Difference between revisions of "2002 AMC 12P Problems/Problem 4"
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== Solution 2== | == Solution 2== | ||
| − | The only tricky part about this equation is the fact that the left-hand side has fractions. Multiplying both sides by <math>b(b+10a)</math> gives us <math>2ab+10a^2+10b^2=2b^2+20ab.</math> Moving everything to the left-hand side and dividing by <math>2</math> gives <math>5a^2-4b^2 -9ab,</math> which factors into <math>(5a-4b)(a-b)=0.</math> Because <math>a \neq b, 5a=4b \implies \frac{a}{b}=0.8</math> giving us our answer of <math>\boxed{\textbf{(E) } 0.8}.</math> | + | The only tricky part about this equation is the fact that the left-hand side has fractions. Multiplying both sides by <math>b(b+10a)</math> gives us <math>2ab+10a^2+10b^2=2b^2+20ab.</math> Moving everything to the left-hand side and dividing by <math>2</math> gives <math>5a^2-4b^2 -9ab,</math> which factors into <math>(5a-4b)(a-b)=0.</math> Because <math>a \neq b, 5a=4b \implies \frac{a}{b}=0.8</math> giving us our answer of <math>\boxed{\textbf{(E) } 0.8}</math> for the AMC 12P and <math>\boxed{\textbf{(C) } 0.8}</math> for the AMC 10P. |
=== Note === | === Note === | ||
Latest revision as of 16:40, 15 July 2024
- The following problem is from both the 2002 AMC 12P #4 and 2002 AMC 10P #10, so both problems redirect to this page.
Contents
Problem
Let 
and 
be distinct real numbers for which
![\[\frac{a}{b} + \frac{a+10b}{b+10a} = 2.\]](http://latex.artofproblemsolving.com/3/0/3/303291e61ebd979d00d2fb126fd207adb3df898b.png)
Find 
Solution 1
For sake of speed, WLOG, let 
. This means that the ratio will simply be because 
Solving for with some very simple algebra gives us a quadratic which is 
Factoring the quadratic gives us 
. Therefore, 
or 
However, since and must be distinct, cannot be 
so the latter option is correct, giving us our answer of 
for the AMC 12P and 
for the AMC 10P.
Solution 2
The only tricky part about this equation is the fact that the left-hand side has fractions. Multiplying both sides by 
gives us 
Moving everything to the left-hand side and dividing by 
gives 
which factors into 
Because 
giving us our answer of for the AMC 12P and for the AMC 10P.
Note
For some unknown reason, the answer choices for the 2002 AMC 10P are different from the answer choices for the 2002 AMC 12P, even though the question is exactly the same. Indeed, 
is the correct answer choice for the 2002 AMC 10P.
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
| 2002 AMC 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
