Difference between revisions of "1956 AHSME Problems/Problem 6"
Megaboy6679 (talk | contribs) (→Solution 1) |
Megaboy6679 (talk | contribs) (→Solution 1) |
||
Line 8: | Line 8: | ||
==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> | + | 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:
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 | |
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.