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

(Solution 3)
(Solution 1)
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Add the equations to get
 
Add the equations to get
 
<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{085hi}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 21:00, 17 September 2020

Problem

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 $k\%$ acid. From jar C, $\frac{m}{n}$ 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 $m$ and $n$ are relatively prime positive integers, find $k + m + n$.

Solution 1

There are $\frac{45}{100}(4)=\frac{9}{5}$ L of acid in Jar A. There are $\frac{48}{100}(5)=\frac{12}{5}$ L of acid in Jar B. And there are $\frac{k}{100}$ L of acid in Jar C. After transferring the solutions from jar C, there will be
$4+\frac{m}{n}$ L of solution in Jar A and $\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}$ L of acid in Jar A.

$6-\frac{m}{n}$ L of solution in Jar B and $\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}$ 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.
\[\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}\] \[\frac{24}{5}-\frac{km}{50n}+\frac{k}{50}=6-\frac{m}{n}\] Add the equations to get \[\frac{42}{5}+\frac{k}{50}=10\] Solving gives $k=80$.
If we substitute back in the original equation we get $\frac{m}{n}=\frac{2}{3}$ so $3m=2n$. Since $m$ and $n$ are relatively prime, $m=2$ and $n=3$. Thus $k+m+n=80+2+3=\boxed{085hi}$.

Solution 2

One might cleverly change the content of both Jars.

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

and Jar B into 1 gallon liquid with $50\%-5(2\%) =40\%$ 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 $\dfrac{2}{3}$ of Jar C will be pour into Jar A.

Thus, $m=2$ and $n=3$.

$\dfrac{30\% + \frac{2}{3} \cdot k\%}{\frac{5}{3}} = 50\%$

Solving for $k$ yields $k=80$

So the answer is $80+2+3 = \boxed{085}$

Solution 3

One may first combine all three jars in to a single container. That container will have $10$ liters of liquid, and it should be $50\%$ acidic. Thus there must be $5$ liters of acid.

Jar A contained $45\% \cdot 4L$, or $1.8L$ of acid, and jar B $48\% \cdot 5L$ or $2.4L$. Solving for the amount of acid in jar C, $k = (5 - 2.4 - 1.8) = .8$, or $80\%$.

Once one knowss that the jar C is $80\%$ acid, use solution 1 to figure out m and n for $k+m+n=80+2+3=\boxed{085}$.

Video Solution

https://www.youtube.com/watch?v=_znugFEst6E&t=919s

~Shreyas S

See also

2011 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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