1959 AHSME Problems/Problem 9

A farmer divides his herd of $n$cows among his four sons so that one son gets one-half the herd, a second son, one-fourth, a third son, one-fifth, and the fourth son, $7$ cows. Then $n$ is: $\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 140\qquad\textbf{(D)}\ 180\qquad\textbf{(E)}\ 240$


The first three sons get $\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{19}{20}$ of the herd, so that the fourth son should get $\frac{1}{20}$ of it. But the fourth son gets $7$ cows, so the size of the herd is $n=\frac{7}{\frac{1}{20}} = 140$. Then our answer is $\boxed{C}$, and we are done.

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