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bearhunter [10]
3 years ago
11

A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure anti

freeze so that the new mixture is 60% antifreeze?
a. 3 quarts
c. 4.28 quarts
b. 2 quarts
d. 2.67 quarts
Mathematics
2 answers:
Nadya [2.5K]3 years ago
7 0

Answer:

d

Step-by-step explanation:

its actually d on edg.

prohojiy [21]3 years ago
3 0

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 .

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