Math, asked by yashsvisinghooszfd2, 11 months ago

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

Answered by nishbiju
0

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

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