Question

A tank contains 5832 litres of water. Each day one-third of the water in the tank is removed and not replaced. How much water remains in the tank at the end of 6 days? [Hint: answer 512 litres] Show me the steps​

Answer 1

Here is your answer.

A tank contains 5832 litres of water.

Total amount of water in tank = 5832 litres.

Each day one-third of the water in the tank is removed and not replaced.

Since one third of the water is removed for six days.

1st day = 1/3rd of 5832 litres is removed

`= (5832)/(3) = 1944`

Amount of water left = 5832 - 1944 = 3888 litres

2nd day = 1/3rd of 3888 litres is removed

`= (3888)/(3) = 1296`

Amount of water left = 3888 - 1296 = 2592 litres

3rd day = 1/3rd of 2592 litres is removed

`= (2592)/(3) = 864`

Amount of water left = 2592 - 864 = 1728 litres

4th day = 1/3rd of 1728 litres is removed

`= (1728)/(3) = 576`

Amount of water left = 1728 - 576 = 1152 litres

5th day = 1/3rrd of 1152 litres is removed

`= (1152)/(3) =384`

Amount of water left = 1152 - 384 = 768 litres

6th day = 1/3rrd of 768 litres is removed

`= (768)/(3) = 256`

Amount of water left = 1768 - 256 = 512 litres

So, 512 litres water remains in the tank at the end of 6 days