Jar A contains 78 litres of milk and water in the respective ratio of

1.	Jar A contains 78 litres of milk and water in the respective ratio of 6 : 7. 26 litres of the mixture was taken out from Jar A. What quantity of milk should be added to jarA, so that water constitutes 40% of the resultant mixture in jar A?1). 8 litres2). 36 litres3). 12 litres4). 14 litres
Answer» Solution Jar A has 78 litres of mixture of milk and water in the respective RATIO of 6 : 7 => Quantity of milk in Jar A = $\frac{6}{13} \times 78 = 36$ litres Quantity of water in Jar A = $78 - 36 = 42$ litres 26 litres of the mixture was taken out from Jar A, i.e., $\frac{26}{78} = (\frac{1}{3})^{rd}$ => Milk left = $36 - \frac{1}{3} \times 36 = 24$ Water left = $42 - \frac{1}{3} \times 42 = 28$ Let milk added to jar A = $x$ litres Acc. to QUES, => $\frac{24 + x}{28} = \frac{60}{40}$ => $\frac{24 + x}{28} = \frac{3}{2}$ => $48 + 2X = 84$ => $2x = 84 - 48 = 36$ => $x = \frac{36}{2} = 18$ litres

Jar A contains 78 litres of milk and water in the respective ratio of 6 : 7. 26 litres of the mixture was taken out from Jar A. What quantity of milk should be added to jarA, so that water constitutes 40% of the resultant mixture in jar A?1). 8 litres2). 36 litres3). 12 litres4). 14 litres

Answer»

Solution

Jar A has 78 litres of mixture of milk and water in the respective RATIO of 6 : 7

=> Quantity of milk in Jar A = $\frac{6}{13} \times 78 = 36$ litres

Quantity of water in Jar A = $78 - 36 = 42$ litres

26 litres of the mixture was taken out from Jar A, i.e., $\frac{26}{78} = (\frac{1}{3})^{rd}$

=> Milk left = $36 - \frac{1}{3} \times 36 = 24$

Water left = $42 - \frac{1}{3} \times 42 = 28$

Let milk added to jar A = $x$ litres

Acc. to QUES, => $\frac{24 + x}{28} = \frac{60}{40}$

=> $\frac{24 + x}{28} = \frac{3}{2}$

=> $48 + 2X = 84$

=> $2x = 84 - 48 = 36$

=> $x = \frac{36}{2} = 18$ litres