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Re: 6 pints of a 20 percent solution of alcohol in water are mixed with 4 [#permalink]
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New solution composition based on alcohol weight = [(6*20%) + (4*10%)] / 10 = 16% alcohol by weight in the new solution.
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Re: 6 pints of a 20 percent solution of alcohol in water are mixed with 4 [#permalink]
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Total solution = 6+4= 10

Total alcohol= 20% of 6= 1.2 + 10% of 4= .4
1.2+.4= 1.6

%= 1.6/10*100= 16

'A' is the answer
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Re: 6 pints of a 20 percent solution of alcohol in water are mixed with 4 [#permalink]
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We can use the concept of weighted averages as the quantities are different.

\(P_{Avg} = \frac{P_1Q_1 \space + \space P_2Q_2}{Q_1 + Q_2}\)

Here Q1 and Q2 are the quantities and P1 and P2 are the percentages.

Substituting the given values, we get \(P_{Avg} = \frac{20*6\space + \space 4*10}{6 + 4} = \frac{120 + 40}{10} = \frac{160}{10} = 16\)%


Option A

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Re: 6 pints of a 20 percent solution of alcohol in water are mixed with 4 [#permalink]
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Re: 6 pints of a 20 percent solution of alcohol in water are mixed with 4 [#permalink]
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