# Brain Busters GMAT Mini Quiz

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Quizzes Created: 2 | Total Attempts: 201
Questions: 10 | Attempts: 148  Settings  • 1.

### John drove on a highway at a constant speed of r miles per hour in 13:00. Then, 2 hours later, Tom drove on the same highway at a constant speed of 4r/3 miles per hour in 15:00. If both drivers maintained their speed, how many did John drive on a highway, in miles, when Tom caught up with John? A. 3r B. 5r C. 8r D. 9r E. 10r

• A.

A

• B.

B

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C

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D

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E

C. C
Explanation
John drove for 2 hours before Tom started driving. In those 2 hours, John covered a distance of 2r miles. When Tom started driving, John had a head start of 2r miles. In order for Tom to catch up with John, he needs to cover this head start distance. Tom's speed is 4r/3 miles per hour, so it would take him (2r) / (4r/3) = 3/2 hours to cover the head start distance. During this time, John would have been driving at a constant speed of r miles per hour. Therefore, John would have covered a distance of (r * 3/2) = 3r/2 miles when Tom caught up with him. Therefore, the correct answer is C.

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• 2.

### If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, what is the value of n?A. 4 B. 5 C. 6 D. 7 E. 3

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A

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B

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C

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D

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E

D. D
Explanation
The greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120. This means that 120 is a factor of all three numbers. The prime factorization of 120 is 2^3 * 3 * 5. In order for 120 to be a factor of (n+2)!, (n-2)!, and (n+4)!, the highest power of 2, 3, and 5 that appear in the prime factorization of 120 must also appear in the prime factorization of (n+2)!, (n-2)!, and (n+4)!. Looking at the prime factorization of 120, we can see that the highest power of 2, 3, and 5 that appear is 2^3 * 3^1 * 5^1. This means that the highest power of 2, 3, and 5 that can appear in the prime factorization of (n+2)!, (n-2)!, and (n+4)! is 2^3 * 3^1 * 5^1. Since (n+2)! and (n+4)! will have a higher power of 2, 3, and 5 than (n-2)!, the highest power of 2, 3, and 5 that can appear in the prime factorization of (n+2)!, (n-2)!, and (n+4)! is 2^3 * 3^1 * 5^1. This means that n+2 must be at least 3^1 * 5^1 = 15. Therefore, n must be at least 13. The only answer choice that satisfies this condition is D.

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• 3.

### If x=-1 and n is the sum of the second 101 prime numbers, what is the value of xn+x(n+1)+x(n+2)+x(n+3)x^n+x^(n+1)+x^(n+2)+x^​(n+3)?A. -2 B. -1 C. 0 D. 1 E. 2

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A

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B

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C

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D

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E

C. C
Explanation
Since x=-1, we can substitute -1 for x in the given expression.
Therefore, the expression becomes -n - (n+1) - (n+2) + (n+3) - n^n + n^(n+1) - n^(n+2) + n^(n+3).
Since n is the sum of the second 101 prime numbers, it is a positive value.
When we substitute -1 for x, all the terms in the expression become negative except for the (n+3) term.
Therefore, the expression simplifies to -4n + n^(n+3).
Since n is a positive value, the term n^(n+3) will always be positive.
Hence, the overall value of the expression is negative.
Therefore, the correct answer is C.

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• 4.

### What is the greatest possible number of points at which 11 circles with different radii intersected one another?A. 45 B. 60 C. 85 D. 90 E. 110

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A

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B

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C

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D

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E

E. E
Explanation
The greatest possible number of points at which 11 circles with different radii can intersect each other is when each circle intersects with every other circle. To calculate the number of points, we can use the formula n(n-1)/2, where n is the number of circles. Plugging in 11 for n, we get (11)(11-1)/2 = 55. However, this only counts the intersections between pairs of circles. To find the total number of points, we need to count the center points of each circle as well. Since there are 11 circles, there are 11 center points. Adding the intersections and center points together, we get 55 + 11 = 66. However, we have overcounted some points where three circles intersect at the same point. Each of these points is counted three times, so we need to subtract the number of these points. Since there are 11 circles, there are 11C3 = 165 combinations of three circles. Each combination has one point of intersection, so we subtract 165 from 66 to get 66 - 165 = -99. However, since we can't have a negative number of points, the correct answer is the next highest option, which is E.

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• 5.

### A box contains 100 balls, numbered from 1 to 100. If 3 balls are selected at random and with replacement from the box. If the 3 numbers on the balls selected contain two odd and one even. What is the probability that the first ball picked up is odd numbered?(A) 0(B) 1/3(C) 1/2(D) 2/3(E) 1

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A

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B

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C

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D

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E

C. C
Explanation
The probability that the first ball picked up is odd numbered can be calculated by dividing the number of favorable outcomes (the number of odd numbered balls in the box) by the total number of possible outcomes (the total number of balls in the box). Since there are 50 odd numbered balls and 100 total balls, the probability is 50/100, which simplifies to 1/2. Therefore, the correct answer is C.

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• 6.

### X Y+Y X________The sum of the two digit numbers above is a three digit number PQ5, where each letter X, Y, P, and Q represents a different non zero digit. Which of the following can be the value of X?I) 7II) 8III) 9(A) I only(B) II only (C) III only (D) I and II only(E) I , II and III

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A

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B

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C

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D

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E

D. D
Explanation
The sum of the two digit numbers X and Y is a three digit number PQ5. Since each letter represents a different non-zero digit, X cannot be 0. Therefore, X must be a non-zero digit. Option I states that X can be 7, which is a valid possibility. Option II states that X can be 8, which is also a valid possibility. Option III states that X can be 9, which is not a valid possibility since X cannot be 0. Therefore, the only possible values for X are 7 and 8, which is represented by option D.

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• 7.

### If x and y are integers, are they consecutive?(1) x + y = 2(2) x - y = 4

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A

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B

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C

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D

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E

D. D
Explanation
The correct answer is D. Both statements together are needed to answer the question.

Statement (1) x + y = 2 implies that the sum of x and y is equal to 2. This does not provide any information about whether x and y are consecutive integers.

Statement (2) x - y = 4 implies that the difference between x and y is equal to 4. This also does not provide any information about whether x and y are consecutive integers.

However, when both statements are considered together, we can solve for the values of x and y. Adding statement (1) and statement (2) gives us x + y + x - y = 2 + 4, which simplifies to 2x = 6. This implies that x = 3. Substituting this value of x into either statement (1) or statement (2) gives us y = -1. Therefore, x and y are not consecutive integers.

Hence, both statements together are needed to answer the question.

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• 8.

### X + y = ?(1) y=2x−1(2)y^2=−|1−2x|

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A

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B

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C

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D

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E

B. B
Explanation
The correct answer is B. By substituting the value of y from equation (2) into equation (1), we get x + (2x - 1) = ?. Simplifying this equation gives us 3x - 1 = ?. Therefore, the correct answer is B.

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• 9.

• A.

A

• B.

B

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C

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D

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E

D. D
• 10.

### If |3x| > |4y|, is x > y?(1) x > 0(2) y > 0

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A

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B

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C

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D

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E

A. A
Explanation
Statement (1) x > 0 provides information about the value of x, but does not provide any information about the value of y. Therefore, we cannot determine if x > y based on this statement alone. However, we can determine that x is positive, which means that |3x| will always be greater than 0. This implies that |3x| > |4y| will always be true, regardless of the value of y. Therefore, x > y is not necessarily true based on statement (1) alone.

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