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, let $m \le n$.

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 $\fbox{(D) three}$

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 37
Followed by
Problem 39
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