1971 AHSME Problems/Problem 2

Problem

If $b$ men take $c$ days to lay $f$ bricks, then the number of days it will take $c$ men working at the same rate to lay $b$ bricks, is

$\textbf{(A) }fb^2\qquad \textbf{(B) }b/f^2\qquad \textbf{(C) }f^2/b\qquad \textbf{(D) }b^2/f\qquad \textbf{(E) }f/b^2$

Solution

We can use a modified version of the equation $\text{Distance} = \text{Rate} \times {\text{Time}}$ , which is $\text{Work Done} = \text{Rate of Work} \times{ \text{Time Worked}}$ . In this case, the work done is the number of bricks laid, the rate of work is the number of men working, and the time worked is the number of days. With these definitions, we see from the first equation that $f = b \cdot c$ . If we let $X$ be the number of days it will take $c$ men working at the same rate to $b$ bricks, then we have the equation $b = c \cdot x$ . So, $x = \frac{b}{c}$ . The first equation says that $\frac{f}{c^2} = \frac{b}{c}$ , which leads to $x = \frac{f}{c^2}$ . This doesn't match any of our answer choices though, so we have to fiddle around a bit before we realize that $c = \frac{f}{b}$ , a substitution we can make to see that $x = \frac{b^2}{f}$ . Thus, the answer is $\boxed{\textbf{(D) }b^2/f}$ .

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1971 AHSME Problems/Problem 2

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