Problem Solving

Question 1

If  Sqrt((x + 4)^2) = 3, which of the following could be the value of x – 4?
A. -11
B. -7
C. -4
D. -3
E. 5

Answer 1

CAT 1, Q 2. Answer A. Exponents & Roots

Sqrt((x + 4)^2) = 3
(x + 4)^2 = 9
(x + 4) = Sqrt(9)

NOTE: Even exponents hide the sign of the base, so there are two solutions to the equation

(x + 4) = 3
x = -1 
OR
(x + 4) = -3
x = -7

x – 4 = -7 – 4 = -11 OR x – 4 = -1 – 4 = -5

Alternatively, the expression can be simplified to |x + 4|, and the original equation can be solved accordingly.
If |x + 4| = 3, either x = -1 or x = -7

Question 2

A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?
A. 1/9
B. 1/6
C. 2/9
D. 5/18
E. 1/3

Answer 2

CAT 1, Q 8. Answer D. Probability. Combination (order does not matter)

Need to determine (i) number of possible teams of five that can be formed with nine players (ii) number of teams of five that will include John and Peter

(i) nCk = n! / k!(n – k)! = 9! / (5!4!) = (9 * 8 * 7 * 6 * 5) / (5 * 4 * 3 * 2 * 1) = 9 * 2 * 7 = 126
(ii) Reserve two of five spots, which means three spots are left that must be filled by three of the remaining seven players. 7C3 = 7! / (3!4!) = 35

35 of the total possible 126 teams will include John and Peter. Probability that the coach chooses a team that includes both John and Peter is 35/126 = 5/18

Question 3

In the addition shown below, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
  ABC
\+BCB
  CDD
A. 8
B. 10
C. 12
D. 14
E. 18

Answer 3

CAT 1, Q 9. Answer B. Extra problem types. What we know:
B + C = D
B + C is less than or equal to 9
A + B = C

In order to maximize A*B, only consider cases in which B + C = 9

B = 1, C = 8 means A = 7, which results in A*B = 7
B = 2, C = 7 means A = 5, which results in A*B = 10
B = 3, C = 6 means A = 3, which results in A*B = 9
B = 4, C = 5 means A = 1, which results in A*B = 4

Largest possible value is 10

Question 4

Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover the exact distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?
A. 60
B. 72
C. 84
D. 90
E. 120

Answer 4

CAT 1, Q 11. Answer D. Rates & Work.
R(TOM) = 6 mph
R(LINDA) = 2 mph but she leaves one hour before Tom
D(TOM) = 6T
D(LINDA) = 2(T + 1)

Tom has covered Linda’s distance: 6T = 2T + 2, which results in T = 30min
Tom has covered twice Linda’s distance: 6T = 2(2T + 2), which results in T = 120min
Positive difference = 120 – 30 = 90min

Question 5

Each light bulb at Hotel California is either incandescent or fluorescent. At a certain moment, forty percent of the incandescent bulbs are switched on, and ninety percent of the fluorescent bulbs are switched on. If eighty percent of all the bulbs are switched on at this moment, what percent of the bulbs that are switched on are incandescent?
A. 22 (2/9)%
B. 16 (2/3)%
C. 11 (1/9)%
D. 10%
E. 5%

Answer 5

CAT 1, Q 15. Answer D. Percents. Although there are two pairs of mutually exclusive classifications – incandescent vs. fluorescent, and on vs. off – this is not best approached as an overlapping set problem. the question is only concerned with the number of light bulbs that are turned on, and be approached as a mixtures problem

I = # of Incandescent light bulbs
F = # of Fluorescent light bulbs
I + F = Total

I(ON) = 0.4I 
F(ON) = 0.9F
I+F(ON) = 0.8(I + F)

.8I + 0.8F = 0.4I + 0.9F
F = 4I

% of Incandescent bulbs that are switched on = 0.4I / (0.8I + 0.8F) = 0.4I / (0.8I + 0.8(4I) = 10%

Question 6

Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?
A. 2
B. 5
C. 6
D. 8
E. 9

Answer 6

CAT 1, Q 23. Answer C. Overlapping Sets. Group 1 (Poetry): 22 
Group 2 (History): 27
Group 3 (Writing): 28 

The easiest way to answer this overlapping sets questions is to use the following formula:
Total = Group 1 + Group 2 + Group 3 – (People in 2 Groups) – 2(People in All 3 Groups) + None
59 = 22 + 27 + 28 – 6 – 2x + 0
59 = 71 – 2x
x = 6

*NOTE: People in all 3 groups are counted 3 times, once for each group that they are in. As a result, they are double counted twice, and we must adjust for this double counting

Question 7

If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5

Answer 7

CAT 1, Q 36. Answer B. Consecutive Integers. RULE: The product of three consecutive integers will always be a multiple of three.

If x, x – 1, and x – k are consecutive integers, their product must be divisible by 3. Note that x and x – 1 are consecutive, so the three terms would be consecutive if (x – k) is either the lowest of three (k = 2) or the greatest of the three (k = -1).

Note that the difference between k = -1 and k = 2 is 3. Every third consecutive integer would serve the same purpose in the product x(x – 1)(x – k): periodically serving as the multiple of three in the list of consecutive integers. Thus, k = -4 and k = 5 would also give us a product that is always divisible by three.

When k = -1, x(x – 1)(x – k) is not evenly divisible by three

Question 8

There is a number line that ranges from 0 to 1. The segment from 0 to 1 has been divided into fifth, and also into sevenths. What is the least possible distance between any two of the tick marks?
A. 1/70
B. 1/35
C. 2/35
D. 1/12
E. 1/7

Answer 8

Lecture # 2, Q 162. Answer B. Find the least common denominator (35). Draw a number line and accurately divide the 0 to 35/35 segment into fifths (7, 14, 21, 28, 35) and into sevenths (5, 10, 15, 20, 25, 30, 35). In doing so, we can see that the least possible distance between any two of the tick marks is 1/35 (e.g. 15/35 – 14/35).

Question 9

At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?
A. 1/2
B. 2/5
C. 3/5
D. 4/5
E. 5/8

Answer 9

OG13, Lecture # 2, Q151. Answer E. Draw a table and choose Smart Numbers.

Columns: Day, Night, Total
Rows: Boxes, Workers, Total

Boxes: Day 4 per, Night 3 per
Workers: Day 5, Night 4
Total: Day 20, Night 12

Day Crew / Total = 20/32 = 5/8

Question 10

If (0.0015 * 10^m) / (0.03 * 10^k) = 5 * 10^7, then m – k =
A. 9
B. 8
C. 7
D. 6
E. 5

Answer 10

Answer A. Roots & Exponents
(0.0015 / 0.03) * (10^m / 10^k) = 5 * 10^7
(15 / 3) * (10^-4 / 10^-2) * (10^m / 10^k) = 5 * 10^7
5 * 10^-2 * 10^(m-n) = 5 * 10^7
10^(-2 + m – n) = 10^7
-2 + m – k = 7
m – k = 9

Question 11

At a certain school, the ratio of the number of second graders to the number of fourth graders is 8 to 5, and the ratio of the number of first graders to the number of second graders is 3 to 4. If the ratio of the number of third graders to the number of fourth graders is 3 to 2, what is the ratio of the number of first graders to the number of third graders?
A. 16 to 15
B. 9 to 5  
C. 5 to 16
D. 5 to 4
E. 4 to 5

Answer 11

OG13, HW2, Q66. Answer E. Ratios. Because the ratio involves four different grades, using variables will generally just make this kind of problem harder. Instead we can put all of the information into a table and then use that table to merge the ratios by finding common terms.

1st : 2nd : 3rd : 4th
        8                : 5	
3    : 4 
                   3     : 2
The next step is to start merging rows by multiplying the ratios in order to get the numbers in each column to match up. This works because the ratio can be thought of as fractions, and it does not change the value of the fraction to multiply both the numerator and denominator by the same number.

3*4 = 12
8*2 = 16
4*4 = 16
3*5 = 15
5*2 = 10
2*5 = 10

1st : 2nd : 3rd : 4th
12 : 16 : 15 : 10

1st : 3rd = 12 : 15 = 4 : 5

Question 12

On a recent trip, Cindy drop her care 290 miles, rounded to the nearest 10 miles, and used 12 gallons of gasoline, rounded to the nearest gallon. The actual number of miles per gallon that Cindy’s car got on this trip must have been between
A. 290/12.5 and 290/11.5
B. 295/12 and 285/11.5
C. 285/12 and 295/11.5
D. 285/12.5 and 295/11.5
E. 295/12.5 and 285/11.5

Answer 12

OG13, HW2, Q142. Answer D. Rates & Work. To optimize the value of mpg in both directions, we should consider extreme values of d and g. 285 ≤ distance

Question 13

The value of (2^-14 + 2^-15 + 2^-16 + 2^-17)/5 is how many times the value of 2^-17?
A. 3/2
B. 5/2
C. 3
D. 4
E. 5

Answer 13

OG13, Lecture # 3, Q 230. Answer C. Exponents & Roots
(2^-14 + 2^-15 + 2^-16 + 2^-17)/5 = x * 2^-17
x = (2^-14 + 2^-15 + 2^-16 + 2^-17) / (5 * 2^-17)
x = 2^-17 * (2^3 + 2^2 + 2^1 + 1)/(5 * 2^-17)
x = (2^3 + 2^2 + 2^1 + 1)/5
x = 3

Question 14

Sqrt[(16)(20) + (8)(32)]
A. 4√20
B. 24
C. 25
D. 4√20 + 8√2
E. 32

Answer 14

OG13, Lecture # 3, Q 35. Answer B. Exponents & Roots. 
= Sqrt[(16)(20) + (8)(16)(2)]
= Sqrt[(16)(20 + 16)]
= Sqrt[(16)(36)]
= Sqrt(16) * Sqrt(36)
= 4 * 6
= 24

Question 15

If s equals the multiplication of the numbers 100 through 200, inclusive, and t equal the multiplication of the numbers 100 through 201, inclusive, then what is the value of 1/s + 1/t?
A. (201^2) / t
B. (202)(201) / t
C. 201 / t
D. 202 / t
E. (202)(201) / (t^2)

Answer 15

OG12, Lecture # 3, Q 89. Answer D. Algebraic Translations
t = 201 * s
s = t / 201

1/s + 1/t
= 1/(t/201) + 1/t
= 201/t + 1/t
= 202/t

Question 16

If 4 is one of the solutions of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?
A. -7
B. -4
C. -3
D. 1
E. 6

Answer 16

OG13, HW 3, Q 45. Answer A. Quadratic Equation. We are told that one solution is 4, which we can plug into the equation to find the value of k. Once we know the value of k, we can factor the quadratic. In order to get positive 3x, we need a positive 7, which results in -7 as the other solution to the equation.

x^2 + 3x + k = 10
4^2 + 3(4) + k = 10
16 + 12 + k = 10
k = -18

x^2 + 3x – 18 = 10
x^2 + 3x – 28 = 0
(x – 4)(x + 7) = 0
x = 4 OR x = -7

Question 17

If n = 3^8 – 2^8, which of the following is NOT a factor of n?
A. 97
B. 65
C. 35
D. 13
E. 5

Answer 17

OG13, HW 3, Q 117. Answer C. Quadratic Equations. The question asks for the answer choice that is NOT a factor of n. Thus, 4 of the 5 answer choices must be factors of n. In general, the best way to find the factors of a number is to break it down into its prime factors. Calculating 3^8 – 2^8 in less than two minutes without a calculator would be very difficult, and finding the factors of a large number is typically difficult. Both of these factors point to there being a shortcut built into the problem.

This a difference of squares problem.
3^8 – 2^8 = (3^4 – 2^4)(3^4 + 2^4) = (81 – 16)(81 + 16) = 65 * 97

We can eliminate answer choices A and B, but there are still two answer choices to go. Although 97 is prime, 65 can be further broken down into 13 * 5, which eliminates answer choices D and E.

Question 18

Given that x + y = a and x – y = b, what is the value of 2xy?
A. (a^2 - b^2)/2
B. (b^2 - a^2)/2
C. (a - b)/2
D. (ab)/2
E. (a^2 + b^2)/2

Answer 18

OG12, Lecture # 4, Q 227. Quadratic Equations. Answer A. Set the problem up as two systems of equations OR PLUG IN.
x + y = a
+(x – y) = b
2x = a + b

x + y = a
-(x – y) = b
2y = a - b
y = (a - b)/2

2xy = (a+b)(a-b)/2 = (a^2 – b^2)/2

Takeaway:

• PLUG IN when variables appear in the answer choices
• ALWAYS run your chosen numbers through all five answer choices

Question 19

If d > 0 and 0  0
II. c/d  1
A. I only
B. II only
C. I and II only
D. II and III only
E. I, II and III

Answer 19

Quant2, Lecture # 4, Q 134. Fractions. Answer C. Split the compound inequality into:
c/d 0 is TRUE. Given that d > 0 and c/d > 0, c must be positive.
II. c/d 1 is FALSE. Plug in values to prove insufficiency
0.2^2 + 0.4^2 > 1
0.04 + 0.16 > 1 IS NOT TRUE

Takeaways:

• Always split compound inequalities
• Plug in with the goal of proving insufficiency

Question 20

A store currently charges the same price for each towel that it sells. If the current price of each towel were to be increased by $1, 10 fewer of the towels could be bought for $120, excluding sales tax. What is the current price of each towel?
A. $1
B. $2
C. $3
D. $4
E. $5

Answer 20

OG 13, Lecture # 5, Q 203. Algebraic Translation. Answer C. There are two approaches to this problem (1) direct algebra (2) back solve. Either way, because we are given categories, we should set up a table. Note that under both the current and future scenarios, total cost is $120.

Current: P * Q = 120
Future (P + 1) * (Q – 10) = 120

The two equations are not linear, so solving for the variables will involve quadratics, which will take a fair amount of time. Switch gears and back solve
Test the middle number first, $3. Current Q would equal 40. Future Q would equal 4(Q – 10) = 120, or Q = 40. This is the answer.

Takeaways:

• Any time the GMAT categorizes (e.g. current vs future), then use a table
• Back solve when (1) problem solving question (2) values are given in the answer choices (3) questions asks for a value, not a relationship (e.g. how much more does Joe earn than Susan)

Question 21

Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had Train X traveled when it met Train Y?
A. 37.5
B. 40.0
C. 60.0
D. 62.5
E. 77.5

Answer 21

Quant2, Lecture # 5, Q 119. Rates & Work. Answer A. There are two approaches to this problem. Alternative # 1 is to set up a table that tracks the progress of each train and the total combined progress at intervals in time.

0hr: X = 0, Y = 0, Total = 0
1hr: X = 20, Y = 33.3, Total = 53.3
2hr: X = 40, Y = 66.6, Total = 106.6…which means the train must have traveled for less than two hours since the total distance of the route is only 100 miles. Thus, train X has traveled less than 40 miles, which tells us that Answer A is Correct.

Alternative # 2 is to set up a RTD table and add the rates since the trains are moving in opposite directions. Thus, X has traveled D = 20 * (15/8) = 37.5 miles
X: 20mph * 5hrs = 100mi
Y: 100/3mph *  3hrs = 100mi
X + Y: 160/3mph * T = 100mi
T = (100 * 3) / 160 = 15/8 

Takeaways:

• Objects moving in opposite directions (toward or away) – ADD RATES
• Objects moving in the same direction – SUBTRACT RATES (imagining that the first guy is at d = 0)

Question 22

Car A is 20 miles behind Car B, which is traveling in the same direction along the same route as Car A. Car A is traveling at a constant speed of 58 miles per hour and Car B is traveling at a constant speed of 50 miles per hour. How many hours will it take for Car A to overtake and drive 8 miles ahead of Car B?
A. 1.5
B. 2.0
C. 2.5
D. 3.0
E. 3.5

Answer 22

OG 13, Lecture # 5, Q 207. Rates & Work. Answer E. There are three approaches to this problem. Alternative # 1 is to set up a table that tracks the progress of each car and the difference at intervals in time.

0hr: A = 0, B = 20, Difference = - 20
1hr: A = 58, B = 70, Difference = -12
2hr: A = 116, B = 120, Difference = -4
3hr: Difference = +4
4hr: Difference = + 12…which means it takes car A between 3 and 4 hours to drive 8 miles ahead of Car B. Thus, the answer is 3.5 hours

Alternative # 2 is that to use the following formula for relative rates:
(Rate A – Rate B) * T = Difference in Distance
(58 – 50) * T = 28
8T = 28
T = 28/8 = 3.5 hours

Alternative # 3 is to set up a RTD table and use substitution.
A: 58T = D + 28
B: 50T = D
Using substitution, 58T = (50T) + 28, which means T = 3.5 hours

Question 23

Working alone, Printers X, Y, and Z can do a certain printing job, consisting of a large number of pages, in 12, 15, and 18 hours, respectively. What is the ratio of the time it takes Printer X to do the job, working alone at its rate, to the time it takes Printers Y and Z to do the job, working together at their individual rates?
A. 4/11
B. 1/2
C. 15/22
D. 22/15
E. 11/4

Answer 23

Quant2, Lecture # 5, Q 130. Rates & Work. Answer D. The best way to solve this problem is to set up a RTW table and pick a Smart Number. 
X: 15 * 12 = 180
Y: 12 * 15 = 180
Z: 10 * 18 = 180
Y + Z: 22 * 180/22 = 180

Time: X / (Y + Z) = 12 / (180/22) = (12 * 22) / 180 = (6 * 2 * 2 * 11) / (6 * 2 * 15) = 22/15

Takeaways:

• Save time by picking SMART NUMBERS!
• Working together – ADD rates

Question 24

A case contains c cartons. Each carton contains b boxes, and each box contains 100 paper clips. How many paper clips are contained in 2 cases?
A. 100bc
B. 100b/c
C. 200bc
D. 200b/c
E. 200/bc

Answer 24

OG13, HW 5, Q 4. Answer C. Algebraic Translations. Two alternatives – direct algebra or plug in.
Alternative # 1 – Direct Algebra.
1 case = c cartons
1 carton = b boxes
1 box = 100 paper clips
1 case = bc boxes = bc(100 paper clips) = 100bc paper clips
2 cases = 200bc paper clips

Alternative # 2 – Plug In. Let c = 3 and b = 2. Each case must contain 32 = 6 boxes, and each box contains 100 paper clips for a total of 6100 = 600 paper clips. Therefore, 2 cases contain 1,200 paper clips. Plug c = 3 and b = 2 into each answer choice to find 1,200
A. 100bc = 100(2)(3) = 600 INCORRECT
B. 100b/c = (100*2)/3 = 200/3 INCORRECT
C. 200bc = 200(2)(3) = 1,200 CORRECT

Question 25

On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $0.60 per glass on the first day, what was the price per glass on the second day?
A. $0.15
B. $0.20
C. $0.30
D. $0.40
E. $0.45

Answer 25

OG 13, HW 5, Q 60. Answer D. Algebraic Translation. Two alternatives – Direct Algebra or Ratios.
Alternative # 1. Use a chart in order to help translate this problem.
Day: Amount of OJ + Amount of Water = Amount of Orangeade
Day 1: j + j = 2j
Day 2: j + 2j = 3j

We are told that the revenue from selling the orangeade was the same for both days, and the price per glass on day 1 was $0.60. Although we do not know the number of glasses sold on day 1 or day 2, we do know the total amount of orangeade sold on both days. We can use 2j to represent the number of items sold on day 1, and 3j to represent the number of items sold on day 2. Finally, let p represent the price per glass on day 2.
Day 1: Revenue = 0.6 * 2j = 1.2j
Day 2: Revenue = p * 3j = 3jp

1.2j = 3jp
p = $0.40 

Alternative # 2. (3/2) = (0.60/p) = 3p = 1.2. Thus, p = 0.4

Question 26

A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase?
A. 10
B. 11
C. 12
D. 13
E. 14

Answer 26

OG13, HW 5, Q 64. Answer B. Algebraic Translations. A + B = ? Note the integer constraint. We know that the values of A and B must be integers. The way to solve this problem is to start listing possibilities and stop as soon as we find one that works because only one answer can be the correct one.
0.70A + 0.50B = 6.30
7A + 5B = 63
Apples, Cost of Apples, Amount Remaining to Purchase Bananas, # Bananas
1, 7, 63 – 7 = 56, 56 is not divisible by 5
2, 14, 56 – 7 = 49, 49 is not divisible by 5
3, 21, 49 – 7 = 42, 42 is not divisible by 5
4, 28, 42 – 7 = 35, 35 IS divisible by 5

Thus, (4)(0.70) + (7)(0.5) = 2.80 + 3.50 = 6.30 works out to be 11 pieces of fruit

Question 27

Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?
A. x/(x + y)
B. y/(x + y)
C. xy/(x + y)
D. xy/(x – y) 
E. xy/(y – x)

Answer 27

OG 13, HW 5, Q 81. Answer E. Rates & Work. There are two alternatives – direct algebra and plug in.
Alternative # 1. Set up a Rate Chart and solve algebraically
R * T = W
Machine A: 800/y * y = 800
Machine B: 800/b * b = 800
Together: 800/x * x = 800

We know that A & B’s combined rate is the sum of A’s and B’s separate rates, so we can compute Machine B’s rate by subtracting A’s rate from A and B’s combined rate
800/b = 800/x – 800/y 
800/b = 800y/xy – 800x/xy
800/b = 800(y – x)/xy
800(y – x)b = 800xy
b = xy / (y – x)

Alternative # 2. Let x = 8 and y = 10 and plug into the answer choices looking for a value of 40.
Machine A: 80 * 10 = 800
Machine B: 20 * 40 = 800
Together: 100 * 8 = 800
C. xy/(x + y) = 8*10 / 18 = NOT INTEGER
D. xy/(x – y) = 80 / (8 – 10) = NOT NEGATIVE
E. xy/(y – x) = 80 / (10 – 8) = 40 CORRECT

Question 28

At his regular hourly rate, Don had estimated the labor cost of a repair job as $336 and he was paid at that amount. However, the job took 4 hours longer than he had estimated and, consequently, he earned $2 per hour less than his regular hourly rate. What was the time Don had estimated for the job, in hours?
A. 28
B. 24
C. 16
D. 14
E. 12

Answer 28

OG13, HW 5, Q 137. Answer B. Algebraic Translations. There are two alternatives – Direct Algebra and Back Solve. 
Alternative # 1 – Back solve. We know 336 = hr AND 336 = (h + 4)(r – 2). Test the answers until you find one that matches the rate equation. 
Estimated Hours (h), Actual Hours (h + 4), Estimated Hourly Rate (r), Actual Hourly Rate (r – 2)
28, 32, 336/28 = 12, 336/32 = 10 something -- NO MATCH
24, 28, 336/24 = 14, 336/28 = 12 -- CORRECT
16, 20, 336/16 = 21, 336/20 = not an integer -- NO MATCH
14, 18, 336/14 = 24, 336/16

Question 29

During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine’s average speed for the entire trip?
A. (180 – x)/2
B. (x + 60)/4
C. (300 – x)/5
D. 600/(115 – x)
E. 12,000/(x + 200)

Answer 29

OG13, HW 5, Q 162. Answer E. Rates & Work. We can make this difficult average rates problem more approachable by choosing Smart Numbers either (1) for the total distance traveled or (2) for x

Smart Number for Total Distance = 100
1st Leg: 40 * 40/x = x
2nd Leg: 60 * (100 – x)/60 = 100 – x
Total: ? * (x/40) + (100 – x)/60 = 100

R = 100 / [(x/40) + (100 – x)/60] = 100 / [(30x/120) + (200 – 2x)/120] = 100 / (x + 200)/120 = 100*120 / (x + 200) = 12,000 / (x + 200)

Smart Number for x = 40
1st Leg = 40 * 1 = 40
2nd Leg = 60 * 1 – 60
Total = ? * 2 = 100

R = 50 – now plug in to the answer choices

Question 30

10% of GMAT test takers take the exam more than once and have a high score of 650 or greater, and 20% of GMAT test takers who take the only once have a score of 650 or greater. If 40% of GMAT test takers take the exam more than once, what percent of GMAT test takers have a high score of less than 650?
A. 10.5%
B. 22%
C. 30%
D. 70%
E. 78%

Answer 30

MGMAT, Online Lecture #6. Answer E. Overlapping Sets. Choose a smart number for the total (100) and set up a double-set matrix in order to calculate the answer.

Score 650: More than Once = 10, Once Only = ?, Total = ?
Total: More than Once = 40, Once Only = ?, Total = 100

Score 650: More than Once = 10, Once Only = 0.2(60) = 12, Total = 22
Total: More than Once = 40, Once Only = 60, Total = 100

Question 31

In a group of 68 students, each student is registered for at least one of three classes – History, Math, and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?
A. 13
B. 10
C. 9
D. 8
E. 7

Answer 31

MGMAT, Section #6. Answer B. Overlapping Sets. Set up a simple Venn diagram in order to help visualize the problem and then use the standard 3-set equation.

Total Actual # = Group 1 + Group 2 + Group 3 – (People in Exactly 2 Groups) – 2(People in All 3 Groups) + None

Group 1 + Group 2 + Group 3 = 84
-(People in Exactly 2 Groups) = -x
-2(People in All 3 Groups) = -2(3) = -6
\+People in Neither Group = 0
= Actual Total Number of People = 68

68 = 84 – x – 6 
x = 10

Question 32

Of the 50 researchers in a workgroup, 40 percent will be assigned to Team A and the remaining 60 percent to Team B. However, 70 percent of the researchers prefer Team A and 30 percent prefer Team B. What is the lowest possible number of researchers who will NOT be assigned to the team they prefer?
A. 15
B. 17
C. 20
D. 25
E. 30

Answer 32

OG13, HW 6, Q 25. Answer A. Overlapping Sets. Note that, although we want to maximize the number of people in the top-left and middle boxes, we are constrained by the totals. Because only 15 people prefer B, the maximum number of people who can be assigned to the middle box is 15. A hidden constraint is at work here: the number of people in any box cannot be negative. Similarly, only 20 people are assigned to A, so 20 is the maximum number of people who can be in the top-left box.
Prefer A: Assigned to A = 20, Assigned to B = 15, Total = 35
Prefer B: Assigned to A = 0, Assigned to B = 15, Total = 15
Total: Assigned to A = 20, Assigned to B = 30, Total = 50

Question 33

Of the 300 subjects who participated in an experiment using virtual-reality therapy to reduce their fear of heights, 40 percent experienced sweaty palms, 30 percent experienced vomiting and 75 percent experienced dizziness. If all of the subjects experienced at least one of these effects and 35 percent of the subjects experienced exactly two of these effects, how many of the subjects experienced only one of these effects?
A. 105
B. 125
C. 130
D. 180
E. 195

Answer 33

OG13, HW 6, Q 178. Answer D. Overlapping Sets. Draw a Venn Diagram in order to visualize the three overlapping sets in this problem. Utilize the three set formula to determine the number of people who experienced only one of these side effects.

Actual Total = Group 1 + Group 2 + Group 3 – (People in 2 Groups) – 2(People in All 3 Groups) + None
300 = 120 + 90 + 225 – 105 – 2(x) + 0
300 = 330 – 2x
2x = 30
x = 15

People in ONLY one group = Actual Total – People in 2 Groups – People in All 3 Groups = 300 – 105 – 15 = 180

Takeaway:

• Memorize the overlapping set formulas; given 3 groups (x, y, z)
  (1) Act Tot = Only(X) + Only(Y) + Only(Z) + Only(XY) + Only(YZ) + Only(XZ) + XYZ
  (2) Act Tot = Tot(X) + Tot(Y) + Tot(Z) – (XY + YZ + XZ) – 2(XYZ)

Question 34

Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest possible piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?
A. 82
B. 118
C. 120
D. 134
E. 152

Answer 34

OG13, HW 6, Q 183. Answer D. Statistics. Given Mean = 68 and Median = 84. First, the sum is equal to 68 * 7 = 476. Second, if the median length is 84, then the 4th longest length is 84. Call the shortest length S and the longest length L. We are told that L = 4S + 14, which we want to maximize. List out the numbers in the set:

S, a, b, 84, c, d, 4S + 14

The 7 pieces of rope have are limited by the fixed sum of 476. If we want to maximize the length of the longest piece, we need to minimize the lengths of the first six pieces. Since c and d have to be at least as long as the median, Minimum(c, d) = 84. By a similar logic, Minimum(a,b) = S. Rewrite the numbers in the set and set the sum equal to 476 in order to solve for S.

S + S + S + 84 + 84 + 84 + 4S + 14 = 476
7S + 266 = 476
7S = 210
S = 30
L = 4(30) + 14 = 134

Question 35

Of the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from 
A. 20 to 50
B. 40 to 70
C. 50 to 130
D. 110 to 130
E. 110 to 150

Answer 35

OG13, HW 6, Q 222. Answer D. Overlapping Sets. We must determine the minimum and maximum number of students who could be majoring in both chemistry and biology. Set up a double-set matrix

Minimum
Bio: Chem ≥ 110, No Chem ≤ 40, Total = 150
No Bio: Chem ≤ 20, No Chem ≥ 30, Total = 50
Total: Chem = 130, No Chem = 70, Total = 200

Maximum
Bio: Chem ≤ 130, No Chem ≥ 0, Total = 150
No Bio: Chem ≥ 0, No Chem ≥ 30, Total = 50
Total: Chem = 130, No Chem = 70, Total = 200

Question 36

Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X?
A. 10%
B. 33 1/3%
C. 40%
D. 50%
E. 66 2/3%

Answer 36

OG13, HW 6, Q 224. Answer B. Percents. Set up a mixture table in order to determine what equation we need to write.

0.4X + 0.25Y = 0.3(X + Y)
0.4X + 0.25Y = 0.3X + 0.3Y
0.1X = 0.05Y
X/Y = 1/2

Mixture X as % of Total Weight = X/(X + Y) = 1/3 = 33.3%

Question 37

Simplify the following expression: √1 / [(3^0) + (3^-1) + (3^-2) + (3^-3) + (3^-4)]
A. 1/11
B. 9/11
C. 11/9
D. 11
E. 3^10

Answer 37

CAT 2, Q4. Answer B. Exponents & Roots. There are three approaches to this problem.
(1) Pull out a common term (my first intuition)
= √[1 / (3^-4(3^0 + 3^1 + 3^2 + 3^3 + 3^4))]
= √[1 / (3^-4 * 121)]
= √(81 / 121)
= 9/11

(2) Get rid of the negative exponents by multiplying by 3^4 / 3^4
= √[1 / (3^0 + 3^-1 + 3^-2 + 3^-3 + 3^-4) * (3^4 / 3^4)]
= √[3^4 / (3^0 + 3^1 + 3^2 + 3^3 + 3^4)]
= √(81/121)
= 9/11

(3) Get a common denominator 
= √[1 / (1 + 1/3 + 1/9 + 1/27 + 1/81)]
= √[1 / ((81 + 27 + 9 + 3 + 1) / 81)]
= √[1 / ((81 + 27 + 9 + 3 + 1) / 81)]
= √[81 / (81 + 27 + 9 + 3 + 1)]
= √(81/121)
= 9/11

Question 38

Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joseph, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?
A. 1/3x
B. x/(x – 3)
C. (x – 1)/3x
D. x/(x – 1)
E. (x – 1)/x

Answer 38

CAT 2, Q6. Answer C. Rates & Work. There are two approaches to this problem
(1) Plug In (preferred). Let x = 4, which means Lindsay can paint 1/4 of the room in 1 hour and her rate is 1/4. This implies that Joseph can pain 3/4 of the room in one hour. At that rate, he could paint 1/4 of the room in 20 minutes. Only Answer Choice C equals 1/4 when you plug in 4 as the value for x: (4 – 1) / (3 * 4) = 1/4

(2) Direct Algebra (more cumbersome) 
1/x + 1/y = 1
1/y = 1 – (1/x)
Rate = 1/y = (x – 1)/x 
Work = (1/3)(x – 1)/x = (x – 1) / 3x

Question 39

Each book on a certain shelf is labeled by a single category. For every 2 history books, there are 7 fantasy books and for every 3 fantasy books, there are 5 reference books. If the proportion of history to reference books is doubled, while the proportion of fantasy to reference books is maintained, which of the following could be the number of history books if there are fewer than 60 fantasy books on the shelf after the changes?
A. 6
B. 21
C. 24
D. 35
E. 36

Answer 39

CAT 2, Q11. Answer C. Ratios. This ratios problem asks for the value that COULD be the number of fantasy books on a certain shelf. The ratios both contain fantasy books. To combine the two ratios, make the number of fantasy books consistent across the two equations. Books must come in integer units, so the smallest combined number of fantasy books is the least common multiple of 7 and 3, or 21. Multiply the first ratio by 3 and the second ratio by 7:
H:F = 6:21
F:R = 21:35
The combined ratio is:
H:F:R = 6:21:35

The proportion of history to reference books is doubled but the proportion of fantasy to reference books is unchanged. Don’t change the fantasy or reference numbers. Double the history component from 6 to 12. The new ratio is:
H:F:R = 12:21:35

21x

Question 40

Six mobsters have arrived at the theater for the premiere of the film “Goodbuddies.”  One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him.  How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?
A. 6
B. 24
C. 120
D. 360
E. 720

Answer 40

CAT 2, Q12. Answer D. Combinatorics. Ignoring Frankie’s requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie’s requirement, the six mobsters could be arranged in 720/2 = 360 different ways

Question 41

The ACME company manufactured x brooms per month from January to April, inclusive. On the first of each month, during the following May to December, inclusive, it sold  x/2 brooms. At the beginning of production on January 1st, the ACME company had no brooms in its inventory. If storage costs were $1 per month per broom, approximately how much, in terms of x, did the ACME company pay for storage from May 2nd to December 31st, inclusive?
A. $x
B. $3x
C. $4x
D. $5x
E. $14x

Answer 41

CAT 2, Q13. Answer E. Algebraic Translations. Since this problem includes variables in both the question and the answer choices, we can try solving by plugging in smart numbers. For x, we want to choose a multiple of 2 because we will have to take x/2 later. Let’s say that ACME produces 4 brooms per month from January to April, so x = 4. The total number of brooms produced was (4 brooms x 4 months), or 16 brooms.

ACME sold x/2 brooms per month, or 2 brooms per month (because we chose x = 4). Now we need to start figuring out the storage costs from May 2nd to December 31st.

Since ACME sold 2 brooms on May 1st, it needed to store 14 brooms that month, at a cost of $14. Following the same logic, we see that ACME sold another two brooms June 1st and stored 12 brooms, which cost the company $12. We now see that the July storage costs were $10, August were $8, September $6, October $4, November $2, and for December there were no storage costs since the last 2 brooms were sold on December 1st.

So ACME’s total storage costs were 14 + 12 + 10 + 8 + 6 + 4 + 2 = $56. Now we just need to find the answer choice that gives us $56 when we plug in the same value, x = 4, that we used in the question. Since 14 x 4 = 56, $14x must be the correct value.

Question 42

If n is a non-negative integer such that 12^n is a divisor of 3,176,793, what is the value of n^12 – 12^n?
A. -11
B. -1
C. 0
D. 1
E. 11

Answer 42

CAT 2, Q16. Answer B. Exponents & Roots. Since n must be a non-negative integer, n must be either a positive integer or zero. Also, note that the base of the exponent 12^n is even and that raising 12 to the nth exponent is equivalent to multiplying 12 by itself n number of times. Since the product of even integers is always even, the value of 12^n will always be even as long as n is a positive integer. For example, if n = 1, then 121 = 12; if n = 2, then 122 = 144, etc.

Since integer 3,176,793 is odd, it cannot be divisible by an even number. As a result, if n is a positive integer, then 12^n (an even number) will never be a divisor of 3,176,793. However, if n is equal to zero, then 12^n = 120 = 1. Since 1 is the only possible divisor of 3,176,793 that will result from raising 12 to a non-negative integer exponent (recall that all other outcomes will be even and thus will not be divisors of an odd integer), the value of n must be 0.

012 – 120 = 0 – 1 = -1

Question 43

A driver paid n dollars for auto insurance for the year 1997. This annual premium was raised by p percent for the year 1998; for each of the years 1999 and 2000, the premium was decreased by 1/6 from the previous year’s figure. If the driver’s insurance premium for the year 2000 was again n dollars, what is the value of p?
A. 12
B. 33 1/3
C. 36
D. 44
E. 50

Answer 43

CAT 2, Q18. Answer D. FDPs. First, translate the information into equations. The 2000 premium = (5/6)(5/6)(1 + p)(n) = n. From here, there are two methods.

(1) Pick Smart Numbers. Note that you can pick smart numbers for n but not for p. The value of p must match one of the answer choices. Let the 1997 premium be n = 100 and back solve.
D. (5/6)(5/6)(1.44)(100) = (5/6)(120) = 100

(2) Direct Algebra (not easy)
n = (5/6)(5/6)(1 + p)(n) 
1 = (25/36)(1 + p)
36/25 = 1 + p
P = 36/25 – 25/25 = 11/25 = 44/100

Question 44

The number x is the average (arithmetic mean) of the positive numbers a and b, and the reciprocal of the number y is the average (arithmetic mean) of the reciprocals of a and b. In terms of a and b, x – y = ?
A. (a^2 + b^2) / 2(a + b)
B. (a – b)^2 / 2(a + b)
C. (a + b) / 2
D. (a – b) / 2
E. 0

Answer 44

CAT 2, Q21. Answer B. Statistics. To solve this problem, find the expressions for x and y in terms of a and b, and perform the indicated subtraction (ie we are looking for the value of x – y).

x = (a + b)/2

1/y = (1/a + 1/b) / 2
2 = y (1/a + 1/b)
y = 2ab / (a + b)

x – y = (a + b)/2 – 2ab/(a + b)
x – y = (a + b)^2 / 2(a + b) – 4ab / 2(a + b)
x – y = (a^2 + 2ab + b^2 – 4ab) / 2(a + b)
x – y = (a^2 – 2ab + b^2) / 2(a + b)
x – y = (a – b)^2 / 2(a + b)

Question 45

ABCD is a square with two diagonal lines dividing the square into three regions of equal area, including two isosceles triangles and a central region.  If AB = 3, what is the approximate length of w, the perpendicular distance between the two diagonal lines?
A. 0.8
B. 1.2
C. 1.4
D. 1.8
E. 2.1

Answer 45

CAT 2, Q22. Answer A. Triangles & Diagonals. If AB = 3, then the area of square ABCD is 9, so each of the three regions has area 3. Using the 45−45−90 triangle ratios, the diagonal BD of the square has length 3√2. Label the rest of this diagonal:

Let Q equal the height of each isosceles triangle. w + 2Q = 3√2, so w = 3√2 – 2Q. To find Q, note that the two triangular regions can be put together to form a square with area 6 (because each of the regions has an area of 3). The area of the square is equal to s^2, so s = √6.

Using the 45−45−90 triangle ratios again, the diagonal of this smaller square (2Q) would be (√6)(√2) = √12 = 2√3. 2Q = 2√3, so Q = √3.

The value of w is thus 3√2 – 2Q = 3√2 – 2√3. Since the approximate value of √2 is 1.4 and the approximate value of √3 is 1.7, this approximates to 3(1.4) – 2(1.7) = 0.8.

ALTERNATIVE – EDUCATED GUESS. The diagram is drawn to scale, so the value of w can be estimated visually. The length of AB is 3; in the diagram, w appears to be around one-fourth of that length, so w should be somewhere around 3/4. Answer choice A, 0.8, is the closest value. The second closest answer, 1.2 (answer choice B), would require that w be 40% of the length of AB. Given the diagram, this seems rather unlikely.

Takeaway:
-PS questions are drawn to scale unless otherwise stated, data sufficiency questions are NOT

Question 46

If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =
A. (18 – 3y) / y^3
B. 18 / y^2
C. 18 (y^2 + 3y)
D. 9 / y^2
E. 36 / y^2

Answer 46

CAT 2, Q23. Answer D. Quadratic Equations. There are two approaches.

(1) Direct Algebra. The equation x^2 * y^2 = 18 – 3xy is really a quadratic, with the xy as the variable.

X^2 * y^2 + 3xy – 18 = 0
(xy + 6)(xy – 3) = 0
xy = 3 or -6

However, we are told that x and y are positive so xy must equal 3. Therefore, x = 3/y and x^2 = 9/y^2.

(2) Plug In. Alternatively, this is a VIC and can be solved by plugging numbers. If we plug a value for y and find the corresponding value of x, we can check the answers to see which one matches the value of x. Looking at the values 3 and 18 in the equation, a y value of 3 makes sense.
x^2 * (3)^2 = 18 – 3(x)(3) 
9x^2 = 18 – 9x
9x^2 – 9x + 18 = 0
X^2 – x + 2 = 0
(x + 2)(x – 1) = 0
x = 1, -2
But since x cannot be negative, x = 1. If we plug y = 3 into each of the answer choices, C and D both give an x value of 1. We must now plug in another value of y to decide between C and D

Question 47

A store sells a certain product at a fixed price per unit. At the product's current price, q units cost a total of exactly $300. If the price were lowered by $5 from its current value, then q + 2n units would cost exactly $300; if the price were raised by $5, then q – n units would cost exactly $300.  What is the value of q?
A. 10
B. 15
C. 20
D. 25
E. 30

Answer 47

CAT 2, Q31. Answer C. Algebraic Translations. Translate the question into three equations and then back solve (the algebra is FAR too cumbersome to do in two minutes.

(1) P*Q = 300
(2) (P – 5)(Q + 2n) = 300
(3) (P + 5)(Q – n) = 300

C. Given Q = 20, use equation (1) to determine that P = 15. Next, use equation (2) to determine that n = 5. Finally, use equation (3) to verify that the equation holds true. The result is 300, therefore we have a match!
D. Given Q = 25, use equation (1) to determine that P = 12. Next, use equation (2) to determine that n ~ 9. Finally, use equation (3) to verify that the equation holds true. The result is approximately 272, which is close but no match!

E. Given Q = 30, use equation (1) to determine that P = 10. Next, use equation (2) to determine that n = 15. Finally, use equation (3) to verify that the equation holds true. The result is 225, which is further away from the desired result!

Question 48

Jacob drove from Town A to Town B at an average rate of x miles per hour, then returned along the same route at y miles per hour.  If he then drove back to Town B at z miles per hour along the same route, what was Jacob’s average rate of speed for the entire trip, in miles per hour?
A. (x + y + z) / 3
B. 3xyz / (xy + yz + zx)
C. xyz / (x + y + z)
D. (xy + yz + zx) / (x + y + z)
E. 3(x + y + z) / xyz

Answer 48

CAT 2, Q32. Answer B. Rates & Work. This question asks for the average rate, which is total distance divided by total time. We don’t know the distance between the two towns, but we do know that the distance was the same for each leg of the journey. Let’s call it n miles. That means the total distance traveled is 3n.

Now we need to calculate the total number of hours that the trip took. For each leg of the trip, we know the distance traveled (n) and the rate at which Jacob traveled. Time = Distance / Rate.

The first leg of the trip took Jacob n/x hours. The second leg took n/y hours, and third leg took n/z hours. The total time is the sum of the three times, or n/x + n/y + n/z. Dividing these, we get:

Average Rate = Total Distance / Total Time
= 3n / (n/x + n/y + n/z)
= 3 / (1/x + 1/y + 1/z) Divide the numerator and denominator by n
= 3xyz / (yz + xz + xy) Multiply the numerator and denominator by xyz

Takeaway:
-Do not plug in when the answer choices are complicated

Question 49

Two sides of a triangle have lengths x and y and meet at a right angle.  If the perimeter of the triangle is 4x, what is the ratio of x to y?
A. 2:3
B. 3:4
C. 4:3
D. 3:2
E: 2:1

Answer 49

CAT 2, Q33. Answer B. Triangles & Diagonals. There are two approaches to this problem
(1) Back Solve (easiest). Plug answer choices back into the problem and check the truth of the final condition (i.e., the perimeter of the triangle must work out to 4x). Given a ratio x : y, we can select two specific values x and y having that ratio, use the Pythagorean theorem to find the hypotenuse of the triangle, add up all three sides to find the perimeter, and, finally, check whether the perimeter is indeed equal to 4x as required.
A. Let x = 2 and y = 3. Using Pythagorean’s Theorem, the hypotenuse of the triangle is √13. These values yield a perimeter of 5 + √13, which is not equal to 4x = 8
B. Let x = 3 and y = 4. Using Pythagorean’s Theorem, the hypotenuse of the triangle is 5. These values yield a perimeter of 12, which is equal to 4x = 12. We have a match!

(2) Direct Algebra. We know that the perimeter of the triangle equals 4x = x + y + c, where c is the hypotenuse, c = 3x – y

x^2 + y^2 = (3x – y)^2
x^2 + y^2 = 9x^2 – 6xy + y^2
0 = 8x^2 – 6xy
0 = 2x(4x – 3y)
x = 0 OR 4x – 3y = 0, which means x/y = 3/4

Question 50

Health insurance Plan A requires the insured to pay $1,000 or 50% of total cost, whichever is lower. Plan B requires the insured to pay the initial $300, but then pays 80% of the cost over $300. Which of the following is a cost level for which both insurance plans pay out the same amount?
A. $600
B. $1,000
C. $3,800
D. $5,300
E. $6,200

Answer 50

CAT 2, Q37. Answer C. Algebraic Translations. There are two approaches to this problem, but first translate the given information. Let C equal total cost, which is the variable for which we are solving.
Plan A Cost to Insured = min(1000, 0.5C)
Plan B Cost to Insured = 300 + 0.2(C – 300)

(1) Back Solve. When the two calculated Plan amounts are equal, we have found our answer
A. C = 600. Plan A Cost to Insured = 300. Plan B Cost to Insured = 300 + 0.2(600 – 300) = 360
B. C = 1000. Plan A Cost to Insured = 500. Plan B Cost to Insured = 300 + 0.2(1000 – 300) = 440
C. C = 3800. Plan A Cost to Insured = 1000. Plan B Cost to Insured = 300 + 0.2(3800 - 300) = 1000
D. C = 5300. Plan A Cost to Insured = 1000. Plan B Cost to Insured = 300 + 0.2(5300 - 300) = 1300
E. C = 6200. Plan A Cost to Insured = 1000. Plan B Cost to Insured = 300 + 0.2(6200 - 300) = 1480

(2) Direct Algebra. Because there are two possible payment structures for Plan A, so we need to set up 2 equations. We are looking for the cost level for which both plans pay out the same amount, so we can set the two plans equal to each other. The two equations are:

0.8(x – 300) = x – 1,000
x = 3800

0.8(x – 300) = .5x
x = 800

$800 isn’t one of our options, but $3,800 is.

Question 51

What is the area of a trapezoid with two parallel sides equal to 2 and 5 and the line perpendicular to these two sides equal to 12?
A. 39
B. 40
C. 42
D. 45
E. 46.5

Answer 51

OG12, Lecture # 7, Q 145. Answer C. Triangles & Diagonals. There are six possible approaches to this problem.

(1) Area of Trapezoid: h * (Base1 + Base2)/2 = 12 * (5 + 2)/2 = 42
(2) Area of Rectangle + Area of Triangle: (12)(2) + (1/2)(12)(3) = 42
(3) Area of Two Triangles: (1/2)(12)(5) + (1/2)(2)(12) = 42
(4) Area of Two Triangles: (1/2)(2)(12) + (1/2)(12)(5) = 42
(5) Negative Space: (12)(5) – (1/2)(3)(12) = 42
(6) Mirror Image: (12)(7)/2 = 42

Takeaway:

• If you are stuck on geometry, come up with creative approaches!
• Negative space and mirror images are great approaches to geometric areas

Question 52

In a circle, chord PQ is parallel to diameter OR, and OR has a length of 18. Additionally, the diagonal line from P to R forms a 35 degree angle at PRO. What is the length of minor arc PQ?
A. 2pi
B. (9pi)/4
C. (7pi)/2
D. (9pi)/2
E. 3pi

Answer 52

OG11, Lecture #7, Q 206. Answer A. Circles & Cylinders. From the diagram, we know that angle PRO equals 35. Draw a line between P and the center, point C. Because angle PCO opens up to the same arc as angle PRO, we know that angle PCO is two times the degree of angle PRO. Thus, angle PCO equals 70.

We now know that the central angle that open up minor arc PQ = 180 – 2(70) = 40. Thus, the length of minor arc PQ = (40/360)(18pi) = 2pi

Question 53

In a five pointed star with tip angles v, w,  x, y, and z, what is the sum of these five angles?
A. 45
B. 90
C. 180
D. 270
E. 360

Answer 53

OG13, HW 7, Diag 10. Answer C. Lines & Angles. Label the interior angles of the polygon and use the formula to find the sum of the interior angles.
a + b + c + d + e = 180(n – 2) = 180(5 – 2) = 540

Take advantage of symmetry to make this problem easier. The right answer must be the same for all cases, including the symmetrical one, so we know that we can find the answer this way with less work. Make each interior angle equal to 540/5 = 108. With this, we can also find the angles for their supplements. Every supplementary angle will be equal to 180 – 108 = 72

Now that we have the values of all the supplementary angles, we can find the values of v, w, x, y, and z. We know that 72 + 72 + v = 180. This means that v = 36. And all five angles will be the same, because they are all the third angle in the triangle with two angles of 72. The sum is therefore 5 * 36 = 180.

Question 54

A carpenter constructed a rectangular sandbox with a capacity of 10 cubic feet. If the carpenter were to make a similar sandbox twice as long, twice as wide, and twice as high as the first sandbox, what would be a the capacity, in cubic feet, of the second sandbox?
A. 20
B. 40
C. 60
D. 80
E. 100

Answer 54

OG13, HW 7, Q 18. Answer D. Polygons. The new capacity will be eight times the original capacity of 10 cubic feet, or 80 cubic feet.

Volume(old) = lwh
Volume(new) = (2l)(2w)(2h) = 8(lwh)

Question 55

In a rectangular solid, three sides have areas 12, 15, and 20, respectively. What is the volume of the solid?
A. 60
B. 120
C. 450
D. 1,800
E. 3,600

Answer 55

OG13, HW 7, Q 78. Answer A. Polygons. We know the surface area of three of the sides. Fortunately, it is irrelevant which side is length, which side if width, and which side if the height.
12 = lw
15 = lh
20 = wh

lw * lh * wh = 12 * 15 * 20
(l^2)(w^2)(h^2) = 12 * 15 * 20
lwh = √(12 * 15 * 20)
lwh = √(3 * 4 * 3 * 5 * 4 * 5)
lwh = 2 * 4 * 4
lwh = 60

Alternatively, we could have multiplied the numbers. 12 * 15 * 20 = 60^2

Question 56

A ladder of a fire truck is elevated to an angle of 60 degrees and extended to a length of 70 feet. If the base of the ladder is 7 feet above the ground, how many feet above the ground does the ladder reach?
A. 35
B. 42
C. 35√3
D. 7 + 35√3
E. 7 + 42√3

Answer 56

OG13, HW 7, Q 92. Answer D. Triangles & Diagonals. Let a equal the distance from the base of the ladder to the imaginary wall. Let b equal the height from the base of the ladder to the top of the ladder. We are looking for the value of 7 + b. The fact that the ladder is at a 60 degree angle provides us with the information necessary to determine that the triangle formed is a 30-60-90 special right triangle.
SL : LL : Hypotenuse
x : x√3 : 2x

The longest side (2x) is the hypotenuse, or c = 70. The short leg (x) is opposite the smallest angle (30 degrees) and is represented by a. The height of the ladder is 7 + 35√3
c = 2x = 70
a = x = 35
b = x√3 = 35√3

Question 57

In a rectangular lot, there is a triangular section in one of the corners that represents a flower bed. One of the legs equals x and the other equals y. The hypotenuse of the triangular flower bed equals z. If the area of the bed is 24 square yards and x = y + 2, then z equals
A. √13
B. 2√13
C. 6
D. 8
E. 10

Answer 57

OG13, HW 7, Q 165. Answer E. Triangles & Diagonals. Area = (1/2)(x)(y) = 24. x = y + 2.

24 = (1/2)(y)(y + 2) 
48 = y^2 + 2y
y^2 + 2y – 48 = 0
(y + 8)(y – 6) = 0
y = -8 OR y = 6

Using Pythagorean’s theorem or a Pythagorean triplet, we know that 36 + 64 = z^2. Thus, z = 10

Question 58

In a rectangular coordinate system, the line y = x is the perpendicular bisector of segment AB (not shown), and the x-axis is the perpendicular bisector of segment BC (not shown). If the coordinates of point A are (2, 3), what are the coordinates of point C?
A. (-3, -2)
B. (-3, 2)
C. (2, -3)
D. (3, -2)
E. (2, 3)

Answer 58

OG13, HW 7, Q 202. Answer D. Coordinate Plane. If y = x is the perpendicular bisector of AB, then the line is perpendicular to AB, and intersects AB at its midpoint. This means the distance from point A to the line is equal to the distance from the line to point B. We know that AB will have a slop equal to the negative reciprocal of y = x, so the line AB can be written as y = -1x + b. Plugging in the coordinates for point A yields y = -x + 5. The midpoint of AB will be –x + 5 = x, or (2.5, 2.5). B will have the coordinates (3, 2).

If the x-axis is the perpendicular bisector of BC, we can also say that point C is the reflection of point B through the x-axis. Thus, point C will have the coordinates (3, -2)

Alternatively, using our reflection process, we can estimate the correct answer. With this analysis, we know that point C must be in quadrant IV and to the right of point A, which eliminates all answers except D.

Question 59

Simplify √[(√96) + 2/(5 + 2√6)]
A. √(10 + 8√6) 
B. √5
C. √10
D. √11
E. √(6/17)

Answer 59

MGMAT Challenge Problem. Answer C. Exponents & Roots. Two methods: (1) simplify by breaking apart the expression into pieces (2) use estimation to strategically guess

Piece 1: 2/(5 + 2√6) = 2/(5 + 2√6) * (5 – 2√6)/(5 – 2√6) = 10 - 4√6
Piece 2: √96 = √(16 * 6) = 4√6
Together: √[(√96) + 2/(5 + 2√6)] = √[(4√6) – (4√6) + 10] = √10

OR use estimation.

Piece 1: 2/(5 + 2√6) = 2/(5 + 5) ~ 2/10
Piece 2: √96 ~ 9.8
Together: √[(√96) + 2/(5 + 2√6)] = √(9.8) + (2/10) = √10

Question 60

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?
A. 10
B. 12
C. 14
D. 16
E. 18

Answer 60

OG13, Lecture # 8, Q 116. Answer C. Divisibility & Primes. We need to determine how many 3’s are factors of p, which equals 30!

The factor foundation rule tells us that each multiple of 3 from 1 to 30 will contribute factors of 3 to the overall product. The non-multiples of 3 will not contribute factors of 3 and can thus be ignored. Go through the list below and add up the number of 3’s that appear
3 = 1 * 3…(1)
6 = 2 * 3…(1) 
9 = 3 * 3…(2)
12 = 2 * 2 * 3…(1)
15 = 5 * 3…(1)
18 = 2 * 3 * 3…(2)
21 = 7 * 3…(1)
24 = 2 * 2 * 2 * 3…(1)
27 = 3 * 3 * 3…(3)
30 = 2 * 5 * 3…(1)

We can see that, collectively, the multiples of 3 between 1 and 30, inclusive, contribute 14 factors of 3 to p. Therefore, 14 is the answer

Question 61

For the positive integers a, b, and k, a^k ǀǀ b means that a^k is a divisor of b, but a^(k+1) is not a divisor of b. If k is a positive integer and 2^k ǀǀ 72, then k is equal to
A. 2
B. 3
C. 4
D. 8
E. 18

Answer 61

OG13, HW 8, Q 110. Answer B. Divisibility & Primes. In order to figure out how many 2’s there are among 72’s prime factors, use a factor tree to find the prime factorization of 72.

72 = 3^2 * 2^3

There are three 2’s. That means that 72 is divisible by 2 three times. Thus, k must equal 3.

Question 62

If 3 < x < 100, for how many values of x is x/3 the square of a prime number?
A. Two 
B. Three
C. Four
D. Five
E. Nine

Answer 62

OG13, HW 8, Q 127. Answer B. Divisibility & Primes. A good way to solve this problem is just to construct and list out the possible values of x. We should not have too many numbers to the list because the largest choice is nine.

Prime = 2, Prime^2 = 4, x = 12
Prime = 3, Prime^2 = 9, x = 27
Prime = 5, Prime^2 = 25, x = 75
Prime = 7, Prime^2 = 49, x = 147 TOO BIG

Only three primes match the description: 2, 3, and 5.

Question 63

If y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer, then y must be 
A. 2
B. 5
C. 6
D. 7
E. 14

Answer 63

OG13, HW 8, Q 155. Answer E. We must first find the prime factors of 3,150 to determine which ones are “single.” 3,150 = 2 * 3 * 3 * 5 * 5 * 7 = 2 * 3^2 * 5^2 * 7. We need the prime box of 3,150y to contain only pairs of primes if 3,150y is to be a perfect square. Single primes are not allowed. This means that y must provide at least one 2 and one 7. Also, because we are looking for the smallest possible y, those two factors should be all that y provides. This leads to y = 2 * 7 = 14. This is the answer

Question 64

How many of the integers that satisfy the inequality (x + 2)(x + 3) / (x – 2) ≥ 0 are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5

Answer 64

OG13, HW 8, Q 229. Answer D. Positives & Negatives. A good way to solve this problem is to start testing integers that are less than 5, starting with the number 4. There cannot possibly be more than five of these integers because the largest answer choice is five.

4: (4 + 2)(4 + 3) / (4 – 2) = 21 (WORKS)
3: (3 + 2)(3 + 3) / (3 – 2) = 30 (WORKS)
2: Does not work because it makes the denominator zero
1, 0, -1: do not work because they all make the numerator positive, but the denominator negative
-2: (-2 + 2)(-2 + 3) / (-2 – 2) = 0 (WORKS)
-3: (-3 + 2)(-3 + 3) / (-3 – 2) = 0 (WORKS)
-4: and any number less than -4 will not work because they make the numerator positive and the denominator negative, resulting in a negative overall value

So there are a total of four integers: 4, 3, -2 and -3

Question 65

If s and t are positive integers such that s/t = 64.12, which of the following could be the remainder when s is divided by t?
A. 2
B. 4
C. 8
D. 20
E. 45

Answer 65

OG13, HW 8, D 13. Answer E. Divisibility & Primes.
12/100 = Remainder/t
0.12t = R

Approach # 1: Back solve
A. If 0.12t = 2, then t will not be an integer
B. If 0.12t = 4, then t will not be an integer
C. If 0.12t = 8, then t will not be an integer
D. If 0.12t = 20, then t will not be an integer
E. If 0.12t = 45, then t = 375 CORRECT

Approach # 2: Divisibility Theory
12/100 = R/t
3/25 = R/t
3t = 25R
Because 25 has no 3’s in it, the integer on the right must contain a 3 (the prime factors on each side must match, because each variable is an integer). So the remainder, which is an integer, must be a multiple of 3. Of the answer choices, only 45 is a multiple of 3

Question 66

An operation • is defined by the equation a • b = (a – b) / (a + b), for all numbers a and b such that a ≠ -b. If a ≠ -c and a • c = 0, then c = ?
A. –a
B. -1/a 
C. 0
D. 1/a
E. a

Answer 66

OG13, HW 4, Q 134. Formulas, Answer E. We are told that (a – c)/(a + c) = 0. Note that the fraction equals 0 if its numerator equals zero. For example, 0/5 = 0. Therefore, we must have a – c = 0, or a = c, which is the answer. It is important to note that a fraction will NOT be equal to zero when the denominator is equal to zero.

Takeaways:
-Dividing by zero results in an undefined expression

Question 67

What is the value of Sqrt(Cbrt(0.000064))?

Answer 67

Since 0.000064 has six decimal places, its cube root will have one-third as many, or two decimal places. Thus, cbrt(0.000064) = 0.04. Finally, Sqrt(0.04) = 0.2

Alternative Method: Use the powers of 10
Sqrt[Cbrt(64) * Cbrt(10^-6)] = Sqrt(4 * 10^-2) = Sqrt(4) * Sqrt(10^-2) = 2 * 10^-1 = 0.2

Question 68

A certain junior class has 1,000 students, and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A. 3 / 40,000
B. 1 / 3,600
C. 9 / 2,000
D. 1 / 60
E. 1 / 15

Answer 68

OG12, Lecture # 9, Q 217. Answer A. Probability. There are two parts of this question that make it difficult. First, what exactly is a sibling pair? A sibling pair is a pair of two students that are related. Thus, if there are 60 sibling pairs, then there are 120 students in total (60 juniors, 60 seniors. Second, what is the probability that, after selecting the junior sibling, you pick his or her matching sibling from the senior class?

P(Junior Sibling Chosen) = 60 / 1,000
P(Senior Match Chosen) = 1 / 800

P(Sibling Pair) = (60/1,000) * (1/800) = 6 / 80,000 = 3 / 40,000

Question 69

Two integers will be randomly selected from the sets below, one integer from set A and one integer from set B. What is the probability that the sum of the two integers will equal 9?
A = {2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
A. 0.15
B. 0.20
C. 0.25
D. 0.30
E. 0.33

Answer 69

OG13, Lecture # 9, Q 68. Answer B. Probability. We can solve this problem by either of the two fundamental approaches to probability: by basic formula or by combining probabilities of component events.

Basic Probability Formula: Since set A has four elements and set B has five, the total number of outcomes is 4 * 5 = 20. Of these outcomes, exactly four are successful in summing to 9: (2 + 7, 3 + 6, 4 + 5, 5 + 4)

Probability = # of successful outcomes / total # of outcomes = 4 / 20 = 1 / 5 = 0.20

Domino Effect: multiply the successive independent probabilities. First, we calculate, separately, the probability of each of the four events listed above. For each of these cases, the chance of picking the “correct” first number (from set A) is 1/4, and the chance of picking the “correct” second number (from set B) is 1/5. Therefore, the probability of each of these outcomes is (1/4)(1/5) = 1 / 20. Since the four outcomes are mutually exclusive (that is, no two of them can happen at the same time), we add together the four probabilities to get 4/20 = 0.2

Question 70

The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails?
A. 1/8
B. 1/2
C. 3/4
D. 7/8
E. 15/16

Answer 70

OG13, Lecture # 8, Q 193. Answer D. Probability. Two approaches: list all the possibilities for the three tosses or use the 1 – x principle.

HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
Out of the 8 possible outcomes, 7 of them show at least one tails. Thus, the probability of getting at least one tails in three tosses is 7/8

Another approach to this problem is to use the 1 – x principle. What is the probability of NOT showing ALL heads?
P(at least one tails) = 1 – P(HHH) = 1 – (1/2)^3 = 1 – 1/8 = 7/8

Question 71

Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?
A. m
B. 10m/7
C. 10m/7 – 9/7
D. 5m/7 + 3/7
E. 5m

Answer 71

CAT #3, Q 2. Answer C. Statistics. In all that follows, the numbers in set S are assumed to be in increasing order (so that the 1st value is the lowest and the 7th value is the highest). Also, if we want the highest possible average, then we need to make every member of Set S as large as possible.

The median of S is the 4th value in the set. For the median to equal m, the 4th value must equal m.

We must choose the highest possible value for every member of the set. The maximum value of any member of the set is 2m, so the 7th value is 2m. Because every member of the set is an integer, the next highest possible value is 2m – 1, which is the 6th value. The 5th value is the next highest possible number, which is 2m – 2.

By a similar logic, the 3rd value is m – 1, the 2nd value is m – 2, and the 1st value is m – 3.

S = {m – 3, m – 2, m – 1, m, 2m – 2, 2m – 1, 2m}

The sum of the elements of the set is 10m – 9. The average of the values in this set is 10m/7 – 9/7.

Question 72

A certain square is to be drawn on a coordinate plane.  One of the vertices must be on the origin, and the square is to have an area of 100.  If all coordinates of the vertices must be integers, how many different ways can this square be drawn? 
A. 4
B. 6
C. 8
D. 10
E. 12

Answer 72

CAT #3, Q 10. Answer E. Coordinate Plane. Each side of the square must have a length of 10. If each side were to be 6, 7, 8, or most other numbers, there could only be four possible squares drawn, because each side, in order to have integer coordinates, would have to be drawn on the x- or y-axis. What makes a length of 10 different is that it could be the hypotenuse of a Pythagorean triple, meaning the vertices could have integer coordinates without lying on the x- or y-axis.

For example, a square could be drawn with the coordinates (0,0), (6,8), (-2, 14) and (-8, 6). (It is tedious and unnecessary to figure out all four coordinates for each square).

If we label the square abcd, with a at the origin and the letters representing points in a clockwise direction, we can get the number of possible squares by figuring out the number of unique ways ab can be drawn.

a has coordinates (0,0) and b could have the following coordinates: (-10,0); (-8,6); (-6,8); (0,10); (6,8); (8,6); (10,0); (8, -6); (6, -8); (0, 10); (-6, -8); (-8, -6). There are 12 different ways to draw ab, and so there are 12 ways to draw abcd.

Question 73

Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value? 
A. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33

Answer 73

CAT #3, Q 11. Answer C. Probability. The chance of getting AT LEAST one pair of cards with the same value out of 4 dealt cards should be computed using the 1-x technique. That is, you should figure out the probability of getting NO PAIRS in those 4 cards (an easier probability to compute), and then subtract that probability from 1.

First card: The probability of getting NO pairs so far is 1 (since only one card has been dealt).

Second card: There is 1 card left in the deck with the same value as the first card. Thus, there are 10 cards that will NOT form a pair with the first card. there are 11 cards left in the deck. Probability of NO pairs so far = 10/11.

Third card: Since there have been no pairs so far, there are two cards dealt with different values. There are 2 cards in the deck with the same values as those two cards. Thus, there are 8 cards that will not form a pair with either of those two cards. There are 10 cards left in the deck. Probability of turning over a third card that does NOT form a pair in any way, GIVEN that there are NO pairs so far = 8/10.

Fourth card: Now there are three cards dealt with different values. There are 3 cards in the deck with the same values; thus, there are 6 cards in the deck that will not form a pair with any of the three dealt cards. There are 9 cards left in the deck. Probability of turning over a fourth card that does NOT form a pair in any way, GIVEN that there are NO pairs so far = 6/9.

Cumulative probability of avoiding a pair on the second card AND on the third card AND on the fourth card = cumulative product = (10/11) (8/10) (6/9) = 16/33.

Thus, the probability of getting AT LEAST ONE pair in the four cards is 1 − 16/33 = 17/33.

Question 74

Two musicians, Maria and Perry, work at independent constant rates to tune a warehouse full of instruments. If both musicians start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Perry were to work at twice Maria’s rate, they would take only 20 minutes. How long would it take Perry, working alone at his normal rate, to tune the warehouse full of instruments?
A. 1 hr 20 min
B. 1 hr 45 min
C. 2 hr
D. 2 hr 20 min
E. 3 hr

Answer 74

CAT #3, Q 12. Answer E. Rates & Work.

Organize the given information in a RTW chart, using m and p to represent Maria’s and Perry’s normal rates, and w to represent work (to tune 1 warehouse full of instruments).

Maria & Perry: m + p * 45 minutes = w
Maria & Perry (hyp): m + 2m * 20 minutes = w

3m(20) = w
m = w / 60

(w/60 + p)(45) = w 
(3/4)w + 45p = w
45p = (1/4)w
180p = w
180 = w/p

Perry’s time = w/p = 180 minutes = 3 hours

Question 75

If a rectangle with length √a units and width √b units is inscribed in a circle of radius 5 units, what is the value of a + b?
A. 10
B. 20
C. 25
D. 50
E. 100

Answer 75

CAT #3, Q 14. Answer E. Circles & Cylinders. The problem doesn’t indicate exactly how to inscribe the length and the width in the circle. Whatever way this is done, though, it will still be the case that the diameter of 10 will represent the diagonal of the rectangle. This diagonal also creates a right triangle (all rectangles have angles of 90°). The values of the sides a and b, then, can be represented using the Pythagorean Theorem. (√a)^2 + (√b)^2 = 10^2 or a + b = 10

Question 76

If 4^(4x) = 1600, what is the value of 4^(x–1)^2?
A. 40
B. 20
C. 10
D. 5/2
E. 5/4

Answer 76

CAT #3, Q 17. Answer D. Exponents & Roots. Manipulate the given equation in order to determine the value 4^(2x–2).
4^(4x) = 1600
[4^(4x)]^(1/2) = (1600)^(1/2)…Square root each side
4^(2x) = 40
4^(2x) / 4^2 = 40 / 4^2…Divide by 4^2
4^(2x–2) = 40 / 16 = 20 / 8 = 10 / 4 = 5 / 2

Question 77

Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50%. What is his current speed?
A. 6
B. 8
C. 10
D. 12
E. 16

Answer 77

CAT #3, Q 22. Answer B. Rates & Work. The best way to solve this problem is to Back Solve.

A. 4mph slower: speed = 6 – 4 = 2, time = 96/2 = 48
50% faster: speed = 6 * 1.5 = 9, time = 96/9 = 10 2/3
Difference = more than 16

B. 4mph slower: speed = 8 – 4 = 4, time = 96/4 = 24
50% faster: speed = 8 * 1.5 = 12, time = 96/12 = 8
Difference = 16 (CORRECT. Matches the 16 hours from the question)

C. 4mph slower: speed = 10 – 4 = 6, time = 96/6 = 16
50% faster: speed = 10 * 1.5 = 15, time = 96/15 = 6.4
Difference = 9.6

Question 78

A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated, creating trees that are part Fuji and part Gala. The number of his trees that are pure Fuji plus the number that are part Fuji and part Gala totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?
A. 22
B. 33
C. 55
D. 77
E. 88

Answer 78

CAT #3, Q 24. Answer B. Overlapping Sets. This problem can be solved using a set of four equations with four unknowns. The fourth variable will represent the total number of apple trees. We’ll add two equations together to solve for the number of Gala apples relatively simply:

Let F = the number of Fuji trees
Let C = the number of cross pollinated trees
Let G = the number of Gala trees
Let T = the total number of trees

Equation (1): By definition, F + C + G = T
Equation (2): The pure Fuji trees plus the cross pollinated trees total 187, so F + C = 187
Equation (3): 75% of his trees are Fuji, so F = 0.75T
Equation (4): 10% of his trees cross pollinated, so C = 0.1T

We can add these two equations together to show that 85% of the trees are Fuji or cross pollinated: F + C = 0.85T

Substituting from Equation (2), we get 0.85T = 187, or T = 220

Plugging back into the original “total” equation, we get the following:

F + C + G = T
(F + C) + G = 220
187 + G = 220
G = 33

Question 79

If ¡n! = (n!)^2, then ¡17! - ¡16! =
A. ¡1!
B. (¡16!)(¡4!)(2)
C. (¡16!)(12)(2)
D. 17^2
E. (¡16!)(12^2)(2)

Answer 79

CAT #3, Q 25. Answer E. Formulas. The two numbers are close to identical, but ¡17! contains two additional factors: 17 and 17. Factor out a ¡16! from the two terms:
¡17! – ¡16! 
¡16!(17*17 – 1)
¡16!(289 – 1)
¡16!(288)
¡16!(12^2)(2)

Question 80

Refer to picture (CAT # 3, Q 27). If BE ǀǀ CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?
A. 12
B. 18
C. 24
D. 30
E. 48

Answer 80

CAT #3, Q 27. Answer B. Triangles & Diagonals. Since BE ǀǀ CD, triangle ABE is similar to triangle ACD (parallel lines imply two sets of equal angles).  We can use this relationship to set up a ratio of the respective sides of the two triangles:
AB
AB/AC = AE/AD
3/6 = 4/AD
AD = 8

We can find the area of the trapezoid two different ways:

Approach #1: Area Trapezoid = Area Triangle CAD – Area Triangle BAE
Area Trapezoid = (1/2)(6)(8) – (1/2)(3)(4) = 24 – 6 = 18

Approach #2: Take the mirror image of the shape.
Area Trapezoid = [Area Large Rectangle – 2(Area Small Triangle)] / 2 = [(6)(8) – (3)(4)] / 2 = 18

Question 81

If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?
A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72

Answer 81

CAT #3, Q 29. Answer D. Formulas. We are told that p = (km^3) / q^2. Because the problem asks us for the greatest value, we will need to compare the five answer choices. We don’t necessarily need to find all 5 values, though; let’s give the choices a quick scan to see which are the most likely to produce the largest values, given our equation. Answer A is the only one in which m is larger than q, but the actual numbers are quite small; this isn’t likely to produce the largest value overall. Answers B and C have the same values for m and q; of the two, the 20/20 pair is likely going to result in the larger value – though we have to check to make sure. Answers D and E are large enough to look annoying; let’s not start with either of them. Answer C looks like the best place to start.

C. 20^3 / 20^2 = 20
D. 30^3 / 36^2 = (5^3 * 6^3) / (6^4) = 125/6 = 20 5/6
E. 36^3 / 72^2 = 36^3 / (2^2 * 36^2) = 36/4 = 9

Question 82

Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?
A. 1 / (5^4)
B. 1 / (5^3)
C. 6 / (5^4)
D. 13 / (5^4)
E. 17 / (5^4)

Answer 82

CAT #3, Q 31. Answer E. Probability. There are two ways that Leila can succeed on at least 3 of the throws—she can succeed on exactly 3 throws or on all 4. We can find the total probability by determining the chance of each of these two distinct outcomes and adding them together.

Let’s start with the chance of succeeding on all 4 throws. Since Leila’s chance of success is 1/5, we can simply multiply out her chance of succeeding 4 times in a row:
1/5 × 1/5 × 1/5 × 1/5 = (1/5)^4

Note that we can now eliminate answer B; it can’t be correct, because we haven’t yet added in the possibility of succeeding on exactly 3 throws.

If Leila succeeds on exactly 3 of the throws, then she must have 1 missed throw. The chance of missing a throw is 1 – 1/5 = 4/5, so we can multiply the individual probabilities to get the total probability:
1/5 × 1/5 × 1/5 × 4/5 = 4/5^4

However, this is the probability of only one specific outcome: hit, hit, hit, miss. Since she doesn’t need to make her throws in this specific order, we need to consider the other possibilities:
hit, hit, miss, hit
hit, miss, hit, hit
miss, hit, hit, hit

There are 4 different ways that Leila can get 3 hits and 1 miss. (Note that this wasn’t an issue when we calculated the chance of succeeding on all 4 throws, because there is only one way to do that.) Therefore, her chance of successfully making exactly 3 out of 4 throws is 4(4/5^4)=16/5^4.

We can now find the probability that Leila succeeds on at least 3 of the throws:
1/(5^4) + 16/(5^4) = 17/(5^4)

Question 83

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of √(288kx)?
A. 24k√3
B. 24√k
C. 24√3k
D. 24√6k
E. 72√k

Answer 83

CAT #3, Q 34. Answer B. Exponents & Roots. The value √(288kx) can be simplified to 12√(2kx). Given that x is divisible by 6, for the purpose of solving this problem x might be restated as 6y, where y may be any positive integer. The expression could then be further simplified to 12√(12ky) or 24√(3ky). The following are possible values

y = 1, solution = 24√(3k)
y = 2, solution = 24√(6k)
y = 3, solution = 72√(k)
y = k, solution = 24k√(3)

Question 84

100 people are attending a newspaper conference. 45 of them are writers and more than 38 are editors. Of the people at the conference, x are both writers and editors and 2x are neither. What is the largest possible number of people who are both writers and editors?
A. 6
B. 16
C. 17
D. 33
E. 84

Answer 84

CAT #3, Q 35. Answer B. Overlapping Sets. This Overlapping Sets problem asks us to determine the maximum value of x, the number of people who are both writers and editors. We can set up a Double Set Matrix using the information given in the question to tackle this problem. From the matrix, we find that the number of people who are not writers is 100 – 45 = 55, and therefore the number of people who are editors and not writers is 55 – 2x. We can continue to populate our matrix.

Because the total number of editors must be more than 38, we can set up and solve the following inequality:

x + (55 – 2x) > 38
55 – x > 38
x