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08 Mar 2012, 13:27
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Baker's Dozen: A challenging set of GMAT Problem Solving (PS) questions that will put your quantitative reasoning skills to the test. Sharpen your mathematical abilities as you tackle thirteen thought-provoking problems.
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000
2. If \(y=\frac{(3^5-3^2)^2} {(5^7-5^4)^{-2}}\), then y is NOT divisible by which of the following?
A. 6^4
B. 62^2
C. 65^2
D. 15^4
E. 52^4
3. For the past \(k\) days, Liv has baked an average (arithmetic mean) of 55 cupcakes per day. Today, Bibi helped Liv, and together they baked 100 cupcakes, raising the average to 60 cupcakes per day. What is the value of \(k\)?
A. 6
B. 8
C. 9
D. 10
E. 12
4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441
5. In how many ways can we select 8 shirts from a closet containing 7 distinct red shirts and 5 distinct blue shirts, such that at least one shirt of each color remains in the closet, if the order of selection does not matter?
A. 460
B. 490
C. 493
D. 455
E. 445
6. A swimming pool has two water pumps, A and B, and one drain, C. Pump A can fill the empty pool by itself in \(x\) hours, while pump B can do so in \(y\) hours. The drain C can empty the entire pool in \(z\) hours, with \(z > x\). When both pumps A and B are operating simultaneously and drain C is open until the pool is filled, which of the following represents the fraction of the total pool volume added by pump A until the pool is filled?
A. \(\frac{yz}{x+y+z}\)
B. \(\frac{yz}{yz+xz-xy}\)
C. \(\frac{yz}{yz+xz+xy}\)
D. \(\frac{xyz}{yz+xz-xy}\)
E. \(\frac{yz+xz-xy}{yz}\)
7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?
A. $60,000
B. $90,000
C. $100,000
D. $120,000
E. $180,000
8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215
9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y
10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?
A. 7
B. 11
C. 13
D. 17
E. 19
11. In the infinite sequence 1, 3, 9, 27, ... , each term after the first is three times the preceding term. What is the positive difference between the sum of the 13th and 15th terms and the sum of the 12th and 14th terms of the sequence?
A. 10*3^11
B. 20*3^11
C. 10*3^12
D. 40*3^11
E. 20*3^12
12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
A. 12
B. 20
C. 24
D. 29
E. 33
13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)?
A. 0
B. 6
C. 7
D. 12
E. 14
The solutions can be found below on this page.
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000
2. If \(y=\frac{(3^5-3^2)^2} {(5^7-5^4)^{-2}}\), then y is NOT divisible by which of the following?
A. 6^4
B. 62^2
C. 65^2
D. 15^4
E. 52^4
3. For the past \(k\) days, Liv has baked an average (arithmetic mean) of 55 cupcakes per day. Today, Bibi helped Liv, and together they baked 100 cupcakes, raising the average to 60 cupcakes per day. What is the value of \(k\)?
A. 6
B. 8
C. 9
D. 10
E. 12
4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441
5. In how many ways can we select 8 shirts from a closet containing 7 distinct red shirts and 5 distinct blue shirts, such that at least one shirt of each color remains in the closet, if the order of selection does not matter?
A. 460
B. 490
C. 493
D. 455
E. 445
6. A swimming pool has two water pumps, A and B, and one drain, C. Pump A can fill the empty pool by itself in \(x\) hours, while pump B can do so in \(y\) hours. The drain C can empty the entire pool in \(z\) hours, with \(z > x\). When both pumps A and B are operating simultaneously and drain C is open until the pool is filled, which of the following represents the fraction of the total pool volume added by pump A until the pool is filled?
A. \(\frac{yz}{x+y+z}\)
B. \(\frac{yz}{yz+xz-xy}\)
C. \(\frac{yz}{yz+xz+xy}\)
D. \(\frac{xyz}{yz+xz-xy}\)
E. \(\frac{yz+xz-xy}{yz}\)
7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?
A. $60,000
B. $90,000
C. $100,000
D. $120,000
E. $180,000
8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215
9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y
10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?
A. 7
B. 11
C. 13
D. 17
E. 19
11. In the infinite sequence 1, 3, 9, 27, ... , each term after the first is three times the preceding term. What is the positive difference between the sum of the 13th and 15th terms and the sum of the 12th and 14th terms of the sequence?
A. 10*3^11
B. 20*3^11
C. 10*3^12
D. 40*3^11
E. 20*3^12
12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
A. 12
B. 20
C. 24
D. 29
E. 33
13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)?
A. 0
B. 6
C. 7
D. 12
E. 14
The solutions can be found below on this page.
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13 Mar 2012, 03:53
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7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?
A. $60,000
B. $90,000
C. $100,000
D. $120,000
E. $180,000
Fritz owns \(\frac{2}{3}\) of the shares owned by the other three shareholders. In other words, Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\) of all the shares. To put it more simply, Fritz owns 2 parts, while the other three shareholders together own 3 parts. Thus, Fritz owns 2 parts out of a total of 5 parts.
Luis owns \(\frac{3}{7}\) of the shares owned by the other three shareholders. In other words, Luis owns \(\frac{3}{3+7}=\frac{3}{10}\) of all the shares.
Alfred owns \(\frac{4}{11}\) of the shares owned by the other three shareholders. In other words, Alfred owns \(\frac{4}{4+11}=\frac{4}{15}\) of all the shares.
Combined, these three shareholders own \(\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}\) of all the shares. This implies that Werner owns the remaining shares, which amounts to \(1-\frac{29}{30}=\frac{1}{30}\). Therefore, from the total of $3,600,000, Werner receives \($3,600,000 * \frac{1}{30} = $120,000\).
Answer: D
A. $60,000
B. $90,000
C. $100,000
D. $120,000
E. $180,000
Fritz owns \(\frac{2}{3}\) of the shares owned by the other three shareholders. In other words, Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\) of all the shares. To put it more simply, Fritz owns 2 parts, while the other three shareholders together own 3 parts. Thus, Fritz owns 2 parts out of a total of 5 parts.
Luis owns \(\frac{3}{7}\) of the shares owned by the other three shareholders. In other words, Luis owns \(\frac{3}{3+7}=\frac{3}{10}\) of all the shares.
Alfred owns \(\frac{4}{11}\) of the shares owned by the other three shareholders. In other words, Alfred owns \(\frac{4}{4+11}=\frac{4}{15}\) of all the shares.
Combined, these three shareholders own \(\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}\) of all the shares. This implies that Werner owns the remaining shares, which amounts to \(1-\frac{29}{30}=\frac{1}{30}\). Therefore, from the total of $3,600,000, Werner receives \($3,600,000 * \frac{1}{30} = $120,000\).
Answer: D
13 Mar 2012, 04:05
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13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)?
A. 0
B. 6
C. 7
D. 12
E. 14
Apply \(a^2-b^2=(a-b)(a+b)\): \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3\).
Next, \(\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38\).
Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^2-38\) will have 00-38=62 as the last digits: 6*2=12.
Answer: D.
A. 0
B. 6
C. 7
D. 12
E. 14
Apply \(a^2-b^2=(a-b)(a+b)\): \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3\).
Next, \(\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38\).
Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^2-38\) will have 00-38=62 as the last digits: 6*2=12.
Answer: D.
