Difference between revisions of "1970 AHSME Problems/Problem 35"
(→Solution) |
(→Solution) |
||
(One intermediate revision by the same user not shown) | |||
Line 18: | Line 18: | ||
<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> | ||
− | Because both sides have <cmath>k^2x</cmath>, they cancel out. Similarly, the second equation will become <math>bk^2 = q^2 + 2qk\sqrt{ | + | 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>. |
== See also == | == See also == |
Latest revision as of 04:42, 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 years more, his pension would have been
dollars greater, whereas had he served
years more
, his pension would have been
dollars greater than the original annual pension. Find his annual pension in terms of
and
.
Solution
Note the original pension as , where
is the number of years served. Then, based on the problem statement, two equations can be set up.
Square the first equation to get
Because both sides have
, they cancel out. Similarly, the second equation will become
Then,
can be multiplied to the second equation and
can be multiplied to the first equation so that the left side of both equations becomes
. Finally, by setting the equations equal to each other,
Isolate
to get
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.