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Let the total capacity of the tank be 96 units (LCM of 24 and 32) . So, the efficiency of tap 'Q' = (96/24) = 4 units/minute So, the efficiency of tap 'R' = (96/32) = 3 units/minute So, the efficiency of tap 'S' = 4 X 1.75 = 7 units/minute Increased efficiency of tap 'R' = 3 X (5/3) = 5 units/minute So, combined efficiency of taps 'R' and 'S' = 7 + 5 = 12 units/minute Therefore, required time = (96/12) = 8 minutes
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