In the addition shown, each of $P, Q$ and $R$ is a digit.

$$
\begin{array}{r}
P Q R \\
+\quad Q R \\
\hline 1012
\end{array}
$$

The value of $P+Q+R$ is
(A) 12
(B) 15
(C) 13
(D) 22
(E) 20
Solution 1

Beginning with the units (ones) column, the units digit of the sum $R+R$ is 2 .

Thus, $R=1$ or $R=6$ are the only possibilities $(1+1=2$ and $6+6=12$ have units digit 2 ).

If $R=1$, then there is no 'carry' from the units column to the tens column.

In this case, the units digit of the $\operatorname{sum} Q+Q$ is 1 .

This is not possible since $Q+Q=2 Q$ which is an even number for all possible digits $Q$.

Thus, $R=6$, which gives $R+R=12$, and so the 'carry' from the units column to the tens column is 1 .

In the tens column the units digit of the sum $Q+Q+1$ is 1 , and so the units digit of $Q+Q$ is 0 . Thus, $Q=0$ or $Q=5$ are the only possibilities $(0+0$ and $5+5$ have units digit 0$)$.

If $Q=0$, then there is no 'carry' from the tens column to the hundreds column.

This is not possible since the sum of the hundreds column is 10 and $P$ is a digit and so cannot be 10 .

Thus, $Q=5$, which gives $Q+Q+1=11$, and so the 'carry' from the tens column to the hundreds column is 1 .

The sum of the hundreds column is 10 , and so $P+1=10$ or $P=9$.

The value of $P+Q+R$ is $9+5+6=20$.

Solution 2

$Q R$ is a 2-digit number and thus less than 100 .

The sum of $P Q R$ and $Q R$ is greater than 1000, and so $P Q R$ must be greater than 900 which means that $P=9$.

Since the sum of the hundreds column is 10 , the 'carry' from the tens column must be 1 .

Considering the tens and units (ones) columns together, we see that the result of $Q R+Q R$ has units (ones) digit 2, tens digit 1, and hundreds digit 1 (since there is a 'carry' of 1 to the hundreds column). That is, $Q R+Q R=112$ and so $Q R=56$.

The value of $P+Q+R$ is $9+5+6=20$.

ANSWER: (E)