Can someone help me with this math problem?

Question

how many quarts of pure anti-freeze must be added to 2 quarts of a 20% anti-freeze solution to obtain a 40% anti-freeze solution?-

(if you get the answer,plzz work the problem out so i can know how you got it)

Answer 1

Let x = number of quarts of pure anti-freeze.
You are given that 2 = number of quarts of 20% anti-freeze
The mixture contains both the pure anti-freeze and the 20% anti-freeze; therefore,
x+2 = number of quarts of the mixture of 40% anti-freeze.
Note that pure anti-freeze is 100% anti-freeze.
Insert your percentages.
100x + 20(2) = 40(x+2)
Now solve for x.
100x + 40 = 40x + 80
Subtract 40x from each side
60x + 40 = 80
Subtract 40 from each side
60x = 40
Divide both sides by 60
x = 40/60
Reduce
x = 2/3 quarts of pure anti-freeze

Answer 2

Ok, so you want to start by determining how much water you have first.

80% of 2 quarts is 1.6 quarts. So whatever your total volume is after you add the new antifreeze is added. The 1.6 q will now become 60% of the total volume, since you're trying to achieve 40% anti-freeze.

To find the total volume, you would divide 1.6 q by it's new percentage, because to find a volume equivalent of a percentage, you multiply the total volume by the percentage, so we're just reversing here. Follow me?

So 1.6 q will be 60% of 2.66666666... q (or 2 2/3 q)

Subtract your original amount from this new total,
2 2/3 - 2, and you get 2/3 q

So the total amount you would have added would have been 2/3 q.

Follow?

The equation would be:

where t is the new total volume in quarts

t = 1.6 + 0.4t (0.4t is the same as saying 40% of t, 1.6 is the original volume of water, which was 80% of 2 quarts)

t - 0.4t = 1.6 + 0.4t - 0.4t subtract t from both sides

0.6t = 1.6

(0.6t)/0.6 = (1.6)/0.6 divide both sides by 0.6, which is 60 percent, to solve for t

t = 2.666666666... or 2 2/3 q

This is your new total volume. Since what you are trying to find is the amount you have to add of antifreeze and you're adding a pure amount of it with no water, you can simply subtract the original volume to get the remainder, which is the total amount of antifreeze added.

2/3 q must be added to make this solution a 40% mixture.

Answer 3

1 will indicate the pure 100%
2 will indicate the 20%
3 will indicate the 40%

You have to mix a certain volume of 1, say V1 with a certain volume of 2, say V2 and you need to get the 40% with a volume of V3.

So, you know that V2 = 2 quarts and you know that V1 + V2 = V3

V1 + 2 = V3

The percent indicates mass per volume of solution so if you multiply volume times percent the unit will be:

[Volume]*[Percent] = [Volume]*[Mass/Volume] = Mass.

So let's say you take V1 from the 100%, the mass, M1 will be

M1 = V1 * 100% = V1 * 1
M2 = V2 * 20% = 2 * 0.2
M3 = V3 * 40% = V3 * 0.4

Now M1 + M2 = M3
So
V1 + 2*0.2 = V3 * 0.4

We know that V3 = 2 + V1
V1 + 0.4 = 0.8 + 0.4 V1
V1 - 0.4 V1 = 0.8 - 0.4
0.6 V1 = 0.4

V1 = 2/3 quart