In 66kg of adulterated milk, the ratio of water to milk was found to be 1 $:$ 5. Water was added to it to make the ratio 3 $:$ 5. The quantity of water added is?

15L

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17L

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19L

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20L

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22L

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Solution

The correct option is E
22L

$F_{1} = Fraction of milk in the adultered milk = \frac{5}{6} F_{2} = Fraction of milk in water = 0 F_{3} = Fraction of milk in the new mixture = \frac{5}{8}$

$∴ F_{1} : F_{2} = \frac{5}{8} : \frac{5}{24} = 3 : 1$

$If we have 66 kg of adulterated milk, water = \frac{66}{3} \times 1 = 22 l i t r e s$

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In 66kg of adulterated milk, the ratio of water to milk was found to be 1:5. Water was added to it to make the ratio 3:5. The quantity of water added is?

The correct option is E 22L F1=Fraction of milk in the adultered milk=56F2=Fraction of milk in water=0F3=Fraction of milk in the new mixture=58 ∴F1:F2=58:524=3:1 If we have 66 kg of adulterated milk, water=663×1=22 litres

In 66kg of adulterated milk, the ratio of water to milk was found to be 1 $:$ 5. Water was added to it to make the ratio 3 $:$ 5. The quantity of water added is?