Time and work questions are easy - just follow this easy math trick. To donate to the tecmath channel: paypal.me/tecm... To support tecmath on Patreon: / tecmath To buy tecmath mechandise: teespring.com/...
I’m not a fan of trick without explaining the logic of why we take the product A*B and divide by the sum A+B. I would explain it like this: Adam’s rate is 1/12 job per day while Bob’s rate is 1/6 or 2/12. So their combined rate is 3/12 job per day. We want the answer in day/job so we invert the resulting ratio to get 12/3 days per job. Edit: with the above logic, let’s prove why the “trick” in the video works the way it was shown. Let A and B be the number of days Adam and Bob work per job, respectively. Therefore, their combined rate in term of job/day is: 1/A + 1/B = (A + B)/(A*B) job per day. To get the answer in days/job we invert the resulting ratio to get AB/(B+A) as shown in the video.
The key for solving this kind of problem is coming up with a common unit of measurement, which would be the fraction of the job done in one day. For Adam, this is 1/12, and for Bob, it's 1/6. Adding the two fractions yields 3/12, which simplifies to 1/4, the portion of the job they get done in one day working together. This means they would complete the job in four days together.
I have no idea what all the calculations are, all I did in my head was: Adam finishes 1/12th of the job per day, Bob finishes 1/6th of the job per day, which is 2/12th, so they both finish 3/12th of the job per day together and if you finish 3/12 per day, you get 12/12th after 4 days.
You are correct. This terrible video uses a magic recipe. The problem with that approach is that you need a magic different recipe for each and every type of problem.
Yeah, I think this is a much better way to explain it. But you don't need to use 1/12 or 1/6. You can use 3 days as a measure for example, because both 12 and 6 can be divided by 3. So Adam does 1/4 of his job in 3 days, Bob does 1/2 of his job in 3 days. When you combine those two, you get Adam and Bob did 3/4 of their job in 3 days. Assuming that they work at a constant rate in 4 days they'll finish 4/4 of the work.
6 days as long as adam stays out of bobs way and doesn't mess anything up 😂 adam isn't what you call good help, and guys like that can actually slow you down
I perfectly understand how to execute the math-step now. I do wonder why it works though. Why do we multiply the times of their efforts together and shrink them back down? Who came up with that?
@@zalida100 thanks, I'll check it out. I'm realizing that my education - while certainly not poor - did not advertise just how awesome mathematics can be in real life situations. It's difficult know where to start with my renewed love for numbers. There's the skeletal framework of how to execute formulas, but I have a limited understanding of the concepts. The concepts would be immensely helpful. The 'whys'.
@@zalida100how is that a harmonic mean? You’re suggesting in the first example that we have 72 of something, what is this thing? And we’re not dividing by the sum of the reciprocals of these things, unless you’re saying 12 and 6 are the reciprocals of those things, in which case I don’t know where the 72 comes from.
For these kinds of problems you're essentially finding the equivalent resistance of a network of parallel resistors. However in these cases the resistor values are given in days rather than Ohms. Consider the first problem: 1/Req = 1/R1 + 1/R2. In this case 1/Req = 1/12 + 1/6 = 3/12 = 1/4. Therefore, Req = 4 days. ◼
Adam makes 1/6 of the job per day. Bob makes 1/12 of the job per day. Working together they do. 1/6+1/12 =3/12=1/4. So if working together make 1/4 per day it Will take 4 days to do the complete job
A long time ago, it would've been impossible for me to solve questions like these on my own. Now that I've revisited math and studied other things as well, I could answer all of them correctly! Not by using this trick, but by reasoning through it. Thank you! :)
Each day, Adam does 1/10 of the job. Bob does 1/15 and Connie does 1/12. Altogether they do 1/10+1/15+1/12 =6/60+4/60+5/60 = 15/60 = 1/4 of the job per day. So it would take them 4 days to finish the job.
@@jmmainat Thank you for sharing! Yes, our method is the same, there's just a small difference in the presentation. I wrote it out as an equation: (1/10+1/15+1/12)*x=1, where x is the number of days. I've picked 120 as the common denominator, but 60 is even better, because it's the lowest common denominator! :) Lesson learned: when I've found a common denominator and it's divisible by two, I start dividing it by 2 until I get the lowest common denominator. Thanks again!
As an engineer who did a lot of work in production years ago, it's NEVER as simple as you imply. In each separate case, you need to answer the basic questions: What kind of job is it? Do Adam and Bob both do the same thing, independently, or are their individual efforts dependent on what the other does? Is it the kind of job where Adam must wait for Bob to finish his part before he can begin? Every job is different, especially jobs that require a number of steps that must be completed in a specified order. What is the division of labor involved - is Adam good at some things, not so at others? Ditto, Bob. The answers to these questions, and many others, will definitely affect the calculations which are nowhere near as simplistic as this. BTW - in the Real World, in any multi-operation serial process, the slowest worker or the slowest operation sets the pace. Any Industrial Engineer will tell you that.
While your statement is correct, we do have to start teaching the math somewhere, and many people (if not most) are intimidated enough by math as it is. (To be honest, I don't particularly like the "trick" used in this video. I prefer to use rates for these types of questions.)
This is the only mathematical way to calculate it: In one day Adam finishes 1/10 of the job, Bob 1/15 and Connie 1/12. Altogether they do 1/10+1/15+1/12=1/4 of the job per day. Consequently, it will take them 4 days to finish the job. Understood?
Following problems also can be done by taking LCM example A does job in 12 days B does same job in 4 days let's assume the job as LCM of both that is 12 so 12 will be job now take A and B efficiency so A's efficiency will be 12/12=1 and B's will be 12/4=3 now add both efficiency that will be 1+3=4 now same job / total efficiency 12/4=3 days it is
Ok, the method for the third case is not the same as the first two, and all three are overcomplicated. 🙂 And, to be fair, I'm a simpleton so I have to look for simple mechanisms. This is a time and velocity problem. The workers have work velocities. The job is the integration of velocity over time. So, pick the third scenario, and choose a worker at random. Let's go with Bob and declare he works at a rate of one unit per day. Since he completes the job in fifteen days, a job is 15 units. Velocities over an equal distance are in inverse proportion, so Adam's velocity is Bob's time:Adam's time, or 15/10 (1.5). Connie's velocity is 15/12 (1.25) So, job/total velocity=time. 15/(1+1.5+1.25) or 4. But even that is not really general. What if 3.2 Adams could complete the job in 10 days, Bob and his twin could do it in 15, and the Connie triplets could do the job in 12 days? How long would it take two of each to do the job? Let's use Connie as our unit. 3c*12 = job. Set our reference with Connie's velocity = 1, so the job is 36 units. 2b*15=job=36, so Bob's velocity is half of 36/15, or 36/30. 3.2a*10=job=36, so Adam's velocity is 36/32. The total velocity is two of each, or 2*1 + 2*36/30 + 2*36/32. For convenience, call that 2(1+36/30+36/32). Connie defined our units and set the job at 36, so 36/2(1+36/30+36/32) = 5.41 days for two of each worker. The method works for any number of workers, any number of each worker, and any grouping for the solution.
First is helps to see that together they can do the job in less time than either can do it alone. So the answer has to be less than 12 or 6 days... and indeed... less than 6 days. You can see that if the both can do the job in 6 days, the job will take 3 days. Since Adam is slower than Bob the job will take more than 3 days, and less than 6 days. Along the lines of what Tzukova posted...the average of 3 and 6 is 4.5. Thus,4 and 1/2 days would be a reasonable estimate. But you can do better if you understand how to weight the average.
It's very wise to understand first the formula before teaching the shortcut method. My approach to the problem is much easier and very understandable. It would be: 1/12 + 1/6 = 1/X 1 represents the number of their work and X represents if they work together. Now solve it: 1/12 + 1/6 =1/x 3/12 = 1/x 1 times 3 divide 12 is 4. (ratio rotation method). 4=X.
For the first kind of problem (where we have Adam doing job in 12 days and Bob doing job in 6 days) it can be done another way. Take 3 days as a measure. For 3 days, Bob will do 1/2 of all the job; while Adam will do 1/4 of all the job. If you add 1/4 to 1/2 you'll get 3/4... That means that Adam and Bob in 3 days would do 3/4 of their job, and that means for 4 days they'd do 4/4 of their job that is all the job.
You don't need a formula. Just find the number of jobs each can do in the same amount of time. Adam does 1 job in 12 days. Bob does 2 jobs in 12 days. Adamn and Bob do 1+2 = 3 jobs in 12 days. i.e. 1 job in 4 days. Adam does 1 job in 30 days. Adam and Bob do 5 jobs in 30 days. Meaning Bob does 4 of those jobs in the 30 day period. i.e. Bob does 1 job in 7.5 days. The least common multiple (lcm) of the days is 60. Adam does 6 jobs in 60 days. Bob does 4 jobs in 60 days. Connie does 5 jobs in 60 days. Adam, Bob, Connie do 6+4+5 = 15 jobs in 60 days. i.e. 1 job in 4 days.
The way I did these in my head was combining the rate of work (jobs/day) and then the reciprocal would give the rate of days per job. So in example 1 Adam works at a rate of 1 job per 12 days and Bob at a rate of 1 job per 6 days: ((1/12)+(1/6))^-1 = ((1/12)+(2/12))^-1 = (3/12)^-1 = 12/3 = 4 ((1/a)+(1/b))^-1 = ((b/ab)+(a/ab))^-1 = ((a+b)/(a*b))^-1 = ((a*b)/(a+b))
It's faster to observe that, given Bob works twice as fast as Adam, if they team up they will work 3 times as fast as Adam alone, and the job will be done in (12/3) = 4 days.
This is a trick and doesn't really promote thinking the problem through. This is all fine as long as the student understands at some point why this trick works. To try and make sense of the problems I would think as follows: Q1: If Bob takes 6 days to complete 1 job then he can do 2 jobs in 2*6=12 days. So, working together Adam and Bob can do 1+2=3 jobs in those 12 days, or 1 job in 12/3=4 days. Q2: Together Adam and Bob can do 1 job in 6 days or 5 jobs in 5*6=30 days. So, since Adam can do 1 job in those 30 days Bob must do 5-1=4 jobs in the same 30 days, or 1 job in 30/4=7.5 days. Q3: Adam can do 1 job in 10 days, or 6 jobs in 6*10=60 days. Bob can do 1 job in 15 days, or 4 jobs in 4*15=60 days. Connie can do 1 job in 12 days or 5 jobs in 5*12=60 days. So together they can do 6+4+5=15 jobs in those 60 days, or 60/15=4 jobs in 1 day. The approach used in each case is to look for a common number of days that they can complete work in, then you can simply divide by the number of jobs that can be completed in that time to find the number of days it takes to complete 1 job.
My theory is, 12 + 6 = 18 / 2 = 9 / 2 = 4,5days. So i made an average task time for both which is 9 days for one person which i divided by 2 bcs they are going to be 2 times faster together.
The job will really take 18 days. Adam will get pissed off with Bob for being a suckhole and do everything in his power to slow him up. The Project Manager will get pissed off and fire the both of them.
I looked at it as Adam works half the speed of Bob. This suggests that (ignoring the complexities of sharing a job) that Bob will get 2/3s of the work done and Adam 1/3. If you ask yourself how long it would have taken Adam to do a job 1/3 of the size well that's just 1/3 X 12 = 4. Just to confirm, give Bob a job 2/3s of the size and that's 2/3 X 6 = 4. So the job is done in 4 days. But how did I work out that Bob would be doing 2/3s of the work? Well I assumed that Adams rate was the standard and called that 1. Now Bob gets the work done in half the time so his rate is twice that of Adam so I call his rate as 2. So I imagine the two of them working together meaning we have a total work rate of 1+2=3. Adam is doing a rate of 1 so his contribution is 1/3 and Bob's is 2/3.
Meanwhile, back on Planet Real, it will take 8 days. Bob will be doing his best to get the job done in time but he'll have to keep stopping while Adam shifts his lazy but out of the way.
what if Adam slows Bob down, cos Adam isnt quite as experienced as Bob, if this was the case, it would likely take 8 days if the work together and would cost more than if you just hired Bob.
I mean obviously 13 days. They spend the first day arguing about technique, and at the end of the first day Bob storms off frustrated with Adam’s inefficiency, leaving Adam to start the job over and take 12 days.
How much job can Adam do in 1 day? 1/12 jobs How much job can Bob do in 1 day? 1/6 jobs. How much job they can both do in 1 day? 1/12+1/6 = 3/12 = 1/4. They do the job in 4 days. Now generalyze A takes X days, so he makes 1/X job in a day B takes Y days, so he makes 1/Y job in a day. They both do the job in 1/X + 1/Y. Let's multiply both members by 1, in the form of XY/XY XY/X²Y + XY/XY² = Y/XY + X/XY = X+Y/XY. This is the inverse of the time, so the answer in how many days will be XY/(X+Y)
Here's a simple way. If A is Adam-days, and B is Bob-days, then A/12 + B/6 = 1. Assuming you want both Adam and Bob to work the same number of days, you want A = B. Therefore, A/12 + A/6 = 1, or 3A = 12, or A = 4 = B. 4 days.
Bob will have 1 sixth of the job done in 1 day. Adam will do 1 twelfth. So in 1 day, they will do 3 twelfth. 3 twelfths multiple by 4 days is 12 twelfth.
It would be nice to see the logic leading to your method, complete with units of jobs and days. In the first case I like a different method. Convert everything to Bob-units. 12 Adam = 6 Bob = 1 job, so Adam equals Bob/2. A day's work for both is then 1.5 Bob. X days of 1.5 Bobs equals one job, or 6 Bob. Divide each side by 1.5 and you get four days. In the second case, j=30A=6(A+xA). In other words, 30 days of Adam or 6 days, each of which is one Adam plus one Adam scaled to equal one Bob. Divide both 30A and 6(A+xA) by 6, you get 5A=A+xA. Subtract A from each side for 4A=xA, or x=4. One Bob-day equals 4 Adam-days. The job, 30 Adams divided by 4 to convert to Bob-days, says Bob alone gets the job done in 7.5 days. In the third instance, one job equals 10 Adam, 15 Bob, or 12 Connie. Bob = 2/3A, Connie equals 5/6A. So 10 Adam (one job) equals x days(Adam+2/3Adam+5/6Adam) or 10 Adam = x(2.5 Adam). Divide each side by 2.5 Adam to reveal x = 4, the number of days all working together would take. I like your method but don't yet see the derivation. The workers are values in units of job/days, so multiplying workers yields units of job^2/days^2. Dividing by a sum of job/days gets you back to units of job/days, so that makes sense.
Great stuff…but please make sure you are giving an explanation as to why you do things…would be much more helpful in understanding the problem and solution better with having an explanation as to why you do something. Thanks!
I'm trying to remember this so can you do more problems like the 2nd problem. I missed a step so I need it broken down alittle better please and thank you.
Abigail can have a baby in 270 days Bethany can have a baby in 270 days How long will it take for them to have a baby if they work together? 135 days, apparently
@@silverhammer7779 This is a funny rabbit trail. The average production rate is a baby per month. They just plop out in 9 month cycles. So if you only measure at the end of a 9 month cycle, it works (except for the recovery time [ 18 months recommended ]) The real rate is a baby per mother per 27 months (or so). So 9 women can average a baby every 3 months. Good luck making that work.
The job still takes 6 days, Bob paints his half of the house in 3 days then spends the next 3 days chatting up the hot house wife, while adam finishes his half of the house.
1:00 Real world maths. Answer 6 days. Explanation: Adam is a Union man. If Bob is to work with him Bob must join the Union and work at the Union pace and no faster. If Bob does not join the union Adam downs tools and Bob continues his work alone.
Adam is no dill, he has been scamming this job for some time. He can see Bob is going to make him look bad. He will fix it so something breaks and stops everything for some days. Bobs tools will also mysteriously go missing. Total length of job 12+6 of course.
Why? I understand how to do this as 1/(12 hours) + 1/(6 hours) = 1/(4 hours), that makes intuitive sense to me, but the ab/(a+b) I'm not getting why/how it works.
The problem with your solution is that nobody knows why you are doing the math you presented. It works, yes, but why. Here is my way of formulating the problem: Ar is Adam's work rate; He can perform a job in 12 days so Ar = J/12. Br (Bob's work rate, 1 Job in 6 days) = J/6. The time, T for Adam and Bob to work together. Therefore, J = T* Ar + T* Br = T*J/12 + T * J/6 = T*J(1/12 + 2/12) = 3/12 = 1/4. J's cancel out. T = 4.
One man can dig a post hole in 10 minutes. How long will it take 10 men to dig a post hole? Answer -- 30 minutes or more. In the REAL world, things don't work with mathematical precision. Sure, if someone gets this sort of question on a math test, you can use this method to arrive at an answer. But you'd be a fool to rely on the answer you'll arrive at if you need to apply the answer to the real world.
This is a TERRIBLE way to teach maths, but sadly schools are full of incompetent maths teachers like this one. He uses a recipe and then plugs in the numbers. Since there is no understanding, pupils must learn a different recipe for each type of problem. Le Phan provides the correct method to solve the problem.
