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Bunuel
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If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol 26 Nov 2019, 02:03
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If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol solution, what percentage of the final solution is alcohol?

(1) \(5x + 3y = 280\)
(2) \(2x + 7y = 460\)


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nick1816
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If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol 26 Nov 2019, 03:08
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z is the percentage of resultant solution
z= 50x+30y/80

z=5x+3y/8

Statement 1= We know the value of 5x+3y
Sufficient

Statement 2-
We know the value of 2x+7y

We can't find 5x+3y using only 1 equation

Insufficient


Bunuel
If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol solution, what percentage of the final solution is alcohol?

(1) \(5x + 3y = 280\)
(2) \(2x + 7y = 460\)


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exc4libur
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If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol 26 Nov 2019, 04:54
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Bunuel
If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol solution, what percentage of the final solution is alcohol?

(1) \(5x + 3y = 280\)
(2) \(2x + 7y = 460\)
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(50x%+30y%)/80=z*100
([50x+30y]/100)/80=z*100
([50x+30y]/100)/80=z*100
[50x+30y]/80=z
5x+3y/8=z

(1) \(5x + 3y = 280\) sufic

(2) \(2x + 7y = 460\) insufic

Ans (A)
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If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol 26 Nov 2019, 04:56
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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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If 50 mL of an x% alcohol solution is mixed with 30 mL of a y% alcohol 20 Apr 2026, 05:15
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