2006 CEMC Fermat Problems/Problem 8

Problem

If Corina had added the numbers $P$ and $Q$ correctly, the answer would have been $16$ . By mistake, she subtracted $Q$ from $P$ . Her answer was $4$ . What is the value of $P$ ?

$\text{ (A) }\ 4 \qquad\text{ (B) }\ 5 \qquad\text{ (C) }\ 8 \qquad\text{ (D) }\ 10 \qquad\text{ (E) }\ 16$

Solution 1

We can set up equations using $P$ and $Q$ :

$P + Q = 16$

$P - Q = 4$

From the second equation, we have:

$P = 4 + Q$

$Q = P - 4$

Plugging that into the first equation, we get:

$P + P - 4 = 16$

$2P - 4 = 16$

$2P = 20$

$P = \boxed {\textbf {(D) } 10}$

We can also check our answer:

If $P = 10$ , then $Q = P - 4 = 10 - 4 = 6$

Plugging this into the equations, we have:

$10 + 6 = 16 = 16$

$10 - 6 = 4 = 4$

This works out, so our answer for $P$ is correct.

~anabel.disher

Solution 1.5

We can use elimination to solve the equations:

$P + Q = 16$

$P - Q = 4$

We can notice one equation has $Q$ while the other has $-Q$ , so we can add them:

$P + Q + P - Q = 16 + 4$

Simplifying both sides, we get: