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Thomas bought 12 pencils and erasers altogether and paid $13

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by Max@Math Revolution » Mon Dec 16, 2019 11:57 pm
[GMAT math practice question]

Thomas bought 12 pencils and erasers altogether and paid $132 for them. He bought at least one pencil and at least one eraser. The price of a pencil is $3 more than the price of an eraser, and the number of pencils is greater than the number of erasers. What are the numbers of pencils and erasers?

A. 7, 5
B. 8, 4
C. 9, 3
D. 10, 2
E. 11, 1
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by Max@Math Revolution » Thu Dec 19, 2019 6:07 am
=>

Assume p and e are the numbers of pencils and erasers, respectively and a and b are the prices of a pencil and an eraser, respectively.
Then we have p + e = 12, a = b + 3 and p > e.

We have ap + be = (b + 3)p + b(12 - p) = bp + 3p + 12b – bp = 3p + 12b = 132 or p + 4b = 44.

Since p + e = 12 and p > e, we have p > 12 – p or 6 < p ≤ 12.
Since p = 44 – 4b = 4(11 - b), p is a multiple of 4.
If p = 8, then we have e = 4, b = 9 and a = 12.
If p = 12, then we have e = 0 which does not satisfy the condition.
Therefore, p = 8 and e = 4 is the unique solution.

Therefore, B is the answer.
Answer: B
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by Scott@TargetTestPrep » Wed Jan 01, 2020 6:45 pm
Max@Math Revolution wrote:[GMAT math practice question]

Thomas bought 12 pencils and erasers altogether and paid $132 for them. He bought at least one pencil and at least one eraser. The price of a pencil is $3 more than the price of an eraser, and the number of pencils is greater than the number of erasers. What are the numbers of pencils and erasers?

A. 7, 5
B. 8, 4
C. 9, 3
D. 10, 2
E. 11, 1
We can let p = the number of pencils and c = cost of each pencil. Therefore, there are (12 - p) erasers, and the cost of each eraser is (c - 3). We can create the cost equation:

pc + (12 - p)(c - 3) = 132

pc + 12c - 36 - pc + 3p = 132

12c + 3p = 168

4c + p = 56

p = 56 - 4c

p = 4(14 - c)

We see that p, the number of pencils, must be a multiple of 4, so choice B is the correct answer.

Answer: B

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