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PS Expert - Math Revolution: Ask Me Anything about GMAT PS 10 Nov 2020, 10:33
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Que: A chemical evaporates out of a beaker at the rate of x liters for every y minutes. If the chemical costs 25 dollars per liter, what is the cost, in dollars, of the amount of the chemical that will evaporate in z minutes?

(A) \(\frac{25}{yz}\)

(B) \(\frac{xz}{25q}\)

(C) \(\frac{25y}{xz}\)

(D) \(\frac{25xz}{y}\)

(E) \(\frac{25yz}{x}\)
Show more


Solution: Amount of chemical evaporated in ‘y’ minutes: x liters

Therefore, the amount of chemical evaporated in 1 minute = \(\frac{x}{y}\) liters

Hence, the amount of chemical evaporated in z minute = \(\frac{(xz)}{y}\) liters

=> Cost of 1 liter of chemical: $25

Thus, cost of the chemical evaporated in z minutes = $25 * \(\frac{(xz)}{y}\) = \(\frac{(25xz)}{y}\)

D is the correct answer.

Answer D
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Que: Suzy saves $30 per month. In each of the next 30 months, she saved $30 more than he saved in the previous month. What is the total amount she saved during the 30-month period?

(A) $900
(B) $2,800
(C) $13,950
(D) $9,300
(E) $12,000
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PS Expert - Math Revolution: Ask Me Anything about GMAT PS 12 Nov 2020, 12:44
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(Functions): If 2f (x) + 3f (−x) = 2x − 4, what is the value of 5f (1)?

A) 14/5
B) -2
C) 14
D) -14/5
E) -14


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: 2f(x) + 3 f(-x) = 2x - 4 --------- equation (1)

Substituting x = 1 in equation (1)

2f(1) + 3f(-1) = 2(1) – 4 =2 – 4= -2

2f(1) + 3f(-1) = -2--------- equation (2)

Substituting x = -1 in equation (1)

2f(-1) + 3f(1) = 2(-1) – 4 = -2 – 4 = -6

2f(-1) + 3f(1) = -6--------- equation (3)

Multiplying equation (2) by ‘2’ we get,

4f(1) + 6f(-1) = -4--------- equation (4)


Multiplying equation (3) by ‘3’ we get,

9f(1) + 6f(-1) = -18--------- equation (5)

Subtracting Eqn(5) - Eqn(4)

=> 5f(1) = -14

=> f(1) = \(\frac{-14}{5}\)

D is the correct answer.

Answer D
Show more


Since the question asks for the value of 5f (1), the correct answer is E) -14
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Que: Suzy saves $30 per month. In each of the next 30 months, she saved $30 more than he saved in the previous month. What is the total amount she saved during the 30-month period?

(A) $900
(B) $2,800
(C) $13,950
(D) $9,300
(E) $12,000
Show more

Solution: Sum of first ‘n’ positive integers = \(\frac{[n ( n +1)] }{ 2}\)

The total amount she saved during the 30-month period is $30 + 2 * $30 +…+ 29 * $30 + 30 * $30 = $30(1 + 2 + … + 29 + 30).

Since the sum of first 30 positive integers 1 + 2 +…+29 + 30 = \(\frac{(30 * 31)}{2}\) = 15 * 31 = 465, we get total amount saved in 30 months

=> $30(1+2+…+29+30) = $30 * 465 = $13,950

Therefore, C is the correct answer.

Answer C
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PS Expert - Math Revolution: Ask Me Anything about GMAT PS 14 Nov 2020, 00:03
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Que: A volcanic lava laterally moves at the rate of \(\frac{15}{4}\) feet per hour. How many days does it take the lava to move \(\frac{9}{2}\) miles? (1 mile = 5,280 feet)

(A) 205
(B) 264
(C) 730
(D) 80
(E) 528
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PS Expert - Math Revolution: Ask Me Anything about GMAT PS Updated on: 14 May 2021, 03:27
Originally posted by MathRevolution on 16 Nov 2020, 04:47.
Last edited by MathRevolution on 14 May 2021, 03:27, edited 1 time in total.
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Que: A volcanic lava laterally moves at the rate of \(\frac{15}{4}\) feet per hour. How many days does it take the lava to move \(\frac{9}{2}\) miles? (1 mile = 5,280 feet)

(A) 205
(B) 264
(C) 730
(D) 80
(E) 528
Show more


Solution: Distance covered by the lava in 1 hour = \(\frac{15}{4}\)feet

=> \(\frac{15}{4}\) feet : 1 hour

=> \(\frac{15}{4}\) (\(\frac{1}{5280}\) mile) : 1 hour (since 1 foot = \(\frac{1}{5280}\) mile)

=> \(\frac{15 }{ (4 * 5280)}\) miles : 1 hour

=> \(\frac{15 }{ (4 * 5280)}\) miles : 1 hour = \(\frac{9}{2}\) miles : x hours

=> (x hours) * [\(\frac{15 }{ (4 * 5280)}\) miles] = (1 hour)(9/2 miles)

=> x[\(\frac{15}{(4 * 5280)}\)] = (1)(\(\frac{9}{2}\))= \(\frac{9}{2}\) (cancel out (hours)(miles) from both sides)

=> x = \((\frac{9}{2}) / [\frac{15 }{ (4 * 5280)}]\)= \(\frac{(9 * 4 * 5280) }{ (2 * 15)}\)= 6,336.

Therefore, hours needed to cover \(\frac{9}{2}\) miles = 6,336 hours, and days needed to cover \(\frac{9}{2}\) miles = \(\frac{6,336 }{ 24}\) = 264

B is the correct answer.

Answer B
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Que: The trader first decreased the selling price of a bicycle by 25 percent and then increased by 25 percent. Which of the following represents the final percent change in the selling price of the bicycle?

(A) 6.25% less
(B) 0%
(C) 6.25% more
(D) 93.75% more
(E) 100% more
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PS Expert - Math Revolution: Ask Me Anything about GMAT PS Updated on: 14 May 2021, 03:28
Originally posted by MathRevolution on 26 Nov 2020, 03:05.
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Que: The trader first decreased the selling price of a bicycle by 25 percent and then increased by 25 percent. Which of the following represents the final percent change in the selling price of the bicycle?

(A) 6.25% less
(B) 0%
(C) 6.25% more
(D) 93.75% more
(E) 100% more
Show more


Solution: Let us apply the IVY approach to solve the question. As we are dealing with a percentage, then let the original price of a bicycle be $100.

Price after price reduced by 25%: (100 - 25)% of $100

=> (\(\frac{75}{100}\)) * $100

=> $75

Price after the new price increased by 25%: (100 + 25)% of $75
=> (\(\frac{125}{100}\)) * $75

=> $93.75

As the base price is $100, the final price would be 93.75% of the base price.

Since Percent change = \(\frac{(After - Before)}{Before}\)*100(%) and $100 (Before) -> $93.75 (After),

We get Percent change = \(\frac{($93.75-$100)}{$100}\) * 100(%) = \(\frac{(-6.25)}{100}\) * 100(%)= -6.25%

Therefore, A is the correct answer.

Answer A
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Que: A juice manufacturer has 1,200 liters of mango pulp in stock, 25 percent of which is water. If the manufacturer adds another 400 liters of mango pulp of which 20 percent is water, what percent, by volume, of the manufacturer’s mango pulp contains water?

(A) 21.50%
(B) 23.75%
(C) 33.33%
(D) 35.00%
(E) 37.50%
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PS Expert - Math Revolution: Ask Me Anything about GMAT PS Updated on: 30 Mar 2022, 02:58
Originally posted by MathRevolution on 01 Dec 2020, 01:43.
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Que: A juice manufacturer has 1,200 liters of mango pulp in stock, 25 percent of which is water. If the manufacturer adds another 400 liters of mango pulp of which 20 percent is water, what percent, by volume, of the manufacturer’s mango pulp contains water?

(A) 21.50%
(B) 23.75%
(C) 33.33%
(D) 35.00%
(E) 37.50%
Show more


Solution: The resultant percentage will be the weighted average of the percentages of the above two stocks.

=> \(\frac{[(1,200 * 0.25) + (400 * 0.20)]}{ [(1,200 + 400)]}\) * 100

=> \(\frac{(300 + 80) }{ (1,600)}\) * 100

=> (\(\frac{380}{1,600}\)) * 100 => 23.75%

Therefore, B is the correct answer.

Answer B
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Que: An instructor gave the same test to three groups: P, Q, and R. The average (arithmetic mean) scores for the three groups were 64, 84, and 72, respectively. The ratio of the numbers of candidates in P, Q, and R groups was 3: 5: 4, respectively. What was the average score for the three groups combined?

(A) 72
(B) 75
(C) 77
(D) 78
(E) 80
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Que: An instructor gave the same test to three groups: P, Q, and R. The average (arithmetic mean) scores for the three groups were 64, 84, and 72, respectively. The ratio of the numbers of candidates in P, Q, and R groups was 3: 5: 4, respectively. What was the average score for the three groups combined?

(A) 72
(B) 75
(C) 77
(D) 78
(E) 80
Show more

Solution: Let the number of candidates in groups P, Q, and R be 3k, 5k, and 4k, respectively, where ‘k’ is a constant of proportionality.

The average (arithmetic mean) scores for the three groups were 64, 84, and 72, respectively. The combined average for three groups is

=> \(\frac{[ (64 * 3k) + (84 * 5k) + (72 * 4k)]}{ (3k + 5k + 4k)}\)

=> \(\frac{[(192k + 420k + 288k)]}{ (12k)}\)

=> \(\frac{900k }{ 12k}\) = 75

Therefore, B is the correct answer.

Answer B
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Que: Few coins are put into 7 boxes such that each box contains at least two coins. At the most '3' boxes can contain the same number of coins, and the remaining boxes cannot contain an equal number of coins. What is the minimum possible number of coins in the 7 boxes?

(A) 18
(B) 20
(C) 24
(D) 27
(E) 30
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Que: Few coins are put into 7 boxes such that each box contains at least two coins. At the most '3' boxes can contain the same number of coins, and the remaining boxes cannot contain an equal number of coins. What is the minimum possible number of coins in the 7 boxes?

(A) 18
(B) 20
(C) 24
(D) 27
(E) 30
Show more

Solution: Since each box contains at least two coins, we have 7 * 2 = 14 coins minimum.

We know that at the most 3 boxes can have the same number of coins. Since we need to minimize the total number of coins, we must have as many boxes having the same number (minimum possible number, i.e. 2 coins) of coins as possible. Thus, for each of the 3 boxes containing an equal number of coins, we have ‘2’ coins. Thus, number of coins in the 3 boxes = 2 × 3 = 6.

Since each of the remaining 4 boxes have a different number of coins, let us put in 3, 4, 5, and 6 coins in those boxes. Thus, the total number of coins = 6 + (3 + 4 + 5 + 6) = 24.

Therefore, C is the correct answer.

Answer C
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PS Expert - Math Revolution: Ask Me Anything about GMAT PS 05 Dec 2020, 23:58
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Que: A color “code” is defined as a sequence of three dots arranged in a row. Each dot is colored either “red” or “black.” How many distinct codes can be formed?

(A) 4
(B) 5
(C) 6
(D) 8
(E) 1
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PS Expert - Math Revolution: Ask Me Anything about GMAT PS Updated on: 30 Mar 2022, 03:00
Originally posted by MathRevolution on 08 Dec 2020, 09:40.
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Que: A color “code” is defined as a sequence of three dots arranged in a row. Each dot is colored either “red” or “black.” How many distinct codes can be formed?

(A) 4
(B) 5
(C) 6
(D) 8
(E) 1
Show more


Solution: Number of ways arranging ‘n’ objects out of which ‘r’ objects are identical is n!/ r!

Case I: Let us assume we use two BLACK and 1 RED dot. Thus, the total way of arranging them is 3!/2! = 3

Case II: Similarly, let us assume we use two RED and 1 BLACK dot. Thus, the total way of arranging them is 3!/2! = 3

Case III: All three BLACK: 3!/3! = 1

Case IV: All three RED: 3!/3! = 1

Total number of codes: 3 + 3 + 1 + 1 = 8

The first dot can use RED or BLACK, the second dot also can use RED or BLACK, and the third dot also can do, so 2*2*2=8

Therefore, D is the correct answer.

Answer D
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Que: On day 1, a shopkeeper increases the price of an item by k%, and on day 2, he decreases the increased price by k%. By the end of day 2, the price of the item drops by $1. On day 3, he again increases the decreased price by k%, and on day 4, he again decreases the increased price by k%. If, at the end of day 4, the price of the item comes to $398, what was the approximate initial price of the item?

(A) $325
(B) $350
(C) $375
(D) $400
(E) $450
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Que: On day 1, a shopkeeper increases the price of an item by k%, and on day 2, he decreases the increased price by k%. By the end of day 2, the price of the item drops by $1. On day 3, he again increases the decreased price by k%, and on day 4, he again decreases the increased price by k%. If, at the end of day 4, the price of the item comes to $398, what was the approximate initial price of the item?

(A) $325
(B) $350
(C) $375
(D) $400
(E) $450
Show more


Solution: Let us apply the IVY approach to solve the question.

As we are dealing with percentage, then let the initial price of the item be 100. Since there is a common k% of increase and decrease, then let us assume k% = 10%

Day 1: Price after price increased by 10%: (100 + 10)% of 100
=> (\(\frac{110}{100}\)) $100 => $110

Day 2: Price after the new price decreased by 10%: (100 - 10)% of 110
=> (\(\frac{90}{100}\)) $110 => $99

Day 3: Price after price increased by 10%: (100 + 10)% of 99
=> $99 * \(\frac{110}{100}\) => $108.9 ≈ $109

Day 4: Price after the new price decreased by 10%: (100 - 10)% of 109
=> $109 * \(\frac{90}{100}\) => $98.1 ≈ 98

For $ 100 = final price is $98
For $x = $398 [given in the question]

=> x = \(\frac{(100 * 398) }{ 98}\) = 406 ≈ 400

Therefore, D is the correct answer.

Answer D
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Que: A number 4p25q is divisible by both 4 and 9; where p and q are the thousands and units digits, respectively. What is the minimum possible value of \(\frac{p}{ q}\)?

(A) \(\frac{1}{8}\)

(B) \(\frac{1}{7}\)

(C) \(\frac{1}{6}\)

(D) \(\frac{2}{5}\)

(E) \(\frac{5}{2}\)
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Que: A number 4p25q is divisible by both 4 and 9; where p and q are the thousands and units digits, respectively. What is the minimum possible value of \(\frac{p}{ q}\)?

(A) \(\frac{1}{8}\)

(B) \(\frac{1}{7}\)

(C) \(\frac{1}{6}\)

(D) \(\frac{2}{5}\)

(E) \(\frac{5}{2}\)
Show more

Solution: A number, here 4p25q 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).

A number, here 4p25q is divisible by ‘9’ if the sum of the digits of the number is divisible by 9.

The sum of the digits = 4 + p + 2 + 5 + q = 11 + p + q. Thus, (11 + p + q) is divisible by 9

If q = 2:

=> 11 + p + q = 11 + p + 2 = 13 + p

13 + p is divisible by ‘9’ when p = 5. (Since, 13 + 5 = 18, which is divisible by 9)

If q = 6:

=> 11 + p + q = 11 + p + 6 = 17 + p

17 + p is divisible by ‘9’ when p = 1. (Since, 17 + 1 = 18, which is divisible by 9)

Therefore, we have two possible combinations:

=> p = 5, q = 2 then \(\frac{p}{q = 5/2}\)

=> p = 1, q = 6 then \(\frac{p}{q = 1/6}\)

Thus, the minimum value of \(\frac{p}{q}\) is \(\frac{1}{6}\).

Therefore, C is the correct answer.

Answer C
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