20% alcohol present in a 15 litres mixture and rest of water. If 3 litres of water be mixed with it, the percentage of alcohol in the new mixture would be?

50/3

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40/3

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20/3

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10/3

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None

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Solution

The correct option is A
50/3

20% present in a 15 litres mixture

So, $\frac{20 x 15}{100} = 3 l i t r e s .$

Water in it = ( 15 - 3 ) = 12 litres .

New quantity of mixture = ( 15 + 3 ) = 18 litres .

Percentage of alcohol in new mix is = $\frac{3 x 100}{18}$

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20% alcohol present in a 15 litres mixture and rest of water. If 3 litres of water be mixed with it, the percentage of alcohol in the new mixture would be?

The correct option is A 50/3 20% present in a 15 litres mixture So, 20x15100=3litres. Water in it = ( 15 - 3 ) = 12 litres . New quantity of mixture = ( 15 + 3 ) = 18 litres . Percentage of alcohol in new mix is = 3x10018

The correct option is A
50/3

20% present in a 15 litres mixture

So, $\frac{20 x 15}{100} = 3 l i t r e s .$

Water in it = ( 15 - 3 ) = 12 litres .

New quantity of mixture = ( 15 + 3 ) = 18 litres .

Percentage of alcohol in new mix is = $\frac{3 x 100}{18}$