Question

Pipes A, B and C can fill a tank in 10, 15 and 18 hours, respectively. A and C are opened for 5 hours, then only pipe A is closed and B is opened at the same time (with C). The total time taken to fill the tank completely is:

• A. $$3\frac{3}{2}$$
• B. $$6\frac{9}{11}$$
• C. $$5\frac{8}{11}$$
• D. $$2\frac{1}{3}$$

Answer 1

Answer:

$$6\frac{9}{11}$$

Solution:

Formula:

Efficiency = Total work done/ total time taken

Calculation:

Let the volume of the tank

⇒ LCM of 5, 10, 15, and 18 = 90 units

Efficiency of A = 90/10 = 9 units/hr

Efficiency of B = 90/15 = 6 units/hr

Efficiency of C = 90/18 = 5 units/hr

Total volume filled by A and C in 5 hours

⇒ (9 + 5) × 5 = 70 units

Remaining volume = 90 - 70 = 20 units

 

Total volume filled by B and C in 1 hour

⇒ 6 + 5 = 11 units/hr

Time taken to fill the remaining volume by B and C

$\frac{20}{11}=1\frac{9}{11}$

Time taken to fill the complete tank $= 5 +1\frac{9}{11}=6\frac{9}{11}$hours

The total time taken to fill the tank  is $6\frac{9}{11}$hours.