Difference between revisions of "1956 AHSME Problems/Problem 6"

(Solution 1)
(Solution 1)
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==Solution 1==
 
==Solution 1==
  
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.<cmath>4x+2y=14+2(x+y)</cmath> <cmath>4x+2y=14+2x+2y</cmath> <cmath>2x=14</cmath> <cmath>x=\boxed{\textbf{(C) }7}</cmath>
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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:
 +
<cmath>4x+2y=14+2(x+y)</cmath> <cmath>4x+2y=14+2x+2y</cmath> <cmath>2x=14</cmath> <cmath>x=\boxed{\textbf{(C) }7}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 16:23, 14 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: \[4x+2y=14+2(x+y)\] \[4x+2y=14+2x+2y\] \[2x=14\] \[x=\boxed{\textbf{(C) }7}\]

See Also

1956 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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