Question

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

Answer 1

Answer:

At t=0, Volume of Water = $V_{0}$ = 3,000 Litre

After t hours , the Volume of water in the tank = $V_{1}$

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

$V_{1}$ = Amount of water after an hour = 3,000 - $3,000 \times \frac{x}{100}$ = 3,000 - 30 x

$V_{2}$ = Amount of water after two hour =(3,000 - 30 x) - (3,000 - 30 x)×$\frac{x}{100}$= (3,000 - 30 x)$(\frac{100 - x}{100})$

Also, $V_{2} = 2881.2$

⇒$(3000 - 30 x)(100 -x)$ = 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, $V_{1} = k ^1 V_{0}$

k = $\frac{V_{1}}{V_{0}}$

$V_{1} = 3,000 - 30 \times 2 = 3,000 - 60= 2,940$

k = $\frac{2940}{3000}$= 0. 999.......(non terminating repeating)

k= 0.999 (approx)