Question

A can contains liquids L1 and L2 in the ratio 11 : 13. Draw 24 L of mixture from this Can and refill it by liquid L2. If the can has now L1 and L2 in the ratio 2 : 3, how much quantity of L1­ was in the can initially?

(a) 14 2/7L

(b) 86 3/7 L

(c) 187 1/7 L

(d) 55 5/7 L

Answer 1

The quantity of L1 initially in the can was 86 3/7 L.

• The ratio of liquids in the can L1 and L2 is 11:13.

• Let the common ratio be x.

• Therefore, quantity of L1 = 11x and quantity of L2 is 13x.

• Total amount of mixture = 11x + 13x = 24x.

• Now, 24L of the mixture from the can is drawn from the mixture and has added 24L of liquid L2 in the mixture.

• Amount of liquid left in the can is (24x-24) litres.

• Now , the amount of liquid L1 in the can= 11x - (24/24x)  × 11x and the amount of liquid L2 left in the can = 13x - (24/24x)  × 13x + 24

• Now, the ratio of the liquids become 2:3.

$\frac{L1}{L2}$ = $\frac{11x - 11}{13x - 13 + 24}$

$\frac{2}{3}$ = $\frac{11x - 11}{13x+11}$

26x + 22 = 33x - 33

55 = 7x

x = 55/7

• Now, the amount of L1 in the mixture is 11x

Amount of L1 = 11 × 55/7 = 605/7

Amount of L1 = 86 3/7 L