Which terms must be removed from the sum 12+14+16+18+110+112 if the sum of the remaining terms is to equal 1?

1985 AMC 12 Problem 5

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Which terms must be removed from the sum

\frac12+\frac14+\frac16+\frac18+\frac1{10}+\frac1{12}

if the sum of the remaining terms is to equal

1?

\frac14

and

\frac18

\frac14

and

\frac1{12}

\frac18

and

\frac1{12}

\frac16

and

\frac1{10}

\frac18

and

\frac1{10}

Compute how much larger the full sum is than

1

Use a common denominator of

120

to compare the candidate pairs

The full sum is

\begin{aligned} \frac{\substack{60+30+20\\{}+15+12+10}}{120} &=\frac{147}{120}. \end{aligned}

The removed terms must therefore total

\frac{27}{120}=\frac{9}{40}.

Since

\frac{1}{8}+\frac{1}{10}

=\frac{5}{40}+\frac{4}{40}

=\frac{9}{40},

those are the required terms.

Thus the correct answer is E.