# 1970 AHSME Problems/Problem 35

## Problem

A retiring employee receives an annual pension proportional to the square root of the number of years of his service. Had he served $a$ years more, his pension would have been $p$ dollars greater, whereas had he served $b$ years more $(b\ne a)$, his pension would have been $q$ dollars greater than the original annual pension. Find his annual pension in terms of $a,b,p$ and $q$. $\text{(A) } \frac{p^2-q^2}{2(a-b)}\quad \text{(B) } \frac{(p-q)^2}{2\sqrt{ab}}\quad \text{(C) } \frac{ap^2-bq^2}{2(ap-bq)}\quad \text{(D) } \frac{aq^2-bp^2}{2(bp-aq)}\quad \text{(E) } \sqrt{(a-b)(p-q)}$

## Solution

Note the original pension as $k\sqrt{x}$, where $x$ is the number of years served. Then, based on the problem statement, two equations can be set up. $$k\sqrt{x+a} = k\sqrt{x} + p$$ $$k\sqrt{x+b} = k\sqrt{x} + q$$

Square the first equation to get $$k^2x + ak^2 = k^2x + 2pk\sqrt{x} + p^2.$$ Because both sides have $$k^2x$$, they cancel out. Similarly, the second equation will become $bk^2 = q^2 + 2qk\sqrt{x}.$ Then, $a$ can be multiplied to the second equation and $b$ can be multiplied to the first equation so that the left side of both equations becomes $abk^2$. Finally, by setting the equations equal to each other, $$bp^2 + 2bpk\sqrt{x} = aq^2 + 2aqk\sqrt{x}.$$ Isolate $k\sqrt{x}$ to get $\fbox{D} = \frac{aq^2-bp^2}{2(bp-aq)}$.

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