1985 AMC 12 Problema 5

Question

¿Qué términos deben eliminarse de la suma

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

para que la suma de los términos restantes sea

1

?

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

y

\frac18

\frac14

and

\frac18

\frac14

y

\frac1{12}

\frac14

and

\frac1{12}

\frac18

y

\frac1{12}

\frac18

and

\frac1{12}

\frac16

y

\frac1{10}

\frac16

and

\frac1{10}

\frac18

y

\frac1{10}

\frac18

and

\frac1{10}

Calcula cuánto excede de

1

la suma completa

Compute how much larger the full sum is than

1

Usa un denominador común de

120

para comparar los pares candidatos

Use a common denominator of

120

to compare the candidate pairs

La suma completa es

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

Por lo tanto, los términos eliminados deben sumar

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

Como

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

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

=\frac{9}{40},

esos son los términos requeridos.

Así, la respuesta correcta es E.

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.

El Problema 5 en otros años