Most US College Students Cannot Solve This Basic Math Problem. The Working Together Riddle
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Most US College Students Cannot Solve This Basic Math Problem. The Working Together Riddle


Hey, this is Presh Talwalkar. Alice and Bob can complete a job in two hours. Alice and Charlie can complete the same job in three hours. Bob and Charlie can complete the same job in four hours. How long will the job take if Alice, Bob, and Charlie work together? Assume each person works at a constant rate whether working alone or working with others. This problem has been asked to students in US colleges. To the professor’s surprise, many of the students set up the wrong equations and could not solve this problem. Can you figure it out? Give this problem a try and when you’re ready keep watching the video for the solution. Before I get to the solution, let me go over a common mistake in how students get to the wrong answer. They read the first sentence, that Alice and Bob can complete a job in two hours, and translate the names and the numbers into an equation. They say this must mean that A + B=2. They look at the second sentence, that Alice and Charlie can complete the job in three hours, and they similarly convert it to A + C=3. The third condition, that Bob and Charlie can complete the job in four hours, gets converted to the equation B + C=4. The question of how long it will take for all three of them working together gets translated into the question
of “what is A + B + C =?” So to solve this system of equations… They want to solve for A + B + C, so they can add up all the equations together. We end up getting two terms of A, two terms of B, and two terms of C, to equal 2 + 3 + 4. If we group the factors [summands], we get
2A + 2B + 2C=9. We then divide by two and that gets us to A + B + C=9/2=4.5 (four and a half). So evidently this will be the answer that many students get. They would say that it takes 4.5 hours for all three of them, when working together. But let’s think about: does this answer make any sense? We know that Alice and Bob take two hours, Alice and Charlie take three hours and Bob and Charlie take four hours, but somehow when all three are working together they take four and a half hours? This makes no sense. When three people work together it should take less time then when only two people work together! But 4.5 hours is more time, so this answer must be wrong. Not only were the equations set up incorrectly, but any student who submits this answer is not thinking about whether the answer makes any logical sense. So how do we solve this problem? We need to set up the equations in the correct method. We know that Alice and Bob can complete a job in two hours, so how do we translate this into an equation? Well, if they complete the job in two hours, that means the percentage of the job that Alice does in two hours plus the percentage of the job that Bob does in two hours equals 100%, or that equals 1. Now since they work at a constant rate, we can say the amount of the job that Alice does in two hours is 2 times the amount of the job that she does in one hour. And the same thing goes for Bob. So we now have a natural choice for our variables we can say the percentage of the job that Alice does in one hour will be the variable A, and the percentage of the job that Bob does in one hour will be the variable B. This leads to the equation that 2A + 2B=1. And that’s how we can translate this we can group this out to be 2 times (2x) the quantity A + B=1. So we can now translate the second sentence. We have Alice and Charlie completing the job in three hours. This will translate into 3 times the quantity A + C=1, where C is the percentage of the job that Charlie completes in one hour. We also have that Bob and Charlie can complete the job in four hours, so that would mean 4 times the quantity B + C=1. Now we want to figure out… What would happen if they all three work together? So we are needing to solve for… the time (t) x (A + B + C)=1. We need to solve for this variable “t”. So how do we do that? Well, we can similarly add up all the equations But we have different quantities of each of these variables, so in order to get the same number of each variable, we’re going to do a little trick. We’re going to multiply each equation so that there’s a leading coefficient of 12, which is the lowest common multiple of 2, 3 and 4, so the first equation will multiply by six. This will get 12 times the quantity A + B to be equal to 6. The second equation will multiply by four and the third equation will multiply by three. We can now add up all of these equations We’ll end up with 12A (2 times), 12B (2 times), and 12C (2 times), and this will be equal to 6 + 4 + 3. We can factor out the 24 of each of these variables, and that will be equal to 6 + 4 + 3, which equals 13. We now divide by 13… and we end up with 24 divided by 13 times the quantity A + B + C=1, and THAT is what we wanted to figure out. So we go back to our setup and we can see that we get to the answer of 24 over 13. So the job will take 24 over 13 hours or about 1 hour and 51 minutes. And this is a sensible answer because it takes less time than any pair working together. Did you figure this out? Thanks for watching this video; please subscribe to my channel. I make videos on math. You can catch me around around my blog, mindyourdecisions, that you can follow on Facebook, Google+, and Patreon. You can catch me on Social Media @preshtalwalkar and if you like this video please check out my books! (There are links in the video description.)

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100 thoughts on “Most US College Students Cannot Solve This Basic Math Problem. The Working Together Riddle

  1. Unless Alice, Bob and Charlie are robots,
    The correct answer is: Undetermined.
    Reason: Human factor
    Same trick question with: If a person is running 4 mph constantly, how many total miles the person would cover if he ran for 24 hours straight?

  2. Ok so Im tsking a wild guess becuse my mobile data is so slow I can only see the thumbnail. so if alice does 30 minites of work, and bob takes hour 30, whih means charlie taked 2hours 30 and for some reason adding more peopke decreases efficiency so im saying 4 hours 30 final awnser

  3. It would have been at least 4.5 hrs. Clearly, Charlie makes mistakes and Alice and Bob have to fix them as well as do their own. Dammit Charlie.

  4. One can use the method of finding total resistance in a parallel circiut too:
    (1/a + 1/b)^(-1) = 2
    (1/a + 1/c)^(-1) = 3
    (1/a + 1/c)^(-1) = 4
    (1/a + 1/b + 1/c)^(-1) = t
    t = 24/13
    Finding the total resistance in a parallel circuit and this is mathematicly the same question as this problem.

  5. There’s possibility that someone is pulling back so that despite the fact 3 people work together but take longer time to done it

  6. We learned to use (1/A + 1/B = 1/X) and (1/A + 1/C = 1/Y) and (1/B + 1/C = 1/Z) and solve for H in (1/A + 1/B + 1/C = 1/H) in mid. school (i think 6th grade) in Iran.

  7. So the key to understand the problem is notice the relation of the hours and individual work with the constant rate, so proper equations could be formed.

  8. Here’s my take, Alice can complete her tasks in .5 hours, bob can complete in 1.5 hours, and Charlie in 2.5, it adds up to 4.5 hours

  9. I knew it was percentage based but I didn’t know how to convert what I was thinking into a formula and I was always thing a bit wrong is saying that each person does a certain percentage of the job and that was a fixed number (like Alice always did 60% of the job or something)

  10. (1/rA + 1/rB)= 1/2
    (1/rA+1/rC) = 1/3
    (1/rB+1/rC) =1/4
    2*(1/rA+1/rB+1/rC)= 13/12
    1/rA + 1/rB + 1/rC = 13/24 meaning that they accomplish about 54.17% of the work in 1 hr by working together so they can finish the work in 24/13 hrs

  11. Not a common answer but the job may be quick but they argue a lot.
    Alice and bob could argue less than alice and charlie. And bob and charlie argue more than alice and bob do. Bob likes alice better than charlie. Charlie likes alice more than bob. They could just get along with each other and all do it together in an hour.

  12. 360 boxes need to be moved.
    A+B MOVE 1/2 OF THEM / HOUR, OR 180 / HOUR
    A+C MOVE 1/3 OF THEM / HOUR, OR 120 / HOUR
    B+C MOVE 1/4 OF THEM / HOUR, OR 90 / HOUR
    A MOVES 105 / HOUR, B MOVES 75 / HOUR, C MOVES 15
    105 + 75 + 15 = 195 / HOUR
    1.846 HOURS / 360 BOXES
    1.846 HOUS = 1:51

  13. Great math but an employer will have Charlie fired or work by himself monitor productivity and fire him if needed
    Do the cost analysis : Paying Charlie for two hrs only gets you 18 mins of productivity! hit the road Jack ! Throwing more people at some jobs makes it take longer and/or less productivity
    fire the supervisor/ manager if it happens again
    That’s the solution

  14. Great math but an employer will have Charlie fired or work by himself monitor productivity and fire him if needed
    Do the cost analysis : Paying Charlie for two hrs only gets you 18 mins of productivity! hit the road Jack ! Throwing more people at some jobs makes it take longer and/or less productivity
    fire the supervisor/ manager if it happens again !
    Another way to look at it is : if Charlie is their instead of bob he gives a half hearted effort so Alice doesn’t have to do all the work herself . But put Charlie with bob and he will let bob do all the work! Fire Charlie , That’s the solution

  15. i ended up converting job to distance and the combinations of people into vehicles going at a certain speed. To make it ez, i made the distance 100km, so alice and bob are a vehicle going at 50km/h, alice and charlie one going at 33.33km/h and bob and charlie one going at 25. Then you make it so each individual person represents a certain ammount of speed, so sa + sb = 50km/h and so on. Then its just a simple equation system and dividing 100 by whatever speed you get for sa + sb + sc and then converting whatever you get to hours and minutes. This will give you the 1h 51 minutes travel time.

  16. This is the first and easiest problem in 7th grade mathematics book in india and even college students can't solve it.
    Shame

  17. a + b = 2 etc happens when you have a dysfunctional educational system. Ok, I haven't done any real maths in like 8 years, so nobody can fault me, but if people who ARE a part of that system have this exact same problem…time to change the system.

    edit: my most common math question: Why is this important, where will I use it, seems useless.

  18. I've solved this very simply. Let:
    w = work done
    a = work rate Alice
    b = work rate Bob
    c = work rate Charlie
    where work rate = work done per hour.
    Problem's statement is: w = 2(a + b) = 3(a + c) = 4(b + c)
    We're searching x such that w = x(a + b + c) => x = w / (a + b + c). We have:
    2a + 2b = 3a + 3c => a – 2b + 3c = 0
    2a + 2b = 4b + 4c => a – b – 2c = 0
    3a + 3c = 4b + 4c => 3a – 4b – c = 0
    Solving these equations we get: a = 7c, b = 5c
    Thus a + b = 12c => a + b + c = 13c
    Since w = 2(a + b) = 24c we get x = 24/13 hours.

  19. Presh' answer makes sense in his model world where every person works at a constant rate AND DOES NOT INTERFERE IN THE OTHERS' WORK. Both assumptions do not necessarily apply outside the mathematical model world. In the real world, a coworker may not only work slowly, but may interfere in the other's work by engaging the other in conversation, asking questions, making the other wait for the first to finish a stage before handing the work to the faster worker, checking his/her iphone for messages, texting, bathroom breaks, arguing about how the work should be done, criticizing the other's work product, sabotaging the other's work in order to look better to the boss, etc.

  20. am i tbe only one who finish a work faster alone than with buddies?
    ..u know…a drink or two in between, roasting each other,

  21. As a college student who has knowingly turned in incorrect answers, it's not that we don't think about whether the answer makes sense, it's that we don't know how to get the real answer and it's better to put something on the paper in the hopes of partial credit, even if it's wrong.

  22. I approach these type of questions using velocity-displacement-time formula, i.e v=s/t. Where, v can be used as the rate at which a person does the work, s be the work and t be the time taken.
    Here
    Va+Vb=W/2
    Va+Vc=W/3
    Vb+Vc=W/4,
    And
    Va+Vb+Vc=W/t
    t can easily be calculated by adding the first three eqns and dividing the eqn by 2 on both sides.

  23. A + B in one hour complete 1/2 job
    B + C in one hour complete 1/3 job
    C + A in one hour complete 1/4 job
    So in one hour 2(A + B + C) complete 1/2 + 1/3 + 1/4 = 13/12 job
    => A + B + C in one hour complete 13/24 job
    So to complete the job they need 24/13 hour.

  24. The real question is what were Alice and Charlie doing in that extra hour? Specifically, did they use a condom? And what beer joint did Bob and Charlie go to before they started work?

  25. I got one hour and 27.5 minutes.
    I reasoned that:
    A + B complete 1/2 a job in 1 hour
    A + C complete a 1/3 in 1 hour
    B + C complete 1/4 in 1 hour

    If duplicate people were possible:
    (A + B) + (A + C) + (B + C) = 2(A + B + C) would complete 13/12 job in one hour
    meaning A + B + C would complete 13/24 job in one hour.
    To complete the remaining 11/24, 11/24 * 1 hour would be needed or 27.5 minutes.

    EDIT: Derp!
    If 13/24 of the job is completed in 1 hour then it takes 11/13 hours to complete the remaining 11/24 of the job which is 51 minutes. I should have never doubted you!

  26. Too complicated for non-matematicians. Use W=V*T i.e. work = speed * time instead of (percentage per time) * (time). For many people speed is more natural term.
    A + B = 1/2 has meaning that speed of Alice and Bob equals to half job per hour and etc.

  27. If my friend does his work in 2 hours and I do the same work in 1 hour that doesn't mean that we both combined can do it in a specific time limit. We can finish it in 0.5 hours also if we are determined and work together with coordination and we can finish it in 4 hours also if we don't do it with proper teamwork and determination. So, the question is pointless. You can't predict anything. Questions asking that if a car travels at 50 kmph stops after 10 hours how much distance would it have covered are pointless too coz it's not necessary that motion is uniform and even if it is, it's unpredictable until many conditions are specified. We just think of everything as ideal and therefore the questions which we solve can't be applied on real situations. Like , neglecting Air resistance would not work in freely falling body questions.

  28. Law of diminishing returns in economics, one of the few practical things I learned at school in Economics classes that I still remember.

  29. its quite easy, easier than what u have done
    work done by a+b in 1 hour = 1/2
    work done by a+c in 1 hour = 1/3
    work done by b+c in 1 hour = 1/4
    work done by a+b+a+c+b+c in 1 hour = 1/2+1/3+1/4 => 2(a+b+c)= 13/12
    work done by a+b+c in 1 hour = 13/24
    total time for them to do the work= 24/13= 1 hour 51 mins

  30. I think "basic" is the wrong choice of words. "Common" fits much better.

    In management, this sort of math is routine. We'd assign manpower value to each employee to make sure we have enough to do the task. On smaller scales, most small businesses estimate this type of math intuitively.

    If we think of Alice as the veteran employee, Bob the average employee, and Charlie the new hire; we assign an estimated manpower value to each with Bob as the value of 1.
    Alice would receive a higher value due to her proficiency.
    Charlie would be a fractional in the above example to keep the math simple. Although, typically I'd assign a negative value due to trainees pulling manpower to be taught.

  31. If Alice, Bob and Charlie are getting paid for this work, I'd say fire Charlie, he only helps to get the job done 9 minutes faster than if Alice and Bob worked alone! What a slacker.

  32. They need to make it clear that this task can be completed in parallel, for example painting a room or shoveling a sidewalk.

  33. Why is this Question considered hard
    We used to solve same type of Q in class 7 and we could figure it out even at that time

  34. Here is the more easier method ☺️
    Let's suppose 'T' time taken by all to complete work when working together.
    Then part of work done by (A+B) = T/2 ……..eqn(1)
    Similarly part of work done by (A+C) = T/3……. eqn(2)
    AND
    Part of work done by pair (B+C) = T/4…….(eqn)
    Since once we add above parts of work we get twice of work done because
    (A+B)+(A+C)+(C+B)=2(A+B+C)
    So on adding parts of work done by individual pairs in T time we got following equation
    (T/2) + (T/3) + (T/4) = 2
    T × (13/12) = 2
    T = 2×12/13
    T = 24/13
    T = 1Hour 51 min.
    Where A: Alice ; B: Bob ; C: Charlie

  35. We can think this problem in another corner. Like Take the work as distance , and the
    2(speed of A + speed of B) = S
    3(speed of A + speed of C) = S
    4(speed of B + Speed of C) = S.
    By solving this we can get the answer. This is my method. It actually looks like your method.
    Actually I Really proud of my self.
    🇱🇰 I'm a Sri lankan.

  36. I think an easier way to work this out is set out the equations as:
    a+b=1/2
    a+c=1/3
    b+c=1/4
    And then the answer is the reciprocal of a+b+c

  37. I mean, i just did trial and error to sum it all up, figured bob would be 1.30 and alice .30, so 4 – 1.30 =2.30, so .30+1.30+2.30 = 4.30

    Yes .30 as in half of an hour

  38. I think that's why physics is so helpful when facing math problems.
    This was basically a no-brainer if you understand how speed and time for example are connected.

  39. Guess it was sort of incorrect You can immidiately see 3 names begining with letters a b and c. Having 3 formulas means we can solve for all three of them

  40. Let’s make the amount of a job Alice can do in an hour be A, and B for Bob and C for Charlie. Then we take A+B, which is the sum of his much Alice and Bob can do in an hour. If they do 1 job in 2 hrs, they do 1/2 job in 1 hr. So A+B=1/2. So for A+C, since they do it in 3 hrs, they do 1/3 job in an hr. A+C=1/3. Then obviously B+C=1/4, applying the same rule. So A+B-1/6=A+C, B-1/6=C. Now we have B+C=1/4, B+B-1/6=1/4, 2B=5/12, B=5/24. So C=5/24-1/6=1/24. So A+1/24=1/3, A=7/24. So we have A+B+C=7/24+5/24+1/24=13/24 of a job per hr. So to find hrs for one job, we do (13/24)/1=1/x, cross-multiply to get (13/24)x=1, x *24/13 hrs*. Boom. How do college kids get this wrong, I got this in less than 3 minutes and I’m a sophomore in high school. Also Bruh Charlie sucks ass

  41. 🙁 I wasl ike the answer doesn't make sense… then I was like.. I need ot know J.. and find out T and it has to be less than 2 hours. I mean I would of guessed one hour… but I set it up the wrong way… It knowing what you can do toa question that reveals it. I didn't know I could assign 1 as 100% for J. Which would of led me down the correct path… the notion of thoguth stating.. hey. 2A+2B=1 one representing the missing theing for meaning 100% of the job takes. Thus equal to one… A B are going ot be a varrible or fraction/ percentage of it. Hwoever you wish to word it. Logic states to me that A works the fastest, B mid, nad C slowest.. so I was doing other nonsense maths, lol.. but.. if you don't know something then work with what is given. So regardless fo who works faster, simply put that both take 2 hours to complete the job, and the variable is there somewhere… there first method works if YOU know how long the Job takes and thus you should subjtract from the job… So I had something like A + B – J = 2 hours. lol.. That wasn't getting me anywhere.. lol…

  42. Hooray! I got it (because I caught the trick: it's about rates, i.e., quotients). Therefore, it's about a harmonic mean, not a geometric or arithmetic mean.

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