Difference between revisions of "1959 AHSME Problems/Problem 9"
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== Problem == | == Problem == | ||
| − | A farmer divides his herd of <math>n</math>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, <math>7</math> cows. Then <math>n</math> is: <math>\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 140\qquad\textbf{(D)}\ 180\qquad\textbf{(E)}\ 240</math> | + | A farmer divides his herd of <math>n</math> 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, <math>7</math> cows. Then <math>n</math> is: <math>\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 140\qquad\textbf{(D)}\ 180\qquad\textbf{(E)}\ 240</math> |
== Solution == | == Solution == | ||
| − | The first three sons get <math>\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{19}{20}</math> of the herd, so that the fourth son should get <math>\frac{1}{20}</math> of it. But the fourth son gets <math>7</math> cows, so the size of the herd is <math>n=\frac{7}{\frac{1}{20}} = 140</math>. Then our answer is <math>\ | + | The first three sons get <math>\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{19}{20}</math> of the herd, so that the fourth son should get <math>\frac{1}{20}</math> of it. But the fourth son gets <math>7</math> cows, so the size of the herd is <math>n=\frac{7}{\frac{1}{20}} = 140</math>. Then our answer is <math>\fbox{\textbf{(C) }140}</math>. |
| + | |||
| + | ==See also== | ||
| + | {{AHSME 50p box|year=1959|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 11:08, 21 July 2024
Problem
A farmer divides his herd of 
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, 
cows. Then is: 
Solution
The first three sons get 
of the herd, so that the fourth son should get 
of it. But the fourth son gets cows, so the size of the herd is 
. Then our answer is 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
