SOLUTION: Pump A can empty a swimming pool in 5 hours, pump B can empty it in 8 hours, and pump C can empty it in 10 hours. If all three of these pumps are used together, how long will it t

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Pump A can empty a swimming pool in 5 hours, pump B can empty it in 8 hours, and pump C can empty it in 10 hours. If all three of these pumps are used together, how long will it t      Log On


   

Question 1196421: Pump A can empty a swimming pool in 5 hours, pump B can empty it in 8 hours, and pump C can empty it in 10 hours. If all three of these pumps are used together, how long will it take to empty the pool?
Found 3 solutions by ikleyn, greenestamps, math_tutor2020:
Answer by ikleyn(52903) About Me  (Show Source):
You can put this solution on YOUR website!
.

Pump A makes  1%2F5  of the job per hour;

Pump B makes  1%2F8  of the job per hour;

Pump C makes  1%2F10  of the job per hour.


Working together, the three pumps make

    1%2F5+%2B+1%2F8+%2B+1%2F10 = 8%2F40+%2B+5%2F40+%2B+4%2F40 = 17%2F40

of the job per hour.


So, in  40%2F17 = 2 6/17 = 2.353 hours = 2 hours and 22 minutes (rounded) the job will be complete.    ANSWER

Solved.

On the way, you learned the notions of individual rate of work, combined rate of work
and how to apply them to solve this joint work problem.

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

It is a standard and typical joint work problem.

There is a wide variety of similar solved joint-work problems with detailed explanations in this site.  See 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.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic
"Rate of work and joint work problems"  of the section  "Word problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.



Answer by greenestamps(13215) About Me  (Show Source):
You can put this solution on YOUR website!


Here is an alternative to the standard method for solving the problem shown by the other tutor.

The three times are 5, 8, and 10 hours. Consider the least common multiple of those times, which is 40 hours.

In 40 hours...
Pump A could empty the pool 40/5 = 8 times;
Pump B could empty it 40/8 = 5 times;
Pump C could empty it in 40/10 = 4 times

So in 40 hours, the three pumps together could empty the pool 8+5+4 = 17 times.

Therefore, the number of hours it would take the three pumps to empty the pool one time is 40/17.

ANSWER: 40/17 hours

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

LCM = lowest common multiple
GCF = greatest common factor

LCM formula:
LCM of x and y = (x*y)/(GCF of x and y)

The LCM of 5 and 8 is 5*8/1 = 40, where the GCF is 1
The LCM of 40 and 10 is 40*10/10 = 40, this time the GCF is 10

Overall, the LCM of the set {5,8,10} is 40
Let's multiply that by 1000
40*1000 = 40,000
so that we have a possibly realistic gallon amount for the pool

Consider a pool with 40,000 gallons
Pump A working alone can empty the pool in 5 hours
The rate of this pump is (40,000)/5 = 8,000 gallons per hour

Pump B can empty it in 8 hours
The unit rate here is (40,000)/8 = 5,000 gallons per hour

Pump C can empty it in 10 hours
Unit rate = (40,000)/(10) = 4,000 gallons per hour

Add the three unit rates
A+B+C = 8000+5000+4000 = 17,000 gallons per hour

x = number of hours to empty the pool if the three pumps are used together

(unit rate)*(number of hours) = total amount drained
(17000 gallons per hour)*(x hours) = 40000 gallons
17000x = 40000
x = 40000/17000
x = 40/17
x = (34+6)/17
x = (34/17)+(6/17)
x = 2+(6/17)

It takes 2 whole hours, plus an additional 6/17 of an hour, to get the job done if the three pumps work together.

6/17 of an hour = (6/17)*60 = 21.17647 minutes approximately

If rounding to the nearest whole minute, then we get 22 minutes

So it takes about 2 hours, 22 minutes if the pumps work together (and no single pumps hinder any others).

Side notes:
2 hr + 22 min = 2*60+22 = 120+22 = 142 min
(40/17) hr = (40/17)*60 = 141.17647 = 142 min