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[GMAT math practice question]

How many triples (a,b,c) of even positive integers satisfy a3 + b2 + c = 50?

A. one
B. two
C. three
D. four
E. five

=>

Consider the variable a first.
Since 4^3 > 50, we can only have a = 2.
If a = 2, then b^2 + c = 42 since a^3 = 2^3 = 8.
Since 8^2 = 64 > 42, we can only have b = 2, 4 or 6.
If b = 2, then b^2 + c = 2^2 + c = 4 + c = 42 and c = 38.
If b = 4, then b^2 + c = 4^2 + c = 16 + c = 42 and c = 26.
If b = 6, then b^2 + c = 6^2 + c = 36 + c = 42 and c = 6.
Thus, there are three possible triples: ( 2, 2, 38 ), ( 2, 4, 26 ) and ( 2, 6, 6 ).

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

If 123,456=123a+87 and 234,567=123b+6, how many multiples of 123 lie between 123,456 and 234,567?

A. a
B. b
C. a+b
D. a-b
E. b-a

=>

Since 123,456 = 123*1003 + 87, we must have a = 1003.
Since 234,567 = 123*1907 + 6, we must have b = 1907.
The multiples of 123 between 123,456 and 234,567 are 123*1004, 123*1005, …, 123*1907.
Thus, 1907 – 1004 + 1 = 1907 – 1003 = b – a multiples of 123 lie between 123,456 and 234,567.

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

How many 4-digit integers have the form abcd, where b is even and d >= 2b?

A. 900
B. 1200
C. 1520
D. 2400
E. 2700

=>

Suppose abcd is a 4-digit number.
There are 9 possible values for a: a = 1, 2, …, 9.
There are 10 possible values of c: c = 0,1,2,…,9.
Since b is even, b can take on the values 0,2,4,6 and 8.
However, the condition d >= 2b limits the possible values of b to 0, 2 and 4.

Case 1 : b = 0 => d = 0, 1, … , 9
The number of possible values of d is 10.
There are 10 * 9 * 10 = 900 4-digit integers with b = 0.

Case 2: b = 2 => d = 4, 5, …, 9
The number of possible values of d is 6.
There are 6 * 9 * 10 = 540 4-digit integers with b = 2.

Case 3: b = 4 => d = 8, 9
The number of possible values of d is 2.
There are 2 * 9 * 10 = 180 4-digit integers with b = 4.
In total, there are 900 + 540 + 180 = 1520 possible 4-digit integers of this form.

Therefore, the answer is C.

Answer: C
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[GMAT math practice question]

The median of 5 numbers is 50, and their range is 40. If the median of the 3 smallest numbers is 40, which of the following could not be the range of the 3 largest numbers?

I. 0
II. 20
III. 40

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

=>

Suppose a, b, c, and d satisfy a ≤ b ≤ 50 ≤ c ≤ d. Since the median of the 3 smallest numbers is 40, we must have b = 40.

The range of the 3 largest numbers is d – 50.

Since we are told that d – a = 40, the maximum range of the 3 largest numbers occurs when a = b = 40.
Then d = 80, and the maximum range is d – 50 = 80 – 50 = 30.

The minimum range of the 3 largest numbers occurs when a = 10 and b = 40.
Then we have c = 50 and d = 50, and the range is d – 50 = 50 – 50 = 0.

The range of the 3 largest numbers lies between 0 and 30, inclusive.
Thus, 0 and 20 are the only possible values.

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

What is the remainder when 7^{100} is divided by 50?

A. 0
B. 1
C. 7
D. 21
E. 49

=>

The remainder when 7^{100} is divided by 50 depends only on the units and tens digits.

The units digits of 7^n cycle through the four values 7, 9, 3, and 1.
The tens digits of 7^n cycle through the four values 0, 4, 4, and 0.

We have the following sequence of units and tens digits for 7^n:

7^1 = 07 ~ 07
7^2 = 49 ~ 49
7^3 = 343 ~ 43
7^4 = 2401 ~ 01
7^5 = 16807~ 07


So, 7^{100} = (7^4)^}{25} has the same units and tens digits as 7^4, that is, 01.
Thus, the remainder when 7^{100} is divided by 50 is 1.

Therefore, B is the answer.

Answer : B
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[GMAT math practice question]

The alarm of a certain clock rings every 15 minutes. If the alarm first rings at 12:15, when will it ring for the 24th time?

A. 17:30
B. 18:00
C. 18:30
D. 19:00
E. 19:30

=>
It takes 1 hour = 4 x 15 minutes for every fourth alarm to ring. The time period starts at fifteen minutes before 12:15, or 12:00. So:

The 4th alarm rings at 13:00.
The 8th alarm rings at 14:00.
The 12th alarm rings at 15:00.
The 16th alarm rings at 16:00.
The 20th alarm rings at 17:00.
The 24th alarm rings at 18:00.

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

If 3 persons are selected at random from 8 persons for 3 positions present, vice-present, and secretary, how many such ways are possible?

A. 56
B. 70
C. 240
D. 336
E. 1680

=>

8P3 = 8 x 7 x 6 = 336

Ans: D
Shouldn't the answer be A?

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[GMAT math practice question]

John invested a total of $100,000 in two accounts. The first account had an annual interest rate of 4%, and the second account had an annual interest rate of 6%. If his investment earnings at the end of the first year were at least $5,000, what was the smallest possible amount that he could have invested at an annual interest rate of 6%?

A. $40,000
B. $42,000
C. $45,000
D. $50,000
E. $54,000

=>

Let x be the principal invested at the 6% interest rate.
Then we must have x*0.06 + ( 100,000 – x )*0.04 ≥ 5,000.
Now,
x*0.06 + ( 100,000 – x )*0.04 ≥ 5,000
=> x*0.06 + ( 4,000 – x *0.04 ) ≥ 5,000
=> x*0.02 + 4,000 ≥ 5,000
=> x*0.02 ≥ 1,000
=> x ≥ 50,000

Therefore, the smallest amount he could have invested at 6% was $50,000.
Therefore, D is the answer.

Answer: D
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[GMAT math practice question] 5.29

The operation # is defined to be multiplication. Which of the following must be true?

Ⅰ. x#1=x
Ⅱ. x#0=0
Ⅲ. x#y=y#x

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

=>

If # is the multiplication operation, then we must have x#1 = x, x#0 = 0 and x#y = y#x.

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

If 3 persons are selected at random from 8 persons for 3 positions present, vice-present, and secretary, how many such ways are possible?

A. 56
B. 70
C. 240
D. 336
E. 1680

=>

8P3 = 8 x 7 x 6 = 336

Ans: D
Shouldn't the answer be A?

Sent from my MotoG3-TE using GMAT Club Forum mobile app

We have 8 choices for president.
After choosing president, 7 people are left.
We have 7 choices for vice-president.
After choosing vice-president, 6 people are left.
We have 6 choices for secretary.
Thus 8*7*6 is the number of the cases
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[GMAT math practice question]

Attachment:
6.1.png
6.1.png [ 8.01 KiB | Viewed 1888 times ]

If the distances between each pair of consecutive ticks are equal, what is the value of x?

A. 2^12
B. 3(2^10)
C. 5(2^10)
D. (2^13)
E. 3(2^12)

=>

The distance between each pair of consecutive ticks is 2^11 – 2^10 = 2*2^10 – 2^10 = 2^10.
Thus, x = 2^11 + 2^10 = 2*2^10 + 2^10 = 3(2^10).

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

When x and n are positive integers, if x^{2n}>(3x)^n, which of the following must be true?

A. x>3
B. n>1
C. x=3
D. n=3
E. x=n

=>

x^{2n}>(3x)^n
=> x^{2n} > 3^nx^n
=> x^n > 3^n
=> x > 3

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

The test scores of students at a school have a distribution that is symmetric about the mean m. 68 percent of the test scores lie within one standard deviation d of the mean. If the average score of students at the school is 65 and the standard deviation is 5 points, what percent of the students at the school score more than 70 points?

A. 10%
B. 12%
C. 14%
D. 16%
E. 18%

Attachment:
6.5.png
6.5.png [ 27.47 KiB | Viewed 1830 times ]

=>

50% – ( 68% / 2 ) = 16%

Therefore, D is the answer.

Answer: D
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[GMAT math practice question]

If the sum of the minimum and the maximum of 7 consecutive odd integers is 114, what is the average (arithmetic mean) of the seven integers?

A. 19
B. 38
C. 57
D. 76
E. 114

=>

Let the 7 consecutive odd integers be 2n-5, 2n-3, 2n-1, 2n+1, 2n+3, 2n+5, and 2n+7. Then their minimum is 2n-5 and their maximum is 2n+7. The sum of the maximum and minimum is ( 2n – 5 ) + ( 2n + 7 ) = 4n + 2 = 114 and so n = 28.
Since the average and the median of consecutive odd integers are equal, the average is 2n + 1 = 57.

Therefore, the answer is C.

Answer: C

Another way to solve this problem is to note that the average of the consecutive odd integers is equal to the average of the two end points. That is, 114 / 2 or 57.
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[GMAT math practice question]

How many integers between 50 and 100, inclusive, are divisible by 2 or 3?

A. 35
B. 37
C. 42
D. 47
E. 52

=>

Let A be the set of integers between 50 and 100 that are divisible by 2.
Let B be the set of integers between 50 and 100 that are divisible by 3.
Let C be the set of integers between 50 and 100 that are divisible by both 2 and 3. This is the same as the set of integers between 50 and 100 that are divisible by 6. Then
A = { 50, 52, …, 100 }
B = { 51, 54, …, 96, 99 }
C = { 54, 60, …, 96 }

The number of elements of the set A is |A| = ( 100 – 50 ) / 2 + 1 = 26.
The number of elements of the set B is |B| = ( 99 – 51 ) / 3 + 1 = 17.
The number of elements of the set C is |C| = ( 96 – 54 ) / 6 + 1 = 8.
Using a Venn diagram, we can see that we need to find |A| + |B| - |C| as the integers in the intersection of sets A and B are counted twice.

Attachment:
6.11.png
6.11.png [ 5.62 KiB | Viewed 1818 times ]

|A| + |B| - |C| = 26 + 17 – 8 = 35.


Therefore, the answer is A.

Answer : A
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[GMAT math practice question]

A college has a soccer club, a tennis club and a basketball club. Students can enroll in only one of these three clubs. The ratio of the number of soccer club members to the number of tennis club members is 2:3. The ratio of the number of tennis club members to the number of basketball club members is 4:5. A total of 350 students have joined one of these three clubs. How many students are enrolled in the soccer club?

A. 80
B. 90
C. 100
D.120
E. 150

=>

Let S, T and B be numbers of members in the soccer club, the tennis club and the basketball club, respectively.
Since S:T = 2:3 = 8:12 and T:B = 4:5 = 12:15, we have S:T:B = 8:12:15.
Let S = 8k, T = 12k and B = 15k.
Then S + T + B = 8k + 12k + 15k = 35k = 350,
and k = 10.
The number of members of the soccer club is
S = 8k = 80.

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

If n is the sum of the first 50 positive integers, what is the greatest prime factor of n?

A. 3
B. 5
C. 7
D. 17
E. 51

=>

1 + 2 + 3 + … + 50 = 50*(50 + 1)/2 = 25*51 = 52*3*17
17 is the greatest prime factor of n.

Therefore, the answer is D.
Answer: D
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