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Pipe A is 40% more efficient then pipe E, Pipe B requires 4 minutes less to fill up the tank alone then pipe E, ratio of time taken by Pipe C and pipe A is 5 : 6, pipe C requires 15 minutes to empty the whole tank, if all the pipes are opened together then it requires 48/ 11 minutes to fill up the whole tank. Find the ratio of time taken by pipe B and pipe C to fill up the tank alone.

Pipe A is 40% more efficient then pipe E, Pipe B requires 4 minutes less to fill up the tank alone then pipe E, ratio of time taken by Pipe C and pipe A is 5 : 6, pipe C requires 15 minutes to empty the whole tank, if all the pipes are opened together then it requires 48/ 11 minutes to fill up the whole tank. Find the ratio of time taken by pipe B and pipe C to fill up the tank alone. Correct Answer 8 : 5

Given:

Time taken by pipe A to fill the tank alone = time taken by E × 40% less

Pipe B = Pipe E = 4

Pipe C: Pipe A = 5 : 6

Total time = 48/ 11 minutes

Pipe D = 15 minutes

Formula:

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

Calculation:

Let time taken by pipe E = X

⇒ Time taken by pipe A = X × (1 – 0.40)

⇒ Time taken by pipe A = 0.60 × X

⇒ Time taken by pipe D: Times taken by pipe A = 5 : 6

⇒ Time taken by pipe D = 0.6X × 5/6

Now, placing them in formula

⇒ 11/ 48 = (1/X) + (1/ 0.6X) + (2/X) – (1/15)

⇒ 11/ 48 = (45 + 75 + 90 – 3X)/45X

⇒ 12780 = 639X

⇒ X = 20

⇒ Time taken by pipe A = 20 × 0.60

⇒ Time taken by pipe A = 12 minutes

⇒ Time taken by pipe B = 20 – 4

⇒ Time taken by pipe B = 16 minutes

⇒ Time taken by C = 0.6 × 20 × 5/ 6

⇒ Time taken by C = 10 minutes

Ratio required = 16 : 10

∴ Ratio required is 8 : 5

C:A = 5:6 (given in question) and C takes 15 mins to empty the tank the same has been reversed in the solution D;A = 5:6 and D takes 15 mins to empty the tank....?

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