A certain mixture is made up of compounds A, B, and C. Compound B acco

Let the total volume of the mixture be x and the individual compounds volume be A,B and C

A+B+C = x

B=0.7x

Hence A+C = 0.3x

Statement 1:

Given that if 3A is added to the mixture then 50% of the mixture will be compound A, implies that

4A=\(\frac{(x+3A)}{2}\)

Solving it we get

x=5A

We still cannot deduce the additional litres required to make the mixture 25% with out knowing the value of A

Statement 2:

half of compound B is removed implying only 50% of the compound B is left in the mixture

\(\frac{0.7x}{2}\)= 175

x=500 litres

Now we know x=500, B=0.7*500=350

So A+C=150

But we still don't know that value of C to determine the additional litres of A

Hence Statement B is insufficient

Statement 1 and 2 combined

we have x=5A, x=500, B=350 and A+C=150

we can find A using x=5A and x=500

A=100

And using A+C=150, we can find that C=50

So the current composition of A is 100 litres or \(\frac{(100}{500)}\) 20% of the mixture

Now we can find the additional A required. There is no point going ahead to solve for the exact number of litres required. You can conclude that both the statement together are sufficient. Hence OA is C

For those who want to know the final answer here it is

Let y be the additional litres of compound A

25% = \(\frac{(100+y)}{(500+y)}\)

Solving it

500+y=400+4y

100=3y

y=\(\frac{100}{3}\)

Hence we need to add 33.33 litres of the compound A to attain 25% composition in the mixture

Kudos if it helps!