# The bathtup-Problem

• September 8th 2011, 01:20 PM
Benjy
The bathtub-Problem
Hello!

I've encountered a difficult problem here and would like some help.
It would be great for my understanding If you could help me solve it with fractions.

A bathtub is filled in 5 minutes and emptied in 10 minutes when the plug is pulled out. How much time will it take to fill the bathtub if you forget to insert the plug?
• September 8th 2011, 01:43 PM
skeeter
Re: The bathtup-Problem
Quote:

Originally Posted by Benjy
Hello!

I've encountered a difficult problem here and would like some help.
It would be great for my understanding If you could help me solve it with fractions.

A bathtub is filled in 5 minutes and emptied in 10 minutes when the plug is pulled out. How much time will it take to fill the bathtub if you forget to insert the plug?

rate tub is filled ... $\frac{1 \, tub}{5 \, min}$

rate tub is emptied ... $\frac{1 \, tub}{10 \, min}$

combined rate ... $\frac{1 \, tub}{5 \, min} - \frac{1 \, tub}{10 \, min} = ?$
• September 8th 2011, 02:21 PM
Vingar
Re: The bathtup-Problem
Quote:

Originally Posted by Benjy
Hello!

I've encountered a difficult problem here and would like some help.
It would be great for my understanding If you could help me solve it with fractions.

A bathtub is filled in 5 minutes and emptied in 10 minutes when the plug is pulled out. How much time will it take to fill the bathtub if you forget to insert the plug?

I'm not sure whether there is a specific answer for this question, here is a try:

If x is defined as the volume of the bathtub, then x/5 is the speed of water inflow per minute and x/10 is the speed of water outflow per minute. Then you have a net inflow rate of y=(x/5-x/10).

Hope this helps :)
• September 8th 2011, 02:25 PM
skeeter
Re: The bathtup-Problem
Quote:

Originally Posted by Vingar
I'm not sure whether there is a specific answer for this question ...

then you better have another look at post #2
• September 8th 2011, 02:27 PM
TheBoss
Re: The bathtup-Problem
Quote:

Originally Posted by Benjy
Hello!

I've encountered a difficult problem here and would like some help.
It would be great for my understanding If you could help me solve it with fractions.

A bathtub is filled in 5 minutes and emptied in 10 minutes when the plug is pulled out. How much time will it take to fill the bathtub if you forget to insert the plug?

This question is very simple. If the time it takes to empty is double the time it takes to fill, it should fill in 10 minutes, since its rate of fill was cut in half. Correct?
• September 8th 2011, 02:51 PM
Benjy
Re: The bathtup-Problem
Quote:

Originally Posted by TheBoss
This question is very simple. If the time it takes to empty is double the time it takes to fill, it should fill in 10 minutes, since its rate of fill was cut in half. Correct?

yes It's correct

Thanks
• September 8th 2011, 02:55 PM
Benjy
Re: The bathtup-Problem
Quote:

Originally Posted by skeeter
rate tub is filled ... $\frac{1 \, tub}{5 \, min}$

rate tub is emptied ... $\frac{1 \, tub}{10 \, min}$

combined rate ... $\frac{1 \, tub}{5 \, min} - \frac{1 \, tub}{10 \, min} = ?$

Thank you

I understand you

I first tried to fit it into the "distance=speed*time"-Formula but it didn't work for me
• September 8th 2011, 03:03 PM
Benjy
Re: The bathtup-Problem
Quote:

Originally Posted by Vingar
I'm not sure whether there is a specific answer for this question, here is a try:

If x is defined as the volume of the bathtub, then x/5 is the speed of water inflow per minute and x/10 is the speed of water outflow per minute. Then you have a net inflow rate of y=(x/5-x/10).

Hope this helps :)

Thank you