1965 AMC 12 Problem 32

Question

An article costing

C

dollars is sold for

\$100

at a loss of

x

percent of the selling price. It is then resold at a profit of

x

percent of the new selling price

S'.

If the difference between

S'

and

C

is

1\dfrac19

dollars, then

x

is:

undetermined

\dfrac{80}{9}

10

\dfrac{95}{9}

\dfrac{100}{9}

The first loss makes

C=100+x

The resale condition gives

S'=\frac{10000}{100-x}

A loss of

x\%

of the

\$100

selling price means

C=100+x.

On resale,

S'-100=(\frac{x}{100})S',

so

S'=\frac{10000}{100-x}.

Therefore

\begin{gathered} \frac{10000}{100-x}-(100+x)\\ =\frac{x^2}{100-x} =\frac{10}{9}. \end{gathered}

Thus

9x^2+10x-1000=0,

whose positive root is

x=10.

Therefore, the correct answer is C.

Problem 32 in other years