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 litres) to meet the demand?

a)10 b)30 c)40 d)50

• in the period of first 3 hrs, the demand of water = 10000.
  but the water poured by 10000 per hour.
  Thus the end of 3 hours it will poured 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
  
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