A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely ? Solve the word problem
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Here is your answer,
Let the volume of the tank be V
Together two taps take 2 hours to fill it completely.
Rate of both the taps together = V/2
Let the two taps be A and B.
Time taken by tap A = t hours
So, rate = V/t
Time taken by tap B = (t + 3) hours
So, rate = V/(t + 3)
Combined rate = V/t + V/(t + 3)
We already know that combined rate = V/2
⇒ V/t + V/(t + 3) = V/2 ..................(1)
Dividing this equation by V, we get.
⇒1/t + 1/(t + 3) = 1/2
Taking LCM of the denominators and then solving it.
⇒ (t + 3 + t)/t(t + 3) = 1/2
Now, cross multiplying.
⇒ 2t + 6 + 2t = t² + 3t
⇒ t² + 3t - 4t - 6 = 0
⇒ t² - t - 6 = 0
⇒ t² - 3t + 2t - 6 = 0
⇒ t(t - 3) + 2(t - 3) = 0
⇒ (t + 2) (t - 3) = 0
⇒ t = - 2 or t = 3
Since, time cannot be negative.
So, time taken by tap A is 3 hours
And, time taken by tap B = 3 + 3 = 6 hours.
Hope it helps you!
Here is your answer,
Let the volume of the tank be V
Together two taps take 2 hours to fill it completely.
Rate of both the taps together = V/2
Let the two taps be A and B.
Time taken by tap A = t hours
So, rate = V/t
Time taken by tap B = (t + 3) hours
So, rate = V/(t + 3)
Combined rate = V/t + V/(t + 3)
We already know that combined rate = V/2
⇒ V/t + V/(t + 3) = V/2 ..................(1)
Dividing this equation by V, we get.
⇒1/t + 1/(t + 3) = 1/2
Taking LCM of the denominators and then solving it.
⇒ (t + 3 + t)/t(t + 3) = 1/2
Now, cross multiplying.
⇒ 2t + 6 + 2t = t² + 3t
⇒ t² + 3t - 4t - 6 = 0
⇒ t² - t - 6 = 0
⇒ t² - 3t + 2t - 6 = 0
⇒ t(t - 3) + 2(t - 3) = 0
⇒ (t + 2) (t - 3) = 0
⇒ t = - 2 or t = 3
Since, time cannot be negative.
So, time taken by tap A is 3 hours
And, time taken by tap B = 3 + 3 = 6 hours.
Hope it helps you!
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