Difference between revisions of "1956 AHSME Problems/Problem 6"
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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{( | + | 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{(B) }7}</cmath> | ||
==See Also== | ==See Also== |
Latest revision as of 12:31, 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:
Solution 1
Let there be cows and
chickens. Then, there are
legs and
heads. Writing the equation:
See Also
1956 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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