Filling a tank. Two pipes are connected to the same tank. Working together they can fill the tank in 4 hrs. The larger pipe working alone can fill the tank 6 hrs less than the smaller one, HOw long would the smaller one take, working alone, to fill the tank?

Question

Filling a tank. Two pipes are connected to the same tank. Working together they can fill the tank in 4 hrs. The larger pipe working alone can fill the tank 6 hrs less than the smaller one, HOw long would the smaller one take, working alone, to fill the tank?

Answer 1

Let time taken by smaller pipe be t hrs

then time takes by larger pipe is t-6 hrs

rate of smaller pipe = 1/t
rate of larger pipe = 1/(t-6)
combined rate = 1/t + 1/(t-6)
= (2t-6)/(t(t-6))

time at combined rate = 1/ [ (2t-6)/(t(t-6)) ]
= t(t-6)/(2t-6)

then t(t-6)/(2t-6) = 4
t^2 - 6t = 8t - 48
t^2 - 14t + 48 = 0
(t-8)(t-6) = 0
t = 8 or t = 6 , but t = 6 would make the rate of the larger pipe undefined, so we reject that answer

So it would take 8 hours to fill with only the smaller pipe.

Answer 2

To solve this problem, we can assign variables to the rates at which the pipes fill the tank. Let's say the smaller pipe fills the tank at a rate of x tanks per hour, and the larger pipe fills the tank at a rate of y tanks per hour.

From the information given, we know that when the two pipes work together, they can fill the tank in 4 hours. This means that the combined rate of the two pipes is 1 tank per 4 hours. So we can write the equation:

1/4 = x + y

We also know that the larger pipe working alone can fill the tank 6 hours less than the smaller one, which means its rate is slower. So we can write another equation:

y = x - 6

Now we have a system of two equations with two variables. We can solve this system to find the values of x and y.

First, let's substitute the second equation into the first equation:

1/4 = x + (x - 6)

Simplifying this equation gives us:

1/4 = 2x - 6

Next, let's isolate the variable x by adding 6 to both sides of the equation:

1/4 + 6 = 2x

Combining the fractions and simplifying further gives us:

25/4 = 2x

Dividing both sides by 2 gives us:

25/8 = x

So the smaller pipe fills the tank at a rate of 25/8 tanks per hour.

Now that we have the rate of the smaller pipe, we can determine how long it would take for it to fill the tank alone. Since the rate is given in tanks per hour, the time it takes (t) to fill the tank is the reciprocal of the rate:

t = 8/25

Therefore, the smaller pipe would take 8/25 hours to fill the tank working alone.