Questions on Word Problems: Rate of work, PAINTING, Pool Filling answered by real tutors!

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Question 1210372: If painter A can paint a wall in one hour and painter B can paint the same wall in two hours, how long does it take for painter A and B to paint the same wall together?

Found 4 solutions by math_tutor2020, josgarithmetic, mccravyedwin, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

Let's say the wall has an area of 300 square feet.
This value can be changed to anything you want, and the final answer will remain the same.

When working alone painter A can paint this wall in one hour, so his or her rate is 300/1 = 300 sq ft per hour.
The formula to use would be: speed = amountDone/time
Painter B has a rate of 300/2 = 150 sq ft per hour when working alone.

When the two work together, assuming neither worker slows down the other, we add their unit rates.
A+B = 300+150 = 450
Their combined rate is 450 sq ft per hour.

rate*time = amountDone
(450 sq ft per hr)*(x hours) = 300 sq ft
450*x = 300
x = 300/450
x = (2*150)/(3*150)
x = 2/3
It takes 2/3 of an hour if the two work together.
This is equivalent to 40 minutes because (2/3)*60 = 40.

Answer: 40 minutes

Answer by josgarithmetic(39613)   (Show Source):
You can put this solution on YOUR website!

PAINTER RATE TIME WORK A 1 1 1 B 1/2 2 1 A&B 1+1/2 ? 1

x, the time for A and B together to do the one job of "paint the wall"

or about 40 minutes

Answer by mccravyedwin(405)   (Show Source):
You can put this solution on YOUR website!

Another way to look at it is what they COULD do in the same time. Painter A can paint a wall in 60 minutes and painter B can paint the same wall in 120 minutes. Therefore, painter A COULD paint 2 walls in 120 minutes while painter B COULD paint only 1 wall in that same 120 minutes, so in 120 minutes together they COULD paint 3 walls, so to paint only 1 wall it would take them only 1/3 of 120 minutes or 40 minutes. Edwin

Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.
If painter A can paint a wall in one hour and painter B can paint the same wall in two hours,
how long does it take for painter A and B to paint the same wall together?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

This problem is so simple that I will solve it mentally using reasoning, ONLY.

Working together, they will spend equal time. Since painter A works twice as fast as painter B, painter A will make twice greater job than painter B. Hence, painter A will make 2/3 of the job, while painter B will make 1/3 of the job. To make 2/3 of the job, painter A needs 2/3 of an hour, i.e. 40 minutes. Hence, working together, both painters will complete the job in 2/3 of an hour, or in 40 minutes.

Solved.

Simple problems should be solved with simple methods, isn't it ?

///////////////////////////////////

This problem is interesting and deserves a thorough discussion.

From one hand side, it is a typical joint-work word problem.

But a typical joint-work word problem usually tells about small pieces of a whole job
that can be done by participants in a certain time interval.

Therefore, been a typical problem, this given problem, from the other hand side, is unusual,
because it tells about grand pieces of a whole job, like the whole wall, which can be painted in one hour.

And if a student is non-experienced, this immediately puts the student in a dead end.

Some tutors may come explaining that a student can work with grand pieces exactly in the same
way as he/she works with small pieces.

But I do not believe that an average/typical beginner student will be satisfied and will accept such an explanation.
Such explanations can be good for tutors or for teachers (because formally they complete their job doing this way),
but a student will unlikely be satisfied and unlikely will accept such a solution.

Again, explanations with grand pieces are good for teacher and for tutors, especially if they want to shock a student.
Yes, the shocking method is sometimes a good method for teaching, especially for advanced students.
But an average young student will remain unsatisfied. His soul will ask for a more standard solution/explanation
in terms he/she is accustomed to: in terms of "small pieces".

Here is an example on how to build/(to create) such a solution/explanation for this given problem.

Let's make our reasoning not in terms of hours, but in terms of minutes.

Then the problem can be equivalently re-phrased this way

If painter A can paint a wall in 60 minutes and painter B can paint the same wall in 120 minutes, how long does it take for painters A and B to paint the same wall together?

Surely, now this modified problem looks like an absolutely standard even for a beginner.
Now we pronounce a standard mantra

painter A makes 1/60 of the job in one minute, painter B makes 1/120 of the job in one minute. And working together, they will make + = + = = of the job per minute. From it, we conclude immediately that the necessary time to complete the job working together is 40 minutes.

This way, the problem is solved within the usual/traditional framework.

I wrote these explanations to show that sometime it is useful
to take a look to a problem from the other/unexpected point/angle of view.

XXXXXXXXXXXX   S e v e r a l   d a y s   l a t e r   XXXXXXXXXXXX

And now I invite you to laugh with me at the AI solution to a similar problem.

Today, 8/13/2025, I posted this similar problem to Google

If painter A can paint a wall in half an hour and painter B can paint the same wall in half an hour, how long does it take for painters A and B to paint the same wall together?

Below is copy-paste of the AI response

            - - - - - - - S T A R T - - - - - - - - -

AI Overview

If painter A can paint a wall in 30 minutes (1/2 hour) and painter B can paint the same wall in 30 minutes (1/2 hour),
together they would take 15 minutes (1/4 hour) to paint the wall.

Explanation:

Rate of painting: Painter A paints 1/2 of the wall in 1 minute, and painter B also paints 1/2 of the wall in 1 minute.

Combined rate: Together, they paint 1/2 + 1/2 = 1 wall in one minute.

Time to paint the whole wall: Since they paint 1 wall in 1 minute, they can paint the whole wall together in 1/(1/2 + 1/2) = 1/1 = 15 minutes.

https://www.google.com/search?q=If+painter+A+can+paint+a+wall+in+half+an+hour+and+painter+B+can+paint+the+same+wall+in+half+an+hour%2C+how+long+does+it+take+for+painter+A+and+B+to+paint+the+same+wall+together%3F&rlz=1C1CHBF_enUS1071US1071&oq=If+painter+A+can+paint+a+wall+in+half+an+hour+and+painter+B+can+paint+the+same+wall+in+half+an+hour%2C+how+long+does+it+take+for+painter+A+and+B+to+paint+the+same+wall+together%3F&gs_lcrp=EgZjaHJvbWUyBggAEEUYOdIBCTE5MDZqMGoxNagCCLACAfEFFpskuz7iUCjxBRabJLs-4lAo&sourceid=chrome&ie=UTF-8

            - - - - - - - F I N I S H - - - - - - - - -

8/13/2025

Notice that the answer of the AI is correct, but his intermediate reasoning and calculations are wrong.

In other words, in this case, the AI, without having a prototype of the solution / (without having a TEMPLATE)
behaves exactly like a dumb student: he can't figure out what to do and how to get the answer correctly.
This only confirms my reasoning above about the average beginner student.

The Google AI and such a student behave identically and suffer identically.
Proved experimentally.

Ha - ha - ha.

///////////////////////////////////////////

This story has interesting continuation.

On September 2, 2025, I submitted this problem to Google AI again.

This time it produced correct, right, perfect and ideal solution.

It tells me that AI needs very much to have an adequate, attentive and qualified teacher/tutor/mentor.

Then the AI learns and increases its level and becomes adequate.

I am glad to see this progress.

Question 1119: Sally can paint a house in 4 hours and Joe can paint the same house in 6 hours. If they work together how long will it take them to paint the house?
thanks
Found 2 solutions by timofer, KtipsayFoozeBamps:
Answer by timofer(104)   (Show Source):
You can put this solution on YOUR website!
Sally, 1 house in 4 hours
Joe, 1 house in 6 hours

Both together, the combined rate is .

In hours per 1 house, this is
or 2 hours 24 minutes.

Answer by KtipsayFoozeBamps(1)   (Show Source):
You can put this solution on YOUR website!
12/5

Question 1182648: The total energy required to move a 10-N block up a 25-degree incline for 10 meters is 500 Joule. A frictional force of 10-N opposes the motion of the block. Determine force (parallel the inclined) required to move the block, the rate of work done if the work was done in 10 seconds, and the initial velocity of the block if its velocity at 10 meters is 30 m/s.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Here's the solution:
**1. Force parallel to the incline:**
* **Work done against gravity (W_g):**
W_g = mgh = mg * d * sin(θ)
Where:
* m = mass of the block (weight = mg = 10 N, so m = 10 N / g, where g ≈ 9.8 m/s²)
* h = vertical height gained = d * sin(θ)
* d = distance moved along the incline = 10 m
* θ = angle of incline = 25°
W_g = (10 N / 9.8 m/s²) * 9.8 m/s² * 10 m * sin(25°)
W_g ≈ 42.3 J
* **Work done against friction (W_f):**
W_f = frictional force * distance
W_f = 10 N * 10 m = 100 J
* **Total work done (W_total):** This is given as 500 J.
* **Work done by the applied force (W_applied):**
W_total = W_g + W_f + W_applied
500 J = 42.3 J + 100 J + W_applied
W_applied ≈ 357.7 J
* **Force parallel to the incline (F):**
W_applied = F * d
357.7 J = F * 10 m
F ≈ 35.77 N
**2. Rate of work done (Power):**
Power = Work done / Time
Power = 500 J / 10 s = 50 W
**3. Initial velocity:**
We can use the work-energy theorem to find the initial velocity. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.
* **Change in kinetic energy (ΔKE):**
ΔKE = (1/2)mv² - (1/2)mv₀²
Where:
* v = final velocity = 30 m/s
* v₀ = initial velocity (what we want to find)
* **Net work done (W_net):** The net work is the total work minus the work done against friction.
W_net = W_total - W_f = 500J - 100J = 400 J
* **Apply the work-energy theorem:**
W_net = ΔKE
400 J = (1/2) * (10 N / 9.8 m/s²) * (30 m/s)² - (1/2) * (10 N / 9.8 m/s²) * v₀²
400 J ≈ 459.2 J - 0.51v₀²
0.51v₀² ≈ 59.2 J
v₀² ≈ 116.1
v₀ ≈ √116.1 ≈ 10.77 m/s
**Summary:**
* Force parallel to the incline: ≈ 35.77 N
* Rate of work done: 50 W
* Initial velocity: ≈ 10.77 m/s

Question 1182652: The total energy required to move a 10-N block up a 25-degree incline for 10 meters is 500 Joule. A frictional force of 10-N opposes the motion of the block. Determine the force (parallel the inclined) required to move the block, the rate of work done if the work was done in 10 seconds, and the initial velocity of the block if its velocity at 10 meters is 30 m/s.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
Here's the solution:
**1. Force parallel to the incline:**
* **Work done against gravity (W_g):**
W_g = mgh = mg * d * sin(θ)
Where:
* m = mass of the block (weight = mg = 10 N, so m = 10 N / g, where g ≈ 9.8 m/s²)
* h = vertical height gained = d * sin(θ)
* d = distance moved along the incline = 10 m
* θ = angle of incline = 25°
W_g = (10 N / 9.8 m/s²) * 9.8 m/s² * 10 m * sin(25°)
W_g ≈ 42.3 J
* **Work done against friction (W_f):**
W_f = frictional force * distance
W_f = 10 N * 10 m = 100 J
* **Total work done (W_total):** This is given as 500 J.
* **Work done by the applied force (W_applied):**
W_total = W_g + W_f + W_applied
500 J = 42.3 J + 100 J + W_applied
W_applied ≈ 357.7 J
* **Force parallel to the incline (F):**
W_applied = F * d
357.7 J = F * 10 m
F ≈ 35.77 N
**2. Rate of work done (Power):**
Power = Work done / Time
Power = 500 J / 10 s = 50 W
**3. Initial velocity:**
We can use the work-energy theorem to find the initial velocity. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.
* **Change in kinetic energy (ΔKE):**
ΔKE = (1/2)mv² - (1/2)mv₀²
Where:
* v = final velocity = 30 m/s
* v₀ = initial velocity (what we want to find)
* **Net work done (W_net):** The net work is the total work minus the work done against friction.
W_net = W_total - W_f = 500J - 100J = 400 J
* **Apply the work-energy theorem:**
W_net = ΔKE
400 J = (1/2) * (10 N / 9.8 m/s²) * (30 m/s)² - (1/2) * (10 N / 9.8 m/s²) * v₀²
400 J ≈ 459.2 J - 0.51v₀²
0.51v₀² ≈ 59.2 J
v₀² ≈ 116.1
v₀ ≈ √116.1 ≈ 10.77 m/s
**Summary:**
* Force parallel to the incline: ≈ 35.77 N
* Rate of work done: 50 W
* Initial velocity: ≈ 10.77 m/s

Question 1188303: For a given mass of gas in a piston, 𝑝𝑉^1.5 = 400 where p is the pressure in N/m^2, and V is the volume in m^3. Suppose the pressure increases at a constant rate of 3 N/m^2 per minute. Find the rate at which the volume is changing at the instant when the pressure is 50 N/m^2.
Answer by artographer(1)   (Show Source):
You can put this solution on YOUR website!
𝑝𝑉1.5 = 400
Solving the equation for V,
V = 266.6667 * p^-1
Rate the volume dV/dt,
p is increasing at the rate of dp/dt = 3N/m^2, p = 50N/m^2
dV/dt = 266.6667 * p^-2 * dp/dt
Put dp/dt = 3, p = 50
dV/dt = (266.6667 * 3) / 50^2 = 0.32m^3/minute

Question 1209440: Sliding at 18 cm/min instead of 16 cm/min, a slug saves 25 seconds crossing a cabbage leaf. How long is the leaf, in cm?
CC12#5
Found 2 solutions by ikleyn, josgarithmetic:
Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.

It was solved at this forum many years ago.

See the solution under this link

https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.1172769.html

https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.1172769.html

Answer by josgarithmetic(39613)   (Show Source):
You can put this solution on YOUR website!

(cm/min) (min) (cm) SPEED TIME DISTANCE 18 d/18 d 16 d/16 d 25/60

lowest denominator to use is

Question 1209017: A bathroom tub will fill in 15 minutes with both faucets open and the stopper in place. With both faucets closed and the stopper removed, the tub will empty in 20 minutes. How long will it take for the tub to fill if both faucets are open and the stopper is removed?
Found 3 solutions by math_tutor2020, greenestamps, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

Answer: 1 hour

Explanation

Consider a 300 gallon tub.
I arrived at this value by multiplying the given numbers 15 and 20.
There's nothing special about 300 since we can change it to any other value to get the same answer at the end.

If the tub starts off empty, and takes 15 minutes to fill up with stopper in place, then the two faucets have a combined unit rate of 300/15 = 20 gallons per minute.
rate = amountDone/time

After the tub is completely filled up and the faucets turned off, the drain is now opened.
It takes 20 minutes to fully empty.
The drain rate is 300/20 = 15 gallons per minute.

Now consider when both faucets are on and the drain is open at the same time.
In one minute, the faucets add 20 gallons while the drain takes away 15 gallons.
The tub will have 20-15 = 5 gallons.
Meaning that in this tug-of-war of faucets vs drain, the faucets win out to give a net unit rate of 5 gallons per minute.
The tub will ultimately fill up although a bit more slowly of course.

Lets determine how long it takes for the tub to fill up.
rate*time = amountDone
time = amountDone/rate
time = (300 gallons)/(5 gallons per min)
time = (300/5) minutes
time = 60 minutes
time = 1 hour

Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!

The response from tutor @ikleyn shows a standard formal algebraic solution and a quick mental solution. While you might require a formal algebraic solution, you should not ignore the fact that solving problems like this mentally is excellent brain exercise.

Here is solution by another method that can be done mentally and quickly. This method can be used to solve many similar "working together" problems.

Consider the least common multiple of the two given times. The two faucets can fill the tub in 15 minutes; the drain can empty the tub in 20 minutes. The least common multiple of those two times is 60 minutes

Now consider what the faucets and drain could accomplish in 60 minutes.

The faucets could fill the tub 60/15 = 4 times; the drain could empty the tub 60/20 = 3 times.

So in 60 minutes with the faucets and drain both operating, the overall result would be filling the tub 4-3 = 1 time.

So the time to fill the tube with both faucets and the drain both operating is 60 minutes, or 1 hour.

ANSWER: 1 hour

Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.
A bathroom tub will fill in 15 minutes with both faucets open and the stopper in place.
With both faucets closed and the stopper removed, the tub will empty in 20 minutes.
How long will it take for the tub to fill if both faucets are open and the stopper is removed?
~~~~~~~~~~~~~~~~~~

In this problem, there are two opposite processes. One process is filling with the rate of of the volume per minute (two faucets combined). Another process is draining with the rate of the volume per minute. So, when both facets are open and the stopper is removed, the net filling rate is the difference - = - = of the volume per minute. It means that the filling process will take 60 minutes, or 1 hour. ANSWER

Thus, the problem is just solved (mentally).

If you want to write an equation, you should formalize this reasoning.

Let "t" be the time to fill, in minutes. Then in t minutes, the two facets will fill part of the volume, while through the stopper hole, the part of the volume will be removed. The volume will be fully filled when - = 1 (the whole volume). Write with common denominator and find t - = 1, = 1, t = 60*1 = 60 minutes. ANSWER. The required time is 60 minutes, or 1 hour.

Solved using equation.

Question 1209016: An oil tanker can be emptied by the main pump in 4 hours.An auxiliary pump can empty the tanker in 9 hours. If the main pump is started at 9 AM, when should the auxiliary pump be started so that the tanker is emptied by noon?
Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.
An oil tanker can be emptied by the main pump in 4 hours.
An auxiliary pump can empty the tanker in 9 hours.
If the main pump is started at 9 AM, when should the auxiliary pump be started
so that the tanker is emptied by noon?
~~~~~~~~~~~~~~~~

The main pump is scheduled to work 3 hours, from 9 am to noon. In 3 hours, the main pump will make 3/4 of the job; the remaining 1/4 part of the job should be done by the auxiliary pump. The auxiliary pump needs = = 2 hours. Here is the rate of work of the auxiliary tank. Hence, the auxiliary pump should start 2 hours before noon, i.e. at 9:45 am. ANSWER. The auxiliary pump should start at 9:45 am.

Solved.

Question 1208959: Patrice, by himself, can paint four rooms in 10 hours. If he hires April to help, they can do the same job together in 6 hours. If he lets April work alone, how long will it take her to paint four rooms?
Let A = length of time it will take April to paint four rooms

Here is my equation:
(1/10) + (1/A) = 1/6
Is this correct?

Found 3 solutions by math_tutor2020, greenestamps, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

You have the correct set up.
To solve you can multiply everything by the LCD 30A and you should get 3A+30 = 5A.
Which at this point should be fairly easy to solve.
The result is A = 15.

Another option is to solve like so

Rewrite each fraction in terms of a common denominator

Cross multiply

I'll let the student take over from here.

--------------------------------------------------------------------------

Another approach.

Let's say the job is to paint 600 square feet of wall and ceiling area.
I arrived at this number by multiplying the given values 10 and 6.
Then tacking on a 0 at the end to make the area a bit more realistic.

This number isn't particularly important since it can be changed to anything else. You will get the same result at the end.
Also, the "four rooms" can be changed to any number of rooms to get the same answer. So we'll ignore this figure.

Patrice can paint all 600 sq ft in 10 hours when working alone.
His rate is 600/10 = 60 sq ft per hour.
rate = amountDone/time

When Patrice and April work together, assuming neither person hinders the other, they paint all 600 sq ft in 6 hours.
rate = 600/6 = 100 sq ft per hour.

April's unit rate must be 100-60 = 40 sq ft per hour since the two people's individual rates add to the combined rate 100 sq ft per hour.

Let's see how long it takes her to do the job by herself.
rate*time = amountDone
time = amountDone/rate
time = (600 sq ft)/(40 sq ft per hr)
time = (600/40) hours
time = 15 hours

Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!

Your setup is fine, using the method shown in most references for solving "working together" problems. Patrice alone can do the job in 10 hours, so the fraction of the job he does in 1 hour is 1/10. If A is the number of hours it takes April alone to do the job, then the fraction she does in 1 hour is 1/A. And since it takes the two of them 6 hours to do the job together, the fraction they do together in 1 hour is 1/6.

Then, adding the fractions of the job each of them does alone in 1 hour gives you the fraction of the job they do together in 1 hour:

To solve the equation, multiply everything by the least common denominator, 30A.

The number of hours it takes April alone to do the job is 15.

ANSWER: 15 hours

While you might be expected to set up and solve the problem by this method, there is an alternative method that many students find easier.

Consider the least common multiple of the two given times, which is 30 hours.

In 30 hours, Patrice could do the job 30/10 = 3 times.

Working together, the two of them could do the job 30/6 = 5 times.

That means that April alone can do the job 5-3 = 2 times.

And that means the number of hours it takes her to do the job alone is 30/2 = 15.

ANSWER (again, of course): 15 hours

Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.

This setup is correct.

Here "1" in the setup equation represents one whole job, which is painting 4 rooms.

So, your equation (= your setup) is correct. Please accept my congratulations.

But it is not whole solution to the problem.

To complete the solution, you should find A from the setup equation.

---------------------

To see many other similar (and different) problems on joint work,
solved with complete explanations to teach you, look into the lessons

    - Using Fractions to solve word problems on joint work
    - Solving more complicated word problems on joint work
    - Selected joint-work word problems from the archive

in this site.   Read them and get be trained in solving joint-work problems.

Consider these lessons as your textbook, handbook, guidebook, tutorials and (free of charge) home teacher,
which is always with you to help, to feed you with ideas and to inspire.

Question 1208958: It takes an HP LaserJet 1300 laser printer 10 minutes longer to complete a 600-page print job by itself than it takes an HP LaserJet 2420 to complete the same job by itself. Together, the two printers can complete the job in 12 minutes. How long does it take for each printer to complete the print job alone? What is the speed of each printer?

Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.
It takes an HP LaserJet 1300 laser printer 10 minutes longer to complete a 600-page print job
by itself than it takes an HP LaserJet 2420 to complete the same job by itself.
Together, the two printers can complete the job in 12 minutes.
How long does it take for each printer to complete the print job alone? What is the speed of each printer?
~~~~~~~~~~~~~~~~~~~~~~~

Let x be the required time for the fast printer, in minutes. Then the time for the slow printer is (x+12) minutes. The rate of work of the fast printer, of the job per minute, plus rate of work of the slow printer, of the job per minute, is of the job per minute. So, we write this rate of work equation + = . (1) To solve it, we multiply all the terms by 12x*(x+10). It gives 12(x+10) + 12x = x*(x+10). Simplify 12x + 120 + 12x = x^2 + 10x, 24x + 120 = x^2 + 10x x^2 - 14x + 120 = 0. Factor left sides (x-20)*(x+6) = 0. The roots are x= 20 and x= -6. We want the positive root (the time). So, we take x = 20. ANSWER. Fast printer will complete the job in 20 minutes working alone. Slow printer will complete the job in 20+10 30 minutes working alone. CHECK. I will check the validity of equation (1). + = + = = = . Precisely correct !

Solved.

Question 1208894:
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!

Question 1208470: A certain job can be done by 72 men in 100 days. There were 80 men at the start of the project but after 40 days, 30 of them had to be transferred to another project. How long will it take the remaining workforce to complete the job?
Found 5 solutions by Edwin McCravy, math_tutor2020, josgarithmetic, greenestamps, ikleyn:
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!

I think this type problem should be considered as a problem in VARIATION. Common sense tells us that: Time (t) increases as the amount of job-doing (j) increases. Time (t) decreases as the number of workers (w) increases. Therefore, time (t) varies directly as the amount of job-doing (j) and inversely as the number of workers (w). That is: -----------------------------------------
A certain job [THAT'S 1 JOB] can be done by 72 men in 100 days.
Solve for k So
There were 80 men at the start of the project
but after 40 days,....
So for those first 40 days, So the amount of job-doing during those first 40 days was 4/9ths of a job. So that means there is still 5/9 of a job left to be done.
.....30 of them had to be transferred to another project.
So the amount of job-doing left for the remaining 50 men to do was 5/9 of a job. Answer: 80 days. Edwin

Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

Answer: 80 days

Explanation

Consider a warehouse with 72000 boxes.
There are 72 men who can move these boxes in 100 days.
Assume that none of the workers hinder one another.
Each man would move 72000/72 = 1000 boxes.
Each man has a unit rate of 1000/100 = 10 boxes per day.

1 man can move 10 boxes per day.
80 men can move 80*10 = 800 boxes per day.
Over the course of 40 days, that would be 800*40 = 32000 boxes.
There are 72000-32000 = 40000 boxes remaining.

The 80-man crew drops to 50 since 30 people transfer.
x = number of days to move 40,000 boxes
1 man can move 10 boxes per day
50 men can move 50*10 = 500 boxes per day
500x = number of boxes to move
500x = 40000
x = 40000/500
x = 80

Answer by josgarithmetic(39613)   (Show Source):
You can put this solution on YOUR website!

N, how many workers
R, work rate for any one worker
T, time
J, how many jobs

------------------------

First sentence

Second sentence
x for the unknown number of days
80 men for 40 days and then 50 men for x days
The work to be done is 1 job.

Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!

The job requires 72 men working 100 days; the number of man-days required for the job is 72*100 = 7200.

80 men working for the first 40 days perform 80*40 = 3200 man-days of labor. The number of man-days remaining to complete the job is 7200-3200=4000.

The remaining 4000 man-days of labor, performed by 80-30=50 men, requires an additional 4000/50 = 80 days.

ANSWER: 80 days

Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.
A certain job can be done by 72 men in 100 days. There were 80 men at the start of the project
but after 40 days, 30 of them had to be transferred to another project.
How long will it take the remaining workforce to complete the job?
~~~~~~~~~~~~~~~~~~~~~~

Since the job can be done by 72 men in 100 days, it means that the entire job is 72*100 = 7200 man-days. 80 men in 40 days completed 3200 man-days of the job; the remaining job is 7200-3200 = 4000 man-days. The remaining workforce is 80-30 = 50 men. They will complete the remaining part of the gob in 4000/50 = 80 days. ANSWER

Solved.

========================

If you read my solution and the solution by @greenestamps attentively, then with no doubts
you will see that they are very similar. We use similar words and similar construction of
the solution.

It is because problems of this kind are very old: they trace their lineage back to the times
of Ancient Egypt.

During many years and many centuries, people rolled their solutions in the same way
as the surf rolls stones, making them ideally smooth.

This form of solutions, which @greenestamps and me used in our posts, were formed verbally
in the last 300 years.

So, when @greenestamps and me re-tell this classic solution, we use the same words, the same form
and the same ideas as in ancient times or in the last 300 years.

This is a tribute to the tradition of retelling old solutions for old problems in an old wrapper - as a standard classic mantra.

It is good if you retell it to your teacher in the same words - the teacher will see that you
not only know the solution, but know and follow traditions, too.

May be, in 20 years later you will re-tell it to your children and in 50 years later will re-tell to your
grandchildren. In this sense, it is good to know the tradition and to pass it to the next generations.

Question 1208469: Working alone, A can complete a task in ‘a’ days and B in ‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together?

Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.
Working alone, A can complete a task in ‘a’ days and B in ‘b’ days.
They take turns in doing the task with each working 2 days at a time.
If A starts they finish the task in exactly 10 days.
If B starts, they take half a day more.
How long does it take to complete the task if they both work together?
~~~~~~~~~~~~~~~~~~~~~~

So, their rates of work are 1/a for A and 1/b for B. In the first scenario, A works 6 days; B works 4 days. So, first equation is + = 1 (the entire job) (1) In the second scenario, A works 4.5 days; B works 6 days. So, second equation is + = 1 (the entire job) (2) To solve, multiply equation (1) by 3; multiply equation (2) by 2 and subtract the second from the first. You will get then - = 3 - 2, or = 1. Hence, a = 9, which means that A makes the entire job in 9 days working alone. Then from (1) = 1 - = 1 - = . Hence, b = 12, which means that B makes the entire job in 12 days working alone. Working together, they will complete the job in = = 5 days = 5 days. ANSWER

Solved.

Question 455508: The number of hours h that it takes m men to assemble x machines varies directly as the number of machines and inversely as the number of men. If four men can assemble 12 machines in four hours, how many men are needed to assemble 36 machines in eight hours?
Answer by mccravyedwin(405)   (Show Source):
You can put this solution on YOUR website!

Time required varies directly with the number of jobs and inversely with the number of workers. I found this problem from long ago, which is worded for combined variation, to show the other tutors that combined variation is probably what we are supposed to use on these type problems. Notice how the problem is worded.

The number of hours h that it takes m men to assemble x machines varies directly
as the number of machines and inversely as the number of men.
It really isn't necessary for the problem to tell us that because its easy for any student to see from plain common sense. In other words: 1. It's easy to see that if the number of men is held constant, the more machines there are, the more hours will be required, Also, the fewer machines, the fewer hours will be required. That is DIRECT variation. and 2. It's easy to see that if the number of machines is held constant, the more men there are, the fewer hours will be required. And the fewer men there are, the more the hours that will be required. That is INVERSE variation. If four men can assemble 12 machines in four hours, how many men are needed to assemble 36 machines in eight hours? Multiply by 3m It will require 6 men. Edwin

Question 1208171: If 8 men take 12 days to assemble 16 machines, how many days will it take 15 men to assemble 50 machines
Found 5 solutions by mccravyedwin, Edwin McCravy, greenestamps, josgarithmetic, ikleyn:
Answer by mccravyedwin(405)   (Show Source):
You can put this solution on YOUR website!

Here is what I now believe is the best way to do this kind of problem, and how to think it out, I believe it should be taught this way: I will use COMBINED VARIATION, defined as follows: Combined variation describes a situation where a variable depends on two (or more) other variables, and varies directly with some of them and varies inversely with others (when the rest of the variables are held constant). We are asked for TIME REQUIRED, so let's see how TIME varies with each of the other two variables, the number of workers and the number of jobs. Time required varies DIRECTLY with the number of jobs IF the number of workers remains constant. (The more jobs, the more time required. The less jobs, the less time required. Obvious!) Time required varies INVERSELY with the number of workers IF the number of jobs remains constant. (The more workers, the less time required. The less workers, the more time required. Obvious!) Therefore, when we let everything vary, we have a case of combined variation So to state the combined variation involved: Time required varies directly with the number of jobs and inversely with the number of workers. Let T = time required, J = number of jobs, W = number of workers.
8 men take 12 days to assemble 16 machines
Solve for k: Substitute 6 for k:
how many days will it take 15 men to assemble 50 machines?
answer: 20 days. Edwin

Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!

You can also use the worker-time-job formula, which is: where W₁ = the number of workers in the first situation. T₁ = the number of time units (days in this case) in the first situation. J₁ = the number of jobs in the first situation. W₂ = the number of workers in the second situation. T₂ = the number of time units (days in this case) in the second situation. J₂ = the number of jobs in the second situation. W₁ = 8 W₂ = 15 T₁ = 12 T₂ = the unknown quantity J₁ = 16 J₂ = 50 Answer: 20 days Edwin

Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!

You have, so far, two responses showing very different methods for solving the problem.

There are numerous other methods; below is a solution by the method I find easiest.

The problem asks for how many days it will take. So start with the given number of days, 12, and multiply it by factors based on how the values of the other parameters change.

The given scenario has 8 men; the new scenario has 15 men. More men means fewer days, so the number of days changes by a factor of 8/15.

The given scenario has 16 machines; the new scenario has 50. More machines to make means more days, so the number of days changes by a factor of 50/16.

The calculation is then

Cancel some common factors before performing the multiplication:

ANSWER: 20 days

Answer by josgarithmetic(39613)   (Show Source):
You can put this solution on YOUR website!
Simple formula without explanation,

Quick setup

---------twenty days

Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.
If 8 men take 12 days to assemble 16 machines,
how many days will it take 15 men to assemble 50 machines
~~~~~~~~~~~~~~~~~~~~~~~

Let x be the number of days in the second scenario. The rate of work in the first scenario is = = of the machine per man per day. The rate of work in the second scenario is = of the machine per man per day. The rate of work is assumed to be the same, so we write this proportion = . From this proportion, find x x = = 10*2 = 20 days. ANSWER. 20 days for 15 men to assemble 50 machines.

Solved.

----------------

To see many other similar (and different) problems, solved by the same method, look into the lesson
    - Rate of work problems
in this site.

This lesson is written specially for you, to make your horizon WIDER.

Question 1207799: I have been answering textbook math questions for over 20 years. It is terribly embarrassing for me to admit that in all those years solving word problems have been a nightmare. I am sure most students agree that the wording is word problems are not always clear and precise.
Here is one:

Two pumps of different sizes, working together, can empty a fuel tank in 5 hours.The larger pump can empty this tank in 4 hours less than the smaller one. If the larger pump is out of order, how long will it take the smaller one to do the job alone?

Together two pumps can do the job in 5 hours.
I understand this:
Pump 1 + pump 2 = 5 hours.
Is this right?
Smaller pump = x
Larger pump = x - 4

Is this right?

Honestly, I now find myself guessing my way to an equation.

Here it is:

(1/x) + 1/(x - 4) = 5

If my equation and reasoning is wrong, explain why it is wrong.
Thanks

Found 5 solutions by ikleyn, mccravyedwin, MathTherapy, timofer, greenestamps:
Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.

In opposite to your lamentations about word problems, this concrete word problem in your post
is absolutely clear standard school Math problem, and it is written in a classic clear Math language.

Every word is correct, all the words are in their proper places, no excessive words
and all the words are connected in the right order, expressing very clear description and idea.

Every time, when Math problem is written by a professional, it is clear.
As a rule, such problems are brilliant pieces of art.
Many of them were developed and polished during hundreds of years,
in different countries and in different cultures
by people who cared about the mathematical education of subsequent generations.

If a wording Math problem is not clear, it means (with the probability of 129% ),
that the problem is written by a non-professional.

If somebody does not understand clearly written Math problems, it only means
that this person has no enough adequate training in solving/understanding such problems.

It doesn't mean anything else.

Those who have enough adequate training in solving Math problems, write the standard setup equation
immediately and without errors - - - because they are trained.

            By the way, it is a simple criterion if the student
            knows the subject in the school elementary Math.

        The necessary and sufficient condition is that the student
        writes the setup equation immediately and without errors.

Answer by mccravyedwin(405)   (Show Source):
You can put this solution on YOUR website!

Word problems having more than 1 unknown are easier to set up using a separate letter for each unknown, and a separate equation for each sentence. They are not necessarily easier to solve that way, but they're easier to set up that way. Look at the first sentence:
Two pumps of different sizes, working together, can empty a fuel tank in 5 hours.
If you write the equation first in words, it's pretty easy to see that that sentence translates into this equation: Suppose it takes the large pump L hours to empty the tank. Then the large pump's rate is Suppose it takes the small pump S hours to empty the tank. Then the small pump's rate is [Notice that S will be the final answer!] So now we have this for the first sentence's equation: Taking out the words, we just have the equation. Now we look at the second sentence:
The larger pump can empty this tank in 4 hours less than the smaller one.
That is simply or, removing the words, So now you have two equations in 2 unknowns: Then you just substitute S-4 for L in the first equation and you have Multiply through by LCD S(S-4) Those values are approximately, using the +, 12.4 hours and using the -, 1.6 hours, and certainly we reject the second one, so the answer is , approximately 12.385 hours. Edwin

Answer by MathTherapy(10549)   (Show Source):
You can put this solution on YOUR website!

I have been answering textbook math questions for over 20 years. It is terribly embarrassing for me to admit that in all those years solving word problems have been a nightmare. I am sure most students agree that the wording is word problems are not always clear and precise. Here is one: Two pumps of different sizes, working together, can empty a fuel tank in 5 hours.The larger pump can empty this tank in 4 hours less than the smaller one. If the larger pump is out of order, how long will it take the smaller one to do the job alone? Together two pumps can do the job in 5 hours. I understand this: Pump 1 + pump 2 = 5 hours. Is this right? Smaller pump = x Larger pump = x - 4 Is this right? Honestly, I now find myself guessing my way to an equation. Here it is: (1/x) + 1/(x - 4) = 5 If my equation and reasoning is wrong, explain why it is wrong. Thanks NOT QUITE! , the 2 COMBINED/SUMMED hourly-rates of the pumps, is correct. However, both pumps working together, can empty the tank in 5 hours, or of the tank in 1 hour So, we get: , and NOT:

Answer by timofer(104)   (Show Source):
You can put this solution on YOUR website!
large pump rate 1/(x-4)

small pump rate one tank in uknown time, 1/x

both pumps combined 1/(x-4)+1/x=1/5, because when both tanks are working at the same
time, their rates are summed. This equation shows the described, 1 tank in 5 hours.

The property of adding in any order means also can say 1/x+1/(x-4)=1/5

Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!

Together two pumps can do the job in 5 hours.
I understand this:
Pump 1 + pump 2 = 5 hours.
Is this right?

We don't know what it means, so perhaps we don't know if it is right.

But in fact we know it can't be right, because the equation literally says the sum of two things is some number of hours. If you are adding two things and the sum is some number of hours, then the two things you are adding must be hours. Pumps are not hours.

Smaller pump = x
Larger pump = x - 4
Is this right?

Again we don't know, because we don't know what it means. Are "x" and "x-4" the names of the two pumps? I doubt it.

No matter how much experience you have working math problems, you are always potentially in trouble if you don't start with clear and precise definitions of the variables and expressions you are going to use.

The standard method for solving "working together" problems is to write an equation that says the sum of the fractions of the job that each worker does in some amount of time is equal to the fraction of the job that they do together in that time.

The problem says that the larger pump can do the job in 4 hours less than the smaller pump. So the proper definitions of "x" and "x-4" are

Let x = # of hours the smaller pump takes to do the job alone
Then x-4 = # of hours the larger pump takes to do the job alone

(Compare that to the "definitions" you show for x and x-4....)

Next, from there (as you tried to do in your setup),

Then 1/x = fraction of job done by smaller pump in 1 hour
And 1/(x-4) = fraction of job done by larger pump in 1 hour

Our equation is going to say "fraction done by one pump in 1 hour, plus fraction done by other pump in 1 hour, equals fraction done together in 1 hour". Since they can do the job together in 5 hours, the fraction of the job they do together in 1 hour is 1/5.

So, finally,

That completes the setup. Actually solving that leads to a very ugly quadratic equation, so I would use a tool like a graphing calculator to find the answer.

But perhaps what you were looking for was help in doing the setup....

Question 1207751: Patrick, by himself, can paint four rooms in 10 hours. If he hires April to help, they can do the same job together in 6 hours. If he lets April work alone, how long will it take her to paint four rooms?

Let me see.

Patrick = 1/10

April = 1/x

Together = 1/6

(1/10) + (1/x) = (1/6)

Is this the correct equation?
Found 3 solutions by greenestamps, math_tutor2020, ikleyn:
Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!

Here is another way to solve this kind of problem without using fractions.

Consider the least common multiple of the two given times, which is 30 hours.

In 30 hours, Patrick himself could paint the four rooms 30/10 = 3 times; in 30 hours, the two together could paint the four rooms 30/6 = 5 times.

That means in 30 hours April alone could paint the four rooms 5-3 = 2 times.

And that means she could paint the four rooms alone in 30/2 = 15 hours.

ANSWER: 15 hours.

Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

You have the correct set up so far.

I'll show two ways of solving that equation.
Here is one way.
1/10 + 1/x = 1/6
30x(1/10 + 1/x) = 30x(1/6)
3x + 30 = 5x
30 = 5x-3x
30 = 2x
2x = 30
x = 30/2
x = 15 hours is the amount of time April takes when working alone.

Or you can solve it like this.
1/10 + 1/x = 1/6
x/(10x) + 10/(10x) = 1/6
(x+10)/(10x) = 1/6
6(x+10) = 10x
6x+60 = 10x
60 = 10x-6x
60 = 4x
4x = 60
x = 60/4
x = 15 hours

--------------------------------------------------------------------------

Another approach

Let's say hypothetically that the total surface area to paint all four rooms is 600 square feet.

Patrick, when working alone, can paint all 600 square feet in 10 hours.
His unit rate is 600/10 = 60 sq ft per hour.
Formula: rate = amountDone/time

When Patrick and April work together, they get the job done in 6 hours.
I'll assume neither person slows down the other.
Their combined unit rate is 600/6 = 100 sq ft per hour.

Therefore, April's unit rate must be 100-60 = 40 sq ft per hour when she works alone.

Let's determine how long April will take when she does the job alone.
rate*time = amountDone
time = amountDone/rate
time = 600/40
time = 15 hours

The hypothetical 600 sq ft figure I came up with isn't special at all. Turns out you can replace it with any other positive number to get the same final answer marked in red.
I picked 600 since 10*6 = 60, and then I tacked a zero at the end to make the area appear a a bit more realistic pertaining to those four rooms.

Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.

        Yes, this setup equation is correct.

        You can continue and solve it.

        But this problem can be solved without using any equation,
        by manipulating fractions, ONLY.

        See the reasoning below.

Patrick and April can do the job together in 6 hours. Hence, their combined rate of work is of the job per hour. Patrick, working alone, can do the job in 10 hours. Hence, Patrick's rate of work is of the job per hour. Hence, April's rate of work is the difference - = = = = . It means that April makes 1/15 of the job per hour. Hence, April needs 15 hours to complete the job working alone.

At this point, the problem is solved in full, with complete explanations.

==================

So, you can solve this problem in two ways :   manipulating with fractions or using equations.

Both methods have equal rights to exist.

Both methods implement the same idea: making a balance of rates of work.
One method implements it using manipulations with fractions;
other method implements it using equations.
But the idea of the solution is the same in both methods - only the implementation forms are different.

Solving by manipulating fractions is accessible method for 5-th or 6-th grade students.

Solving using equations is accessible method for 7-th grade students, who are just familiar with equations.

To see many other similar (and different) problems on joint work,
solved with complete explanations to teach you, look into the lessons

    - Using Fractions to solve word problems on joint work
    - Solving more complicated word problems on joint work
    - Selected joint-work word problems from the archive

Read them and get be trained in solving joint-work problems.

Consider these lessons as your textbook, handbook, guidebook, tutorials and (free of charge) home teacher,
which is always with you to help and to inspire.

/////////////////////

By the way, this problem was solved at this forum many years ago (perhaps, 20 years ago).
For the solutions, see the links

https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Rate-of-work-word-problems.faq.question.264833.html

https://www.algebra.com/algebra/homework/coordinate/word/Linear_Equations_And_Systems_Word_Problems.faq.question.68586.html#google_vignette

Question 1207767: A 5-horsepower (hp) pump can empty a pool in 5 hours.A smaller, 2-hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool.After two hours, the 2-hp pump breaks down. How long will it take the larger pump to empty the pool?
Found 5 solutions by mccravyedwin, ikleyn, math_tutor2020, mananth, josgarithmetic:
Answer by mccravyedwin(405)   (Show Source):
You can put this solution on YOUR website!
A 5-horsepower (hp) pump can empty a pool in 5 hours.
A smaller, 2-hp pump empties the same pool in 8 hours. The pumps are used
together to begin emptying this pool.

The LCM of 5 and 8 hours is 40 hours, So the 5-horsepower (hp) pump can empty 8 pools in 40 hours. The smaller, 2-hp pump can empty 5 pools in 40 hours. So in 40 hours both pumps cane empty 13 pools, Since 2 hours is 1/20 of 40 hours, in 2 hours together they can empty 1/20 of 13 pools. That's what they've pumped out together in those first 2 hours. So the question becomes: how long does it take the larger pump to pomp out the remaining 7/20 of 1 pool? It can empty 1 pool in 5 hours. So it can empty 7/20 of a pool in 7/20ths of 5 hours, or 35/20ths 7/4ths of an hour, 1 3/4 hours or 1 hour and 45 minutes. Edwin

Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.

This problem was solved at this forum many years ago (perhaps, 20 - 25 years ago).

For the solution, see the link

https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.5399.html

Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

Answer: 1.75 hours

Explanation

The horsepower values are a distraction. Ignore them entirely.

Consider a 4000 gallon pool.
The larger pump, which I'll call pump A, can empty the entire pool in 5 hours when working alone.
A's unit rate is 4000/5 = 800 gallons per hour.
Formula used: rate = (amountDone)/time

The smaller pump B can empty the entire pool in 8 hours when working alone.
B's unit rate is 4000/8 = 500 gallons per hour.

When the two pumps work together, without either pump slowing down the other, their combined rate is 800+500 = 1300 gallons per hour.
The two pumps work together for 2 hours. That drains 1300*2 = 2600 gallons and leaves 4000-2600 = 1400 gallons remaining.

Pump B breaks down at the 2 hour marker.
Pump A now works alone to empty the remaining 1400 gallons of water.
rate*time = amountDone
time = amountDone/rate
time = 1400/800
time = 14/8
time = (2*7)/(2*4)
time = 7/4
time = 1.75 hours is the amount of time needed for the larger pump to finish the job of emptying the pool.

The 4000 value mentioned isn't special. Feel free to change it to any other positive number you want. The final answer will still be the same regardless of the pool volume.
I picked this value based on the LCM of 5 and 8, which is 40. Then I tacked on a few extra zeros to lead to a slightly more realistic pool volume. Even then 4000 is probably on the smaller side of things.

--------------------------------------------------------------------------------

Another approach

one job = emptying the pool
Pump A does one job in 5 hours when working alone
A's unit rate is 1/5 of a job per hour
Pump B does one job in 8 hours when working alone
B's unit rate is 1/8 of a job per hour

Their combined unit rate is 1/8 + 1/5 = 5/40 + 8/40 = 13/40 of a job per hour.
This assumes neither pump slows the other down.

In 2 hours the pumps work together to handle 2*13/40 = 13/20 of the job.
1 - (13/20) = 20/20 - 13/20 = 7/20 of the job remains.

x = amount of extra time, in hours, pump A needs to work alone to finish the job
This is after pump B stops working

rate*time = amountDone
(1/5 of a job per hour)*(x hours) = 7/20 of a job remains
(1/5)x = 7/20
x = 5*7/20
x = 5*7/(5*4)
x = 7/4
x = 1.75 hours

Extra info:
1.75 hours = 60*1.75 = 105 minutes
1.75 hours = 1 hour & 45 minutes since 0.75 hr = 0.75*60 = 45 min

Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!

A 5-horsepower (hp) pump can empty a pool in 5 hours.A smaller, 2-hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool.After two hours, the 2-hp pump breaks down. How long will it take the larger pump to empty the pool?
Consider emptying the pool is 1 job
A 5-horsepower (hp) pump can empty a pool in 5 hours. rate of Emptying = 1/5 of the job
A smaller, 2-hp pump empties the same pool in 8 hours.1/8 of the job
The pumps are used together for two hours
2(1/5 +1/8) of the job is done in two hours =26/40 = 13/20
Balance job = 1- 13/20 = 7/20
1/5 of job the 5Hp does in 1 hour
7/20 of the job it will do in (7/20)/ (1/5)= 35/20 = 7/4 hours
It will take the 5-hp pump an additional 1.75 hours to empty the pool after the 2-hp pump breaks down.

Answer by josgarithmetic(39613)   (Show Source):
You can put this solution on YOUR website!
5 hp pump does one job in 5 hour.
speed,

2 hp pump does one job in 8 hours
speed,

Both together for 2 hours and then only the 5 hp pump for unknown time x. One job to complete.

----------a good starting equation.
Multiply both sides by 40, simplify, and solve for x.