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

If there are $x$ pennies, then there are $2x$ dimes and $3(2x) = 6x$ quarters. Since pennies are $$0.01$ , dimes are $$0.10$ , and quarters are $$0.25$ , the total amount of money in the bag is $$ \left(0.01x+(0.10)(2x)+(0.25)(6x)\right) = $1.71x.$ Therefore, the possible amounts of money are precisely the integer multiples of $$1.71$ .

Since the answer choices are all integer numbers of dollars, we multiply by $100$ to deduce that the answer must be an integer multiple of $$171$ . The only such multiple among the answer choices is $$(2 \cdot 171) = \boxed{\text{(C)} \ $342}$ .

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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