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. | + | 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 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:
Solution 1
Let there be 
cows and 
chickens. Then, there are 
legs and 
heads. Writing the equation:
![\[4x+2y=14+2(x+y)\]](http://latex.artofproblemsolving.com/2/5/6/256a8069593109d909461721ce401183a05f1d87.png)
![\[4x+2y=14+2x+2y\]](http://latex.artofproblemsolving.com/3/c/a/3cace345423e82749b9e357fa6b206599d68edc9.png)
![\[2x=14\]](http://latex.artofproblemsolving.com/5/0/a/50ab895b3d9888e3924262a36a3172e35706b162.png)
![\[x=\boxed{\textbf{(B) }7}\]](http://latex.artofproblemsolving.com/5/0/c/50cba5d8fd2abdb073d0b67462c331ce8762ba59.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
