Question

A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 40% antifreeze?

Answer 1

Answer:

1 3/7 quarts should be drained off and replaced with pure antifreeze.

1 3/7 ≈ 1.4286

Current amount of antifreeze in quarts is -

30/ 100 × 10 = 3

40% ---> 4 quarts

Let the amount drained of and replaced with antifreeze be x-

The amount left after draining off is 10 − x.

The amount of antifreeze  is 30/ 100 (10−x).

30/100(10-x)+x=4

3-3/10x+x=4

3+x(1-3/10)=4

x=1*10/7=1 3/7 quarts

check;

10- 1 3/7 = 8 4/7

=(30/100*8 4/7)+1 3/7

=(3/10 * 60/7) + 10/7

=3*6/7 + 10/7

=28/7

=4

4 liters of pure antifreeze is mixed into 10 quarts.