Tap A can fill 1/4th of tank in D hours and the remaining tank can be filled by tap B in (D + 16) hours. A is 40% more efficient than B. Both the taps are opened to fill the tank, but A was working with 2/7th of its efficiency and B was working with 2/5th of its efficiency, after few hours, they started working with their full efficiency, and remaining tank is filled in 10 hours. Find the time for which taps didn’t work with their full efficiency?

12 hours
8 hours
5 hours
10 hours
None of these

LohaHridoy

5 views

2025-01-04 04:42

1 Answers

AhmedNazir

Option 3 : 5 hours

Given, tap A is 40% more efficient than tab B

∴ The ratio of efficiency of A : B = 7 : 5

Total time taken by A alone to fill the tank = 4D hours

∴ Total time taken by B alone to fill the tank = 7/5 × 4D hours = 5.6D hours

Time taken by B alone to fill 3/4^th of the tank = 4.2D hours

According to the question,

4.2D = D + 16

⇒ D = 5 hours

∴ Total time taken by A alone to fill the tank = 4 × 5 = 20 hours

Total time taken by B alone to fill the tank = 5.6D = 28 hours

Let the total capacity of the tank be LCM of (20 and 28) = 140

Number of units filled by A in 1 hour = 140/20 = 7

Number of units filled by B in 1 hour = 140/28 = 5

Let the number of hours for which A and B didn’t work with their full efficiency be ‘x’ hours

Now, according to the question,

⇒ x + × 10 = 140

⇒ 4x + 10 × 12 = 140

⇒ x = 5 hours

5 views Answered 2022-09-04 04:29

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