1985 AHSME Problems/Problem 4

Problem

A large bag of coins contains pennies, dimes, and quarters. There are twice as many dimes as pennies and three times as many quarters as dimes. An amount of money which could be in the bag is

$\mathrm{(A)\ }$ $ $306 \qquad \mathrm{(B) \ }$ $ $333 \qquad \mathrm{(C)\ }$ $ $342 \qquad \mathrm{(D) \ }$ $ $348 \qquad \mathrm{(E) \ }$ $ $360$

Solution

Let there be $p$ pennies. Therefore, there are $2x$ dimes and $3(2x)=6x$ quarters. Since pennies are $ $0.01$ , dimes are $ $0.10$ , and quarters are $ $0.25$ , the total amount of money is $ $0.01x+(2)(0.10x)+(6)(0.25x)=$ $ $1.71x$ . Therefore, any amount of money must be a multiple of $ $1.71$ .

Since the answer choices are all whole dollar amounts, we multiply by $100$ to get a whole dollar number: $ $171$ . Notice that twice this is $ $342$ , which is answer choice $\boxed{\text{C}}$ .

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1985 AHSME Problems/Problem 4

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