At time t=0 hours a tank contains 3000 litres of water . water leakes from the tank. at the end of every hour there is a x% less water in the tank then at the start of the hour the volume of water in litres, in the tank at time t hours is v1. given that v1=k^1v0 v2=2881.2 work out the value of k and value of x

At t=0, Volume of Water = = 3,000 Litre

After t hours , the Volume of water in the tank =

As it is also given , At the end of every hour there is a x% less water in the tank then at the start of the hour

= Amount of water after an hour = 3,000 - = 3,000 - 30 x

= Amount of water after two hour =(3,000 - 30 x) - (3,000 - 30 x)×= (3,000 - 30 x)

Also,

⇒ = 288120

⇒300000 - 3000 x - 3000 x + 30 x²  = 288120

⇒ 30 x ² - 6,000 x + 11880= 0

Dividing both sides by 30, we get

⇒ x² - 200 x + 396=0

Factorizing the quadratic expression

(x -198)(x-2) = 0

x = 198, x = 2

⇒ Which gives a value of x = 2 as a possible solution of this equation.

Also,

k =

k = = 0. 999.......(non terminating repeating)

k= 0.999 (approx)