A pump can fill a pool in 10hrs. Another pump can fill it in 15hrs. Both pumps together will take? 

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.

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

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.

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.

It's like resistors in parallel, so 1/10 + 1/15 = 1/total = 6 hours.

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.

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!

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).

In 30 hours you would fill 5 pools, therefore you would fill one in 6 hours.

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.

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.

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

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

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.

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)!

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.

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.

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!

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.

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)

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✔️

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.

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)

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.

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

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.

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.

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.

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 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.

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.

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.

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.

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!!

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!

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!!

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.

For two parallel variables, T = AB/(A+B) ; T = 15*10/25 = 6. took longer to type than figure.

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.

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.

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.

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.

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.

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.

Speaking as a chemical engineer, the actual answer would rather depend on the pipework system used, not just the pumps...

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.

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.

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.

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 😊

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.

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.

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.

(10x15)/(10+15)=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

The way I did it was to calculate the gallons per hour each pump can deliver, then just add those two rates together to get the total hourly flow rate. To keep the math easy, I just assigned the pool to be 10 gallons in total. So running the first pump alone at 1 gallon flow per hour, it would take the (given) 10 hours to fill the pool. The second pump would pump at only at 10/15 gallons per hour (2/3 gallons per hour). So the pumps running together have a flow rate of 1+2/3 gallons per hour. (1.666667 gph). A 10 gallon tank would then take 10/1.666667 hours to fill with both pumps running together. Disregarding the previous rounding error, the two pumps would need 6 hours to fill the pool.

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.

you take too long to solve

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.

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 !

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.

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.

Depends. If the source of the water can do that unimpeded or can only service the pump(s) at at low rate.

Will also depend on if they pump through the same fill pipe or separate fill pipes.

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!

If you represent the totall mass in the pool by 1 then: The speed of one pump will be 1/10 and the other 1/15 per hour. The combined speed will be 1/10 + 1/15 = 10/150 + 15/150 = 25/150 = 1/6 per hour. The time it takes to fill will then be 1/(1/6) = 6 hours. I originally took the challenge of solving it entirely in my head by setting up an equation in my head without writing and solving it, which worked. But then I found out that it can be solved better stepwise as depicted. The problem with mathematical tools like equations. is that they often will hide the deeper understanding of a problem for you even though you get the right answar.

I made volume 150(the actual volume doesn't matter, all you need is time it takes to fill) for easy arithmetic (10x15), so pump 1 takes 10hrs so rate is 15, pump 2 takes 15hrs so rate is 10, combined rate for both is 25, 150/25=6hrs

Sensible sentence maths method: Imagine the pool contains 30 tons of water (any other number would work but 30 is chosen because it is divisible by 10 and 15 so all the calculations produce whole numbers which is just nicer). Because pump 1 fills 30 tonne pool in 10 hours then it pumps at 3 tonnes per hour (30/10 = 3). And because pump 2 fills pool in 15 hours then it pumps at 2 tonnes per hour (30/15 = 2). Together the pumps will pump at the rate of 5 tonnes per hour (3 + 2 = 5), therefore to fill the 30 tonne pool will take 6 hours ( 30/5 = 6). This works just fine so up u algebra. But has anyone taken into account that the pumps and hoses from them to the pool also contains water therefore have to fill up before any water goes into the pool? Bet not; but does it make any difference? Yeah it does and because we don't know how much water is needed to fill the pumps and hoses then we do not have enough information to calculate the answer. Then there is the evaporation rate from the pool during the pumping time, as well as ......

The problem is underdefined. Depending on how the pumps are connected to the pool and what characteristics the pumps have, the result may be different. For example, it could be that the 15h pump fills 80% in 7.5h but has very little capacity at the end and needs 7.5h for the last 20%, while the 10h pump has a straight characteristic and fills the last 20% in an hour.

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.

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.

The way I did it in my head, is that I hypothetically made the size of the pool 1500 gallons. So that is a flow rate of 100 gallons per hour for the 15 hour pump and a flow rate of 150 gallons per hour for the 10 hour pump. so with both of them thats a flow rate of 250 gallons per hour. So I divide 1500 by 250, and get 6 hours as the answer.

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.

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.

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...

I got there using dumbo mental arithmetic, where the pool capacity is fixed at an imaginary 15L, so those two pumps could be accomodated that way. In a real life situation a proper math approach like in the video would definitely have been preferable : D

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.

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!

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.

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

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

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!

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.

13:28 You did say, earlier, that the LCD was an important piece for solving with fractions When we got on toward the second method, tho, even though you still made a fraction, I didn’t think the LCD would be involved, this time Which is funny, having remarked that, because I’m starting to remember my grade school lessons - but in those, we didn’t multiply all the numbers by the LCD, we actually raised all the denominators to the LCD, so they could be simply added together Thus, the equation becomes 3/30 + 2/30, so if 5x/30=1, x=6 I didn’t learn about matrices until Algebra 2, but then I took no further classes, so all of that knowledge is frayed by a lack of reinforcement

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

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

pump a fills it at a rate of 1/10 per hour and pump b at 1/15 per hour. the smallest common denominator is 30. a = 3/30 b = 2/30, so a+b = 5/30. so it takes them 30/5 = 6 hours.

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 did a rough calcation in my head going by... in 5 hours, the 10hr pump will have filled half and the 15hr will have filled 1/3. Then something in my head said 6. Using the calculator was easy by using a random volume to give a flow rate per pump, as the video explains.

Pretty easy to solve in your head quickly as soon as you see the 3/2 ratio. My mind just thought of it as flow rates.. 2L/s and 3L/s. Using the 10h and 15h, 15x2 = 30L ; 10x3 = 30L. So a 30L bucket would get filled up at 5L per second. 6 hours.

I did it by imagining that there were 150 units of water. Pump 1 fills the pool with 15 (150units/10hrs) units of water/hr Pump 2 is 10 (150/15) units water per hour. Together they are 25 units of water per hour and 150/25 = 6. I guess it would have been easier to do it with an LCD of 30? 150 just seemed more intuitive for some reason.

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 thought of the problem in terms of pump power. 10p = 1 15q = 1 q in terms of p (divide 15 by 3 and multiply by 2). q = (2p/3) Add p + 2p/3 = 5p/3 Find the value of p 10p = 1 p = 1/10 Calculate the hours h(5p/3) = 1 h(5/30) = 1 h(1/6) = 1 h = 6

This is the exact same formula for calculating parallel resistors! A 10-ohm resistor in parallel with a 15-ohm resistor will yield an equivalent resistance of 6 ohms.

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.

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.

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 got it right. My logic was that the second pump must have a water flow rate that was 2/3rd that of the first pump (10/15) so combined the flow of both would be 5/3rds. Inverting this gives 3/5ths which times 10 gives 6 hours.

I looked at it from a different direction, I imagineded a 300 litre pool, calculated the flow rate of each pump based on the 10 and 15 hours provided and then divided the imagined capacity by the combined flow rate to arive at the desired result. I chose 300 litres as it was easily divisible by both 10 and 15, not necessarily the lowest common denominator but as good as. I said 300 litres in 10 hrs is a flow rate of 30l/hr 300 litres in 15 hrs is a flow rate of 20 l/hr Combined is 20+30=50l/hr To pump 300 litres at 50l/hr is 300/50 = 6 hours. It took longer to type than to calculate the answer.

It took me one minute to remember that this is the classic inverse of the sum of the inverses problem that I learned half a century ago in middle school. In this instance, time = 1 / ( 1 / 10 + 1 / 15 ).

Conceptually, the first pump fills the pool in ten hours, so it fills one tenth of the pool per hour. The second pump fills the pool in fifteen hours, so it fills one fifteenth of the pool per hour. If both are on together, they will fill one tenth plus one fifteenth, or one sixth of the pool per hour, and therefore it will take six hours to fill the full six sixths of the pool. I think this is the easiest way for folks that are having trouble with the logic to wrap their heads around it.