Pipe C requires 50% more time then pipe A to fill up the tank when operated alone and ratio of time taken to fill the tank by pipe B and pipe C is 4: 5. It takes 4 minutes to fill up the tank when all the pipes are opened together. What is the total time taken to fill the tank if all three pipes are alternatively opened after every 5 minutes?

Question

Pipe C requires 50% more time then pipe A to fill up the tank when operated alone and ratio of time taken to fill the tank by pipe B and pipe C is 4: 5. It takes 4 minutes to fill up the tank when all the pipes are opened together. What is the total time taken to fill the tank if all three pipes are alternatively opened after every 5 minutes?

11.8 minutes
10.75 minutes
22.5 minutes
10.45 minutes
11.25 minutes

LohaHridoy

7 views

2025-01-04 04:42

1 Answers

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

Option 5 : 11.25 minutes

Given:

Pipe C = 50% + Pipe A

Pipe B: Pipe C = 4 : 5

Pipes opened every 5 minutes

Formula:

If time taken by pipe A to fill tank is ‘a’ minutes and by pipe B is ‘b’ minutes then total time taken to fill the tank t ⇒ (1/t) = (1/a) + (1/b)

Calculation:

Let, time taken to fill the tank by pipe A = X

⇒ Pipe C = 0.5X + X

⇒ Pipe C = 1.5X

⇒ Pipe B: Pipe C = 4: 5

⇒ Pipe B = 4 × 1.5X/5

⇒ 1/ 4 = (1/ X) + (5/6X) + (1/1.5X)

⇒ 1/ 4 = (6 + 5 + 4)/(6X)

⇒ 1/ 4 = 15/ 6X

⇒ 6X = 60

⇒ X = 10 minutes

⇒ Time taken by pipe C = 15 (∵ 1.5 × 10 = 15)

⇒ Time taken by pipe B = 6 × 10/5

⇒ Time taken by pipe B = 12 minutes

Calculating units/ minutes for each pipe

LCM for 10, 12 and 15 = 60

⇒ Pipe A = 6 (∵ 60/10 = 6)

⇒ Pipe B = 5 (∵ 60/12 = 5)

⇒ Pipe C = 4 (∵ 60/15 = 4)

Total quantity to be filled = 60 unit

⇒ Units filled by pipe A = 30 (∵ 6 × 5 min = 30)

Remaining Unit = 60 - 30 = 30

⇒ Units filled by pipe B = 25 (∵ 5 × 5 min= 25)

Remaining Unit = 30 - 25 = 5

⇒ Units filled by pipe C = 5 (∵ 4 × 5/4 min= 5)

⇒ Time taken = 5 + 5 + (5/4)

⇒ Time taken = 11.25 minutes

∴ Time taken to fill up the tank is 11.25 minutes