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

It will take the smaller pipe 12 hours working alone

Step-by-step explanation:

In this question, we are asked to calculate the time it will take for a smaller pipe when working alone to fill a tank if the rate at which a larger one fills the tank is given and the rate at which they both fill when working together is given.

First, let’s represent the the volume to fill with say x cubic feet

When working together, their rate would be x/4

Now, the larger pipe can fill the tank in 6 hours less time. Let’s say the time taken for the smaller y, for the bigger, it definitely would be y-6

Hence rate of bigger will be x/(y-6)

For the smaller, the rate will be x/y

Now when we add both rates together, we have x/4 total rate for both

Let’s do tihis;

x/y + x/(y-6) = x/4

X(1/y + 1/y-6) = x(1/4)

1/y + 1/(y-6) = 1/4

y-6+y/(y(y-6) = 1/4

2y-6/y(y-6) = 1/4

Cross multiply;

4(2y-6) = y(y-6)

8y-24 = y^2 -6y

y^2 -6y -8y+24 = 0

y^2 -14y +24 = 0

y^2 - 2y-12y+24 = 0

y(y-2)-12(y-2) = 0

(y-12)(y-2) = 0

y = 12 or 2

y cannot be 2 because since it take the bigger 6 hours less, 2-6 = -4 is not possible as hours cannot be negative

12 Hours

Step-by-step explanation:

Let the volume of the tank be x cubic feet

When working together, the rate at which the tank would be filled =

If the smaller pipe fills the tank in y hours

Then the rate at which the smaller pipe fills the tank =

The larger tank working alone, can fill the tank in 6 hrs less time than the smaller one. i.e (y-6) hours

Then the rate at which the larger pipe fills the tank =

Next, we add both rates together

Since y has to be bigger than 6 hours, the smaller one  takes 12 Hours to fill the tank.

answer: #1 ads up by 8 and i don't know #2