8 liters of solution is removed from 20% milk solution and 8 liters of

Milk/total solution = 20% initially then 16%;
M=0.2T

16%=[M-(8*20%)]T;
16%=(0.2T-1.6)/T
0.04T=1.6; T=40 IMO B

let x=initial quantity of 20% milk solution
.8x-8(.8)+8=.84x→
x=40 liters
B

total qty of milk ; .2*8 ; 16
milk /total = 20%
milk-1.6/total = 16%
milk = .2*total
.2*total-1.6/total = .16
solve for total = 40
IMO B

When 8 liters of the solution is removed, the 8 liters consists of 0.2 x 8 = 1.6 L of milk, and 6.4 L or water. If we let s = the initial amount, in liters, of the solution, so originally there are 0.2s liters of milk and we can create the equation:

(0.2s - 1.6 + 0)/(s - 8 + 8) = 16/100

(0.2s - 1.6)/s = 4/25

25(0.2s - 1.6) = 4s

5s - 40 = 4s

s = 40

Alternate Solution:

We start with x liters of 20% milk. We remove 8 liters of this solution, so we are removing 8 liters of 20% milk. Then we add back 8 liters of water (which is 0% milk), and the result is x liters of 16% milk. We can summarize this in an equation as:

x(0.20) - 8(0.20) + 8(0) = x(0.16)

0.20x - 1.6 + 0 = 0.16x

0.04x = 1.6

4x = 160

x = 40

Answer: B

we can do witha allegations:

we see we have 80%water solution, 8 litre is removed and is replayed by 100% water solution to make a 84% water solution.
so

80 100
84
16:4
=4:1
So, 1/5 of the total solution was repplaced with 8 litres.
so, total solution is 40

can we do it this way?

20/100 * x = 8 ltrs.
upon solving, we get x = 40 ltrs. ?

(percentage method?)

We are told 8 liters of a solution is removed from a 20% milk solution.

Of the 8 liters removed, 20% consists of milk. Thus, 1.6 liters of the 8 liters consists of milk.

We are then given that the resulting solution consists of 16% milk. From this piece of information, we can create the following formula:

(0.2X - 1.6) / X * 100 = 16

20X - 160 / X = 16
16X = 20X - 160
X = 40

Answer is B.

Please correct me if I'm wrong, but we did not need to know that '8 liters of water is added' to solve this question.

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