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Michael arranged all his books in a bookcase with 10 books

Expert replies
by BTGmoderatorDC » Wed Sep 19, 2018 6:04 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.
(2) Before Michael acquired the 10 additional books, he had more than 24 books.

OA A

Source: Official Guide
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Source: — Data Sufficiency |

by Jay@ManhattanReview » Wed Sep 19, 2018 9:07 pm
BTGmoderatorDC wrote:Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.
(2) Before Michael acquired the 10 additional books, he had more than 24 books.

OA A

Source: Official Guide
Say the bookcase has x numbers of shelves. Thus, the number of books = 10x

After the addition of 10 more books, Michael has 10x + 10 books that can be arranged such that there are 12 books in each shelf.

=> 10x + 10 = 10(x + 1) is completely divisible by 12

Let's take each statement one by one.

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

=> 10x < 96

x: {1, 2, ..., 9}

Since 10(x + 1) is completely divisible by 12, among x: {1, 2, ..., 9}, only x = 5 qualifies.

Thus, Michael had 10x = 50 books before he acquired the 10 additional books. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

=> 10x > 24

x: {3, 4, 5, ..., 11, 12, ...}

Since 10(x + 1) is completely divisible by 12, among x: {3, 4, 5, ..., 11, 12, ...}, many values of x such as 5 and 11 qualify. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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