If two distinct positive divisors of 64 are randomly

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Veritas Prep

If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?

A. 1/3
B. 1/2
C. 5/7
D. 10/21
E. 15/28

OA D

[Brent@GMATPrepNow]

Veritas Prep

If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?

A. 1/3
B. 1/2
C. 5/7
D. 10/21
E. 15/28

OA D

Here are the positive divisors of 64: {1, 2, 4, 8, 16, 32, 64}

We can see that, as long as we DON'T select the 32 or 64, then the sum of the two numbers is guaranteed to be less than 32
So, P(sum is less than 32) = P(neither is a 64 nor the 32 is selected)
= P(no 64 or 32 is selected on the 1st try AND no 64 or 32 is selected on the 2nd try)
= P(no 64 or 32 is selected on the 1st try) x P(no 64 or 32 is selected on the 2nd try)
= 5/7 x 4/6
= 20/42
= 10/21

Answer: D

Cheers,
Brent

[Scott@TargetTestPrep]

Veritas Prep

If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?

A. 1/3
B. 1/2
C. 5/7
D. 10/21
E. 15/28

OA D

The positive divisors of 64 are:

1, 2, 4, 8, 16, 32, and 64

If the sum of two distinct divisors of 64 is less than 32, each has to be less than 32. Therefore, they are 1, 2, 4, 8, and 16. The number of ways to select 2 divisors from these 5 divisors is 5C2 = (5 x 4)/2 = 10.

Since the total number of ways to select 2 divisors from all 7 divisors is 7C2 = (7 x 6)/2 = 21, the probability is 10/21.

Answer: D