'Two Working Together' and other Rate, Time and Distance problems.
A typical worded math problem is the ‘two working together’ problem, which is described here in detail. This is a classic test problem encountered in algebra and both quick and in-depth explanations are provided. Other examples are included.
For this “working together” problem, savvy students will recognize it and think, “Oh, the product over the sum solution”, and write down (or punch in) (3×4)/7, or about 1.7 hours (how many digits you include may depend on your teacher’s demands). This is correct but, what is behind this technique?
The first thing to realize is that this is a “Rate, Time and Distance” type of problem. I always remember that Rate X Time = Distance, or “RTD”. More generally, Rate X Time = Task Complete. Rate is always in ‘units of progress’ over time, such as miles per hour or other things done per time. Example: If you travel at 60 miles per hour for 2 hours, you have gone 60 x 2 = 120 miles. If you didn’t know how long it would take to travel 120 miles at a rate of 60 miles per hour, you would solve 60 X T = 120, or T = 120/60 = 2 hours. Any one of the three terms can be solved for if you know the others.
For the above word problem, first think of Jim’s rate to paint a room as 1 room in 3 hours and denote Rj = 1/3 or 1 room in 3 hours. If you care to divide 1 by 3, you see that he can paint one third of a room in one hour. Tom’s rate, Rt, is 1/4 and he can paint a quarter of a room in one hour.
OK, we figure that the sum of these two rates multiplied by the time working together is equal to 1 (room complete). So, using the RTD format, (Rj + Rt) * T = 1.
[Note: I sometimes use spaces in place of parentheses as it makes typed equations easier to read.]
Now, cross dividing to solve for T, T = 1/(Rj + Rt) = 1/(1/3 + 1/4), or simplifying to a common denominator, T = 1/(4/12 + 3/12) = 1/(7/12); multiplying numerator and denominator by 12/7 we get T = 12/7 / (12/7 * 7/12)), = 12/7! The answer is about 1.7 hours. 12/7 is simply the product over the sum. In the actual effort of painting the room, Jim paints 1.7 * 1/3 = .57– or about 57% of the room and Tom paints 1.7 * 1/4 = .43– or about 43% of the room.
What if Jim’s little brother helps them? It takes him 10 hours to paint a room! OK, T = 1/(1/3 + 1/4 + 1/10) = about 1.46 hours. That helps a bit, but little brother drips a lot of paint. The ‘product over the sum’ technique doesn’t work here (why?) so be careful – it only works for two! For more than two, you have to do it the long way.
Here’s a quick calculator sequence to solve the inverse of such fractional sums for the three workers:
3,1/x,+4,1/x,+10,1/x,=,1/x. The result is 1.46…
Try this other example: You drive half way around a race track at 60 mph and then the remaining half at 50 mph. What is your average speed? Is it 55 mph? No!
Solution: You can’t use the product over the sum technique here, as that only works to solve for time (the T part of RTD), given the other terms. This problem is to solve for the combined rate (R). So, what do we know? Let’s say D is 1 (track length) and the rates are as listed. But we can’t just average the rates and say it’s 55 mph because the times for each half of the track are different. Well, for the first half, T = D/R = .5/60 = .008333— hours and for the second half, .5/50 = .01 hours, for a total of .018333— hours. Finally, solving, R * .018333— = 1 (track), or R = 1/.018333— = 54.5454– mph. This was the first tricky algebra problem to given us in the 8th grade and everyone was mystified that the answer was not 55 mph and unsure how to get the right answer!
About The Author
|Math Expert: Basic Math, Algebra, Geometry, Pre-Ca|
|BSEE, University of Cincinnati, 1979. Well versed in math and the foundation of understanding in both math and science. I have tutored before, while in school and afterwards and have always found it rewarding. I\'m looking forward to sharing knowledge.|