How many liters of water must be added to a 20 liter solution of water

Choice:B
Time taken:1:26
here we go
Since currently acid is 24% in 20 liter mean 4.8 Litre
So
(4.8/20+x)*100=8(The new required level)
480=8(20+x)
480=160+8x
320=8x
x=40

Let's say, we need X liters of water (pure water with no acid). So can make the following equation-

X*0 + 20 * 24% = (X+20) * 8%
=> 20*24/100 = (X+20) * 8/100
=> 20*24 = (X+20) * 8
=> 20*3 = (X+20)
=> X = 60 - 20 = 40

Answer: B

let w=liters of water to be added
.76*20+w=.92(20+w)
w=40
B

Let x be amount of water to be added to the original 20 liters

x is 100% water, so there's an implied 1 before the first x in equation:

.24*20 + x = .08*(20+x)

4.8 + x = 1.6 + .08x
3.2 =.08x
x=40

We are going to mix a 24% acid solution with water which is basically a 0% acid solution. And the result required is 8% acid. So we can apply weighted average concept here.

Ratio in which these two solutions (24% and 0%) should be mixed in order to get an 8% result can be found this way:

24-8 = 16 and 8-0 = 8. So the net result must be 16 steps away from 24 and 8 steps away from 0. OR we can say that ratio in which 24% and 0% solutions should be mixed (in order to get an 8% solution) has to be 8:16 or 1:2.

Since the ratio is 1:2, the amount of 0% solution (water in our case) must be double that of 24% solution.

So the amount of water must be = 2*20 = 40 litres.

Total = 20 liters

Acid = 4.8

Now we have to mix water in order to make acid 8% of the total volume.

let x liters of water must be added to the total volume in order to dilute acid concentration.


4.8 / ( 20 + x ) * 100 = 8

480 = 160 + 8x

8x = 320

x = 40.

B is the correct answer.

Allegation Method:

Water (Add)_____Old Mix (Water's component)

100-------------------------76

_________92_________

16---------------------------8

Which gives, 2:1

Since old mixture was 20l, new mixture will be 20*2 = 40l.

Note: 92% represents the share of water in the new mix.