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Difference between revisions of "1956 AHSME Problems/Problem 6"
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==Solution 1==
 
==Solution 1==
  
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Suppose that there are <math>o</math> cows and <math>h</math> chickens. Then there are <math>4o + 2h</math> legs and <math>o + h</math> heads. Then we have <math>(4o + 2h) = 14 + 2(o + h)</math>. This expands to <math>4o + 2h = 14 + 2o + 2h</math>. Canceling <math>2o + 2h</math> from both sides, we get <math>2o = 14</math>, implying that <math>o = 7</math>. Therefore, the answer is <math>\boxed{\textbf{(C)}}</math>, and we are done.
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Let there be <math>x</math> cows and <math>y</math> chickens. Then, there are <math>4x+2y</math> legs and <math>x+y</math> heads. Writing the equation:
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<cmath>4x+2y=14+2(x+y)</cmath> <cmath>4x+2y=14+2x+2y</cmath> <cmath>2x=14</cmath> <cmath>x=\boxed{\textbf{(B) }7}</cmath>
  
 
==See Also==
 
==See Also==
  
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{{AHSME box|year=1956|num-b=5|num-a=7}}
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{{AHSME 50p box|year=1956|num-b=5|num-a=7}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:31, 15 March 2023

Problem 6

In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 14$


Solution 1

Let there be $x$ cows and $y$ chickens. Then, there are $4x+2y$ legs and $x+y$ heads. Writing the equation: \[4x+2y=14+2(x+y)\] \[4x+2y=14+2x+2y\] \[2x=14\] \[x=\boxed{\textbf{(B) }7}\]

See Also

1956 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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

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