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Re: A 120-milliliter mixture of Chemical X and water [#permalink]
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
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Re: A 120-milliliter mixture of Chemical X and water [#permalink]
1
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.

aishumurali wrote:
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
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Re: A 120-milliliter mixture of Chemical X and water [#permalink]
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.

Amuk wrote:
A 120-milliliter mixture of Chemical X and water contained 40 percent Chemical X. A part of the mixture was removed and replaced with an equal quantity of water. If the resulting mixture contained 10 percent Chemical X, what is the volume of the mixture that was removed?


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Re: A 120-milliliter mixture of Chemical X and water [#permalink]
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