Question

A tanks 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?

Answer 1

Let x and y be the respective times in which the tap fills the tank
Now this is work and efficiency question
Before solving this u need to remember that linear operations can be performed only on work done and the rate of doing work
So if a tap is filling a tank in x time then it is doing one work in a designated amount of time
So,
Rate of doing work=(1/x)
Now as per question if both the taps are open then the tank is filled in 2 hours
So
(1/x)+(1/y)=(1/2)

Now it is further stated as
smaller tap alone takes 3 hours more to fill the tank in the same amount
So it can derived that,

(1/x)+(1/(x-3))=(1/2)
Solving this equation ,
(2x+3)/(x^2 -3x)=(1/2)
Rearranging the equation,
x^2-7x+6=0
The following quadratic equation can be solved as,
(x-1)(x-6)=0
x=1,x=6
But x=1 is not possible as x-3 cannot be negative
So
x=6
y=x-3=6-3=3
So the individual time taken by the taps are 3 hours and 6 hours respectively

Answer 2

Solution :-

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.

Answer.