In a tin can, there is a certain number of pencils, 40 percent without erasers, and 35 percent without points. If of the

[BTGmoderatorDC]

In a tin can, there is a certain number of pencils, 40 percent without erasers, and 35 percent without points. If of the pencils 1/6 have no erasers and no points, what fractional part of the pencils have both points and erasers?

A) 11/12
B) 7/12
C) 5/12
D) 1/3
E) 1/4


OA C

Source: Princeton Review

[Jay@ManhattanReview]

In a tin can, there is a certain number of pencils, 40 percent without erasers, and 35 percent without points. If of the pencils 1/6 have no erasers and no points, what fractional part of the pencils have both points and erasers?

A) 11/12
B) 7/12
C) 5/12
D) 1/3
E) 1/4

OA C

Source: Princeton Review

Given the information, there are 40% without erasers, and 35 percent without point pencils. Thus, 40% + 35% = 75% = 3/4 pencils either do not have erasers or do not have point.

Since we are given that 1/6 of the pencils have no erasers and no points; thus,

Of all the pencils, 3/4 – 1/6 = 7/12 either do not have erasers or do not have point or both.

So, the fractional part of the pencils have both points and erasers = 1 – 7/12 = 5/12

The correct answer: C

Hope this helps!

-Jay
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