Respuesta :
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:
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 =[tex]\dfrac{x}{4}[/tex]
If the smaller pipe fills the tank in y hours
Then the rate at which the smaller pipe fills the tank =[tex]\dfrac{x}{y}[/tex]
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 =[tex]\dfrac{x}{y-6}[/tex]
Next, we add both rates together
[tex]\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[/tex]
Since y has to be bigger than 6 hours, the smaller one takes 12 Hours to fill the tank.
