A pump can fill a pool in 10hrs. Another pump can fill it in 15hrs. Both pumps together will take?
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- čas přidán 13. 07. 2024
- How to solve an algebra work word problem.
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I did it on percentages. Pump 1 fills 10% of the pool per hour, pump 2 fills 6.67% per hour. Together they fill 16.67% per hour, so 100% divided by 16.67 = 6 hours.
I did it taking a fictitious volume that was comfortable to fill with both pumps, say 150 m3
The first pump works with 15 m3/hour and the second with 10 m3/hour.
So both pumps combined pour 25 m3/hour and to fill 150 m3 it will take 6 hours 🙂
That’s how I did it except, I simply did 1/(1/10+1/15), brute force math.
@@64Rossomy unit is 1 pool.
You're absolutely correct! This is how it's supposed to be done! Any number of gallons is turned into 100% And if you know how many gallons the pool holds (eg.950 gallons) you'll also know how many gallons you're pumping per hour, minute, seconds etc. Pump one = 95 gals. per hr and pump two = 63.3333 gals. per hr...Add those two numbers together then divide them into the 950. 95 + 63.33 = 158.33...950/158.33=6
That's what I did as well. 100 gallon pool, Pump 1 pumps at 10 gallons per hour. Pump 2 pumps at 6.6 gallons per hour. together they pump at 16.66 gallons per hour. To fill a 100 gallon pool it would take 6.02409 hours
In 30 hours, pump A fills 3 and pump B fills 2. So In 30 hours they fill a total of 5 pools. 30/5 = 6
Excellent, this is both elegant and rigorous!
I didn't think of it that way. Thanks for the insight.
I solved it by assigning an arbitrary size to the pool. I choose 300 gallons because it's evenly divisible by so many numbers.
300 gallons in 10 hours equals 30 gallons per hour.
300 gallons in 15 hours equals 20 gallons per hour.
Add the two together and one gets 50 gallons per hour.
300 divided by 50 equals 6 hours.
I solved it exactly (nearly verbatim) as @angelleiva36. I think it is the most obvious way to do it.
To be able to add the flows you have to convert both flows to pools per hour and then you normalize to the same hours, add and then invert to get hours.
Nice. I said A runs at 1/10 pools per hour, B runs at 1/15 pools per hour. A + B = 0.1666~ pools per hour. 1 pool / 0.16666~ = 6
I am a Physicist. I encountered a similar exercise about 2 weeks into my first Algebra semester in University and I am happy to say that 40 years later, it took me about 10 seconds to solve this one mentally. I am delighted you have half a million subscribers! Not everyone wants to be an influencer or TikToker!
Maths teacher here. Pro tip: These problems come down to inverting the units. You are given “hours per pool” (hpp), but what you need in order to combine them is “pools per hour”. So 10 hours to fill gives a rate of 1/10 pools per hour and 15 hours to fill gives 1/15 pph.
Now, we need to find a common denominator, which is 30 in this case. 1/10 = 3/30 and 1/15 = 2/30 which add to 5/30. This simplifies to 1/6 pph. Inverting this back gives 6hpp. So with both running, the pool will fill in 6 hours.
It’s the same with all problems of this sort.
Retired maths teacher here. Your method is perfect and exact, just what I would do.
But may be some people will find it difficult to follow this reasoning, so perhaps this will help:
If the pool volume is X m^3 (or another volume unit), in ONE hour the pumps will fill up volumes X/10 and X/15 (m^3), together (5/30)X. In a time T (hours) they will fill up the volume (5/30)X * T, which shall equal X, the whole volume, and we have the equation
(5/30)XT = X or (1/6)T = 1. Multiply by 6 and you have T = 6, 6 hours.
(The volume X disappeared and it is not essential at all, it could be anything!)
By the way it was a good thing to understand, before starting calculating, that the time needed when both pumps are working must be shorter than time needed by the fastest one alone, 10 hours. If you get the answer 25 hours you would know that must be wrong.
I never taught math, although I did teach chemistry. Anyway, the method you described is exactly the way I did it.
Exactly how I did it.
Thank you so much for the explanation. I just couldn't figure out how he'd establish that sum of 2 inverted times (in both explanations he gave) to start with the equation. That was the only thing that got me scratching my head, and the first line of explanation you gave just cleared everything.
@@user-gr5tx6rd4h I was scratching my head when I paused the video, thinking that there was a data missing which is the volume of the pool. You've basically explained why the volume is irrelevant in this situation, since the way the problem is set, we already have the flow of the pumps and the time required to fill that volume no matter how big or small he his.
As a licensed engineer who has designed and installed dozens of industrial pumping systems, I must comment. The actual answer is much more complicated. You need to account for one of the pumps failing, a hose clamp coming off, neighbors complaining about the noise, discovery that the pool is leaking, whether the contractor brought both pumps, if all the hoses and clamps fit, the city inspector saying there is no permits, the city water department shutting you down because they don’t allow pools to be filled during drought and a bunch of other things.
Good point; and you didn't take 15 minutes to explain it 😄
And if you're in Australia, don't forget the fire fighter's 'heli-tanker' hovering over halfway through to suck up a few thousand litres.
NIce. No one should ever confuse an engineer with a mathematician!
As another licensed engineer, you find that the pump contractor actually installed a cheaper pump which takes 9 hours because he got the job by underbidding the job, then telling the owner he can save him money because he has a cheaper pump than the one you designed in his warehouse and the pump you specified has a two year lead time. When you reject the submittal, he calls you whining that he can’t make money because you gold plated the job.
@@marcholland1554 do you have a hidden camera in my office?
Flow rate for faster pump: 1/10 of a pool per hour
Flow rate for slower pump: 1/15 of a pool per hour
Combined flow rate: 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6 of a pool per hour
So you need 6 hours to fill the pool.
Could have aid this in 30 seconds. Instead, he spends all day to explain it.
That's the way I figured it out. I never took algebra just general math. Just have to know how to work with fractions.
@@whiteshadow1771 Yeah, but he's trying to explain the logic plus some of the math.
6.25 hours.
WRONG
I love teaching my students clever problem solving. They know V=RT, and they know they can invent an arbitrary volume for such a problem. If they chose V=150 m^3 then they know P1 pumps 10 m^3/hr to accomplish the same V as P2 pumping 15 m^3/hr for 10 hours. Adding P1+P2 for a total output of 25 m^3/hr. they can simply divide 150 m^3 by 25 m^3/hr to arrive at 6 hrs.
We recently did something very much like this to find the resistor required in a parallel circuit to achieve a specific overall resistance. Such problems come up in most of my classes and I believe that finding and testing clever solutions is a very valuable step toward finding purely algebraic solutions because it helps us to better understand problems. Using 30ml (or 30 mi^3!) rather than 150 would actually yield easier computations, but students often get mixed up using LCMs and GCFs, so simplicity generally outweighs elegance (in this step).
I just tuned in to have a look at how hard a “perfect little challenge for you” can be drawn out for 15 minutes. I’ll come back to this just before I fall asleep. Thanks, Jon. See you later.
It's like resistors in parallel, so 1/10 + 1/15 = 1/total = 6 hours.
This is how I looked at it. Super easy.
That's also how I solved this problem. As a check I figured it's more than 5 hours and less than 7.5 hours.
Of course. I knew it could be done with reciprocals, now I remember where I learned it. Thanks Grahamaus for jogging my memory.
Or capacitors in series
@@whomigazone True, but pumps filling a pool is a lot more analogous to current in a resistor, so it seems like a more intuitive comparison.
In 30 hours you would fill 5 pools, therefore you would fill one in 6 hours.
That was the easiest way to solve it.
Elegant
how did you get 30 hours?
Brilliant solution. Better than mine.
@@olgaa8310 The number 30 is arbitrary here, any number would do. But 30 is the smallest number divisible by both 10 and 15, so it is convenient to use. (If the number disturbs you, "invent" a new unit so that the pool volume is 30 of those new units!)
Raynewport's method is fine for those who prefer using concrete numbers. My method with symbolic equation making may be preferred by others.
This dint need to be a 16min video - I did this quickly and simply as follows:
Pump 1 fills at rate 1/10th of the pool per hour
Pump 2 fills at 1/15th or pool per hour.
Lowest common demoninator of the above 2 fractions is 1/30th
So pump 1 fills at 3/30th per hour and pump 2 fills at 2/30th per hour
Pump1 + pump 2= 3/30th +2/30th = 5/30th = 1/6th per hour
So working together they fill 1/6th of the pool per hour, so it takes 6 hours to fill the whole pool. I wrote this out before he started his solution.
He probably proved this in 2 seconds, but his 15 minutes INTENTIONS was merely to teach dumb headed people like me… I learned a great deal from this video… by the way, parallel connection formula simplified…
12:16 He knows how to confront learners. He teaches, discusses, and hence....
Clearly, you're not in his intended audience.
I did the same way
This problem has been already solved correctly by many commentators below; I am just rewording the simple algebraic equation as follows: Let V be the volume of the pool and T ( hours) be the time needed fill up the pool with both pumps working. Then, T * (V/10+V/15) = V Solving for T we have T ( 1/10+1/15) =1 Then T= 6
My method was to assume a pool size - 300 gallons, and calculate that pump A works at 30g/h, and Pump B at 20g/h - a total of 50g/h, so 300/50=6h.
That’s how I did it. It’s about flow rate.
I used the same method but assumed a pool of 150 ‘units’, with one pump delivering 10 units/hr and the other 15/hr, so 25/hr together. It’s interesting to see how different people approached the same problem.
Yes this is how math works in the real world. No need for fancy BS. It's to bad teachers can't just teach like this.
Same, but i assumed 30 gallons because that was my braindead easy lowest common denominator
@@neverknow69 Though I also solved the problem in the same way I couldn't help but think there was a better way to do it.
I’m not an algebra problem solver at all. But as I scrolled through some of the logic other viewers were using to solve the problem it made sense. I appreciate the effort the author put into trying to explain how he solved the problem, but the more he talked the quicker he lost me. Thanks to all of you for sharing your experience!
The narrator of the video did a awful job, sorry to have to say it, but he did.
I was just pretending I did not know any math, and he never made me feel like I had any idea what he was doing.
He never said why he was doing what he was doing.
And he talked to much without saying anything.
If talking quickly he lost you in 8 minutes, how long would he have taken to lose you if he had spoken 1.5 times as fast?
I usually use the formula RxT=J (R=rate, T=time, J=jobs). Use R = J/T to find the rate of each person/item first, then plug their individual rates into the formula (RT)' + (RT)" = J. I like this setup because I can make adjustments to the time of an individual if they don't run concurrently.
1/10T + 1/15T = 1
5/30T = 1
1/6T = 1
T= 6 hours
If there is a group rate, I use RxWxT = J (where W = # of people working together)
I have always solved these kinds of problems by adding the rates of each "worker" together and moving from there.
For this problem, I would say that the first pump filled 10% of the pool per hour and the second pump filled 6.67% of the pool per hour, so together they fill 16.67% of the pool every hour. Since a full pool is 100%, I divide 100% by 16.67% and I get 6 hours.
Strangely enough, I was first introduced to this kind of word problem by the show Boy Meets World. The protagonist was a kid in school whose teacher gave them a problem that went something like this: Bill can wash a car in 8 minutes, Ted can was a car in 6 minutes. If they work together, how quickly can they wash a car? His answer was the average: 7 minutes, and the teacher marked him wrong.
I remember thinking, "Why is that wrong?" And then either I realized that it wouldn't make sense for Ted to allow Bill to help him if that made him a minute slower, so obviously the average is wrong. I tried a lot of wrong ways to solve this problem. One of my favorites of these was to divide the job in half, so Bill washes half, and it takes him 4 minutes, but Ted finishes his half in only 3 minutes, which means that a whole minute goes by during which Ted could be helping Bill. So it can't be 4 minutes.
It took a while, but I eventually figured out the above way so that both workers can be working the entire time and I can determine how quickly the job is finished. This was way back in the mid '90s, and it was right around when I started to actually like math and even get a little better at it. So, thanks Mr. Feeny (played by William Daniels, who was also the voice for KITT in Knight Rider)!
I don't blame Ted Bill is a fool, he always washes cars from the bottom up.
Thank you!!! That is logical!!!!!!
As said by others we should start with noting that when both work together the work will take shorter than if the quickest one works alone, so less than 6 minutes in your example - we suppose they work seriously and don't disturb each other ;-)
Thus we would know that 7 minutes must be wrong.
Anyone who complains about this videos being too long, and too detailed, relax. This guy is teaching. My first year high school grandson is watching his videos. And learning.
I went about this problem a little differently.. but I see in the comment, lots of people did the same as me.
Pool volume 750gals (any volume will work, but must be consistence)
10hrs flow rate = 75 gals per hr
15hrs flow rate = 50 gals per hr
Add together rate = 125 gals per hr
125 per hr / 750 = 6 hrs
I used the same approach.
@@johnr5252
Moma Always Told Me ~ *"Great Minds Think Alike"* ( ͡❛ ͜ʖ ͡❛)✌
That's how i did it but you sure explained it better than I did.
@@MyRook
Back in College, (late 70's) my very 1st semester, I had a math class that about whipped my ass!!
I was in an Engineering program. This teacher from the 1st day would say every day in class.
"Keep it Simple Stupid" better known as the "KISS Principle" I learn that this phase was NOT new.
However, He explained the theory behind THAT Phase.
So, it "DoesntMatter" if your designing a rocket ship or doing a math problem, KEEP it Simple. It creates less kaos.
His name was Zwicker, years before, he and his wife left Nazi Germany and made it to the American.
Ironically, he helped bring the Nazi's to their knees. To this day, I love WW ll history.
I went to his funnel in 1992. The world lost a great man that day!
how did i get 6.25
If one pump takes 10 hours and another one takes 15 hours this means that the first time that they will both finish filling a whole number of pools together is after 30 hours. In 30 hours the first pump will have filled 3 pools and the second pump will have filled 2 pools, which is 5 pools .. so just divide 30 by 5 and you get 6 hours, telling you that it would take 6 hours if they were filling one pool together.
Yep... Simplest way to go about it. Took me the whole of 45 seconds, then reading the comments I started wondering why people were looking for so many complicated ways to do the same :-)
Thats a good way to do it!
@philipac2gmail Pump 2 will fill the pool in 15 hours. Pump 1 will fill 1,5 pools in 15 hours.
Therefore pump 1 and pump 2 will fill 2,5 pools in 15 hours. Therefore 15 houers divided by 2,5 pools to find the time to fill 1 pool = 6 hours.
6.25 hours.
Great problem. It has been 60+ years since I had high school algebra. Although I struggled a bit, I did ultimately solve the problem before opening the video - although I did chew up my fair share of paper. Thanks so much.
the following method seemed more intuitive:
create a variable to represent the pool capacity and set it to the lowest common denominator of the two pump times, and then determine each pump's rate of flow using the formula "rate = capacity / time".
capacity = 30 ... this is the lowest common denominator of 10 and 15
rate1 = 30 / 10 ... this gives the first pump a flow rate of 3
rate2 = 30 / 15 ... this gives the second pump a flow rate of 2
this formula gets your answer:
answer = capacity / (rate1 + rate2)
I took the same approach and, once the pool volume drops out, the result is the same. Perhaps a science/engineering background serves to over-complicate the whole process. Which, in my case, took a 5 minute problem and stretched it into an hour and 5 sheets of paper - oops.
You can look at it in these terms: the faster pump has a flow rate equivalent to 1.5 times the slower pump. So, if the two pumps work together, it's as if you had (1.5 + 1) = 2.5 of the flow rate of the slower pump. The slower pump takes 15 hrs. Therefore, 2.5 equivalent pumps will only need 15/2.5 = 6 hrs. Or, you can say that the slower pump is equivalent to 0.67 of the faster pump and get the same result: (10/1.67) = 6.
thats how i got my answer, i was just having trouble in how to word it in my head lol
I started with the slow pump equals 0.67 of the fast one. mu brain flipped something and I thought 7.5. ..... Then. Nope, that'll overfill it. Then I realized that 6 gave me a whole number no matter what I did with the numbers while I was actually thinking.... So I plugged it in and voila! It worked.
That is the engineer btw much faster then the video trying to explain😊
Thats faster for mental arithmetic.
Ratios. The first pump is 1.5 times as fast as the second. It will fill 60% of the pool, whilst the slower pump will fill 40%. If the first pump can fill the whole pool in 10 hours, it will take 6 hours to fill 60% of the pool. At the same time, if the slower pump can fill the whole pool in 15 hours, it will take 6 hours to fill 40% of the pool.
Too much bla bla bla!😅😅😅
A good method if you don't have a problem with understanding that if a number shall be split in two parts, one being 1.5 times as big as the other, those two parts must amount to 60 % and 40 % of the number. Perhaps for most people this is easy but may be not for all?
If you buy two things and pay 110 $ total and one of them cost 100 $ more than the other, what were the individual prices? (Not 100 $ and 10 $, of course)
My cheesy estimate came up with 6.25 hours, but watching this re-introduced me to the algebra I've forgotten over the years. Thank you for posting these videos!
If V = Volume to fill then pump rate to fill which we call Pr = V/tf, where tf = time to fill. We can invert equation to give tf = V/Pr. If we have multiple pumps, V stays the same, and we just add the pump rates, so tf = V/(Pr1 + Pr2 + ... + Prn) for n pumps. For the two pumps we know Pr1 = V/10 and Pr2 = V/15, (time is in units of hours). Thus tf = V/(V/10+V/15) = V/(25V/150) = 150/25 = 6 and units is hours so 6 hours to fill pool.
A = pool/10hr
B = pool/15hr
or
A = (1/10) pool/hr
B = (1/15) pool/hr
combined how long for both together?
1/10 pool/hr + 1/15 pool/hr
= (6/60 + 4/60) pool/hr
= 10/60 pool/hr
= 1/6 pool/hr
or a full pool in 6 hours
VERIFY
6hr(A + B) =? 1 pool
6hr (1/10pool/hr+1/15pool/hr)
=? 1pool
3/5pool + 2/5pool =? 1pool
5/5pool =❤ 1pool✔️
Hi Tom @tomtke7351 can you explain where the "60" came from in your solution.
What made you go from
1/10 + 1/15 to
6/60 + 4/60 : what led to your LCD being 60.
Thanks
You need a common denominator so that you are comparing the same thing across both pumps. I would have chosen 30 which is the lowest common denominator 15 * 2 = 30 and 10 * 3 = 30.
@@pollyanna1112
lcd for 1/10 & 1/15
10 = 2×5
15 = 3×5
LCD = 2×5×3
= 30
I errantly doubled it to.60 but
with no harm
pump A fills 1/10 of the pool per hour, pump B fills 1/15th per hour. 1 represents when the pool is filled, x represents the hours to fill the pool. 1= X (1/10 + 1/15) covert 1/10 and 1/15 to a common denominator, 30. 1= X (3/30 + 2/30) 1=X (5/30) so X, the number of hours, equals 6
6.25 hours.
I'm glad you're posting these problems. I love working on them. However, I thought your explanation would be confusing for a beginner. This one over plus one over business. This is a rate problem; volume over(per) time. If you included units into your method I think it would be much easier to follow. Units would cancel out and you're left with 6 hours. To me this seems easier to understand. And, keep it up. These problems are good for everyone.
It's nice how everyone seems to have different mental processes. I took it as pump 1 = 10% per hour, and pump 2 fills at 3/2 the rate. So after 3 hours we have 30% + 20% and are halfway there, 6 hours is the answer.
Wrong. Pump 2 fills at 2/3 of the rate, not at 3/2 of the rate. So 1 + 2/3 = 5/3 faster than pump 1 alone. So 10 hours / 5/3 = 10 x 3/5 = 6 hours.
Although your calculation shows the correct answer, your explanation of how you did it contains an error.
The 2nd pump is slower, it contributes 2/3 of what the 1st pump does (not 3/2).
I.e. when pumps 2 contributes 2/3 of the 1st pumps 30%, you'd get the 20% you showed. (Had you used the incorrect 3/2, then the 2nd pump's contribution would have been 45% - and we know this isn't true :-)
Put a common number with easy results for the pool size, 150G, so one is pumping 15G/H while the other 10G/H, together 25G/H, 6 hours to do 150G, without paper and pencil. You could use 7 and 13 H and without paper and pencil would get more difficult, but the common number would be 7*13 so (7*13)/(7+13) would be the answer. You would blow the mind of the average student by adding something irrelevant like the pools is 14040G.
That's exactly how I did it. Two minutes to figure it in my head, 15 minutes to watch the proper algebraic method.
With both pumps working in parallel ("together" could mean in series):
18Kl in pool (arbitrary)
18K / 10 = 1800l/h
18K / 15 = 1200l/h
1800 + 1200 = 3K
18K / 3K = 6
I love maths but I also live in the real world.
My first thought when reading the question was: "Are both pumps running off the same source? What is the capacity of that source? I have a shower and a washing machine that run off the same source and when the washing machine fills, the shower pressure drops."
I'm not sure that this is a well thought through question!!
The product of 10 times 15 is 150. The sum of 10 and 15 is 25. Divide 150 by 25 and the answer is 6.
I made the size of the pool 150 gallons, which is evenly divisible by both 10 and 15. Next, I figured that 10 pump takes 15 hours, and 15 pump takes 10 hours. Add 10 plus 15, 25. 150 gallons divided by 25 equals 6 hours. So, I did it in my head in 20 seconds.
Me too! I also used 150 gallons, and used the same reasoning as @XtrmiTeez. I don't think it took me more than 20 seconds, but I used the back of an envelope.
I did the same, but made the pool 30 gallons. One pump runs at 3 gal/hr, the other runs at 2 gal/hr. Both are 5 gal/hr, so it takes 6 hours to fill the 30 gallon pool.
Same except without a value for the pool. I did x/10+ x/15 = x/6
That's exactly how I did it. I had the solution before I started the video.
6.25 hours.
The trick of this problem is to find the flow rate of each pump. In electronics it is like finding the resistance of two parallel resistors one being 10 ohms and the other 15 ohms that divides the flow rate of the current. The simple formula for that is R1*R2 / (R1 + R2). So in hours, 10 x 15 / (10 + 15) or 150 / 25 = 6 hours. If you have more than one pump then the formula would be 1 / (1/R1 + 1/R2 + 1/R3) and just keep adding for more pumps.
6.25 hours.
Your formula is the easiest. ❤
@@HappyBuddhaBoyd Check your math. Yes, I know is takes 15 minutes to turn on all the pumps. LOL
I like this method
You are a competent mathematician.
This problem is a day to day applicable and so highly. useful .
Keep it up to sharing math techniques for our brain exercise.
God bless you richly.
My contribution was recognizing that since the size of the pool doesn't matter there's no need to use plausible swimming pool volumes. So a whopping 30 gallon pool. One pump moves 3 gallons/hr and the other one pumps 2/hr. Combined they pump 5 per hour. 30 gallon pool, so it takes 6 hours.
The 2 pumps rate of filling should be added up. I simply added 1/10 to 1/15. 1 represents the singular whole that is the pool's total volume. 1/10 would be the rate of the 10hr hose and 1/15, the 15hr hose. The resulting answer is 5/30 = 1/6 and thus it would take 6 hours.
I solved it this way:
Pump A will fill half the pool in 5 hours, whereas in 5 hours, pump B will only fill one-third of the pool.
One-half plus one-third equals five-sixths.
Therefore both pumps will fill the pool five-sixths full in five hours,
meaning they will completely fill the pool in six hours!!
That's how I did it.
I made the size of the pool 150 gallons, which is evenly divisible by both 10 and 15. Next, I figured that 10 pump takes 15 hours, and 15 pump takes 10 hours. Add 10 plus 15, 25. 150 gallons divided by 25 equals 6 hours. So, I did it in my head in 20 seconds.
Another way to look at it is:
in 15 hours pump 2 gives us one complete pond.
Also in 15 hours pump 1 would give us 1.5 completed ponds.
So working together for 15 hours would give us 2.5 completed ponds.
Since we only want one pond we need to divide the time by 2.5
15/2.5=6
and that's our answer
That was my hillbilly logic method. It works, so what the hell!
Wow you did all that figuring and kept up with the time...Impressive 😂😂@@XtremiTeez
I haven't read thru the comments so perhaps this has already been mentioned. He hasn't mentioned *why* 1/p1 + 1/p2 = 1/x is the formula to use. Think of it like this: how much of the pool will pump 1 fill in just 1 hour? It can fill the pool in 10 hours, so it'll fill 1/10 of the pool in 1 hour. The other pump, therefore, will fill 1/15 of the pool in 1 hour. With both pumps working together, after 1 hour, the pool will now be filled to 1/10 + 1/15 = 1/6 full. It then follows that if the pool is filled 1/6 of the way after 1 hour, it'll be 6 hours until it's full. If there had been a 3rd pump that could fill the pool in 12 hours, it would be able to do 1/12 in 1 hour and the three working together would fill the pool 1/10 + 1/12 + 1/15 = 1/4 in 1 hour. Consequently, the 3 pumps working together will fill the pool in 4 hours.
it just turns it into a decimal percentage of the pool. one divided by anything gives a percentage. for example, one divided by one hundred is point zero one, or one percent. It is one of those 'show your work' problems that used to drive me nuts. "Show my word??? I know how I got the answer, and I got the right answer, but I have no idea what 'the work' looks like!" Hehe!
This is the only A+ I ever got in algebra. I sucked at it. But I guess that would be a very low grade algebra class--and since then I've taken college calculus. The way I thought of it is working out how much of the pool was filled each pump per hour then adding those for a combined rate. Then looking at how long it would take at that rate. I had to think it of it that way for it to make sense.
Speaking as a chemical engineer, the actual answer would rather depend on the pipework system used, not just the pumps...
😂
Don't forget about humidity and evaporation rate, all influenced by the time of day and year.
For it to work as presented, you would have to assume that the pumps are independently piped, which is probably not how it would be in the real world. If you have two centrifugal pumps both feeding or drawing from the same undersized pipe, you might not gain much at all by using both pumps.
Three words: Product over sum. 10X15=150. 10+15=25. If you count on your fingers by 25, just count up to 150. Then, count the number of fingers you have extended. 150/25=6, No calculator necessary, and is MUCH faster, using fewer steps, than finding the common denominator. In the defense of common denominator, it can also be done without the use of a calculator by anyone that knows the basic X table.
If you were to add a 3rd pump, you would use the same formula, by considering the first 2 smaller pumps, one large pump, capable of filling the pool in 6 hours. 6 times the number of hours your 3rd pump could fill it divided by 6 plus the time it would take your 3rd pump to do it. Consider the first 2 pumps the first pump, consider the 3rd pump the second one.
wow thats the shortest method
And to think that I've been doing the old "reciprocal of the sum of the reciprocals" method for over 30 years now (I'm an electronics hobbyist). Geez, all those unnecessary button pushes when calculating parallel resistances all these years for nothing. You bastard, where were you thirty years ago!? :)
6.25 hours.
@@HappyBuddhaBoyd???
The answer is 6 hours. The good pump you paid several hundred dollars for was turned on noon. At 12:15pm, you fired up the $49 Chinese one you got at Harbor Freight. The Harbor Freight pump took a dump 45 minutes in, so you exchanged it for another and it was back running by 1:30pm. At that point, you realized you were behind schedule and the pool party was in doubt, so you got two more Harbor Freight pumps to run at the same time. Those kept tripping the breaker, so in a move Clark Griswold would be proud of, you jumped around the breaker to keep it running. A few minutes later, the house catches fire and the fire department shows up. Turns out, the water from putting out the fire ran off into the backyard and got the pool filled by 6:00pm. Simple... no math required.
My calculation was that it took 90 mins for pump B to fill 10% of the pool. In the same period, pump A would fill 10% and half again = 15%. Add them together and you get 25% in 90 mins. So 100% would take 4 times that = 360 mins = 6 hours.
It took me a couple of minutes, because I'm an English teacher, out of practice with these problems, but I enjoyed it a lot. Especially when I realised I'd got it right.
It's interesting to note in the comment session how many different ways there are to solve it, or at least, to try to explain the solution!
I think the most intuitive approach is to add together the flow rates. Pump 1 takes 10 hours to fill the pump, so its flow rate is 1/10 = 0.1 pools per hour. For pump 2 it is 1/15 = 0.067 pools per hour. Add the flow rates: 0. 1 + 0.067 = 0.167 pools per hour. Therefore to fill one pool, time = 1 / 0.167 = 5.988 hours.
Obviously I rounded the flow rate for pump 2; it is actually 0.06666666...7, which gives us the proper answer of 6 hours.
My reasoning was the same. hrs/pool is confusing, so pools/hrs is more intuitive and can be added!
@@MarcosGallardo1959 Exactly. Convert to flow rates and add.
This is the method I used. Sadly I arrived at the wrong answer because It seems I forgot how to add fractions!!
That was my immediate guesstimation ... I suppose that old adage applies here: "Well, it's close enough for government work..."
That how I did it, except if you keep it in fractions you can add 15/150 + 10/150 = 1/6 (pools/Hr) in your head if there isn't a calculator nearby.
For two parallel variables, T = AB/(A+B) ; T = 15*10/25 = 6. took longer to type than figure.
That's what I did.
6.25 hours.
@@HappyBuddhaBoyd No supporting information here, but regardless, you're wrong. It's 6.0 hours.
Here's how I did it in my head:
One pump fills in 10 hours
One pump fills in 15 hours
Lets make the 15 hour pump be the 100% efficiency pump
That means the 10 hour pump is working at 150% of the first one
Together, both pumps will have an efficiency of 250%
15 hours / 2.5 = 6 hours
it is very interesting to note that in real life, people have an inventive way of figuring it out logically. the percentage and fractions approach were delightful! and so was the arbitration of a certain constant. they did not sound too classroom , or algebraic, simply put, genius and practical !
Well, let's see. A politician can write a piece of legislation in ten hours. A different Pol can do it in 15 hours. How much time does it take both of them together? Answer: Forever, because they can't agree and refuse to compromise on anything.
Well yes. The fact that it takes one woman 9 months to have a baby does not mean that 9 women can do it in one month.
@@russlehman2070 but on average that would be 1 baby a month for a 9 month period.
That's because they are working in opposition rather than in cooperation. If in cooperation you add rates in opposition you subtract rates. The problem as written is similar to two cars traveling at different speeds towards each other, how long till they meet. Your problem is like two cars traveling away from each other at different speeds, they will never meet regardless of speed traveled.
@@josephmalone253No, because the earth is round. If they both travel at the same speed they will meet in about 10,000 miles! LOL.
1500=A x 10
1500=B x 15
1500 ÷ 10=150=A
1500 ÷ 15=100=B
A + B=250
1500 ÷ 250=6=6hours
1500 can be any number,here it represents gallons or liters water to fill the pool.
Btw, I never made it through highschool 😊
Nah man don't worry you took the time out of your day to solve a random problem, you got a good brain
Depends. If the source of the water can do that unimpeded or can only service the pump(s) at at low rate.
I don't know much about algebra but I solved the problem in my head before I clicked on the video. I just designated the pool size to be 100 gallons so the first pump will pump 10 gallons of water per hour and the second pump will pump 6.667 gallons per hour...16. 667 gallons per hour together. At 16.667 gallons per hour the pool will be full in 6 hours.
I did it in my head first.
To make the equation easier to work with, I pretended that the pool size was 10 x 15, which is 150 in volume Litres/Gallons or whatever.
150/10 = 15, so Pump 1 pumps 15 Litres/Gallons per hour.
150/15 = 10, so Pump 2 pumps 10 Litres/Gallons per hour.
Together they pump 15 + 10 Litres/Gallons per hour.
150 / 25 is 6 hours.
This is exactly how I addressed the problem; much easier than racking my brain for an algebraic formula. I also like the way raynewport9395 visualized the two pumps working continuously for 30 hours would fill 5 pools, therefore 1 pool is filled in 6 hours.
I thought this was fairly simple. Maybe unorthodox, but I give the pool a volume, divide each to see how much per hour each, added together divided by total volume was 6 hours. Although I believe this only works if pool volume was divisible by both 10 and 15. Otherwise I think there's a more accurate fractional way. Which I can't do in my head. eg: give pool 30 (divisible by both 10 and 15). 30/10 = 3, and 30/15 = 2. Then (representing both pumps simultaneously) 2 + 3 = 5, and 30/5 = 6... which I can do in my head. To be fair, this was 3rd grade work for me, which I haven't done in over 40 years.
So I wasn't wrong, just went about it a bit differently.
Go with a volume of 360, it's a highly divisible number.
@@JMcMillen Unless there's something I'm missing about 360...
As mentioned above a far better value is 30 as it's not just divisible by both [10 & 15] but the lowest whole number. Like when finding LCD in fractions, since that's sorta what we're doing here (pump gpm/gph), in a round about way. That can be done in your head. Like I said, this was 3rd grade for me so it's not hard off the top of the head.
@@MrTwisted003Best of all: Put the volume = V (or X), make the equation and V will immediately cancel out, so it is not needed to be known. Then you have also showed clearly that the volume is irrelevant.
I wrote it as I heard it to write and solve formulas for the data given as - -
Let p1 = pump 1 flow rate and p2 is that for pump 2.
Let V = volume of pool to be filled, t = filling time in hours. General equation to fill the pool is p x t = V, So for
a) For pump 1 p1 x 10 = V and b) pump 2 p2 x 15 = V
Equating a) and b) we can write p1 x 10 = p2 x 15; rearranging and simplifying this gives c) p1 = 1.5 x p2
Equation for both pumps p1 + p2 filling the pool using our general equation is d) (p1 + p2) x t = V where we want to solve for time t in hours.
We can rearrange d) as t = V/ (p1 + p2) and then substitute for V from b) and p1 from c) to write t = p2 x 15 /( 1.5 x p2 + p2).
The p2 s all cancel out leaving t= 15/(1.5 + 1) = 15/2.5 = 6 hours
Yes I know there are intuitive and much faster ways to get the right answer but I hope this helps those who missed that.
Its easily adapted and solved for say, pump 1 half fills the pool in 10 hours, equation a) becomes p1 x 10 = 0.5 x V, and you can crack it using this process.
It wins hands down if the easy 10 and 15 hours are replaced by say 44.73 and 53.292 hours respectively and things aren't as obvious.
An interesting little puzzle with different ways of tackling it, enjoy!
There was another one you did which was the same thing, only a different kind of work rates, but would be symbolically identical in form. The "Dan and Jon" digging of the hole example is exactly the same form as this one here.
We have 1/x and 1/y as parameters, so let's multiply x and y. So we have 10 and 15 as the parameters. 10*15 = 150 (this is the pool size in arbitrary units). As we'd divide 150 by the other parameter each time to have the time it takes to fill one pool but also to get what part of the pool is filled,, we can just add them together instead: 10+15 = 25 (so each pump fills 25 units per hour). Now take the 150, divide by 25. That makes 6.
6 hours, I did it in my head 😀 it has to be slightly more than half the time of the 10-hour pump and slightly less than half the time of the 15 hour pump the ratio of 10 to 15 is 1 to 1.5 so one hour more than half the 10 hour pump is six hours and 1.5 hours less than half the time of the 15 hour pump is six hours. the answer is 6 hours
Amazing! I'm baffled by the logic of the Work Formula (I'd never heard of such a beast.), but I did more 2 pump problems with different rates and they (of course!) worked out.
I can't overemphasize the importance of making an intuitive guess at the beginning so as to have confidence in the calculated answer, as was done by figuring, "Well, it's gotta be less than 10 hrs." It forces me to understand the problem.
I learned a good deal from this problem, including that the variable x has to be part of the LCD.
Keep doing what you're doing. Good teaching style. Patience is a virtue.
(10x15)/(10+15)=6
@michaelvanhorn643 The same formula as two resistors in parallel (R1 x R2) / (R1 + R2).
you take too long to solve
It’s not about solving a single problem but understanding how to solve all future similar problems.
You can’t spell “too”
If you’re going to criticise someone, especially John, better make sure you can spell. Otherwise unwarranted criticism with come your way.
Then leave the math class and join a spelling class.
There's room for improvement in Jon's spelling record.
In my head in 15 seconds: 6 hours!
Picked a nice pool size that worked with 10 and 15. I used a 300 gallon pool. One pump is 30 gal/hrs and one is 20 gal/hrs. Together, they pump 50 gal/hr.
50x[SIX]=300
I have a degree in accounting, so of course I used rates. I set the volume of the pool to 1000 gallons and calculated the gallon per hour of each pump. Then I summed up the rates, times them by Xhours and equated them to the total volume of the pool. (Rate1 + Rate 2)X =1000.
I calculated it in in my head in like 5 seconds.. by multiplying 10 * 15 just to find the common number of 150 (units = gallons, hectoliters, whatever). Pump A fills it in 10 hours meaning it pumps 15 units/hour, pump B fills it in 15 hours which means it pumps 10 units/hour. Together it's 10 + 15 = 25 units/hour. And in order to fill 150 units of water at rate 25 per hour, it will take 6 hours.
I did the same
I did it to find a generic formula with V=d1·t1=d2·t2=(d1+d2)·t3 we can then replace with d1=V/t1 and d2=V/t2 -> V=(t2V + t1V)/(t1·t2) -> V·t1·t2=V(t2+t1)·t3, we can then simplify by V and we get: t1·t2=(t1+t2)·t3 from which we can extract t3 = t1·t2/(t1+t2) which is quite elegant formula ;) Then we replace with t1 = 10h, t2 = 15h and we get t3 = 10·15/(10+15) = 150/25 = 30/5 = 6 hours 🙂
I done this in my head for 2 pumps by using the Product over the Sum formula.
(10x15)/(10+15)=6 just like two resistors in parallel. You have done the reciprocal way.
I read some of the answers in the comments. Who taught you people, Terrence Howard? Talk about doing stuff the hard way. Just do it the easy way. We have to know how big the tank is. Let's say a nice even number like 1000 gal. 1 pump fill it in 10 hr so 1000/10 = 100 gal an hr. The other in 15hr. 1000/15 =66 gal an hr. Adding the 2 pumps together gives 166 gal an hr. 1000/166=6.02 hr to fill the tank. Now isn't that grade school easy? Some people make things so hard to show how smart they think they are.
I love these problems.. I can work them out in my head. I relate it to the work that I have done over the years..
Will also depend on if they pump through the same fill pipe or separate fill pipes.
Easy. One simple solution to take is this: The slowest pump take 15h, so you can take this as the time it needs for "one unit". When I add the second faster pump I have 2.5 times the mass I can pump (1times the slow pump + 1.5 times the faster pump). So I have a speedup of 2.5 (assuming the pumps have a constant lift of mass, but the exercise does not tell otherwise), so 15h / 2.5 = 6h, which is the result. The charme of that solution is: you can do it quickly in your head.
I always think of these kinds of questions in terms of what will happen in one single hour of work. That standardizes the difference in flow rates into a one hour work scenario.
If it takes 10 hours for pump one to fill the pool, it means that in one hour, a tenth of the pool will be filled.
Similarly, if it takes 15 hours for pump 2 to fill the pool, it means that in one hour, a fifteenth of the pool will be filled.
So every hour consists of a 1/10 element of Pump 1 and a 1/15 element of pump 2.
Together, in that time, 1 hour out of the total hours for the 2 pumps together to fill the pool has passed.
If we let the total hours to fill the pool be x,
Then
1 hour of pump 1 plus
1 hour of pump 2 gives
1 hour out of the total hours (x) for them to fill the pool together
.
Hence
1/10 + 1/15 = 1/x
x=6 hours
Also,
to simplify the above, it means that in one hour, a sixth of the pool will be filled by the 2 pumps
(1/10 + 1/15 = 5/30 = 1/6).
If it takes one hour to fill a sixth of the pool, it will take 6 hours to fill the pool.
I arrived at the same answer - I like the formula but I just found the lowest common multiple (30) treated that as the volume - worked out one pump would work at 2x per hour and other pump worked at 3x ph therefore together they were working at 5x =30 together therefore x = 6. It's exactly the same logic. Did it in my head. I really wished you'd explained the derivation of the formula a little bit better. I understood it but plenty didn't. BTW - an easier way to represent this is to get directly to the answer x = ab/a+b - which is exactly the same formula. 1/a + 1/b = 1/x multiply both sides by ab gets you a+b = ab/x now divide both sides by a+b and you get 1=ab/(a+b)x now multiply both sides by x and you get x = ab/a+b - much more elegant.
I used method 2 and did it in my head. I can still remember being taught this method in high school back in the late 70’s. I don’t know why I remember this actual type of problem in the actual class.
Well to solve that < problem > , you will need one more initial value, that is not indicated, and that is the volume of the pool that is to be filled up, n'est-ce pas?
Thus I set the volume of the pool at 1000 liters and that way came out at 5 hours and 59 minutes for both pumps working all along.
I got 167 liters per hours for both pumps per hour thence got 5,98 after dividing 1000 by 167, which results in 5,98... ...I multiplied 0,98 by 60 and got 58,80 so to be more precise the time is 5 hours 58 minutes and 8 seconds...
Le p'tit Daniel, who hopes to have sorted it out right, if not... ...just leave me a note...
My unique off hand calculations:
2 pump 1 fills a pool in 5 hr.
2 pump 2 fills a pool in 7.5 hours
1 pump 1 + 1 pump 2 = (5 hr + 7.5 hr) ÷ 2 = 6 hr avg.
If the pool volume is V, then the rate of flow of first pump is V/10 cubic metres per unit time
Similarly for second pump the rate of flow is V/15
If it takes time T to fill the pool from both pumps the total combined flow rate must be V/T
Hence V/T = V/10 + V/15
so V/T = V(1/10 + 1/15) and since V cancels out the actual pool volume is irrelevant here
leaving 1/T = 1/10 + 1/15
to solve for T, the common denominator is 10 x15 so we can write
1/T = (10 + 15)/ (10 x 15)
T = (10 x15)/ (10 + 15)
= 150/25
=
As others have pointed out, the mathematics for this problem is identical to two resistors in parallel :)
Slow pump = 2/3 (10/15) flow rate of fast pump (call that f) => flow rate (fast + slow) = f + 2/3f = 5/3f. Since flow rate x time = volume then the combined flow rate for the same volume will yield 3/5 of the original time for the same volume since they are directly proportional. So 3/5 of the original time = 6 hours
The first pump fills 1/10 of the pool per hour. The second fills 1/15 of the pool per hour.
I used the lowest common denominator and made those fractions 3/30 & 2/30 - so between them they fill (3+2)/30 = 5/30 =1/6 of the pool per hour - so together they take 6 hours to fill the pool
I used fractions.
Pump a fills the pool in 10 hrs. Assuming constant flow rate, pump a fills 1/10 of the pool per hour.
Using the same logic on pump B you get the following.
Pump a = 1/10 /hr
Pump B = 1/15 /hr
To add these together you multiply each fraction by the denominator of the other to get like denominatiors.
Pump a = 1/10 *15/15 = 15/150
Pump B = 1/15 * 10/10 = 10/150
Pump a + pump B = 25/150=1/6
Both pumps together pump 1/6 of the pool /hr or to answer the question:
Both pumps working together will fill the pool in 6 hours assuming a constant flow rate.
I was never very good at maths but I remembered my teacher telling me the rule of 3 is important. It never made sense to multiply letters to me and never remembered formula's. To divide it down to 1 I'd need a volume. Pump 2 takes 15 hours, if it was 15 litres it would be 1 L/hour. Pump 1 would be 1.5 L/hour. In 6 hours pump 1 would move 6 x 1,5 L = 9L + 6 x 1 L = 6 L. 9 L + 6L = 15 L (volume of pool)
So many different ways that people have done this!
For me, the pool nominally holds 15 units of water. One pump pumps 1.5 units per hour, the other 1 uph. Together, that's 2.5 uph. 15/2.5=6 hours.
No two pumps can work consistently all the time. "Work together" that's a variable.
Theory in the classroom never works in real life. But it's a good start.
First classroom lesson--
Murphy's law, Anything that can go wrong will go wrong, and at the worst possible time."
Funny to see educators theories go up in smoke when trying to prove them in real situations.
In this case, I would say the pool would fill in just over 5 hours if the pumping goes as good as expected.
The best way to explain it is by thinking about speed of water going out of each pump. One would pool volume divided by 10 and th either is volume divided by 15. Then you can say the speed of the two pumps would be (volume/10 + volume/15) then you can say with new speed of water how long it is going to take to fill up the pool which is [volume/(volume/10+volume/15)]. This gives 6. But here we understand it with speed of water.
The same principle applies about the situation of two cars with different speeds completing a journey. The you can ask how long would it take that the two cars meet if the cars from different end of the trip to meet up somewhere in the middle. Normally the question in these question are provided with time not speed. So again working out the speed is the right way.
Simplest and most intuitive method is as we have learnt in our primary classes :
Working alone in 1 hr 1st pump can fill 1/10th of the tank and the 2nd pump can fill 1/15th of the tank.
Working together in 1 hr they can fill 1/10 + 1/15 or (15+10)/10*15 or 25/150 or 1/6 of the tank.
So working together the can fill the tank in 1/(1/6) or 6 hrs.
Had no idea you could solve this algebarically ;-) Thanks!!
There is a lot more going on in the math question than just knowing the appropriate math formula to apply ... e.g.:
● What are the assumptions to be considered, such as if the pumps can both discharge at the same pressure into the pool
● Are the pumps connected in parallel or in series to the pool?
● How is the formula derived, and why it applies
While it may be difficult for younger minds to comprehend some of such real world considerations, not discussing the latter may teach the wrong thing or lead to more confusion.
As well, maybe the math question should come with a sketch of the pump arrangements to the pool? A picture can explain many things, too!
Cheers!
Great explanation. Thank you!
Simple: 6 hours! We don't need the Volume to be determined. Vol=X, time=T, hours, and we know how many hours each pump takes to fill the pool on its own. Thus using the two flows (X/10 and X/15, respectively), time will be equal: T hours, which we need to find out.
X/10*T+X/15*T=X volume, solve this and get T=6 hours. X cancels each other in this equation and we are left with T=6 hours
(10XT+15XT)/150=X
That's (25/150)XT=X
1/6(XT)=X, cancel X out, we have
1/6T=1
Multiply both sides by 6 and we have the answer:
Thus T=6 in hours
Thank you for this full explanation.
As a retired Firefighter you left out a lot of important information, IE, volume of the pool, the flow rate of each pump in gallons per min/hrs. Supply source of water, tanker truck, pumping from a pond or lake or from a fire hydrant.
Actually the only bit of information that could alter the result is the flowrate over time. And perhaps the temperature. If the pumps speed up or slow down over time we need to account for that. If it's very hot out, some water might evaporate. If it's freezing, we've got a problem too. Don't know why you would want to know all the other things to solve this?
What should not be overlooked is the formula for work is quite similar to a couple of formulas used to sum passive electronic components: resistors in parallel (1/Rt=1/R1+1/R2+1/R3...) and capacitors in series (1/Ct=1/C1+1/C2+1/C3...).
I assigned a flow of 100GPH to the 10 hour pump for a total capacity of a pool at 1,000 gallons. Then divided the 1,000 gallons capacity into 15 hours to get the flow of the 2nd pump at 66 gph. Then it was a matter of adding both flow rates (166 gph) and divide into the 1000 gallons capacity for a result of 6 hrs. To me it makes more sense than following a strict mathematical formulas exercise.
Seasoned Mechanical Engineer using a calculator.
I gave the pool 100 gallons of water to find the gal/hour rating for each pump.
Pump 1 rate: (100 gal/10 hrs) = 10 gal/hr
Pump 2 rate: (100 gal/15 hrs) = 6.67 gal/hr
Pump 1 + Pump 2 = 16.67 gal/hr = Combined pump output
Then work it backwards to find the time needed and lose the gallons in the rate calc. (1 hr / 16.67 gal) x 100 gal = 6 hr 🤔
To explain clearly your formulas:
Let A = rate of first pump(volume/hr), B=rate of 2nd Pump and C= total volume of pool
T=time for both pumps to fill up pool
10*A=C, therefore A=C/10
15*B=C, therefore B=C/15
C =(A+B)T
C=(C/10 + C/15)T
C= (25C/150)T
T=(150C/25C)
T=6
This is the same form as solving resistors in parallel, the formula for which can be transformed for R as R=R1xR2/(R1+R2) which in this case is 10 x 15 / (10 + 15)
which is 150/25 = 6.
Indeed. Good thinking.
I took a little longer route but got the same answer of 6. Interesting how the Volume doesn't need to be calculated or known.
We are given t1=10 t2=15. But volume V is the same for both. V= t1(r1) = t2(r2) That means r2= (10/15)r1 = (t1/t2)r1 .
The volume V is still the same when you use both pumps at once. So V= R(T) = [r1+r2]T Substitute r2= (10/15)r1 then
V=[r1 + (10/15) r1 ] T = r1[1+(10/15)] T = t1(r1) = V from the first equation . r1 cancels out, solve for T so you get
T = t1/[1+(10/15)] Since t1 =10 T = 10/[1+(10/15)] = 6. Yep. Long way around but I skipped no algebra steps.
My answers is more of a proof.
Chat GPT says 😁
To solve this, we can consider the rate at which each pump fills the pool and then combine these rates to find out how long it takes for both pumps to fill the pool together.
The first pump can fill the pool in 10 hours, which means its rate is 110101 of the pool per hour.
The second pump can fill the pool in 15 hours, which means its rate is 115151 of the pool per hour.
When both pumps work together, their rates add up. So, the combined rate is: 110+115101+151
To add these fractions, we need a common denominator, which would be 30 in this case: 330+230=530303+302=305
Simplified, this is 1661 of the pool per hour.
Thus, working together, both pumps can fill the pool in 116=6611=6 hours.
6 hours. I took an example of a 400M^3 pool. 1st pump pumps 40 cub per hour, the second one 26.667 cub per hour. Together 66.667 Cub per hour. 400/66.667 = 1200/ 200 = 6 hours
I did it differently in my head by assigning a number to get gallons per hour per pump and then divided total gallons by the combined gallons per hour. So, I figured that the slower pump would fill a 15 gallon pool at 1 gallon per hour and 15/10 for the faster pump is 1.5 gallons per hour. added together 1.5 gallons per hour + 1 GPH = 2.5 GPH and 15 total gallons divided by the combined 2.5 GPH = 6