Difference between revisions of "1951 AHSME Problems/Problem 2"
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A rectangular field is half as wide as it is long and is completely enclosed by <math>x</math> yards of fencing. The area in terms of <math>x</math> is: | A rectangular field is half as wide as it is long and is completely enclosed by <math>x</math> yards of fencing. The area in terms of <math>x</math> is: | ||
| − | <math>(\mathrm{A})\ \frac{x^2}2 \qquad (\mathrm{B})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}18 \qquad (\mathrm{E})\ \frac{x^2}72</math> | + | <math>(\mathrm{A})\ \frac{x^2}2 \qquad (\mathrm{B})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}{18} \qquad (\mathrm{E})\ \frac{x^2}{72}</math> |
== Solution == | == Solution == | ||
Revision as of 02:26, 8 February 2009
Problem
A rectangular field is half as wide as it is long and is completely enclosed by 
yards of fencing. The area in terms of is:
Solution
By definition of a percent, the answer is 
.
See also
| 1951 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
