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

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== Problem 1 ==
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== 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 <math>k\%</math> acid.  From jar C, <math>\frac{m}{n}</math> 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 <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>k + m + n</math>.
 
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 <math>k\%</math> acid.  From jar C, <math>\frac{m}{n}</math> 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 <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>k + m + n</math>.
  
 
== Solution 1==
 
== Solution 1==
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There are <math>\frac{45}{100}(4)=\frac{9}{5}</math> L of acid in Jar A. There are <math>\frac{48}{100}(5)=\frac{12}{5}</math> L of acid in Jar B. And there are <math>\frac{k}{100}</math> L of acid in Jar C. After transfering the solutions from jar C, there will be
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Jar A contains <math>\frac{11}{5}</math> liters of water, and <math>\frac{9}{5}</math> liters of acid; jar B contains <math>\frac{13}{5}</math> liters of water and <math>\frac{12}{5}</math> liters of acid.
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The gap between the amount of water and acid in the first jar, <math>\frac{2}{5}</math>, is double that of the gap in the second jar, <math>\frac{1}{5}</math>. Therefore, we must add twice as much of jar C into the jar <math>A</math> over jar <math>B</math>. So, we must add <math>\frac{2}{3}</math> of jar C into jar <math>A</math>, so <math>m = 2, n=3</math>.
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Since jar C contains <math>1</math> liter of solution, we are adding <math>\frac{2}{3}</math> of a liter of solution to jar <math>A</math>. In order to close the gap between water and acid, there must be <math>\frac{2}{5}</math> more liters of acid than liters of water in these <math>\frac{2}{3}</math> liters of solution. So, in the <math>\frac{2}{3}</math> liters of solution, there are <math>\frac{2}{15}</math> liters of water, and <math>\frac{8}{15}</math> liters of acid.  So, 80% of the <math>\frac{2}{3}</math> sample is acid, so overall, in jar C, 80% of the sample is acid.
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 +
Therefore, our answer is <math>80 + 2 + 3 = \boxed{85}</math>.
 +
 
 +
~ ihatemath123
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== Solution 2==
 +
There are <math>\frac{45}{100}(4)=\frac{9}{5}</math> L of acid in Jar A. There are <math>\frac{48}{100}(5)=\frac{12}{5}</math> L of acid in Jar B. And there are <math>\frac{k}{100}</math> L of acid in Jar C. After transferring the solutions from jar C, there will be
 
<br> <math>4+\frac{m}{n}</math> L of solution in Jar A and <math>\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}</math> L of acid in Jar A.<br>
 
<br> <math>4+\frac{m}{n}</math> L of solution in Jar A and <math>\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}</math> L of acid in Jar A.<br>
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<br> <math>6-\frac{m}{n}</math> L of solution in Jar B and <math>\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}\right</math> of acid in Jar B.
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<br> <math>6-\frac{m}{n}</math> L of solution in Jar B and <math>\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}</math> of acid in Jar B.
 
<br>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.<br>
 
<br>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.<br>
 
<cmath>\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}</cmath>
 
<cmath>\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}</cmath>
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<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>.
  
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== Solution 2 ==
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== Solution 3 ==
 
One might cleverly change the content of both Jars.  
 
One might cleverly change the content of both Jars.  
  
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So the answer is <math>80+2+3 = \boxed{085}</math>
 
So the answer is <math>80+2+3 = \boxed{085}</math>
  
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== Solution 3 ==
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== Solution 4 ==
 
One may first combine all three jars in to a single container.  That container will have <math>10</math> liters of liquid, and it should be <math>50\%</math> acidic.  Thus there must be <math>5</math> liters of acid.   
 
One may first combine all three jars in to a single container.  That container will have <math>10</math> liters of liquid, and it should be <math>50\%</math> acidic.  Thus there must be <math>5</math> liters of acid.   
  
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Jug A contained <math>45\% \cdot 4L</math>, or <math>1.8L</math> of acid, and jug B <math>48\% \cdot 5L</math> or <math>2.4L</math>.  Solving for the amount of acid in jug C, <math>k = (5 - 2.4 - 1.8) = .8</math>, or <math>80\%</math>.
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Jar A contained <math>45\% \cdot 4L</math>, or <math>1.8L</math> of acid, and jar B <math>48\% \cdot 5L</math> or <math>2.4L</math>.  Solving for the amount of acid in jar C, <math>k = (5 - 2.4 - 1.8) = .8</math>, or <math>80\%</math>.
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Once one knowss that the jar C is <math>80\%</math> acid, use solution 1 to figure out m and n for <math>k+m+n=80+2+3=\boxed{085}</math>.
  
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Once one knows that the jug C is <math>80\%</math> acid, use solution 1 to figure out m and n.
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==Video Solution==
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https://www.youtube.com/watch?v=_znugFEst6E&t=919s
  
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~Shreyas S
  
 
== See also ==
 
== See also ==
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{{AIME box|year=2011|n=I|before=-|num-a=2}}
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{{AIME box|year=2011|n=I|before=First Problem|num-a=2}}
 +
{{MAA Notice}}

Latest revision as of 15:43, 5 February 2022

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

Jar A contains $\frac{11}{5}$ liters of water, and $\frac{9}{5}$ liters of acid; jar B contains $\frac{13}{5}$ liters of water and $\frac{12}{5}$ liters of acid.

The gap between the amount of water and acid in the first jar, $\frac{2}{5}$, is double that of the gap in the second jar, $\frac{1}{5}$. Therefore, we must add twice as much of jar C into the jar $A$ over jar $B$. So, we must add $\frac{2}{3}$ of jar C into jar $A$, so $m = 2, n=3$.

Since jar C contains $1$ liter of solution, we are adding $\frac{2}{3}$ of a liter of solution to jar $A$. In order to close the gap between water and acid, there must be $\frac{2}{5}$ more liters of acid than liters of water in these $\frac{2}{3}$ liters of solution. So, in the $\frac{2}{3}$ liters of solution, there are $\frac{2}{15}$ liters of water, and $\frac{8}{15}$ liters of acid. So, 80% of the $\frac{2}{3}$ sample is acid, so overall, in jar C, 80% of the sample is acid.

Therefore, our answer is $80 + 2 + 3 = \boxed{85}$.

~ ihatemath123

Solution 2

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 $k+m+n=80+2+3=\boxed{085}$.

Solution 3

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 4

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 (Problems • Answer Key • Resources)
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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