2018 AMC 12B Problems/Problem 6

Problem

Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?

$\textbf{(A) } \frac{4DQ}{S} \qquad \textbf{(B) } \frac{4DS}{Q} \qquad \textbf{(C) } \frac{4Q}{DS} \qquad \textbf{(D) } \frac{DQ}{4S} \qquad \textbf{(E) } \frac{DS}{4Q}$

Solution 1

Each can of soda costs $\frac QS$ quarters, or $\frac{Q}{4S}$ dollars. Therefore, $D$ dollars can purchase $\frac{D}{\left(\tfrac{Q}{4S}\right)}=\boxed{\textbf{(B) } \frac{4DS}{Q}}$ cans of soda.

~MRENTHUSIASM

Solution 2

Note that $S$ is in the unit of $\text{can}.$ On the other hand, $Q$ and $D$ are both in the units of $\text{cost}.$

The units of $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},$ and $\textbf{(E)}$ are $\frac{\text{cost}^2}{\text{can}},\text{can},\frac{1}{\text{can}},\frac{\text{cost}^2}{\text{can}},$ and $\text{can},$ respectively. Since the answer is in the unit of $\text{can},$ we eliminate $\textbf{(A)},\textbf{(C)},$ and $\textbf{(D)}.$ Moreover, it is clear that $D$ dollars can purchase more than $S=\frac{DS}{4Q}$ cans of soda, from which we eliminate $\textbf{(E)}.$ Therefore, the answer is $\boxed{\textbf{(B) } \frac{4DS}{Q}}.$

~MRENTHUSIASM

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AMC 12 Problems and Solutions

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2018 AMC 12B Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See Also