In the table, the sum of the numbers in each row, column and diagonal is the same. What is the value of $P+Q+R+S$ ?
(A) 56
(B) 60
(C) 64
(D) 68
(E) 72

| $P$ | 4 | $Q$ |
| :---: | :---: | :---: |
| 10 | 16 | 22 |
| $R$ | 28 | $S$ |
Solution 1

The sum of the numbers in the second row is $10+16+22=48$, so the sum of the numbers in any row, column or diagonal is 48 .

In the first row, $P+4+Q=48$ so $P+Q=44$.

In the third row, $R+28+S=48$ so $R+S=20$.

Therefore, $P+Q+R+S=44+20=64$.

## Solution 2

The sum of the numbers in the second row is $10+16+22=48$, so the sum of the numbers in any row, column or diagonal is 48 .

From the first row, $P+4+Q=48$ so $P+Q=44$.

From the first column, $P+10+R=48$ so $P+R=38$.

Subtracting these two equations gives $(P+Q)-(P+R)=44-38$ or $Q-R=6$.

From one of the diagonals, $R+16+Q=48$ or $Q+R=32$.

Adding these last two equations, $2 Q=38$ or $Q=19$, so $R=32-Q=13$.

Also, $P=44-Q=25$.

From the last row, $13+28+S=48$, or $S=7$.

Thus, $P+Q+R+S=25+19+13+7=64$.

## Solution 3

The sum of all of the numbers in the grid is

$$
P+Q+R+S+10+16+22+28+4=P+Q+R+S+80
$$

But the sum of the three numbers in the second column is $4+16+28=48$, so the sum of the three numbers in each column is 48 .

Thus, the total of the nine numbers in the grid is $3(48)=144$, so $P+Q+R+S+80=144$ or $P+Q+R+S=64$.