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Difference between revisions of "1959 AHSME Problems/Problem 9"

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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>
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== Problem ==
  
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==Solution==
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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>
  
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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>\boxed{C}</math>, and we are done.
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== Solution ==
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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>.
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==See also==
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{{AHSME 50p box|year=1959|num-b=8|num-a=10}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 12:08, 21 July 2024

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

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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