Difference between revisions of "1951 AHSME Problems/Problem 2"

Revision as of 02:31, 8 February 2009

Problem

A rectangular field is half as wide as it is long and is completely enclosed by $x$ yards of fencing. The area in terms of $x$ is:

$(\mathrm{A})\ \frac{x^2}2 \qquad (\mathrm{B})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}{18} \qquad (\mathrm{E})\ \frac{x^2}{72}$

Solution

Let $w$ be the width. Then $l = 2w$ , and the perimeter is $x = 2(2w)+2w = 6w \implies w = \frac{x}6$ . The area is $wl = w(2w) = 2w^2 = 2\left(\frac{x^2}{36}\right) = \frac{x^2}{18}$ , so the answer is $\mathrm{D}$ .

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

@@ Line 5: / Line 5: @@
 == Solution ==
-By definition of a percent, the answer is <math>\mathrm{(B)}</math>.
+Let <math>w</math> be the width. Then <math>l = 2w</math>, and the perimeter is <math>x = 2(2w)+2w = 6w \implies w = \frac{x}6</math>. The area is <math>wl = w(2w) = 2w^2 = 2\left(\frac{x^2}{36}\right) = \frac{x^2}{18}</math>, so the answer is <math>\mathrm{D}</math>.
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Difference between revisions of "1951 AHSME Problems/Problem 2"

Revision as of 02:31, 8 February 2009

Problem

Solution

See also