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