[GMAT math practice question]

When the positive integer n is divided by 8, the remainder is 3, and when n is divided by 5, the remainder is 2. What is the remainder when the smallest possible value of n is divided by 6?

A. 0
B. 1
C. 2
D. 3
E. 4

=>

If n has remainder 3 when it is divided by 8, then n = 8a + 3 for some integer a. The possible positive integer values of n are 3, 11, 19, 27, 35, … .
If n has remainder 2 when it is divided by 5, then n = 5b + 2 for some integer b. The possible positive integer values of n are 2, 7, 12, 17, 22, 27, 32, … .
The smallest value of n that appears in both lists is 27. Therefore, the smallest possible value for n is 27.
27 = 6*4 + 3, 27 has remainder 3 when it is divided by 6.

Therefore, the answer is D.
Answer: D
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[GMAT math practice question]

The terms of the sequence {An} satisfy A2 - A1 = 2, A3 - A2 = 5 and A4 - A3 = 10. Which of the following could be the formula for An+1 - An, in terms of n?

A. n + 1
B. n^2 + 1
C. n^2 - 1
D. n - 1
E. n^2 + 3

=>

We try each possibility until we find a formula that works.
A): If An+1 - An = n + 1, then A2 - A1 = 1 + 1 = 2, which is correct, but
A3 - A2 = 2 + 1 = 3 ≠ 5.
So, A) is not the answer.
B): If An+1 - An = n^2 + 1, then
A2 - A1 = 1^2 + 1 = 2,
A3 - A2 = 2^2 + 1 = 5,
A4 - A3 = 3^2 + 1 = 10.
So, this formula is possible.

Therefore, the answer is B.
Answer: B
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[GMAT math practice question]

If the sequence {An} satisfies An = An-1 - An-2, A1 = 0, and A2 = 1, where n is an integer greater than 2, then what is the sum of the first 100 terms of {An}?

A. 1
B. 2
C. 3
D. 4
E. 5

=>

We determine the period of the sequence by examining its terms:
A1 = 0 and A2 = 1.
A3 = A2 – A1 = 1 – 0 = 1
A4 = A3 – A2 = 1 – 1 = 0
A5 = A4 – A3 = 0 – 1 = -1
A6 = A5 – A4 = -1 – 0 = -1
A7 = A6 – A5 = -1 – (-1) = 0
A8 = A7 – A6 = 0 – (-1) = 1
A9 = A8 – A7 = 1 – 0 = 1
A10 = A9 – A8 = 1 – 1 = 0
…

The sequence has period 6.
The sum of the first six terms is A1 + A2 + … A6 = 0 + 1 + 1 + 0 + (-1) + (-1) = 0.
We apply this fact to determine the sum of the first 100 terms of the sequence:
A1 + A2 + … A100
= ( A1 + A2 + … A6 ) + … + ( A91 + A92 + … A96 ) + A97 + A98 + A99 + A100
= 0 + … + 0 + A97 + A98 + A99 + A100
= A97 + A98 + A99 + A100
= 0 + 1 + 1 + 0 = 2.

Therefore, the answer is B.
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[GMAT math practice question]

If n is the product of 5 different prime numbers, how many factors does n have?

A. 2
B. 4
C. 8
D. 16
E. 32

=>

Since p, q, r, s, t are different prime factors of n, we have n = p*q*r*s*t = p^1q^1r^1s^1t^1.
The number of factors of n is (1+1)(1+1)(1+1)(1+1)(1+1) = 2*2*2*2*2 = 32.

Therefore, the answer is E.
Answer: E
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[GMAT math practice question]

A certain characteristic in a large population has a distribution that is symmetric about the mean m. 68 percent of the distribution lies within one standard deviation d of the mean. The average score of students in a certain class is 72 pts. 84% of students in the class score less than 80 pts. What is the standard deviation of the students’ scores?

A. 1
B. 2
C. 4
D. 8
E. 16

=>



80 – 72 = 8 is the standard deviation of the distribution.


Therefore, D is the answer.
Answer: D
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[GMAT math practice question]

<x> is the largest integer less than or equal to x and [x] is the smallest integer greater than or equal to x. If x is not an integer, which of the following could be the value of [x] - <x>?

A. 1
B. 2
C. 3
D. 4
E. 5

=>

<x> can be defined in the following way:
If m ≤ x < m + 1, where m is an integer, then < x > = m.

[x] can be defined in the following way:
If n – 1 < x ≤ n, where n is an integer, then [x] = n.

Since x is not an integer, there is an integer n such that n – 1 < x < n.
So, <x> = n -1 and [x] = n.
Thus, [x] - <x> = n – ( n – 1 ) = 1.

We can also solve this problem by plugging in numbers. If x = 1.1, then
[1.1] - <1.1> = 2 – 1 = 1.

Therefore, the answer is A.
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[GMAT math practice question]

If n is an integer, which of the following must also be an integer?

I. n(n+1)/2
II. n(n+1)(n+2)/6
III. n(n+1)(n+2)(n+3)/8

A. I only
B. II only
C. III only
D. I and II
E. I, II and III

=>

Statement I
Since n and n + 1 are consecutive integers, n(n+1) is a multiple of 2.
Thus, n(n+1)/2 is an integer.

Statement II
Since n and n + 1 are consecutive integers, n(n+1) and n(n+1)(n+2) are multiples of 2.
Since n, n + 1 and n + 2 are three consecutive integers, n(n+1)(n+2) is a multiple of 3.
Thus, n(n+1)(n+2) is a multiple of 6, and n(n+1)(n+2) / 6 is an integer.

Statement III
Since n and n + 1 are two consecutive integers, n(n+1) is a multiple of 2.
Similarly, (n+2)(n+3) is a multiple of 2.
Also, either n and n + 2 or n + 1 and n + 3 are consecutive even integers. Thus, either (n + 1)(n+3) is a multiple of 8 or n(n+2) is a multiple of 8 since one of them is a multiple of 4.
It follows that n(n+1)(n+2)(n+3) is a multiple of 8, and n(n+1)(n+2)(n+3)/8 is an integer.

Therefore, the answer is E.

Answer: E
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[GMAT math practice question]

If An=An-1/An-2 (n3), A1=1, and A2=-2, then A100=?

A. 1
B. -1
C. 2
D. -2
E. -1/2

=>
A1 = 1, A2 =-2.
A3 = A2 / A1 = (-2)/1 = -2.
A4 = A3 / A2 = (-2)/(-2) = 1.
A5 = A4 / A3 =1/(-2) = -1/2
A6 = A5 / A4 = 1/2 / 1 = -1/2
A7 = A6 / A5 = -(1/2) /-1/2 = 1 = A1.
A8 = A7 / A6 = 1 / (-1/2) = -2= A2.
Thus, every 6th term is the same. That is,
A1 = A7 = A13 = … = 1.
A2 = A8 = A14 = … =-2.
A3 = A9 = A15 = … = -2.
A4 = A10 = A16 = … =1.
A5 = A11 = A17 = … = -(1/2).
A6 = A12 = A18 = … = -(1/2).
Now, 100 = 6 * 16 + 4. Thus,
A100 = A4 = 1.
Therefore, the answer is A.
Answer: A
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[GMAT math practice question]

Alice travels 120 km, the first 40 km at x km/h and the remaining 80 km at y km/h. What is her average speed for the entire trip?

A. 2xy/(2x+y)
B. 2xy/(3x+y)
C. 3xy/(2x+y)
D. xy/(2x+y)
E. xy/(x+2y)

=>

The average speed is (Total Distance) / (Total Time).
The time for the first 40 km is 40 / x and the time for the remaining 80 km is 80 / y. Thus, the total time for the trip is ( 40/x + 80/y ).
Her average speed for the trip is
120 / ( 40/x + 80/y ) = 120xy / ( 40y + 80x ) = 3xy / ( y + 2x ) = 3xy /
(2x +y).

Therefore, the answer is C.
Answer : C
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[Math Revolution GMAT math practice question]

A piece of cardboard measures 30 cm by 40 cm. We can make two different right circular cylinders by rolling the cardboard along its length and by rolling the cardboard along its width. What is the difference in their volumes?

A. 1000 / π
B. 2000 / π
C. 3000 / π
D. 4000 / π
E. 5000 / π

=>



Case 1: 2πr = 30 and h = 40
r = 15/π
The volume is πr^2*h = π(15/π)^2 * 40 = ( 225 * 40 ) / π = 9000 / π.

Case 2: 2πr = 40 and h = 30
r = 20/π
The volume is πr^2*h = π(20/π)^2 * 30 = ( 400 * 30 ) / π = 12000 / π.

Thus, the difference in the two volumes is 12000 / π - 9000 / π = 3000 / π.


Therefore, C is the answer.
Answer: C
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[Math Revolution GMAT math practice question]

There are 3 ways to make the number 12 using products of two positive integers. These are 1*12, 2*6, and 3*4. In how many ways can 2700 be written as the product of two positive integers?

A. 6
B. 12
C. 15
D. 18
E. 36

=>

2700 = 2^2*3^3*5^2
The number of distinct factors of 2700 is (2+1)(3+1)(2+1) = 36.
Since the order of multiplication does not matter (i.e. 30 * 90 = 90*30), the number of pairs of positive integers that multiply to give 2700 is 36/2 = 18.

Therefore, the answer is D.
Answer: D
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[Math Revolution GMAT math practice question]

If x and y are integers such that (xy)^2 + x^2 – 2xy – 2x + 2 = 0, what is the value of y?

A. -2
B. -1
C. 0
D. 1
E. 2

=>

(xy)^2 + x^2 – 2xy – 2x + 2 = 0
=> (xy)^2 – 2xy + 1 + x^2 – 2x + 1 = 0
=> (xy – 1)^2 + (x – 1)^2 = 0
=> xy = 1 and x = 1.
Thus , x = 1 and y = 1.

Therefore, the answer is D.
Answer: D
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[Math Revolution GMAT math practice question]

What is the remainder when 7^8 is divided by 100?

A. 1
B. 2
C. 3
D. 4
E. 5

=>

The remainder when 7^8 is divided by 100 is equal to the final two digits of 7^8.
Now, 7^1 = 7, 7^2 = 49, 7^3 = 343, and 7^4 = 2401.
So, the final two digits of 7^n have period 4:
The tens digits are 0 -> 4 -> 4 -> 0
and the units digits are 7 -> 9 -> 3 -> 1.
It follows that the tens and units digits of 7^8 are 0 and 1, respectively.
Therefore, the remainder when 7^8 is divided by 100 is 1.

Therefore, the answer is A.
Answer : A
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[Math Revolution GMAT math practice question]

The population of state X increases by 10 percent every 10 years. If the population in 1990 was P million, how much larger is the population in 2010 than the population in 1990?

A. 0.1*P
B. 1.1*P
C. 0.01*P
D. 0.11*P
E. 0.21*P

=>

The population in 2010 is (1.1)^2P and the difference between the populations in 2010 and in 1990 is (1.1)^2P – P = ((1.1)^2 – 1)P = (1.1+1)(1.1-1)P = 2.1*0.1P = 0.21P.

Therefore, the answer is E.
Answer: E
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[Math Revolution GMAT math practice question]

If m=-1 and n = 1^2 + 2^2 + … + 10^2, what is the value of m^n+m^{n+1}+m^{n+2}+m^{n+3}?

A. -2
B. -1
C. 0
D. 1
E. 2

=>

m^n+m^{n+1}+m^{n+2}+m^{n+3}
= m^n(1+m^1+m^2+m^3)
= (-1)^n(1+(-1)^1+(-1)^2+(-1)^3)
= (-1)^n* 0 = 0
Whatever the value of n is, the answer is 0.

Therefore, C is the answer.

Answer: C
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[Math Revolution GMAT math practice question]

When 2 numbers are selected from the integers from 1 to 21 inclusive, what is the probability that the 2 selected numbers are prime numbers?

A. 2/15
B. 1/5
C. 1/3
D. 4/15
E. 2/5

=>

There are 8 prime numbers between 1 and 21, inclusive: 2, 3, 5, 7, 11, 13, 17 and 19.
So, there are 8C2 ways of selecting 2 numbers from these 8 prime numbers.
There are 21C2 ways of selecting 2 numbers from the 21 numbers from 1 to 21, inclusive.
Thus, the probability that the 2 selected numbers are prime numbers is 8C2 / 21C2 = ( 8*7 / 1*2) / ( 21*20 / 1*2 ) = 8*7 / 21*20 = 2/15.

Therefore, the answer is A.

Answer: A
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[Math Revolution GMAT math practice question]

If a=2, b=-2, and c=3, what is the value of a^4b^{-3}c^0?

A. -2
B. -1
C. 0
D. 2
E. 3

=>

a^4b^{-3}c^0
= 2^4*(-2)^{-3}*3^0
= -(2^4*2^{-3}*1)
= -2

Therefore, the answer is A.

Answer: A
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[Math Revolution GMAT math practice question]

If (n+2)!/n!=90, then n=?

A. 8
B. 9
C. 10
D. 11
E. 12

=>

(n+2)!/n!= (n+2)(n+1) = 90 = 10*9.
Thus, n + 2 = 10 or n = 8.

Therefore, the answer is A.
Answer : A
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[Math Revolution GMAT math practice question]

If the range of set A is 5 and the range of set B is 11, which of the following CANNOT be the range of sets A and B combined?

A. 10
B. 11
C. 12
D. 13
E. 14

=>

For example, if A = {0,5} and B = { 0, 11 }, then the set A is included in the set B and the smallest possible range of the sets A and B combined is 11, which is the range of the set B.
Thus, any number less than 11 cannot be the range of sets A and B, combined.

Therefore, A is the answer.
Answer: A
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[Math Revolution GMAT math practice question]

n is a positive integer less than 50. When n is divided by 6, the remainder is 2, and when n is divided by 7, the remainder is 6. What is the remainder when n is divided by 8?

A. 1
B. 2
C. 3
D. 4
E. 5

=>

The positive integers less than 50 that have a remainder of 2 when they are divided by 6 are n = 2, 8, 14, 20, 26, 32, 38 and 44.
The positive integers less than 50 that have a remainder of 6 when they are divided by 7 are n = 6, 13, 20, 27, 34, 41 and 48.
The only integer in both lists is n = 20.
20 has a remainder of 4 when it is divided by 8.

Therefore, the answer is D.
Answer: D
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