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Que: Mandy sold 600 apples in 10 days. If the number of apples she sold increased by 4 each day, how many apples did she sell on the \(10^{th}\) day?

A. 36
B. 42
C. 60
D. 78
E. 100
Show more

Solution: We have to find the number of apples did Mandy sell on the \(10^{th}\) day.

Given: Mandy sold 600 apples in 10 days and the number of apples she sold increased by 4 each day

\(1^{st}\) day: x

\(2^{nd}\) day: x + 4

\(3^{rd}\) day: x + 4 + 4 = x + 2 * 4 .
.
.
\(10^{th}\) day: x + 9 * 4

=> x + (x + 4) + (x + 2 * 4) + ….+(x + 9 * 4) = 600

=> 10 x + 4 + 2 * 4 + …..+ 9 * 4 = 600

=> 10x + 4( 1 + 2 + …. + 9) = 600

=> 1 + 2 + 3 + 4 +…..+ n = \(\frac{n(n+1) }{2}\)

=> \(10x + 4[\frac{(9 * 10)}{2}] = 600\)

=> 10x + 180 = 600

=> 10x = 420

=> x = 42

=> ∴ \(10^{th}\) day = x + 9*4 = 42 + 36 = 78

Therefore, D is the correct answer.

Answer D
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Que: John purchased 2,000 bolts for $15 each. He sold 1,000 bolts for $20 each in the first week and sold the rest for $30 each in the second week. What was his total profit from sales?

A. $5,000
B. $10,000
C. $15,000
D. $20,000
E. $25,000
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Que: John purchased 2,000 bolts for $15 each. He sold 1,000 bolts for $20 each in the first week and sold the rest for $30 each in the second week. What was his total profit from sales?

A. $5,000
B. $10,000
C. $15,000
D. $20,000
E. $25,000
Show more


Solution: We have to find the total profit from the sale of 3,000 bolts

Revenues = Total sales; Cost = Purchases

Total cost of 2,000 bolts = 2,000 * 15 = $30,000

Revenue = 1,000 * 20 + 1,000 * 30 = 20,000 + 30,000 = $50,000

=> ∴ Profit = $50,000 - $30,000 = $20,000

Therefore, D is the correct answer.

Answer D
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Que: If x and y are different prime numbers greater than 20 but less than 30, what is the value of x + y?

A. 23
B. 29
C. 50
D. 52
E. 56
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Que: If x and y are different prime numbers greater than 20 but less than 30, what is the value of x + y?

A. 23
B. 29
C. 50
D. 52
E. 56
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Solution: Prime number: An integer that has two positive factors = only 1 and itself => ex) 2, 3, 5, 7…

We have to find the value of x + y

=>If x and y are different prime numbers greater than 20 but less than 30

=> Prime numbers between 20 and 30: 23 and 29 => ∴ x + y = 23 + 29 = 52 .

Therefore, D is the correct answer.

Answer D
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Que: If n is positive integers, what is the unit digit of \((3^{4n+1})(9^{19})\)?

A. 3
B. 7
C. 6
D. 2
E. 8
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Que: If n is positive integers, what is the unit digit of \((3^{4n+1})(9^{19})\)?

A. 3
B. 7
C. 6
D. 2
E. 8
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Solution: Powers that repeat every 4th power:

Ex) Units digit: The digit 3 repeats after every fourth power

=> \(3^1= ~3, ~3^2= ~9, ~3^3 = ~7, ~3^4= ~1, ~3^5= ~3, ... \)

=> Pattern: 3, 9, 7, 1, 3, 9, 7, 1 ….

Ex) Units digit: The digit 9 repeats after every second power

=> \(~9^1 = ~9, ~9^2 = ~1, ~9^3 = ~9, ...\) => Pattern: 9, 1, 9, 1…

We have to find the units digit of \((3^{4n+1})(4^{19})\) if n is positive integers

=> \((3^{4n+1})(9^{19})\)

=> \((3^4 )^n * 3^1 *(~9)\)

=> \((81)^n* 3^1 *(~9) =(~1)* 3^1 *(~9) =(~7)\)

Therefore, B is the correct answer.

Answer B
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Que: How many different prime factors of \(4^5 + 4^6 + 4^7\) are there?

A. 2
B. 3
C. 1
D. 5
E. 6
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Que: How many different prime factors of \(4^5 + 4^6 + 4^7\) are there?

A. 2
B. 3
C. 1
D. 5
E. 6
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Solution: Factors of M are the integers dividing M without a remainder => If M=ab (a and b are positive integers) => a and b are factors of M => When M is expressed as a product of its prime factors only

=> We have prime factorized M =>If we prime factorize a positive integer M => \(M = (p_1)^{t_1}∙(p_2)^{t_2}∙…∙(p_n)^{t_n}\)

=> \(p_i \): Different prime factors and \(t_i\) : Positive integers and the exponents of different prime factors, where
i = 1, 2,….,n

=> number of prime factors = n

We have to find the number of different factors of \(4^5 + 4^6 + 4^7\)

=> \(4^5 + 4^6 + 4^7 = 4^5 + 4^5 * 4^1 + 4^5 * 4^2\)

=> \(4^5 (1 + 4^1 + 4^2)\)

=> \(4^5 (1 + 4 + 16)\)

=> \(4^5 * 21\)

=> \((2^2)^5 * 3 * 7\)

=> ∴ Prime factors 2, 3 and 7

=> ∴ Number of prime factors =3

Therefore, B is the correct answer.

Answer B
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Que: If n is an odd integer, which of the following MUST BE an integer?

I. \(\frac{(n^2 - 1) }{ 2}\)

II. \(\frac{(n - 1) }{ 2}\)

III. \(\frac{n }{ 2}\)


A. I only
B. II only
C. III only
D. I & II only
E. I, II, & III
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Que: If n is an odd integer, which of the following MUST BE an integer?

I. \(\frac{(n^2 - 1) }{ 2}\)

II. \(\frac{(n - 1) }{ 2}\)

III. \(\frac{n }{ 2}\)


A. I only
B. II only
C. III only
D. I & II only
E. I, II, & III
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Solution: Multi-Choice questions: ‘Must be’ VS ‘Could be’.

We have to find the options that must be true

‘MUST BE’: Choose only the options that are absolutely true in ALL cases

=> ‘n’ is an odd integer, Always an integer

=> n = odd = 2m+1 (m is always an integer)

I. \(\frac{(n^2 - 1) }{ 2} = \frac{[(2m+1)^2 – 1]}{ 2} = \frac{[4m^2+4m+1-1]}{2} = 2m^2 + 2m\) => AN INTEGER ∴ Option is TRUE

II. \(\frac{(n - 1) }{ 2} = \frac{(n - 1)}{2 }= \frac{(2m +1 – 1)}{ 2} = m\) => AN INTEGER ∴ Option is TRUE

III. \(\frac{n }{ 2} = \frac{2m+1}{2} = m + 0.5\) => NOT AN INTEGER ∴ Option is NOT TRUE

Only options I and II are TRUE.


Therefore, D is the correct answer.

Answer D
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Que: If the sum of 5 consecutive integers is 0, what is the least of the 5 integers?

A. -2
B. -1
C. 0
D. 1
E. 2
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Que: If the sum of 5 consecutive integers is 0, what is the least of the 5 integers?

A. -2
B. -1
C. 0
D. 1
E. 2
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Solution: Consecutive integers: Integers that are placed next to each other on number line =>…, n-2, n-1, n, n+1, n+2, …

We have to find the greatest integer of 5 consecutive integers if their sum is 0

=> 5 consecutive integers: n-2, n-1, n, n+1, n+2

=> (n-2)+(n-1)+n+(n+1)+(n+2)=0

=> 5n=0 = n=0

=> ∴ The least of the 5 consecutive integers: n-2=0-2=-2

Therefore, A is the correct answer.

Answer A
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Que: A number 42,p5q is divisible by both 4 and 8; where p and q are the thousands and units digits, respectively. What is the minimum possible value of \(\frac{p}{ q}\)?

(A) \(\frac{1}{2}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{2}{3}\)
(E) \(\frac{1}{8}\)
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Que: A number 42,p5q is divisible by both 4 and 8; where p and q are the thousands and units digits, respectively. What is the minimum possible value of \(\frac{p}{ q}\)?

(A) \(\frac{1}{2}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{2}{3}\)
(E) \(\frac{1}{8}\)
Show more

Solution: A number, here 42p5q is divisible by ‘4’ if the number formed by the last two digits of the number is divisible by 4.

Thus, the number 5q is divisible by 4.This is possible when q = 2(52) or 6(56).

Here 42p5q is divisible by ‘8’ if the number formed by the last three digits of the number p5q is divisible by 8.

Thus, the number p5q is divisible by 8

If q = 2:

=> p = 1(152), 3(352), 5(552), 7(752), 9(952) is divisible by 8.

If q = 6:

=> p = 2(256), 4(456), 6(656), 8(856) is divisible by 8.

Therefore, we have following possible combinations:

=> p = 1,3,5,7,9 , q = 2 then minimum \(\frac{p}{q} = \frac{1}{2}\)

=> p = 2,4,6,8, q = 6 then minimum \(\frac{p}{q} = \frac{2}{6} = \frac{1}{3}\)

Thus, the minimum value of \(\frac{p}{q}\) is \(\frac{1}{3}\).

Therefore, B is the correct answer.

Answer B
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Que: A merchant bought 1,000 fans for $15 each. He sold 40 percent of the fans for $20 each and the rest for $30 each. What was the merchant’s average profit per fan?

(A) $11
(B) $15
(C) $18
(D) $21
(E) $26
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(Integers) If m, n, p, and q are distinct positive integers, greater than 1 such that mnpq = 660 and m<n<p<q, how many possible combinations of values exist for m, n, p, and q?

A) Two
B) Three
C) Four
D) Five
E) Seven

Thank you for your replies GMAT Club members. GMAT quant is based on logic, tricks, and quick approaches. Always try to find a quick approach to solve any PS or a DS question. We apply the IVY approach for PS and Variable Approach for DS.


Solution: Let us find the prime factors of 660.

660 can be written as 660 = 2 * 2 * 3 * 5 * 11.

We have an extra ‘2’ and this can be combined with other factors to generate different values.

Also, considering all other factors than ‘2’, we may combine to generate different values for m, n, p, and q.

Attachment:
Possible Combinations.jpg

Therefore, we have ‘4’ possible combinations.

C is the correct answer.

Answer C.
Show more


Apart from the 4 combinations mentioned above, we can also consider one more combination of 2, 3, 5, 11 right?
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Que: A merchant bought 1,000 fans for $15 each. He sold 40 percent of the fans for $20 each and the rest for $30 each. What was the merchant’s average profit per fan?

(A) $11
(B) $15
(C) $18
(D) $21
(E) $26
Show more


Solution: Cost price of 1,000 fans = $ (15 * 1000) = $15,000

=> Number of fans sold at $20 each = \(\frac{40}{100} * 1,000 = 400\)

=> Thus, selling price of these 400 fans = $(400 * 20) = $8,000

=> Number of fans sold at $ 30 each = 1000 – 400 = 600

=> Thus, selling of these 600 fans = $(600 * 30) = $18,000

=> Total selling price of 1,000 fans = $8,000 + $18, 000 = $26,000

=> Total profit: Selling price – Cost price = $26,000 - $15,000 = $21,000

=> Profit per fan: \(\frac{$21,000 }{1,000} = 21\)

Therefore, D is the correct answer.

Answer D
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Que: Box P and Box Q each contain many red balls and white balls. All of the white balls have the same radius. The radius of each white ball is 6 inches less than the average radius of the balls in Box P and 3 inches greater than the average radius of the balls in Box Q. What is the difference between the average (arithmetic mean) radius, in inches, of the balls in Box P and of the balls in Box Q?

(A) 3
(B) 4
(C) 6
(D) 8
(E) 9
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Que: A juice manufacturer has 2,000 liters of mango pulp in stock, 40 percent of which is water. If the manufacturer adds another 1,000 liters of mango pulp of which 50 percent is water, what percent, by volume, of the manufacturer’s mango pulp contains water?

(A) 10%
(B) 16.67%
(C) 26.67%
(D) 43.33%
(E) 65%
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