Error list

Question 1

QUANT: Read questions carefully - why?

Answer 1

Because there may be little changes that go against what you expect to solve

e.g., Which of the following numbers is the THIRD LARGEST –> not asking about the largest number…

Don’t RUSH too much –> focus on understanding Q and approach first

Question 2

RC: Read questions carefully - why?

Answer 2

Because CONTEXT is everything

e.g., A question about the author’s tone will be DIRECTED TOWARDS specific subject matter (so you can FOCUS on the right part of the passage)

Pay attention to what you are comparing and what the QUESTION STEM says is the FOCUS

Question 3

QUANT: when you see decimals in a division format…

e.g., what is 0.4545/45?

Answer 3

DON’T PANIC
> express numerator as a whole number
e.g., (4545 * 10^-4)/45

> then simplify as if it were a FRACTION
e.g., cancel out the 5s first, then 9s,

OR remember that you can SPLIT decimals into a sum to make division easier:
(0.45 + 0.0045)/45
= 0.0101

Question 4

QUANT: |2 - 6| = ?

Answer 4

Don’t mistake ABSOLUTE SIGNS for 1

|2 - 6| = |-4| = 4

1 has a little tip

Question 5

QUANT: What is the greatest prime factor of …
(11! * 10! + 10! * 9!) / 111

Answer 5

Don’t panic –> you know the question is testing FACTORING and FACTORIALS
> Also know that the term MUST simplify into an integration, and 111 must be cancelled out somehow
> start by factoring as much as possible, end up with [10! * (11! + 9!)] / 111
> ** KEEP SIMPLIFYING (you can factor out multiple terms!); Factor out 9!

[10! * 9! (11*10 + 1)] / 111
[10! * 9! (111)] / 111 —> 111 crosses out, so greatest prime factor must be 7

When asked about factors, need a PRODUCT

Question 6

RC: Make you understand the option sets and their meaning carefully –> avoid DISTORTION trap

Answer 6

Focus on the WORDS used and their relationship to each other

BE WARE OF DISTORTED ANSWERS (sound similar but are NOT right)

e.g., “The second paragraph presents a convincing challenge of the validity of evidence discussed in the first paragraph”
> In reality, the second paragraph is not challenging “Validity of evidence” (i.e., evidence that Neanderthals produced pitch for making tools) –> instead, the second paragraph is challenging the CONCLUSION from this evidence

SOMETIMES the difference between two option sets comes down to ONE WORD (e.g., once vs before)

Question 7

QUANT: Data sufficiency
What should you remember to do after you’ve “solved” each statement?

E.g., If x and y are integers, what is the value of x + y?

(1) x(x^2y) = 1
(2) y(y^2x) = 1

Answer 7

Treat the answer to each statement as a CONDITION (esp. in algebra-related Qs that require you to determine the value of something)

> when you evaluate answer C, FIRST think about the ANSWER TO the statements individually (if you’ve managed to narrow down the option set)

Example: (1) tells me that x = y = +/- 1
(2) also tells me than x = y = +/- 1
(3) statement 1 and 2 give me the SAME INFO, so E
> if you set the two equations equal to each other, you might incorrectly conclude that x and y can = 0, but you have to think about the statements as CONDITIONS (cannot equal 0)

Question 8

QUANT: dividing unknowns, what should be triggered?

Answer 8

Make sure that the variable DOES NOT EQUAL 0
> do not assume that the variable =/ 0
> if it COULD = 0, move over to one side of the equation instead of dividing out
> applies also to unknown expressions e.g., x-1

Question 9

QUANT: Data sufficiency
What is the value of r?

(1) rs = 42
(2) r - s = 1

Answer 9

Concept: HIDDEN Quadratic equations –> which can have UP TO 2 SOLUTIONS
> don’t just blindly follow the common saying that if you have 2 variables, you can solve with 2 unique equations

In this case, putting both statements together, we get a quadratic equation
(s-6)(s+7) = 0, so both s and r have two values

E

Question 10

QUANT: When you see in a question, “… is approximately what percent of …”, what should you think of?

Answer 10

Percent = /100

What percent = x/100

Question 11

QUANT: If n =/ 6 and (n+2)/(n-6) = n, what is the value of n^2 -7n - 1?

Answer 11

CONCEPT: Quadratic equations trap answer (look for COMBOs)
> if you are UNABLE to factor a quadratic equation easily, look for COMBOS

In this question, you end up with n^2 - 7n = 2 —> can sub into the question stem to get 2 - 1 = 1

Question 12

QUANT: Be careful with your work

Answer 12

Make sure you write clearly and be careful of:
> signs during algebra and when interpreting ROOTS of quadratic equations
> unclear variables

Question 13

QUANT: If (x+9) is a factor of the expression x^2 - nx - 36, where n is a constant, what is the value of n?

Answer 13

Concept: Quadratic equations and solving for coefficients
> if given a FACTOR or ROOT of a quadratic expression, you can solve for coefficients if you set the expression equal to 0
> be ware of signs on the coefficients

In this case, if x+9 is a factor, then x=-9 is a ROOT –> sub into quadratic expression and set = 0

(-9)^2 - n(-9) - 36 = 0
81 + 9n - 36 = 0
9n = -45
n = -5 (NOT 5)

Question 14

QUANT: Recognize different forms of exponents (powers)

e.g., x^y^2

Answer 14

Recall power rules:

(ab)^c = a^c * b^c ——> and VICE VERSA (sprinkle effect)

If you have the SAME BASE, then you can ADD or SUBTRACT exponents if multiplication or division

a^x * a^y = a^(x+y)
(a^x)/(a^y) = a^(x-y)

Question 15

QUANT: Recognize that you can SPLIT fractions if base is one term and numerator has +/-

e.g., (x-y)/(xy)
e.g., (x-6)/x

Answer 15

Concept: We split fractions if the base is one term to help with simplificaiton

e.g., (x-y)/(xy) = x/xy - y/xy = 1/y - 1/x
e.g., (x-6)/x = x/x - 6/x = 1 - 6/x

Question 16

RC: What should you think of when you read qualifiers like “some”

Answer 16

“Some” is extremely vague and just means “more than one”

So answer choice could just indicate that were were “some” exceptions to what the passage implies, but the conclusion of the passage could still remain TRUE

Question 17

QUANT: If 5a - 3b = c and 2a - b = d, what is the value of a - b?

1) 2a - b = 5
2) 2d = c - 3

Answer 17

4 variables, given 2 equations, asked for COMBO
> whenever you are asked to solve for an EXPRESSION –> alert to combo
> so when you are solving the linear equations by elimination, rearrange in a way that you get to a COMBO
> generally with 4 variables, you need at least 4 different equations, UNLESS you are solving for combo or variables cancel out

Ex:
5a - 3b = c
2a - b = d —> 4a - 2b = 2d
————————————-
subtract: a - b = c - 2d

(1) No way for you to isolate for 2a - b –> NS
(2) you know that c - 2d = 3 = a - b –> S (ans B)

Question 18

QUANT: What are arithmetic sequences?

Answer 18

Sequence / pattern that always has a CONSTANT positive or negative difference between any two consecutive terms
> alternatively named as “equally spaced sequences”

Arithmetic sequences ALWAYS START AT “a1”
n >= 1

Term:
An = a1 + (n - 1)*d
where d is the constant difference (+ or -) between any two consecutive terms

Sum:
Sum from a1 to an = (average * # of terms) —> equally spaced sequences
= (a1 + an)/2 * n

> Average = median = first + last divide by 2

LINEAR growth problems can also be solved as an arithmetic sequence
> e.g., monthly info +/- constant amount
> e.g., height

Question 19

QUANT: If x and y are positive integers such that 1 < x < y AND y / x is an integer, what does this mean for the relationship between x and y?

Answer 19

1) y is a multiple of x (x is a factor of y)
2) y must be BIGGER THAN x
3) y must contain ALL the prime factors of x, AND at least one additional prime factor
> therefore, y cannot have the same number of prime factors as x

Question 20

QUANT: DS, what should you get into the habit of doing

Answer 20

COVERING the other statement (so you don’t accidentally use info from the other statement)

Question 21

QUANT: If A is a positive integer, what is the remainder when A is divided by 6?
(1) A + 4 is divisible by 7
(2) A is divisible by 5

Answer 21

Concept: Remainders / divisibility

Layout the prompt: A/6 = Q + R/6
Trying to find R = ?

(1) A + 4 = 7k
A = 7k - 4
A/6 = (7k - 4)/6

Depending on the value of k, remainder changes
e.g., k = 1, R = 3
But if k = 4, R = 0
NS

(2) A = 5m
A/6 = 5m/6

Depending on the value of m, remainder changes
e.g., m = 1, R = 5
But if m = 6, R = 0

(3) Create COMBINED algebraic equation for A so that you can come up with potential values of A

Recall to create a combined equation = LCM of divisors + smallest possible value

LCM of divisors = 35
Smallest possible value of A = 10

A = 35Q + 10
Possible values of A = 10, 45, 80 etc.

We can see that remainder still changes depending on the value of Q:
If A = 10, R = 4
But if A = 45, R = 3

NS

Question 22

QUANT: If n is a whole number, what is the units digit of n! (i.e., n factorial)?

(1) n < 6
(2) n > 4

Answer 22

Remember: Any factorial >= 5! will have a zero in its units digit (because there’s at least one 2*5 pair, creating a trailing 0)

Once a factorial number ends with 0, then all the factorial numbers larger than it will end with 0 also

(1) n could equal 0, 1, 2, 3, 4, 5
Testing will reveal NS

(2) n! >= 5! –> units digit is always 0

Question 23

QUANT:
What is the greatest common factor of the positive integers x and y?

(1) x = y^20
(2) y = (243)^(1/5)

Answer 23

Concept: GCF of a number and its factor will equal the factor
> if y divides evenly into x, then LCM is x and GCF is y
> HOWEVER since this is a data sufficiency question, you need to FIX THE NUMBER (not just fix the relationship)
> Lesson: Don’t get too excited by the answer –> always see for value question WHAT IS THE VALUE? (and equivalently for y/n question what is the the answer, always y or always n)

(1) x is a multiple of y
so GCF = y, but WE DON’T KNOW THE VALUE OF Y

(2) y = 3
NS because we don’t know the value of x

(3) Sufficient since we know y = 3

Question 24

QUANT: Is the positive integer x a perfect square?

(1) x = t^n, where t is a positive integer and n is odd
(2) x^0.5 = k, where k is a positive integer

Answer 24

CONCEPT: perfect squares, however be aware of special numbers 0 and 1

(1) x = t^n
If t = 1 and n = 1, then x = 1 —> 1 is a perfect square!! so is 0 (Yes)

If t = 2 and n = 3, then x = 8 –> 8 is not a perfect square (No)
NS

(2) x^0.5 = k (squaring both sides)
x = k^2 —> perfect square

Question 25

RC: What is the name of this kind of trap?

Passage: “Moreover, because of its potential to give rise to many additional economic benefits, including export trade growth and job creation, the production of these reactors has recently received rare bipartisan political support, a development suggesting that the future of nuclear power is promising”

Option set: “Reference to bipartisan political support provides an example of the progress that has been made in developing nuclear reactors”

Answer 25

Distortion (similar words, but different meaning)
> “developing” vs “development”

The fact that reactors have “bipartisan political support” IS NOT AN EXAMPLE of progress made in the specific activity of “developing nuclear reactors”

It is merely an example in support of the conclusion that economic hurdles are being overcome

To solve for this next time:
> Be very literal with the option set (recognizing some lee way for synonyms) –> be ware of stretch options too

Question 26

RC: What is the name of this kind of trap?

Passage: “According to the WHO, nuclear power generation is statistically the safest among the eight most common methods of energy production”

Option set: “Nuclear power generation is safer than the vast majority of other methods of energy production”

> you really need to practice going back to the passage to understand what each sentence is talking about (esp. look above markers)

Answer 26

Stretch
> passage says nuclear power generation is the safest among the EIGHT MOST COMMON methods of energy generation, NOT ALL methods

> so you cannot agree with the option that says nuclear power generation is safer than the vast majority of other methods of energy production (beyond the 8, there could be many other methods that are safer than nuclear power generation, just not commonly used)

To solve for this next time:
> Be very literal with the option set (recognizing some lee way for synonyms) e.g., “populace of the US” matches “US residents”; “perception is overstated” matches “understanding has room for improvement”

Question 27

QUANT: If A and B are positive integers, is B divisible by A?

(1) 2B/A is an integer
(2) B^2/A is an integer

Answer 27

Watch out for TEST CASES (always see if you can find a yes and a no)

Concept: Divisibility, factors and multiples

Approach: Test cases

(1) 2B/A = int

Test case 1: Yes
B=2, and A = 2

Test case 2: No
B=1, and A = 2

NS

(2) B^2/A = int

Test case 1: Yes
B=2, A = 2

Test case 2: No
B=2, A=4
** missed this case before because you did not consider than factors of B together can be a factor of A, but individually are not a factor of A

NS

(3) Still Y or N answer if try:
B=2, A=2
B=2, A=4

Question 28

QUANT: If M and N are positive integers greater than 1, does M have more unique prime factors than N?

(1) 2N/M is an integer
(2) N^2/M is an integer

Answer 28

Concept: Unique prime factors

Make sure you understand what you are comparing (in this Q, whether M > N in # of unique prime factors)

e.g., 23 => 2 unique prime factors
e.g., 2^23^2*5^3 =>3 unique prime factors

Approach: Test cases

(1) 2N/M = integer

Test case 1: Yes
N=3, M=2*3 (N does not need to have 2)

Test case 2: No
N=3, M=2

NS

(2) N^2/M = integer

Test case 1: No
N=2*3, M=2

Test case 2: Yes??
> not possible
> because if M has any additional unique prime factors that N does not have, then the division would not be an integer
> So only N can have more unique prime factors than M

S

Question 29

CR:

Newly Elected President: In running for president, I said that I would take a variety of actions, which included cutting taxes, reducing regulations, and increasing spending on infrastructure. Given that a majority of the voters voted to elect me president rather than to reelect the previous president, it must be the case that most voters are in favor of my taking the actions that I talked about during my campaign.

Which of the following is an assumption that the newly elected president made in drawing her conclusion?

(A) If the newly elected president’s platform had included a set of actions different from the set it included, the opposing candidate would have won the election

(B) Using their votes to express dissatisfaction with the previous president’s policies was not the only concern of the vast majority of people who voted for the new president

Answer 29

Ans B

> when you are STUCK between two answers for CR, really try the negation technique and DO NOT create a convoluted justification
Remember to understand the CONCLUSION clearly

Conclusion: Most voters are in favor of the actions PROMISED during her campaign

A is wrong
> Negate: If the newly elected president’s platform had included a different set of actions, the opposing candidates WOULD STILL NOT HAVE WON
> Does conclusion still hold true?
> YES!!!!
> Still true that the evidence that she became president implies that MOST VOTERS are IN FAVOR of her campaign actions

(I originally thought the conclusion was tied to specific actions like cutting taxes, reducing regulation etc.) = “the actions” and if she had still won, it must mean that there was another reason why she was elected (other than “the actions”)

Question 30

QUANT: What should you always test when you see inequalities with squares and/or square roots

e.g., sqrt(y) > x
e.g., y^2 > x^2

e.g., m^a > m^b

Answer 30

FIRST test EXPONENTS using number line (while keeping base the same)
> 0, -1, 1, 2, -2, 1/2, -1/2

Next use number line to test BASES (while keeping exponent the same)
(A) < -1 —> - 2
(B) -1 < x < 0 —> -1/2
(C) 0 < x < 1 –> 1/2
(D) x > 1 —> 2

Particularly don’t forget to test FRACTIONS

sqrt(1/9) = 1/3
sqrt(1/4) = 1/2
sqrt(1/16) = 1/4

Question 31

QUANT - Always remember what you are SOLVING FOR and don’t get distracted by intermediate calculations

e.g., if you are asked to solve for xy, don’t just submit the answer for y

Answer 31

How to address this?
> circle final answer and PUT THE ANSWER right besideit before moving forward

Question 32

QUANT: What is the value of x^2 - 2x - 14?

(1) x < 0
(2) (x - 5)(x + 3) = 0

Answer 32

Quadratic equations and COMBOS

> notice right off the bat that you CANNOT FACTOR x^2 - 2x - 14 –> should be a SIGN that you are looking for COMOB
e.g., If you know x^2 - 2x then you can solve

(1) NS
(2) (x-5)(x+3) = 0
EXPAND this further

x^2 - 2x - 15 = 0
x^2 - 2x = 15

THEREFORE statement 2 is SUFFICIENT

B

Question 33

DS: when you see unknown variables and equations or inequalities, what should you automatically think of?

Answer 33

Be ware of the unknown variable being equal to:
> 0 (cannot divide both sides by 0)
> negative number (for inequalities)

Question 34

What are your top QUANT mistakes?

Answer 34

(1) Fell for trap answer (29%)

(2) Did not understand the concept tested (21%)

(3) Got so excited that I knew how to answer the question that I made a careless error (20%)

(4) Understood the concept tested but failed to properly apply it (16%)

(5) Made a careless math mistake (7%)

(6) Failed to use all of the information provided to me in the stem (4%)

(7) Misread / misinterpreted the question / written work was unorganized (4%)

Question 35

What are your top CR mistakes?

Answer 35

(1) Eliminated the correct answer because it seemed irrelevant (50%)

(2) Chose a trap choice because it seemed to have the effect I was looking for (25%)

(3) Made up a convoluted, unsupported story to support my choice (13%)

(4) Missed important details in the passage

Question 36

What are your top RC mistakes?

Answer 36

(1) Understood the concept but failed to properly apply it (33%)

(2) Eliminated the correct answer because it used words that didn’t resemble the words in the passage (25%)

(3) Chose a trap choice written to appear to match what the passage said (17%)

(4) Other (misread / misinterpreted the Q, did not understand the concept tested, read too quickly to comprehend the passage)

Question 37

QUANT: If m > 0, is n^a / m^b > 1?

(1) n = m
(2) a is a factor of b

Answer 37

Test cases for powers
> think of a NUMBER LINE
> -2
> -1
> -1/2
> 0
> 1/2
> 1
> 2

Ans E because even though we know that m^(a-b) can either be m^0 = 1 or m^negative, m could be a fraction or an integer:

0 < m < 1 or m > 1

e.g., (1/2)^-5 = 2^5
e.g., (2)^-5 = 1/2^5

Question 38

QUANT - convert the following into fractional approximations

0.143
0.833
0.714
0.167
0.222

Answer 38

0.143 = 1/7
0.833 = 5/6
0.714 = 5/7
0.167 = 1/6
0.222 = 2/9

Question 39

QUANT: Sara and Catherine are two of the twenty players on a softball team. If twelve runs were scored in a particular game, how many runs did Catherine score?

(1) The ratio of the number of runs Sara scored to those that Catherine scored was 1 to 3

(2) Had 3 more runs been scored, Catherine’s runs would have represented 2/5 of the total runs scored in the game

Answer 39

Ratios
> made a mistake the first time because I DID NOT UNDERSTAND “had 3 more runs been scored…” —> this means 3 additional runs were added to 12, bringing the total to 15, BUT WE CANNOT ASSUME that Catherine scored 0 of those 3 new runs!!

12 runs = Sara’s runs + Catherine’s runs + other eighteen players’ runs

(1) S/C = 1/3 –> ratio does not tell us anything about the absolute quantity of S or C
NS

(2) Total runs = 15 now

If Catherine contributed 0 of those 3 runs, then:

C/15 = 2/5 —> C = 6

BUT if Catherine contributed 1 of those runs, then:
(C + 1)/15 = 2/5 —> C = 5

In fact, if we assume x = number of new runs contributed by Catherine…

(C+x)/15 = 2/5
C+x = 6 –> x max value is 3 and min value is 0, so C could be 3, 4, 5, 6

NS

(3) knowing statement 1 alongside statement 2 does not help

E

Question 40

QUANT: Be careful with set-matrix questions, sort out ROWS AND COLUMNS

At a particular party with 1000 people, 30 percent of the people like salsa but do not like guacamole. Forty percent of the people who do not like salsa do like guacamole. If 400 people like guacamole, how many of the people who do not like salsa do like guacamole?

Answer 40

Math mistake (did not keep rows and columns in the set-matrix separate)

If you set x = number of people who do not like salsa, then 0.6x + 300 = 600

Not 0.4x = 300

Ans =0.4*500 = 200 people do not like salsa but do like guacamole

Question 41

QUANT DS

When you have statements that present a different condition than the question stem

e.g., if wages were equal to 0, Breck would have paid $15 in taxes

e.g., if tax rate increased by 10%, total profit would have been $30,000

Answer 41

• Don’t assume that the conditions presented in one of the statements is the same as for the broader question

e.g., wages in the actual question stimulus DOES NOT EQUAL 0

e.g., tax rate in the actual question DID NOT increase by 10% and profit DOES NOT equal $30000

Question 42

QUANT PS: When 24 is divided by the positive integer n, the remainder is 4. Which of the following statements about n must be true?
(I) n is even.
(II) n is a multiple of 5.
(III) n is a factor of 20.

Answer 42

Remainder

(1) Always set up the equation

24 / n = int + 4 / n

Rearranged… 24 = nint + 4
20 = nint

(2) List out the possible values of n (unknown)

0 <= R < n
0 <= 4 < n

Therefore, n > 4 ***** don’t forget this condition

From 2^2 * 5 = n*int and n > 4, we know that the possible values of n are: 5, 10, and 20
(originally got this wrong because I did not think of n > 4 condition)

(3) Go through each statement and be careful to test if needed

I) not necessary (e.g., n = 5)
II) yes (since n > 4)
III) yes

Question 43

What is the greatest possible integer value of n such that 9^n is a factor of 43! + 44!?

Answer 43

Ans 10

> first need to express as a PRODUCT = 43!*45
need to PRIME factorize 9^n first –> 3^2n —> figure out how many 3s first
then figure out how many 9s
also notice that the numerator becomes a factorial * # so that involves different approaches

(1) factorial shortcut for figuring out how many 3s are in 43!
43/3 = 14 3’s
43/9 = 4 3’s
43/27 = 1 3’s
Total 3’s in the factorial = 19 3’s

(2) how many 3s are in 45 = 2 3’s

(3) total number of 3’s = 19+2 = 21 3’s

(4) total number of 9’s = at most is 10 (cannot be 11)
> 10 9’s means 20 3’s
> 11 9’s means 22 3’s (not enough)

Question 44

If it takes a certain fish 30 minutes to swim a straight line from one end of a pond to the other, did the fish swim more than 1 mile? (1 mile = 1.6 km)

(1) the fish travels at a constant rate that is less than 4 kilometers per hour

(2) the fish travels at a constant rate that is greater than 3 kilometers per hour

Answer 44

Rate and inequality question

If t = 30 minutes = 0.5 hours, does r*t > 1.6 km?
(BE CAREFUL with ranges when determining if sufficient or not)

(1) r < 4 km/h
rt < 2 km ——> NS to say rt > 1.6

(2) r > 3 km/h
rt > 1.5 km ——-> *** NS to say rt > 1.6

(3) 1.5 < rt < 2 —-> NS to say rt > 1.6 (rt could be 1.9 or 1.51)

Question 45

If x > 0 and 5 - sqrt(5) < sqrt(x) < 5 + sqrt(5), what is the value of x?

(1) x is an even integer
(2) sqrt(x) is an integer

Answer 45

Roots and inequalities
> given a range of values —> figure out the possible range of values for x

FIRST square every term to isolate x (because every term is positive, no need to flip inequality signs), THEN approximate the endpoints:

30 - 10sqrt(5) < x < 30 + 10sqrt(5)
30 - 102.2 < x < 30 + 102.2
8 < x < 52

(1) x is even integer —> in that rate, x could be more than one value
NS

(2) sqrt(x) is an integer –> x is a perfect square

in that range, x could be more than one value ( 9, 16, 25, 36)

NS

(3) x is an even perfect square —> x could be 16 or 36

NS

E

Question 46

What is the value of x?

x - 10 = sqrt(x) + sqrt(10)

Answer 46

Difference of squares involving roots:

x - 10 = (sqrt(x) + sqrt(10))*(sqrt(x) - sqrt(10)

Therefore:
x - 10 = sqrt(x) + sqrt(10)

(sqrt(x) + sqrt(10))*(sqrt(x) - sqrt(10) = (sqrt(x) + sqrt(10))

Since sqrt(x) + sqrt(10) > 0, we can divide terms on both side, leaving:

sqrt(x) - sqrt(10) = 1
sqrt(x) = 1 + sqrt(10)
x = (1+sqrt(10)^2
x = 1 + 2sqrt(10) + 10
x = 11 + 2sqrt(10)

Question 47

If x^102 = 100, which of the following must be the value of x^51?

A) -50
B) -10
C) 10
D) 50
E) None of the above

Answer 47

Watch out for MUST BE TRUE Qs given multiple possible values of X —> does not HAVE TO BE TRUE

x^102 = (x^51)^2 = 100

E —> x DOES NOT have to be -10, x DOES NOT have to be 10

Question 48

The expression sqrt( 7+2sqrt(6)) + sqrt( 7-2sqrt(6)) is equal to which of the following?

A) 2sqrt(6)
B) 3sqrt(7)
C) 8
D) 9
E) 10

Answer 48

OPTION 1) NO PATTERN that you can leverage: sqrt(a + b) + sqrt(a - b)

THEREFORE —> Approximate the value of the expression and match with PS

sqrt(6) = ~2.4
2*sqrt(6) = ~4.8

7+4.8 = ~11.8
7-4.8 = ~2.2

Sqrt(11.8) = ~sqrt(12) —> between Sqrt(9) and Sqrt(16), or between 3 and 4 —> ~3.5

Sqrt(2.2) = ~sqrt(2) —> between sqrt(1) and sqrt(4), or 1 and 2 —-> ~1.4

Therefore ans: 3.5 + 1.4 = ~4.9 —-> closet to A (4.8) vs B (7.8)

OPTION 2) set the Expression = x
> then you can SQUARE both sides
> also notice sqrt(a + b) + sqrt(a - b) —-> has arguments that when multiplied are a DIFFERCE OF SQUARES —-> so need to multiply arguments somehow

Question 49

What is the following expression equal to?

3 + 3 +3 + 3 + 3 + 3 + 3^2 + 3^3 + 3^3 + 3^4 + 3^4 + 3^5 + 3^5 + 3^6 + 3^6 + 3^7 + 3^7

Answer 49

SUM of Exponents with like bases SHORT CUT:
> if you have a sum of power with BASE a, then as long as you have a number of terms added together, you can factor out common term and increase power by 1
(match number of identical terms with the base of the power)
> JUST SKIP TO THE LAST “a” terms and raise it by 1

e.g., base 2 –> need at least two 2’s
2^n + 2^n = 2^n *(1 + 1) = 2^n+1

e.g., base 3 –> need at least three 3’s
3^n + 3^n + 3^n = 3^n*(1 + 1 + 1) = 3^n+1

e.g., base 4 –> need at least four 4’s
4^n + 4^n + 4^n + 4^n = 4^n*(1 + 1 + 1 +1) 4^n+1

Therefore –> in this question, base is 3, so we need at least three identical 3s with the same power
(NOT matching with the exponent)

Start with having three 3^2 —-> (3+3+3) + (3+3+3) = 3^2 + 3^2

Next: (3^2 + 3^2 + 3^2) + 3^3 + 3^3 + 3^4 + 3^4 + 3^5 + 3^5 + 3^6 + 3^6 + 3^7 + 3^7
= (3^3) + + 3^3 + 3^3 + 3^4 + 3^4 + 3^5 + 3^5 + 3^6 + 3^6 + 3^7 + 3^7
= 3^4 + 3^4 + 3^4 + 3^5 + 3^5 + 3^6 + 3^6 + 3^7
= 3^5 + 3^5 + 3^5 + 3^6 + 3^6 + 3^7 + 3^7
= 3^6 + 3^6 + 3^6 + 3^7 + 3^7
= 3^7+ 3^7 + 3^7
= 3^8

Question 50

If m > 0, is n^a / m^b > 1?

(1) n = m
(2) a is a factor of b

Answer 50

Divisibility and exponents

TRAP: m>0 —-> 0 < m < 1 OR m > 1

Is n^a > m^b?

(1) n = m
NS without exponents

(2) a is a factor of b
NS without bases

(3) No: if 1^2 vs 1^2
Yes: if (1/2)^2 vs (1/2)^4

E

Question 51

Are there more than 100 cheerleaders at Hill University?

(1) the ratio of cheerleaders to coaches at Hill University is 13 to 1
(2) the ratio of cheerleaders to teachers at Hill University is 8 to 1

Answer 51

Ratios with integer quantities
> ratio multiplier must be an integer
> can deduce the MINIMUM actual quantities

(1) NS because the actual number of cheerleaders can be any multiple of 13

(2) NS because the actual number of cheerleaders can be any multiple of 8

(3) Combined ratio of cheerleaders to coaches to teachers is…
104x : 8x: 13x

Minimum number of cheerleaders is 104 (when x = 1)

Sufficient (C)

Question 52

If -2 is a solution to the equation x^2 - 6nx = 40, in which n is a constant, which of the following could be a product of n and x?

2
3
6
20
60

Answer 52

Quadratic equations:
> when x = -2, the quadratic equation = 0

(1) solve for n (watching out for negative sign)

x^2 - 6nx - 40 = 0 —–> plug x = -2 in

4 + 12n - 40 = 0
12n = 36
n = 3

(2) Re-write the quadratic equation *** (remember there are TWO solutions for x)

x^2 -18x - 40 = 0
(x + 2)*(x - 20) = 0
x = -2, x = 20

nx = -6
or nx = 60

Question 53

If the product of all two-digit positive integers is divisible by n!, then n could be which of the following?

I. 89
II. 90
III. 99

Answer 53

Factorials - product of CONSECUTIVE INTEGERS is divisible by n!, where n represents the number of consecutive integers in the product

[10, 99] has 90 consecutive integers —> divisible by 90!

BUT ALSO DIVISIBLE by ALL the factors of n!

90! and 89!

I and II

Question 54

Over the span of 3 games, Sara scored x, y, and z goals, respectively. The SD of the number of goals scored per game was n. Was the SD of the number of goals scored over the next three games greater than n?

(1) over the next three games, Sara scored x+k, y+k, z+k goals, respectively

(2) Sara scored two goals in each of the next three games

Answer 54

Statistics: SD
> SD = 0 means the terms in the set are all the same
> adding a constant to each term does not change SD or range (but does increase mean and median by the constant)

(1) SD does not change when there is a transformation involving +k
Therefore no, SD is not greater than n (SD = n)

Sufficient

(2) SD = 0

Question becomes: is 0 > n? —-> always no*****

Sufficient

D

Question 55

For deaths from accidents to represent less than 20 percent of all deaths from the top 6 causes for this age group, the number of deaths from accidents must be decreased by what?

Current number of deaths from accidents = 14,000
Current total number of deaths = 26,500

6600
8800
10,200
11,600

Answer 55

Hypothetical changes to ratio —> when you remove or add to a subgroup, don’t forget to also ADJUST THE TOTALS

Let x = the number of deaths from accidents that must be decreased

14k - x < 0.2*(26.5k - x)

x > 10.75k

ans 11,600

Question 56

If 3c/10 = 10d and d^3 < 0, which of the following must be true?

c+d > 1
d < 2c
2d > 4c
c > 2d
c > d

Answer 56

Inequalities problem solving: TWO variables and equation —> SUB in variable so you have ONE variable

We know:
d < 0
c = 100d/3 —–> SUB into EACH answer choice and evaluate if it is TRUE

a) obviously false since d < 0 and c < 0

b) d < 2*(100d/3) —> divide both sides by d (d < 0)
1 > 200/3 —-> FALSE

c) 2d > 4*(100d/3) —> divide both sides by d
2 < 400/3
1 < 200/3 —> TRUE

d) 100d/3 > 2d —> divide both sides by d
100/3 < 2 —> FALSE

e) 100d/3 > d —> divide both sides by d
100/3 < 1 —> FALSE

Question 57

CR Logic: could the # of jazz listeners increase and proportion of people listening to jazz increase, WHILE at the same time not impacting # of listeners of other genres?

Answer 57

Yes –> If the TOTAL # of listeners increase and become jazz listeners –> does not impact # of listeners of other genres

Mathematically:

a / T < 1 –> (a+c) / (T + c) —> closer to 1

Question 58

CR Logic: Given that students who are significantly older than average in grade school are more likely to attend college, is this statement true?

The proportion of college students significantly older than average for their classes will likely be greater than the proportion of grade school students significantly older than average for their grades

Answer 58

Yes

(older students / total students) * (x/y) where x/y > 1 because of the greater likelihood to attend college

So new proportion > old proportion

Learnings:
> Math in CR –> write out the components involved
> Inference Qs –> must be supported by the passage. COULD be represented in a HIDDEN WAY

ALSO related to the “over-represented” logic
> default - should be same proportion in both groups if “as likely”
> since students who are significantly older than average in grade school are MORE LIKELY to attend college –> expect such students to have higher representation in college

Question 59

Translate the following into equation:

The proportion of college students significantly older than average for their classes

Answer 59

Concept: the proportion or percentage OF X —-> X is the denominator

Quality follows X and is the NUMERATOR

Therefore

college students significantly older than average for their classes / college students

Question 60

QUANT/DI: There are four factories, together having a total of 135 employees. If one employee were randomly selected from those 135 employees, and the selected employee were a line worker, which factory is the employee most likely to work at?

Factory A has ~15 line workers out of ~25 employees
Factory B has ~20 line workers out of ~38 employees
Factory C has ~15 line workers out of ~47 employees
Factory D has ~5 line workers out of ~25 employees

Answer 60

Probability

Need to figure out which factory has the HIGHER NUMBER OF LINE WORKERS

Because: we know the employee IS A LINE WORKER, so DENOMINATOR IS total number of line workers
(NOT calculating individually the probability of selecting a line worker at each factory)

P(A) = 15 / total line workers
P(B) = 20 / total line workers —-> highest probability
P(C) = 15 / total line workers
P(D) = 5 / total line workers

Question 61

If px < 5x, which of the follow must be correct?

p<6
p>6
p>x
p/x < 5/x
p/x > 5x

Answer 61

Inequality –>x^2 is always > 0 so you can multiply and divide (as long not equal to 0) without changing the direction of the inequality sign

p/x < 5/x

Question 62

If x+y = 20 and 2x + 3y < 54, which of the following must be correct?

I. 3x > 17
II. 2y < 30
III. y - x <8

Answer 62

Inequality MUST BE true and multi-variable with equations that you can sub in
> BE CAREFUL WHEN EVALUATING RANGES

After simplifying we get x > 6

I) Therefore we know 3x > 18
Is 3x > 17??
Well, 3x > 18 > 17 —> SO YES this is true ***** (3x will ALWAYS be greater than 17)

II) after simplifying we get y < 14 or 2y < 28
Is 2y < 30??
Well 2y is ALWAYS less than 28 which is less than 30. So YES 2y is ALWAYS less than 30

2y < 28 < 30

III) y - x < 8 —> evaluate as is 12 < 2x ?
or is x > 6? We know this to be true already

I, II and III are true

Question 63

Sqrt[ sqrt(16 / 4 * (x-10) + 4) - 4] / (3x - 75)

For the expression above, which of the following values for x will result in an expression that is a real number?

I. 8
II. 10
III. 25

Answer 63

A real number –> not undefined (denominator cannot equal 0), or argument under square root cannot be negative

WATCH OUT FOR DENOM when you sub in each number

ans None of the above

PEMDAS order of operations —> even though 4*(x-10) has a brackets, we cannot simplify the inside of the brackets any further. So we should go left to right –> 16/4 first = 4 then multiply by x - 10 to get 4x - 40

Denominator =/ 0 —> x =/ 25 (rules out III)

First square root >= 0 —> x >= 9

Second square root —> x >= 13 (rules out I and II)

Question 64

If m is the product of 4 consecutive positive integers, and n is the product of 5 consecutive positive integers, then mn must be divisible by which of the following?

I. 5!
II. 6!
III. 7!

Answer 64

Factorials and PRODUCT of consecutive integers
> a product of n consecutive integers is divisible by n!
> a product of integers is divisible by the PRODUCT OF THEIR FACTORS (mushed together)
> be careful of hidden factors represented by the factorials

m is divisible by 4! and n is divisible by 5!

so mn is divisible by 4! * 5!

I) true
II) ALSO TRUE
6! = 6*5! and 4! is divisible by 6

III) 7! not true because there is no 7 factor

Question 65

Data insights (Graphics Interpretation) –> scatter plot with 2 axis and 2 groups –> watch out for what…

Answer 65

MISTAKE READING VALUES OFF GRAPH:

Aligning vertical points correctly (usually represents the SAME object, just two characteristics)
> don’t trust your eye –> count LEFT TO RIGHT to double check the right vertically aligned pairs

Other common mistakes:
> symbols
> axis
> units

Question 66

Graphics interpretation - watch outs with correlation involving two objects

Answer 66

There are two types of correlation to keep track of:

(1) general correlation between x variable and y variable
> as X increases, Y tends to do what?

(2) Correlation between OBJECTS
> as X increases, Object A does what? Object B does what? —> see if they move together or move in opposite directions

Example: Overall negative correlation, but positive correlation among objects
> As X increases, Object A decreases and Object B decreases

Question 67

Graphics Interpretation general watch outs

Answer 67

1) What you are SOLVING FOR (INTERPRET the question properly)
> venn diagram –> only vs shared segments

2) Watch out for your math accuracy

3) Read charts accurately

Question 68

At a certain repair shop, a flidget costs $15 to replace, a spleener costs $40 to replace, and a bwam costs $100 to replace. If a spleener were twice as likely as a bwam to break, and a flidget were three times as likely as a spleener to break, then the average cost to repair a product at the repair shop would be approximately how many dollars?

Answer 68

RATIOS WORD PROBLEM:

FIRST: translate relationship into algebra
S = 2B —> number of spleener repairs vs Bwam repairs
F = 3S

SECOND: before you can turn into a ratio table, you need to EXPRESS AS RATIOS

S/B = 2/1
F/S = 3/1

THIRD: express as a combined ratio using table and like variable S

F : S : B
6: 2: 1

Total = 9x

Therefore average cost can now be calculated using these part to totals:

avg cost = 6/9$15 + 2/940 + 1/9*100 = $30

Question 69

There are a total of 51 apples and bananas at a fruit stand. How many apples are at the fruit stand?

(1) The number of bananas at the fruit stand is greater than four times the number of apples at the fruit stand

(2) The number of bananas at the fruit stand is less than 42

Answer 69

Word problems - inequalities
> note apples and bananas MUST BE positive integers
> multiple variables with equations –> CAN SUB IN equation into inequality for variable we want

(1) can simplify to a < 10.2
NS

(2) can simplify to a > 9

(3) 9 < a < 10.2 AND a is an integer, we know a = 10

Sufficient
C

Question 70

A certain airline charges a dollars for each passenger’s first 2 checked bags and b dollars for each checked bag thereafter. How much does the airline charge for 10 checked bags?

(1) The airline charges $200 for 6 checked bags
(2) The airline charges $100 for 2 checked bags

Answer 70

Word problems involving price
> “a” dollars for EACH PASSENGER’S first 2 checked bags —> $a for up to two checked bags (not $a per bag)

Total cost = a + b*(10-2)
= a + 8b

(1) 200 = a + 4b
NS

(2) 100 = a
NS without b

(3) from 1 and 2, we can determine value of a and b
Sufficient C

Question 71

If k is an integer greater than 75, (k^2 + 2k)*(k^2 - 1) must be divisible by how many of the following?

I. 4
II. 6
III. 9
IV. 12
V. 18
VI. 20

Answer 71

Divisibility and product of consecutive integers

Notice that (k^2 + 2k)(k^2 - 1)
= (k-1)k(k+1)(k+2)
Is a product of 4 consecutive integers

Must be divisible by 4! and its factors

I) Y
II) Y
III) Not necessarily
IV) Yes
V) not necessarily
VI) not necessarily

e.g., 757677*78

(k > 75 is a red herring)

Question 72

When X is divided by 23, the remainder is 12. When X is divided by 29, the quotient is equal to Q, and the remainder is 12. Q must be divisible by which of the following numbers?

6
12
23
29
52

Answer 72

Remainder theory:
> write algebraic expressions
> integers

X = 23int + 12 (int >= 0)
X = 29Q + 12

Same remainder, DIFFERENT divisors

Set X = X, we know Q must be an integer

23int + 12 = 29Q + 12
23int = 29Q
int = 29Q/23 —-> Q must be a multiple of 23 to make integer

Question 73

A certain candy store charges x dollars for the first pound of candy purchased and y dollars for each additional pound purchased. If Jake purchased 10 pounds of candy, how much did it cost?

(1) The cost of 8 pounds of candy is 220 percent of the cost of 2 pounds of candy

(2) The cost of each additional pound of candy is 1/4 of the cost of the first pound

Answer 73

Word problems

Cost = x + 9y

(1) TRANSLATE –> “is 220 percent OF” (NOT “greater than”)
x + 7y = 2.2*(x + y)
x/y = 4 —-> Ratio does not tell us anything about actual values

(2) Translate:
y = x/4
x/y = 4 —> ratio does not tell us anything about actual values

(3) same info
E

Question 74

If |x+y| = |x| + |y| = 14, which of the following could be equal to x+y?

I. 14
II. 7
III. -14

Answer 74

Multi-variable Absolute values (special rules)
> = means that xy > 0 (if x and y are non zero) OR xy = 0 (if one is 0 and the other is +/-14)

I. x+y could equal 14 –> x = 0, y = 14. Or x = 7, y = 7

II. x+y cannot equal 7 –> |x+y| would not equal 14

III. x+y could equal -14 –> x = 0, y = -14. OR x = -7, y = -7

I and III

Question 75

If |p| - |q| = |p-q|and |p| < 5, which of the following is a possible value of q?

I. -4
II. 0
III. 5

Answer 75

Multi-variable Absolute values (special rules)
> = means pq non zero AND pq > 0 AND |p| >= |q|
so 5 > |p| >= |q|

> OR pq = 0 AND both 0
OR pq = 0 AND p non zero, q zero

I. possible for q = -4 and |q| = |p| = 4

II. possible for q to equal 0 and any value of p will work |p| = |p|

III. since 5 > |p| >= |q| —> not possible for q = 5

Question 76

In how many different ways can the letters of the word SPENCER be arranged?

Answer 76

Permutation with duplicates

7! / 2! ——> NOT equal to 7*6! because need to cross out 2 from top and bottom

Instead = (7*720)/2 = 2520

Question 77

In a particular basket, there are hard candies and soft candies. If 20 hard candies were added to the basket, there would be four times as many hard candies as soft candies. Is the current number of hard candies in the basket greater than the number of soft candies?

(1) The number of soft candies in the basket is greater than 6
(2) The number of hard candies in the basket is less than 9

Answer 77

Word problems with inequality
> INTEGER CONSTRAINTS and RELATIONSHIP
(need to test values… can’t 100% rely on inequality manipulation)
> two variables with INTEGER inequality word problem –> Express as one integer and check values against constant

H >= 0 int
S >=0 int

H + 20 = 4S

Is H > S?

Is 4S - 20 > S?
Is 3S > 20?
Is S > 6.67?
Is S >= 7? (since S and H must be integers)

(1) S > 6 —–> S is AN INTEGER
So S >= 7
SO Always Yes

(2) H < 9 —–> H is AN INTEGER
So H <= 8
Therefore 4S - 20 <= 8
4S <= 28
S <=7
NS (S could be 7 or less than 7)

A

Question 78

What is 10.60824/159.1236 equal to?

1/15
1/17
2/3
1/2
1/16

Answer 78

Messy decimals, round to nearest WHOLE number then long division

11/159 —> 0.069

Closest to 1/15 = 0.066

DON’T round to 10/160 —> incorrectly leads to you choose 1/16 when you see answer choices are very close to each other

Question 79

If n is a positive two-digit integer, how many different values of n allow n^3 - n to be a multiple of 12?

Answer 79

Product of consecutive integers
> to be a multiple of 12, need to be divisible by 3 AND 4
> n^3 - n is (n-1)n(n+1) —-> product of 3 consecutive integers
> n is [10, 99]

Product of 3 consecutive integers is already divisible by 3!

In order for n^3 - n to ALSO be divisible by 4, need at least two factors of 2 ***

CASES OF N:

(1) n is odd –> E * O * E (divisible by 8, so definitely has two factors of 2)
= (99-11)/2 + 1
= 45

(2) n is even (NOT ENOUGH –> n must be EVEN AND MULTIPLE OF 4)
= (96-12)/4 + 1
= 22

Therefore, total possible values of n = 45 + 22 = 67

Question 80

If |2x + 8| = |4x - 4|, what is the value of x?

(1) |x| > x^2
(2) -|x| < -x

Answer 80

Single variable inequality and absolute value –> unravel absolute value signs and SOLVE IT
> based on question stem, either x = 6 (same sign of arguments) OR x = -2/3 (opposite sign of arguments) —–> need to know which solution is it

> boundaries are -4 and 1

(1) |x| > x^2
WE KNOW x cannot equal 0 (otherwise both sides =)

1 > x^2 / |x|
1 > |x| > 0

Satisfies x = -2/3

LONG WAY:
-1 < x < 1 (x=/0)

(2) -|x| < -x

|x| > x
WE KNOW x cannot equal 0

1 > x/|x| —> x/x should equal |1|, so x must be negative and x/|x| must equal -1

Satisfies x = -2/3

D

Question 81

Jake set up a business at a lake conducting jet ski trips for visitors. On each trip, he carries one passenger and each passenger pays him 10 dollars per mile. However, Jake has to pay $100 every 50 miles to refuel his jet ski. If Jake started the day with a full tank of gas and drove a total of 50 visitors an average of 5km each that day, how much profit did Jake make? (Note: 1.6km = 1 mile)

Answer 81

Word problems - profit and loss
> tricky part was “$100 every 50 miles” –> CANNOT BLINDLY ROUND –> need to figure out how many miles Jake needs and determine number of refuels necessary (1 refuel is necessary if total miles exceeds 50; 2 refuels are necessary if total miles exceed 100; 3 refuels are necessary if total miles exceed 150…)
> not always round up to get integers! could round down like in this case…
> in this question, it is easier to SPLIT OUT CALCULATIONS

Total miles = 5/1.6 * 50
= 156.25 —> be comfortable with decimals (long division if necessary)

Total revenue = 10 * 156.25 = 1562.50

Total cost –> number of refuels needed?

Since total miles = 156.25, Jake needs 3 refuels (NOT 4)
So total cost = 100*3 = 300

Total profit = 1562.5 - 300 = 1262.5

Question 82

A gnome who was 16 inches tall swallowed a pill that caused him to grow by a certain factor x every minute for 6 minutes. After 6 minutes, the gnome was 1024 inches tall. He then took another pill that reduced his height by a certain factor y every minute. If this reduced the gnome’s height by twice as much as factor x increased his height, how many minutes did it take the gnome to grow to the maximum height and then shrink to a height less than his original height of 16 inches?

Answer 82

Exponential growth and decay word problem:
> Growth factor = *x
> Decay factor = *(1/y)
e.g., reduced height by factor 4 every minute = *(1/4) = divide by 4

Growth to maximum height takes 6 minutes:
1024 = 16*(x)^6
x = 2

Therefore since decay factor is twice as much as x, decay factor y = 4 (divide by 4 or multiply by 1/4)

Time it takes from max height to reach below 16 inch:
16 > 1024*(1/4)^t
when t = 3, height equals 16, so t = 4 to be below 16

Therefore total time it takes is 6 + 4 = 10 min (NOT 9)

Question 83

There are two types of seats at a concert venue, floor seats and orchestra seats. The floor seats cost $100 per seat and the orchestra seats cost $30 per seat. If a total of 50 tickets were sold on Saturday, how many floor seat tickets were sold that day?

(1) the revenue from the tickets sold on Saturday was between $3680 and $3800

(2) The number of floor seat tickets sold on Saturday was greater than the number of orchestra seat tickets sold

Answer 83

World problem –> INEQUALITY or weighted average
> use inequality since it is more accurate and does not rely on testing
> two variables with INTEGER inequality word problem –> Express as one integer and check values against constant

f = number of floor seats
y = number of orchestra seats
BOTH ARE INTEGERS

f + y = 50
y = 50 - f

f =?

(1) THIS IS AN INEQUALITY

3680 < 100f + 30y < 3800
3680 < 100f + 30*(50-f) < 3800
3680 < 100f + 1500 - 30f < 3800
3680 < 70f + 1500 < 3800 —–> CREATE RANGE FOR f (one variable)
2180 < 70f < 2300
218 < 7f < 230
31.xx < f < 32.xx —-> f must be 32 (sufficient)

(2) f > o
NS

Question 84

Sara purchased both hard candy and chocolate candy at a store. Hard candy costs $0.25 per piece; chocolate candy costs $0.60 per piece. Did Sara purchase more than 1 piece of hard candy? (Assume that a candy is either hard or chocolate but that there is no hard, chocolate candy).

(1) The total cost of the candy was less than $3.60
(2) Sara purchased more than four chocolate candies

Answer 84

Word problem - inequalities
> two variables with INTEGER inequality word problem –> Express as one integer and check values against constant
> notice how we are NOT given any equation –> less likely to be sufficient

Is H > 1?

(1) 0.25H + 0.6C < 3.60

Simplifies to H < 12(6-C)/5

TEST VALUES:
If C =1, H < 12 —> H < 1 or H > 1

NS

(2) C > 4
NS without any info on H

(3) C > 4 —> test C = 5

H < 2.4 —> H < 1 or H > 1

NS

E

Question 85

A dairy sells each case of buttermilk for $90 and each case of ice cream for $20. If the dairy sold a total of 17 cases of buttermilk and ice cream last month, how many cases of buttermilk were sold last month (Assume that fractions of a case cannot be sold)

(1) Had 3 more cases of ice cream been sold, ice cream would have represented 1/2 of the total cases sold last month

(2) Last month the dairy had more than $1000 but less than $1100 in revenue from buttermilk and ice cream sales

Answer 85

Word problem - inequality and equations
> two variables with INTEGER inequality word problem –> Express as one integer and check values against constant

17 = B + I

B = ?

(1) (I + 3) = 0.5*(20) —> know I, and know B

I = 10 - 3
I = 7
B = 10

Sufficient

(2) 1000 < revenue < 1100
1000 < 90B + 20I < 1100
100 < 9B + 2I < 110 —–> SUB IN I = 17 - B
100 < 9B + 2*(17-B) < 110
100 < 9B + 34 - 2B < 110
100 < 7B + 34 < 110
66 < 7B < 76
9.xx < B < 10.xx

B = 10
Sufficient

D

Question 86

In the sequence x0, x1, x2, …, xn, each term from x1 to xk is 3 greater than the previous term, and each term from xk+1 to xn is 3 less than the previous term, where n and k are positive integers and k < n. If x0=xn=0 and if xk = 15, what is the value of n?

5
6
9
10
15

Answer 86

Arithmetic sequences
> Last resort, write out terms according to the rule
> Don’t miss that x0=xn=0

General formula for arithmetic sequences:
an = ak + (n-k)*d

xk = 0 + (k)*3 = 15
k = 5

Therefore:
xn = xk + (n-k)d
xn = 15 + (n-5)(-3) = 0
3n = 30
n = 10

Question 87

If a, b and c are constants, and a > b > c, and x^3 - x = (x-a)(x-b)(x-c) for all numbers x, what is the value of b?

-3
-1
0
1
3

Answer 87

Product of three consecutive integers –> divisible by 3!

(x-a)(x-b)(x-c) is the product of three consecutive integers

BUT NOTICE HOW THERE IS X on BOTH SIDES ———-> WE CAN DETERMINE value of a, b and c

(x-1)(x)(x+1) = (x-a)(x-b)*(x-c)

Since a > b > c, x-a must be smallest and x-b must be largest

x-1 = x-a —> a = 1
x = x-b —> b = 0
x+1 = x-c —-> c=-1

b = 0

Question 88

Car X averages 25.0 miles per gallon of gasoline and car Y averages 11.9 miles per gallon. If each car is driven 12,000 miles, approximately how many more gallons of gasoline will car Y use than car X ?

320
480
520
730
920

Answer 88

Rate word problem:
> misread 12000 as 12

Mpg = distance / gallons

Mpg * number of gallons = miles
so:
Number of gallons = miles / MPG

Car X: 12000 / gallons = 25
gallons = 12000/25 —–> MATH
= (120*100)/25
= 480

Car Y: 12000/gallons = 11.9 ——> ROUND 11.9 to 12
gallons = 12000/12
= 1000

Y - X = 1000-480
= 520

Question 89

If n = 4p , where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

Answer 89

Factors
> we know p must be an odd prime number
> total number of factors (in prime factorized form) = (exponent on prime + 1)*(exponent on prime + 1) …

n=2^2 * p

n has (3*2) or 6 total factors

LIST THEM OUT to determine which ones are EVEN (1 * P, 2 * __, etc., with each number being a factor):

1
4p
2
2p
4
p

Of these 6, 4p, 2, 2p, and 4 are even = 4 even

ALTERNATIVELY, choose any odd prime like 3 –> 2^2 * 3 has 6 factors

1
* 12
* 2
*6
3
*4

Question 90

Sequence an is a geometric sequence with terms a1 = 1, a2 = 2, a3 = 4 and so on. Sequence Sn is a sequence such that its nth term is the sum of the first n terms of the sequence an. That is Sn = a1 + a2 + … + an. Which of the following equations correctly states the relationship between Sn and an?

Sn = 3 + an
Sn = 2an - 1
Sn = 3a - 1
Sn = 2an + 3
Sn = 3an + 3

Answer 90

Geometric sequence and sequence representing SUM of geometric sequence terms

Can solve algebraically

Question 91

How many unique three digit numbers can you create if:

CASE 1)
> even number
> repetition

CASE 2)
> even number
> not repetition

Answer 91

Counting (permutation)

(1) write out possible values for each digit
(2) write out cases for where 0 can be (for without repetition)
(3) start with strictest constraint (units digit eg..)

CASE 1) with repetition = slot method
_ _ _
> first digit [1,9]
> second digit [0,9]
> third digit must be even [0, 2, 4, 6, 8]

9 * 10 * 5 = 450

CASE 2) without repetition = slot method with cases
_ _ _
> first digit [1,9]
> second digit [0,9]
> third digit must be even [0, 2, 4, 6, 8]

Case 1: _ _ 0
= 891 (1 option for last digit, 9 options for middle digit if the last digit is 0, leaving 8 options for first digit)
= 72

Case 2: _ 0 _
= 814 (1 option for middle digit, 4 options for last digit if the middle digit is 0, leaving 8 options for first digit)
= 32

Case 3: _ _ _ (non zeros)
= 784 (4 options for last digit, 8 non-zero options for middle digit, leaving 7 options for first digit)
= 224

TOTAL = 72 + 32 + 224 = 328

Question 92

In sequence an, a1 = 4 and an = 2an-1 for n >=2. Which of the following produces the largest 3-digit number?

a1
a5
a8
a9
a10

Answer 92

Geometric sequence
> we are looking for the LARGEST 3 digit number
> a10 produces a FOUR DIGIT NUMBER —> TRAP
> go from highest n to lowest n to save time

THESE TYPES OF SEQUENCE QUESTIONS –> write out values

Standard explicit form
an = 4*(2^n-1)

a9 = 4(2^8) = 4256 = 1024
a8 = 4(2^7) = 4128 = 512 —-> ANSWER

Question 93

A local skeet shooting tournament had 10 contestants. The maximum number of targets that any contestant could hit was 25. If only one contestant hit 25 targets, was the average (arithmetic mean) number of targets hit less than 20?

(1) Average (arithmetic mean) number of targets hit by the 9 contestants who did not hit the maximum 25 was 19 targets

(2) The median number of targets hit was 20

Answer 93

Statistics
> integer targets hit

We are looking for whether average < 20?
Alternatively whether Sum of the targets scored by the other 9 contestants < 175

(1) sufficient
19 = sum / 9
Always yes

(2) since we KNOW that the actual answer is always yes, let’s save time by finding IF THE MAXIMUM is ALWAYS less than 175

Let 3 contestants score 24 each and 6 contestants score 20 each (to satisfy median = 20)
Sum = 324 + 620
= 192 which is greater than 175 –> so it is still possible to score above 175

NS

A

Question 94

At a certain hedge fund, there are only CFAs and MBAs. The average number of vacation days taken by CFAs is 5 fewer than the average number of vacation days taken per employee at the hedge fund. The average number of vacation days taken by the MBAs is 10 more than the average number of vacation days taken per employee at the hedge fund. What fraction of the hedge fund employees are CFAs?

Answer 94

Weighted average - give 3 averages
> anchor using TOTAL weighted average

Xt = (XaA + XbB)/(A+B)
= Sum of vacation days taken / total number of employees

We also know:
CFAs: Xa = Xt - 5
MBAs: Xb = Xt + 10

Replace Xa and Xb with Xt

Xt = (Xt-5)A + (Xt+10)B / (A+B)

NOTICE how XtA and XtB cancel out:

5A = 10B
A/B = 2/1

so A/(A+B) = 2/3

Question 95

When rounded to the nearest tenth, the standard deviation of the following set is 10.4. How many of the values in the following set are more than 1 standard deviation from the mean:
{60.5, 85.5, 72.5, 68, 68, 80, 68, 85.5, 90, 90, 90}

Answer 95

Statistics

We are looking for value x such that:

x > mean + 10.4
or
x < mean + 10.4

So we NEED TO DETERMINE MEAN
> no special rules –> sum / 11
> try to GROUP to make the math easier

= (903 + 683 + 85.52 + 80 + 60.5 + 72.5)/11
= 858/11
= 78 (using LONG DIVISION) —> 7911 = 869 > 858, but 78*11 = 858

How many terms x are:
x > 88.4
or
x < 67.6

There are 4 terms: 60.5, 90, 90, 90

Question 96

Jeff ordered two types of pizzas for a party, cheese and sausage. If the cheese pizzas cost $10 each, what was the average price he paid per pizza?

(1) Jeff spent a total of $100 on all the pizzas
(2) Jeff ordered cheese pizzas and sausage pizzas at a ratio of 3 to 1

Answer 96

Weighted average / ratio word problem

average price per pizza –> we need to know the price per sausage pizza, and ratio of quantities

Let P = price per sausage pizza
C = number of cheese pizzas = integers
S = number of sausage pizzas = integers

(1) 100 = 10C + PS
NS –> multiple combinations of P, S, and C
100 = 101 + 452
100 = 101 + 2*45

(2) C / S = 3 / 1
NS without prices
C/(C+S) = 3/4
S/(C+S) = 1/4

(3) average pizza = (103/4) + (P)(1/4)

Do we know what P?

Lets sub in C = 3x and S = 1x:
100 = 103x + P1x
100 = 30x + Px
100 = x(30+P)
100 / (30+P) = x (integer) —-> do we know x and P (one value?) ** NOT ONE VALUE

P = 70, x = 1
P = 20, x = 2

NS

E

Question 97

At a cooking competition, 7 contestants had 45 minutes to bake as many cookies as possible. Jenny and Martha were 2 of the contestants. If a total of 69 cookies were baked, did Martha bake more than 8 cookies?

(1) Each contestant baked at least 6 cookies, and none of the contestants baked an equal number of cookies

(2) Jenny baked more than 17 cookies, and Martha baked the 3rd least number of cookies

Answer 97

Multiple variable integer inequality sum question

Key:
> know the constraints and Total
> understand tradeoffs must be made (if you find a perfect combination of values, any + changes must be offset by any - changes)

_ _ _ _ _ _ _ = 69 (integers)
Is M > 8?

(1) all different cookies, minimum 6
6 7 8 9 10 11 18 = 69 —-> M can be any one of these (some greater than 8, some less than or equal to 8)

NS

(2) J > 17 —> J>=18 and Martha baked 3rd least number of cookies

Case 1: 6 7 8 9 10 11 18 = 69 —> M = 8 (no)
Case 2: 5 7 9 9 10 11 18 = 69 –> M = 9 (Yes)
NS

(3) Understand tradeoffs in a restricted sum:
6 7 8 9 10 11 18 = 69 –> this works, and M = 8 (no)

If M were to increase, we need to SUBTRACT from someone else –> cannot subtract from 6 or 7 (minimum 6 and no two identical number of cookies) , also cannot subtract from 9, 10, 11 (end up with same order)

Sufficient C

Question 98

After purchasing 3 liters of cleaning solution with a ratio of 3 parts bleach to 2 parts surfactant, Jorge realizes that the solution contains too much bleach for his needs. He then purchases a cleaning solution whose ratio is 1 part bleach to 4 parts surfactant and replaces some of the original solution until the resulting mixture is 25% bleach.

How many liters of the original solution are replaced?

How many liters of bleach are in the resulting mixture?

Answer 98

Mixture Problem / Ratio – changing the solution to get to desired percentage of substance
> Two mixtures, each with different composition of substances –> need to create SEPARATE EQUATIONS for EACH unique substance (1 for bleach, 1 for surfactant)
> “replacing” solution means TOTAL VOLUME IS THE SAME
> If approaching using mixtures –> easiest to convert ratios into % weights

JOT DOWN INFO:
> we have TWO SUB PARTS (B and S)

ORIGINAL MIXTURE: 3 liters = Liters of Bleach + Liters of Surfactant

And we are given: B/S = 3/2 —-> B/total = 3/5, S/total = 2/5

We can determine the CURRENT QUANTITIES of Bleach and Surfactant

SECOND MIXTURE: B/S = 1/4 –> S = 4B

RESULTING MIXTURE: B / (B+S) = 1/4

Resulting mixture has 0.25 * 3 = 0.75 Liters of bleach

Amount of Bleach in OG: 3 liters (3/5) = 30.6 = 1.8 liters
Amount of Surfactant in OG: 3 liters * (2/5) = 1.2 liters

Jorge removes x liters of this 60% solution and replaces it with x liters of 20% bleach solution, leaving him with 25% bleach solution:

Liters of Bleach = Liters of Bleach

0.75 liters = 1.8 - x(0.6) + x(0.2)
0.75 = 1.8 - 0.4x
0.4x = 1.05
x = 2.625

Question 99

In how many ways can a line of 3 boys and 2 girls be formed so that none of the boys stand next to each other?

Answer 99

Permutation –> Cannot have boys standing next to each other (NOT ONLY consecutively)
> Therefore, boys must be separated by girls

BGBGB is the only arrangement

Slot method for each gender:
32211
= 12

Question 100

In how many ways can the letters of the word OCARINA be arranged if the O, R and N must remain in their original positions?

12 or 24

Answer 100

Anchor Permutation with duplicates

O, R and N are fixed, leaving C, A, I, A to be arranged:

1 * 4 * 3 * 1 * 2 * 1 * 1 —–> need to divide by 2! —–> you forgot to do this in the moment (pay attention to duplicates of LETTERS)

12

Question 101

In how many ways can the word ANTEDILUVIAN be arranged if the T must be placed in the first spot and the U and V must stay together

11!/4 or 10!/4

Answer 101

Permutation
> letters - pay attention to duplicates
> 12 letters
> ignoring the T, we now have 11 letters to arrange, and U and V must stay together –> Link together
> now we have 10 (NOT 11)
> Letters with duplicates: A, N, and I

Ans: 10!/(2! * 2! * 2!) * 2 to arrange U and V

= 10!/4

Question 102

In April 2000, the value of a certain house was $100,000. If the value of the house increased 40 percent per year for 5 years, what was the value of the house in April 2005?

260,192
274,548
300,525
384,160
537,824

Answer 102

Percent
> need to multiply 100,000 by 1.4^5
> did not apply units digit strategy correctly (normally 0* any digit = 0, but what if there were SO MANY digits to the right of the decimal, that the 100,000 just moves the decimal, leaving a units digit that does not equal 0)

Try your best to do the math

100 –> 140 –> 196 –> ~274 –> ~384 –> ~500+

Question 103

The population of city n is 100,000 greater than the population of city m. If both cities increase in population by x percent in a given year, how much greater is the population of city n than the population of city m after the increase?

(1) x = 10
(2) the original population of city m was 10/11 of the original population of city n

Answer 103

Percent word problem
> translate first and try to simplify
> MISREAD THE QUESTION (we are NOT looking for percent change. We are looking for ABSOLUTE CHANGE)

n = 100,000 + m
n(1+x/100)
and m(1+x/100)

We are looking for: n(1+x/100) - m(1+x/100) =?
=(1+x/100)*(n-m)

WE KNOW n-m = 100,000, so we really just need x:
= (1+x/100)*100,000

(1) Sufficient

(2) m = 10n/11
11m = 10n —> we can solve for n and m but don’t know x

NS

A

Question 104

On the xy-plane, point A is in Quadrant II, and point B is in Quadrant IV. If z and m are integers, and if z is the x-coordinate of point A, and m is the y-coordinate of point B, which of the following must be true?

(I). m^3z^2 > m^2z^3
(II). mz < m/z
(III). 1/m < 1/z

Answer 104

Coordinate geometry:

Point A is in quad II –> (negative, positive) = (z, y) —-> z < 0
Point B is in quadrant IV –> (positive, negative) = (x, m) —-> m < 0

Must be true –> looking for at least one counterexample to rule out

(I) Is: (-)(+) > (+)(-) ? NS
SIMPLIFIES TO Is: m > z? We don’t know

(II): is mz < m/z?

is: mz^2 > m?
is: z^2 < 1 —–> not sure (z = -1/2 or z = -4) NS

(III) is: 1/m < 1/z? (reciprocal both negative, change sign)
is m > z ? We don’t know

None of these

Question 105

When a pilot uses an airplane’s normal power setting, a certain transcontinental flight generally takes 24 hours from start to finish. However, for every 1 percent increase above the plane’s normal power setting, the flight time will be reduced by 1/4 of the time taken at the previous power setting. If the plane’s power setting only moves in 1 percent increments, what is the smallest percent increase in power which reduces the flight time to under 10 hours?

Answer 105

Word problems - exponential decay
> “for every 1 percent increase above the plane’s normal power setting, the flight time will be reduced by 1/4 of the time taken at the previous power setting” —-> every 1 percent increase, the flight time is reduced by 25% = decline of 25%
> decay factor is *3/4

LIST OUT in a table: estimate as best as you can

0 = 24 hours
1 percent increase = 24(3/4) = 18
2 percent = 18(3/4) = 54/4 = 27/2 = 13 + 1/2
3 percent = 27/2(3/4) = 81/8 = 10 + 1/8
4 percent = 81/8(3/4) = 243/32 = LESS THAN 10 ** 4 percent

Question 106

At a certain carnival, there is a basketball game in which a participant can make either two-point or three-point shots. If a total of n+3 shots were made during the game and if there were 4 more two-point shots made than three-point shots, which of the following represents the total number of points scored during the game?

5n+11
n+14
(5n+17)/2
(5n+11)/2
(3n+17)/2

Answer 106

Word problem - general (linear equations) with integers
> watch out when substituting expressions INTO expressions (math mistake, write legibly and big)

Let a = number of 2 point shots
b = number of 3 point shots

n+3 = a + b
a = 4 + b
2a + 3b = ? —> need to express in terms of n

a = n+3-b
n+3-b = 4+b
n-1=2b
b=(n-1)/2

Therefore, a = 4 + (n-1)/2 = (7+n)/2

2a+3b
= 2((7+n)/2) + 3((n-1)/2)
= (14+2n + 3n-3)/2
= (5n+11)/2

Question 107

A dance delegation of 4 people must be chosen from 5 pairs of dance partners. if 2 dance partners can never be together on the delegation, how many different ways are there to form the delegation?

Answer 107

“Dance partner”

Combination = permutation / k!

= 10 * 8 * 6 * 4 / 4! ————> first spot has 10 ppl to choose from, second spot has 8 ppl (cannot choose the first person’s partner) etc.
= 80

ALTERNATIVELY:
> first choose which pairs are in the selection
> then multiply by 2^k (each pair can have 2 choices)

= Number of pairs Choose k spots * 2^k
= 5C4 * 2^4
= 5 * 16
= 80

Question 108

A particular trail mix with 6 ingredients is made from nuts, chocolate, and fruit. Five different kinds of chocolate, 6 different kinds of fruit, and 3 different kinds of nuts are available for the mix. If the trail mix must have at least 3 kinds of fruit and at least 2 kinds of chocolate, and at least 2 different types of ingredients must be included in the trail mix, how many different ways are there to make the trail mix?

Answer 108

Trail mix combination - “at least” = cases or Total - X

Mandatory: F F F C C _

Case 1: F F F C C F
= 6C4 * 5C2

Case 2: F F F C C C
= 6C3 * 5C3

Case 3: F F F C C N
= 6C3 * 5C2 * 3C1

Ans: 950

Question 109

There are a number of toys in a toy chest, including 2 toy trucks. If 4 toys, including the 2 trucks, can be selected in 91 ways, how many toys are in the toy chest?

11
14
16
17
20

Answer 109

Combination –> solving for N
> use PS answer choices to help guide you when factoring, but DO NOT EXPECT BOTH factors to be in the answer choice

_ _ Truck1, Truck2

(N-2)C2 = 91 —-> from N-2 toys, choose 2 to create the set of 4 toys including 2 trucks

(N-2)! / [(N-2-2)!(2!)] = 91
[(N-2)(N-3)] /2! = 91
n^2 -5N + 6 = 182
n^2 - 5N - 176 = 0

Two numbers that multiply to 176? (units digit is 6)
1611
> break up factors of 176: 444 = 1144 = 11*16

(n-16)*(n+11) = 0
n= 16 (positive answer)

OR sub in answer choices in the form: 182 = (n-2)*(n-3) –> n=16 works

Question 110

OG: If x and y are positive integers such that y is a multiple of 5 and 3x + 4y = 200, then x must be a multiple of which of the following?

3
6
7
8
10

Answer 110

Multiples –> rearrange FOR X
(if that doesn’t work, QUICKLY rearrange for Y)

X = (200-4y)/3 —-> factor out 4 and 5 from numerator (since we know y is a multiple of 5 and can be expressed as 20*int)

X = 20*(10-int) / 3

Therefore x is an integer and MUST be a multiple of 20 —-> also a multiple of 10

Question 111

OG: Which of the following fractions is closest to 1/2?

4/7
5/9
6/11
7/13
9/16

Answer 111

Approximation
> the fraction that is CLOSEST TO 1/2 will have the SMALLEST DIFFERENCE
(Fraction - 1/2 = smallest difference)

Step 1: Write out what the equivalent 1/2 fraction is for each denominator

3.5/7, 4.5/9, 5.5/11, 6.5/13, 8/16

Step 2: calculate the difference (Fraction - 1/2)
= 0.5/7
= 0.5/9
= 0.5/11
= 0.5/13
= 1/16

Step 3: determine the SMALLEST DIFFERENCE

Between 0.5/13 and 1/16

Is 0.5/13 < 1/16?
Is 8 < 13? Yes

Therefore 7/13 is the closest to 1/2

Question 112

OG: In Western Europe, x bicycles were sold in each of the years 1990 and 1993. The bicycle producers of Western Europe had a 42 percent share of this market in 1990 and a 33 percent share in 1993. Which of the following represents the decrease in the annual number of bicycles produced and sold in Western Europe from 1990 to 1993 ?

Answer 112

Percent word problem –> UNDERSTAND MEANING (get the RIGHT TOTAL)

> x bicycles sold in each year IN WESTERN EUROPE
Western Europe bike producers have a certain share of this total x
Decrease from 1990 to 1993 in the number of bicycles produced/sold BY Western Europe bike producers

42%x - 33%x = 9%x

Question 113

In a certain sequence, a1 = 3 and a2 = -3. Every even-numbered term after a2 is obtained by an = an-3 and every odd-numbered term after a1 is obtained by an = an-1. What is the sum of the terms from the 2nd to the 21st term, inclusive?

Answer 113

Sequences and multiples:
> looking for sum of a2 to a21

LIST OUT some of the terms to find the pattern:
a2 = -3
a3 = -3
a4 = 3
a5 = 3
a6 = -3
a7 = -3
etc.

Pattern: every 4 terms, the pattern repeats [-3,-3,3,3] and the SUM equals 0

How many cycles of 4 are there from 2 to 21?
> Number of terms: (last-first)/increment + 1 = 21-2+1 = 20 terms
> so there are a perfect number of cycles of 4

Sum = 0

If there had been 21 terms (Remainder = 1), then there would be one -3 left over

If there had been 22 terms (Remainder = 2), then there would be, -3 and -3 left over (sum of -6)

Question 114

OG: if x and y are positive numbers such that x+y = 1, which of the following could be the value of 100x + 200y?

I) 80
II) 140
III) 199

Answer 114

Creating RANGES of values for unknown expressions (inequalities)
> we are given x and y are positive numbers
> since x+y = 1, this means x and y are between 0 and 1

CREATE RANGE for what 100x + 200y can be:

100x + 100y < 100x + 200y < 200x + 200y
100 < 100x + 200y < 200

Therefore, II and III are possible

Question 115

If m and p are positive integers and m^2 + p^2 < 100, what is the greatest possible value of mp?

36
42
48
49
51

Answer 115

Perfect squares and inequality:

Max MP by maximizing m and p
> testing values is the fastest approach

m and p both < 10

m < sqrt(100 - p^2)

p = 9, the m < sqrt(19) or m <= 4 —> max mp = 9*4 = 36

p = 8, then m < sqrt(36) or m <=5 —> max mp = 8*5 = 40

p = 7, then m < sqrt(51) or m <=7 —> max mp = 7*7 = 49

p = 6, the m < sqrt(64) or m <=7 —> max mp = 6*7 = 42 (starting to decline now)

if p <= 5, then m <= 8 and max mp <= 40

OR WORK BACKWARDS FROM ANSWER CHOICES:

51:
151 or 317 –> however 3^ +17^2 > 100

49:
149 or 77 –> 7^2 +7^2 = 98 < 100 (WORKS)

Question 116

OG: Robin invested a total of $12,000 in two investments, X and Y, so that the investments earned the same amount of simple annual interest. How many dollars did Robin invest in investment Y ?

(1) Investment X paid 3 percent simple annual interest and investment Y paid 6 percent simple annual interest.

(2) Robin invested more than $1,000 in investment X.

Answer 116

Word problems (simple interest)
> keep track of how many unknowns you have and what you actually need to solve

x(rx%) = (12000-x)(ry%) ——> we just need either x, or rx AND ry

(1) this is SUFFICIENT (we are given rx and ry, and then have only one variable left)

(2) x > 1000 is not sufficient (y can be anything <11000)

Question 117

OG: If 3 < x < 100, for how many values of x is ​x / 3​ the square of a prime number?

Answer 117

Prime numbers and squares
> forgot to only count primes

x = 3*prime^2

Primes: 2, 3, 5, 7

3*prime^2: 12, 27, 75 —> only 3 that are between 3 and 100

Question 118

OG: If a (a + 2) = 24 and b (b + 2) = 24, where a ≠ b , then a + b =

Answer 118

Quadratic equation –> recognize this:

a+b —> no combo here

Need to SOLVE FOR a and SOLVE for b:

a(a+2) = 24 —> a^2 + 2a - 24 = 0
(a+6)(a-4) =0
a = -6 or a=4

Since a and b do not equal each other, let a = -6 and b = 4

a+b = -6+4 = -2

Question 119

OG: If the ratio of the number of teachers to the number of students is the same in School District M and School District P, what is the ratio of the number of students in School District M to the number of students in School District P ?

(1) There are 10,000 more students in School District M than there are in School District P.

(2) The ratio of the number of teachers to the number of students in School District M is 1 to 20.

Answer 119

Ratios
> given Mt/Ms = Pt/Ps
> get comfortable testing cases quickly

We want to know: Ms / Ps =?

(1) Ms = 10000 + Ps
NS
Ms/Ps = (10000+Ps)/Ps

(2) Mt/Ms = 1/20 = Pt/Ps —–> SAME RATIO DOES NOT MEAN EQUAL COUNT

Ms = 20a
Ps = 20b

e.g., Ms = 20 and Ps = 40
or Ms = 20 and Ps = 20
NS

(3) From statement 2 we know 20a / 20b = a/b is what we want

From statement 1 we know:

20a = 10000 + 20b
a = 500 + b —-> still not enough to determine a/b

e.g., b = 100, a = 600
b = 200, a = 700

E

Question 120

OG: In each game of a certain tournament, a contestant either loses 3 points or gains 2 points. If Pat had 100 points at the beginning of the tournament, how many games did Pat play in the tournament?

(1) At the end of the tournament, Pat had 104 points.

(2) Pat played fewer than 10 games.

Answer 120

Two-variable integer
> be ware: individual integer values can equal 0

A+b = ?

Statement 1) 104 = 100 - 3A + 2B
4 = -3A +2B
(4+3A)/2 = B
2 + 3A/2 = B (integer)

A = 0, B = 2, A+B = 2
A = 2, B = 5, A+B = 7
A = 4, B = 8, A+B = 12

NS

Statement 2) A+b < 10
NS

Statement 3) Together, there are still two different possible values
NS

Question 121

OG: A merchant discounted the sale price of a coat and the sale price of a sweater. Which of the two articles of clothing was discounted by the greater dollar amount?

(1) The percent discount on the coat was 2 percentage points greater than the percent discount on the sweater.

(2) Before the discounts, the sale price of the coat was $10 less than the sale price of the sweater.

Answer 121

Multi-variable inequality –> compare signs

Cdiscount on coat vs Sdiscount on sweater → Multi-variable inequality (compare signs)

(1) NS without initial prices of each article of clothing

(2) C = S - 10 → NS without discount percents

PLUG IN VARIABLES and see if the inequality can be deemed true or not:

Is (S-10)(2+d) < Sd?
Is: 2S + SD - 20 - 10d < SD?
Is: 2S - 20 - 10d < 0? → depends on S and d (NS)

ALso we are given too many variables (4) and too few unique equations (2)

IF we knew that C > S and dc >ds, then it is sufficient to say that the dollar amount of the discount was greater for C than for S

Question 122

OG: Is the sum of the prices of the 3 books that Shana bought less than $48 ?

(1) The price of the most expensive of the 3 books that Shana bought is less than $17.

(2) The price of the least expensive of the 3 books that Shana bought is exactly $3 less than the price of the second most expensive book.

Answer 122

Multi-variability inequality –> instead of comparing signs, create RANGES of sum with max or min given

Let 0<a <= b <= c

Is a + b + c < 48?

(1) c < 17 —-> c cannot be equal to or greater than 17

sum: a + b + c < 17 + 17 + 17 = 51

Therefore NS
(Sum could be less than 48 or greater than 48)

(2) a = b - 3 –> plug into expression

Is: b-3 + b + c < 48?
Is: 2b + c < 51? —-> NS (c could be anything)

(3) c < 17, b < 17, a < 14

Sum: a + b + c < 14 + 17 + 17 = 48

Therefore, Sufficient

Question 123

OG: At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?

Answer 123

Mixture:
> Set up your mixture equation properly (especially the TOTALs)
> figure out what CHANGES and what DOESN’T CHANGE (e.g., total count will decrease, but number of apples won’t change)

Average * total count = TOTAL sum

1056 = 40A + (10-A)60
A = 2 and Oranges = 8

Let x = oranges when average price is 52
(2+x)52 = 402 + x*60
X = 3

Therefore she must return 5 oranges

Question 124

OG: Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 4. Three of the sandwiches were evenly divided among the students. Since 4 students did not want any of the fourth sandwich, it was evenly divided among the remaining students. If Carol ate one piece from each of the four sandwiches, the amount of sandwich that she ate would be what fraction of a whole extra-large sandwich?

Answer 124

Fraction word problems

1 sandwich divided by m people means EACH SLICE = 1/m
1 sandwich divided by m-4 people means EACH SLICE = 1/(m-4)

So Carol ate: 3/m + 1/(m-4) slices
= (4m-12)/(m*(m-4)) ——> algebra (clean writing)

Question 125

OG: If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p ?

Answer 125

Factorial short cut –> to determine the number of factors of 3 in 30!

30/3^1 = 10 3’s
30/3^2 = 3 3’s (Every multiple of 9 has another 3 factor)
30/3^3 = 1 3’s (Every multiple of 27 has another 3 factor)

In total, there are 14 3’s in 30!

Key:
> express factor in PRIME NOTATION –> adjust exponent accordingly

e.g., 4^k = 2^2k —> we need to find the number of 2’s first, then solve exponent

> if there are TWO primes, solve for the MOST RESTRICTIVE prime (lower frequency = higher prime)

e.g., between 2 and 5 –> count how many 5s exist

Question 126

OG: A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n , then the least and greatest possible values of n , respectively, are

0 and 13
0 and 14
1 and 10
1 and 9
2 and 8

Answer 126

Word problem max and min involving two equations
> fastest to set up equations and TEST options (starting with smallest value and largest value)

37 teachers = a + b + n
64 classes = 1a + 2b + 3n

We see that n can equal 0 and n cannot equal 14 but can equal 13

ans A

Question 127

OG: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

144
152
160
168
176

Answer 127

Codes with constraints (third digit depends on the second digit) –> need cases
> need to test different cases for 0 (impacts the third digit)
> repetition allowed
_ _ _

Case 1) middle digit is 0
= 8 options * 1 option * 9 options
= 72 options

Case 2) middle digit is 1
= 8 options * 1 option * 10 options (last digit can also be 1 or 0 or anything else)
= 80 options

Together = 152 options

OR

Total number of ways = 8210 = 160
Minus both 0: 811 = 8
= 152

Question 128

OG MOCK: Karina is deciding whether to use Brand A or Brand B motor oil in her car. Brand A costs 50% more per liter than Brand B. But, according to the recommendations for frequency of oil changes, if she uses Brand A, she will be able to drive 2.5 times as many kilometers before she changes her oil as she will if she uses Brand B. The cost per kilometer driven is what percent less when using Brand A than when using Brand B?

Answer 128

Percent Change and Unit Measurement

Looking for the percent CHANGE (decrease) in COST PER KILOMETER

Total Cost Per KM = Cost per liter * Number of Liters per km

B = B’s price per liter
D = B’s distance per oil capacity

NEW A: 1.5B = price per liter and 2.5D = distance per certain number of liters (same for A and B), call it N liters

Cost/km = (1.5B)*N / (2.5D)

PERCENT CHANGE: ((1.5BN/2.5D) - BN/D)/(BN/D)

= -40%

Question 129

OG MOCK: The value of (sqrt(8!) + sqrt(9!))^2 is an integer. What is the greatest integer n such that 2^n is a factor of (sqrt(8!) + sqrt(9!))^2?

Answer 129

Factorial short cut
> requires a numerator that is in PRODUCT FORM
> 2^n is already in prime form

(1) Expand the square of a sum
= 8! + 2sqrt(8!9!) + 9! —–> must be an integer so underneath the square root must be a perfect square
= 8! + 238! + 9!
= 8!(1 + 6 + 9)
= 8!*(16)

(2) Count how many 2s are in the numerator

16 = 2^4 —> 4 twos

8! –> 8/2 = 4 twos
8! –> 8/2^2 = 2 twos
8! –> 8/2^3 = 1 two

(3) add the number of twos
= 4 + 4 + 2 + 1
= 11 twos

Question 130

OG MOCK: Marie wishes to enclose a rectangular region in her backyard using part of her 50-foot-long house as 1 side and a total of 80 feet of fencing for the other 3 sides. If Marie chooses the dimensions of the rectangular region so that the region has the greatest area, what is the length, in feet, of the other side of the rectangular region that is bounded by her house?

10
20
25
40
50

Answer 130

Complex Max/Min problem (quadratic)
> set up equations AND TEST options

OR given this is a quadratic function –> you can take the first derivative of the area equation, set to zero and solve

OR know that max/min occurs at the HALFWAY POINT between roots
> Area (f(x)) = -x/2 * (x-80) ——> area = 0 when x = 0 or x = 80
> so the MAX area occurs when x = 40

WATCH OUT FOR PERIMETER = only 3 sides

Let a = 2 sides of the fence representing the width
Let x = length of the rectangle, where x<=50

Perimeter: 80 = 2a + x
Area: a*x —> MAX

DERIVATIVE METHOD: sub in value for a to create a quadratic equation
Area = (80-x)/2 * x
Area = 40x - x^2/2

Area’ = 40 - x
0 = 40 - x
x = 40

OR SET UP EQUATIONS AND TEST starting with largest value of x:

x = 50, then a = 15, area = 750
x = 40, then a = 20, area = 800 ** MAXIMUM
x = 25, then a = 27.5, area = 687.5
x = 20, then a = 30, area = 600
x = 10, then a = 35, area = 350

Question 131

OG MOCK: A store sold 60 percent of the hats from a shipment of hats at a selling price that was 50 percent greater than the store’s cost for each hat. Then the store reduced the selling price by 66(2/3) percent and sold 70 percent of the remaining hats at the reduced selling price. If the store did not sell any other hats from the shipment, then the store’s gross profit from the sale of the hats from the shipment was what percent of the store’s cost for the hats from the shipment?

Answer 131

Percent word problem / profit
> PERCENT OF total cost = Profit / total cost * 100 (including UNSOLD hats)
> Calculate Total Revenue then Total Cost separately

First batch of hats sold: 0.6N hats at a price of 1.5C each
> Revenue = 0.6N1.5C = 0.9NC

Second batch of hats sold: 0.7 OF THE REMAINING HATS
= 0.7(0.4N) at 1/3(1.5C)
= 0.28N * C/2
= 0.14NC

Total Cost of all N hats = NC

Revenue = 0.9NC + 0.14NC = 1.04NC
Cost = NC
Profit = 1.04NC - NC = 0.04NC

Profit % = 0.04NC / NC * 100 = 4%

Question 132

OG MOCK: The domain of the function f(x) = sqrt(sqrt(x+2) - sqrt(4-x)) is the set of all real numbers x such that …

Answer 132

Domain of function –> must not make the function invalid
> cannot divide by 0
> argument underneath roots must be >= 0
> Tricky part: nested functions or nested calculations

To deal with nested calculations –> solve inside out, then set BIG VARIABLES to determine domain of the largest function
> then combine conditions together

sqrt(x+2) —> x+2 >= 0, so x >=-2
sqrt(4-x) —> 4-x >=0 , so x <= 4

LARGE SQUARE ROOT: let A = x+2 and B = 4-x
sqrt(A) - sqrt(B) >= 0
sqrt(A) >= sqrt(B) *** remember to move so that both sides have a square root term
A >= B (can square because we know sqrt is positive)

x+2 >= 4-x
2x >=2
x >= 1

THEREFORE:
1 <= x <= 4

ALSO can plug in answer choices (at the = part)

Question 133

OG MOCK: User Friendly: 56%
Fast response time: 48%
Bargain Prices: 42%

The table gives three factors to be considered when choosing an Internet service provider and the percent of the 1,200 respondents to a survey who cited that factor as important. If 30 percent of the respondents cited both “user-friendly” and “fast response time”, what is the maximum possible number of respondents who cited “bargain prices”, but neither “user-friendly” nor “fast response time”?

A. 312
B. 336
C. 360
D. 384
E. 420

Answer 133

Triple sets with min/max
> we are trying to MAXIMIZE “Bargain Prices” ONLY
> Don’t assume null = 0 unless told so right off the bat
> since we are given % and counts, start with percents (totaling 100) then convert to count
> to maximize bargain prices only, we need to MINIMIZE THE OVERLAP (all three, FB only and UB only)

100 = U + F + B - UF - UB - FB + all three + null

100 = 56 + 48 + 42 - 30 - UB - FB + all three + null
100 = 116 - UB - FB + all three + null
UB + FB - all three - null = 16 —–> SPLIT INTO “only” components

UB only + all three + FB only + all three - all three - null = 16
UB only + FB only + all three - null = 16

Plus we know Bargain Prices = 42 = B only + FB only + UB only + all three

So we need to MINIMIZE FB only + UB only + all three by making null as SMALL as possible

UB only + FB only + all three = 16 + null —> make null = 0

So B only = 42 - 16 = 26 percent of 1200 = 312

Question 134

If x > y > 0, which of the following must be negative?

I. y - x^3
II. x^2 * y - x^3
III. x^2*y - xy^2

I only
II only
III only
I and II
I and III

Answer 134

Exponents, Signs and inequalities
> don’t forget that x and y could be FRACTIONS

I. y - x^3
Test y < x where y = 1/4 and x = 1/2
1/4 - 1/8 —> POSITIVE
Not necessarily negative

II. x^2 * y - x^3
Need to FACTOR PROPERLY
= x^2*(y - x) —-> y-x is negative and x^2 is positive
YES

III. x^2y - xy^2
Need to FACTOR properly
xy(x - y) —> xy is positive and x-y is positive
Must always be positive, not negative

II only

Question 135

OG: Last school year, each of the 200 students at a certain high school attended the school for the entire year. If there were 8 cultural performances at the school during the last school year, what was the average (arithmetic mean) number of students attending each cultural performance?

(1) Last school year, each student attended at least one cultural performance.

(2) Last school year, the average number of cultural performances attended per student was 4.

Answer 135

Stats - Mean - tickets / attendance trap
> same numerator
> think about TOTAL NUMBER of tickets sold or TOTAL ATTENDANCE (INCLUDES repeat attendees)
> how you cut the pie doesn’t change the TOTAL

Looking for TOTAL attendance / 8
where:
Total attendance = Sum of attendance at each performance = Sum of number of performances each student attended

(1) NS –> can come up with multiple different averages and sums

(2) EQUIVALENT to what we are looking for (800)

Sufficient

Question 136

If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n ?

(1) 69 + n and 94 + n are the squares of two consecutive integers.

(2) 94 + n and 121 + n are the squares of two consecutive integers.

Answer 136

Squares and consecutive integers
> big question is whether n is unique

“Squares of two consecutive integers” e.g., 1 and 4 are squares of two consecutive integers (1 and 2)

> list out perfect squares (ORDERED): 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196 …
(notice how the perfect squares have a gap that is consecutive odd integers: +3, +5, +7, +9, +11, +13…)

(1) 69+n and 94+n are squares of two consecutive integers —> will be a pair of the squares listed together above
> NOTICE HOW there is a DIFFERENCE: 94+n - 69-n = 25 —> matches with 144 and 169 where n = 75
> since the difference between the ordered set of perfect squares is INCREASING (by +2), this is the only time the gap is 25 so n is a unique number

Sufficient

(2) 94+n and 121+n —> difference is 27, which matches with 196 and 169
> n = 75
> since the difference between the ordered set of perfect squares is INCREASING (by +2), this is the only tie the gap is 27 so n is a unique number

Sufficient

D

Question 137

What are all the factors of 4p, if p is a prime number greater than 2?

Answer 137

4p = 22p —-> 6 factors (1<=k<=n)

Coming up with factors is all about USING ALL OF THE factors
> two products method is the fastest to orderly come up with factors

1 and 4p (1 and itself)
2 and 2p
p and 4

Question 138

OG: List T consists of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum of the 30 decimals, E, is defined as follows. Each decimal in T whose tenths digit is even is rounded up to the nearest integer, and each decimal in T whose tenths digit is odd is rounded down to the nearest integer; E is the sum of the 30 resulting integers. If ​one third​ of the decimals in T have a tenths digit that is even, which of the following is a possible value of E − S ?

−16
6
10

Answer 138

Inequality RANGES with decimals
> focus on MAX and MIN changes, as well as the NET EFFECT of rounding
> since there are two effects, we want to ADD INEQUALITIES to know the range of E - S (positive contributions - negative contributions)

e.g., 20 integers each changed by this amount, 10 integers each changed by this amount

EVEN TENTHS DIGIT: FOCUS ON MAX AND MIN DIGITS ONLY
> Smallest even tenths digit = 0 = biggest rounding up (will set the max)
> Increase after rounding < 1 (for each number)
0.01 to 1.00 = 0.99
» For 10 numbers, increase after rounding < 110 = 10
> Largest even tenths digit = 8 = smallest rounding up (will set the min)
> Increase after rounding > 0.1
0.89 to 1.00 = 0.11
» For 10 numbers, increase after rounding > 0.110 = 1

ODD TENTHS DIGIT:
> Smallest odd tenths = 1 = smallest rounding down (will set max value)
> Decrease after rounding > 0.1 (<-0.1)
1.10 to 1= -0.1
» For 20 numbers, decrease after rounding < -2
> Largest odd tenths = 9 = largest rounding down (will set min value)
> Decrease after rounding < 1 (>-1)
1.99 to 1 = -0.99
» For 20 numbers, decrease after rounding > -20

TOGETHER:

1 < Even contribution < 10
-20 < Odd contribution < -2
ADD INEQUALITIES:
-19 < Even and odd contribution < 8

I and II work

Question 139

OG: Max purchased a guitar for a total of $624, which consisted of the price of the guitar and the sales tax. Was the sales tax rate greater than 3 percent?

(1) The price of the guitar that Max purchased was less than $602.

(2) The sales tax for the guitar that Max purchased was less than $30.

Answer 139

Inequality word problem with non-integers
> sub inequation into inequality to evaluate

624 = P(1+t/100)
Or
624 = P + P(t/100)

P < 602 → SUB IN EQUATION (P=624/(1+t/100)
624/(1+t/100) < 602 → solve for t inequality
624/602 < 1+t/100
(624/602 - 1)*100 < t (use calculator)
T > 3.65 (sufficient)

P(t/100) < 30 → SUB IN EQUATION (P=624/(1+t/100)
(624/(1+t/100)(t/100) < 30
T < 5.05 (NS)

Question 140

OG: A company makes and sells two products, P and Q. The costs per unit of making and selling P and Q are $8.00 and $9.50, respectively, and the selling prices per unit of P and Q are $10.00 and $13.00, respectively. In one month the company sold a total of 834 units of these products. Was the total profit on these items more than $2,000.00 ?

(1) During the month, more units of P than units of Q were sold.

(2) During the month, at least 100 units of Q were sold.

Answer 140

2 variable integer inequality
Key concept: need to re-express inequality question stem as ONE VARIABLE vs constant, then use statements to help evaluate inequality

Is: 2P + 3.5Q > 2000? Given that P+Q = 834 → Sub in P = 834 - Q
Is: 2*(834-Q) + 3.5A > 2000?
Is: Q >= 222?

(1) P>Q means that P>834/2 and Q<834/2 → Q<417
NS

(2) Q>=100 also is NS

(3) together 100<=Q<417, still NS

Question 141

A gumball machine contains 21 gumballs, of which 8 are red, 7 are blue, and 6 are green. If a child is to choose 4 gumballs randomly, what is the probability that he will choose exactly 2 blue gumballs and exactly 1 red gumball?

Answer 141

Probability
> even though we have identical balls, for probability it doesn’t matter (but for CASES it does matter)

TWO APPROACHES:
Chain method –> DON’T FORGET you need to multiply by cases (permutation for the ORDER of selecting colours)
P(BBRG) = (6/20)(8/19)(6/18) * 4!/2!
= 16/95

Combinatorics:
Total # ways to select 4 balls (assuming each ball is unique): 21C4

favorable ways based on colours (counting method): 7C2 * 8C1 * 6C1

Question 142

If there are a certain number of students in a classroom, what is the probability that at least 3 of them have birthdays in the same month?

(1) There are no students born in the month of December
(2) There are 25 students in the classroom

Answer 142

Birthday Probability Q
> “same month” –> try to see if you can SPREAD the students in as many different months as possible

Looking for: how MANY students have birthdays in the same month / total number of students

(1) No students born in December –> 11 months
> could have every single student born in January = 100% probability that at least 3 of the students have birthday in the same month
> OR could have 3 students, each in different months = 0% probability

(2) N = 25 students –> know the denominator

IF we spread these students as far as part as possible
> 12 students in different months
> another 12 students in different months
> 1 student MUST be in the same month as 2 other students

100% probability

B

Question 143

The product of all the prime numbers less than 20 is closest to which of the following powers of 10?

10^8
10^7
10^6

Answer 143

Approximation utilizing powers of 10
> try to create intuitive GROUPINGS (close to powers of 10)
> watch out for your multiplication (3*7 = 21 NOT 210)

Product of prime numbers less than 20 = 2357111317*19

25 = 10
37 = ~20 = 210
1119 = ~200 = 210^2
1317 = ~200 = 2*10^2

Product = 10210210^2210^2
= 10^6 * 8 (8 million)
= 10^7 (closest to 10 million)

ALTERNATIVELY:
= (25)(37)(1113)(1719)
= (10)(21)(143)(323)
= (10)(~20)(~150)(~300)
= 102101510310^2
= 10^5(303)
= 10^5(90)
= 10^6*9 ——> closer to 10^7 than 10^6
=10^7

Question 144

From the consecutive integers −10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

(−10)^20
(−10)^10
0
−(10)^19
−(10)^20

Answer 144

Consecutive integers (minimize product) and number properties (positive/negative)
> even number of negative values = positive
> odd number of negative values = negative

REMEMBER YOUR POTENTIAL INTEGERS:
> repetition allowed
> [-10, 10]
> picking 20 integers (range is NOT -20 to 20)

Least possible value of product = MOST NEGATIVE VALUE

-10 * 10^19
= -10^20
= -(10^20)

USE ANSWER CHOICES TOO:
> a and b are obviously positive
> c is 0
> d and e are the only negative options

Question 145

A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

4
5
6
7
8

Answer 145

Complex Combinatorics involving inequality (min)
> need to PLUG IN NUMBER if you cannot solve the quadratic equation

Let N = number of letters used
> “pair of distinct letters written in alphabetical order” means if A and B are used, only ONE valid combo is AB, not BA = COMBINATION (not a permutation)

12 <= N + NC2

12 <= N + (N*N-1)/2

e.g., N = 2 —> 2 unique single letter codes + 1 pair code = 3

THIS IS NOT FACTORABLE –> plug in 4 –> does not work

Plug in 5 –> satisfies the inequality

Question 146

A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer’s initial cost. What was the dealer’s approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?

7% loss
13% loss
7% profit
13% profit
15% profit

Answer 146

Profit/Loss word problem
> always write out profit/loss equation

Profit is generally = PriceQuantity - VariableCostquantity - FixedCost
Profit % = (PriceQuantity - VariableCostquantity - FixedCost) / (VariableCost*Quantity + FixedCost)

ans= profit 13%

Question 147

A car is traveling on a straight stretch of roadway, and the speed of the car is increasing at a constant rate with respect to time. At time 0 seconds, the speed of the car is v0 meters per second; 10 seconds later, the front bumper of the car has traveled 125 meters and the speed of the car is v10 meters per second.

Select values of v0 and v10 that are together consistent with the information provided. Make only two selections, one in each column.

5
18
20
36
72

Answer 147

DI - rate and arithmetic sequence question

Given an arithmetic sequence of SPEED, with first term = V0 and last term = V10
> there are 11 terms
> median is located as position 6 = V5 (NOT V6)

Average Speed = Total Distance / Time = 125 / 10 = 12.5 m per second

Average = median = V5 = (V0 + V10)/2 = 12.5

V0+V10 = 25

Must be 5 and 20

Question 148

The cost of a mixture that contains candy-coated chocolate pieces and salted peanuts in the ratio of 2 pounds of chocolate pieces to 3 pounds of peanuts is $3.80 per pound. If the chocolate pieces cost $2.00 more per pound than the peanuts, how much do the peanuts cost per pound?

Answer 148

Mixture with price expressed in an equation and volume expressed as a ratio

C = pounds of chocolate
P = pounds of peanuts

C/P = 2x/3x where x is the ratio multiplier >= 1 (int)

Pc = 2 + Pp

Price per pound * Pounds

PcC + PpP = 3.8(C+P)
(2+Pp)(2x) + Pp*3x = 3.8(5x) —> ratio multiplier x cancels out from every term

4 + 2Pp + 3Pp = 19
5Pp = 15
Pp = 3

Question 149

A group consisting of several families visited an amusement park where the regular admission fees were ¥5,500 for each adult and ¥4,800 for each child. Because there were at least 10 people in the group, each paid an admission fee that was 10% less than the regular admission fee. How many children were in the group?

(1) The total of the admission fees paid for the adults in the group was ¥29,700.

(2) The total of the admission fees paid for the children in the group was ¥4,860 more than the total of the admission fees paid for the adults in the group.

Answer 149

Inequality with prices and integers
> be careful with inequality (may not be sufficient)

C + A >= 10
48000.9 = 4320 is the price for each child
55000.9 = 4950 is the price for each adult

C = ?

(1) 4950*A = 29700
A = we know
But without total N, we cannot know value of C

(2) 4320C - 4950A = 4860
We could plug in a value for A into C+A >=10
But we end up with an inequality again
Multiple value of C work here

(3) Together from 1 we know A, and from 2 we know C

Sufficient

Question 150

A committee recently held a planning meeting for a craft fair. The committee decided that the fair would occur on a Tuesday and that a public discussion about the fair would be held sometime after, also on a Tuesday.

In the table, select for Fair a number of days between the planning meeting and the fair and select for Discussion a number of days between the planning meeting and the public discussion that would be jointly consistent with the given information, Make only two selections, one in each column.

260
271
296
306
311

Answer 150

Multiples
> given weekday intervals (every 7 days it repeats the same day of the week)
> “number of days BETWEEN A and B” = Number of days not including end points

Let X = number of days between Planning Meeting and the Fiar
Let X + 7int = number of days between Planning Meeting and the Public Discussion
> Fair and Public Discussion have 7int - 1 days in between
> Planning and Public Discussion must have x + 1 + 7int - 1 = x + 7int days in between

Looking for X and X+7*int

Only 271 and 306 work together (separated by 35 days which is a multiple of 7)

Question 151

Kylie invested a certain amount of money at r% yearly interest compounded at the end of each year and the same amount of money at (r + 2)% yearly interest compounded at the end of each year. What was the amount of money that Kylie invested at the (r + 2)% yearly interest rate?

(1) At the end of 1 year, the investment at the (r + 2)% yearly interest rate earned $200 more than the investment at the r% yearly interest rate.

(2) r = 3

Answer 151

Compound interest
> over ONE period, the amount of compound interest = Starting Value * (rate per period)

Looking for investment amount , given t = 1 and n =1

(1) Interest at (r+2)% = Interest at r% + 200
X(r+2)/100 = X(r/100) + 200
X(r+2)/100 = (Xr/100) + 20000/100
(Xr+x2)/100 = (Xr + 20000)/100
Xr + X2 = Xr + 20000 —–> Xr cancels out
2X = 20000
X = 10000 (Sufficient)

(2) r = 3 —> NS without interest amount

A

Question 152

Each of the 20 people working in a certain office contributed either $9, $10, or $11 toward an office party. What was the average (arithmetic mean) amount contributed per person in the office?

(1) The number of people who contributed $9 was the same as the number of people who contributed $11.
(2) The number of people who contributed $9 was more than the number of people who contributed $10.

Answer 152

Word problems
> remember - always express as many equations as possible
20 = A + B + C (Hidden C trap)

Looking for (9A + 10B + 11C)/20

(1) A = C
So 20 = A + B + A = 2A + B
and (20A + 10B)/20
= 10*(2A+B)/20 —-> CAN SUB IN 2A + B= 20
= 10
Sufficient

(2) A > B
NS without actual quantities
e.g., A=5, B=1, C = 14
or A = 5, B=2, C=13 —-> produces different numerators and therefore different averages

A

Question 153

Roses: $1 each
Daisies: $0.50 each

Kim and Sue each bought some roses and some daisies at the prices shown above. If Kim bought the same total number of roses and daisies as Sue, was the price of Kim’s purchase of roses and daisies higher than the price of Sue’s purchase of roses and daisies?

(1) Kim bought twice as many daisies as roses.

(2) Kim bought 4 more roses than Sue bought.

Answer 153

Complex weighted average inequality
> two sums are EQUAL to each other, but relative proportion of each component might not be equal
> this affects both the total cost as well as average cost

Strategy: reflect the shared TOTAL SUM as “N” and sub in for two of the variables (will cancel out)

LOGIC:
> if one person bought more Roses, then R% > D%
> if one person bought more Roses than the other person, then their weighted average cost per item will ALSO INCREASE (since total N is the same, total cost of the basket will also increase)
> DEFAULT: assume equal number of roses and daisies were bought and see if this changes

(1) not helpful because we don’t know anything about Sue’s purchases relative to Kim’s purchases
> Sue could have bought the same number of roses and daisies or fewer roses and more daisies than Kim

(2) Means that weighted average price per item in Kim’s purchase is GREATER than Sue’s purchase –> Kim’s purchase is more expensive than Sue’s

Question 154

Of the 66 people in a certain auditorium, at most 6 people have their birthdays in any one given month. Does at least one person in the auditorium have a birthday in January?

(1) More of the people in the auditorium have their birthday in February than in March.

(2) 5 of the people in the auditorium have their birthday in March.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.

Answer 154

Birthday month questions - testing dispersion of people ACROSS MONTHS
> with 66 people, they can ALL AVOID having birthdays in January if each of the 11 other months have 6 people

(1) since F > M, this means that March must have < 6 people
> means January must have at least one person (given all other months are maxed out)

Sufficient

(2) similar reasoning to (1)

D

Question 155

A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. If n is a positive integer, for which of the following values of k is ​25 10 to the nth power + k10 raised to the 2n power​ divisible by 9 ?

9
16
23
35
47

Answer 155

Divisibility rules
> need to plug in answer choices
> also need to assume n can be equal to 2 (or any other positive integer)

Lets assume n = 2 then factor:
10^2 (25 + k10^2)
2+5+k digits must be a multiple of 9

Ans 47
(2+5+4+7 = 18)

Question 156

What is the greatest positive integer n such that 5^n divides 10! – (2)(5!)^2?

Answer 156

Factors –> always need to make number a PRODUCT (not sum or difference)

10! – (2)(5!)^2
= 5!(109876 - 25!)
=5!3!(10987 - 254)
=5!3!4(10927 - 25)
= 5!3!425(927-1)
= 5!3!425(125)
= 5!3!425(125)
= 5!3!425(5^3)
=5!5^43!8

How many 5s are in this product?
= 1 + 4
= 5 5’s

(no need for factorial shortcut cause we can count the number of 5’s pretty easily)

Question 157

Half of a large pizza is cut into 4 equal-sized pieces, and the other half is cut into 6 equal-sized pieces. If a person were to eat 1 of the larger pieces and 2 of the smaller pieces, what fraction of the pizza would remain uneaten?

Answer 157

Fraction word problems:

Let 1 = 1 pizza

1/2 pizza is cut into 4 equal-sized pieces: 1/2 divide by 4
= 1/2 * 1/4 = 1/8 per slice

1/2 pizza is cut into 6 equal-sized pieces: 1/2 divide by 6
= 1/2 * 1/6 = 1/12 per slice

If a person eats: 1/8 pizza + 2/12 pizza = (3+4)/24 = 7/24 pizza would be eaten

So 1 pizza - 7/24 pizza remains
= 17/24 pizza remains

Question 158

A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when nm is divided by 18?

Answer 158

Remainder theory
> formula way is not very helpful since we cannot get a clean remainder **
> so should instead MULTIPLY n and m then find the remainder when the product is divided by 18

n = 9int + 6
m = 12int + 4

nm = (9int + 6)(12*int + 4) —-> THEN EXPAND
nm = 108int + 36int + 72int + 24
nm/18 = int + int + int + 24/18
nm/18 = int + int + int + 1 + 6/18

so the remainder is 6

Question 159

A clothing retailer used to sell only “fast-fashion” pieces, which were low priced and had a profit markup of 50 percent of the per-item cost (including, for example, the costs of wholesale purchase and marketing). On average, each customer spent $850 annually on around 65 such pieces from the retailer. Now the retailer wishes to double its total profits by selling only “classic” pieces. It plans to double its percentage profit markup per item and generate more revenue per customer while leaving unchanged the company’s total costs. The plan assumes that for each classic piece, on average, customers will pay five times what they paid for each fast-fashion piece, and that the total number of customers for the retailer’s clothing products will remain the same.

Statements: Customers paid an average of _____1_____ dollars (rounded to the nearest dollar) for each of the retailer’s fast-fashion pieces. The retailer will need to sell an average minimum of ______2_____ classic pieces per person (rounded to the nearest whole number) to achieve its profit goals for classic pieces.

Select values for 1 and for 2 that create the statements that are most strongly supported by the information provided and in accordance with the retailer’s plan. Make only two selections, one in each column.

Answer 159

DI TPA - profit
> need to recognize that the SELLING PRICES AND UNIT COSTS ARE DIFFERENT
> keep fractions and don’t use calculator until the last step

Profit = (Price - Unit cost)*Quantity sold

Part 1: each customer spends $850 on 65 pieces
> means price per unit = 850/65 = ~$13

Part 2: Goal is to DOUBLE PROFITS
> “double its percentage profit markup PER ITEM” means New Price = 100% markup from cost
New Price = 2New Unit Cost
> “for each classic piece, customers will pay five times what they paid for each fast-fashion piece”
New Price = 5(850/65)

Fast-fashion unit cost? Price = 1.5C
C = P/1.5 = (850/65)*(2/3)

Classic unit cost? Price = 2C
C = P/2 = 5(850/65)0.5

Fast-Fashion PROFIT = (Price-Unit cost)Quantity per person
= (850/65)(1/3)*65 units

DESIRED PROFITS = 2(850/65)(1/3)*65 units

EQUATION with unknown new quantity:
2(850/65)(1/3)65 = 5(850/65)(0.5)Q —-> THINGS CANCEL OUT

Leaving: (265)/3 = 50.5Q
Q = (213*2)/3
Q = 52/3
Q = ~17

Question 160

Yesterday Bookstore B sold twice as many softcover books as hardcover books. Was Bookstore B’s revenue from the sale of softcover books yesterday greater than its revenue from the sale of hardcover books yesterday?

(1) The average (arithmetic mean) price of the hardcover books sold at the store yesterday was $10 more than the average price of the softcover books sold at the store yesterday.

(2) The average price of the softcover and hardcover books sold at the store yesterday was greater than $14.

Answer 160

Inequality word problem
> when you get to evaluating both statements, REWRITE WHAT YOU KNOW you don’t forget

S = 2H

Rs > Rh? → Plug in
PsS > PhH
Ps2H > PhH → H’s cancel
2Ps > Ph?

(1) Ph = 10 + Ps —> plug into question stem
Is: 2Ps > 10 + Ps?
Is: Ps > 10? → NS

(2) (Rs + Rh)/(3H) > 14 → bring in prices and quantity

(PsS + PhH) > 42H
Ps*2H + PhH > 42H → H’s cancel out
2Ps + Ph > 42
2Ps > 42 - Ph → not sufficient to know whether 2Ps > Ph
E.g., Ph = 40 and 2Ps = 4 (ans No)
E.g., Ph = 2 and 2Ps = 100 (ans Yes)

(3) 2Ps > 42 - Ph
Ph = 10 + Ps → rewrite what you know from statements 1 and 2
Plug in

2Ps > 42 - 10 - Ps
3Ps > 32
Ps > 10 + ⅔ → ans Yes based on statement 1 is Ps > 10

Question 161

What is the probability that Lee will make exactly 5 errors on a certain typing test?

(1) The probability that Lee will make 5 or more errors on the test is 0.27.

(2) The probability that Lee will make 5 or fewer errors on the test is 0.85.

Answer 161

Probability DS involving overlap

(1) NS
> cannot isolate P(exactly 5 errors)

(2) NS
> cannot isolate P(exactly 5 errors)

(3) TOGETHER we know from complements
1 = P(at most 5 errors) + P(at least 5 errors) - P(overlap = exactly 5 errors)

Sufficient

Question 162

The figures above show a hexagonal nut that has a width of 1 5/16 inches and a wrench that, in order to fit the nut, must have a width of at least 1 5/16 inches. Of all the wrenches that fit the nut and have widths that are whole numbers of millimeters, the wrench that fits the nut most closely has a width of how many millimeters?

Note 1 inch = ~25.4 millimeters

(FIGURE shows Nut = 1 + 5/16 inch)

30
31
32
33
34

Answer 162

Decimal and UNIT CONVERSION question (NOT approximation question)
> “wrench that fits the nut most closely has a width of how many millimeters?”
> Need to determine the width of the NUT in millimeters and check answers that can FIT THE NUT (closest distance to 0 but has to be BIGGER than the nut)

CONSTRAINT: Width of wrench >= Nut’s width in mm

1) Nut’s width in mm? —> Don’t waste time being afraid of decimal operations … JUST DO IT using speed math
= 25.4 + 5/16(25.4)
= 25.4 + 5/8(12.7)
= 25.4 + 63.5/8
= 25.4 + 7.93xx
= 33.33 mm

2) compare to width of wrenches
33, 32, 31, 30 are all TOO SMALL

34 is the only other answer

Question 162

Rita and Sam play the following game with n sticks on a table. Each must remove 1, 2, 3, 4 or 5 sticks at a time on alternate turns, and no stick that is removed is put back on the table. The one who removes the last stick (or sticks) from the table wins. If Rita goes first, which of the following is a value of n such that Sam can always win no matter how Rita plays?

7
10
11
12
16

Answer 162

Hard “game” question –> related to MULTIPLES
> didn’t know what this was testing

So when Rita starts, Sam can complement her move each time (5-1, 4-2, 3-3, 2-4, 1-5) and leave her with 6 sticks at the end if the total number of sticks is a multiple of 6. There is only one multiple of 6 in the options.

NEED TO UNDERSTAND THE “MUST BE TRUE” CONDITION FOR “winning”
> So the GOAL of the game is to leave 6 sticks for your opponent –> guarantees your win (they can be 1 through 5 sticks, leaving <=5 left over for you to win)

7 –> R could pick 1 and leave you with 6 –> you don’t win
10 –> R could pick 4 and leave you with 6 –> you don’t win
11 –> R could pick 5 and leave you with 6 –> you don’t win

12 –> R could pick 1 and leave you with 11, then you pick 5 and R ends up with 6 –> You WIN
16 –> R could pick 5 and leave you with 11, then you pick 5 and R ends up with 6 –> YOU WIN

BETWEEN 12 and 14: see if a counter case exists

16 –> R could pick 4 and leave you with 12, then you pick 5 and leave R with 7, then R picks 1 and leaves you with 6 –> YOU LOSE

12 –> R could pick 5 and leave you with 7, you pick 1 and leave R with 6 –> YOU WIN
R could pick 4 and leave you with 8, you pick 2 and leave R with 6 –> YOU WIN
R could pick 1 and leave you with 11, you pick 5 and leave R with 6 –> YOU WIN

Regardless of what R picks, you can always win

Question 163

Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array?

Answer 163

Counting x factors x even and odd

Looking for integer solutions to n^2 - k^2 = 48

Equivalent to:
(n+k)*(n-k) = 48

ACRONYM: CFE

1: Prime factorize 48 = 86 = 2^3 * 23 = 2^4 * 3
2: number of factors = 5*2 = 10 factors
3: list out factors
1, 48
2, 24
3, 16
4, 12
6, 8

4: Use even/odd properties to shortlist potential integer pairs ***
n+k and n-k have the same properties, and 48 needs at least 1 even factor
> must be even * even
> short list: 2x24, 4x12, 6x8

5: Test to check

n+k = 24
n-k = 2
——-
2n-2 = 24
n = 13 and k = 11 (satisfies constraint)

n+k = 2
n-k = 24
—–
2n-24 = 2
2n = 26
n=13 and k = - (not possible)

n+k = 12
n-k = 4
—-
2n - 4 = 12
2n = 16
n = 8 and k = 4 (satisfies constraint)

n+k=8
n-k = 6
—
2n - 6 = 8
2n = 14
n = 7 and k = 1 (satisfies constraint)

n+k=6
n-k=8
—-
2n-6=8
2n= 14
n=7 and k = - (not possible)

So in total there are only 3 possible solutions

Question 164

A certain manufacturer uses the function C(x) = 0.04x2 – 8.5x + 25,000 to calculate the cost, in dollars, of producing x thousand units of its product. The table above gives values of this cost function for values of x between 0 and 50 in increments of 10. For which of the following intervals is the average rate of decrease in cost less than the average rate of decrease in cost for each of the other intervals?

x, C(x)
0, 25000
10, 24919
20, 24846
30, 24781
40, 24724
50, 24675

Options:
A. From x = 0 to x = 10
B. From x = 10 to x = 20
C. From x = 20 to x = 30
D. From x = 30 to x = 40
E. From x = 40 to x = 50

Answer 164

Functions
> Average Rate of Change = SLOPE BETWEEN TO POINTS
> doesn’t matter if asking for Average Rate of DECREASEor INCREASE (still treat as “CHANGE”)

Average Rate of Decrease in Cost “less than” the average rate of decrease in cost for other intervals = looking for MOST SHALLOW SLOPE over that interval (closest to 0 in value or smallest absolute value of slope)

Since the intervals are all the same (10), just look at the Change in C(x)

Interval from 40 to 50 has the smallest decline = smallest absolute value of slope

Question 165

If C is the temperature in degrees Celsius and F is the temperature in degrees Fahrenheit, then the relationship between temperatures on the two scales is expressed by the equation 9C = 5(F – 32). On a day when the temperature extremes recorded at a certain weather station differed by 45 degrees on the Fahrenheit scale, by how many degrees did the temperature extremes differ on the Celsius scale?

Answer 165

Functions
> given a CHANGE in one of the variables, asked about the corresponding CHANGE in the other variables

You can use smart numbers here because we just need two temperatures that differ by 45 degrees
e.g., F1 = 55 and F2 = 100

This works because we end up with
Change C = 5/9*(45) = 25

OR you can set up equation: F2 - F1 = 45 and then solve for C2 - C1

Question 166

A certain truck traveling at 55 miles per hour gets 4.5 miles per gallon of diesel fuel consumed. Traveling at 60 miles per hour, the truck gets only 3.5 miles per gallon. On a 500-mile trip, if the truck used a total of 120 gallons of diesel fuel and traveled part of the trip at 55 miles per hour and the rest at 60 miles per hour, how many miles did it travel at 55 miles per hour?

Answer 166

Hard PS
> Fuel efficiency = Miles Per Gallon Consumed

Since you are looking for miles, set that as the variable
(instead of first calculating hours then calculating distance r*T)

Number of Gallons = Miles / mpg

120 = A/4.5 + (500-A)/3.5 ——> deal with decimals in denominator by multiplying top and bottom by powers of 10

A = 360 miles

Question 167

A merchant paid $300 for a shipment of x identical calculators. The merchant used two of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?

A. 24
B. 25
C. 26
D. 28
E. 30

Answer 167

Hard PS
Other Profit questions
> when asked to calculate QUANTITY, you can plug in answer choices instead of going through complex factoring

Revenue = 120 + Cost
Revenue = 120 + 300
Revenue = 420

Revenue = Price * Quantity
420 = (5 + 300/x) * (x - 2)

Quantity must be an integer, so Quantity - 2 must also be an integer

Therefore 5 + 300/x must also be an integer so 300/x must be an integer

Ans E

Question 168

A car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city. If the car traveled 6 fewer miles per gallon in the city than on the highway, how many miles per gallon did the car travel in the city?
A. 14
B. 16
C. 21
D. 22
E. 27

Answer 168

Hard PS
> Fuel efficiency = Miles Per Gallon Consumed

We are looking for MILES PER GALLON = Miles / Tank * Tank / Gallons

Let 1 tank = x gallons —> rate becomes 1 tank / x gallons

WATCH OUT FOR CITY VS HIGHWAY NUMBERS:

City’s mpg = Highway’s mpg - 6
336/x = 462/x - 6
x = 21

City’s mpg = 336/21 = 16

Question 169

The annual stockholders’ report for Corporation X stated that profits were up 10 percent over the previous year, although profits as a percent of sales were down 10 percent. Total sales for that year were approximately what percent of sales for the previous year?

Answer 169

Complex percents and ratios
> whenever you are given TIME PERIODS, write it out in a table

Looking for “PERCENT OF SALES”
= (Sales for year)/(Sales for previous year) * 100

A1 = Profit Previous Year
A2 = Profit This Year
B1 = Sales Previous Year
B2 = Sales This Year

A2 = A1*(1.1)

(A2/B2) = 0.9*(A1/B1)

What is B2 / B1 * 100 =?

ALGEBRA
= 1.1*(10/9)
= 11/9
= 1 + 2/9
= 1.22

Ans 122%

Question 170

A certain brand of house paint must be purchased either in quarts at $12 each or in gallons at $18 each. A painter needs a 3-gallon mixture of the paint consisting of 3 parts blue and 2 parts white. What is the least amount of money needed to purchase sufficient quantities of the two colors to make the mixture?
(4 quarts = 1 gallon)
A. $54
B. $60
C. $66
D. $90
E. $144

Answer 170

Ratios x unit conversion x integer
> Cannot buy mixture of paint (need to buy paint separately)
> Cannot buy fractional gallons or fractional quarts –> need to round up to nearest gallon or quart

Since we are trying to MINIMIZE the amount of money spent, we need to evaluate BOTH OPTIONS (is it cheaper to buy a certain quantity of Blue paint at $12 per quart or $18 per gallon)

(1) Determine quantities of blue and white paint needed (ratio component)
3x + 2x = 5x

5x = 3 gallons, so x = 3/5

Blue paint needed = 9/5 gallons = 1 gallon + 4/5 gallon
White paint needed = 6/5 gallons = 1 gallon + 1/5 gallon

(2) Determine lowest cost options for EACH COLOUR SEPARATELY

Blue paint: 1 gallon –> $18 or $48 for 4 quarts —> $18
4/5 gallon —> $18 for 1 gallon or $48 for 4 quarts –> $18

White paint: 1 gallon –> $18
1/5 gallon –> $18 for 1 gallon or $12 for 1 quart –> $12

So total cost = 18 + 18 + 18 + 12 = $66

Question 171

Feb = +10%
March = -15%
April = +20%
May = -10%
June = +5%

The table above shows the percent of change from the previous month in Company X’s sales for February through June of last year. A positive percent indicates that Company X’s sales for that month increased from the sales for the previous month, and a negative percent indicates that Company X’s sales for that month decreased from the sales for the previous month. For which month were the sales closest to the sales in January?
A. February
B. March
C. April
D. May
E. June

Answer 171

Percent change –> but not given a starting number –> ASSUME STARTING number is 100 (Smart Numbers)

Feb: 1001.1 = 110
March: 1100.85 = 93.5
April: 93.51.2 = 112.2
May: 112.20.9 = 100.98 ** closest to 100
June: 100.981.05 > 100.98

Question 172

What is the following expression equal to?

2/(3x) - 2/(x+y)*[ (x+y)/(3x) - x - y) ] / (x-y)/x

Answer 172

PEMDAS and being careful when cancelling like terms in the numerator and denominator

ans: 2x/(x-y) —–> NOT 2/(x-y)

Question 173

What is the following expression equal to?

1/(1-x) + 1/(1+x) + 2/(1+x^2) + 4/(1+x^4)

Answer 173

Fractions simplification using quadratic identities in denominator
> NOTICE some patterns
(1) difference of square in the first two terms
= (1-x)*(1+x) = 1 - x^2

(2) Notice powers on the x term increase by 2 —> ANOTHER DIFFERENCE OF SQUARE
= “inching” approach

= 2/(1-x^2) + 2/(1+x^2) + 4/(1+x^4)
= (2 + 2x^2 + 2 - 2x^2)/(1 - x^4) + 4/(1+x^4)
= 4/(1-x^4) + 4/(1+x^4)
= (4 + 4x^4 + 4 - 4x^4)/(1-x^8)
= 8/(1-x^8)

Question 174

If a + 1/a = 3, what is the following expression equal to: a^2 / (a^4 + 1)?

Answer 174

Recognize special quadratic identity: square of a sum of reciprocals

(a + 1/a)^2 = a^2 + 2 + 1/a^2 = 9
a^2 + 1/a^2 = 7

BUT MOST IMPORTANTLY, know that you can set the original fraction as T and then calculate 1/T
> it is always easier to deal with fractions where the denominator is a SINGLE TERM than a sum

SOLVE FIRST: 1/T = (a^4 + 1)/a^2
= a^2 + 1/a^2
= 7

So 1/T = 7 and T = 1/7

Question 175

A merchant purchased a jacket for $60 and then determined a selling price that equaled the purchase price of the jacket plus a markup that was 25 percent of the selling price. During a sale, the merchant discounted the selling price by 20 percent and sold the jacket. What was the merchant’s gross profit on this sale?

Answer 175

Profit/Loss
> TRUST the GMAT’s wording (“SELLING PRICE that equaled the purchase price of the jacket plus a markup that was 25 percent of the SELLING PRICE”) ****

P = 60 + 0.25P
0.75P = 60 —> CAN SOLVE FOR P
P = 80

Revenue = 80*0.8 = 64
Cost = 60

Gross Profit = 4

Question 176

A factory assembles Product X from three components, A, B, and C. One of each component is needed for each Product X and all three components must be available when assembly of each Product X starts. It takes two days to assemble one Product X. Assembly of each Product X starts at the beginning of one day and is finished at the end of the next day. The factory can work on at most five Product Xs at once. If components are available each day as shown in the table above, what is the largest number of Product Xs that can be assembled during the three days covered by the table?

TABLE: Monday
A = 3
B = 6
C = 4

Tuesday
A = 6
B = 3
C = 7

Wednesday
A = 3
B = 4
C = 4

Answer 176

Constrained resource production problem
> solve ONE DAY AT A TIME

“One of each component is needed for each Product X and all three components must be available when assembly of Product X starts” —> need 1 A, 1 B and 1 C to produce 1 X

“It takes two days to assemble one Product X. Assembly of each Product X starts at the beginning of one day and is finished at the end of the next day” —> Two days to assemble (so a product that starts production on Monday won’t be finished until Tuesday night)

“The factory can work on at most five Product Xs at once” —> max number of simultaneous production “machines” is 5

On Monday:
> 3 parts A can produce up to 3 units of X
> 6 parts B can produce up to 6 units of X
> 4 parts C can produce up to 4 units of X
> therefore, on Monday the factory can assemble 3 units of X which will be finished by Tuesday night
> leaves 2 machines, 3 parts B, and 1 part C

On Tuesday:
> we have more than enough resources A, B, and C, however we only have 2 machines
> so at most, we can produce 2 units of X that will be finished by Wednesday night

Wednesday:
> anything that starts today won’t be finished in time

So max number of units produced over the 3 days = 3 + 2
= 5 units of X

Question 177

How many positive integers n have the property that both 3n and n/3 are 4-digit integers?

Answer 177

“How many” = CFE acronym

Constraints —> n must be a multiple of 3 **** and be a 4 digit integer
E = even/odd or MULTIPLES

Statement 1: 1000 <= 3n <= 9999
and
Statement 2: 1000 <= n/3 <= 9999 —–> n MUST be a multiple of 3

From statement 2 we can create:
3000 <= n < 30000 —> n must be a multiple of 3 so let’s change the range to be inclusive

3000 <= n <= 29,997

From statement 1 we can also get an inequality for n:
1000/3 <= n <= 3333

With two equations for n, create a NEW INEQUALITY with SHARED REGIONS:

3000 <= n <= 3333 —> n must be a multiple of 3

How many multiples of 3 exist in this range?

[3000, 3333]= (3333-3000)/3 + 1 = 111 + 1 = 112

Question 178

If Whitney wrote the decimal representations for the first 300 positive integer multiples of 5 and did not write any other numbers, how many times would she have written the digit 5?

Answer 178

Counting digits - count in each place value (units digit, tens digit, hundreds digit)

[5, 10 …, 1500]

Units digit: alternates between 0 and 5
= 300 positive integers / 2
= 150 5’s in the unit digit

Tens digit: 50, 55 –> twice in every 100s
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 –> 10* 2= 20 5’s in the tens digit

1000s - 0, 1, 2, 3, 4 –> 5 * 2 = 10 5’s in the tens

ALTERNATIVELY …
> [50, 150, 250 ..1450] = (1450-50)/100 + 1 = 15 integers
> [55, 1455] = 15 integers
> 30 5’s in the tens

Hundreds digit: 5xx
500-595 –> 595-500 / 5 + 1 = 20 5’s in the hundreds

1500 –> 1 5’s in the hundreds ** missed this one

TOTAL number of 5’s = 150 + 20 + 10 + 20 + 1 = 201

Question 179

The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7?

Answer 179

Digit place values counting: CFE
Remember –> abc = 100a + 10b + c
> need to create a PRODUCT of FACTORS in order to determine multiple of 7 ****

100a + 10b + c - 100c - 10b - a = 7int
99a - 99c = 7int
99(a-c) = 7int
911(a-c) = 7*int

Therefore a-c must be 7 , where a and c [1,9 digits]

a-c could be:
9-2
8-1
(a, b and c are NON ZERO)

Therefore, abc could be: 9b2 and 8b1 –> what can b be?

9b2 = 9 options for b (a, b and c are NON ZERO)
8b1 = 9 options for b
——-
18 options

Question 180

Let S be the set of all positive integers having at most 4 digits and such that each of the digits is 0 or 1. What is the greatest prime factor of the sum of all the numbers in S ?

Answer 180

Counting

List out ALL the numbers: {1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111}
To find the greatest prime factor of the SUM → need to determine the SUM and then prime factorized
SUM equals 8888
= 8*1111
= 2^3 * 11 * 101 → ans: 101

Prime numbers beyond 100 –> 101, 103, 107, 109, 113, 127, 131, 137
> how to tell? see if divisible by primes 13, 17 and 19
> if these primes skip them, then it is a prime

13: 13, 26, 39, 52, 65, 78, 91, 104, 117, 130, 143
17: 17, 34, 51, 68, 85, 102, 119, 136
19: 19, 38, 57, 76, 95, 114

Question 181

A certain online form requires a 2-digit code for the day of the month to be entered into one of its fields, such as 04 for the 4th day of the month. The code is valid if it is 01, 02, 03, …, 31 and not valid otherwise. The transpose of a code xy is yx. For example, 40 is the transpose of 04. If N is the number of valid codes having a transpose that is not valid, what is the value of N?

Answer 181

Counting: CFE
> MISREAD QUESTION (looking for N = number of codes that are NOT VALID *****)

Valid codes: 01 - 09 —-> 10, 20, 30 are valid
10-19 —-> 01, 11, 21, 31 are valid
20-29 —-> 02, 12, 22 are valid
30, 31 —> 03, 13 are valid

12 are valid
31 - 12 = 19 are NOT VALID

Question 182

If x < y < z and y − x > 5, where x is an even integer and y and z are odd integers, what is the least possible value of z − x?

A. 6
B. 7
C. 8
D. 9
E. 10

Answer 182

LEAST possible value
> even/odd properties (x even, y odd, z odd)

z - x must be ODD
> MINIMIZE by minimizing the value of Z and maximizing the value of X (use y as the connector)

z = y + 2
x = y - 1

z - x = y + 2 - y + 1 = 3 (HOWEVER y - x > 5 –> minimum gap is 5)

SO
z = y + 2
y - x = 7 (odd - even = odd)
so x = y - 7

THEREFORE: z - x = y + 2 - y + 7 = 9

Question 183

Five integers between 10 and 99, inclusive, are to be formed by using each of the ten digits exactly once in such a way that the sum of the five integers is as small as possible. What is the greatest possible integer that could be among these five numbers?

A. 98
B. 91
C. 59
D. 50
E. 37

Answer 183

Max integer

[10, 99] –> five integers

“using each of the tens digits exactly once” —> which tens digit? MISREAD ****
> “TEN” digits = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Minimize sum —> smallest tens, largest unit digit:
WANT LARGEST possible integer

MINIMUM SUM = 180 occurs when 1x, 2x, 3x, 4x, and 5x as the tens and 6, 7, 8, 9, 0 as units digit –> doesn’t matter where we place the single digit integers ** should have checked that the minimum DOESN’T CHANGE

1x, 2x, 3x, 4x, 5x
19, 28, 37, 46, 50

What about: 10, 26, 37, 48, 59 (also equals 180!)

so 59 is the largest such number (NOT 50)

Question 184

In an auditorium, 360 chairs are to be set up in a rectangular arrangement with x rows of exactly y chairs each. If the only other restriction is that 10 < x < 25, how many different rectangular arrangements are possible?

Answer 184

how many = CFE = counting
> integers

xy = 360 = 252^23^2 = 2^33^25

360 has a total of 432 = 24 factors

10 < x < 25 is another constraint: so START HERE TO SAVE TIME

12 and 30
15 and 24
18 and 20
20 and 18
24 and 15

5 possible arrangements

Question 185

Last year ​three fifths​ of the members of a certain club were males. This year the members of the club include all the members from last year plus some new members. Is the fraction of the members of the club who are males greater this year than last year?

(1) More than half of the new members are male.

(2) The number of members of the club this year is ​six fifths​ the number of members last year.

Answer 185

Fraction ratio and group composition
> we care about the COMPOSITION OF THE NEW MEMBERS –> are > 60% of the new members men?
> related to weighted average logic (if the composition of the new members is at least 60%, then the fraction of members who are males will be greater than or equal to 60%)

PROOF:
(0.6L + x)/(L+N) > 0.6?
Becomes:
x > 0.6N? (Is the number of new male members greater than 0.6 of the new additions)?

(1) x > 0.5N (NS)

(2) does not tell me anything about the composition of the new members

(3) together still not sufficient
E

Question 186

At a certain company, 30 percent of the employees live in City R. If 25 percent of the company’s employees live in apartments in City R, what is the number of the employees who live in apartments in City R?

(1) Of the employees who live in City R, 6 do not live in apartments.

(2) Of the employees, 84 do not live in City R.

Answer 186

Sets - misread one condition “25% of the company’s employees live in apartments in City R”

(1) 0.05N = 6 → N we know and therefore we know X (Sufficient)

(2) 0.7N = 85 → N we know and therefore we know X (Sufficient)

Question 187

Each Monday through Friday (a workweek), Avinash will bring either exactly one apple or exactly one banana with him to his workplace for an afternoon snack. To avoid having to decide which to bring, each morning Avinash will toss a coin with a face on exactly one side that is equally likely to land faceup or facedown. If the coin lands faceup, then he will bring an apple, and if the coin lands facedown, then he will bring a banana. Avinash correctly determined the probability that, for a given workweek, either he would bring an apple on at least 4 consecutive days or he would bring a banana on at least 4 consecutive days. This probability was m divided by n.

Select for m and for n values jointly consistent with the given information. Make only two selections, one in each column.

Answer 187

Probability TPA
> “at least 4 CONSECUTIVE DAYS” –> LINK METHOD

AAAAB or BAAAA are the only two orders (not 5 cases)

Probability(at least 4 consecutive days he would bring an apple)
= P(AAAAB + BAAAA + AAAAA)
= (½)^5 * 3

Probability(at least 4 consecutive days he would bring an orange)
= = P(BBBBA + ABBBB + BBBBB)
= (½)^5 * 3

SUM = 6/32

Question 188

OGMOCK: Kesha paid a sales tax of x percent on her purchase. If the sales tax had been only 2 percent, she would have paid $27 less in sales tax on her purchase. What was the value of x if the total amount Kesha paid for her purchase, including sales tax, was $486?

8
7
6
5
4

Answer 188

Word problems: sales tax and price
> “$27 less in sales tax if the sales tax had only been 2 percent”

2% sales tax = X% sales tax - 27

End up with: 486 = (2700/(x-2)) * (1 + x/100)

APPROACH 1: Solve for x algebraically (it’s OK that x is in the denominator, don’t be afraid! Cross multiply)
> then use long division

APPROACH 2: Plug in answers for x, starting from the top to finish

Ans 8

Question 189

OGMOCK: How many different positive two-digit integers are there such that the sum of the two digits is a prime number greater than 11?

4
5
6
7
8

Answer 189

Digits and number properties
> whenever you see digits write [0,9] or [1,9]
> read carefully what you are being asked to solve (digits vs integer)

in this case, we have a two-digit integer: ab, and want to know how many ab exist such that a+b > 11 and is a prime number

Ans 8

> max sum a+b is 9+9 = 18
min sum a+b is 1+0 = 1

Question 190

OGMOCK: Roy sells homegrown cucumbers and peppers at the local farmer’s market. He charges one price for each cucumber and another price for each pepper. If Roy charges $6.00 for 3 cucumbers and 8 peppers and $3.20 for 2 cucumbers and 4 peppers, how much does he charge for 3 cucumbers and 5 peppers?

Answer 190

Linear equations (unknown represents PRICE)
> no short cut with combos, have 2 unknowns and 2 equations, so we can solve for price of cucumbers and price of peppers

Eq’n 1: 6 = 3C + 8P
Eq’n 2: 3.2 = 2C + 4p —> becomes 6.4 = 4C + 8P

Elimination: eq’n 2 minus eq’n 1
0.4 = C
Therefore P = 0.6

So 3C + 5P = 30.4 + 50.6 = 1.2 + 3 = $4.20

Question 191

OGMOCK: A certain wire with a constant mass-to-length ratio has a mass of x grams per y centimeters of its length. Which of the following is the mass of this wire, in kilograms, per meter of length?

Answer 191

Unit conversions –> line up the units and unit conversion factors
> COMMON KNOWLEDGE: 100 cm = 1 meter and 1000 grams = 1 kg
> Other common unit conversion factors: 1000 meters = 1 kilometer, 1 liter = 1000 ml, 1 meter = 1000 mm

x grams / y centimeters * 100 centimeters / 1 meter * 1 kg / 1000 grams
= 100x / 1000y
= x/10y

Question 192

OGMOCK Given that 1/(n(n+1)) - 1/((n+1)(n+2)) = 2/((n)(n+1)(n+2)), which of the following is equal to (229)/(567) + (229)/(678) + (229)/(789) + … + (229)/(282930)?

Answer 192

Complex sequence patterns –> USE THE EQUATION and try to match to the numbers (probably something will cancel out *******

Notice that we have a 2 in the numerator and three terms in the denominator —> re-express each term as a subtraction

BUT FIRST factor out 29

= 29[ 1/(56) - 1/(29*30)]
= 29/30 - 1/30
= 28/30
= 14/15

Question 193

OGMOCK: What amount, in dollars, invested for one year at an interest rate of 2 percent compounded semiannually would produce the same final balance at the end of the year as $10,000 invested for one year at an interest rate of 4 percent compounded quarterly?

Answer 193

Compound interest
> be careful with multiplication of decimals –> properly do long multiplication!

A = unknown amount of initial investment

Semi annual compounding:
n = 2 periods in a year
2% —> 1% per period

Quarterly compounding:
n = 4 periods in a year
4% –> 1% per period

Final balances are equal:
10000(1.01)^4 = A(1.01)^2
A = 10000*(1.01)^2

1.01^2 = 1.011.01 = 1.0201
> either do long multiplication (101101*10^-4)
> or square of sum (1 + 0.01)^2

therefore:
A = 10201

Question 194

OGMOCK: If n is a positive integer, which of the following could be equal to (n+1)^3 - (n^3 + 1)?

(7^3)(6^3 - 1)
(7^3)(6^3 + 1)
22627
33132
0

Answer 194

“Could be equal to” –> looking for NUMBER PROPERTIES
> watch out for factoring cubes

(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

Therefore:
(n^3 + 3n^2 + 3n + 1) - n^3 - 1
= 3n^2 + 3n
= 3n*(n+1) —> two consecutive integers (must be EVEN) and a MULTIPLE OF 3

Leaves:
> 22627 and 33132

3n(n+1) matches 33132 better (a product of two consecutive integers TIMES 3) *******
> n = 31 and n+1 = 32

Question 195

OGMOCK: The value of the expression (20!10!)/(25!*5!) is an integer that ends in how many zeros?

Answer 195

Number of trailing zeros depends on the number of complete 2*5 pairs

First: simplify into one product
> easiest to break 10! first before 20!

20! * 126

Next, to determine the number of complete 2*5 pairs, we need to focus on the CONSTRAINED number (5 - there are fewer 5s than there are 2s in a factorial)

How many 5s are in 20!?
20/5 = 4 5’s

How many 5s are in 126? Zero

Therefore there are at most 4 trailing zeros

Question 196

OGMOCK: The revenue of a small company increased from x dollars in 2000 to z dollars in 2002. If the percent increase in the company’s revenue from 2000 to 2001 was equal to the percent increase in the company’s revenue from 2001 to 2002, what was the company’s revenue in 2001 in terms of x and z?

Answer 196

Percent change and equations
> since we have 3 time periods, WRITE THEM OUT CLEARLY

Let y = percent increase between any consecutive year

Let R = revenue in 2001

Equation 1: x(1+y) = R
Equation 2: R(1+y) = z

We WANT value of R: –> sub in (1+y)

(1+y) = R/x

Therefore: R*(R/x) = Z
R^2 = xz
R = sqrt(xz) —> R >0

Question 197

OGMOCK When fully burned, natural gas produces approximately 1000 BTU of heat per cubic foot of fuel, and propane produces at 2500 BTU of heat per cubic foot of fuel. For a furnace in which either natural gas or propane may be burned as fuel, the efficiency for a given fuel is the usable heat energy produced when the fuel is burned in the furnace, expressed as a percentage of the total heat energy produced when the fuel is fully burned.

Kaisa will purchase a furnace whose efficiency with respect to either natural gas burned alone or propane burned alone is 90%. In the table, select for Natural Gas Cost Per Cubic Foot and for Propane Cost Per Cubic Foot the values that are jointly consistent with the given information for which the fuel cost per BTU of usable heat energy produced by this furnace would be approximately the same for each fuel burned alone.

$0.0035
$0.0070
$0.0110
$0.0175
$0.0350

Answer 197

TPA - unit measurement conversion

NG: 1000 BTU / 1ft^3 or 900 usable BTU / 1ft^3
P: 2500 BTU/ft^3 and 2250 usable BTU / 1 ft^3

UNIT CONVERSION CHAIN:
$/ft^3 * ft^3 / usable BTU = $ / usable BTU

Therefore: “jointly consistent” - N = natural gas cost per cubic foot; P = propane cost per cubic foot
N(1/900) = P(1/2250)
P/N = 2.5

TEST METHODOLOGICALLY –> P = 2.5*N
> so plug in from top to bottom values of N then see if corresponding value of P is also in the table

When N = $0.0070 and P = $0.0175 satisfies this ratio

Question 198

OGMOCK: The difference between Larry’s height and John’s height is twice the difference between Larry’s height and Ken’s height. If Larry is the tallest of these 3 men, what is the average (arithmetic mean) height of these 3 men?

(1) Ken’s height is 180 centimeters
(2) Larry’s height is 190 centimeters

Answer 198

Absolute value (distance)
|L - J| = 2*|L - K|
BUT since Larry is the tallest of these 3 men, we know L > J and L > K and L-J > 0 and L-K > 0

So equation becomes: L-J = 2*(L-k) —> SIMPLIFY
L-J = 2L-2K
L = 2K-J

Arithmetic mean = SUM / 3
= (L+J+K)/3
= (2K - J + J + K)/3
= 3K / 3
= K

(1) K = 180
Therefore L = 360 - J

MEAN = (360 - J + J + 180) / 3
= SUFFICIENT

(2) L = 190
190 = 2K-J
J = 2K - 190

MEAN = (190 + J + K)/3
= (190 + 2K - 190 + k)/3
= 3K/3
= K (not sufficient)

A

Question 199

In a certain learning experiment, each participant had three trials and was assigned, for each trial, a score of either –2, –1, 0, 1, or 2. The participant’s final score consisted of the sum of the first trial score, 2 times the second trial score, and 3 times the third trial score. If Anne received scores of 1 and –1 for her first two trials, not necessarily in that order, which of the following could NOT be her final score?

-4
-2
1
5
6

Answer 199

Word problems:
> let a = first score, b = second score, c = third score where each score can be {+/-2, +/-1, 0}

Final score = a + 2b + 3c

Anna received eitehr:
1 - 2 + 3c or -1 + 2 + 3c as her total score

3c - 1 or 3c + 1 —-> multiple of 3 minus or plus 1

THEREFORE, her score CANNOT BE a multiple of 3 —> 6

Question 200

In a certain sequence, each term after the first term is one-half the previous term. If the tenth term of the sequence is between 0.0001 and 0.001, then the twelfth term of the sequence is between …

Answer 200

Sequences and decimal estimation (inequality)

an = 1/2 * an-1

0.0001 < a10 < 0.001
Therefore a11 = 1/2(a10)

0.0001/2 < a11 < 0.001/2
0.0001/4 < a12 < 0.001/4

0.0001/4 = (110^-4)/4 = 0.25 * 10^-4 (4 leading 0s) = 0.000025
0.001/4 = (110^-3)/4 = 0.25 * 10 ^-3 (3 leading 0s) =0.00025

Question 201

The United States mint produces coins in 1-cent, 5-cent, 10-cent, 25-cent, and 50-cent denominations. If a jar contains exactly 100 cents worth of these coins, which of the following could be the total number of coins in the jar?

I. 91
II. 81
III. 76

Answer 201

Word problems with coins
> tricky dealing with multiple variables
> however usually these questions are testing MULTIPLES as well as just feasibility of making a valid combo
> strategy: start with determining the number of the smallest denomination coin, then consider the next ones

100 cents = 1p + 5n+ 10d + 25q+50h
1p = 5n + 10d + 25q + 50h - 100
p = 5*(n + 2d to 5q + 10 h - 20) —-> number of pennies is a multiple of 5

I. is 91 coins possible?
> could have 90 pennies and 1 dime
> works

II. is 81 coins possible?
> try 80 pennies and 1 20 cent coin (not possible)
> try 75 pennies and 6 coins with a value of 25 cents (next smallest denomination is 5 and 56 = 30 cents) (not possible)
> try 70 pennies and 11 coins with a value of 30 cents (next smallest denomination is 5 and 511 = 55 cents) (not opssible)
> gap keeps increasing
> Not possible

III. is 76 coins possible?
> could have 70 pennies and 6 nickels coins

Question 202

AB
+ BA
——
AAC

In the correctly worked addition problem shown, where the sum of the two-digit positive integers AB and BA is the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the integer AAC?

9
6
3
2
0

Answer 202

Digits [0,9] - all different digits
> units digit of integer AAC –> what is the value of C

USE VALUE FORM:
10A + B + 10B + A = 100A + 10A + C —-> come up with a FACTOR FORM for C
11B = 99A + C
11B - 99A = C

C = 11(B - 9A) —-> C is a multiple of 11 but cannot be greater than 9. So C must be equal to 0

Question 203

A survey of employers found that during 1993 employment costs rose 3.5 percent, where employment costs consist of salary costs and fringe-benefit costs. If salary costs rose 3 percent and fringe-benefit costs rose 5.5 percent during 1993, then fringe-benefit costs represented what percent of employment costs at the beginning of 1993?

Answer 203

Complex Percent Change problems (within a year but with sub parts)
> E = employment costs (E1 and E2)
> S = salary costs (S1 and S2)
> F = fringe benefit costs (F1 and F2)

TOTAL EQUAITONS:
E1 = S1 + F1
E2 = S2 + F2

We are looking for F1 / E1 (“at the beginning of 1993”)
= F1 / (S1 + F1)

Replace E2 = S2 + F2 with E1, S1 and F1…

1.035E1 = 1.03S1 +1.055F1 —> replace E1 = S1 + F1 —-> get a ratio of S1 to F1

1.035S1 + 1.035F1 = 1.03S1 + 1.055F1
0.005S1 = 0.02F1
S1/F1 = 0.02/0.005 —> multiply top and bottom to get rid of decimals
S1/F1 = 20/5
S1/F1 = 4/1
S1 = 4F1

Therefore: F1 / (S1+ F1)
= F1 / (4F1 + F1)
= 1 / 5
= 20%

Question 204

Lee is planning a trip and estimates that, rounded to the nearest 5 kilometers (km), the length of the trip will be 560 km and that, rounded to the nearest ​one fourth​ hour, the driving time for the trip will be 7 hours. If these estimates are correct, then Lee’s average driving speed during the trip will be between __x__ kilometers per hour and _y__ kilometers per hour, where x < y.

From the values given in the table, select for x and for y the values that complete the statement in such a way that the interval between the selected values includes all possible average speeds for Lee’s trip and y – x is minimal. Make only two selections, one in each column.

73
76
79
82
85

Answer 204

TPA involving Rounding and inequalities

Remember:
Rounded Value - Increment < actual value < Rounded Value + Increment

Distance: increment is 5/2 = +/- 2.5
Time: increment is 1/8 = +/- 0.125

557.5 < Actual distance < 562.5

6.875 < Actual time < 7.125

Average rate = Distance / T
> max rate = max distance / min time
> min rate = min distance / max time

557.5 / 7.125 < Average Rate < 562.5 / 6.875
78.25 < Average Rate < 81.81

NEED TO PICK THE SMALLEST RANGE THAT has the calculated range WITHIN IT:

76 < average rate < 82

Question 205

According to a prominent investment adviser, Company X has a 50% chance of posting a profit in the coming year, whereas Company Y has a 60% chance of posting a profit in the coming year.

Select for Least probability for both the least probability, compatible with the probabilities provided by the investment adviser, that both Company X and Company Y will post a profit in the coming year. And select for Greatest probability for both the greatest probability, compatible with the probabilities provided by the investment adviser, that both Company X and Company Y will post a profit in the coming year. Make only two selections, one in each column.

Answer 205

Probability –> actually testing OVERLAPPING SETS
> expected probability is P(A)*P(B)
> but we want LEAST and GREATEST probability that X AND Y post a profit

MIN OVERLAP (spread sets as far as possible) = 10%

MAX OVERAP (smaller set is inside the larger set) = 50%

Question 206

OGMOCK What is the product of all the solutions of (x+2)^2 = |x+2|?

Answer 206

Absolute value equation –> take off the absolute value sign and consider signs then evaluate possible solutions (don’t double count)
> cannot divide both sides by |x+2| because x+2 could be equal to 0

(x+2)^2 = x+2
(x+2)^2 - (x+2) = 0
(x+2)(x+2 - 1) = 0
(x+2)(x + 1) = 0
x = -2, or -1

OR

(x+2)^2 = -(x+2)
(x+2)^2 + (x + 2) = 0
(x+2)(x+2 + 1) = 0
(x+2)(x+3) = 0
x = -2, or -3

SOLUTIONS {-3, -2, or -1}

PRODUCT = -3-2-1 = -6

Question 207

OGMOCK of the 190 bicycle stores in a certain region, 90 percent repair bicycles and 40 percent rent bicycles. If the integer n denotes the number of bicycle stores in the region that both repair and rent bicycles, which of the following indicates all possible values of n?

Answer 207

Overlapping sets - min and max of OVERLAP
> could set up table or just do logic

LOGIC:

MAX overlap is when every “rent” bicycle store ALSO “repairs” bicycles = 40% of 190 = 76

MIN overlap must be 30% of 190 = 57

Question 208

OGMOCK One year ago a window washing service charged $100 for setup and an additional $30 per hour for on-site washing. This year the company charges $20 for setup and an additional $50 per hour for on-site washing. Which of the following is equivalent to the percentage change from last year to this year that the company charges for setup and x hours of on-site washing?

Answer 208

Pricing structure

One year ago: 100 (fixed flat rate) + 30*x hours (P1)

This year: 20 (fixed flat rate) + 50*x hours (P2)

Percent change (P2 - P1)/P1 * 100

= (20 + 50x - 100 - 30x) / (100 + 30x) * 100
= (20x - 80) / (100 + 30x) * 100
= (20(x - 4)) / (10(10+3x)) * 100
= (200*(x-4))/(3x + 10) %

Question 209

A garden center sells a certain grass seed in 5 pound bags at $13.85 per bag, 10 pound bags at $20.43 per bag, and 25 pound bags at $32.25 per bag. If a customer is to buy at least 65 pounds of the grass seed, but no more than 80 pounds, what is the least possible cost of the grass seed that the customer will buy?

$94.03
$96.75
$98.78
$102.07
$105.36

Answer 209

Least Possible Cost (give unit costs) —> buy more of the product with the LOWEST COST PER POUND
> buy in bulk –> more of the 25 pound bag
> still need to test a few cases to guarantee your answer is the minimum

65 <= Pounds <= 80 —-> NOTE: based on multiples, number of pounds must be a multiple of 5 (65, 70, 75, 80 only)

Test: buy 3 25-pound bags (75 pounds)
Cost = 32.25*3 = $96.75

Test: buy 2 25-pound bags plus 1 10-pound bag plus 1 5-pound bag (65 pounds)
Cost = 32.25*2 + 20.43 + 13.85
= 64.50 + 20.43 + 13.85 > 97 > 96.75

Test: buy 2 25-pound bags plus 2 10-pound bags (70 pounds)
Cost = 32.252 + 20.432
= 64.50 + 40.86 > 100 > 96.75

Therefore $96.75 is the lowest cost

Question 210

Each of the nine digits 0, 1, 1, 4, 5, 6, 8, 8, and 9 is used once to form 3 three-digit integers. What is the greatest possible sum of the 3 integers?

Answer 210

Digits - maximize sum by placing the greatest digits in the greatest column, then the next greatest digits in the next column etc.

= 961 + 851 + 840
= 2652

(even if you change the order of the tens and units place, sum remains 2652) — BECAUSE all that matters is WHICH DIGITS ARE IN WHICH PLACE VALUE
> in this case, since we are maximizing sum, we want the GREATEST DIGITS are in the GREATEST COLUMN and the NEXT GREATEST DIGITS are in the NEXT GREATEST COLUMN (order doesn’t matter) and the LEAST GREATEST DIGITS are in the SMALLEST COLUMN

9, 8, 8 MUST be in the hundreds place
6, 5, 4 MUST be in the tens place
0, 1, 1 MUST be in the ones place

Question 211

The toll T, in dollars, for a truck using a certain bridge is given by the formula T = 1.50 + 0.50(x – 2), where x is the number of axles on the truck. What is the toll for an 18-wheel truck that has 2 wheels on its front axle and 4 wheels on each of its other axles?

Answer 211

Functions –> must properly use the function (x-2 not x)

18 wheels = 2 wheels + 4 wheels * N other axels
4 = N other axels
Therefore there are 5 axels in total

T = 1.5 + 0.5(5-2)
= 1.5 + 0.53
= 1.5 + 1.5
= $3.00

Question 212

Pat is reading a book that has a total of 15 chapters. Has Pat read at least ​one third​ of the pages in the book?

(1) Pat has just finished reading the first 5 chapters.

(2) Each of the first 3 chapters has more pages than each of the other 12 chapters in the book.

Answer 212

Linear equations and sub parts
> need to come up with really good test cases that encompass extremes (including extreme DIFFERENCES between chapters)
> change one thing at a time when doing test cases

(1) Not sure how long each chapter is in terms of pages
(2) Not sure how many pages Pat has finished reading
(3) Pat has finished reading the first 5 chapters → do we know how many pages she has read?
DIDN’T FIND TWO OPPOSITE ANSWERS

BECAUSE the next two chapters do NOT need to have the same number of pages as the rest of the book

E.g., first three chapters each has 10 pages and the next two chapters have 1 page each and the remaining 10 chapters have 9 pages each
Total # of pages = 30 + 2 + 90 = 122
Pat has read (30 + 2) = 32 pages < ⅓ of 122

E.g., first three chapters each has 10 pages and other chapters has 1 page
Total # of pages = 30 + 9*1 = 39
Pat has read (30 + 2) = 32 pages > ⅓ of 39

Question 213

If each of the stamps Carla bought cost 20, 25, or 30 cents and she bought at least one of each denomination, what is the number of 25-cent stamps that she bought?

(1) She spent a total of $1.45 for stamps.

(2) She bought exactly 6 stamps.

Answer 213

3+ integer variable and equations
> be comfortable subbing in variables when there are 3 variables too

(1) 145 = 20A + 25B + 30C
29 = 4A + 5B + 6C
NS (24 + 5 or 14 + 15)
(2) A + B + C = 6
NS

(3) SUB IN A = 6 - B - C
End up with:
5 - 2C = B → Still not sufficient (B can be multiple values)

C=1, B = 3, A = 2
C = 2, B = 1, A = 3

Question 214

What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?

x <= -1
-1 < x <= 0
0 < x <= 1
x > 1
All real numbers

Answer 214

Inequality operations
> LOOK AT THE ANSWER CHOICES –> we want to determine the RANGE OF X that satisfies the inequality and equation
> need to get rid of y because answer choices don’t have y

(1) solve for y in terms of x using the equation
xy = x+y
xy - y = x
y(x -1) = x
y = x/(x-1) —>x=/1 (which must be true for the equation to be true anyways)

(2) sub in equation for y into the inequality to solve for range of x
x*(x)/(x-1) > 0
x^2 / (x-1) > 0 —-> sewing approach checking roots x = 0 and x=1

This inequality is satisfied when x > 1

ALTERNATIVELY - plug in answer choices (focus on end points) into the inequality and equation to see if it works

e.g., x=-1, then -y = -1 + y –> y = 1/2 and xy = -1/2 which is not greater than 0

e.g., x = 0, then xy = 0 which is not greater than 0

e.g., x = 1, then y = 1 + y –> which is not true

Cannot be all real numbers (since we just showed that x cannot be equal to -1, 0 and 1)

Question 215

At a certain factory, 4 processes—A, B, C, and D—are carried out 24 hours per day, 7 days per week. Process A operates on a 60-hour cycle; that is, Process A takes 60 hours to complete and immediately begins again when it is completed. Likewise, Process B operates on a 24-hour cycle, Process C operates on a 27-hour cycle, and Process D operates on a 9-hour cycle. On Monday, all 4 processes began together at 10:00 in the morning.

On the basis of the information provided, select for Processes A, B, and D the day of the week on which Processes A, B, and D will next begin together at 10:00 in the morning. Also select forAll 4 processes the day of the week on which Processes A–D will next begin together at 10:00 in the morning. Make only two selections, one in each column.

Answer 215

TPA: Multiples and time problem

Processes will begin together again when TIME ELAPSED EQUAL LCM OF CYCLES

A’s time elapsed = 60 h per cycle * Number of complete cycles
…

Part 1: LCM of 60, 24 and 9 is 360 hours

360 hours / 168 hours per week = 2 weeks + 24 hours = Tuesday

Part 2: LCM of 60, 24, 27 and 9 is 1080 hours

1080 hours / 168 hours per week = 6 weeks + 72 hours = Thursday

Question 216

How many members of a certain legislature voted against the measure to raise their salaries?

(1) ​one fourth​ of the members of the legislature did not vote on the measure.

(2) If 5 additional members of the legislature had voted against the measure, then the fraction of members of the legislature voting against the measure would have been ​one third​.

Answer 216

Ratios
> test cases can be quick and easy
> know that total must be a multiple of 12 (when considering both statements together)

(1) Clearly not sufficient

(2) (A + 5) = ⅓ *(Total) → NS too many variables
E.g., Total = 30, then A = 5 or Total = 60 then A = 15

Together?
¼total = Neither
¾total = A + F

First statement is not helpful and we still have too many variables (all we know is that TOTAL must be divisible by 12) → two cases would prove this to be true

Question 217

Machines K, M, and N, each working alone at its constant rate, produce 1 widget in x , y , and 2 minutes, respectively. If Machines K, M, and N work simultaneously at their respective constant rates, does it take them less than 1 hour to produce a total of 50 widgets?

(1) x < 1.5

(2) y < 1.2

Answer 217

Work and reciprocal inequality

SET UP THE EQUATION:

t < 60 mins?

50 / (1/x + 1/y + 1/2) < 60?

5/6 < 1/x + 1/y + 1/2 ?

1/3 < 1/x + 1/y?

(1) x < 3/2 —–> 1/x > 2/3 > 1/3 (sufficient!)

(2) y < 6/5 ——> 1/y > 5/6 > 1/3 (sufficient)

Question 218

If nth day of August and 2nth day of October of the same year fall on the same day of the week, then how many values of n are possible ?

0
1
2
3
4

Answer 218

Multiples and days of the week
> nth day of August and 2nth day of October will fall on the SAME DAY OF THE WEEK IF 2nth day of October occurs 7*int days after nth day of August

NEED TO DETERMINE NUMBER OF DAYS BETWEEN nth day of August and 2nth day of October first , including one end point (e.g., 2nth day of Oct occurs X days after nth day of August)

Remaining days in August: 31 - n
Days in Sept: 30
Days in Oct: 2n

Therefore, there are 31 - n + 30 + 2n = 61 + n days after nth day of August

AND CONSTRAINT: 2*n <= 31 days of October

n = 2, 9

Only 2 values of n

Question 219

If x = –| w |, which of the following must be true?

x = – w
x = w
x^2 = w
x^2 = w^2
x^3 = w^3

Answer 219

Must be True: inequality and absolute values

x <0, unknown sign of w
|x| = |w| —> x^2 = w^2

|w|^2 = w^2

Question 220

For what value of x between –4 and 4, inclusive, is the value of x^2 – 10x + 16 the greatest?

–4
–2
0
2
4

Answer 220

Quadratic functions

Since a > 0, we know the quadratic function OPENS UP like a U
> minimum occurs halfway between the roots

Roots x = 2 and x = 8 –> minimum occurs at x = 5

Therefore, function has higher values of y at the extremes —> in the range provided x = -3 produces the greatest value of x^2 - 10x + 16

Question 221

The number ​sqrt(63 -36 sqrt(3))​ can be expressed as ​x +y*sqrt(3)​ for some integers x and y . What is the value of xy ?

–18
–6
6
18
27

Answer 221

Roots operations
> “can be expressed” means the same thing as EQUAL
> probably need to square both sides of an equation

sqrt(63 -36 sqrt(3))​ = x +y*sqrt(3)​ —–>square both sides

63 -36sqrt(3) = x^2 + 2xysqrt(3) + y^2*3 —-> MATCH THE TERMS ON EITHER SIDE (we are looking for xy afterall)

-36sqrt(3) = 2xysqrt(3)
xy = -18

Why does this work?
> because irrational numbers like sqrt(3) cannot be expressed as a fraction of two integers

Question 222

Jorge’s bank statement showed a balance that was $0.54 greater than what his records showed. He discovered that he had written a check for $x.yz and had recorded it as $x.zy , where each of x , y , and z represents a digit from 0 though 9. Which of the following could be the value of z ?

2
3
4
5
6

Answer 222

Digits operations

Translate the word problem first: “Jorge’s bank statement showed a balance that was $0.54 GREATER than what his records showed. “
> Bank Balance > Record Balance
> Bank Balance = $0.54 + Record Balance

Jorge RECORDED $x.zy but he WROTE A CHECK for $x.yz ———————-> “writing a check” REDUCES THE BANK BALANCE

DETERMINE WHAT IS THE RIGHT AMOUNT vs WRONG AMOUNT:
> RIGHT AMOUNT: what he wrote the check for = $x.yz
> WRONG AMOUNT: what he recorded = $x.zy (must be greater than $x.yz because it shrunk the recorded balance more than the actual bank balance)

Beginning Balance - $x.yz = 0.54 + Beginning Balance - $x.zy
Becomes: $x.zy - $x.yz = $0.54

THE FOLLOWING IS WRONG: (assumes depositing money)
$x.yz - $x.zy = $0.54

THEN NEED TO CONVERT decimal digits (1) into integers then (2) into value form

$xzy - $xyz = 54
100x + 10z - y - 100x - 10y - z = 54
9z - 9y = 54
z - y = 6 ———–> z = 6+y —> y = 0, z = 6

ans E

Question 223

A certain manufacturer sells its product to stores in 113 different regions worldwide, with an average (arithmetic mean) of 181 stores per region. If last year these stores sold an average of 51,752 units of the manufacturer’s product per store, which of the following is closest to the total number of units of the manufacturer’s product sold worldwide last year?

10^6
10^7
10^8
10^9
10^10

Answer 223

Approximation

Equation becomes: units sold
= 51752 * 181 * 113 —-> round to the nearest power of 10

= 50000 * 200 * 100
= 5210^4 * 10^2 * 10^2
= 10*10^8
= 10^9

Question 224

Three balls numbered 1, 2, and
3 are placed in a bag. A ball is drawn from the bag and the
number is recorded. The ball is then returned to the bag. After
this has been done three times, what is the probability that the
sum of the three recorded numbers is less than 8?

Answer 224

Probability with number operations
> need to figure out the RANGE OF POSSIBLE SUMS FIRST
> with replacement = independent events

3 <= sum <= 9

Probability sum is less than 8 = 1 - probability sum is >= 8

> =8 sum occurs:
(1) 3+3+3 = 9
(2) ** 3+3+2 = 8 (don’t forget this case)

P(three 3s in a row) = (1/3)^3 = 1/27
P(two 3s and one 2) = (1/3)^3 * 3 = 3/27

Therefore probability sum is < 8 = 1 - 1/27 - 3/27
= 23/27

Question 225

While designing a game involving chance, Desmond noticed that the probability a fair coin lands faceup exactly 2 times when the coin is tossed 3 times is equal to the probability that a fair coin lands faceup exactly m times when the coin is tossed n times, where n > 3.

Select for m and for n values consistent with the given information. Make only two selections, one in each column.

1
2
3
4
5
6

Answer 225

TPA independent probability
> fair coin —–> probability is ALWAYS (1/2)^n tosses * cases
(not m times)

First calculate the desired probability (land faceup two times out of 3)
= (1/2)^3 * 3 cases
= 3/8

THEN TESTING METHODOLOGICAL TESTING, given n>3, so start with n = 4 and M=1

n=4 and m=1 probability = (1/2)^4 * 4!/3!*1!
= 1/16 * 4
= 1/4 (not 3/8)

n=4 and m=2 probability = (1/2)^4 * 4!/(2!*2!)
= 1/16 * 6
= 3/8 (MATCHES we can stop)

Question 226

Giulia is planning to sell her car, which is fueled by gasoline (petrol) and averages 20 miles per gallon (mpg), and purchase a diesel-fueled car that averages 30 mpg. She estimates that her future cost per gallon of diesel fuel will be 5% higher than is her present cost per gallon of gasoline. She wishes to estimate (1) the annual cost of fuel for her new car if she maintains her present annual total miles driven and (2) the annual total miles she can drive her new car if she maintains her present annual expenditure on fuel.

Let x represent Giulia’s present annual cost per gallon of gasoline in US dollars, and let y equal her present annual total of miles driven. Select for Cost an appropriate expression for Giulia’s estimate of (1) above, and select for Miles an appropriate expression for her estimate of (2) above. Make only two selections, one in each column.

Answer 226

TPA
> get comfortable with variables (we can use both x and y for both expressions)
> unit measurements

(1) ANNUAL COST OF FUEL FOR DIESEL at present miles driven (y)

= Cost per gallon of diesel * number of gallons of diesel

Cost per gallon of diesel = 1.05(cost per gallon of gas) —-> sub in x
= 1.05(x)

Number of gallons of diesel * miles per gallon of diesel = miles
Number of gallons of diesel = miles / 30 —-> sub in y
= y / 30

THEREFORE ANNUAL COST FOR DIESEL = (1.05xy)/30

(2) ANNUAL TOTAL MILES at present annual expenditure on fuel

Present cost per gallon of gas = x
Present number of gallons = y miles / 20 mpg
Present annual cost on gas = x * number of gallons of gas = x * (y/20)

Cost of gallons of diesel that is purchasable = cost per gallon of diesel * gallons of diesel

(xy)/20 = 1.05x * gallons of diesel
Gallons of diesel = y/(20*1.05)

Miles = Miles per gallon of diesel * Number of gallons of diesel
= (30) * y/(201.05)
= 3y/(21.05)

Question 227

In a two-month survey of shoppers, each shopper bought one of two brands of detergent, X or Y, in the first month and again bought one of these brands in the second month. In the survey, 90 percent of the shoppers who bought Brand X in the first month bought Brand X again in the second month, while 60 percent of the shoppers who bought Brand Y in the first month bought Brand Y again in the second month. What percent of the shoppers bought Brand Y in the second month?

(1) In the first month, 50 percent of the shoppers bought Brand X.

(2) The total number of shoppers surveyed was 5,000.

Answer 227

Sets matrix

Columns: X or Y (first month)
Rows: X or Y (second month)

Let A = number of shoppers who bought brand X in the first month
Let B = number of shoppers who bought brand Y in the first month

0.9A = number of shoppers who bought brand X in both months
0.6B = number of shoppers who bought brand Y in both months

Want to know Number of shoppers who bought brand Y in the second month / Total

(1) 0.5Total = A
0.5Total = B

Therefore: 0.9A = 0.45Total and 0.6B = 0.3Total

ALSO: A - 0.9A = 0.1A = 0.05T and B - 0.6B = 0.4B = 0.2T

So we know ALL the inner cells as a function of T and can calculate ratios

Sufficient

(2) Total = 5000 but we are not given any info about the split between A and B or X and Y etc.

A

Question 228

OGMOCK: from a class of 20 students whose names are listed in alphabetical order, a teacher will choose one group of 3 students to represent the class in a student congress. If the teacher will not choose a group of 3 students whose names are in 3 consecutive positions on the list, how many different groups of 3 students could be chosen by the teacher?

1120
1122
1135
1137
1140

Answer 228

Challenging Combination Q
> helpful to draw out 20 lines

Combination: 3 students _ _ _
EXCLUDING options where students are positioned consecutively on an alphabetically ordered list of names

Total number of possible groups: 20C3 = (20!)/(17! * 3!) = (201918)/(32) = 20193
= 6019
= 1140

LESS: groups where students are positioned consecutively
> 18 such groups (18th group will include the last three students on the list)

1140-18
= 1122

Question 229

OGMOCK: A certain store purchased grills for $50 each and lawn chairs for $5 each and then sold each grill and each chair. The store’s gross profit on each grill was 30 percent of its purchase price, and the store’s gross profit on each chair was 50 percent of its purchase price. If the store sold 5 times are many chairs as grills and if the store’s total gross profit on the grills and chairs was $550, what was the store’s total revenue from the sale of the grills and chairs?

$1530
1800
2050
2100
23600

Answer 229

Profit/Loss (pretty lengthy process)
> always write out the equation
Profit = PQ - VQ - FC

Revenue = PQ + PQ = ** PROFIT + COST

Grills:
> Cost = $50 each
> Profit = 0.3*(50) = $15 each
> THEREFORE PRICE = Profit + Cost = 15+50 = $65 each

Chairs:
> Cost = $5 each
> Profit = 0.5*(5) = $2.50 each
> THEREFORE PRICE = Profit + Cost = 2.5+5 = $7.50 each

QUANTITY…
C = 5G

TOTAL GROSS PROFIT:
$550 = Profit from Grills + Profit from Chairs
$550 = 15G + 2.5C
550 = 15G + 2.5(5G)
550 = 15G + 12.5G
550 = 27.5G
G = 550/27.5 —> LONG DIVISION
G = 5500/275
G = 20
Therefore C = 100

Revenue = Profit + Cost
= 550 + 5020 + 5100
= 550 + 1000 + 500
= 2050

Alternatively:
REVENUE = PQ + PQ
= 6520 + 7.5100
= 1300 + 750
= 2050

Question 230

OGMock: Which of the following is equal to (12 + sqrt(28))/sqrt(9+7)?

Answer 230

Roots
> NOTICE that 9+7 = 16 —> no variables at all
> if you do multiply top and bottom by sqrt(16), make sure to distribute to both terms in the numerator

= (12 + 2*sqrt(7))/4
= 3 + sqrt(7)/2

Question 231

OGMock: The string of digits 3691215…300 is formed by merging together the decimal representations of the first 100 positive integer multiples of 3. Counting from the left, what is the 72nd digit of this string of digits?

0
1
6
8
9

Answer 231

Digits x multiples
> Testing ability to count effectively what is the 72nd digit in this string

Each one digit multiple of 3 contributes 1 digit
Each two digit multiple of 3 contributes 2 digits
Each three digit multiple of 3 contributes 3 digits

There are:
> Three one digit multiples of 3 = length of 3 digits
> [12,99] = (99-12)/3 + 1 = 30 two digit multiples of 3 = length of 60 + 3 = 63 digits so far
> THEN MANUAL COUNTING for three digit multiples of 3:

102, 105, 108

72nd digit is 8

Question 232

The string of digits 135791113…999 is formed by merging together the decimal representations of the odd integers from 1 through 999. Counting from left, what is the 110th digit of this string of digits?

0
3
5
7
9

Answer 232

Digits x multiples
> Testing ability to count effectively what is the 110th digit in this string

Each one digit odd integer contributes 1 digit
Each two digit odd integer contributes 2 digits
Each three digit odd integer contributes 3 digits

There are:
> Five one digit odd integers = length of 5 digits
> [11,99] = (99-11)/2 + 1 = 45 two digit odd integers = length of 90 + 5 = 95 digits so far
> THEN MANUAL COUNTING for three digit odd integers:

101, 103, 105, 107, 109
110th digit is 9

Question 233

OGMOCK: How many different numbers can be obtained as the product of exactly 3 different numbers in the set {2, 3, 5, 7, 11}

6
10
12
15
20

Answer 233

Hidden combination problem involving factors
abc = unique number IF factors are different (grouping matters, not the order of the multiplication)

5C3
= (543)/3!
= 10

Question 234

OGMOCK: For each positive integer n, let an = n(-1)^n(x-1). If the sum of a1 through a20 is equal to 60, what is the value of x?

Answer 234

Sequences and patterns

Write out a few of the terms a1 through a3:
a1 = -1(x-1)
a2 = 2(x-1)
a3 = -3(x-1)
…
Even n –> n(x-1)
Odd n –> -n(x-1)

So Sum equals = (x-1)*(-1 + 2 - 3 + 4 - 5 … - 19 + 20) = 60

Notice how EVERY two terms the sum = 1
20 terms there are 10 pairs so sum equals 10

x-1 = 60 / 10
x - 1 = 6
x = 7

Question 235

OGMock: The value of (8^4 + 8^16) / (4^8 + 16^8) is?

less than 0.00005
greater than 0.00005 and less than 0.05
greater than 0.05 and less than 50
greater than 50 and less than 50,000
greater than 50,000

Answer 235

Exponents
> convert all terms to the SAME BASE (2)

= (1+ 2^36) / (2^4 * (1 + 2^16)) ———> 1 is really insignificant so we can exclude
= ~2^36 / (2^4 * 2^16)
= 2^36 / 2^20
= 2^16 —> approximately 64k which is Greater than 50,000

OR 2^16 = 2^10 * 2^6 = 1024 * 64 = ~64000

2^10 = 1024
2^11 > 2000
2^12 > 4000
2^13 > 8000
2^14 > 16000
2^15 > 32000
2^16 > 64000

Question 236

OGmock: If x^2 = x+1, which of the following is equal to x^3?

x+2
x+3
2x
2x+1
4x+2

Answer 236

Recognizing patterns and manipulating equations
> notice how every answer choice has x
> factoring won’t help

MULTIPLY BOTH SIDES OF THE EQUATION BY X

x^3 = x*(x+1)
x^3 = x^2 + x ——–> we know x^2 = x+1
x^3 = x+1+x
x^3 = 2x + 1

Question 237

A certain computer program reorders the letters of any seven-letter sequence, and the position of a letter in the new order depends only on its position in the original order. The first run of the program changes the initial input ABCDEFG to the output DABCGEF. If the input to each subsequent run is the output from the preceding run, after how many runs will the output first equal the initial input ABCDEFG?

6
7
12
14
24

Answer 237

Test MULTIPLES and LCM

Write out a few positions based on the RULE:

RULE:
Position 1 –> 2
2 –> 3
3 –> 4
4 –> 1
5 –> 6
6 –> 7
7 –> 5

A will take 4 runs of the program to return back to the right position (then the pattern continues, EVERY 4 RUNS will return back to the right position)
B will take 4 runs
C will take 4 runs
D will take 4 runs
E will take 3 runs
F will take 3 runs
G will take 3 runs

SO all the letters will be back at the right position after LCM of the cycles = 12 runs

Question 238

OGmock: How many positive integers less than 500 have a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3?

20
21
24
26
72

Answer 238

Remainder problems
> for unknown number and different divisors –> form COMBINED EQUATION FOR N
> then to determine “how many positive integers less than 500” could n be, remember constraints (e.g., n must be a multiple of something) but just DO COUNTING (last - first/increment +1)
> figure out the first and last possible value of n in the range

Combined equation for n becomes n = 21*int + 8
> remember that the Int could be equal to 0

WRITE OUT A FEW POTENTIAL VALUE OF N

[8, 29, 50 … 491] → then count how many integers are there
= (491-8)/21 + 1 = 23 + 1 = 24 integers

Question 239

How many positive integers less than 100 have the remainder of 2 when divided by 13?
A. 6
B. 7
C. 8
D. 9
E. 13

Answer 239

Remainder problems
> counting the number of integers is easiest when you set up the RANGE of possible values of n and treat as an EQUALLY SPACED sequence

n = 13*int + 2

[2, 15, …93] *** don’t miss 2 (when int = 0)

number of possible values of n that are less than 100?

= (last-first)/increment + 1
= (93 - 2)/13 + 1
= 7+1
= 8

Question 240

Beginning at noon yesterday, water was removed from a partially filled water tank at a constant rate of 300 liters per hour. When there were 600 liters of water left in the tank, no more water was removed from the tank. Were there more than 1000 liters of water in the tank at noon yesterday?

(1) There were 600 liters of water in the tank at 2:00 yesterday afternoon.

(2) There were more than 650 liters of water in the tank at 1:00 yesterday afternoon.

Answer 240

Tank capacity question with a twist
> “when there were 600 liters of water left in the tank, NO MORE WATER WAS REMOVED” —> need to know when this happens

Initial water - 300*hours = Ending water

initial water > 1000?

(1) tells us that water STOPS getting removed either AT 2pm or BEFORE 2pm

if at 2pm: 600 = initial - 300*2 hours —-> initial =1200 (ans Yes)

if stop at 1pm: 600 = initial - 300*1 hour –> initial = 900 (ans No)

NS

(2) water > 650 at 1 pm —> water is still being removed at 1pm
Initial water - 300*1 > 650
initial water > 950

Also NS (could be >1000 or <1000)

(3) Initial water > 950
Water stops being removed after 1pm

Case 1: 700 liters of water at 1pm –> initial water = 1000 (ans No)
Case 2: 710 liters of water at 1pm –> initial water = 1010 (ans Yes)

NS
E

Question 241

Beginning at noon yesterday, water was added to a partially filled water tank at the constant rate of 500 gallons per hour. When there was a total of 2,000 gallons of water in the tank, no more was added. Was there more than 1,000 gallons of water in the tank at noon yesterday?

(1) There were 1,700 gallons of water in the tank at 1:00 yesterday afternoon.
(2) There were 2,000 gallons of water in the tank at 3:00 yesterday afternoon.

Answer 241

Tank capacity question with a twist
> “when there were 2000 gallons of water in the tank, NO MORE WATER WAS ADDED” —> need to know when this happens

Water level = Initial level + 500*hours

Initial level > 1000?

(1) tells us exactly how much water is in the tank after 1 hour

1700 = Initial level + 500*1
Initial level = 1200 > 1000 (Sufficient)

(2) we don’t know when the water reached 2000

If reached at 3pm:
2000 = initial + 500*3
Initial = 500 (ans No)

If reached at 1pm:
2000 = initial + 500*1
Initial = 1500 (ans Yes)

NS

A

Question 242

A furniture manufacturer produced at least 3 cabinets each day for the past 30 working days. What was the median number of cabinets produced daily by the manufacturer for the past 30 working days?

(1) For the past 30 working days, the manufacturer produced at most 6 cabinets each day

(2) For the past 30 working days, the manufacturer produced a total of 104 cabinets

Answer 242

Statistics Median
> minimum number of cabinets = 3 per day
> don’t get overwhelmed by the high number of days –> test cases still (making the two middle terms equal to the median)

(1) Max 6 cabinets per day
Median could be 3 or 6
NS

(2) SUM 104 —> need TEST CASES

Median =3 (position 15 and 16)?

104 = (16 days)(3 per day) + (14 days)(4 per day) —> works

Median = 7?

104 = (16 days)*(7 per day) —-> doesn’t work (already exceed the sum of 104)

Median = 6?
104 = (16 days)(6 per day) + 14 daysx per day
8 = 14 * x –> won’t be an integer and will be less than 3

INTEGERS

104 = 16median + 14x
52 = 8median + 7x —> only one valid case

Sufficient

Alternative answer: FIXED sum and tradeoffs
Initially, our set of daily productions looks like this if we assume 3 cabinets are produced on all 30 days, resulting in a minimum sum of 90 cabinets:

{3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3}

Since the sum is actually 104 cabinets, we need to account for an extra 14 cabinets and observe the changes to the middle values (15th and 16th terms) in this set. Even if we distribute these additional cabinets in the most balanced way possible, by adding 1 cabinet to 14 of the days, the set becomes:

{3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4}

In this configuration, the 15th and 16th terms, which determine the median, remain unchanged at 3. Therefore, regardless of how the additional 14 cabinets are distributed, the median number of cabinets produced daily remains 3.

Question 243

Kelly invested in two different funds, Fund F and Fund M. For each $100 invested in Fund F, Kelly earned $8.50 in interest the first year. For each $100 invested in Fund M, Kelly earned $7.60 in interest the first year. If Kelly earned 8.2 percent in interest from the two investments that year, what dollar amount was invested in Fund M?

(1) The amount invested in Fund F was $20,000 more than 50 percent of the total amount invested

(2) The amount invested in Fund F was $80,000

Answer 243

Interest
(assume simple interest)

Interest rate = income / investment amount

Fund F’s interest rate = 8.5%
Fund M’s interest rate = 7.6%

8.2%(F + M investment) = 8.5%F + 7.6%*M ———-> we KNOW this gives a RATIO between F and M but no actual values, also it is ONE EQUATION

M?

(1) F = 20000 + 0.5*(F + M) —-> another equation. Now we have two variables, two different equations

F = 20000 + 0.5F + 0.5M
0.5F - 20000 = 0.5M

Sufficient

(2) F = 80,000

Since we are given a relationship between F and M, knowing F will allow us to know M
Sufficient

D

Question 244

When the Manor Apartments building was constructed, the parking spaces for the building were all in a single row and were numbered sequentially, from 1 through 8. Each space was assigned to the apartment with the same number; there were no other apartments in the building. A year later, however, the parking space assignment for Apartment 2 was exchanged with that for another apartment, and then a year after that, the parking space assignment for Apartment 2 was exchanged again, this time with that for yet another apartment. No other exchanges have been made. Currently the parking space assigned to Apartment 2 is adjacent on one side to the space assigned to Apartment 6 and on the other side to the space assigned to Apartment 5.

Select two apartment numbers such that the one selected for First could have been the number of the apartment whose assignment was exchanged with that of Apartment 2 after the first year, and the one selected for Second could have been the number of the apartment whose assignment was exchanged with that of Apartment 2 after the second year, jointly consistent with the information provided.

1
3
4
5
6
8

Answer 244

TPA - pure logic
> keyword: “the parking spaces for the building were ALL IN A SINGLE ROW and were numbered SEQUENTIALLY”

Looking for two exchanges of APARTMENT 2’s location

Original line up: 1 2 3 4 5 6 7 8

END UP WITH either 5 2 6 or 6 2 5

Need to exchange spots with apartments that are NEAR 5 and 6 —> try 4 and 7 (not one of the answer choices)

If exchange with 4 —> becomes 1 4 3 2 5 6 7 8

Then exchange with 5 –> becomes 1 4 3 5 2 6 7 8 —> WORKS!

ans: 4 and 5

Question 245

Ken bought a shirt at a price of d dollars, to which a sales tax of p percent was added. He paid with a 20-dollar bill and received less than c dollars in change. Was the price of the shirt, without sales tax, more than 15 dollars?

(1) p = 6
(2) c = 5

Answer 245

Sales price and tax word problem

20 - d*(1+p%) < c

Want to know if d > 15?

Can first rearrange the inequality in terms of d > __

d > (20-c)/(1+p/100) > 15?

(1) p = 6 becomes
20 - d*(1.06) < c —> without c, don’t know

(2) c = 5 becomes
20 - d(1+p%) < 5
15 < d(1+p%) –> without p also don’t know

(3)

15 < d*(1.06)
d > 14.15 —–> still NS (d = 15 ans No or d = 18 ans Yes)

E

Question 246

Printing presses, Press A and Press B, are used to print the daily edition of a certain newspaper. Working together at their individual constant rates, how many hours did it take the 2 presses to print the daily edition of the newspaper yesterday?

(1) Yesterday, Press A printed newspaper at a rate of 4000 newspapers per hour

(2) Yesterday, Press B printed newspapers at 75% the rate that Press A printed

Answer 246

Work

Need to know (combined rate)*time = amount of work

> key trap: don’t know how many newspapers is the “daily edition of the newspaper”

(1) Missing Press B’s rate

(2) Missing absolute rates and work

(3) even together all we know is that Ra = 4000 newspapers/hour and Rb = 3000 newspapers/hour

However we don’t know the Amount of work!

E

Question 247

Economists work with supply and demand curves that show the price P of goods as a function of the quantity Q of those goods supplied or demanded. For a certain product, the supply curve is P = aQ + b and the demand curve is P = k/Q, where a, b, and k are nonzero constants. The point at which these curves intersect in the (Q, P) coordinate plane is referred to as the equilibrium point, and for this product the equilibrium point is (10,5). For this product, what are the values of a, b, and k?

(1) the point (8, 1) is on the supply curve
(2) the point (25, 2) is on the demand curve

Answer 247

Supply and demand (coordinate plane)
> need two points to know the equation of the line

(10, 5) is on BOTH curves (Q, P)

Supply curve: 5 = a*10 + b

Demand curve: 5 = k/10 —> k = 50 (we know k)

We need to know a and b (supply curve) –> one other point on the supply curve will be sufficient

(1) sufficient
(8,1) and (10,5) are two points creating two equations with two unknowns –> sufficient

(2) on the demand curve is NOT that helpful since we already know the demand curve

A only

Question 248

Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was $120,000. What was the median price of the three houses?

(1) The price of Tom’s house was $110,000.
(2) The price of Jane’s house was $120,000.

Answer 248

Statistics: mean and median
> given average, know sum is 120k*3 = 360k
> asked about the median
> test cases
> note that we are given AVERAGE –> one case is when all data points are equal to the average

(1) two cases
> 110 110 140
> 110 120 130
NS

(2) could be all three equal to 120k, but if you want to change ANY single salary, one other salary MUST ALSO CHANGE, leaving 120k as the median

S

B

Question 249

The 50 participants of a management training seminar ate dinner at a certain restaurant. They had 3 choices for their meal: vegetarian lasagna for $12, blackened catfish for $15, or stuffed pork chops for $18. Each participant ordered exactly 1 meal and the total cost of the meals ordered by the participants was $810. How many participants of the management training seminar ordered blackened catfish?

(1) Six more people ordered catfish than lasagna.
(2) Twice as many pork chop meals were ordered as catfish meals.

Answer 249

Linear equations
> write out the equations
> remember: need at least n different equations for n different unknowns to be sufficient UNLESS there are exceptions (like integer constraint)

50 = L + C + P

12L + 15C + 18P = $810 —> simplifies to 4L + 5C + 6P = $270

Want to know: C = ?
> int >= 1

(1) C = 6 + L –> now we have 3 equations and 3 variables, SUFFICIENT

(2) P = 2C –> now we have 3 equations and 3 variables, SUFFICIENT

D

Question 250

TTP: Nick and JP both paint houses. Working alone, JP can paint an entire house in 12 hours. Nick can do the same job alone in 15 hours. Nick and JP begin painting a house together and work for 2 hours, at which time Brent joins them. The three of them complete the job 4 hours later. How long would it take Brent to paint 2 houses alone?

20 hours
36 hours
40 hours
62 hours
80 hours

Answer 250

Combined Workers problem (given hours):
> when dealing with different time (e.g., one person works for x hours, then leaves and then another joins for y hours), it is faster to set up in terms of WORK CONTRIBUTION BY EACH WORKER
(instead of calculating work and time in steps)

Work Contribution By Each Worker = Individual Rate * Time Spent Working

Sum of Individual Work Contributions = Total Work

JP’s rate is 1 house / 12 hours and he works for 2+4 = 6 hours in total
N’s rate is 1 house / 15 hours and he works for 2+4 = 6 hours in total
B’s rate is 1 / t hours and he works for 4 hours in total

Total work = 1 house = (1/12)6 + (1/15)6 + (1/t)*4

t = 40 hours

So it takes Brent 40 hours to paint 1 house and it will take him 80 hours to paint 2 houses alone

Question 251

TTP: Jon and his twin sister, together with their 3 younger brothers, are wrapping presents. Each of the younger brothers can wrap presents at 1/4 of the rate that Jon and their older sister can each wrap presents. The time it takes all 5 children working together to wrap 10 presents is what fraction of the time it takes Jon and his twin sister, working together, to wrap 10 presents?

Answer 251

Work:
> combined worker
> relative rates (we are not given actual individual rates)

There are 5 people –> 3 younger brothers, Jon and his twin sister
> Jon and his twin sister have the SAME WORK RATE (“1/4 of the rate that Jon and their older sister can EACH wrap presents”)

Rb = 1/4(Rj) = 1/4(Rs)
OR: 4Rb = Rj

Looking for: Time it takes when 5 children work together / Time it takes when Jon and his twin sister work together
> work = 10 presents

(1) Create expression for the Time it takes 5 children working together to wrap 10 presents

Combined Rate = Total Work / Time
3Rb + 2Rj = 10 / Time x —> substitute variables representing relative rates**

3Rb + 8Rb = 10 / Time x
Time x = 10/ 11Rb

(2) Create expression for the Time it takes Jon and his twin sister working together to wrap 10 presents

Combined Rate = Total Work / Time y
2*Rj = 10 / Time y
8Rb = 10 / Time Y
Time Y = 5 / 4Rb

(3) Put together to calculate ratio
Time X / Time Y
= 8/11

Question 252

Tom and Jerry are filling up 500 water balloons apiece. It takes Tom 2 hours longer than Jerry to fill up 500 balloons. Jerry can fill 50% more balloons per hour than Tom. How many water balloons per hour can Jerry fill?

Answer 252

Work
> Relative rates (given two equations, one for time and one for individual rates)

Tt = 2 + Tj (when work equals 500 balloons)
Rj = 1.5Rt

Looking for Rj

Given variables for time, we can create equation linking rates and time:

Rate * time = Work

Rj * Tj = 500 and Rt * Tt = 500

CHOOSE ONE EQUATION (e.g., Tt = 2 + Tj) and replace variables with the other one
> in this case, replace all time variables with rates
> then, use relative rates again to express in terms of ONE VARIABLE

500/Rt = 2 + 500/Rj —-> then use Rj = 1.5Rt OR 2/3Rj = Rt
750/Rj = 2 + 500/Rj
Rj = 125 balloons per hour

Question 253

TTP: If 9 men, each working at the same constant rate, can build a house in x days, how many days would it take y men, working at that same constant rate, to build the same house?

(1) One of the 9 men can build the house on his own in 36 days
(2) y = 2x

Answer 253

Combined Worker problem

Given: Combined rate of 9 men = 1 / x
Therefore, individual rate per man = 1 / 9x

Looking for 9x / y (given that y * 1/9x * t = 1)

(1) Individual rate per man = 1/36 = 1/9x —> x = 4
However, we still don’t know y
NS

(2) once you substitute y=2x into the question expression, x cancels out, leaving 9/2 = 4.5 days

Question 254

TTP: a brand x copy machine can copy three times as many pages per minute as a brand y copy machine can. If both machines worked alone at their individual constant rates to copy the same number of pages, how long did it take copy machine y to copy those pages?

(1) when working at a constant rate, it takes a brand x copy machine 20 minutes to copy the pages
(2) if both machines had worked together at their individual constant rates, they would have copied the pages in 15 minutes

Answer 254

Worker problem
> relative rates

Looking for TIME it took machine y to copy “those pages” (can assume equal to z pages)

Rx = 3*Ry
Tx, Ty it takes to copy z pages, where Ty = z / Ry

(1) Rx = z / 20 = 3*Ry
Ry = z / 60
Therefore, Ty = z / z / 60 = sufficient

(2) (Rx + Ry)15 = z pages –> substitute Rx = 3Ry in
(4Ry)*15 = z pages

Ry = z / 60
Therefore, Ty = z / z / 60 = sufficient

Question 255

If x and y are positive integers and 1/x + 1/y < 2, which of the following must be true?

x+y > 4
xy > 1
x/y + y/x < 1
(x-y)^2 > 0
None of the above

Answer 255

Algebraic expressions - didn’t know how to solve

x>0 and y>0

COUNTEREXAMPLES is probably the fastest (find counterexample for ALL answers EXCEPT for 1)

Is it possible for x+y > 4 to be false?
x = 2 and y = 2, but x+y =4

Is it possible for xy > 1 to be false?
Always true

Is it possible for x/y + y/x < 1 to be false?
x=2 and y=2 but x/y + y/x = 1

Is it possible for (x-y)^2 > 0 to be false?
x=2 and y =2 but (x-y)^2 = 0

Question 256

A drain pipe can empty a bucket in 4 hours. On a rainy day, with the bucket full, the drain pipe manages to empty the bucket in 6 hours instead. If during the rainfall, the rain adds 3 liters of water each hour, what is the capacity of the bucket?

9 liters
18
27
36
45

Answer 256

Work/rate

Capacity of bucket in liters?
> WATCH OUT FOR SIGNS (e.g., make rate inflow negative)
> different units (let X = capacity of the bucket)

Rate outflow - rate inflow = Net Rate outflow

(X liters)/4 - 3 = (X liters)/6

X = 36 liters

Question 257

Two pumps, A and B, are used to fill a swimming pool. Pump A, when working alone, can fill the pool in 4 hours, whereas Pump B, when working alone, can fill it in 6 hours. If Pump B starts filling the pool by itself for 1 hour before Pump A joins in to work together with Pump B, how long will it take to fill the pool completely?

Answer 257

Work (combined worker problem)
> be mindful what the VARIABLES you set actually mean
> Always reread the question to make sure you know what you are solving for

Pump A: 1 pool/4 hours
Pump B: 1 pool/6 hours

Total work = sum of individual work

1 pool = 1/61 + (1/4+1/6)t —–> total time = t + 1

Question 258

If a cube with a side length of 4cm is cut into smaller cubes, each with a side length of 1cm, what is the percentage increase in the total surface area of the resulting smaller cubes?

Answer 258

Percentage increase with some geometry
> SA of cube = 6 sides * area of one side

Number of smaller cubes = Volume of larger cubes / volume of smaller cubes
= 4^3 / 1^3 = 64 smaller cubes

Original surface area = 6 * 16 = 96 cm^2
New surface area = 6 * 1 * 64 cubes = 6 *64 = 384 cm^2

Percentage increase in SA: (384-96)/96 * 100
= 288/96 * 100
= 300%

Question 259

If q^2 + pr = 10 and r^2 + pq = 10, and r=/q, what is the value of p^2 + q^2 + r^2?

10
15
20
25
30

Answer 259

Algebra
> plug in answers if don’t know what to do

If you SUM both equations, you get:
q^2 + r^2 + pr + pq = 20
q^2 +r^2 + p(r+q) = 20

Nothing is stopping us from assuming p(r+q) = p^2, so if we assume this, then q^2 + r^2 + p^2 = 20

LONG WAY has algebraic proof using condition r=/q:

(1) subtract equations
q^2 + pr - r^2 - pq = 0

(2) factor
q^2 - r^2 + pr - pq = 0
(q + r)(q - r) - p(q - r) = 0
(q - r)*(q + r - p) = 0

(3) since r=/q, q-r =/0 so q + r - p must = 0
q + r - p = 0
q + r = p

(4) once again, plug back into summed equation
r^2 + q^2 + p(r + q) = 20
r^2 + q^2 + p(p) = 20 —–> Long proof that r + q = p

Question 260

On a farm, there are chickens, cows, and sheep. The total number of chickens and cows is three times the number of sheep. If there are more cows than either chickens or sheep, and the combined total of heads and feet of chickens and cows is 100, how many sheep are on the farm?

5
8
10
14
17

Answer 260

Word problem:
Chickens = A
Cows = B
Sheep = C

A + B = 3C
B > A
B > C
A + B + 2A + 4B = 100 —> 3A + 5B = 100

C =?

*didn’t know how to manipulate inequality (3 variables, only 2 equations)
* when in doubt, PLUG IN ANSWER CHOICES
* ALSO leverage number properties (units digits, even/odd)

First simplify the equations:
3*(3C - B) + 5B = 100
9C - 3B + 5B = 100
9C + 2B = 100

Solution 1: Plug in answer choices and verify that it satisfies all conditions
C = 5, B = not integer

C = 8, B = 14, A = 10, B > A, B > C (ANSWER)

Solution 2: Leverage number properties

3A + 5B = 100 —> A must be a multiple of 5

Test A = 5, 10, 15 … –> A = 10, B = 14 and C = 8 works

Question 261

If x = sqrt(10) + cube root(9) + fourth root(8) + fifth root(7) + sixth root(6) + seventh root(5) + eight root(4) + ninth root(3) + tenth root(2), then which of the following must be true?

x > 12
10 < x < 12
8 < x < 10
6 < x < 8
x < 6

Answer 261

Roots
> Concept: root of any number > 1 will always be > 1
e.g., 1000th root of 2 > 1
> try to do precise rounding for square and cube roots

sqrt(10) –> greater than 3
cube root(9) –> greater than 2
Seven numbers greater than 1

3 + 2 + 7 —> greater than 12

Question 262

Sqrt(7 + sqrt(48)) - sqrt(3)

Answer 262

Sum/difference involving ROOTS

Always simplify the argument into the most simplified form (so sqrt(48) becomes 4sqrt(3))
> either square both sides of the equation (insert = x)
> OR try to MATCH terms together on both sides of the equation (insert x)

e.g., Sqrt(7 + 4sqrt(3)) - sqrt(3) = x
sqrt(7 + 4sqrt(3)) = x + sqrt(3) —–> square both sides
7 + 4sqrt(3) = x^2 + 2xsqrt(3) + 3
4 + 4sqrt(3) = x^2 + 2xsqrt(3) ——> match terms

4 = x^2 and 4sqrt(3) = 2xsqrt(3)

x = 2

Question 263

If a fraction of a 50% alcohol solution is replaced with 25% alcohol solution, resulting in a final solution with 30% alcohol, what is the fraction of the original solution that was replaced?

3%
20%
66%
75%
80%

Answer 263

Mixture (replacement)
> replacing SAME amount of liquid so the TOTAL VOLUME DOESN’T CHANGE

Substance: alcohol
> Let A = original volume
> Let R = replaced volume

Looking for: R / A

0.5(A) - 0.5(R) + 0.25(R) = 0.3(A)
0.2A = 0.25R
R/A = 0.2/0.25
R/A = 4/5 = 80%

Question 264

In a blind taste competition, there are 3 types of tea: Type A, Type B, and Type C. Each type is presented in 3 cups, for a total of 9 cups. If a contestant tastes 4 different cups of tea at random, what is the probability that the contestant does not taste all 3 types of tea?

Answer 264

Combinatorics

A: 3 types
B: 3 types
C: 3 types

P(not taste all 3 types of tea) = 100% - P(taste all 3)

P(taste all 3) = ABCA or ABCB or ABCC (categories only, still need to account of the types of sub-teas)

Denominator: 9C4 = 9!/(5!4!) = 9876/4*3! = 126

Numerator: 3C23C13C1 * 3 = 27 * 3 = 81

P(not taste all 3 types of tea) = 100% - 81/126
= 45/126
= 5/14

Alternatively:
> not tasting all 3 types of tea means you taste 2 types of tea (cannot just taste one type of tea)

Numerator: 3 teas choose 2 * 6 sub-teas choose 4
Denominator: 9 sub-teas choose 4

Question 265

If a/b = x/y and a/y = b/x, where a, b, x and y are non-zero integers, which of the following must be true?

I. x/y = -1
II. x = y
III. |x| = |y|

Answer 265

Must be true (algebra)
> counterexample to eliminate

FIRST SIMPLIFY THE QUESTION STATEMENT: try to LINK the two ratio equations using a/b
(Note how the answer choices show the relationship between x and y only, so we need to get in terms of x and y)

Eq’n 1: a/b = x/y
Eq’n 2: a/b = y/x
Eq’n 1 = Eq’n 2 = a/b = x/y = y/x

Therefore: x^2 = y^2
(x+y)*(x-y) = 0
x=y or x=-y

I) x/y = -1?
Does NOT have to be true (x=y makes it false)

II) x=y?
Does NOT have to be true (x=-y makes it false)

III) |x| = |y|
Must be true
(taking the square root of both sides of x^2 = y^2 equals |x| = |y|)

Question 266

Which of the following expressions has the greatest value?

999^12
10^30
777^10
(-20)^24
(sqrt15)^40

Answer 266

Comparing exponents:
(1) change any PAIR of numbers to the SAME EXPONENT, and compare bases
or
(2) reduce / increase the exponents on any PAIR of numbers by raising by LCM or reducing exponent by GCF and evaluating values

Question 267

The equation x^2 + mx - n = 0, where x is a variable and m and n are constants has equal roots. One of the roots of another equation y^2 + my + 15 = 0, where y is a variable and m is a constant, is 3. What is the value of n?

Provide 2 solutions

Answer 267

Quadratic equations and roots
> A root is a value subbed in for x (or another variable) that makes the equation = 0
> since the equation is set = 0 too, the value of x that works MUST be a solution
> also must use special root formulas for a quadratic equation in the form ax^2 + bx + c:
x1 + x2 = -b/a
x1 * x2 = c/a

SOLUTION 1: using special root formulas
(1) find the value of m using the second equation and plugging in y = 3

(2) find the value of n using special root formulas, where x1 = x2 = x

2x = -(-8)/1
x = 4

(3) find the value of n with x = 4
n=-16

SOLUTION 2 (FASTER): Using Discriminant (recall when there is only 1 solution, the discriminant b2 - 4ac = 0)

64 - 4(1)(-n) = 0
64 + 4n = 0
n = -16

SOLUTION 3: setting quadratic equation equal to a quadratic identity
> “equal roots” means the quadratic function can be expressed as the Square of the Difference

(x-k)^2 = x^2 - 8x - n
x^2 - 2xk + k^2 = x^2 - 8x - n ———> MATCH TERMS

Question 268

If b = a+4, for which of the following values of x is the expression (x - a)^2 + (x - b)^2 the smallest?

a-1
a
a+2
a+3
a+5

Answer 268

Min/Max problems
> easiest way to solve is by PLUGGING IN answer choices

(1) sub in b = a+4
(x - a)^2 + (x - a - 4)^2

(2) Minimize the sum of squares by making the constant as small as possible
> also notice that with every answer choice a cancels out

Ans: a+2
(sum equals 8)

Question 269

How many integers are there between 1 and 1000, inclusive, that are not divisible by either 11 or 35?

Answer 269

Counting / Divisibility / Multiples:

Total number of integers - number of integers that are divisible by 11, 35 or both

Total number of int = 1000

Number of integers divisible by 11: [11, 990]
= (990-11)/11 + 1
= 90 —–> 11*90 = end range

Number of integers divisible by 35: [35, 980]
= (980-35)/35 + 1
= 28 ——> 35*28 = end range

Number of integers divisible by both: LCM of 11 and 35 = 385, 770
= 2 integers

Ans = 1000 - (90 + 28 - 2)
= 1000 - 116
= 884

Question 270

How many integers are there between 324,700 and 458,600 that have a 2 in the tens place and a 1 in the ones place?

Answer 270

Counting / Divisibility / Multiples:
> range is 6 digits, with two digits being 2 1

_ _ _ _ 2 1

> fastest approach is to do “cycle method”: every 100 integers, there is another _ _ _ _ 2 1

Number of integers between 324,700 and 458,600, not inclusive: 458,600 - 324,700
= 133,900 integers (it is okay not to subtract 1 e.g., 800-600 = 200/100 = 2 cycles of 100)

How many cycles of 100? 133,900/100 = 1339

Question 271

How many roots does the equation sqrt(x^2 + 1) + sqrt(x^2 + 2) = 2 have?

0
1
2
3
4

Answer 271

Equations involving exponents and roots:
> a “root” is the same thing as a “solution” (value for a variable that satisfies the equation)

In this case, we need to test values for x, knowing that x^2 will always be >= 0

So test x = 0 (minimum value of x^2) to get sqrt(1) + sqrt(2) on the LHS, which equals 1 + ~1.4 = ~2.4 > 2

so there are no real x that satisfies the equation

Other concepts:
> Number of solutions —> testing discriminant but needs quadratic equation in the form ax^2 + bx + c
> when you see a bunch of radicals, you MUST SQUARE both sides of the equation and KEEP SQUARING until radicals are gone

Question 272

On September 19th, 1987, it was a Saturday. If 1988 was a leap year, what day of the week was it on September 21st, 1990?

Solve via two methods

Answer 272

Multiples/Remainders:
> Day of the week depends on multiple and remainders of 7 —> if the number of days AFTER a certain day is a multiple of 7, then it will land on the SAME day of the week
> Date relates to either every 365 or 366 days
> or can solve via shift method

Shift method: 365 days means the same date in the following year would fall ONE DAY LATER IN THE WEEK; 366 days means the same date in the following year would fall TWO DAYS LATER IN THE WEEK
> Sept 19, 1988 has a shift of +2 days
> Sept 19, 1989 has a shift of +1 day
> Sept 19, 1990 has a shift of +1 day
> Sept 21, 1990 has a shift of +2 days

Total shift = 6 days from Saturday = Friday

Question 273

The value of 1/2 + (1/2)^2 + (1/2)^3 + … + (1/2)^20 is between?

1/2 and 2/3
2/3 and 3/4
3/4 and 9/10
9/10 and 10/9
10/9 and 3/2

Answer 273

Estimation
> recognize that this is a Geometric Series

Sum = a1*(1 - r^n) / (1 - r)
where:
a1 = 1/2
r = 1/2

Sum = (1/2)*(1 - 1/2^20) / (1 - 1/2)
= ~1/2 / 1/2
= ~1 —-> closest is 9/10 and 10/9

Question 274

There are two bars made of gold-silver alloy. The first bar contains 2 parts of gold and 3 parts of silver by weight, while the second contains 3 parts of gold and 7 parts of silver by weight. If both bars are melted and combined to form a single 8-kilogram bar with a gold-to-silver ratio of 5:11 by weight, what was the weight of the first bar?

Answer 274

Mixture
> given two substances (gold and silver)

Let A = the weight of the first bar
Let B = the weight of the second bar

** it is RIGHT to make A + B = 8

SHORT WAY: A + B = 8kg (melting two bars, so combined bar will have a weight EQUAL to the sum of the individual bars)

Gold: 4A + 3*(8 - A) = 25 ——> 4A + 24 - 3A = 25
A = 1

LONG WAY: set separate variables for the first and second bars
Gold: (2/5)(A) + (3/10)(B) = (5/16)(8)
Silver: (3/5)A + (7/10)(B) = (11/16)8

Now we have two unknowns and two equations

Gold: 4A + 3B = 25
Silver: 6A + 7B = 55

B = 7 and A = 1

Question 275

If x is an integer and |1-x| < 2, then which of the following must be true?

x is not a prime number
x^2 + x is not a prime number
x is positive
Number of distinct positive factors of x + 2 is a prime number
x is not a multiple of an odd prime number

Solve via two methods

Answer 275

Absolute value / inequality: must be true
> try to find counterexample and eliminate
> also don’t forget to use the TRUTH

|1-x| < 2 —–> EXPAND AND DETERMINE RANGE OF VALUES FOR X

Case 1: 1-x >= 0 —-> 1 >= x, therefore x > -1
Combined … -1 < x =< 1
(don’t forget = sign)

Case 2: 1-x =< 0 —> 1 =< x, therefore x < 3
Combined … 1 =< x < 3

So the potential values of integer x are: [0, 1, 2]

Counterexamples exist for all except for D:
A - x can be a prime number e.g., x = 2
B - x^2 + x can be a prime number e.g., x = 1
C - x can be negative or equal to 0
E - x can be a multiple of an odd prime number e.g., x = 0, zero is a MULTIPLE of every integer *****

D must be true because the potential values of x [0, 1, 2] and the number of distinct positive factors of x+2 is always a prime number

Visual approach:
|1-x| represents the DISTANCE between 1 and x

If that distance < 2 and x must be an integer, then x can only be [0, 1, 2]