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Option 3 : 560 ml

Suppose the initial quantity of alcohol having 40% concentration is ‘x’ ml and the quantity replaced is y ml;

∴ 0.4(x – y) + 0 × y = 0.3x

⇒ 0.1x = 0.4y

⇒ x = 4y      ---- (1)

Now he mixes ‘y’ ml alcohol having 40% concentration with 260 ml alcohol having 20% concentration,

And thus a mixture having 27% concentration of alcohol is prepared;

∴ y × 0.4 + 260 × 0.2 = (y + 260) × 0.27

⇒ 0.4y + 52 = 0.27y + 70.2

⇒ 0.13y = 18.2

⇒ y = 140

From equation (1)

x = 560 ml

∴ Initial quantity of alcohol having 40% concentration = 560 ml
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