1955 AMC 12 Problem 14

Attempt Problem 14 of the 1955 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1955 AMC 12 solutions, or check the answer key.

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14.

The length of rectangle

R

10

percent more than the side of square

S.

The width of the rectangle is

10

percent less than the side of the square. The ratio of the areas,

R:S,

is:

99:100

101:100

1:1

199:200

201:200

Let the square’s side length be

s

The rectangle’s area is

(1.1s)(0.9s)

If the square side is

s,

then the rectangle has dimensions

1.1s

and

0.9s.

Therefore

\begin{aligned} \frac{[R]}{[S]} &=\frac{(1.1s)(0.9s)}{s^2}\\ &=0.99=\frac{99}{100}. \end{aligned}

Thus, the correct answer is A.

Problem 14 in other years