Question
Tap A is 50% more efficient than Tap B. Tap A can fill
one-third of tank in D hours and the remaining tank can be filled by tap B in (D + 12) hours. If both taps are opened to fill the tank, but due to partial closing of valve, tap A was working with one-third of its efficiency and tap B was working with one-half of its efficiency, after few hours, they started working with their full efficiency, and remaining tank is filled by them in 8 hours, then find the number of hours for which taps didn’t work with their full efficiency?Solution
Given, tap A is 50% more efficient than tap B. So, ratio of efficiency of tap A : tap B = 3 : 2 Total time taken by tap A alone to fill the tank = 3D hours So, total time taken by tap B alone to fill the tank = (3/2) × 3D = (9D/2) hours Time taken by tap B alone to fill 2/3 of the tank = (2/3) × (9D/2) = 3D hours According to question, 3D = D + 12 2D = 12 D = 6 hours Therefore, Total time taken by tap A alone to fill the tank = 3 × 6 = 18 hours Total time taken by tap B alone to fill the tank = (9D/2) = (9 × 6)/2 = 27 hours Let the total capacity of tank be LCM (18 and 27) = 54 units Number of units of water filled by tap A in one hour = 54 ÷ 18 = 3 units Number of units of water filled by tap B in one hour = 54 ÷ 27 = 2 units Let the number of hours for which tap A and tap B didn’t work with their full efficiency be ‘x’ hours. Therefore, according to question, [(1/3) × 3 + (1/2) × 2]x + (3 + 2) × 8 = 54 2x + 8 × 5 = 54 2x = 54 – 40 x = 7 hours
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