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Water is continuously poured from a reservoir to a locality at the steady rate of 10,000 liters per hour. When delivery exceeds demand the excess water is stored in a tank. If the demand for 8 consecutive three-hour periods is 10000, 10000, 45000, 25000, 40000, 15000, 60000 and 35000 liters respectively, what will be the minimum capacity required of the water tank (in 1000 liters) to meet the demand?

A10

B30

C20

D40

Answer: Option (Login/Signup)

Show Explanation

In the period of first 3 hrs, the demand of water = 10000.

But the water poured is 10000 per hour.

Thus at the end of 3 hours, it will pour 30000 ltr, then the excess of water = 20000.

Similarly at the end of 2nd 3 hr period the excess of water 20000+20000 = 40000.



Third, 3 hrs water poured = 30000

already we have 40000 of excess water and demand 45000.

Then the excess water at the end of 3rd 3 hrs

= 40000+30000-45000

= 25000

Fourth 3 hrs. water = 25000+30000–25000

= 30000

Fifth = 30000+30000–40000

= 20000

Sixth = 20000+30000–15000=35000

Seventh = 35000+30000–60000

= 5000

Eighth = 5000+30000–35000

= 0

By analyzing the above results, we have 40000 ltrs is the highest measure of excess water.

So the minimum capacity of water tank = 40000 litres = 40

i.e. = 40

Asked In :: iGate TCS NQT

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