1972 AHSME Problems/Problem 15


A contractor estimated that one of his two bricklayers would take $9$ hours to build a certain wall and the other $10$ hours. However, he knew from experience that when they worked together, their combined output fell by $10$ bricks per hour. Being in a hurry, he put both men on the job and found that it took exactly 5 hours to build the wall. The number of bricks in the wall was

$\textbf{(A) }500\qquad \textbf{(B) }550\qquad \textbf{(C) }900\qquad \textbf{(D) }950\qquad  \textbf{(E) }960$


Every hour, the contractors lay $x/9$ and $x/10$ bricks, assuming a total of $x$ bricks in the wall. Together, $x/5$ bricks are laid per hour: \[x = 5(x/9 + x/10 - 10)\] There are a total of $x=900$ bricks, and the answer is $\textbf{(C)} \quad 900$.

-edited by coolmath34

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