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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