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:

d

Step-by-step explanation:

its actually d on edg.

Answer 2

ANSWER

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

$= \frac{30}{100}$

Now write 60% in the decimal form

$= \frac{60}{100}$

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

The quantity of  antifreeze is

$= \frac{30\times (10 - x)}{100}$

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

than the equation becomes

$= x + \frac{30\times ( 10 - x )}{100} = \frac{60\times10}{100}$

solyving

100x +300 - 30x = 600

70x = 300

$x = \frac{300}{70}$

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 .