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Difference between revisions of "1970 AHSME Problems/Problem 35"

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== Solution ==
 
== Solution ==
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Note the original pension as <math>k\sqrt{x}</math>, where <math>x</math> is the number of years served. Then, based off of the problem statement, two equations can be set up.  
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Note the original pension as <math>k\sqrt{x}</math>, where <math>x</math> is the number of years served. Then, based on the problem statement, two equations can be set up.  
  
 
<cmath>k\sqrt{x+a} = k\sqrt{x} + p</cmath>
 
<cmath>k\sqrt{x+a} = k\sqrt{x} + p</cmath>
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<cmath>k^2x + ak^2 = k^2x + 2pk\sqrt{x} + p^2.</cmath>
 
<cmath>k^2x + ak^2 = k^2x + 2pk\sqrt{x} + p^2.</cmath>
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Because both sides have <cmath>k^2x</cmath>, they cancel out. Similarly, the second equation will become <math>bk^2 = q^2 + 2qk\sqrt{X}.</math> Then, <math>a</math> can be multiplied to the second equation and <math>b</math> can be multiplied to the first equation so that the left side of both equations becomes <math>abk^2<cmath>. Finally, by setting the equations equal to each other, </cmath>bp^2 + 2bpk\sqr{x} = aq^2 + 2aqk\sqrt{X}.</math><math> Isolate </math>k\sqrt{x}<math> to get </math>\fbox{D} = \frac{aq^2-bp^2}{2(bp-aq)}<math>.
+
Because both sides have <cmath>k^2x</cmath>, they cancel out. Similarly, the second equation will become <math>bk^2 = q^2 + 2qk\sqrt{x}.</math> Then, <math>a</math> can be multiplied to the second equation and <math>b</math> can be multiplied to the first equation so that the left side of both equations becomes <math>abk^2</math>. Finally, by setting the equations equal to each other, <cmath>bp^2 + 2bpk\sqrt{x} = aq^2 + 2aqk\sqrt{x}.</cmath> Isolate <math>k\sqrt{x}</math> to get <math>\fbox{D} = \frac{aq^2-bp^2}{2(bp-aq)}</math>.
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</math>\fbox{D}$
 
  
 
== See also ==
 
== See also ==

Latest revision as of 03:42, 2 June 2021

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

See also

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

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