sorry this sum is easy for some but i dont understand.

The mixture first contains 40% chemical x , 40% of 120 is 48 litres
Then the mixture contains 10% chemical x , which is equal to 12 litres

48 - 12 = 36 litres

Hi,

I think you might be lost on wordings.

So total volume of mixture $= 120$ml

Qty of $X = 40$% of $120 = 48$ml

A part of the mixture was removed and replaced with an equal quantity of water.
Let's say $p$ ml is removed and replaced with water. Now $p$ will contain $40$% of $X$ as well.

So, after removing $p$ ml
qty of $X = 48 - 0.4 p$

If the resulting mixture contained $10$% of $X$
Now the resulting mixture is $120$ml so qty of $X = 12$ml

$X = 48 - 0.4 p = 12$

$p = 90$ ml

Please ask if the doubt remains.

To solve for the removed part of mixture (M), we consider that for any part removed, the concentration of M will be remained. Hence, $\frac{0.4*(120-M)}{120}=0.1$, $48-0.4M=12$, $M=90$ is the answer.

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