two taps an together fill a tank in 2 8/11 hrs the tap of the smaller diameter 1 hour more than the larger one to fill the tank separately find the time in which each tap can separately fill the tank
Answers
Answer:
Time taken by the larger tap to fill the tank separately = 5 hours
Time taken by the smaller tap to fill the tank separately = 5 + 1 = 6 hours
Step-by-step explanation:
Time taken by two taps together = 2 8/11 = 30/11 hours
Time taken by larger tap to fill the tank separately = x hours
Time taken by smaller tap to fill the tank separately = x+1 hours
Now,
1/x + 1/x+1 = 11/30
x+1 + x/ x^2 + x = 11/30
2x + 1/ x^2 + x = 11/30
Cross multiplying,
30(2x + 1) = 11(x^2 + x)
60x + 30 = 11x^2 + 11
11x^2 - 49x - 30 = 0
11x^2 - 55x + 6x - 30 = 0
11x(x - 5) + 6(x - 5) = 0
(11x + 6) (x-5) = 0
Therefore,
x = 5 and x = -6/11
Since time cannot be negative, we cancel out -6/11; So,
Time taken by the larger tap to fill the tank separately = 5 hours
Time taken by the smaller tap to fill the tank separately = 5 + 1 = 6 hours