2011 AIME I Problem 1

Question

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.

Answer

Answer: 85

Concepts:mixture percentage

Difficulty rating: 1950

Small hint:

Imagine pouring all three jars together: the result is

10

liters that must be

50

% acid, which determines

k

Big hint:

With

k

known, let

x

be the amount added to jar

A

and set jar

A

’s acid equal to half of its new volume

Solution:

If all three jars were combined, the result would be

10

liters of

50

% acid, since both final jars are

50

% acid. The total acid is therefore

5

liters, so

4(0.45) + 5(0.48) + 0.01k = 5,

which gives

k = 80.

Now let

x

be the number of liters poured from jar

C

into jar

A

. Jar

A

then holds

4 + x

liters containing

1.8 + 0.8x

liters of acid, so

1.8 + 0.8x = 0.5(4 + x),

giving

0.3x = 0.2,

so

x = \frac{2}{3}.

Thus

m + n = 2 + 3 = 5,

and

k + m + n = 80 + 5 = 85.