help please - MATH?

Question

An automobile radiator contains 16 liters of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that there will be 50% antifreeze?


Please explain so I learn.

Answer 1

Here's how to think about it. In the end you will be mixing so many liters of 30% mixture and so many liters of 100% mixture.

Let a be the amount of 30% mixture.
Let b be the amount of 100% mixture.

Added together these must equal 16 liters:
a + b = 16

Also, you want the mixture to equal 50%.
30% x a + 100% x b = 50% x 16

You have two equations and two unknowns, so this should be solvable.

Let's solve for a:
a = 16 - b

Substitute this into the second equation:
30% x (16 - b) + 100% x b = 50% x 16

You could do this all with decimal amounts (30% = 0.30, 50% = 0.50) etc. But it is easier to just do them as whole amounts (30, 100, 50). The answer will work out either way.

Distribute through:
30 x 16 - 30b + 100b = 50 x 16

Simplify:
480 - 30b + 100b = 800

Combine like terms:
480 + 70b = 800

Subtract 480 from both sides:
70b = 800 - 480
70b = 320

Divide both sides by 70:
b = 320 / 70
b = 32 / 7
b = 4 4/7 liters

So you will need to drain 4 4/7 liters of the 30% mixture, then add 4 4/7 liters of the 100% mixture.

As a double check:
a = 11 3/7
b = 4 4/7

30% x 11 3/7 + 100% x 4 4/7 =? 50% x 16
30% x 80/7 + 32/7 =? 8
24/7 + 32/7 =? 8
56/7 = 8 <-- check

So the correct answer is you must drain 32/7 liters, or 4 4/7 liters (as a mixed fraction).

As a decimal, the answer is approximately 4.571 liters.

Answer 2

Of the current 16 liters, 30% is Antifreeze, 70% is water.
16 * .30 = 4.8 liters of antifreeze
16 * .70 = 11.2 liters of water

You need to have 50% Antifreeze, 50% water. SO you need a total of 8 liters of water, and 8 of antifreeze. To get 8 liters of water, you need to get rid of (11.2 - 8) = 3.2 liters.

When you drain it, you're draining the MIXTURE of only 70% water, so for every 1 liter you drain, you drain .7 liters of water.

SO to find how many liters of mixture to drain, you divide 3.2 by .7 = approx 4.571428, or 4.57.

If you want to check this out, subtract 4.57 from 16, which is 11.43, then multiply your .70 * 11.43 and you've got 8 (well 8.001 since 4.57 had to be rounded).

This may not be the best way to solve it, but that's how I'd do it.

Answer 3

amount of antifreeze initially
=30*16/100
=4.8
let x l of water is drained,
but amount of antifreeze is same
now vol.is 16-x
so
50*(16-x)/100=4.8
16-x=2*4.8
solve this for x