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 60% antifreeze?

a. 3 quarts
c. 4.28 quarts
b. 2 quarts
d. 2.67 quarts

Answer 1

Answer: C 4.28

Explanation:

Answer 2

Find out the how much fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze .

To proof

let us assume that the fluid should be drained and replaced with pure antifreeze be x .

As given

A radiator contains 10 quarts of fluid, 30% of which is antifreeze.

Now write 30% in the decimal form

`= (30)/(100)`

Now write 60% in the decimal form

`= (60)/(100)`

After drained and replaced the fluid with pure antifreeze than the quantity becomes = ( 10 - x )

The quantity of antifreeze is

`= (30* (10 - x))/(100)`

The fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze .

than the equation becomes

`= x + (30* ( 10 - x ))/(100) = (60*10)/(100)`

solyving

100x +300 - 30x = 600

70x = 300

`x = (300)/(70)`

x = 4.28 quarts (approx)

Hence 4.28 quarts (approx) fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze .

Therefore the option (c.) is correct .