1970 AMC 12 Problem 35

Attempt Problem 35 of the 1970 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1970 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

35.

A retiring employee receives an annual pension proportional to the square root of the number of years of his service. Had he served aa years more, his pension would have been pp dollars greater, whereas, had he served bb years more (b≠ab\ne a), his pension would have been qq dollars greater than the original annual pension. Find his annual pension in terms of a,a, b,b, p,p, and q.q.
p2−q22(a−b)\dfrac{p^2-q^2}{2(a-b)}
(p−q)22ab\dfrac{(p-q)^2}{2\sqrt{ab}}
ap2−bq22(ap−bq)\dfrac{ap^2-bq^2}{2(ap-bq)}
aq2−bp22(bp−aq)\dfrac{aq^2-bp^2}{2(bp-aq)}
(a−b)(p−q)\sqrt{(a-b)(p-q)}
Answer: D
Concepts:ratio and proportionradicalsystem of equationsalgebraic manipulation
Small hint:
Let the current pension be X=knX=k\sqrt n and write the two hypothetical pensions
Big hint:
Square X+p=kn+aX+p=k\sqrt{n+a} and X+q=kn+bX+q=k\sqrt{n+b}, then eliminate both nn and k2k^2
Solution:
Let the current pension be X=kn.X=k\sqrt n. Since X+p=kn+aX+p=k\sqrt{n+a} and X+q=kn+b,X+q=k\sqrt{n+b}, squaring and using X2=k2nX^2=k^2n gives the first line below. Multiplying its equations by bb and a,a, respectively, and subtracting gives the second: 2pX+p2=k2a,2qX+q2=k2b,2X(bp−aq)=aq2−bp2. \begin{gathered} 2pX+p^2=k^2a,\\ 2qX+q^2=k^2b,\\ 2X(bp-aq)=aq^2-bp^2. \end{gathered} Hence X=aq2−bp22(bp−aq).X=\dfrac{aq^2-bp^2}{2(bp-aq)}.
Therefore, the correct answer is D.
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