1952 AHSME Problems/Problem 38

Problem

The area of a trapezoidal field is $1400$ square yards. Its altitude is $50$ yards. Find the two bases, if the number of yards in each base is an integer divisible by $8$ . The number of solutions to this problem is:

$\textbf{(A)}\ \text{none} \qquad \textbf{(B)}\ \text{one} \qquad \textbf{(C)}\ \text{two} \qquad \textbf{(D)}\ \text{three} \qquad \textbf{(E)}\ \text{more than three}$

Solution

Let us denote $8m$ and $8n$ to be our bases. Without loss of generality, $m \le n$

Thus, $50 * \frac{8m + 8n}{2} = 1400$ $4m + 4n = 28$ $m + n = 7$

Since $m$ & $n$ are integers, we see that the only solutions to this equation are $(1,6)$ , $(2,5)$ , and $(3,4)$ . Therefore, the answer is $\fbox{(D) three}$

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 37	Followed by Problem 39
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1952 AHSME Problems/Problem 38

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