Question

4. A solution of 60% fertilizer is to be mixed with a solution of 21% fertilizer to form 234 liters of a 43% solution. How many liters of the 60% solution must be used?
SHOW YOUR WORK

Answer 1

Given:

A solution of 60% fertilizer is to be mixed with a solution of 21% fertilizer to form 234 liters of a 43% solution.

To find:

The quantity of the 60% solution in the mixture.

Solution:

Let x be the quantity of the 60% solution and y be the quantity of the 21% solution.

Quantity of mixture is 234. So,

$x+y=234$

$x=234-y$                  ...(i)

The mixture has 43% fertilizer. So,

$\dfrac{60}{100}x+\dfrac{21}{100}y=\dfrac{43}{100}\times 234$

Multiply both sides by 100.

$60x+21y=10062$            ...(ii)

Using (i) and (ii), we get

$60(234-y)+21y=10062$

$14040-60y+21y=10062$

$-39y=10062-14040$

$-39y=-3978$

Divide both sides by -39.

$\dfrac{-39y}{-39}=\dfrac{-3978}{-39}$

$y=102$

Putting this value in (i), we get

$x=234-102$

$x=132$

Therefore, 132 liters of the 60% solution must be used.