Kumail, a noted sneaker enthusiast, has a collection

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Kumail, a noted sneaker enthusiast, has a collection consisting of sneakers made by Brand A and Brand B. 5/7 of the Brand A sneakers are low-tops, and 1/3 of the Brand A low-tops are running shoes. If Kumail owns 7 more pairs of Brand B sneakers than Brand A sneakers, what is the minimum possible number of pairs of sneakers in his collection?

A. 21
B. 28
C. 36
D. 40
E. 49

Can some experts show me the best solution in this problem?

OA E

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by ErikaPrepScholar » Fri Feb 23, 2018 10:20 am
We'll start from the smallest possible group we know about and work our way up:

We know that 1/3 of the Brand A low-tops are running shoes. For this to work, the number of Brand A low-tops must be divisible by 3.

We know that 5/7 of the Brand A sneakers are low tops. For this to work, the number of Brand A low-tops must be divisible by 5, and the number of Brand A sneakers must be divisible by 7.

Finally we know that Kumail has 7 more pairs of Brand B sneakers than Brand A sneakers. This doesn't tell us anything about the number of Brand A sneakers we need to have.

Okay, we know that the number of Brand A low-tops must be divisible by 3 and 5. The lowest common multiple of 3 and 5 is 15. If 15 is 5/7 of the number of Brand A low-tops, then there must be 21 total Brand A sneakers. Then, Kumail must have 21 + 7 = 28 Brand B sneakers, for 21 + 28 = 49 sneakers in total.
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by Scott@TargetTestPrep » Fri Jun 14, 2019 2:48 pm
BTGmoderatorDC wrote:Kumail, a noted sneaker enthusiast, has a collection consisting of sneakers made by Brand A and Brand B. 5/7 of the Brand A sneakers are low-tops, and 1/3 of the Brand A low-tops are running shoes. If Kumail owns 7 more pairs of Brand B sneakers than Brand A sneakers, what is the minimum possible number of pairs of sneakers in his collection?

A. 21
B. 28
C. 36
D. 40
E. 49

We are given that 5/7 of the Brand A sneakers are low-tops and 1/3 of the Brand A low-tops are running shoes. Thus, the number of Brand A sneakers must be a multiple of 7 x 3 = 21. Since the smallest positive multiple of 21 is 21, the least number of Brand A sneakers is 21.

Since Kumail owns 7 more pairs of Brand B sneakers than those of Brand A, the minimum number of Brand B sneakers is 28. Thus, the minimum possible number of pairs of sneakers in his collection is 21 + 28 = 49.

Answer: E

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by Scott@TargetTestPrep » Fri Jun 14, 2019 2:48 pm
BTGmoderatorDC wrote:Kumail, a noted sneaker enthusiast, has a collection consisting of sneakers made by Brand A and Brand B. 5/7 of the Brand A sneakers are low-tops, and 1/3 of the Brand A low-tops are running shoes. If Kumail owns 7 more pairs of Brand B sneakers than Brand A sneakers, what is the minimum possible number of pairs of sneakers in his collection?

A. 21
B. 28
C. 36
D. 40
E. 49

We are given that 5/7 of the Brand A sneakers are low-tops and 1/3 of the Brand A low-tops are running shoes. Thus, the number of Brand A sneakers must be a multiple of 7 x 3 = 21. Since the smallest positive multiple of 21 is 21, the least number of Brand A sneakers is 21.

Since Kumail owns 7 more pairs of Brand B sneakers than those of Brand A, the minimum number of Brand B sneakers is 28. Thus, the minimum possible number of pairs of sneakers in his collection is 21 + 28 = 49.

Answer: E

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

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