Question 1201150

A certain automobile cooling system contains 12.0 L of coolant that is 15% antifreeze.

How many liters of mixture must be removed so that, when it is replaced with pure antifreeze, a mixture of 23% antifreeze will result?

Initialquantity = 12L
strength = 15%

remove x litres
Solution remaining = (12-x)
You add pure antifreeze to replace x
Final antifreeze 23%  12 L

(12-x)*15% +100%*x = 23%*12
Multiply by 100 the equation
(12-x)*15 +100x=23*12
180-15x+100x=276
85x =96
x =96/85
=1.13 L has to be removed