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Que: Item A attracts a sales tax rate of $0.54 per $25. What is the sales tax rate, as a percent, for item B that attracts four times as much as the rate for item A?

(A) 216%
(B) 86.4%
(C) 8.64%
(D) 2.16%
(E) 0.135%
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Que: Item A attracts a sales tax rate of $0.54 per $25. What is the sales tax rate, as a percent, for item B that attracts four times as much as the rate for item A?

(A) 216%
(B) 86.4%
(C) 8.64%
(D) 2.16%
(E) 0.135%
Show more

Solution: Tax paid on $25 = $0.54

Thus, a tax, which is four times as much as the above, would be $(0.54 × 4) = $2.16 on $25.

Thus, the tax rate = \(\frac{$2.16}{$25}\)=0.0864

Thus, this tax, expressed as a percentage = 0.0864 = 8.64 × \(\frac{1}{100}\) = 8.64%.

Therefore, C is the correct answer.

Answer C
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Que: A trader buys a batch of 120,000 computer chips for $3,600,000. He sells \(\frac{2}{5}\) of the computer chips, each at 25 percent above the cost per computer chip. Later, he sells the remaining computer chips at a price per computer chip equal to 25 percent less than the cost per computer chip. What was the percent profit or loss on the batch of computer chips?

(A) Loss of 1%
(B) Loss of 5%
(C) Loss of 7.50%
(D) Profit of 10%
(E) Profit of 22.22%
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Que: A trader buys a batch of 120,000 computer chips for $3,600,000. He sells \(\frac{2}{5}\) of the computer chips, each at 25 percent above the cost per computer chip. Later, he sells the remaining computer chips at a price per computer chip equal to 25 percent less than the cost per computer chip. What was the percent profit or loss on the batch of computer chips?

(A) Loss of 1%
(B) Loss of 5%
(C) Loss of 7.50%
(D) Profit of 10%
(E) Profit of 22.22%
Show more


Solution: Total cost of the 120,000 computer chips = $3,600,000

=> Cost of \(\frac{2}{5}\) of the above computer chips = $ 3,600,000 * \(\frac{2}{5}\) = $1, 440, 000.

These were sold at a 25% higher than the cost price. Thus, the selling price of the above computer chips

=> \(\frac{125}{100}\) *$1,440,100 = $1, 800, 000

Cost of the remaining computer chips = $(3, 600, 000 − 1, 440, 000) = $2, 160, 000

Later, these remaining computer chips were sold at a 25% lower than the cost price. Thus, the selling price of the above computer chips

=> \(\frac{75}{100}\) * $2, 160, 000= $1, 620, 000

Thus, total selling price = $(1, 800, 000 + 1, 620, 000) = $3, 420, 000.

Since total selling price (= $3,420,000) < total cost price (= $3,600,000), there is a loss

Thus, percent loss = \(\frac{[(Cost price − Selling price)]}{ Cost price}\) × 100 (%)

=> \(\frac{(3, 600, 000 − 3, 420, 000) }{ 3, 600, 000}\) × 100 (%) = -5% (Loss)

Therefore, B is the correct answer.

Answer B
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Que: As per the previous year's data, $q can buy p number of items. If the average cost of each item increased by 25 percent this year, then the number of items can be bought with $6q equals?

(A) 4.8p
(B) 2.50p
(C) 8p
(D) 6p
(E) 7.50p
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Que: As per the previous year's data, $q can buy p number of items. If the average cost of each item increased by 25 percent this year, then the number of items can be bought with $6q equals?

(A) 4.8p
(B) 2.50p
(C) 8p
(D) 6p
(E) 7.50p
Show more

Solution: Cost of p items = $q

Since the cost increases by 25%, the new cost of p items = \(\frac{125}{100}\) * $q = \(\frac{($5q)}{(4)}\)

Thus, with a budget of \(\frac{($5q)}{(4)}\), 'p' items can be bought this year.

Thus, with a budget of $6q, the numbers of items that can be bought

=> \(\frac{6qp}{\frac{5q}{4}}\) (Since \(\frac{($5q)}{(4)}\) : p = $6q : x, we get x = \(\frac{6qp}{\frac{5q}{4}}\)

=> x = 4.8p

Therefore, A is the correct answer.

Answer A
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Que: In a stadium, the Royal Challengers team had the support of 24,500 spectators from natives and 10 percent of spectators from other than natives. If S is the total number of spectators in the stadium and 40 percent belonged to natives, in terms of S, which of the following represents the number of supporters for the Royal Challengers team?

(A) 0.6S + 12, 250
(B) 0.28S + 12, 250
(C) 0.28S + 24, 500
(D) 0.06S + 24, 500
(E) 0.6S + 24, 500
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Que: Two water pumps, working simultaneously at their respective constant rates, took exactly 4 hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate?

(A) 5
(B) 16/3
(C) 11/2
(D) 6
(E) 20/3
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Que: In a stadium, the Royal Challengers team had the support of 24,500 spectators from natives and 10 percent of spectators from other than natives. If S is the total number of spectators in the stadium and 40 percent belonged to natives, in terms of S, which of the following represents the number of supporters for the Royal Challengers team?

(A) 0.6S + 12, 250
(B) 0.28S + 12, 250
(C) 0.28S + 24, 500
(D) 0.06S + 24, 500
(E) 0.6S + 24, 500
Show more

Solution: Total number of spectators = S

Since 40% of the spectators belonged to natives, percent of other than native spectators = (100 − 40) = 60%.

Thus, the number of other than native spectators = 60% of S = 0.6S

Of the above, Royal Challengers had the support of 10%.

Thus, the number of other than native supporters = 10% of 0.6S = 0.06S [Part becomes the whole]

Number of native supporters = 24,500

Thus, total number of supporters = 0.06S + 24, 500

Therefore, D is the correct answer.

Answer D
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Que: If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between the average of the odd integers and the average of the even integers in set N?

(A) −1
(B) \(\frac{1}{2} \)
(C) 0
(D) 1
(E) 2
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Que: If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between the average of the odd integers and the average of the even integers in set N?

(A) −1
(B) \(\frac{1}{2} \)
(C) 0
(D) 1
(E) 2
Show more

Solution: Since the set starts with an odd number (1) and has an odd number of integers, the set would end with an odd number, too.

Let’s see the set. Set N: {1, 2, 3, 4, 5, ..., (2n + 1)}, where n is a positive integer.

=> Number of odd terms is one more than the number of even terms.

Thus, the number of odd terms = (n + 1) and we derive a new set from set N, that is {1,3,…, 2n+1}.

Then we get the average of the odd integers = \(\frac{[1+3+…+(2n+1)]}{(n+1)}\) and since 1+3+…+(2n+1)=\((n+1)^2\), we get \(\frac{[(1+3+…+(2n+1)]}{(n+1)} = \frac{[(n+1)^2]}{(n+1)} = n+1\)

The number of even terms = n and we derive a new set from set N, that is {2,4,…, 2n}.

Then we get the average of the even integers=\(\frac{(2+4+…+2n)}{n}\)=\(\frac{2(1+2+…+n)}{n}\) and since 1+2+…+n = \(\frac{[n(n+1)]}{2}\), we get \(\frac{(2 + 4 + …+ 2n)}{n}=\frac{2(1+2+…+n)}{n}\) = \(\frac{n(n+1)}{n}\) = n+1

Therefore, we get the difference of the average of the odd integers and the average of the even integers in set N=(n + 1) - (n + 1)=0

Say there are only three terms in the set

S = {1, 2, 3}

=> X = \(\frac{(1 + 3) }{ 2}\) = \(\frac{4}{2} = 2\)

=> Y = 2

=> X – Y = 0

Again, say there are five terms in the set

S = {1, 2, 3, 4, 5}

=> \(X = \frac{(1 + 3 + 5) }{ 3}= \frac{9}{3} = 3\)

=> \(Y = \frac{(2 + 4) }{ 2} = \frac{6}{2} = 3\)

=> X – Y = 0

Therefore, C is the correct answer.

Answer C
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Que: To make a certain color, a paint dealer mixes 3.4 liters of red color to a base that is 68 liters. The paint manufacturer recommends mixing 0.7 liters per 10 liters of the base to make that color. By what percent should the mixing be increased to bring it to the recommendation?

(A) 10%
(B) 33.33%
(C) 40%
(D) 66.66%
(E) 72%
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Que: To make a certain color, a paint dealer mixes 3.4 liters of red color to a base that is 68 liters. The paint manufacturer recommends mixing 0.7 liters per 10 liters of the base to make that color. By what percent should the mixing be increased to bring it to the recommendation?

(A) 10%
(B) 33.33%
(C) 40%
(D) 66.66%
(E) 72%
Show more


Solution: The mixing for 68 liters of the base was 3.4 liters of red color.

The recommended mixing for every 10 liters of the base was 0.7 liters of red color.

Thus, as per the recommendation, the amount of red color required for 68 liters of base = \(\frac{0.7 }{ 10} * 68 = 4.76\) liters

Percent change: \(\frac{(After – Before)}{ Before}\) * 100(%) [After: 4.76 ; Before: 3.4]

=> \(\frac{(4.76 – 3.4) }{ 3.4}\) * 100(%)

=> \(\frac{1.36}{3.4}\) * 100(%) = 40%

Therefore, C is the correct answer.

Answer C
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Que: A merchant bought 2,400 fans for $30 each. He sold 60 percent of the fans for $40 each and the rest for $35 each. What was the merchant’s average profit per fan?

(A) $6
(B) $8
(C) $9
(D) $10
(E) $12
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Hey, got this problem on one of my practice tests and cannot figure it out or understand the explanation given... Can someone break this down? Cheers and happy studying

† and ¥ represent nonzero digits, and (†¥)² - (¥†)² is a perfect square. What is that perfect square?

(a) 121
(b) 361
(c) 576
(d) 961
(e) 1089
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33,56,65 is a triplet. So, (65)^2 - (56)^2 = (33)^2 will be the answer. If you carefully notice, the numbers in (65)^2 - (56)^2 are reversed, i.e 56 and 65 which satisfies the conditions given in the question.
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Que: A merchant bought 2,400 fans for $30 each. He sold 60 percent of the fans for $40 each and the rest for $35 each. What was the merchant’s average profit per fan?

(A) $6
(B) $8
(C) $9
(D) $10
(E) $12
Show more

Solution: Cost price of 2,400 fans = $ (30 * 2400) = $72,000

Number of fans sold at $40 each = \(\frac{60}{100} * 2400 = 1,440\)

Thus, selling price of these 1,440 fans = $(1,440 * 40) = $57,600

Number of fans sold at $ 35 each = 2400 – 1440 = 960

Thus, selling of these 960 fans = $(960 * 35) = $33,600

Total selling price of 2,400 fans = $57,600 + $33, 600 = $91,200

Total profit: Selling price – Cost price = $91,200 - $72,000 = $19,200

Profit per fan: \(\frac{$19,200 }{2,400} = 8\)

Therefore, B is the correct answer.

Answer B
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Que: Box X and Box Y each contain many yellow balls and green balls. All of the green balls have the same radius. The radius of each green ball is 4 inches less than the average radius of the balls in Box X and 2 inches greater than the average radius of the balls in Box Y. What is the difference between the average (arithmetic mean) radius, in inches, of the balls in Box X and of the balls in Box Y?

(A) 4
(B) 10
(C) 7
(D) 8
(E) 6
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Que: Box X and Box Y each contain many yellow balls and green balls. All of the green balls have the same radius. The radius of each green ball is 4 inches less than the average radius of the balls in Box X and 2 inches greater than the average radius of the balls in Box Y. What is the difference between the average (arithmetic mean) radius, in inches, of the balls in Box X and of the balls in Box Y?

(A) 4
(B) 10
(C) 7
(D) 8
(E) 6
Show more

Solution: Let the radius of each green ball = x inches.

Each green ball is 4 inches less than the average radius of the balls in Box X.

Thus, the average radius of balls in Box X = (x + 4) inches.

Also, each green ball is 2 inches greater than the average radius of the balls in Box Y.

Thus, the average radius of balls in Box Y = (x − 2) inches.

Thus, the required difference = ((x + 4) − (x − 2)) = 6 inches.

Therefore, E is the correct answer.

Answer E
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Que: A quiz consists of X questions, each of which is to be answered either “Yes” or “No.” What is the least value of X for which the probability is less than \(\frac{1}{500}\) such that a participant who randomly guesses the answer to each question will be a winner?

(A) 8
(B) 9
(C) 10
(D) 200
(E) 500
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Que: A quiz consists of X questions, each of which is to be answered either “Yes” or “No.” What is the least value of X for which the probability is less than \(\frac{1}{500}\) such that a participant who randomly guesses the answer to each question will be a winner?

(A) 8
(B) 9
(C) 10
(D) 200
(E) 500
Show more


Solution: Total questions: X

Total options for each question: 2 [Yes or No]

Thus, the probability of randomly guessing an answer and getting it correct = \(\frac{1}{2}\)

Thus, the probability of randomly guessing answers to all X questions and getting them correct:

=> \(\frac{1}{2}\) * \(\frac{1}{2}\) * ….. X times

=> \((\frac{1}{2})^x\)

=> \((\frac{1}{2})^x < \frac{1}{500}\)

=> \(\frac{1}{(2^x)} < \frac{1}{500}\)

=> \(2^x > 500\)

For x = 9, \(2^9 = 512\), which just exceeds 500.

Therefore, B is the correct answer.

Answer B
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