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 $\fbox{D} = \frac{aq^2-bp^2}{2(bp-aq)}$ .

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 34	Followed by Problem 35
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1970 AHSME Problems/Problem 35

Problem

Solution

See also