A mixture contains wine and water in the ratio 3 : 2 and another mixture contains them in the ratio 4 : 5. How many litres of the latter must be mixed with 3 litres of the former so that the resulting mixture may contain equal quantities of wine and water?
Solution
Correct Answer: Option A
Solution A ⇒ 3 litres solution (3 : 2 ∷ wine : water)
Solution B ⇒ x litres solution (4 : 5 ∷ wine : water)
According to question Solution A is mixed with Solution B and forms
⇒ (3 + x) litres solution (1 : 1 ∷ wine : water)
For Wine
Wine in Sol A + Wine in Sol B = Wine in net Solution
(3/5) × 3 + (4/9) × x = 1/2 × (3+x)
⇒ (4x/9) – (x/2) = (3/2) - (9/5)
⇒ (8x/18) – (9x/18) = (15/10) – (18/10)
⇒ 3/10 = x / 18
⇒ x = 5.4 litres = \(5\frac{2}{5}\) litre
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