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

Revision as of 15:30, 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

@@ Line 5: / Line 5: @@
 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.
-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>.
+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>.
 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.

2011 AIME I (Problems • Answer Key • Resources)
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