Asked by marathe khemraj
                Two taps of various or different diameter can fill a tank in 9 and 3/8 hours. The tap of larger diameter takes 10 hrs less than the smaller one  to fill the tank separately.Find the time in which each tap can separately fill the tank. 
            
            
        Answers
                    Answered by
            Reiny
            
    time taken by larger diameter tap --- x hrs
time taken by smaller diameter tap ---x + 10 hrs
combined rate = 1/x + 1/(x+1) = (2x+10)/(x(x+10))
 
(2x+10)/(x(x+10)) = 1/(75/8) = 8/75
8x^2 + 80x = 150x + 750
8x^2 - 70x - 750 = 0
(x-15)(8x + 50) = 0
x = 15 or x = some negative, which we'll reject
Time for the larger tap is 15 hrs
time for the smaller tap is 25 hrs
check:
combined rate = 1/15 + 1/25 = 8/75
time at combined rate = 1/(8/75) = 75/8 or 9 3/8 hrs
    
time taken by smaller diameter tap ---x + 10 hrs
combined rate = 1/x + 1/(x+1) = (2x+10)/(x(x+10))
(2x+10)/(x(x+10)) = 1/(75/8) = 8/75
8x^2 + 80x = 150x + 750
8x^2 - 70x - 750 = 0
(x-15)(8x + 50) = 0
x = 15 or x = some negative, which we'll reject
Time for the larger tap is 15 hrs
time for the smaller tap is 25 hrs
check:
combined rate = 1/15 + 1/25 = 8/75
time at combined rate = 1/(8/75) = 75/8 or 9 3/8 hrs
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