A box contains apple and pears.

An equal number of apples and pears are rotten.

$\frac{2}{3}$ of all of the apples are rotten.

$\frac{3}{4}$ of all of the the pears are rotten.

What fraction of the total number of pieces of fruit in the box is rotten?
(A) $\frac{17}{24}$
(B) $\frac{7}{12}$
(C) $\frac{5}{8}$
(D) $\frac{12}{17}$
(E) $\frac{5}{7}$
Solution 1

Since $\frac{2}{3}$ of the apples are rotten, $\frac{3}{4}$ of the pears are rotten, and the number of rotten apples and pears are equal, then we could try 6 rotten pieces of each type of fruit. (We choose 6 as it is a multiple of the numerator of each fraction.)

If there are 6 rotten apples, then the total number of apples is $\frac{3}{2}(6)=9$.

If there are 6 rotten pears, then the total number of pears is $\frac{4}{3}(6)=8$.

Therefore, there are $9+8=17$ pieces of fruit in total, of which $6+6=12$ are rotten.

Thus, $\frac{12}{17}$ of the fruit are rotten.

## Solution 2

Suppose there are $a$ apples and $p$ pears in total.

Since the number of rotten apples and rotten pears are equal, then $\frac{2}{3} a=\frac{3}{4} p$, so $p=\frac{4}{3}\left(\frac{2}{3} a\right)=\frac{8}{9} a$.

Therefore, the total number of pieces of fruit is $a+p=a+\frac{8}{9} a=\frac{17}{9} a$.

Also, the total number of rotten fruit is $2\left(\frac{2}{3} a\right)=\frac{4}{3} a$, so the fraction of the total amount of fruit that is rotten is $\frac{\frac{4}{3} a}{\frac{17}{9} a}=\frac{4}{3} \cdot \frac{9}{17}=\frac{12}{17}$.

ANSWER: (D)