Difference between revisions of "2011 AIME I Problems/Problem 1"
Ewcikewqikd (talk | contribs) (→Solution) |
AlphaMath1 (talk | contribs) m (→Solution) |
||
Line 2: | Line 2: | ||
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 == | + | == Solution 1== |
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 | 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 | ||
<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> | ||
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 == | == Solution 2 == |
Revision as of 17:06, 20 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 1
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. Unknown error_msg) 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