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Bunuel
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sixfivefive
Can someone provide a more clear solution?
Methanol (M) is 80% of Total (T) and Ethanol (E) is 20% of Total (T)

M = 0.8T
E = 0.2T

When 10 litres are removed from T => 8 litres of M, and 2 litres of E are removed

We are told that, \(\frac{E - 2 }{ T} = 15\)%

=> \(\frac{0.2T - 2 }{ T} = \frac{15 }{ 100}\)

=> \(20T - 200 = 15T\)

=> \(T = 40\\
\)
Answer C.

Hope it helps.
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sixfivefive
Can someone provide a more clear solution?

Let's call the total volume of the mixture x.

We remove 10 liters and then add in 10 liters so the value of x is the same before and after.

Solution 1 has 20% or 2/10 or 1/5 ethanol.

If we take out ten liters, then 1/5 of that is ethanol. 10 * 1/5 = 2 liters of ethanol and 8 liters of methanol removed. We have one side of the equation now:
(1/5x - 2).

We are then told that the overall ethanol concentration after adding in 10 liters of methanol is 15% or 15/100 or 3/20.

So...

1/5x -2 = 3/20x

x = 40

C
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