A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 1800 m³. The emptying capacity of the tank is 10 m³/min higher than its filling capacity, and the pump needs 6 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump in m³/min?
- AOption (a)
18
- BOption (b)
27
- COption (c)
50
- DOption (d)
65
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Correct answer
50
Why this answer is correct
Shortcut Trick
Let filling capacity be x. Then emptying capacity is x + 10.
Time difference formula: [Capacity / (Smaller Rate)] − [Capacity / (Larger Rate)] = Time Difference
1800/x − 1800/(x + 10) = 6
Test Option (3): x = 50 ⇒ 1800/50 − 1800/60 = 36 − 30 = 6 minutes. Matches exactly.
∴ The correct answer is 50 m3/min.
Alternate Method
Given: Total Tank Capacity = 1800 m3, Difference in capacity = 10 m3/min, Difference in time = 6 minutes.
Formula Used: Time = Capacity ÷ Rate
⇒ Let Filling Capacity = x m3/min
⇒ Emptying Capacity = (x + 10) m3/min
⇒ 1800/x − 1800/(x + 10) = 6
⇒ 1800 [(x + 10 − x) / (x(x + 10))] = 6
⇒ 1800 × 10 / (x2 + 10x) = 6
⇒ 3000 = x2 + 10x
⇒ x2 + 10x − 3000 = 0
⇒ (x + 60)(x − 50) = 0
⇒ Since capacity cannot be negative, x = 50 m3/min.
∴ The correct answer is 50 m3/min.
Additional Information
Time and Work Relation
Rate of work is inversely proportional to the time taken. If capacity increases, time decreases: Time = Work / Rate.
Quadratic Shortcut
For the equation x2 + (a−b)x − ab = 0, the roots are b and −a. In physical problems like capacity, always pick the positive root.
Pipes and Cisterns
Filling is considered positive work while emptying is considered negative work in net efficiency problems.