A bottle is full of dettol. One-third of it is taken out and then an e

1.	A bottle is full of dettol. One-third of it is taken out and then an equal amount of water is poured into the bottle to fill it. This operation is done four times. Find the final ratio of dettol and water in the bottle (a) 13 : 55 (b) 20 : 74 (c) 16 : 65 (d) 10 : 48
Answer» (c) 16 : 65 Let the original quantity of dettol in the bottle be x litres. Then, quantity of water in the bottle = 0 litres After the 1st operation : Quantity of dettol in the bottle = \(\big(x-\frac{x}{3}\big)\) litres = \(\frac{2x}{3}\) litres Quantity of water in the bottle = \(\frac{x}{3}\) litres After the 2nd operation : Quantity of dettol in the bottle = \(\big(\frac{2x}{3}-\frac13\times\frac{2x}{3}\big)\) litres = \(\frac{4x}{9}\) litres ∴ Quantity of water in the bottle = \(\big(\frac{x}{3}+\frac{2x}{9}\big)\) litres After the third operation : Quantity of dettol in the bottle = \(\big(\frac{4x}{9}-\frac13\times\frac{4x}{9}\big)\) litres = \(\frac{8x}{27}\) litres ∴ Quantity of water in the bottle = \(\big(\frac{x}{3}+\frac{2x}{9}+\frac{4x}{27}\big)\) litres After the fourth operation : Quantity of dettol in the bottle = \(\big(\frac{8x}{27}-\frac13\times\frac{8x}{27}\big)\) litres = \(\frac{16x}{81}\) litres Quantity of water in the bottle = \(\big(\frac{x}{3}+\frac{2x}{9}+\frac{4x}{27}+\frac{8x}{81}\big)\) litres = \(\big(\frac{27x+18x+12x+8x}{81}\big) \) litres = \(\frac{65x}{81}\) litres ∴ Required ratio = \(\frac{16x}{81}\): \(\frac{65x}{81}\) = 16 : 65

A bottle is full of dettol. One-third of it is taken out and then an equal amount of water is poured into the bottle to fill it. This operation is done four times. Find the final ratio of dettol and water in the bottle (a) 13 : 55 (b) 20 : 74 (c) 16 : 65 (d) 10 : 48

Answer»

(c) 16 : 65

Let the original quantity of dettol in the bottle be x litres.

Then, quantity of water in the bottle = 0 litres

After the 1st operation :

Quantity of dettol in the bottle = \(\big(x-\frac{x}{3}\big)\) litres = \(\frac{2x}{3}\) litres

Quantity of water in the bottle = \(\frac{x}{3}\) litres

After the 2nd operation :

Quantity of dettol in the bottle = \(\big(\frac{2x}{3}-\frac13\times\frac{2x}{3}\big)\) litres = \(\frac{4x}{9}\) litres

∴ Quantity of water in the bottle = \(\big(\frac{x}{3}+\frac{2x}{9}\big)\) litres

After the third operation :

Quantity of dettol in the bottle = \(\big(\frac{4x}{9}-\frac13\times\frac{4x}{9}\big)\) litres = \(\frac{8x}{27}\) litres

∴ Quantity of water in the bottle = \(\big(\frac{x}{3}+\frac{2x}{9}+\frac{4x}{27}\big)\) litres

After the fourth operation :

Quantity of dettol in the bottle = \(\big(\frac{8x}{27}-\frac13\times\frac{8x}{27}\big)\) litres = \(\frac{16x}{81}\) litres

Quantity of water in the bottle = \(\big(\frac{x}{3}+\frac{2x}{9}+\frac{4x}{27}+\frac{8x}{81}\big)\) litres

= \(\big(\frac{27x+18x+12x+8x}{81}\big) \) litres

= \(\frac{65x}{81}\) litres

∴ Required ratio = \(\frac{16x}{81}\): \(\frac{65x}{81}\) = 16 : 65