Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

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 =$\dfrac{x}{4}$

If the smaller pipe fills the tank in y hours

Then the rate at which the smaller pipe fills the tank =$\dfrac{x}{y}$

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 =$\dfrac{x}{y-6}$

Next, we add both rates together

$\dfrac{x}{y-6}+\dfrac{x}{y}=\dfrac{x}{4}\\\dfrac{yx+x(y-6)}{y(y-6)}=\dfrac{x}{4}\\\dfrac{x(y+y-6)}{y(y-6)}=\dfrac{x}{4}\\\dfrac{2y-6}{y^2-6y}=\dfrac{1}{4}\\y^2-6y=4(2y-6)\\y^2-6y=8y-24\\y^2-6y-8y+24=0\\y^2-12y-2y+24=0\\y(y-12)-2(y-12)=0\\(y-12)(y-2)=0\\y=12,2$

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