Answer:
x = 2; K = 0.98
Step-by-step explanation:
At time, t = 0, Volume of Water = 3,000 litres
After t hours , the Volume of water in the tank = Vt
At the end of every hour there is a x% less water in the tank then at the start of the hour
After 1 hour, volume of water left Vt = V1
V1 = 3,000 - 3000 × (x/100)
V1 = 3,000 - 30x
After 2 hours, Volume of water left is V2
V2 = (3,000 - 30 x) - (3,000 - 30 x) × (x/100)
V2 = (3,000 - 30 x) - (3000x - 30x²)/100
V2 = {100(3000 - 30x) - x(3000 - 30x)} /100
V2 = {(3000 - 30x)(100 - x)}/100
Recall, V2 = 2881.2
Therefore
{(3000 - 30x)(100 - x)}/100 = 2881.2
(3000 - 30x)(100 - x) = 288120
300000 - 3000 x - 3000 x + 30 x² = 288120
30 x ² - 6000 x + 11880 = 0
Dividing both sides by 30
x² - 200 x + 396=0
Factorizing:
(x -198)(x-2) = 0
x = 198 or x = 2
Since the percentage by which the water reduces cannot be greater than 100, x = 2 is chosen.
Step 2: Solving for k
Given that Vt=KtV0
K = Vt/V0
At t = 1
V1 = 3000 - 30x
V1 = 3000 - 30 × 2
V1 = 3000 - 60
V1 = 2940
K = 2940/3000
K = 0.98
x+3