1959 AHSME Problems/Problem 9

Problem

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: $\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 140\qquad\textbf{(D)}\ 180\qquad\textbf{(E)}\ 240$

Solution

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 $\fbox{\textbf{(C) }140}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1959 AHSME Problems/Problem 9

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