Thread: A leak

Josh has two leaking pipes in his basement. While waiting for the plumber to come, josh puts a bucket under each leak. The two buckets have the same capacity. The bucket uder the first leak fills in 20 minutes. The bucket under the second leak fills in 35 minutes.

Josh's brother takes away one of teh buckets and places the one bucket under the two leaks. About how long will it take for the one bucket to fill completely?

I need to come up with an equation that has "variables on both sides" to represent this situation. I figured out it takes about 12 minutes to fill up but cannot get the equation.

Originally Posted by Jman115

Josh has two leaking pipes in his basement. While waiting for the plumber to come, josh puts a bucket under each leak. The two buckets have the same capacity. The bucket uder the first leak fills in 20 minutes. The bucket under the second leak fills in 35 minutes.

Josh's brother takes away one of teh buckets and places the one bucket under the two leaks. About how long will it take for the one bucket to fill completely?

I need to come up with an equation that has "variables on both sides" to represent this situation. I figured out it takes about 12 minutes to fill up but cannot get the equation.

1. Let b denotes the volume of one bucket.
Then the amount of water from the 1st leak is and the amount of water from the 2nd leak is

2. Let t denotes the time which is necessary for both leaks to fill one bucket:



Solve for t.

Spoiler:

You should come out with 12 min 43,6

That seems like a complicated equation for a 7th grade word problem. Is there no easier way to set it up? All the other problems have been something like:

2x – 4 = 6x + 7

This just seems way over their heads. You have 3 different variables. The chapter is on having the same variable on two different sides of the equation.

I know the first leak is filling up 1/20th of the bucket per minute and the second leak is filling 1/35th of the bucket per minute. Combined they are filling 11/140th of the bucket per minute and 11 goes into 140 12.72 times.



So I know on one side of the equation I have to add 1/20v + 1/35v but the other side of the equation is escaping me. I know I will end up with 11/140v and have to multiply the other side by 140 and divide by 11. Just don't know what is over there.
I just don't know how to get the equation they are looking for. While the one above works, I don't think it is what they had in mind.

Is that a no?

Since my goal is to know how long it would take to fill the bucket, and the topic of the paper is the same variables on both sides of the equation, I need a common thread. Since my goal is time it takes to fill the bucket my guess is the common thread is supposed to have something to do with that.

So 1/20v + 1/35v represents how much time would represent one minute of time....

1 min = 1/20 + 1/35 = 7/140 + 4/140 = 11/140 (of bucket)

11/140 takes 1 min; 1 (the whole bucket!) takes 1 / (11/140) = 140/11 = 12 8/11 minutes.

That's the way it works...handle it "your way" if you wish; matters not: still 12 8/11 minutes !

The issue I am having is I need to come up with an equation that has the same variable on both sides. I know how to get the answer. The equation is what I am troubled with. This isn't my way, this is the expectation for class.

Mais voyons donc!

x = time to fill 1st bucket (20)
y = time to fill 2nd bucket (35)

1 / (1/x + 1/y)
= 1/[(x + y) / (xy)]
= xy / (x + y)
That's it, c'est tout!

Substitute the values:
20*35 / (20 + 35) = 700 / 55 = 12 8/11

If (another case) times are 25 and 40 minutes:
25*40 / (25 + 40) = 1000 / 65 = 15 25/65 = 15 5/13

OK?

Not really. This is a problem on a 7th grade worksheet. This is supposed to be something 7th graders can figure out.

All of the problems in this section in the book and the worksheets all have ONE variable per equation. You have an x and a y. These equations as shown before look like the following:

2x + 4 = 7x – 8


Everything I have done to figure out the problem and everything I have seen here is WAY over their heads and doesn't emulate the lessons taught in this section. I have contacted the company about it, but I doubt they will be prompt in contacting me.

I do appreciate the responses just please keep in mind that these equations and explanations are meant for 7th grade minds (12 years old). They are capable of a lot, but it just seems weird that the company would throw in this random problem that has nothing to do with the section. So far that has never happened on any of the other sheets I have done.

Well, seems like you got a problem.

One variable CAN be used here: x for 1st bucket, x+15 for 2nd;
handling similarly as I've shown gives x(x + 15) / (2x + 15) = 700/55

Still not easy for 7th grade...

According to the section I would need x on both sides of the equation. But that is closer to what I would need. I guess I'll just wait until the company responds. After staring at this problem for 2 hours and reading the responses from you guys I am pretty sure they flubbed up by putting this problem on this sheet.

Code:

  |       |            || ............them's the pipes!
@7:20   @4:35        @11:x
|  |    |  |          |  |
|  |    |  |          |  |
----    ----          ----
140     140           140

The LCM of 20 and 35 being 140, then assign 140 as the buckets' size.

Easy to see that 1st pipe has "speed" of 7 per minute, 2nd of 4 per minute.

Combining the pipes then means a combined speed of 7+4 = 11 per minute.
We let x = time this will take.
11x = 140
x = 140 / 11 = 12 8/11

May be a good/easy way to at least keep the 7th graders interested,
with you at the blackboard describing the above...

As you can see, a real easy equation comes out.
Perhaps that was the problem's intention?

That will work wonderfully. Thank you very much for all the time you have put into this.