1952 AMC 12 Problem 16

Attempt Problem 16 of the 1952 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 1952 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

If the base of a rectangle is increased by

10\%

and the area is unchanged, then the altitude is decreased by:

9\%

10\%

11\%

11\dfrac19\%

9\dfrac1{11}\%

The new altitude must be divided by the same factor by which the base was multiplied

10\%

base increase multiplies the base by

\frac{11}{10}

To keep the area fixed, the altitude is multiplied by

\frac{10}{11}.

Its fractional decrease is

1-\frac{10}{11}=\frac1{11},

which is

\frac{100}{11}\%=9\dfrac1{11}\%.

Thus, the correct answer is E.

Problem 16 in other years