Pipe A and B can fill a tank in 20 and 40 hours respectively. An outlet pipe C can empty it in 60 hours. Just before the opening of the taps, some dust enters in the inlet pipes which reduce the efficiency of both A and B pipes like pipe A works with 1/2 of its efficiency and pipe B works with 1/3 of its efficiency. Find the total time taken to fill the whole tank which is already half filled.?

Question

Pipe A and B can fill a tank in 20 and 40 hours respectively. An outlet pipe C can empty it in 60 hours. Just before the opening of the taps, some dust enters in the inlet pipes which reduce the efficiency of both A and B pipes like pipe A works with 1/2 of its efficiency and pipe B works with 1/3 of its efficiency. Find the total time taken to fill the whole tank which is already half filled.?

40 hours
50 hours
70 hours
55 hours
30 hours

LohaHridoy

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2025-01-04 04:42

1 Answers

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AhmedNazir · Accepted Answer · 2022-09-04T04:29:57+06:00

Option 5 : 30 hours

Given:

Pipe A can fill the tank in 20 hours

Pipe B can fill the tank in 40 hours

Pipe C can empty the tank in 60 hours

A works with 1/2 of its efficiency

B works with 1/3 of its efficiency

Concept used:

Work = Time taken × Efficiency

Total efficiency = Efficiency of Filling pipes – Efficiency of emptying pipes

Calculation:

Let the total work be x

Efficiency of pipe A = x/20

Efficiency of pipe B = x/40

Efficiency of pipe C = x/60

Now, after dust enters Pipes A and B

Efficiency of pipe A = x/20 × (1/2)

⇒ Efficiency of Pipe A = x/40

Efficiency of pipe B = x/40 × (1/3)

⇒ Efficiency of pipe B = x/120

Total efficiency = Efficiency of Filling pipes – Efficiency of emptying pipes

Now, Total efficiency of A + B + C = x/40 + x/120 – x/60

⇒ Total efficiency of A + B + C = 2x/120 = x/60

Part of tank which is already filled = x/2

Tank to be filled = x – x/2 = x/2

Time taken to fill the remaining part of tank = x/2 ÷ (x/60)

⇒ 30 hours

∴ Time taken to fill the whole tank if it is already half filled is 30 hours

Alternate solution:

A can fill the tank in 20 hours

B can fill the tank in 40 hours

C can empty the tank in 60 hours

L.C.M of 20, 40 and 60 = 120

Efficiency of A = 6

Efficiency of B = 3

Efficiency of C = 2

Now, after dust enters Pipes A and B

Efficiency of A = 6 × 1/2 = 3

Efficiency of B = 3 × 1/3 = 1

Efficiency of C = 2

Total efficiency = Efficiency of Filling pipes – Efficiency of emptying pipes

Total Efficiency of A + B + C = 3 + 1 – 4

⇒ Total Efficiency of A + B + C = 2

Part of tank which is already filled = 120/2 = 60

⇒ Part of tank which is to be filled = 120 – 60 = 60

Time taken to fill the remaining tank = 60/2

⇒ 30 hours

∴ Time taken to fill the whole tank if it is already half filled is 30 hours