# Difference between revisions of "2011 AIME I Problems/Problem 1"

AlphaMath1 (talk | contribs) (→Solution) |
Ewcikewqikd (talk | contribs) (→Solution) |
||

Line 12: | Line 12: | ||

<cmath>\frac{42}{5}+\frac{k}{50}=10</cmath> Solving gives <math>k=80</math>. | <cmath>\frac{42}{5}+\frac{k}{50}=10</cmath> Solving gives <math>k=80</math>. | ||

<br>If we substitute back in the original equation we get <math>\frac{m}{n}=\frac{2}{3}</math> so <math>3m=2n</math>. Since <math>m</math> and <math>n</math> are relatively prime, <math>m=2</math> and <math>n=3</math>. Thus <math>k+m+n=80+2+3=\boxed{085}</math>. | <br>If we substitute back in the original equation we get <math>\frac{m}{n}=\frac{2}{3}</math> so <math>3m=2n</math>. Since <math>m</math> and <math>n</math> are relatively prime, <math>m=2</math> and <math>n=3</math>. Thus <math>k+m+n=80+2+3=\boxed{085}</math>. | ||

+ | |||

+ | |||

+ | == Solution 2 == | ||

+ | One might cleverly change the content of both Jars. | ||

+ | |||

+ | Since the end result of both Jars are <math>50\%</math> acid, we can turn Jar A into a 1 gallon liquid with <math>50\%-4(5\%) = 30\%</math> acid | ||

+ | |||

+ | and Jar B into 1 gallon liquid with <math>50\%-5(2\%) =40\%</math> acid. | ||

+ | |||

+ | Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so <math>\dfrac{2}{3}</math> of Jar C will be pour into Jar A. | ||

+ | |||

+ | Thus, <math>m=2</math> and <math>n=3</math>. | ||

+ | |||

+ | <math>\dfrac{30\% + \frac{2}{3} \cdot k\%}{\frac{5}{3}} = 50\%</math> | ||

+ | |||

+ | Solving for <math>k</math> yields <math>k=80</math> | ||

+ | |||

+ | So the answer is <math>80+2+3 = \boxed{85}</math> |

## Revision as of 20:53, 19 March 2011

## Problem 1

Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is acid. From jar C, liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that and are relatively prime positive integers, find .

## Solution

There are L of acid in Jar A. There are L of acid in Jar B. And there are L of acid in Jar C. After transfering the solutions from jar C, there will be

L of solution in Jar A and L of acid in Jar A.

L of solution in Jar B and $\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}\right$ (Error compiling LaTeX. ! Missing delimiter (. inserted).) of acid in Jar B.

Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution.

Add the equations to get
Solving gives .

If we substitute back in the original equation we get so . Since and are relatively prime, and . Thus .

## Solution 2

One might cleverly change the content of both Jars.

Since the end result of both Jars are acid, we can turn Jar A into a 1 gallon liquid with acid

and Jar B into 1 gallon liquid with acid.

Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so of Jar C will be pour into Jar A.

Thus, and .

Solving for yields

So the answer is