1986 AMC 12 Problem 20

Attempt Problem 20 of the 1986 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 1986 AMC 12 solutions, or check the answer key.

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20.

Suppose

x

and

y

are inversely proportional and positive. If

x

increases by

p\%,

then

y

decreases by

p\%

\frac{p}{1+p}\%

\frac{100}{p}\%

\frac{p}{100+p}\%

\frac{100p}{100+p}\%

An increase by

p\%

multiplies

x

\frac{100+p}{100}

Inverse proportionality divides

y

by that factor; compare the new value with the old one

The new value of

x

\frac{x(100+p)}{100}.

Hence the new value of

y

y\cdot\frac{100}{100+p}.

The fractional decrease is

1-\frac{100}{100+p}=\frac{p}{100+p}.

Expressed as a percentage, this is

\frac{100p}{100+p}\%.

Thus the correct answer is E.

Problem 20 in other years