Let X be the capacity of the container.

% of Liq A left / % of A originally present = (1- (Vol of soln replaced/vol of container)) ^ n

[Note this formula is applied for repeated iterations such as these and 'n' = Number of iterations(in this case 2)]

9/25 = (1-15/x)^2 ...... [(9/9+16) / 1(Liq A present originally)]

3/5 = 1 - (15/X)

X=37.5

Hey Ashwin
Good use of the formula. But how would u do this without alligation? Using standard ratios and equations

Wasnt able to format the reply.. See attached pic

Could u please explain me this -(1-15/x)*15?
Ä° didnt get why u multiplied by 15/ I mean why u use multiplication, not substraction?

p.s. sorry for a stupid question

This will be calculation intensive esp. if one does'nt remember the formula.
So to do it within 2-3 mins we will have to use subsitution of option:

Since ratio( A:B) is 9:16 so let A be 9x and B be 16x
Total Volume : 25x

So the answer would probably be either B or C ( 25 or 37.5)

Let try for 25:
Intial : A25 B0
First mixing : A10 B15
Second mix : A4 B21 ( On removing 15lt B will be 6 then adding 15)
Final ratio 4:21 ( Incorrect)

Let try for 37.5:
Intial : A37.5 B0
First mixing : A22.5 B15 ( 22.5:15 is 3:2, so removing 15lt will remove A9 B6)
Second mix : A13.5 B24 ( On removing 15lt B will be 9 then adding 15)

Final Ratio : 13.5:24 => 9:16 (Bingo!!)

Answer : 37.5

@LalaB

Could u please explain me this -(1-15/x)*15?

I think you meant [(X-15)/X] * 15

When you do the replacement for the first time, its straightforward, (You remove 15 Lts of A and replace it by 15 Lts of B)

When this is done the second time, while removing the 15 Lts, you remove some quantity of A and some of B.

To find quantity of A remaining, we need to find the amount of A removed in the 2nd step.

For ease of explanation, consider concentration of A to be 20% (remaining is B 80%) in a 100 Ltr solution. Now if we remove 20 Lts of this solution, quantity of A removed could be found by;

(20/100 )* 20 = 4 Lts (Quantity of A/Total Quantity) * Amount Removed.

Now, [(X-15)/X] * 15 is the same thing as above

X-15 - Quantity of A
X - Total quantity
15 - Amount removed.

Hope it makes sense

shankar.ashwin wrote:Let X be the capacity of the container.

% of Liq A left / % of A originally present = (1- (Vol of soln replaced/vol of container)) ^ n

[Note this formula is applied for repeated iterations such as these and 'n' = Number of iterations(in this case 2)]

9/25 = (1-15/x)^2 ...... [(9/9+16) / 1(Liq A present originally)]

3/5 = 1 - (15/X)

X=37.5

I understood the part % of Liq A left = 9. But I am not sure how % of A originally present = 25. Isn't 9+16=25 the solution mixture after the two replacements? I am jinxed.

I posted an alternate approach here:

https://www.beatthegmat.com/a-ps-from-gm ... tml#426860

msr4mba wrote:
I understood the part % of Liq A left = 9. But I am not sure how % of A originally present = 25. Isn't 9+16=25 the solution mixture after the two replacements? I am jinxed.

Yeah you're right 'replacement' is replacing a certain quantity of 'A' with equal quantity of 'B'. (You don't add B increasing quantity)

The total quantity at the end of each step remains the same. So initial A will be 9+16

Just for fun. A different take on the problem.

If we carefully observe, we are always replacing the same fraction of component from available A each time we remove 15 litres and add 15 litres of B.

lets say each time we replace p fraction of A
initially lets say we have x litres of A component
after first replacement the A component will be x.p - (1)
after second replacement the A component will be x.p^2
we know xp^2/x = 9/25 (nothing but final amt of A/total amt of solution)
p = 3/5 -(2)

so we are removing a fraction of 3/5 each time we are removing 15 litres and adding 15 litres of B
so use material balance after first replacement:
we know that amount of A present after first replacement is x-15
and also we know it is x.p = 3x/5 from (1)&(2)
=> 3x/5 = x-15 => 2x/5 = 1 5=> x = 37.5

user123321