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Set S consists of all the positive multiples of 5 that are l

Expert replies
by BTGmoderatorDC » Thu Feb 14, 2019 8:26 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty

Set S consists of all the positive multiples of 5 that are less than K, and K is a positive integer not divisible by 5. The mean of Set S is not divisible by 5. Let N be the number of members of the set. N is not divisible by 5. What does N equal?

(1) N < 52
(2) K/5 > 48

OA C

Source: Magoosh
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Source: — Data Sufficiency |

by Jay@ManhattanReview » Fri Feb 15, 2019 1:36 am
BTGmoderatorDC wrote:Set S consists of all the positive multiples of 5 that are less than K, and K is a positive integer not divisible by 5. The mean of Set S is not divisible by 5. Let N be the number of members of the set. N is not divisible by 5. What does N equal?

(1) N < 52
(2) K/5 > 48

OA C

Source: Magoosh
Given:

1. Set S consists of all the positive multiples of 5 such that they are less than K, and K is a positive integer not divisible by 5.

For example say, K = 21, then set S:{ 5, 10, 15, 20}; however, if say K = 52, then set S:{ 5, 10, 15, 20, 25, 30, 35, 40, 45, 50}

2. The mean of Set S is not divisible by 5.

Since Set S is a multiple of 5 (equally spaced set), its median and mean would be equal. Given that all the terms of the set are multiple of 5; however mean (median) is not, we must have an even number of terms is the set. For example say, K = 21, then set S:{ 5, 10, 15, 20}; its median = average of 10 and 15 = 12.5, not a multiple of 5.

Thus, N is even.

We have to get the value of N.

Let's take each statement one by one.

(1) N < 52

N can be any even, non-multiple of 5 number such as 48, 46, 44, 42, 38, 36, etc. It will depend on the value of K. Since the value of K is not known, we can't get the unique value of K. Insufficient.

(2) K/5 > 48

=> N can be 48, 52, 54, 56, 58, 62, etc. No unique va;ue fo N. Insufficient.

(1) and (2) together

From (1) and (2), we have only one common value of N, i.e., N = 48. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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