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,
[tex]x+y=234[/tex]
[tex]x=234-y[/tex] ...(i)
The mixture has 43% fertilizer. So,
[tex]\dfrac{60}{100}x+\dfrac{21}{100}y=\dfrac{43}{100}\times 234[/tex]
Multiply both sides by 100.
[tex]60x+21y=10062[/tex] ...(ii)
Using (i) and (ii), we get
[tex]60(234-y)+21y=10062[/tex]
[tex]14040-60y+21y=10062[/tex]
[tex]-39y=10062-14040[/tex]
[tex]-39y=-3978[/tex]
Divide both sides by -39.
[tex]\dfrac{-39y}{-39}=\dfrac{-3978}{-39}[/tex]
[tex]y=102[/tex]
Putting this value in (i), we get
[tex]x=234-102[/tex]
[tex]x=132[/tex]
Therefore, 132 liters of the 60% solution must be used.
