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Re: If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol [#permalink]
Amount of Alcohol in 50ml solution = \(\frac{50x}{100}\)
Amount of alcohol in 30ml solution = \(\frac{30y}{100}\)

Alcohol in mix = \(\frac{50x}{100}\)+\(\frac{30y}{100}\)=\(\frac{(5x+3y)}{10}\)
Total amount of mix =80

To find the percentage of alcohol in new solution we have to divide alcohol in mix by total amount of mix

1) Directly gives what we are looking for -- sufficient
2) From 2x+7y=460 we cant get the value of 5x+3y -- INsufficient

IMO
Ans: A
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Re: If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol [#permalink]
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Re: If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol [#permalink]
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