Quant Gmat

Front 1

Circular permutations = (n-1)!

Back 1

How many ways can six friends be arranged around a circular dinner table?

Circular permutations can be calculated with the formula (n - 1)!, where n = the number of items to be arranged. So in our case, the formula is (6-1)! = 5! = 5 * 4 * 3 * 2 * 1 = 120.

Front 2

Sum of consecutive integers

Back 2

n(n-1)/2

Front 3

sum of odd consecutive integers!!!!!!!!!!!!

?????????

Back 3

??????

Front 4

Given that N=a(ˆ3) * b(ˆ4) * c(ˆ5)

where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?

Back 4

To make N a perfect square, the powers of a, b and c must be even; to make N a perfect cube, the powers must be multiples of 3; to make N a perfect fifth power, they must be multiples of 5. This implies that each of the powers of a, b and c must be a multiple of 30 (because 30 is the LCM of 2, 3 and 5)

Front 5

Finding number of factors

Back 5

Number of factors of 72.

1) Express the number as the product of prime numbers. 72 = 2(ˆ3) ∗ 3(ˆ2)

2) Forget about the bases and concentrate on the exponents. Add one to each exponent (making them 4 and 3 in this case)

3) Multiply the exponents and you'll have your number of total factors. (3*4 = 12).

Thus, 72 has 12 factors.

Front 6

During the 31-day month of May, a tuxedo shop rents a different number of tuxedos each day, including a store-record 55 tuxedos on May 23rd. Assuming that the shop had an unlimited inventory of tuxedos to rent, what is the maximum number of tuxedos the shop could have rented during May?

Back 6

A. That means that the range of consecutive integers would be 25 through 55 (a total of 31 individual values).

Using the "sum of evenly spaced sets" principle:Sum = (# of values)(median value)You can calculate this as (31)(40) = 1240.

Front 7

FLASHCARDS GMAT CLUB

Back 7

:)

Front 8

Divisibility rules

= Is 5632 divisible by 4?

= Is 2658 divisible by 6?

= Is 396 divisible by 9?

Back 8

All yes!

Divisibility by 4 – last 2 digits is a multiple of 4

Divisibility by 6 – the sum of digits is a multipleof 3 and the last digit is even

Divisibility by 9 – the sum of digits is a multipleof 9

Front 9

Is integer (𝑥^2)*(𝑦^4) divisible by 9?

A) x is an integer divisible by 3

B) xy is an integer divisible by 9

Back 9

You could think the answer is B since xy/9 then (𝑥^2)*(𝑦^4) would also, because its just the numbers divisible to some power. Beware!!

tip1= (𝑥^2)*(𝑦^4) is an integer = it doesn't mean x and y are.

x= 81 and y= 1/9

then x is div by 3, xy is an integer divisible by 9, but (𝑥^2)*(𝑦^4) = 1 and is not divisible by 9.

Answer = E

Front 10

Which of the following integers represents a sumof 3 consecutive even integers?

200

303

400

554

570

Back 10

To answer the question, check which integer is both divisible by 3 (since there are 3 integers) and is even (because you can only make an even number with sum of 3 evens).

The only number that falls into both of those 2 categories is 570.

Correct answer is E

Front 11

How many distinct integers arethere between 1 and 21 inclusive?

Back 11

formula inclusive = (21-1) + 1

Front 12

2/5 = 40%

1/20 =5%

1/8 = 12.5%

1/6 = 16.67%

Back 12

1/12 =8.33%

7/8 = 87.5%

5/6 = 83.33%

Front 13

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

2^7 = 128

Back 13

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

Front 14

4^2 = 16

4^3 = 64

4^4 = 256

Back 14

5^2 = 25

5^3 = 125

Front 15

11^2 = 121

12^2 = 144

13^2 = 169

14^2 = 196

15^2 = 225

16^2 = 256

18^2 = 324

19^2 = 361

Back 15

sqroot(2)=1.4

sqroot( 3) = 1.7

sqroot(625) =25

sqroot(169) =13

Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

cube root (125) = 5

fifth root (243) = 3

Front 16

What is the 3- step approach to solvingequations and inequalities with absolutevalue?

Back 16

Step1: Open modulus and set conditions.To solve/open a modulus, you need toconsider 2 situations to find all roots: Positive and Negative

Step 2: Solve new equations from Step 1

Step 3: Check conditions for eachsolution from Step 2

Front 17

Absolute values

Is M<0?

A) −𝑀 = |𝑀|

B) 𝑀2 =9

Back 17

Statement (1) by itself is not sufficient. Fromstatement (1), M is either a negative number orzero. If M=-3, then –(-3)=|-3| or 3=3, which is notsufficient.

Statements (1) and (2) combined are sufficient. Ifwe combine both statements, then M=-3.

Front 18

Number of factors:

prime = 2

sq of prime = 3

prime x prime = 4

product of normal numbers = 5 or more

Back 18

.

Front 19

What is the ratio of a to d if:

2a = 3b

(1/2)*b = 2c

3c=d

Back 19

With questions like these, unless you can spot ashortcut right away, the easiest way to solve is toplug numbers: let’s pick a=6, since there is a 2 and3 involved

2*6 = 3b; b = 4

0.5*4 = 2c; c = 1

3*1 = d; d = 3

Therefore, the ration of a to d is 6:3 which is 2:1

Front 20

Lena’s first test score was at the 80th percentile ina class of 120 students. On another test, 24 out of200 students scored better than Lena. If nobodyhad Lena’s score, what is Lena’s percentile afterthe two tests?

Back 20

Take each test separately.

test 1 = 120 * 80/100 = 96 (she scored more than 96 people)

test 2 = 200 - 24 = 176 ( she was better than 176 people)

sum results = 96+176 = 272

sum total = 120 + 200 = 320

answer = 272/320 = 85 percentile

Front 21

What is the sum of all members ofthe set 9,12,15,18,21,24 ?

Back 21

The sum of elements of evenly spaced set is givenby the formula

Sum= ( (𝑎1+𝑎𝑛) × 𝑛) / 2

n= number of numbers. there are 6 numbers in the set.

Therefore, ((9+24) × 6) / 2 = 99

Front 22

Express 0.393939... in a fraction format

Back 22

1. identify the recurring

2. if the last dig rec. is 9, use this number the amount decimals rec. (0.39 = 2 rec. so 9 and 9 = 99)

Example #1: Convert 0.3939... to a fraction

1: The recurring number is 39

2: 39/99 - the number is 2 digits so two nines are added 99

3: Reducing it to lowest terms: 33

Front 23

There is a set {67,32,76,35,101,45,24,37}. If wecreate a new set that consists of all elements ofthe initial set but decreased by 17%, what is thechange in standard deviation?

Back 23

A decrease in all elements of a set by a constant percentage will decrease the standard deviation of the set by the same percentage (the average is decreased by 17% as well as the difference between average (mean) and all elements or their squares. Thus the decrease in standard deviation is 17%.

Front 24

Venn Diagram formulas

Group1 + Group2 + Neither – Both = Total

G1 + G2 + G3 - (two groups) - (2 * all three)

Back 24

.

Front 25

If a price was increased by x% and thendecreased by y%, is the new price higher thanthe original?

A) x>y

B) x=1.2y

Back 25

Answer E

A is obviously NS

B is also. If y=100 and x=120 then the new price = 0. Is y=10 and x= 12 the new price is greater than the old.

Front 26

Tip - finding area of triangle

If you know 2 sides of a triangle but not its height,you can add an equally sized triangle to create asquare/rectangle/rhombus and find its area (may beeasier). Remember to divide your result by 2.

Back 26

Memorize these triangles:

a) 3, 4, 5

b) 6, 8, 10

c) 5, 12, 13

d) 12, 16, 20

e) 7, 24, 25

Front 27

Equilateral triangles:

height = a/2 * sqroot(3)

Back 27

.

Front 28

Combinations:

A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as: nCk

Now we have to exclude all arrangements of k objects(k!) and remaining (n‐k) objects ((n‐k)!) as the order ofchosen k objects and remained (n‐k) objects doesn'tmatter.

n! / ( (k!) * (n-k)! )

Back 28

Permutations:

A permutation is an ordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as : nPk

Now we have to exclude all arrangements ofremaining (n‐k)! Objects

n! / ( n-k )!

Front 29

When to use Permutation or Combination?

Use Permutation when the sequence of choice matters (meaning a group ABC is different from BAC or CBA). Classic example is choosing nominees for 3 specific positions from a pool of 10 candidates

Back 29

Use combination when the order of selection has no impact and once a smallgroup is formed, it does not matter how theyarrived there. Classic example is picking 3 marblesfrom a bag of 10

Front 30

Circular combinations

If six business partners are having a dinner ata round table, how many seatingarrangements are possible?

Back 30

Formula: (n-1)!

(6-1)! = 5! = 120

Front 31

If there are 5 chairs in a room and Bob andRachel want to sit so that Bob is always left ofRachel, in how many ways this seatingarrangement be achieved?

Back 31

First, identify it is a permutation and not a combination because order matters.

2 - This condition is called symmetry because iteliminates half of the possibilities (Rachel can sitonly left or right of Bob). Therefore, the number of ways that Bob is left of Rachel is 1/2 of the total ways.

5! / (5-2)! = 20

dividing by 2 the answer is 10.

Front 32

In how many different ways can a group of 8people be divided into 4 teams of 2 peopleeach?

Back 32

The solution to this problem is the number of combinations. First we get one team out of 8 . The number of ways to do this would be 8C2. The next combination is 2 out of 6 or 6C2, and so on. Having all four combinations multiplied, we need to divide the total number by the number of ways the teams can be chosen , since we are not interested if the team with two certain people is chosen first, second or third. Therefore, the answer is found by the following:

( 8C2 * 6C2 * 4C2 * 2C2 ) / 4! = 105