Question:easy
It takes 6 hours for pump A, used alone, to fill a tank of water. Pump B used alone takes 8 hours to fill the same tank. A, B and another pump C all together fill the tank in 2 hours. How long would pump C take, used alone, to fill the tank?
It takes 6 hours for pump A, used alone, to fill a tank of water. Pump B used alone takes 8 hours to fill the same tank. A, B and another pump C all together fill the tank in 2 hours. How long would pump C take, used alone, to fill the tank?
Show Hint
Add the rates of A, B and C to equal the combined rate of 1/2 tank per hour, then solve for C's rate.
Updated On: Jul 15, 2026
- 4.8
- 6
- 5.6
- 3
Show Solution
The Correct Option is A
Solution and Explanation
This can also be solved using the LCM-of-capacity (unit work) method instead of fractions.
Assume the tank holds a convenient number of units equal to the LCM of 6, 8 and 2, which is 24 units.
Pump A's rate = $\frac{24}{6} = 4$ units per hour.
Pump B's rate = $\frac{24}{8} = 3$ units per hour.
Together A, B and C fill the tank in 2 hours, so their combined rate = $\frac{24}{2} = 12$ units per hour.
Rate of C alone = combined rate - rate of A - rate of B = $12 - 4 - 3 = 5$ units per hour.
Time taken by C alone to fill the full 24-unit tank = $\frac{24}{5} = 4.8$ hours.\[\boxed{4.8 \text{ hours}}\]
Assume the tank holds a convenient number of units equal to the LCM of 6, 8 and 2, which is 24 units.
Pump A's rate = $\frac{24}{6} = 4$ units per hour.
Pump B's rate = $\frac{24}{8} = 3$ units per hour.
Together A, B and C fill the tank in 2 hours, so their combined rate = $\frac{24}{2} = 12$ units per hour.
Rate of C alone = combined rate - rate of A - rate of B = $12 - 4 - 3 = 5$ units per hour.
Time taken by C alone to fill the full 24-unit tank = $\frac{24}{5} = 4.8$ hours.\[\boxed{4.8 \text{ hours}}\]
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