Russian Elementary Math Problem.

Discussion in 'Community Discussion' started by Frisco, Sep 29, 2011.

1. Frisco macrumors 68020

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#1
From another website forum by Zea mays: Just curious to see who can get it. I have no clue, but will post the answer from the website.

"I recently came across this transportation/math problem for Russian elementary school students (ages ≈ 12 years). I've changed the problem somewhat  miles for kilometers, etc.):

Two Russian cities, A and B, are 500 miles apart. At the crack of dawn, Person A leaves city A and drives to city B at a constant speed. At the same time (sunrise), Person B leaves city B and drives to city A at a constant speed. They take the same highway and pass each other at 12 noon without stopping.

Person A arrives at city B at 4:00 pm. Person B arrives at city A at 9:00 pm.

What time was sunrise?

I would bet that 75% of American six-graders would have trouble with this problem. We are falling behind in math and science, folks. More than half of the undergraduate physics majors at the University of Chicago are from south-east Asia, and I'm sure that is true of other science universities in the U.S."

2. Sep 29, 2011
Last edited: Sep 29, 2011

RITZFit macrumors 65816

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#2
3 a.m.?? assuming same velocity...I still get an off time for B XD

3. fireshot91 macrumors 601

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Northern VA
#3
I got 8 AM.

Since it takes one person 4 hours to travel half the distance. (From noon to Destination B), it must take him 4 hours to get from Position A to midpoint (At noon).

That means he started at 8 AM.

And since he started at sunrise, sunrise is therefore at 8 AM.

Assuming constant velocity.

I think most of the confusion is that people (Like me at first) assume that A is West-er than B.

If you treat it like this:

B--------------*NOON*-----------A

And sunrise starts at A then moves to B

then person B would reach A at 9 PM (Long after the sun has gone past A)
And person A would reach B at 4 PM (Around where the sun is, at that time).

4. munkery macrumors 68020

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Dec 18, 2006
#4
Not good at word problems but I'm going to say 6am.

Assuming that time zones are not relevant to the question.

Let the trip of person A be referred to as taking "a" hours.

Let the trip of person B be referred to as taking "b" hours.

The time required for both travellers to meet each other is the same given that both leave at sunrise. Let this time until the travellers meet be referred to as taking "c" hours.

a = c+4
b = c+9

c = a-4
c = b-9

a-4 = b-9

Using trial and error (there is probably some mathematical way to avoid trial and error):

a-4 = b-9
10-4 = 15-9
6 = 6
c = 6 hours

Known time is noon. So,

12 (noon) - 6 (hours) = 6am

5. Sep 30, 2011
Last edited: Sep 30, 2011

(marc) macrumors 6502a

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#5
Correct! If you don't want to guess, you define x as the distance from city A to the point where persons A and B meet, y as the time of sunrise, vA as the speed of person A and vB as the speed of person B. Then you get the equations vA = 500/(16-y), vB = 500/(21-y), vA * (12 - y) = x, vB * (12 - y) = 500 - x. Now you have 4 linear independent equations, so you'll be able to solve for all variables (x = 300, y = 6, vA = 50, vB = 100/3).

I think the problem might be a bit hard for 6th graders. I'd say it fits 9th or 10th grade better. Someone with a high-school education definitely should be able to solve it. But the students from Western countries prefer to take artsy pictures of themselves from "original" angles and post them on their Facebook account rather than think about a math problem for longer than a few seconds. After all, being bad at maths makes you hip and cool!

6. Sep 30, 2011
Last edited: Sep 30, 2011

munkery macrumors 68020

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#6
Thanks.

I knew using the speed was required but got lazy after drawing a diagram for the problem because it was obvious the answer was 6am.

7. munkery macrumors 68020

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Dec 18, 2006
#7
I would like to see how users solve the following equations:

16/4[(8-6)2]+1=

16/4((8-6)2)+1=

What method did you used to solve the equations?

8. fireshot91 macrumors 601

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#8
I think we had a thread on this. And the general answer was that the question was very ambiguous and the writer of the question should use more grouping symbols.

I solve it as 16/4[4] + 1 = 4 [4] + 1 = 17.

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Dec 18, 2006
#9
10. munkery macrumors 68020

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Dec 18, 2006
#10
I think that it is because in algebra brackets are used in place of parentheses if parentheses are contained within parentheses while brackets are also used to denote inclusion in relation to intervals.

For Wolfram alpha to support the inclusive function, this error in relation to algebraic equations is introduced.

But, I could be wrong.

Any other ideas? Is this explained somewhere?

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#11

12. munkery macrumors 68020

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#12
Hey, at least I tried to explain why this occurs.

Is there an already available explanation?

13. (marc) macrumors 6502a

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#13
Wolfram Alpha does it correctly.

14. munkery macrumors 68020

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Dec 18, 2006
#14
OK, what is the technical difference between the brackets and parentheses?

It would seem that for algebraic equations where parentheses can be arbitrarily supplanted with brackets for the sake of clarity that this may not be correct.

http://www.purplemath.com/modules/orderops.htm

The two equations presented above are functionally the same but provide different answers.

The error must come from a limitation in Wolfram Alpha. A rational explanation is that this error is introduced by there being multiple but independent uses for square brackets in mathematics.

15. ender land macrumors 6502a

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Oct 26, 2010
#15
This problem is imo considerably more difficult than most math questions on the ACT, and American high school students... do not do the best on that section of the ACT.

I'd be amazed if anywhere CLOSE to 25% of sixth graders could get this problem correct in the USA. There are a multitude of ways to do it incorrectly.

For example, another way to do this problem is to realize the total travel time is 13 hours (since after passing each other, one person took 4 hours to go some of the distance and the other person took 9 to do the remainder) and just subtract that from the end time of the person you are interested in. Which gives 8:00 am. This of course incorrectly assumes the same speed for both cars - which is not stated, only that each car itself does not change speed.

It is also not 100% clear to me that both vehicles leave at the same absolute time. This is because there is a phrase "at the same time" which appears to contradict the different descriptions of the actual time, one being "crack of dawn" and the other being "sunrise." Depending on how you interpret this, you might not have any idea how to solve the problem to begin with. A fair number of people might not even say the two cars left at the same time.

munkery, I'm... interested in seeing why you could not have ended up with, say, 11-4 = 16-9 and gotten 5:00am instead

Very few people would make a setup of equations and solve this correctly like (marc) did. I'd be surprised to see more than 25% of high school students in the USA doing this correctly. Especially if it ended up being a "less guessable" answer

----------

Also, let's not argue about the "do you do parenthesis in the strict mathematical sense or the likely intent of the writer sense."

:\

16. munkery macrumors 68020

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Dec 18, 2006
#16
Because that would be wrong.

Are both sides equal? No.

The only reason I did it this way is because the answer was obvious after drawing a diagram of the equation.

I was too lazy to bother figuring out the more appropriate method given that the question was easy enough not to require it and this is just a forum on the internet.

17. munkery macrumors 68020

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Dec 18, 2006
#17
The are multiple uses of square brackets in mathematics.

The inclusive property of square brackets, in the strict mathematical sense, doesn't apply when used in equations such as those above.

18. ender land macrumors 6502a

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#18
Wait, what?

11-4 = 7
16-9 = 7

19. Sep 30, 2011
Last edited: Sep 30, 2011

munkery macrumors 68020

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#19
Whoops, haha.

Didn't notice both sides were changed. Only noticed the 11.

Honestly, I didn't use the method shown in my post to solve the equation.

When starting to solve it, I first drew a diagram and it was obvious that the answer was 6am.

I also did calculate the traveller's speeds but opted not to do the extra work to use the proper method when the answer was obvious.

I can't remember how I did it now but I did determine that one of the traveller's trip time was 10 hours. Then, I worked from there using the method in my post.

20. Sep 30, 2011
Last edited: Sep 30, 2011

munkery macrumors 68020

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#20
Whoops, made another mistake or maybe not. Don't know? So confused?

Need to practice my math skills.

21. Sep 30, 2011
Last edited: Sep 30, 2011

Staff Member

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#21
I got 6 am, although it took me longer to work it out than it should have.

I don't understand this, or rather I think the original poster didn't understand it because the answer is the same regardless of whether it's 500 miles or 500 km.

22. ender land macrumors 6502a

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#22

I think the only way to solve this for real is the way (marc) did.

If it happens to be 5:00am it would just be coincidence Your way just meant you could get any arbitrary time depending on how you arranged it.

23. munkery macrumors 68020

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Dec 18, 2006
#23

In working through it with the more appropriate method, I made an error that lead to the wrong answer. Forgot to carry a sign if that is even the right terminology.

I haven't done this in so long that I wanted to get some practice.

I am also trying to figure out how I got 10 hours for one of the trip times using the speed the first time I solved the question.

I threw out the sheet I was working on and can't recall exactly what I did.

Using Marc's equations as a starting point, nothing I generate looks similar to what I did to find 10 hours.

But, I defined everything differently and didn't use 24 hour time in my calculations.

Everything was from zero and treated not in relation to 24 hour time values. If that makes any sense.

24. chown33 macrumors 604

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#24
Then use parsecs. And one of the cities is named "Kessel".

25. Oct 1, 2011
Last edited: Oct 1, 2011

munkery macrumors 68020

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#25
Found my working sheet and figured out my method used with trial and error.

Given constant speed of travellers and same start time leading to same amount of time travelling until meet, there should be equivalence in the ratios of the unknown travel time with known travel time across the two travellers.

So,

4/c = c/9

Furthermore,

a/b = vA/vB = 4/c = c/9

This is why 6am is the obvious answer when looking at a diagram of the problem.

2/3 is the ratio that is common across the problem.