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Question 1

Do not restate value questions as yes/no questions and vice versa

Answer 1

Good choices for strategic numbers : positive integers, positive proper fractions, zero, negative proper fractions and negative integers.

Question 2

The two statements presented in a DS question will never contradict each other.
For all values and signs of a,b,c and d : a/b + c/d = (ad+bc)/bd and a/b - c/d = ad - bc/db

Answer 2

Simple fraction : consists of two whole numbers : numerator and denominator and denominator can’t be 0
Proper fraction : numerator is less than the denominator
Improper fraction : numerator is greater than the denominator
Mixed number : whole number followed by proper fraction
Two fractions a/b and c/d are equivalent if a x d = b x c

Question 3

10011 - 7677?

Answer 3

A pretty neat way of doing this is writing 10011 as 9999 + 12. This way subtracting 7677 from 9999 becomes smooth since there’s no carry over etc. and each digit can be calculated in a straightforward way.
Further, you don’t even need to round 7677 for this.
This is also better than rounding 10011 to 10000

Question 4

1) If I have a fraction and I add k to both numerator and denominator. What happens to the value of the fraction (increases/decreases) ?
2) If I have a fraction and I subtract k from both numerator and denominator. What happens to the value of the fraction (increases/decreases) ?
3) If I add k to numerator and m to denominator and k > m then?
4) If k < m?
5) What’s the general rule for addition?

Answer 4

GENERAL RULE FOR ADDITION : adding a to the numerator and b to the denominator moves the resultant fraction closer to the fraction a/b. If x/y < a/b, moving the starting fraction close to a/b will make it bigger. If x/y > a/b, moving the starting fraction close to a/b will make it smaller

There’s no general rule for subtraction beyond the case where you subtract the same number from both num and denom since num and denom can weirdly get negative etc.

1) If fraction is less than 1, it moves closer to 1 (increases) and if greater than 1, it moves closer to 1 (decreases) (moves closer to 1 in both less than and greater than 1 case)
2) If fraction is less than 1, it moves away from 1 (decreases) and if greater than 1, moves away from 1 (increases) (moves away from 1 in both less than 1 and greater than 1 case)
3) and 4) See general rule

Question 5

What digits can a square never end with?

What digits can a cube never end with?

Answer 5

1) All even powers can never end with 2,3,7,8

2) Odd powers can end with ANY digits. Hence no such digits exist.

Question 6

Simplify : 4x12 / 3(18 + 5(3+1)^2 + 4 / 2)^0.5

Answer 6

Being neutral between choosing among * and / will get you the wrong result.
Remember that PEMDAS says to apply / and * from left to right. This is because division is not commutative.
Correct answer : 160

PEMDAS :
Parentheses : OPERATIONS WITHIN parentheses. Includes (within) absolute value bards and root symbols/radicals.
Exponents
Multiplication and Division from left to right
Addition and subtraction : from left to right
TREAT 3(10) as 3 x 10 and give priority equal to division and not same as parentheses

Question 7

1) 5 + 3/4?

2) 5 + 9/4?

Answer 7

1) Just attach em : 5 3/4 (mixed fraction)

2) Just attach em : 5 9/4 (basically true for both proper and improper fraction)

Question 8

1) How do you know if a fraction is terminating or not?
2) Is 22/7 repeating?
3) a> c and b>d, is a/b > c/d?

Answer 8

1) If in it’s SIMPLIFIED form, the denominator can be written as 2^x * 5^y ONLY then is it terminating.
NOTE : x, y >=0 (they can be ZERO also)

2) All fractions with whole number numerator and denominators are either terminating or repeating. Hence, yes 22/7 is repeating. It is very close to pi which is non repeating non terminating.
3) No

Question 9

10.60824/159.1236?
Options:
1/15
1/17
2/3
1/2
1/16

Answer 9

It’s obvious that it’s close to 1/16. But it is greater since 10/159 is 1/15.9 which is greater than 1/16 further, we still add something greater to numerator than denominator further cementing this.
Alternate neater approach :
Estimate 10.6 as 11 and 159.1 as 159 to get 11/159.
Not convert the answer options to something which has either the same numerator or denominator :
11/165
11/187
11/16.5
11/22
11/176
The one that is closest is the answer : 11/165
Best approach :
2(5.30412)/3(53.0412) = 2/30 = 1/15

Question 10

What are the values of the following fractions?
1/5
1/6 5/6
1/7
1/8 3/8 5/8 7/8
1/9
1/11
1/12

Answer 10

Ans :
.2
.1666...  | .8333333
.142857 
.125 .375 .625 .875
.1111...
.090909....
.083333

Question 11

What are the values of the following?
3!
4!
5!
6!
7!
8!

Answer 11

Ans :

3: 6
4: 24
5: 120
6: 720
7: 5040
8: 40320

Question 12

For solving 2 linear eqn in 2 variables, consider using the combination method when neither equation can easily be solved for one of the variables and consider using the substitution method when one can eqn can easily be solved for one of the variables.
If you need to find the value of only 1 of the two variables, substitute the for the value you don’t want.
I.e. if you want x, find y = some equation of x and substitute that values of y.

Answer 12

Some cases when one method is obviously superior to the other :

1) If adding or subtracting will eliminate one variable straight away then use combination
2) If the coefficients of any one of the two variables is 1 or can be brought to 1 by simple divisions, use substitution
3) If the coefficients of any one variable across the two equations are simple multiples ie. 2x + 3y = 10, 6x + 11y = 20, then use combination (2 and 6 multiples)

Question 13

For equations with fractions, multiply both sides of the equation with the LCM of the denominators to remove the fractions.

Answer 13

Until I’m proven wrong, converting fractional coefficients to whole numbers proves faster and cleaner

Question 14

‘what is x in terms of y’ simply means find the equation of the form x = some equation involving only variable y (and no other variables)

Answer 14

Given, ax + by = c, dx + ey = f
if a/d = b/e = c/f, then we have inf solutions and the lines are coincident
if a/d = b/e =/= c/f then 0 solutions, lines are parallel
if a/d =/= b/e then unique solution, line intersects

Question 15

if x and y are integers, what is the value of x+y

1) x(xy) = 1
2) x/y = 1

Answer 15

Answer is C
1) Gives two solutions (1,1) AND (-1,1)
If x and y are integers and xy = 1, then (1,1) and (-1,-1) are the only two possible solutions.
When you apply this to the above question, you get possible solution according to 1) as (1,1) but here 1 = x^2 so x can be both 1 and -1, so two possible solutions according to data provided by 1) (x,y) = (1,1) or (-1,1)

Question 16

How to find number of +ve int solutions to equations like 2x + 3y = 763?

Answer 16

1) Find any 1 point that satisfies the equation, helps to start from the lower end. using x = 1 doesn’t give int y. using x=2 gives y = 253 (int) so we start with point (2,253). Can also start with y instead of x. BETTER TO START with the higher coefficient variable i.e. y in this case since testing divisibility by the smaller coefficient is easier.
2) Now x will increase by coefficient of y and y will DECREASE by that of x. So, x will increase by 3 and y will decrease by 2. (y decreases coz we have + between 2x and 3y so if one positive no. increases, the other has to decrease, if it was - as in 2x - 3y, both will INCREASE hence lead to inf solutions)
3) You can find the last possible value by finding the starting pt. of the variable you DID NOT choose in step 1 i.e. y. y= 1 gives x=380. So, we start from (2,253) and end up in (380,1)
4) Now that you know step sizes (3 and 2 respectively), you can find number of terms using AP formula.

(You cannot use this if there are additional constraints like x,y should be between 1 to 9 etc. In such cases you need to brute force with intuition)
(You can start with any solution for x and y and use step size logic to find starting solutions)

Question 17

No. of positive int solutions to eqn (finite or inf) :

1) 2x + 3y = 763
2) 2x - 3y = 763
3) 2x -3y = -763
4) -2x - 3y = 763

Answer 17

1) finite
2) inf
3) inf
4) finite (0)
How? : Sketch line to understand. 1) will cut y axis in the first quadrant 2) will not hence inf solutions

Question 18

If you have ax + by + cz = d and you need to find +ve integral solutions. Eg : 2x + y + 11z = 42, pick the highest coefficient variable i.e. z (coz of 11) and solve for every possible case of z. Use step size logic for solving for x and y. Finally add all the solutions across values of z.

Answer 18

If you have a range [x,y] and step size > y-x, then it is possible that we have a solution in given range. However, if step size <= y-x then we necessarily will have a solution in the given range.

Question 19

If you have 3b + 7s + f = 120, 4b + 10s + f = 164.5 and we need b + s+ f. how we do dis?

Answer 19

Assume multiplier to be m and n. Multiply first equation by m and second by n. now 3m + 4n = 1 and 7m + 10n = 1. Now solve for m and n OR for ration of m to n (something like 2m = -3n, now put m = -3 and n = 2). Note here we want value of new EQUATION and not b, s and f.

Question 20

Given roots a and b, what’s the quadratic equation?

Answer 20

x^2 - (a+b)x + ab
When dealing with quadratic equations, remember that if the roots are equal, you have ONLY ONE solution/possible value for x.

Question 21

1) 462 = (20 - 2x)(25 -2x)(x). Find x

2) x^2 = 4 + (4-x)^0.5

Answer 21

1) Observe that 20 -2x and 25 -2x are 5 units apart. Keep this in mind while splitting or finding factors. Split 462 to 11 x 7 x 3 x 2. See that 11 and 6 are 5 apart. so x =7
2) Before squaring both sides, reorganize the equation in a way that RHS has only the underroot equation (4-x)^0.5. This makes squaring both sides less messy and prevents an ugly middle term.

Question 22

Easy way to find root after middle term split.
This method hinges on the sign of ‘a’ ONLY
1) if eqn is of the form ax^2 + bx + c and you split it to ax^2 + dx + ex + c, the roots are -d/a and -e/a
2) if eqn is of the form -ax^2 + bx + c and you split it to -ax^2 + dx + ex + c, then roots are d/a and e/a

Answer 22

If roots of ax^2 + bx + c are d and e, the roots of cx^2 + bx + a are 1/d and 1/e

a^n + b^n is divisible by a+b if n is odd

a^n - b^n is divisible by a-b always and by a + b if n is even.
a^n - b^n = (a-b) (a^n-1 + (a^n-2).b + …b^n-1)

Question 23

If your count starts with 1:

1) how many even numbers till x if x is even ?
2) how many odd no.s till x if x is even?
3) how many even till x if x is odd?
4) how many odd till x if x is odd?

Answer 23

1) x/2
2) x/2
3) (x-1)/2
4) (x +1)/2

Question 24

Odd or Even? :

1) Sum of 2n odd nos. ?
2) Difference of 2n odd no.s?
3) Sum/Difference of 2n +1 odd no.s?
4) Sum/Diff of 2n or 2n+1 even no.s?

Answer 24

1) Even
2) Even
3) Odd
4) Even

Question 25

Find value of (99^99 - 98^98)/(99^98 + 98^98)?

1) 1
2) <1
3) >1
4) 0

Answer 25

You can either apply the a^n - b^n expansion and proceed OR :
Try for smaller values like 4 and 3. If it holds true for 99 and 98, it should hold true for 4 and 3 respectively

Question 26

How many pairs of +ve int satisfy 1/m + 4/n = 1/12 if n is an odd int < 60

Answer 26

Here n is constrained, hence write m in terms of n (the constrained variable). Now use the constraints to try out all possible values of n and find corresponding m. This way you only have to cycle through limited no. of possible values

Question 27

If you need to simplify expressions like (x-1)(x-3)(x-4)(x-8), then multiply in a way that you get the middle term as equal, getting the product of constants equal will NOT help you in any way.

Answer 27

Find equation whose roots are 2 more than the roots of the equation x^2 -6x +4 =0
Ans : Let the root of this equation be a and b. the new equation will have root x = a + 2. Using this a = x -2. We know a is a root of given equation. Hence put x -2 in the given equation to get required equation.
Similarly if it was 2 less than a and b, we would have put a = x +2

Question 28

1) No. of factors of x?

2) No. of factors of a perfect square? (Odd/ Even)

Answer 28

1) Prime factorize x into a ^ b x c^d x e^f …
The no. of factors = (b+1)(d+1)(f+1)..
NOTE : YOU CAN ONLY DO THIS IF YOU KNOW THAT a,c,e… are PRIMES. DON’T APPLY THIS IF ANY ONE OR MORE OF THEM CAN BE COMPOSITE.
MADE AN ERROR ON DS QUESTION ASSUMING THEY ARE PRIME WHEN THEY WERE ACTUALLY NOT.
2) Odd, use the above formula

Question 29

No. of terms between 2 no.s a and b (both included)?
What if :
1) both odd
2) both even
3) 1 odd 1 even
4) the smaller number is 0

Answer 29

b - a + 1 irrespective of any scenario

Question 30

((5+4)(5^2+4^2)(5^4+4^4)(5^8+4^8) + 4^16)^1/3 ?

Answer 30

Multiply the inner expression by (5-4) which is fine since it is equal to 1 to get ((5-4)(5+4)(5^2+4^2)(5^4+4^4)(5^8+4^8) + 4^16)^1/3
Now repeatedly apply (a+b)(a-b) formula

Question 31

1) Find no. of odd factors of a number. Eg : 900

2) Find no. of even factors of a number. Eg : 900

Answer 31

1 )Prime factorize 900 = 2^2 x 3^2 x 5^2
Ignore even terms and calculate no. of factors using only the odd terms : (2+1)(2+1) = 9
Odd factors = 9
2) Even factors = total no. of factors - odd factors

Question 32

Find sum of factors. Eg : 100

Answer 32

Prime factorize 100 : 2^2 * 5^2

Sum of prime factors : (2^0 + 2^1 + 2^2) (5^0 + 5^1 + 5^2) (you go up till max power in prime factorization)

Question 33

If x and y are positive integers and 5x + 8y = 55, then x = ?

Answer 33

Instead of brute forcing, see that 5x and 55 have 5 in common. So, 8y = 5(11-x)
y = 5(11-x)/8
Now, 8 cannot divide into 5, so 11-x has to be a multiple of 8. Given the constraints, x can only be 3 and so y = 5.

In a similar vein : A store sells card a and card b. card a is for 5 and card b is for 7. a customer spent 60 buying at least 1 of each kind. What are the total no. of cards bought.

Here, if you use data as is, you get 5a + 7b = 60 (with a and b >=1 )and you can take 5a to the RHS and proceed.
HOWEVER, if you re-write it as a 5(a+1) + 7(b+1) = 60 where a and b can be >=0, you run into trouble as it simplifies to 5a + 7b = 12 and nothing is in common anymore. So, don’t be oversmart.

Question 34

Q : ‘Sara purchased m almonds and n peanuts……’
Now, whenever there is an action mentioned in the question stem like purchased, got, lent, bought etc. You can assume that zero is not allowed.

Answer 34

When cancelling terms OR EXPRESSIONS, always make sure that they cannot be zero.
Also, be on the lookup that multiplying by a negative number will change the inequalities up.
This changes a little if it’s across numerator and denominator. In that case, you can cancel even if you’re not sure if it’s zero or not AS LONG AS it’s present in both the numerator and denominator.

Question 35

In general (mostly for algebra questions), if you're unable to understand how to solve the question in 15-20 seconds, START PLUGGING VALUES
Eg : 
(25 - x^9)/(5 - x^3) is equal to which of the following :
1) 5 - x^3
2) 5 + x^3
3) (5 + x)^3
4) (5 - x)^3
5) None

Plugging values would’ve given you None very quickly

Answer 35

If the product of two integers or two expressions of integers which doesn’t convert them to non-integers is equal to 1 then either both are 1 or both are -1.
NOTE : if one of the expressions is an even power, it can be both -1 and 1 as long as the remaining expression is 1.
(Made silly mistake on the above)

Question 36

1) 0^0 =?
2) Is 0 equal to its OPPOSITE?
3) Is 0 a multiple of all the numbers?
4) Is 0 is the only number equal to all its multiples?
5) Is 0 a factor of any number?
6) 0 : even or odd?
7) Is every number a factor of 0?
8) Is 0 a perfect square

Answer 36

1) undefined.
2) 0 is the only number equal to its OPPOSITE (0 = -0)
3) Yes
4) Yes
5) 0 is not a factor of any number EXCEPT itself
NOTE : ‘a’ is a factor of ‘b’ if b = a x k (for some
integer k)
6) 0 is an even number
7) Yes
8) Yes

Question 37

1) Is 1 the only number with exactly 1 factor?

2) Is 1 prime?

Answer 37

1) Yes

2) No, reason in (1)

Question 38

If m and n are integers and mn + n^2 is odd, what can be said about :

1) mn
2) n^2 + m
3) n x m^2

Answer 38

Something that helps is factorizing it to a product of expressions since applying odd even logic to products is easier than applying it to sum/difference (for product, if one subexpression is even, the entire expression is even, for the entire expression to be odd, all the subexpressions have to be odd)
Hence, write the given expression as n(m + n). Now is this is odd then n has to be odd and so does m +n. Hence, m has to be even.
Use this to solve further.

Question 39

even/even can be odd or even
even/odd will be even
odd/odd will be odd

Answer 39

The absolute value of a number is it’s distance from 0 on the number line.
Signed numbers can be positive or negative.

Question 40

When we add two or more negative numbers, the result is always a smaller negative number.

When we add two numbers that have opposite signs, the sum is always between the two addends.

The opposite of a number is the number with the opposite sign.
Opposite of 8 is -8.
Opposite of -5 is 5.

In the expression a + b, ‘a’ is called minuend and ‘b’ is called subtrahend.

Answer 40

When a nonzero base is raised to an even exponent, the result will always be positive.

When a nonzero base is raised to an odd exponent, it retains it’s original sign.

What this means is that, if we know the sign of the result of an expression with even exponent, we CANNOT find the sign of the original expression.
But, if we know the sign of the result of an expression with an odd exponent, we can find out the sign of the original expression.

Question 41

2 3 5 7 11 13 17 19 23 29 31 37
41 43 47 53 59 61 67 71 73 79 83 89
97

2^5 = 32
2^10 = 1024

Answer 41

Number of prime factors and the number of unique prime factors refer to different things.
For 32 = 2^2^2^2^2 (i.e. 2^5), the total number of prime factors is 5 while the total number of unique prime factors is 1
For 12, total = 3, unique = 2

Question 42

To find LCM of 3 or more no.s :
A prime factor is repeated in when that prime factor is shared by at least two no.s in the set. That is, it does not need to be shared by all of the numbers to be considered a repeated prime number.
For the non shared/repeated factors, use them as is with their exponent intact.
Using the above flow, it’s easy to understand why the LCM of x and y is xy if they share no prime factors
Eg : LCM of 15, 18 and 24
15 = 3 x 5
18 = 2 x 3^2
24 = 2^3 x 3
Now 2 is shared across only 2 terms. But we still consider it in the LCM calculation with the highest power. LCM : 2^3 x 3^2 x 5

Eg : LCM of 8, 9, 10
8 = 2^3
9 = 3^2
10 = 2 x5
Common = 2, max power = 3, so pick 2^3. 3^2 and 5 are not shared, pick them as is.
LCM = 2^3 x 3^2 x 5 = 360

Answer 42

For a set of positive integers, the LCM will always be greater than or equal to the largest number in the set and the GCF will always be less than or equal to the smallest number in the set.

If y divides evenly into x, GCF(x, y) = y and LCM(x, y) = x

To find GCF of 3 or more no.s :
Here, for a prime factor to be repeated, it needs to be present IN ALL of the no.s. For the repeated root, find the lowest exponent and pick that. Do this for all the repeated roots and then multiply them all.

GCF(a, b, c) = GCF(GCF(a, b), c)

If the LCM of x and y is p and the GCF of x and y is q, then xy = pq.

The LCM of a set of positive integers provides us with all the unique prime factors of the set.
Thus, it provides all the unique prime factors of the product of the no.s in the set.
It also tells us the max power of the prime factors in any of the set’s no.s
I.e if LCM of x, y and z is a^b x c^d x e^f. Then if c divides x, the max power of c in x will be <=d

Question 43

Euclid Algo for GCD
Find the GCF (182664, 154875, 137688) or GCF (GCF(182664, 154875), 137688)
First we find the GCF (182664, 154875)

182664 - (154875 * 1) = 27789

154875 - (27789 * 5) = 15930

27789 - (15930 * 1) = 11859

15930 - (11859 * 1) = 4071

11859 - (4071 * 2) = 3717

4071 - (3717 * 1) = 354

3717 - (354 * 10) = 177

354 - (177 * 2) = 0

So, the the greatest common factor of 182664 and 154875 is 177.

Now we find the GCF (177, 137688)

137688 - (177 * 777) = 159

177 - (159 * 1) = 18

159 - (18 * 8) = 15

18 - (15 * 1) = 3

15 - (3 * 5) = 0

So, the greatest common factor of 177 and 137688 is 3.

Therefore, the greatest common factor of 182664, 154875 and 137688 is 3.

Go till the RHS in 0 and pick the previous RHS.

Answer 43

‘x is a dividend of y’ means that y is a factor of x
‘y divides into x evenly’ : same meaning as above.

If z is divisible by both x and y, z must also be divisible by the LCM of x and y

Question 44

An item was initially being sold for $x where x is a whole number. During a sale, it was sold for 16% of it’s initial selling price. What can you comment about the sale price wrt divisibility?

Answer 44

sale price should be divisible by 4.
Reason : let n be the sale price
n = 16/100 * x = 4/25 *x
(25/4)*n = x
Since we know that x is a whole number, n must be divisible by 4.

Question 45

Number is divisible by 6 if?
Number is divisible by 4 if?
Number is divisible by 8 if?
Number is divisible by 9 if?
Number is divisible by 11 if?
Number is divisible by 12 if?

Answer 45

It is divisible by 2 and 3.
If the last 2 digits are divisible by 4
If the last 3 digits are divisible by 8
If sum of all digits is divisible by 9
If sum of odd numbered place digits - sum of even numbered place digits is divisible by 11 (numbering starts with 1)
If number is divisible by 4 and 3

The divisibility rules apply EVEN IF THE NUMBERS ARE IN DECIMAL FORM

Question 46

Number is divisible by 7 if?

Eg : Test divisibility of 458409

Answer 46

We first take the last digit and multiply it by 2. So,(9 × 2 = 18). Subtract 18 with the rest of the number, which is 45840. So, 45840 -18 = 45822. We are not sure if 45822 is a multiple of 7.
We repeat the same process again with 45822. Multiply the last digit by 2. So, (2 × 2 = 4). Subtract 4 with the rest of the number, which is 4582. So, 4582 - 4 = 4578. We are not sure if 4578 is a multiple of 7.
Let us repeat the process again with 4578. Multiply the last digit by 2. So, (8 × 2 = 16). Subtract 16 with the rest of the number, which is 457. So, 457 - 16 = 441. We are not sure if 441 is a multiple of 7.
Let us repeat the process again with 441. Multiply the last digit by 2. So, (1 × 2 = 2). Subtract 2 with the rest of the number, which is 44. So, 44 - 2 = 42. 42 is the sixth multiple of 7. Therefore, we can confirm that 458409 is divisible by 7.

Question 47

Product of ANY n consecutive integers is always divisible by n!
Also applies for expressions like : (x+6)(x+7)(x+8) or (x-1)(x-2)(x-3) or x^2 - x or x^2 + x or x^5 - 5x^3 + 4x

Answer 47

Product of ANY n consecutive even integers will always be divisible by (2^n) * n!

No special rule for Odd consecutive integers.

IMPORTANT : You can use these two rules in conjunction :
If you know that you have 3 consecutive no.s and 2 of them are even. Then in addition to being divisible by 3! = 6, it will also be divisible by (2^2)*2! = 8

Question 48

When n is divided by 15, the remainder is 2. When n is divided by 17, the quotient is Z and the remainder is 2. Z must be divisible by all of the following except?
1
3
5
15
30

Answer 48

Super smart way : it can only be 30. If it was 15, it would’ve been 1, 3 and 5 also. It cannot be 3 and not 15/30 and so on.
Normal way : n/15 = q + 2/15 : n = 15q + 2
n = 17z +2
i.e. 17z = 15q. Since q is an integer : 17z/15 is an integer. Hence, z is div by 15.
I was getting stuck since I was thinking only z is an integer and writing z = 15q/17 which was leading me nowhere.

Question 49

If you’re given that some integer x when divided by some integer y gives 9.48, that itself is not sufficient to find the remainder. Why?

It’s not sufficient to find the dividend, divisor or remainder. Only quotient you can find.

Answer 49

Because, all we know is x/y = 9.48 =9 + 48/100.
Now we cannot say for certain that it is 48/100 (which means remainder =48 and y=100). It can also be 4800/10000 (r=4800 y=10000 which is an entirely different division altogether). It can also be 24/50. So, an infinite number of remainders are possible.

All we know is that since the simplest form of 48/100 is 12/25, whatever the remainder is, it is divisible by 12.

To find r, we need to know the value of x or y or both.

Question 50

To find the remainder for a division such as : 9/5 = 1.8,
we can multiply the decimal part by the denominator :
.8 *5 = 4. Thus the remainder is 4
NOTE : we are able to produce an exact remainder since WE KNOW the denominator (y = 5)

Answer 50

Remainders IN FRACTIONAL FORM can be multiplied but we need to correct for excess remainders in the end.

Suppose we need to find remainder when 12 x 13 x 17 is divided by 5 :
12 = 2 2/5
13 = 2 3/5
17 = 3 2/5
ONLY the remainders can be multiplied to get 232 = 12.
Fraction = 12/5 (Don’t straight up multiply the fractions, you’ll get 12/125 in that case. Only take the numerators)
After correcting for excess remainder (remove the excess in the numerator by dividing by the denominator and removing multiples of the denominator until we get a proper fraction), the remainder is 2/5

(Check : 121317 is 2652 which when divided by 5 gives remainder 2)

If the result is negative (eg : 12-1317 => 2/5 , -3/5, 2/5) : -12 then correct for negative remainder. -2 should be corrected to 3. Remainder is 3.

Question 51

X = 500 * 600 * 700. What is the remainder when X is divided by 8.

500 = 8*62 + 4/8 [DON'T SIMPLIFY 4/8 to 1/2. WE NEED TO RETAIN DENOMINATOR AS 8 SINCE THAT IS THE DIVISOR]
600 = 8*75 + 0/8
700 = 8*87 + 4/8

Product of remainders = 4 * 0 * 4 = 0
In fraction = 0/8. Hence, remainder = 0

Answer 51

REMAINDERS can be added or subtracted as well. We need to correct for excess remainder in the case of addition and negative remainder in the case of subtraction. (A remainder must be a non negative integer less than the divisor)

Eg : Remainder when 12 + 13 + 17 is divided by 5
Remainders : 2/5 , 3/5 and 2/5
Adding them : 2 + 3 + 2 (as in the case of multiplication of remainders, only take the numerators) = 7. Fraction = 7/5
After correction : 2/5. Hence, remainder is 2

Eg : Remainder when 17 - 13 is divided by 5
Remainder = 2/5 and 3/5. Subtracting : -1. Fraction : -1/5
After correction (add denominator to numerator until it becomes positive) : 4/5
Hence, remainder is 4

Consider the timing payoff between simply adding the numbers and then finding remainder vs doing using these methods.

Question 52

Any factorial >= 5! will always have 0 as its unit digit

Answer 52

(5x2) pairs give us the number of ZEROES only.
for the digits to the left of the zeroes, you need to see if there are any other factors left, then take their product and pad with the zeroes (from 5x2 pairs) to get final number.
Else, USE 1 and pad 1 with the zeroes. DO THIS CAREFULLY

Question 53

Leading zeros are any zeros that appear after the decimal point and before the first nonzero digit. 0.002 has 2 leading zeros and 0.7310 has no leading zeros.

Answer 53

1) If X is an integer with k digits, and if X is NOT A perfect POWER of 10, then 1/X will will k-1 leading zeros
Eg : 1/10001 will have 4 leading zeros in its decimal form.

2) If X is an integer with k digits, and if X is a perfect POWER of 10, then 1/X will have k-2 leading zeros.

Question 54

To determine the largest number of a PRIME NUMBER x that divides into y! , we perform the following steps:
1) Divide y by x, x^2, x^3 etc and keep track of the quotients while ignoring any remainders. We can stop once the quotient becomes zero
2) Add the quotients from the previous divisions.
The sum represents the number of prime number x in the prime factorization of y!

Answer 54

To determine the largest number of a NON-PRIME NUMBER x that divides into y! (if the prime factorization of x = a x b x c …. where all of a, b, c … have power 1) , we perform the following steps:

1) Break x into prime factors
2) Using the largest prime factor of x, apply the factorial divisibility shortcut to determine the quantity of that prime factor. The quantity determined represents the largest number of x that divides into y!

Question 55

To determine the (largest) value of a non-prime number x (where x = p^k, p is a prime and k is an integer greater than 1) that divides into y!, we perform the following steps:

1) Express x = p^k
2) Apply the factorial divisibility shortcut to determine the quantity of p in y!. Let that be ‘a’. Then create and simplify an inequality : ‘kb <= a’ to determine ‘b’ which is the largest number of x that divides into y!

Answer 55

if 100!/8^n is an integer, what is the largest possible value of integer n.
Expression : 100!/(2^3)^n.
We determine the quantity of 2 in 100! = 97
Now we solve the inequality 3n <=97 (n has to be int)
n = 32. Greatest possible value of n is 32

Question 56

To determine the (largest) value of a non-prime number x (where x = p^k, p is a NON PRIME and k is an integer greater than 1) that divides into y!, we perform the following steps:

1) Express x = q^k * r^s… where q,r… are all primes.
2) Apply previous methods to find largest possible value of q^k, r^s… in y! by solving inequalities etc.
3) Now pick the minimum value among the largest possible values of all the primes in y!

Answer 56

A perfect square OTHER THAN 0 and 1, is a number such that all of its prime factors have even exponents.

A perfect cube OTHER THAN 0 and 1, is a number such that all of its prime factors have exponents that are divisible by 3

Question 57

The decimal equivalent of a fraction will terminate iff the denominator of THE REDUCED FRACTION has a prime factorization that contains only 2s OR 5s OR both.
If the prime factorization of the REDUCED FRACTION’s denominator contains anything other than 2s or 5s, the decimal equivalent will not terminate.

Answer 57

The GCF of two consecutive integers is always 1.

When a whole number is divided by 10, the remainder will be the units digit of the dividend.
When by 100, the last two digits of the dividend will be the remainder and so on.

When integers with the same units digit are divided by 5, the remainder is constant.
I.e remainder when 1{7} is divided by 5 is same as when 342332{7} is divided by 5 since both have the same units place.

Question 58

When the original quantity of x is REDUCED by y%, the remaining portion becomes (1-(y/100))*x NOT yx/100
WASTED TIME ON THIS:
if the original mass of x grams decayed by 94%, the remaining mass can be :
27
28
29
31
32
Kept writing 94x/100 = y again and again instead of 6x/100 = y

Answer 58

If we know the GCF and the LCM of 2 no.s, that is NOT enough to find the 2 no.s, we can only find their product with this information.
HCF won’t even give you all the factors of the no.s, it’ll only give you the factors that it has in common with the lowest possible power among them.

If there are two fractions and both can be expressed as a terminating decimal, then their sum can be expressed as a terminating decimal.

If there are two fractions and one can be expressed as a terminating decimal while the other can’t, then their sum can’t be expressed as a terminating decimal.

It there are two fractions and both can’t be expressed as terminating decimal, then their sum MAY BE a terminating decimal.

Question 59

If (500! x 500! x 500!)/500^n is an integer, what is the largest possible value of integer n?
120
121
122
123
124

Answer 59

Very TRICKY
if you solve normally, 500 factorizes to 5^3 * 2^2.
Now you find the max power of 5 in 500 which comes to 124.
NOW IF YOU SOLVE 3x <= 124, you get 41 and 41 x3 =123 WHICH IS WRONG
Instead understand that the numerator is (500!)^3. So the inequality to solve is 3x <= 3*124
So, x is 124
HENCE, always find the final expression in numerator BEFORE solving the equality.

Question 60

The LCM of ‘n’ no.s is the FIRST number they all divide into at the same time.

Answer 60

ALWAYS think of 1 as a factor when you factorize (especially during word problems)

Question 61

If q = 40! + 1, which of the following cannot be a prime factor of q?
11
19
37

Answer 61

Two consecutive integers cannot share the same factors since GCF = 1. So, 40! and 40! + 1 will not have any common factors except 1.
And since we know 11, 19 and 37 divide 40!, they will NOT divide 40! + 1

Question 62

If you know that the pattern starts from ‘n’th element and has a repeating set of size ‘m’ and you need to find what the kth term will be. No. of terms : k-n+1. Now find the nearest multiple of m to k-n+1. That term will be where the last repeating set ENDS. you can find out the required element from there.

Simple example : pattern starts from 5th element, size is 4, we want 12th element.
12 - 5 +1 = 8 and 8/4 is 0 that means a pattern ends there. so pick the last element of the pattern.
if 11th element, 11-5+1 = 7 and 7/4 is 1 with 3 remainder. so 1 pattern ends and the third element of the next iteration of the pattern is what we need. Hence, pick the 3rd MEMBER OF THE PATTERN.

Answer 62

If you want to find no. of multiples of x between y and z (both inclusive), find nearest multiple of x >=y and nearest multiple of x<=z let them be p and q respectively.
Then, the no. of multiples of x is : ((q-p)/x) + 1

The only integers with exactly 3 factors are squares of prime no.s.

Question 63

If a and b are integers, is a < 0?
1) b^2 - a^2 < 0
2) b^2 - a^3 > 0
Check if it’s C or E

Answer 63

It’s C because, if add the inequalities, you get a^3 < b^2 < a^2
solution interval for a : 0<a></a>

Question 64

Positive integer n is a multiple of 18, and positive integer m is a multiple of 30. Is mn a multiple of 120?

1) n is a multiple of 12
2) m is a multiple of 40

Answer 64

Using 1) you can find that n is a multiple of 36.

Now, there is a nuance to this. What you did was ‘if a number is divisible by a (18) and b (12), it is divisible by its LCM’. Now, you can say that if a number is divisible by m and n, it must be divisible by the LCM of 36 and 30 but that’s not what you want here.
Had you been asked about the characteristics of a number divisible by m and n, you would use the LCM to draw conclusions.
Here, you are straight up given the number which is mn. Hence, mn will be 36p x 30q = 1080pq. While it is obvious that it will be divisible by 180, it is also evident that it is divisible by 120.
The point is don’t just find the LCM of 36 and 30 and since that’ll be 180 conclude that mn need not be divisible by 120. THAT IS WRONG.
That conclusion would have been right if you had not been told that the no. being considered is mn. Since you know the number (mn), you can draw additional conclusions.

Question 65

x has only one unique prime factor DOES NOT mean that x is a prime.
It just means that it can be expressed as x = p^n where p is prime

Answer 65

We know that for a number to be a perfect square, it has to be of the form n^2.
Note that if n is a perfect square in itself, then n^k where k is >= 0 will always be a perfect square so you NEED NOT HAVE THE CONDITION THAT k SHOULD BE EVEN

Question 66

When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x+ y is divided by 21.

1) x^2 divided by 7 leaves a remainder of 4
2) y - 4 is divisible by 3

Answer 66

In questions like this, put values quickly.
x will be 2,5,8,11…
y will be 4,11,18,25,32….

A and B are easy to eliminate since we won’t know anything about x + y since each concern with only x or y

C ) Now, x = 2 and 5 satisfy condition 1 and y = 4, 25 satisfies 2)
x + y = 6, 9, 27, 30. Divided by 21, these leave diff remainders hence E.

Using algebra to solve this would’ve taken ages.

Question 67

If s and k are prime no.s, what’s the GCF of s and k?

Answer 67

NOT ALWAYS ONE, if s = k, GCF(s,k) = 1
MADE A SILLY MISTAKE ON THIS FFS

ALWAYS THINK 0? EQUAL? INTEGER? NEGATIVE? FRACTION? OPPOSITE?

FZONE : Fraction (both +ve and -ve)
               Zero (one or more)
               Opposite
               Negative (one or more)
               Equal (both +ve and -ve)

Question 68

The square roots of non-negative no.s come in pairs except for 0.
BUT, when we use the radical sign, we’re referring to the principal square root which is always positive.

Answer 68

The square root of a variable squared is equal to the absolute value of that variable which is ALWAYS positive except in the case of 0.
(x^2)0.5 = |x|

Question 69

A radical’s index is simply the number found on the top left of the radical. Basically the n is (x)^1/n
Further, (x^n)^1/n = |x| IF n IS EVEN
(x^n)^1/n = x IF n IS ODD

Answer 69

512 and 729 are perfect cubes of 8 and 9 respectively.
Approx values : 
(2)^0.5 = 1.4
(3)^0.5 = 1.7
(values for 5 to 8 are in gaps of .2)
(5)^0.5 = 2.2
(6)^0.5 = 2.4
(7)^0.5 = 2.6
(8)^0.5 = 2.8
(10)^0.5 = 3.15

Question 70

How to find (80)^1/4?

Answer 70

Find closest 4th power immediately less and greater than the given number and apply crude interpolation.

3^4 is 81, 2^4 is 16. (80)^0.5 must be very close to 3 but a smidge less

(9)^0.25 : 1^4 =1 , 2^4 = 16. 9 must be somewhere in between tending towards 2, so 1.6 (actually 1.7)

Question 71

(a)^1/m * (b)^1/m = (ab)^ 1/m
(ab)^ 1/m = (a)^1/m * (b)^1/m
If m is even, a and b MUST BE NON-NEGATIVE

((a)^1/n)/((b)^1/n) = (a/b)^1/n and vice versa. If n is even, a has to be non negative and b has to be positive

Answer 71

Two or more radicals are like radicals if they both have the same root index and the same radicand (expression under the radical)

A simplified fraction will never have a radical in the denominator.

Question 72

IF AN EQUATION INVOLVES A RADICAL SIGN, THEN AFTER SOLVING IT, ALWAYS PLUG BACK THE VALUES TO CHECK IF THE VALUES ARE ACTUALLY A SOLUTION OR NOT

Answer 72

If a =/= -1, 0 or 1, and a^x = a^y, then x = y.
Also, if a =/= -1, 0 or 1, and a^x * a^y = a^z, then x + y = z

x^a * y^a = (xy)^a
If y =/= 0, x^a/y^a = (x/y)^a

Question 73

If a and b are prime and x, y, z and w are int, if a^x * b^w = a^y * b^z, then x = y and w = z

Answer 73

How to compare : (3)^1/2, (4)^1/3, (5)^1/4, (7)^1/6?
Compute LCM of powers : 2,3,4 and 6 => 12, raise them all to the 12th power : 3^6, 4^4, 5^3 and 7^2. Compare these no.s, and their order will give you the order of the initial terms

How to compare : 3^100, 4^75, 5^50, 6^25
Compute GCF of the powers : 100, 75,50 and 25 => 25
Raise all the terms to their 1/GCF th power to get 3^4, 4^3, 5^2 and 6. Compare these no.s, and their order will give you the order of the initial terms.

Question 74

2^n + 2^n = 2^n+1
3^n + 3^n + 3^n = 3^(n+1)
4^n + 4^n + 4^n + 4^n = 4^(n+1)

Answer 74

If base is greater than 1 and exponent is a positive proper fraction => result is smaller
If base is positive proper fraction and exponent is a positive proper fraction => result is larger

Question 75

If a positive integer ending with zeros is a perfect square, it MUST HAVE AN EVEN NUMBER OF ZEROS to the right of its final nonzero digit.
The square root of such a perfect square will have exactly half as many trailing zeros as the perfect square.

Extrapolating this, the square of a number with trailing zeroes will have TWICE AS MANY trailing zeroes

Answer 75

If a decimal with a finite number of decimal places is a perfect square, its square root will have EXACTLY HALF the number of decimal places.
Thus, a perfect square decimal MUST HAVE AN EVEN NUMBER OF DECIMAL PLACES.
NOTE : We’re talking about decimal places NOT DIGITS here.

Extrapolating this, the square of a decimal number will have TWICE AS MANY DECIMAL PLACES

Question 76

The cube root of a perfect cube has exactly ONE THIRD the number of zeros to the right of the final nonzero digit as the original perfect cube.

Extrapolating this, the cube of a number with trailing zeroes will have 3 TIMES AS MANY trailing zeroes

Answer 76

The cube root of a perfect cube decimal has exactly one third the number of decimal paces as the original perfect cube.

Extrapolating this, the cube of a decimal number will have 3 TIMES AS MANY DECIMAL PLACES

Question 77

What’s the decimal value of (0.000005)^2?

Answer 77

Square of 5 is 25.
The number of decimal places is 6, so the square will have 12.
So, fill enough zeroes so that the total no.s of digits is 12 including ‘25’ so basically 10 zeros
0.000000000025

Question 78

What is (7 + 2(6)^0.5)^0.5 + (7 - 2(6)^0.5)^0.5

Answer 78

1) Estimate if the options are discernably apart

2) Square it and simplify to get exact answer. (Rationalization is not always the only approach)

Question 79

After the decimal, if there are n digits with the nonzero number at the end as y, you can convert it to y * 10^-n.

Eg : .0023 = 23 x 10^(-4) (Since there are 4 digits after the decimal)

Answer 79

9^(-x) * 16^(-10y) = 18^-30
You factorize both LHS and RHS to common prime factors on both sides to get :
3^(-2x) * 2^(-40y) = 3^(-80) * 2(-40)
NOW, you can equate -2x to -80 and -40y to -40 ONLY IF YOU KNOW THAT x and y are integers.
Otherwise, we will have infinitely many solutions.
For eg : set x to 0 and y will be approx 4.17 etc.

Question 80

When graphing a variable x that is greater than B, use open circle at point B to indicate that B is not a possible value for x and extend the solution line to the right of B.
Use closed circle if x is greater than or equal to B.
Similar for less than and less than or equal to.

Using a number line to compare relative sizes of variables expressed in multiple inequalities is surprisingly error proof and handy.
Eg : m > n and n<p> q? 1) q >n 2) q >p (Ans : E)</p>

Answer 80

When the signs face the same direction, multiple inequalities can be added in the same way that we add multiple equations.

A compound inequality, or a three part inequality, such as a<= x + 5 <= b can be manipulated in the same way as an equation can. Make sure that whatever you do to one part of the compound inequality, you do to all the parts. When a compound inequality is multiplied or divided by a negative number, both of the inequality signs must be reversed.

Question 81

If a<=x<=b and c<=y<=d, to find the maximum and minimum value of xy, we evaluate ac, ad, bc and bd. The largest of these 4 quantities will be the max value of xy and the smallest will be the minimum value of xy.

Answer 81

If two absolute values/equations are equal, it must be true that the expressions within the absolute value bars are either equal or opposite

Question 82

They (RHS and LHS) can be equal only if :

1) One or both of a and b are 0 OR
2) Both a and b are of the same sign

This can be extended to more than 2 variables.
For eg : for a, b and c :
1) At least 2 of a, b and c are 0 OR
2) All a, b and c are of the same sign

a + b | <= | a | + | b |

Answer 82

They (RHS and LHS) can be equal only if :

1) the second quantity (b) is 0 OR
2) Both quantities are of the same sign and the absolute value of the first quantity is greater than or equal to the absolute value of the second quantity i.e. | a | >= | b |

a - b | >= | a | - | b |

Question 83

Solve | 4x + 6 | < 8

For this, mechanically you do 4x + 6 = 8 and 4x + 6 = -8.
The understanding is that 4x + 6 can either be positive or negative.
If it is positive, it will be 4x + 6 and the max value will be 8 so 4x + 6 = 8
Similarly, if it is negative, it will be -(4x + 6) since modulus always gives positive and the max value will be 8 so -(4x + 6) = 8 which we write as 4x + 6 = -8

ALWAYS USE THE ABOVE METHOD INSTEAD OF MECHANICAL METHOD.
EG OF FUCKUP WITH MECHANICAL :
Solve | 2x - 5 | <6
mechanically did 2x - 5 < 6 and 2x - 5 < -6
The second equation (2x - 5 < -6 ) is WRONG
Correct way :
when positive : 2x - 5 < 6
when negative : -(2x -5) < 6 => -2x + 5 < 6 => -2x < 1
=> x > -1/2

The mechanical approach would’ve given x < 1/2 which is WRONG

Answer 83

If an absolute value equation has a variable on both sides of the equation, WE MUST CHECK the solutions of the equation for extraneous solutions (another case where you need to plug and check is when radical sign is involved).
The extraneous solutions happen because for some solutions, the equation becomes of the form | ax + b | =-k where k is a +ve constant. the above equation form has no solutions. Hence, we need to check to prevent this.

Eg: | 2x + 3 | = x
We solve as normal,
2x + 3 = x and -(2x + 3) = x
x = -3 and x = -1
Plugging them back into the initial equation, we see that both do NOT satisfy the equation. This is because, we have modulus on one side and negative values on the RHS which will obviously not have any solutions.

Question 84

If p and q are the largest and smallest values respectively that k can take and -6 < k < 12, p and q are 11 and -5 respectively.
NOT 12 and -6 YOU IDIOT

Answer 84

a /(b+c) = b/(c+a) = a/(b+c) = r. What values can r take?

When you put smart values, try more than 1 for each case to make sure that it holds for them all.
Case 1 : r = 1/2
a = b = c = 1
1/2 = 1/2 = 1/2 = r = 1/2
Case 2 : r = -1
Now, solving using r = -1 gives you a + b+ c =0
What I did wrong : Take 1, 0 and -1 and see that it does not satisfy the initial equation hence -1 cannot be a value.
What I did not notice : If I take -1,0 and -1, then r does not have a consistent value of -1 in the first place.
What I could have done : Take another value 2, -1, -1 and see that it satisfies all the required equations.

Question 85

For a number line to INCLUDE/ENCOMPASS an inequality, it has to include ALL the values of the inequality.
I.e. if the inequality is |x| >=16
Options : 
1) x >= 30 and x <= -30
2) x >= 10 and x <=-10

PICK OPTION 2 since it includes ALL the points of the given inequality. Option 1 misses out on a few. It is fine that option 2 has extra points that the given inequality doesn’t cover since we just want it to include our given inequality.

Answer 85

If after solving an equality equation (any order) you get the roots are (1, 3) and (3,5) for each of the CASES respectively, consider 3 as the same root and not a repeated root.
I.e. the sum of the roots will be 1 + 3 + 5 and NOT 1 + 3 + 3 +5.
Similarly, product will be 1 * 3 5 and NOT 1335

Question 86

If x > y > 0, is y = 5?

1) 3x + y = 21
2) x + 4y = 23

1) Gives us x = (21 - y)/3
Nifty trick :
put x = (21 - y)/3 in x > y > 0 (from stem)
21-y > 3y > 0 => 21/4 > y => 5.25 > y
This tells us y can be equal or less than 5

2) Gives us x = 23 - 4y
Nifty trick :
23 - 4y > y > 0
23/5 > y > 0 => 4.6 > y
Hence, we know y cannot be 5

Answer 86

If x > y > 0, is x > 3?

1) 2/x + 1/y = 11/15
2) 2x + 3y = 6

It’s easy to see that 2) is enough
For 1, we solve to get y = 15x/11x-30 and we put it into the given inequality (x>y)
Now, here’s the neat part, since it is given that x and y are positive, we know that y is positive and 15x is positive. So, 11x - 30 HAS TO BE positive.
So, x > 15x/11x-30. Now, you can cancel x since you know it is positive and MULTIPLY by 11x-30 without altering the inequality sign SINCE YOU KNOW IT IS POSITIVE TOO.
You get 11x - 30 > 15
x > 45/11
1) is enough too
Answer is D

Question 87

Unit conversion works like this : THIS IS SILLY MISTAKE PROOF :
suppose I have area as 1200 m^2 and I need to convert it to km^2, then I do the following :
Conversion factor = 1 km = 1000 m
Since I’m dealing with m^2, I square the above,
1 km^2 = 10^6 m^2
Now, to convert m^2 to km^2, I need to use the conversion factor in a way that the unit I want to convert (m^2) comes in the denominator.
So, my conversion factor is 1 km^2 / 10^6 m^2
1200 m^2 * 1 km^2 / 10^6 m^2
m^2 cancels from numerator and denominator
to get (1200/10^6) km^2 which is the final answer.

Answer 87

Tricky question :
A certain airplane travels 26 miles per gallon of gasoline. if the plane carries 100 cubic feet of gasoline, how many miles can it travel. 1 gallon = 0.13 cubic feet.

Note that the conversion factor is ALREADY IN CUBIC FEET.
so (100 * 1/0.13 )26 is the answer
I was INCORRECTLY doing :
(100 * (1/0.13)^3)26

Question 88

Sometimes in questions where you’re buying things in packages, you won’t get fractional packages. So you’ll need to round it upwards.
MADE MISTAKE ON THIS SINCE BOTH THE FRACTIONAL ANSWER AND THE WHOLE NUMBER WAS THERE AND I CHOSE THE FRACTIONAL.

Answer 88

For age problems, start with a variable for present day age and then add/subtract from it as needed.

Question 89

A pilot was required to fly a certain number of hours one day. In the morning, she flew 1/4 of the day’s required hours plus 2 more, in the afternoon, she flew 2/5 of the day’s remaining required hours. Then in the evening, she flew the remaining 4 hours. What was the daily requirement of flying hours?

Answer 89

I did the following mistake :
Morning : x/4 + 2
Afternoon : (2/5)(x - x/4) THIS IS WRONG
It should be : (2/5)(x - x/4 - 2)

Question 90

The denom of a fraction is 2 more than 3 times the numerator. IF 7 were added to both the numerator and denom, the new fraction would be equal to 3/5. What is the original fraction.

Answer 90

If you solve normally, you would get original fraction as 2/8
DO NOT SIMPLIFY this to 1/4
Reason, 2+7/8+7 = 9/15 = 3/5
BUT 1+7/4+7 = 8/11 =/= 3/5 
VERY IMPORTANT

Question 91

Simple interest occurs when the interest is based on the original amount of principal borrowed or lent. Compound interest occurs when the interest builds on both the original principal and accumulated interest

Answer 91

Simple Interest = principal x rate x time.
Note : If rate 5%, write as 5/100. Convert rate and time to same terms. Month-month or year-year.

Compound Interest : A = P(1 + r/n)^(nt)
A :future value
P : prinicpal
r : ANNUAL INTEREST RATE (if 5%, write as 5/100)
t : time IN YEARS
n : NUMBER OF COMPOUNDING PERIODS PER YEAR

Question 92

Language : An investment of $100 pays 10% interest semiannually for 1 year. What is the value of the investment after one year.

Answer 92

HERE 10% IS THE ANNUAL INTEREST. IT IS JUST THAT THE INTEREST IS BEING PAID SEMI ANNUALLY

Question 93

Constant growth formula : Final value = initial value + (constant increase rate x number of periods of growth)

Fn = kn + p

Answer 93

Without knowing the growth driver/growth patter, we cannot find the nth element of a series.

Eg : 2, 4, 6, 8, ?
We CANNOT say 10 until we are told that this follows an evenly spaced pattern etc.

Similarly,
200, 400, 800, ?
We CANNOT say 1600 until WE ARE TOLD that there is exponential growth of factor 2 going on

Question 94

When exponential decay occurs, the amount of the decay decreases with each successive decay period. The GREATEST decay occurs during the first period of decay.

Answer 94

A mixture matrix helps a lot in a wet mixture question.

Eg : Alcohol solution 1 contains 60 percent alcohol and solution 2 contains 20 percent alcohol. If 1 and 2 are mixed together to produce 80 L of solution of 50 percent alcohol, how many L of 2 were used?

Mixture matrix looks like this :
| Concentration | volume | Weight of compound
A | .6 | x | .6 * x
B | .2 | (80-x) | .2 * (80 - x)
Final | 0.5 | 80 | .5 * 80

Now solve the last column to get x

The above can be extended to any number of mixtures.

Question 95

Word inequalities

up to        |  <=
at least    |  >=
exceed    |  >
no more  |  <=
at most    |  <=
as few as |  >=

Answer 95

If one is the mth person in a line counted from the beginning and nth person counted from the end, the number of people in the line is (m + n - 1)

Be careful when 2 or more people are involved and you are not told who comes first.

Eg : Adam and Ben are standing in a line, if Adam is 10th from beginning and Ben is 15th from the end and there are 5 people between Adam and Ben, there can be BOTH 30 AND 18 PEOPLE IN THE LINE DEPENDING ON WHO COMES FIRST BETWEEN Ben and ADAM

i.e 3 B 5 A 8 = 18
or 9 A 5 B 14 = 30

Question 96

A pilot on normal plane setting takes 24 hrs for a trip. If he increases power setting by 1%, the flight time is reduced by 1/4 of the previous time taken. Power only moves in 1% increments.
What percent above the normal power settings must the pilot set for the flight time to be less than 10 hrs.

Answer 96

What I did : 24 x (3/4)^n < 10 : (3/4)^n = 5/12 and got stuck
Better solution : Calculate iteratively, manually.
1% increase : 24 - 1/4x24 = 18
2% increase : 18 - 1/418 = 13.5
3% increase : 13.53/4 = 10.125
4% increase will make it sub 10

Question 97

Jenny deposits $d into an account that earns 8% annual interest compounded semi annually. After 6 months, she add another d dollars. How much money will she have after 1 year?

Answer 97

In questions like these, you can decouple the individual payments/installments.
1st installment : d(1 + .08/2)^2
2nd installment : d(1 + .08/2)^1
The sum of the two installments will be total money at the end of the year.

YOU NEED NOT DO : Find money at 6 months, add d and then calculate the final for that amount after another 6 months. This gives some complex numbers.

Question 98

John and Maria are waiting in a line. Counting from the beginning of the line, John is the 5th person; counting from the end of the line, Maria is the 5th person. Including John and Maria, are there more than 10 people waiting in the line?

1) John is ahead of Maria
2) There is 1 person in between John and Maria

Answer 98

I mistakenly selected (B). The answer is (C)
The reason is that there can be more than one person in between John and Maria even when Maria is in front of John while keeping the 5th position of both intact.

Eg : | | M | J | | or in extreme case M | | | J
I mistakenly thought only | | | M J | | | is possible else Maria and John would not be 5th from their respective ends.
If John is before Maria, the arrangement is | | | | J x M | | | | where x can be any number of people including 0.

Question 99

William drove from A to B and then drove along the same road from B to A. Was William’s average speed for the entire trip at least 100 miles per hour?

1) From A to B, William drove at an average speed of at least 10 miles per hour.
2) From B to A, William drove at an average speed of no more than 50 mph.

Answer 99

Answer is B)

Easy to see that (1) alone is not sufficient
In the case of 2, see that if the distance from A to B is 100 (can be anything for that matter).
Then average speed is 200/{(100/y) + (100/50)} where y is the speed from A to B
=> 200/{100/y + 2} => Hence, no matter what y is, the fraction will be less than 100.
Hence, (2) alone is sufficient

Question 100

For work problems, rate x time = work (as always match units)

Answer 100

If it’s y hours, then it is 3600y seconds. NOT y/3600 seconds facepalm

Question 101

For rate problems, it is important to understand this :
If it is given that A does a job in 2 hours, his rate is 1 job/ 2 hours i.e. 1/2 job per hour.
Further, if you take a variable r for the rate, don’t take the rate as 1/r, keep it as r itself (I actually did 1/r once ffs). Now, key insight is that time to complete 1 job would be 1/r.

Answer 101

Useful rate equality :

x workers/combined rate of x workers = y workers/ combined rate of y workers

Question 102

VERY VERY NEAT
Father and son shovel snow together in 24 mins. Son alone takes 20 mins more than father. How long does father take to do it alone?
1) 12 mins
2) 40
3) 72
4) 96
5) 120

Answer 102

let the time taken be f and f + 20. so, the rates are 1/f and 1/(f+20). further, 1/f + 1/(f+20) = 1/24
Simplify this to get : t^2 - 28t - 480
Here comes the beautiful part.
Factorizing the above is time taking. Obviously one root will have to be negative and the other positive.
Check the options, if you pick 12, it’ll have to be +12 so in middle term split, -12x. So, the other root will have to be positive which won’t work since we want sum as -28

pick 40, it’ll have to -40x, the other root can be +12x which works perfectly, sum is -28x and product is -480.

The other options will need a big positive term to get sum to -28x which will blow up the product term hence they won’t work.

Use this instead of time taking factorization

Question 103

When Jefferson works alone, he can mow n lawns in x minutes. When Robert works alone, it takes him 20 mins longer than it takes Jefferson to mow n lawns. If Jefferson and Robert both work together at their own individual constant rates to mow n lawns in t minutes, how many minutes would it take Jefferson to mow n lawns by himself?

1) n = 10
2) The two men work together for 24 minutes

Answer 103

Answer is B)

Important things:
1) Understand the rate variable you create. You write rate as n/x and n/(x+20). THIS MEANS that to mow n lawns, he needs x mins and x+20 mins respectively.
SO, TO ANSWER THIS QUESTION, YOU JUST NEED x NOT n.

2) Further, when you get a quadratic equation like : x^2 - 28x -480 which is what you get here, note that this might give you ONLY 1 possible solution since we have a real variable like time for x which can only take positive values. Solving the above gives x = 40 and -12 and obviously only 40 can be an actual value for x.

Question 104

Machines A and B are both used to make shoes. Working alone, A can make 500 shoes per hour. Machine B, alone, can make 800 shoes per hour. If only one machine can run at a time and the two machines are used for different periods of time to make 4000 shoes, did machine A work longer than machine B?

1) It took more than 6 hours to produce 4000 shoes.
2) It took fewer than 7 hours to produce 4000 shoes.

Answer 104

Before putting smart values, become smart. See that you’ll need to test more than 6 and fewer than 7 both. SO PICK VALUES BETWEEN THE TWO, THAT WAY YOU WON’T NEED TO CREATE MORE VALUE AGAIN AND AGAIN.
Pick 6.1 and 6.9 for example.

Question 105

If you’re given something like ‘m is divisible by 20’, just don’t take it as face value. Use the facts like ‘if div by 20, then div by 10, 4,5,2 etc’ and so last two digits div by 4, even, last digit 0 etc etc.

Answer 105

We know that if we have a fraction a/b and we add k to both a and b, we move it closer to 1.
For subtraction of k, think of it this way.
We know that :
1) if a/b is less than 1, adding k to both num and denom makes it closer to 1 hence greater. So, a+k/b+k > a/b
Observe that a/b is k subtracted from both num and denom of a+k/b+k and since we know that a/b is smaller than a+k/b+k so WE KNOW THAT SUBTRACTING k from both num and denom reduces the number.
2) if a/b is greater than 1, adding k to both num and denom makes it closer to 1 hence smaller. so a+k/b+k < a/b. Observe that a/b is k subtracted from both num and denom of a+k/b+k and since we know that a/b is larger than a+k/b+k so WE KNOW THAT SUBTRACTING k from both num and denom increases the number.

NOTE : WE ASSUME THAT EVEN AFTER SUBTRACTING, THE NUMBER IS STILL POSITIVE

Question 106

Equivalent ways of expressing ratios :
x/y
x:y
x to y

Answer 106

If the ratio of A to B is a to b and x is the ratio multiplier, then ax represents the actual amount of quantity A and bx represents the actual amount of quantity B

Question 107

A pasta sauce recipe calls for 2 cups of a, 3 of b and 7 of c. If I have only 5 cups of c, how much of pasta sauce can I make?

Answer 107

Realize that if you add 2, 3 and 7, you would get 12 CUPS of sauce by default. You WON’T get only 1 CUP of sauce by mixing multiple CUPS of sauce.
Answer: (5/7)*12

Question 108

If x : y = 3:4 and x:z = 7:11
Then, to make it a 3 part ratio, convert it so that the common variable (x in this case) is 1.
So, x : y = 1 : 4/3 and x : z = 11/7
Now, SINCE THE COMMON VARIABLE IS EQUAL FOR BOTH THE RATIOS, WE CAN COMBINE THEM
x:y:z = 1 : 4/3 : 11/7
You can also normalize using LCM of 3 and 7 = 21
21 : 28 : 33

Answer 108

Two equal ratios are known as proportion.

If a/b = c/d then we have a proportion and it must be true that ad = bc

Question 109

Over the course of a certain day, pumpkin and blueberry muffins are baked and sold at the diner. Did the number of blueberry muffins for sale decrease? (Only blueberry and pumpkin muffins exist)

Answer 109

Here, DO NOT ASSUME THAT MUFFINS CANNOT INCREASE OVER THE COURSE OF THE DAY. IT IS GIVEN THAT THEY ARE BAKED AND SOLD OVER THE DAY WHICH MEANS THEY CAN INCREASE IF MORE MUFFINS ARE BAKED THAN ARE SOLD.
Made mistake by assuming the above FFS

Question 110

When we have direct variation relationship between two variables x and y, they can be related by the equation y = kx where k is a positive constant.
When x increases or decreases by any factor, y will increase or decrease by the same factor.
Sample question wording : ‘height of a tree varies directly with the tree’s age’

Answer 110

Direct variation can be between higher order terms as well. If y varies directly with the square of x, we write y = k * x^2 where k is a constant. If x^2 [NOT x] increases or decreases by any factor, y will increase or decrease by the same factor

Question 111

When we have inverse variation relationship between two variables x and y, they can be related by the equation : y = k/x where k is a positive constant.
When x increases by a factor, y will decrease by the same factor.
When x decreases by a factor, y will increase by the same factor.

Answer 111

Inverse variation can be between higher order terms as well. If y varies inversely with the square of x, we write y = k / x^2 where k is a constant.

Question 112

A single variable can vary directly with one variable and inversely with another. if y varies directly with x and inversely with z then y = kx/z where k is a constant.

Answer 112

A single variable can vary directly with two other variables simultaneously. if y varies directly jointly with x and z then y = kzx
It can also vary inversely with two, y = k/zx

Language : a varies inversely jointly with the square of B and fourth root of c : a = k/(b^2 * c^(1/4))

v varies directly jointly with j and square of r. v = kj(r^2)

Question 113

For ratio questions, something that helps a lot is if ratio of a to b is given as 5 : 3, write it as a = 5x and b = 3x and now you can use it like actual numbers/quantities.

Answer 113

A pack has 12 A cards and 8 B cards. What is the lowest number of A cards that would have to be removed from the pack so that no more than 40 percent of the pack would be A cards.

In such questions where you have terms like ‘no more, less than, up to’ etc. , it’s always better to put in the corresponding inequality while solving.
(12 - x)/(20 - x) [<=] 2/5
Solving this would automatically give you the correct inequality to pick the answer : 6.666 <= x
The ‘<=’ is what I’m talking about.

Normally you would put equal to 2/5 and solve to get x = 6.66 and think whether x should be greater than 6.66 (7) or less than (6). This step is eliminated by solving in the way mentioned above.

Question 114

Are there more than 100 cheerleaders in HU?

1) The ratio of cheerleaders to coaches at HU is 13 to 1
2) The ratio of cheerleaders to teachers at HU is 8 to 1

Answer 114

Tricky
(C)
Cheerleader : coach :: 13x : x
Cheerleader : teacher :: 8y : y
It is clear that the number of cheerleaders would have to be a multiple of LCM of 13 and 8 which is 104.
Hence, it is clear that there are more than 104 cheerleaders at HU

Usually when given only two ratios, we cannot find the total number of entities. But here, we don’t need the total, we just need to know if it is greater than 100

Question 115

Sample 3 part ratio solving problem :
In a mix, ratio of F to M to S is 10 : 5 : 2. To improve this mixture, the ratio of F to S was halved, and the ratio of F to M was tripled. If the final mix contained 72 ounces of sugar, how many total ounces were in the mix?

Answer 115

Start off with the usual amount of F, M and S being 10x, 5x and 2x
Ratio of F to S is halved, (10x/2x) x 1/2 = 5x/2x
Ratio of F to M is tripled, (10x/5x) x 3 = 6x/x
DON’T CANCEL x IN THE PREVIOUS STEPS
NOTICE THAT F is common to both the fractions/rations. Make the numerator/denominator corresponding to F same in both the ratios and combine to form 3 part ratio. You can also make the F num/denom equal to 1. This is what we’ll do here.
New ratio of F to M to S is 1 : 1/6 : 2/5
i.e. 30x 5x and 12x. Given, S is 72, x = 6
So, Total : 47 x 6 = 282

Question 116

x percent of y is (x/100)y
5 percent of a is (5/100)a
1 million percent of y is (1,000,000/100)*y
1 thousandth percent of z is (1/1000/100) * z = z/100,000

Answer 116

You can use the above method for stuff like ‘100 is what percent of 50’ etc. => x/100 * 50 = 100, x will be the answer is percentage terms.

The percent figure will always be larger than its equivalent decimal figure (5% > 5/100 since 5 > 0.05 :P )

Question 117

‘Marked down x percent’ => ‘reduced/discounted by x percent’

Answer 117

3 percent greater than z = 103 percent of z
450 percent greater than z = (100+450)/100 z = 5.5z = 550 percent of z
100 percent greater than z = 2z = 200 percent of z
800 percent greater than z = 9z = 900 percent of z

Question 118

Pay close attention to the wording :
p percent of x is (p/100)x
z percent greater/more than x is ((100+z)/100)x
z percent less than x is ((100-z)/100)*x

Answer 118

I purchased 10 shares of a stock for x dollars each. I later sold the 10 shares for y dollars each and paid z percent tax on the profit. After tax, what was the profit I made?

1) y - x = 5000
2) If I had paid 10 percent more tax, my profit would have been 30000 dollars

For 2) If the tax you paid before was z/100, now IT WOULD BE 1.1z/100 and NOT (z+10)/100
To make it (z+10)/100, the language would say something like ‘percentage points/basis points of the tax were increased by 10%/were 10% greater’

Question 119

In 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?

1) 260,192
2) 274,548
3) 300,525
4) 384,160
5) 537,824

Answer 119

Super Smart way
What you need is 100x(1.4)^5
Realize that the power of 4 will follow cycle of 4,6,4… and the only option that satisfies this is 5)

Question 120

Fractional change in price = (Final - Initial)/Initial

Percent change = (Fractional change) x 100

Answer 120

Use set matrix to solve overlapping set problems :
| A | Not A | Total
B | Both A and B| Only B | Total B
Not B | Only A | Neither | Total not B
Total | Total A | Total not a | Total

Also the table headers (A and B) need not be single character long, you can use full words or 2 or more characters also for clarity :P

Question 121

Number of A only + Number of B only + Number of both + Number of neither = Total number of unique elements
Number of A + Number of B - Number of both + Number of neither = Total number of unique elements.

Answer 121

Simplified overlapping set setup,
Draw three horizonal lines, draw three vertical lines intersecting the 3 horizontal lines
Also the table headers (A and B) need not be single character long, you can use full words or 2 or more characters also for clarity :P

     | A | A' | Total B       | B'      | Total |

Question 122

De Morgan’s laws come handy for overlapping set questions :
(A or B)’ = A’ and B’
(A and B)’ = A’ or B’

Answer 122

For triple overlapping sets, we need 3 intersecting circle Venn diagram and a 4th circle depicting ‘neither a nor b nor c’ group.

Total number of unique members : A only + B only + C only + A and B only + B and C only + C and A only + A and B and C + Neither A nor B nor C (last term v important)

Total number of unique members : A + B + C - A and B only - B and C only - C and A only - 2(A and B and C) + Neither A nor B nor C (Easy to understand if you visualize)

Pick which of the above two formulas to use based on the data given. Further, if you don’t have the two pairs (combos of 2 of A, B and C) separately, you can group them into one group of ‘exactly two of 3’ or ‘two of 3’ based on which formula you use.

NOTE : if you are not given how many are A and B etc. then subtract the triple overlap from the double overlap numbers to get only double overlap numbers and then proceed from there. Else, draw Venn diagram and proceed.

Question 123

At a certain car dealership with 3000 cars, there are 1200 red cars and 1400 cars with a sunroof. If at least 500 of the cars at the dealership are not red and do not have a sunroof, what is the greatest number of cars at the dealership that could be red and have a sunroof?

Answer 123

Notice that it is ‘at least 500’, take as 500 + x and solve, answer is 1200.

Question 124

Out of 72 rock stars, 20 play guitar, 22 play drums and 30 sing. If 5 play guitar and sing, 4 play drums and guitar, 3 play drums and sing, and 2 do all three, how many do neither of the above 3 activities?

Answer 124

Ans : 10
NOTE : If it is given that 5 play guitar and sing, then IT INCLUDES THE PEOPLE WHO DO ALL THREE since it is not ‘5 ONLY play guitar and sing’

Question 125

Out of 120 people who play either the piano or the drums or the tuba, 50 play the drums, 70 play the piano and 60 play the tuba. If 30 play exactly 2 of the instruments. How many people do not play all three instruments?

Answer 125

Very important
Find the number of people who play all 3 instruments and subtract it from the total (120) so it’s 105.
The number of people who play neither of the 3 instruments here is 0.

Question 126

Very important
The number of people who do not play exactly 2 of the three instruments = Total - total number of people who play exactly 2 out of the 3 instruments = 120 - 30 = 90
It makes sense to include the people who play all 3 in the count

Answer 126

A or B = A only + B only + both A and B

A or B = A + B - (A and B)

If you have trouble visualizing : A or B = (not A and not B)’
A or B = Total - (not A and not B)

Question 127

VERY VERY VERY IMPORTANT : In the case of set matrix DS problems, draw diagonal lines in each box to separate stem 1 and stem 2 data . This helps a lot when you’re considering 1 stem in isolation and prevents info leakage.

This works for all set matrix DS question forms. -> Trust me I checked

Answer 127

HACK : Make the diagonals as you go and not at the beginning. That way it gets easier when you want to apply both 1 and 2 together.
If you’re not able to understand, just shut up and do it.

Question 128

At a particular school, 60 male students play sports. How many male students attend the school?

1) At the school, 3/4 of the female students play sports.
2) At the school, 1/4 of the students do not play sports.

Answer 128

Super tricky questions.
If you use both S1 and S2 (stem 1 and stem2), you’ll get the equation T - F = 80 (T : variable for total students and F variable for females). Now, I thought this is not enough to find male students and did E but if you look carefully, you’ll see that T-F is the number of male students hence the answer is C. Be more alert!!!!

Answer 129

An evenly spaced set is one in which the numbers in the set increase or decrease by a fixed amount.
The elements of an evenly spaced set share a common difference.

Question 130

When the question says ‘inclusive’, it instructs us to include both of the numbers that define the endpoints of the set.

Answer 130

To count the number of consecutive integers in a set that includes the first and last numbers, use the formula : highest number - least number + 1 (works for all combos : 1) negative 2) some positive, some negative, 0 etc. as well).

If only one of the endpoints is included, no. of consecutive integers : highest - lowest..

If both are excluded i.e. we need to find no. of consecutive integers BETWEEN two numbers : highest - lowest - 1

Question 131

To count the number of consecutive multiples of a given number in a range of values that are inclusive of the first and last numbers, use the formula :

((Highest number div by given number - lowest number div by given number)/Given number) + 1

Answer 131

You can use simple AP formula to find number of multiples.
An = A1 + (n-1)d
Since the terms are 1-indexed, n is the total number of terms or in this case, multiples.
A1 is the lowest number div by given number
An is the highest number div by given number
d is the number whose multiples we want to count

Question 132

You can use the above formula/AP approach to find number of odd/even no.s as well

Find no. of odd nos. between 5 and 55 inclusive.
5 + (n-1)2 = 55 => n = 26
26 odd no.s counting both 5 and 55.

Answer 132

Average of an evenly spaced set of numbers.
Method 1 : Add first (lowest) and last (highest) number and divide by 2
Method 2 : It is equal to the midpoint of the set i.e. median. (the set has to be sorted for this to work)

Note that in case you have an odd number of terms in your evenly spaced set, the average exists in the set but if you an even number of terms, the average DOES NOT exist in the set.

Question 133

The number of multiples of A or B = no. of multiples of A + no. of multiples of B - number of multiples of LCM(A, B)

Answer 133

The number of multiples of A or B but NOT BOTH = no. of multiples of A + no. of multiples of B - 2 x (number of multiples of LCM(A, B))

Question 134

Weighted average : {(freq of data pt 1 x data point 1) + (freq of data pt 2 x data pt 2) … }/(sum of frequencies)

Answer 134

The weighted average of two points will be closer to the data point with the greater no. of observations or with the greater weighted percentage.

Question 135

In case of weighted average questions, don’t try to intuitively find the weights, think in terms of formulas.
Eg : If you’re given at 50 kmph, mileage is 4 km/l and at 60 kmph, mileage is 5 km/l and you travel 100 km at 50 kmph and 300 at 60 kmph, what is the average mileage?

DON’T WONDER HOW TO WEIGHT THE MILEAGES USING DISTANCE ETC.
YOU KNOW MILEAGE = DISTANCE/FUEL CONSUMED
JUST USE THIS.
Total distance covered : 100 + 300 = 400
Total fuel consumed : 100/4 + 300/5 = 85
Average mileage = total distance/ total fuel

Answer 135

If a set of numbers has n terms where n is odd, the median is the value at the (n+1)/2 position when the numbers are in numerical order.

If n is even, the median is the average of the numbers are n/2 and {(n/2) + 1} position

In an evenly spaced set, the mean of the set is equal to the median of the set.

Question 136

There can be more than 1 ‘mode’ for a set.

Eg : mode of {1,4,6,7,4,8,7,7,4} is both 4 and 7 since they both appear 3 times.

Answer 136

IMP : If each number in the data set occurs the same number of times as the others, then there is no mode.

Question 137

Range = Highest number in a set - lowest number in a set.

Range is a measure of spread of data set. Greater the range, more the numbers in the data set are spread out.

Answer 137

Standard deviation measures how far the data points are from the mean of the data set.

Range based on mean and no. of standard deviations apart the points are :
High = mean + n(sd)
Low = mean - n(sd)

Question 138

If you add or subtract the same amount to or from each term in a data set, the standard deviation does not change. Hence, you can have different means but the same standard deviation.

Answer 138

If we multiply or divide the elements of a data set by a constant amount, the standard deviation will also be multiplied or divided by that amount.

Question 139

The least possible standard deviation of a set is zero.

One way to guarantee that a positive standard deviation will decrease is to add elements that equal the mean to a set.

Answer 139

IMPORTANT : To compare the standard deviation among data sets that have an equal number of data points, perform the following steps :
1) Determine the mean of each set
2) For each individual set, determine the abs difference between the mean of that set and each data point in the set
3) Sum the differences obtained from each individual set
The set with the greater sum has the greater standard deviation.

Question 140

Actual formula for SD : [ {sum of (mean - point)^2} for all n points in the set/n ]^0.5

Answer 140

IMPORTANT : Instead of brute forcing to find the mean, check if they are evenly spaced. If so, you have wayyy quicker methods to find the mean

Question 141

When we are given that either the smallest value of a set is equal to the largest value or a set or the range of a set is equal to zero, we know that all the data points are the same. Thus, the SD is equal to zero.

Answer 141

If we are given that either the largest value or the smallest value in a data set is equal to the mean, all the data points in the set are the same, and thus the sd must be zero.

Question 142

In ANY of the following cases, the SD is not equal to 0:

1) Data points of a particular set are not all the same
2) The range is not equal to 0
3) The largest or the smallest data point in the set is not equal to the mean

Answer 142

On the first even day of the month, Steve saves $5 and on each following even day, he saves $5 more than he saved the previous even day of the month. If there were 30 days in the certain month, how much money did Steve save by the end of the month?

So on 2nd, he saves $5 and last day is 30th.
Calculate the number of days : (30-2/2) + 1 = 15
Now, IMPORTANT : find average (DON’T FIND SUM OF FIRST AND LAST NUMBER AND FIND HOW MANY SUCH PAIRS EXISTS AND SUM THEM AND THEN ADD MIDDLE TERM, DONT DO THIS). Last term : 5 + (15-1)5 = 75
Average = 75+5/2 = 40 since evenly spaced
Sum is simply : average x number of terms : 40*15 = 600
This is much simpler, using pairs and finding middle term etc. is much more cumbersome and error prone.

Question 143

Dollar value of bonds to stocks to metals is 7 : 9 : 4
After some transactions, the ratio of bonds to metals is cut in half and the ratio of metals to stocks is doubled. What is the new ratio of bonds to stocks to metals?

Answer 143

I always fumble in this. Understand the core funda
Initial ratio : 7 : 9 : 4
Dollar value = 7x, 9x and 4x
Now, the bonds to metals is cut in half : 7/4 * 1/2 = 7/8
metals to stonks : 4/9 * 2 = 8/9
Now, find what is common to the two ratios : metals is common.
MAKE THE COEFFICIENT OF METALS (COMMON THING) TO BE THE SAME ACROSS BOTH THE RATIOS.
here it is already the same (8). So, you can simply convert this to 3 part ratio :
7 : 9 : 8

Question 144

Combinations are used when the order in which a task is completed does not matter.
Permutations are used when the order in the which a task is completed does matter.

Answer 144

Fundamental principle of counting : If there are m ways to perform task 1 and n ways to perform task 2 and the tasks are independent, then there are m x n ways to perform both the tasks together.

Question 145

If different items are being treated as uniform, you don’t need to make cases, just consider them as uniform/of the same kind.

Eg : A treasure chest contains 5 R, 4 E and 3 D. If a pirate picks 5 jewels of which 3 are rubies, how many possible ways are there to pick?

Answer 145

For rubies, it’s 5C3.
Now, for the remaining two jewels, you can make cases : 1 E and 1 D, 2 E 0 D or 0 E 2 D. But note that, you don’t need to differentiate between the jewels as long as you’re picking two. So, just do 7 (4 E + 3 D) C2
Saves a lot of time.

Question 146

For circular arrangements of k items : (k-1)!
Note that in circular arrangements, if the sequence is same clockwise/counterclockwise, it’s treated as the same arrangement.
Better way to understand : If we count a sequence, it is equivalent to all it’s combos formed by a clockwise or an anticlockwise movement of one or more of it’s elements.

Eg : A,B,C if circular arranged can be arranged in only 2 ways : A, B ,C and A, C ,B
A, B, C = A, B, C | B, C, A | C, A, B
A, C, B = A, C, B | C, B, A | B, A, C

Answer 146

If you know that events are mutually exclusive and completely exhaustive, you can use ‘total - something’ to get required amount as long as ‘something’ is easier to compute that ‘required amount’.
This ends up saving a lot of time.

Question 147

If you end up using the approach ‘x people are available for this slot and then y for the next….’ stream of logic and it is a COMBINATION question, DIVIDE IT BY n! WHERE n IS THE NUMBER OF SLOTS.

Question 148

Need to plant 4 unique apple and 4 unique orange trees. How many ways are there such that no apple tree is adjacent to another apple tree and no orange tree is adjacent to another orange tree?

Answer 148

Realize that there are two possible high level arrangements : one starting with orange and one with apple :
1) OAOAOAOA
2) AOAOAOAO
Ans : 1152

Question 149

The formula for dependent events is P(A and B) = P(A) x P(B | A)

For mutually exclusive events : P(A or B) = P(A) + P(B)

If they are not mutually exclusive : P(A or B) = P(A) + P(B) - P(A and B)

Answer 149

When the probabilities are not equiweighted, you can calculate the prob this way :
Actual prob = no. of outcomes producing the event x prob of 1 outcome

Eg : prob of rain on a given day : 2/5. What is the prob of rain for 3 out of 4 days?

Now, if you calculate no. of possible success events and divide by total events, that will be WRONG.
Instead, do this : no. Of success events = 4!/3! = 4 (arrangements of R R R NR)
Prob of 1 success event = 2/5 x 2/5 x 2/5 x 3/5
Multiply : 4 x 24/625 = 96/625

Question 150

For ‘MISSISSIPPI’, 11!/4!4!2! INCLUDES the fact that the SS can be interchanged since they are equal and it treats them as equal. So, 11!/4!4!2! is the number of unique combos where the non similar characters are moved around. I.E. it treats MIS1S2ISSIPPI as equal to MIS2S1ISSIPPI

Answer 150

Further, the previous concept can be extended to circular arrangements as well :
The only thing that changes is in the numerator, instead of n!, we put (n-1)!. Denom will still be repeated characters’ repetition counts.

Question 151

6 people a,b,c,d,e and f are seated randomly on a circular table. what is the prob that a,c and e will be sitting in a sequence of 3 i.e. together?

Answer 151

b,d,f,x => 3! where x = (a,c,e). We do 3! and not 4! since it is circular arrangement
Total : 5! (and NOT 6!)
Prob = 3! x 3!/5!

Question 152

5 people a,b,c,d and e are lining up for a photo. What is the prob that none of the a,b,c will stand next to one another?
MADE AN ERROR ON THIS FFS

Answer 152

Here, DO NOT USE prob that none of the three stand next to each other = 1 - the three stand next to each other. This will be wrong since we need to remove cases where only a and b are together but c is not together which DOES NOT get removed in the above equation.
The correct equation would be : required prob = 1 - (a is next to b + b is next to c + c is next to a) - 2(a is next to b is next to c)

INSTEAD, use the simpler A X B Y C approach :
X can be filled in 2 ways and Y in 1. Total success : 2 x (3!) = 12
Total : 5! = 120
Final answer : 1/10 => Much easier

Question 153

If I generate a 4 length code using all the alphabets where there can be repetition, AABB is just one of the cases (Don’t count it 4!/2!2! cases).
So, prob of AABB is 1/26^4

Answer 153

1 = P(A) + P(B) - P(A and B) + P(neither A nor B)
To decide how to choose A and B, create a set matrix form and use the column header as A and row header as B or vice versa

Question 154

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

1) There are no students born in the months of December
2) There are 25 students in the classroom.

Answer 154

Note that while 1) will change the answer, it is not necessary to answer on a standalone basis.
2) alone will allow you to answer (you will include Dec as well since only 1 tells you to ignore it)
Hence, the answer is B and NOT C.

Question 155

Out of 100 fishermen, 41 caught Bass, 44 caught Trout, 46 caught Salmon. 16 caught bass and trout, 12 caught bass and salmon, 18 caught trout and salmon and 7 caught all 3. How many fishermen did not catch any of these 3 fish?

Answer 155

NOTE : To apply formula you need double overlap numbers on a pure double overlap sense and not including triple overlap areas.
Hence, convert the double overlap to double overlap ONLY.
bass and trout only = 16 - 7
bass and salmon only = 12 - 7
salmon and trout only = 18 - 7
Now, you can apply formula
100 = 41 + 44 + 46 - 9 - 5 - 11 - 2x7 + x
solve for x to get 8

Question 156

Martha can guess height with success rate of 2/3 and weight with a success rate of 3/5. If Martha meets 3 people, what is the prob that she correctly guesses a person’s weight and height at least once.

Answer 156

Wasted a lot of time on this idk why.
Realize that guessing weight and height are independent.
Prob of guessing height and weight correctly : 2/3 x 3/5 = 2/5
Prob of missing = 3/5 (This includes missing just one of weight/height and missing both)
Hence, required prob = 1 - (3/5)^3

Question 157

Very important : extrapolate this to apply in all multi-event questions
There are a number of red and white beads in a jar. If Mike were to simultaneously pick two beads from the jar, what is the prob that he will not pick any red beads?
1) The prob of selecting 2 red beads in 8/45
2) The prob of selecting 2 white beads is 1/45

Answer 157

In these types of questions, always consider that you can change one of the two variables (beads) to cause a failure.
1) Trap choice: 1 - Prob(2 red beads) is NOT prob of not picking any red beads since we are not accounting for cases like picking 1 red and 1 white which is also not a success event.
2) Sufficient since this is what we want.
Answer : B

Question 158

A certain computer program generates 4 digit even numbers using the following numbers : 0,1,2,3,4 and 5 and the digits can be repeated. What is the prob that a number formed by this program will be a multiple of 4?

Answer 158

Now you know that for a number to be a multiple of 4, the last 2 digits need to be div by 4. After you find the suitable combos satisfying this, DON’T FORGET TO MULTIPLY BY THE POSSIBLE CASES FOR THE HUNDRED’S AND THOUSAND’S PLACE SINCE THOSE TWO PLACES’ DIGITS CAN BE ANYTHING SINCE WE ONLY CARE ABOUT THE TENS AND ONES PLACE DIGITS. this will increase the number of multiples.
I FORGOT TO DO THIS AND MADE AN ERROR.

A quicker way to do this would be realizing that only the last two digits matter. So final prob = suitable last two digits/total combos of LAST TWO DIGITS only.
We can practically ignore the remaining 2 digits (thou and hundreds) since they will be common across num and denom.

Question 159

An unmarked dreidel with 4 sides is spun. After it lands on a particular side, that side is marked with a yellow highlighter. What is the prob that after 4 spins, every side of the dreidel will have been marked with the yellow highlighter?

Answer 159

My approach was give each face a name of A, B, C ,D so we want prob of A, B, C, D in any combo. So 4!/4^4
(4^4 since sides can repeat)

Alternate neat solution which can be used for a surprisingly large number of questions is :
Prob that a unmarked side comes up in 1st spin =1
in 2nd spin : 3/4
3rd spin : 1/2
4th spin : 1/4
Total : 1 x 3/4 x 1/2 x 1/4

Question 160

56 D and 44 R. If two people are randomly selected from among the D and R, what is the prob that the 2 will be Rs?

Answer 160

Even though it seems like the selection is happening as a single event, we can still use the turn by turn approach as highlighted in previous card :
For 1st R, prob = 44/100
For 2nd R, 43/99
Final : 44/100 x 43/99

Question 161

All lines are straight, they have no curvature
ray = half line
line segment = segment

A line is named by any two points on it (applies to lines, line segments and rays). They are differentiated by the symbol on top of their name :
Line has :
Line segment has : —-
Ray has : —->
Note that when you name a ray by two points, the first point must be the endpoint.

Answer 161

An angle is the union of two rays sharing a common endpoint. The common endpoint is called the vertex of the angle.
The two rays are called the sides of the angle.

Obtuse angle : more than 90 but less than 180
Straight angle : angle whose measure is exactly 180

For an angle, the vertex is the middle of the three points and must be in the middle of the angle’s designation.

Question 162

All lines are straight, they have NO curvature
ray = half line
line segment = segment

A line is named by any two points on it (applies to lines, line segments and rays). They are differentiated by the symbol on top of their name :
Line has :
Line segment has : —-
Ray has : —->
Note that when you name a ray by two points, the first point must be the endpoint.

Answer 162

An angle is the union of two rays sharing a common endpoint. The common endpoint is called the vertex of the angle.
The two rays are called the sides of the angle.

Obtuse angle : more than 90 but less than 180
Straight angle : angle whose measure is exactly 180

For an angle, the vertex is the middle of the three points and must be in the middle of the angle’s designation.

Diagrams are not necessarily drawn to scale on GMAT.
Whenever a degree measure is used on the GMAT, it is accompanied by the degree symbol.
If the symbol is not present then it not a measure, it is instead a name for that angle/angle measure.

To distinguish a straight angle from a straight line, we put a semicircular arc on the straight line.

We put a full circle to show a 360 rotation about end point.

Question 163

When n lines intersect through a common point, the sum of all the angles created by those n lines at that point in 360 degrees.

Angles are supplementary if their measures add up to 180 degrees.

Answer 163

When a transversal cuts a pair of parallel lines, ANY acute angle + ANY obtuse angle will sum to 180 degrees.
Vertical angles are equal when a transversal cuts a pair of parallel lines.
Vertical angles are angles that are diagonally oriented to each other. a/b : a & d and b & c
c/d

Question 164

Even when a pair of lines intersect a point, the vertical angles are equal.

When two lines are cut by a transversal, corresponding angles are formed. Corresponding angles are two non adjacent angles that are on the same side of the transversal but one is inside the parallel lines while one is outside the parallel lines.

Answer 164

When two parallel lines are cut by a transversal, corresponding angles are equal.

When a pair of parallel lines are cut by a NON PERPENDICULAR transversal, both obtuse and acute angles are necessarily created.

Question 165

For a triangle, each interior angle must measure more than 0 degree and the three interior angles must sum up to 180 degrees.

In a triangle, the largest angle is always opposite the longest side or the triangle. The smallest angle is opposite the shortest side of the triangle. In addition, equal sides will always be opposite equal angles.

Answer 165

Given any polygon, when taking one exterior angle at each vertex, the sum of the measures of the exterior angles will always equal 360 degrees.

VERY VERY IMPORTANT : Exterior angle is the angle created by ONE SIDE OF A TRIANGLE and the EXTENSION of AN ADJACENT SIDE. Any angle that follows this definition will sum up with all the other exterior angles chosen in any combo to sum up to 360 (as long as it is one angle per vertex).
It doesn’t matter if you pick the angles clockwise/anticlockwise/whatever. They will ALWAYS sum up to 360 as long as they SATISFY the definition of the exterior angle.

Question 166

When you say that an angle x is remote from an angle y, it means that angle x is on the opposite side of the triangle from angle y.

An exterior angle of a triangle is equal to the sum of its two remote interior angles.

Answer 166

To determine the height of a triangle, we might need to draw ALTITUDE lines which may be inside/outside the triangle and perpendicular to the base/extension of the base of the triangle.

The difference of lengths of any two sides of the triangle is LESS than the length of the third side.
Even if either the sum or the difference property holds, we can draw a triangle. We do not need to satisfy both conditions (One implies the other anyway).

Question 167

If we classify triangles by sides, we have scalene, equilateral and isosceles.
Scalene : no sides equal, correspondingly no two angles are equal
Isosceles: exactly two sides are of the same length and two angles are of the same measure.
Equilateral : All 3 sides are of the same length and all three angles are of the same measure

If we classify triangles by angles : acute, obtuse and right
Acute triangle : each of its angles measures less than 90 degrees. It can be scalene, isosceles or equilateral
Obtuse : One angle is greater than 90 degrees but less than 180 degrees. It can be scalene of isosceles.
Right : One angle is 90. It can be scalene or isosceles.

Answer 167

If a triangle with side A, B and C has C^2 = A^2 + B^2 then angle opposite side C must measure 90 degrees and thus the triangle must be a right triangle.

3-4-5 Right Triangle : If we know that the two smaller sides of a triangle are of the form 3x and 4x, then we know that it is a right triangle and the third side it the hypotenuse of length 5x. This is true for all x >= 1.

5-12-13 Right Triangle : If we know that the two smaller sides of a triangle are of the form 5x and 12x, then we know that it is a right triangle and the third side it the hypotenuse of length 13x. This is true for all x >= 1.

Question 168

The sides of an isosceles right triangle are in the ratio : x : x : x*((2)^0.5)

The sides of a 30-60-90 triangle are in the ratio of x : x*((3)^0.5) : 2x

The above can be applied in multiple ways, if we know it is a right triangle and the sides follow the above ratio, we can say that the angles follow the above and vice versa too.

Answer 168

Area of a equilateral triangle : ((s^2) x (3)^0.5)/4

Dropping an altitude from upper vertex to the base of an equilateral triangle produced two identical 30-60-90 triangles.

Question 169

If one triangle is simply an enlargement of another triangle, the triangles are similar.
There are three general ways in which two triangles can be similar :
1) Three angles of one triangle are the same measure as three angles of another triangle
2) The three pairs of corresponding sides have lengths in the same ratio
3) An angle of one triangle is the same measure as an angle of another triangle and the sides surrounding these angles are in the same ratio.

The above statements imply equivalence i.e. if angles are equal, sides will be in the same ratio and vice versa. If angle is same and enclosing sides are in same ratio then all angles are same and all sides are in ratio and so on.

Answer 169

Parallelogram : Opposite sides are equal in length, opposite angles are equal in measure, the diagonals bisect each other and each diagonal divides the parallelogram into 2 congruent triangles.
Any two consecutive angles within a parallelogram are supplementary.
Area : base x height

Square : The diagonals of a square are perpendicular to each other (not the case in rectangles)

VV Important : Given a rectangle with a fixed perimeter, the rectangle with the max area will be a square.

Question 170

I would like to maximize the area of the dance floor used for my party. I’m choosing between floors X and Y both of which have the same perimeter. Is the area of the dance floor X greater than the area of floor Y?

1) X is a rectangle but not a square
2) Y is a square

Answer 170

Tricky, I wrongly chose B
If I choose B, i do not know the shape of X. If X is a rectangle then X cannot have an area greater than Y but if X is a circle, it can.
MAKE SURE OF THE SHAPE BEFORE YOU APPLY LOGIC.

IMPORTANT :
A circle gives the maximum area for a given perimeter.

So the triangle that gives the maximum area for a given perimeter is an equilateral triangle. The quadrilateral that gives the maximum area for a given perimeter is a square. The pentagon that gives the maximum area for a given perimeter is a regular pentagon. The n-gon that gives the maximum area for a given perimeter is a regular n-gon.

Additionally, as the number of sides, n, grows in a regular n-gon, the area will increase for a given perimeter. Thus of a square and an equilateral triangle with the same perimeter, the square will have more area. But of a square and a regular pentagon with the same perimeter, the pentagon will have the larger area.

Question 171

VV IMPORTANT : Given a rectangle of fixed area, the rectangle with the minimum perimeter is a square.

It’s an inverse relation, for a given perimeter, a regular figure gives max area,
for a given area, regular gives minimum perimeter.

Answer 171

A trapezoid is a quadrilateral in which one pair of opposite sides called the bases are parallel (never of equal length) but the other pair of opposite sides are not parallel. If the two non parallel sides are equal in length, the trapezoid is referred to as an isosceles trapezoid. In that case, the angle they make with the bases are equal.
Area of trapezoid = (base1 + base2)height/2

Question 172

Sum of interior angles of a polygon = (n - 2) x 180

The sum of exterior angles of any polygon, when taking one exterior angle at each vertex, add up to 360.

Answer 172

In a regular polygon, all the SIDES are equal in length along with all the ANGLES.

Question 173

Area of a regular hexagon : {3((3)^0.5)/2}s^2 where s is the length of any one side (6 x formula for area of equilateral triangle of side ‘s’)
A regular hexagon can be divided into 6 equilateral triangles.

Answer 173

A chord is a line segment that connects ANY two points on the circle.
If a chord passes through the center of the circle, it is called a diameter of the circle.

Question 174

A central angle is any angle at the center of the circle that is formed by two radii.
An arc is a portion of the circumference of a circle. Arc length is measured using measures of length. However an arc can be measured in degrees as well.

A sector of a circle is the region of a circle that is defined by two radii and their intercepted arc

Answer 174

central angle/360 = arc length/circumference = area of sector/area of circle

Very important : If points A and B are two points on a circle and arc AB is not a semicircle, arc AB refers to the shorter portion of the circumference between A and B. This shorter portion is also known as the minor arc.
The longer portion of the circumference between A and B is known as the major arc but is usually referred as arc ACB, C representing some third point on this longer portion of the circumference.

Question 175

An inscribed angle is an angle formed by two chords of a circle with the vertex of the angle on the circumference.
The degree measure of an inscribed angle is equal to half of the degree measure OF THE ARC that it intercepts. Conversely, when an inscribed angle shares the same endpoints as the central angle, the degree measure of the central angle is twice the degree measure of the inscribed angle.

The Central Angle Theorem is always true unless the vertex of the inscribed angle lies on the minor arc instead of the major arc. In this case, the inscribed angle is the supplement of half the central angle. As a formula: angle = 180 −central angle/2

Answer 175

A triangle is inscribed in a circle when all the three of the triangle’s vertices lie on the circle’s circumference.
When a triangle is inscribed in a circle, if one side of the triangle is also the diameter of the circle, then the triangle is a right triangle with the 90 degree angle opposite the diameter.

When an equilateral triangle is inscribed in a circle, the center of the triangle coincides with the center of the circle. A line segment drawn from the center of the triangle to a vertex of the triangle would bisect that interior angle. The triangle divides the circumference into three arcs of equal length.

Question 176

If a circle is inscribed in an equilateral triangle, the points at which it touches the triangle are the midpoints of those sides. Further, the line to the midpoints is perpendicular to those sides as well. When joined to the vertices of the triangle, we get a 90-60-30 triangle.
(The above is when each side of the triangle is tangential to the circle and it touches the triangle at exactly 3 points and the circle is the largest circle that can fit inside the triangle)

Note : a circle can be inscribed in any triangle but it won’t always touch the midpoint unless it’s equilateral triangle

Answer 176

When a rectangle is inscribed in a circle, a diagonal of the rectangle is also a diameter of the circle.

When a circle is inscribed in a square, each point at which the circle touches the square is the midpoint of the side of the square
(The above is when each side of the square is tangential to the circle and it touches the square at exactly 4 points and the circle is the largest circle that can fit inside the square)

When a regular polygon is inscribed in a circle, the polygon divides the circle into arcs of equal length. So they subtend equal angles at the center equal to 360/n where n is the number of sides

Question 177

If a shape is inside another shape touching all the sides, the inside one is inscribed and the outside one is circumscribed.

The area of an inscribed square will be smallest when the vertices of that square are located at the midpoints on the respective edges of the circumscribed square. Furthermore ,the area of such an inscribed square will be half of the area of the circumscribed square.

Answer 177

When a rectangle is inscribed in a semicircle and a radius is drawn from one of the vertices of the rectangle to the center of the circle, a right triangle is created in which the base of the triangle is equal to half the side of the rectangle.

Question 178

When a frame of uniform width surrounds something, the area of the frame/border does NOT include the area of what it encloses.

The longest line segment that can be drawn within a rectangular solid is (l^2 + b^2 + w^2)^0.5

Answer 178

On the GMAT, cylinder = right circular cylinder. Other forms of cylinder are elliptical cylinder (base is not circular) and oblique cylinder (bases (top and bottom) are not aligned)

Surface area of right circular cylinder : 2pirh + 2pir^2
Volume of a right circular cylinder : pir^2*h

Question 179

To determine the rate at which a liquid will rise within a 3d object, we must know the rate at which the liquid flows into the figure and the exact dimensions of the figure.
To determine the number of smaller objects of known volume that will fit within a larger object of known volume, we must know the exact dimensions of both the smaller objects and the larger object.

Answer 179

In inscribed figure questions, it is sometimes easier and more error proof to calculate the diameter rather than the radius

Don’t assume that the border is always of equal width.

Inscribed angle : Angle formed by two chords of a circle with the vertex of the angle on the circumference.

VERY VERY IMPORTANT : The degree measure of an inscribed angle is equal to half of the degree measure of the arc that it intercepts. This is very important. This concept is independent of the central angle principle.

The Central Angle Theorem is always true unless the vertex of the inscribed angle lies on the minor arc instead of the major arc. In this case, the inscribed angle is the supplement of half the central angle. As a formula: angle = 180 −central angle/2

An extension of the above is : When an inscribed angle shares the same endpoints as the central angle, the degree measure of the central angle is twice the degree measure of the inscribed angle.

Question 180

If two circles intersect, then the common area is divided equally by a common chord only if the radii are equal. In that case the two isosceles triangles will be congruent.

Answer 180

Just knowing the volume or the TOTAL SURFACE AREA of an object does not tell us anything about its dimensions/shape unless we know it is a regular object (in which case, all sides will be equal in length).
Even if we know it’s a rectangular cube (not regular cube), that still does not tell us anything about the dimensions. We can have wonky dimensions. I knew about volume but wasn’t sure about total surface area

Question 181

The x coordinate and the y coordinate together make up a coordinate pair. Also referred to as an ordered pair. => (x, y)
The quadrants are numbered anticlockwise. 2|1
3|4
There is a difference between points in a quadrant and points on the axes (check).

A line segment includes the end points.

Slope = delta y/delta x = Rise/Run

The slope of a line can be zero (a horizontal line), undefined (a vertical line), positive (a line that rises when the line moves from left to right), or negative (a line that falls when the line moves from left to right). Note that it is UNDEFINED and not INF

Answer 181

Lines with positive slopes always intersect quadrants 1 and 3. A positively sloped line intersects quadrant 4 only if its x intercept is positive and it intersects quadrant 2 only if x intercept is negative. If its x intercept is 0, then the line will intersect only quadrants 1 and 3.

Lines with negative slopes always intersect quadrants 2 and 4. A negatively sloped line intersects quadrant 1 only if its x intercept is positive and it intersects quadrant 3 only if x intercept is negative. If its x intercept is 0, then the line will intersect only quadrants 2 and 4.

We use x intercept to apply logic in the above and not y intercept since y intercept can be theoretically +inf/-inf.

A horizontal line intersects quadrants 1 and 2 if the line’s y intercept is positive and intersects quadrants 3 and 4 if the line’s y intercept is negative. If the y-intercept of a horizontal line is zero, the line is coincident with the x axis and does not intersect any of the quadrants.

A line with an undefined slope i.e. vertical line intersects quadrants 1 and 4 if the x intercept of the line is positive and intersects quadrants 2 and 3 if the x intercept of the line is negative. If the x intercept of a line with undefined slope is zero, the line is coincident with the y axis and DOES NOT INTERSECT ANY OF THE QUADRANTS

Question 182

IMPORTANT : The larger the absolute value of the slope of a line, the steeper the line

Answer 182

In y = mx + b, b is the y intercept.

Standard/General form of equation of a line : Ax + By = C where A, B and C are all constants.

You can define a line if you know a point on that line in addition to one of the following :

1) The slope of the line, or the slope of a line that is parallel or perpendicular to the line
2) A second point on the line

Parallel lines have the same slope and different y intercepts and as a result, they never intersect.

The slopes of two perpendicular lines are negative reciprocals, negative reciprocals multiply to -1

Question 183

The reflection of the point (x, y) over the x axis is (x, -y)
The reflection of the point (x, y) over the y axis is (-x, y)
The reflection of the point (x, y) over the origin is (-x, -y)

Answer 183

To reflect a line segment over the x axis, y axis or origin, reflect the endpoints of the line segment. Once we have the two new endpoints, we can draw a line segment connecting them, and this new line segment is the reflection of the original.

Question 184

To reflect a polygon over the x axis, the y axis or the origin, we can reflect each vertex of the polygon, and then connect the reflected vertices to form the reflected polygon

Answer 184

The reflection of (x, y) : Just memorize this, tough to do it logically under time constraints

1) over the line y = x is (y, x)
2) y = -x is (-y, -x)
3) y = b is (x, 2b -y)
4) x = a is (2a -x, y)
5) (x = a, y = b) is (2a-x, 2b-y) -> IMPORTANT

Question 185

The coordinates of the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) are (xm, ym) = ((x1 + x2)/2, (y1+y2)/2)

The equation of a circle centered at point (a, b) is (x - a)^2 + (y - b)^2 = r^2

Answer 185

Before sketching inequalities, rewrite to get y isolated on LHS before proceeding.

‘Greater than’ inequality (y > mx + c) graph has a dashed line to signify that the points on the line DO NOT satisfy the equation. The shaded region lies ONLY ABOVE THE LINE. -> VERY IMPORTANT

In ‘less than’ equality (y < 6x + 8), the shaded region lies ONLY BELOW THE LINE.

‘Greater than or equal to’ inequality (y >= mx + c) graph has a solid line to signify that the points on the line DO satisfy the equation. The shaded region lies ABOVE THE LINE and INCLUDES THE POINTS ON THE LINE.

Question 186

NICE HACK for graphing :
I have two inequalities :
1) y <= 6x + 8
2) y >= 2x + 1

After you graph the two and find the regions, one step that is taken is finding where they’ll intersect (if they do) and graphing that. To help in that, you can observe that the slop of 1) is higher than 2) and both are positive i.e. both will increase as you move left to right. Hence 2) will never catchup with the y coordinate of 1) due to it’s slope. Hence, they’ll intersect in the leftward direction (3rd quadrant in this case.)

Answer 186

If you have one point (x,y) and slope is for p/q and you are given that the points of a certain figure are integers then to try out points you can simply do (x + q, y + p), (x + 2q, y + 2p)….

Equation of line is y = (-W/Z)*x + C and WZC =/= 0. If (a, b) is a point on this line, what is the value of b?
1) -W/Z = 5/2
2) y intercept is 4
Now even though by a,b it seems like you know the value of a and b, it is not so. If you’re asked value, it cannot be in terms of ‘a’ or something else unless explicitly mentioned. So, E

Similar example : if point (x, y) is on line L. what is the slope of L given that (8x, 6y) also lies on L.
Ans : We cannot determine 5y/7x even though IT MAY SEEM that x and y are given as fixed values.

IMPORTANT : Circle X has center at (10, 10) and diameter d. Circle Y has center at (5, -8) and radius r. Does Y intersect X?
If one circle is inside the other, THAT DOES NOT MEAN THAT IT IS INTERSECTING. MADE A MISTAKE ON THIS FFS. INTERSECTION MEANS THERE HAS TO BE CROSSOVER. x^2 + y^2 =2 does NOT INTERSECT x^2 + y^2 = 4
Even if the radius of one circle is larger than the distance between the centers of the two circles (which is a good metric to use), is is possible that one circle has a radius so small that it is contained in the other and doesn’t intersect.

Question 187

Stationary point :
When a figure is reflected over another line, which is called the line of reflection, if there are any stationary points, they must lie on the line of reflection.

A line that intersects a circle only once is referred to as a tangent. Remember that it does INTERSECT.

Answer 187

Ratio of area of similar triangles = (side of triangle 1/side of triangle 2)^2

Use hero’s formula to understand the derivation of the above

Question 188

In f(x) = y, the variable x is called the independent variable and y is called the dependent variable.

If a graph is the graph of a function, then any vertical line drawn can only intersect the graph at exactly one point or at no points. This is called the vertical line test.

Hence, the graph pertaining to x = 4 or x^2 + y^ 2 = 81 are NOT graphs of any functions -> IMPORTANT

Answer 188

Every sequence has a rule or a formula that dictates how the sequence works. Without knowing this rule, we cannot make any conclusions about the value of any term in the sequence. If the first few terms of a sequence are given, but the rule is not given, we cannot determine any further terms in the sequence.

Example : If x is the fifth term of some sequence an, what is the value of x?

1) The first four terms of an are 2,4,16 and 256
2) Each term of an from the 2nd term on is obtained by squaring the term preceding it.

ANSWER IS ‘C’ even though the logic of the sequence is obvious on using 1), IT IS NOT ‘A’

Question 189

When the formula of a sequence is given in terms of n, an = 2n -1, it is called an explicit formula

If it is given are an = 3a(n-1) then it is recursive formula.

Answer 189

VERY IMPORTANT : If the formula of a sequence is given explicitly, we can find any term of the sequence without knowing the term or terms prior to it. However, if the formula is given recursively, we must know ALL the terms prior to it.

AP = an = a1 + (n-1)*d
Sn = (a1 + an)n/2

GP = an = a1 * r^(n-1)

You can use some heuristic based methods to calculate mid elements i.e. if we have GP with terms as an then a4 will be (a2*a6)^0.5
if AP then a4 = (a2 + a6)/2

Question 190

OG Notes :

1) If in a geometry question, you feel like you’re lacking info, just drop a perpendicular and proceed.
https: //gmatclub.com/forum/in-the-figure-point-d-divides-side-bc-of-triangle-abc-into-segments-126934.html

2) In geometry inequality questions, SKETCHING IS INVALUABLE, manually, you can miss out on points but sketching is idiot(you)-proof.
Q : https://gmatclub.com/forum/in-the-xy-plane-region-r-consists-of-all-the-points-x-y-102233.html

3) Average speed does not IMPLY the vehicle was travelling at the same speed at mid point, this comes up in relative speed questions.
Q : https://gmatclub.com/forum/stations-x-and-y-are-connected-by-two-separate-straight-parallel-77177.html

4) Always try to bound problems if counting is not working out/tough.
Q : https://gmatclub.com/forum/list-t-consist-of-30-positive-decimals-none-of-which-is-an-integer-131755.html

Answer 190

5) Know when to put values. THIS IS A SKILL I CONSTANTLY GET TRIPPED UP ON. And before putting values, think of an optimized strategy and then start.
6) In counting questions, take a second to check if it will be easier to count normally on a case by case basis or if it would quicker to do total - complement. Sometimes, counting misses out on edge cases which the latter approach captures.
7) In word problems, when factorizing a number, always note that 1 IS ALSO A FACTOR, say you’re dividing 10 toffees among n kids such that each gets an equal number, the possibilities of the no. of toffees one gets is 5, 2 AND 10!!!! DON’T FORGET 10 which happens when there is only 1 kid.
8) In DS, before putting values, LOOK AT BOTH THE OPTIONS TO PICK VALUES SMARTLY.
9) When you have a uniform border, the length and breadth increase by 2x and NOT x.

Question 191

in two similar triangles, the ratio of their areas is the square of the ratio of their sides

in an obtuse triangle, a^2 + b^2 < c^2. Then in an acute triangle, a^2 + b^2 > c^2

Tricky, reverses the usual flow of problem solving : https://gmatclub.com/forum/if-4-7-x-3-which-of-the-following-must-be-true-168681.html

Nice trick : https://gmatclub.com/forum/if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html

Answer 191

In some questions, use mathematical solving for some parts and put values for the rest, THIS IS VVVVV HELPFUL. Could have saved tons of time on this : https://gmatclub.com/forum/if-d-0-and-0-1-c-d-1-which-of-the-following-must-be-true-168429.html

This one straight up fucked me up : https://gmatclub.com/forum/the-table-above-shows-the-results-of-a-survey-of-100-voters-who-each-89187.html