Get Help

Turn on suggestions

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

- Community
- :
- PTC Mathcad
- :
- PTC Mathcad
- :
- WORD PROBLEM, BET U CANT DO IT

Options

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Mute
- Printer Friendly Page

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Notify Moderator

10-02-2000
03:00 AM

10-02-2000
03:00 AM

WORD PROBLEM, BET U CANT DO IT

Two taps turned on together can fill a tank in 15 minutes. By themselves, one takes 15 minutes longer than the other to fill the tank. find the time taken to fill the tank by each tap alone.

Labels:

2 REPLIES 2

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Notify Moderator

10-02-2000
03:00 AM

10-02-2000
03:00 AM

WORD PROBLEM, BET U CANT DO IT

The faster takes 7.5x(1+sqrt(5)), about 24 min 16.23 seconds. The slower, quite obviously, takes 15 minutes more than that.

If you would like to present a challenge, it may be more appropriate in the "StudyWorks Challenge" section at the bottom.

If you would like a few more details of the solution, you can suggest some things you have tried and we will be glad to clarify whatever you like.

If you would like to present a challenge, it may be more appropriate in the "StudyWorks Challenge" section at the bottom.

If you would like a few more details of the solution, you can suggest some things you have tried and we will be glad to clarify whatever you like.

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Notify Moderator

10-03-2000
03:00 AM

10-03-2000
03:00 AM

WORD PROBLEM, BET U CANT DO IT

On 10/2/00 5:21:24 PM, Stormkill wrote:

>Two taps turned on together

>can fill a tank in 15 minutes.

>By themselves, one takes 15

>minutes longer than the other

>to fill the tank. find the

>time taken to fill the tank by

>each tap alone.

Let A be the fraction of the tank filled by the first tap. Let B be the fraction filled by the second tap. Thus, A+B represents the entire take, so it is 1. A+B=1.

Let t represent the time taken to fill the tank together. In this case, t=15 since we want t to be the time taken to fill the tank together. Let R and S be the rates at which both takes are filled.

Rt=A St=B A+B=1

Those are the equations I have so far. R is the rate at which it is filled. ie, rate = (fraction of tank) / (time to fill it). since the rate is calculated by the time taken to fill the tank, rate = 1 / time. Let T and U be the times required for each tap to fill it alone.

R=1/T S=1/U

Rt+St=1 combine the first set of equations

t/T+t/U=1 or 1/T+1/U=1/t

The formula is well known and is easily generalized. if n taps working together can perform the task in t1, t2, t3, ..., tn time, the time taken if they work together is

1/t=1/t1+1/t2+1/t3+...+1/tn.

This formula applies in many occasions. (eg, people mowing lawns)

Back to your problem.

Because one tap is 15 minutes faster, let U=T+15. Of course, t=15.

1/T+1/(T+15)=1/15

multiply both sides by the denominators to remove the fractions.

15(T+15)+15T=T(T+15)

15T+225+15T=T^2+15T

225+15T=T^2

T^2-15T-225=0

Solve this and the results is as Nathan said.

Craig

>Two taps turned on together

>can fill a tank in 15 minutes.

>By themselves, one takes 15

>minutes longer than the other

>to fill the tank. find the

>time taken to fill the tank by

>each tap alone.

Let A be the fraction of the tank filled by the first tap. Let B be the fraction filled by the second tap. Thus, A+B represents the entire take, so it is 1. A+B=1.

Let t represent the time taken to fill the tank together. In this case, t=15 since we want t to be the time taken to fill the tank together. Let R and S be the rates at which both takes are filled.

Rt=A St=B A+B=1

Those are the equations I have so far. R is the rate at which it is filled. ie, rate = (fraction of tank) / (time to fill it). since the rate is calculated by the time taken to fill the tank, rate = 1 / time. Let T and U be the times required for each tap to fill it alone.

R=1/T S=1/U

Rt+St=1 combine the first set of equations

t/T+t/U=1 or 1/T+1/U=1/t

The formula is well known and is easily generalized. if n taps working together can perform the task in t1, t2, t3, ..., tn time, the time taken if they work together is

1/t=1/t1+1/t2+1/t3+...+1/tn.

This formula applies in many occasions. (eg, people mowing lawns)

Back to your problem.

Because one tap is 15 minutes faster, let U=T+15. Of course, t=15.

1/T+1/(T+15)=1/15

multiply both sides by the denominators to remove the fractions.

15(T+15)+15T=T(T+15)

15T+225+15T=T^2+15T

225+15T=T^2

T^2-15T-225=0

Solve this and the results is as Nathan said.

Craig