Can you solve this math problem without algebra?
This is something of a rant.
I came across this problem in a book I am using for tutoring. The solution given is really ugly. I can see teaching the use of algebra for these types of problems, but the intuitive solution should be included as well.
A person runs two laps on a track. The first lap is done at 5 mph and the second at 4 mph. If the total time for the two laps is 54 minutes, how long did it take to run each of the two laps?
Solution given by book.
Let t = time for five minute lap. Note that 54 minutes is .9 hours.
The person covers 5t for the first lap and 4(t – .9) for the second. Since the two laps have the same distance, we get
5t = 4(.9 – t)
9t = 3.6
t = .4 hour = .4 x 60 = 24 minutes.
I don’t have the book with me. I don’t recall if they evaluated (.9 – t) for the second lap (I sure hope not). The second lap is obviously 54 – 24 = 30 minutes.
YUK!!
This can be done without algebra and without conversion between minutes and hours. The numbers work out simply enough that it can be done in your head. Do you see how? While you are at it, figure out the length of the track.
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11 Answers
With your clues I can do it as fractions. Is this without algebra?
1) TIME
Total time = 54 minutes = 9/10 hours
Ratio of velocities is 4/5
Thus time will be split 4/9 + 5/9
4 * 1/9 hours = 4 * 6 minutes = 24 minutes
5 * 1/9 hours = 5 * 6 minutes = 30 minutes
2) DISTANCE
4 miles/hour * 5/10 hours = 20/10 miles = 2 miles
5 miles/hour * 4/10 hours = 20/10 miles = 2 miles
54 is divisible by 9.
Then you get neat chunks of 6 minutes. That by 5 and 4, and you get 30 and 24.
Yeah, well, I’ve done a lot of math problems and seem to have unusually high mathematical aptitude, I like sensible rate-time-distance problems, and I have extremely high suspicion and skepticism for story problems, so I intuitively guessed that the author contrived a nice round number and felt the relationship between 5 and 4 probably corresponded to a 30 minute lap and a 24 minute lap. There are some easy tests to see if that’s the case, such as:
30 minutes is the rounder number and would be for the slower speed since it took longer, so if at 4mph it takes half an hour, it’d take an hour at 2mph, which literally says the track is two miles long (really, story-problem author? that’s a freakin’ long track!) if so, then at 5mph, does that in fact take 24 minutes? Well, at 1 mph, 2 miles would take 2 hours, which is 120 minutes, divided by 5 for going 5 mph is like dividing by 5 and multiplying by 2, which is 24, so yes, my math/fractions/story-problem intuition turns out to be the answer.
@Call_Me_Jay , You did it right, but there is no need to work with hours. Just take 4/9 of 54 minutes and 5/9 of 54 minutes.
I am finding quite a number of problems in the book could be solved without algebra. Here is another one that is typical of a certain type of problem.
A person has 8 more dimes than quarters. If the total amount of the quarters and dimes is $5.35, how many dimes are there?
It’s a silver (supposedly) 10 cent piece.
If I didn’t use algebra for the dime & quarter problem, I would brute force it, starting with an estimate, calculating the answer, and adjusting up or down for the next guess. Which is against my nature.
You are actually close to the solution. Take a guess. It does not matter how close you are. Then figure out from the guess how you can get the exact solution.
Here is a way of solving the second problem without algebra.
Give an estimate, for example 8 dimes and no quarters.
8 dimes and no quarters is 80 cents. How far off are we? $5.35 – ,80 = $4.55.
We can improve the estimate by adding a quarter and a dime, preserving the difference in the number of coins, and getting 35 cents closer to $4.55. How many times do we need to do that? $4.55/.35 = 13. We need to add 13 more quarters and 13 more dimes, giving 13 quarters and 8+ 13 = 21 dimes.
Let’s look at the algebraic approach. Let d = number of dimes. 10d + 25(d – 8) = 535. 35d = 535 + 200. d = 21
If we use the estimation approach with 0 dimes and -8 quarters, we do exactly the same arithmetic.
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