How many liters of pure alcohol must be added to 15 liters of a 20% solution to obtain a mixture which is 30%?

Question

How many liters of pure alcohol must be added to 15 liters of a 20% solution to obtain a mixture which is 30% alcohol? can u show me the solution?=)

Answer 1

2.1

Answer 2

The previous answer does not take into consideration that adding more alcohol to the solution will increase the volume. If only add 1.5 l of alcohol then you will get 27.2% because the volume is now 16.5 l. The correct answer is 2.1428l

because
first find water content
0.8*15=12

x/(x+12)=0.3 This says that the soution is 30% alcohol containing 12 liters of water

some algebra simplifies x to be 5.1428. Since we already have 3 liters of alcohol present (.2*15=3) we must add 5.1428-3= 2.1428

Answer=2.1428

Answer 3

15 litres of a 20% solution contains 15* 20/100 = 3 litres of pure alcohol.
Let the volume of pure alcohol needed to boost the solution to 30% of pure alcohol be x litres.
The following equation holds:
Total volume of new solution = 15+ x
Total volume of pure alcohol in this solution = (3 + x) lit
Therefore, (15+x) *30/100 = 3 + x
Solving for x, we have 7x = 15, or x = 15/7 = 2.143 litres

Answer 4

If you already have 20% alcohol solution and 15 litres of it and you want 30%.

First find what 30% of 15 litres is = 4.5 litres

you have 20% which is 3 litres (of pure alcohol already in.) Therefore if you add another 1.5 litres (4.5 - 3) of 'pure' alcohol, it will be a 30% alcohol solution.

Answer 5

You start with 3 litres of pure alcohol in 15 litres. this is 3/15 or 20%. You add x litres of alcohol the proportion you now have is
(x+3)/(x+15), this has to equal 0.3.
Rearranging this equation gives x+3=0.3(x+15)
This simplifies to 0.7x=1.5
x=2.14 litres

Answer 6

15/20 = (x + 15)/30
450 = 20x + 300
x = 7.5 L

Answer 7

2.142857...litres
i already solved it for someone just see...