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midgic
Joined: 14 Feb 2004
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 Posted: Sat Oct 10, 2009 6:38 am Post subject: |
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There are two islets A and B, one people on each islet.
Two people set off at the same time and are going to swim two lengths of the distance between A nd B.
ex. A->B->A or B->A->B
The first time they meet at 700m from A, the second time they meet at 400m from B.
What is the distance between A and B? |
I thought of an alternate solution for this problem. Suppose swimmer B is much faster than swimmer A. Swimmer B meets swimmer A 700 m from islet A. Then swimmer B catches swimmer A before swimmer A even reaches islet B. The math is as follows (I think):
Let x be the distance swimmer B completed while swimmer A completed 700 m.
(x) over (700) = (x + 1000) over (x - 400)
Then, x^2 - 1100x - 700,000 = 0
Solving the quadratic gives: x = 1551.2
So, the distance between A and B is 700 + 1551.2 = 2251.2 meters.
Can anyone verify this solution? |
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Panda
Joined: 25 Oct 2008
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| Posted: Sat Oct 10, 2009 6:43 am Post subject: |
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Ontheway: you must be some mutant genius.
Midgic: wow, I really love that 0.2 meters. Thanks for the great job. |
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ontheway
Joined: 24 Aug 2005
Location: Somewhere under the rainbow...
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| Posted: Sat Oct 10, 2009 8:19 am Post subject: |
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| Panda wrote: |
Ontheway: you must be some mutant genius.
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I joined the thread from where it had stopped previously as I generally avoid going back to the beginning of things I have scanned before. You, I just noticed, posted your new problems as changes to your OP. I didn't look there for your problems. So, it was "guess the problem that results in this answer." |
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ontheway
Joined: 24 Aug 2005
Location: Somewhere under the rainbow...
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| Posted: Sat Oct 10, 2009 9:25 am Post subject: |
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| midgic wrote: |
| Quote: |
There are two islets A and B, one people on each islet.
Two people set off at the same time and are going to swim two lengths of the distance between A nd B.
ex. A->B->A or B->A->B
The first time they meet at 700m from A, the second time they meet at 400m from B.
What is the distance between A and B? |
I thought of an alternate solution for this problem. Suppose swimmer B is much faster than swimmer A. Swimmer B meets swimmer A 700 m from islet A. Then swimmer B catches swimmer A before swimmer A even reaches islet B. The math is as follows (I think):
Let x be the distance swimmer B completed while swimmer A completed 700 m.
(x) over (700) = (x + 1000) over (x - 400)
Then, x^2 - 1100x - 700,000 = 0
Solving the quadratic gives: x = 1551.2
So, the distance between A and B is 700 + 1551.2 = 2251.2 meters.
Can anyone verify this solution? |
Nice alternative solution.
I checked your solution after I first solved it this way:
let L be the total distance from A to B.
700/(L - 700) = (L - 1100)/(L + 300)
Solving the quadratic equation: L = 2251.25 meters
double check. |
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Shiktang
Joined: 10 May 2009
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| Posted: Sat Oct 10, 2009 10:01 am Post subject: |
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What about the other problems. Anyone do the 6 digit thingy. Haven't got a pencil right now, but it is a typical problem.
Koreans learn the short cut methods, and they are able to apply these methods to more difficult problems when in highschool. Come exam time they have to solve their problems in the quickest way possible, so the mathematics programs in the elementry schools are geared to teaching quick more efficent ways of solving problems. Most lamebrains brought up in the western way of doing things( that includes most of us) will be forever stuck doing it the longwinded and timecosuming algebraic way.
Their approach to teaching mathematics leaves the methodology employed in your basic western school in the dust.
All elementry school students are held to the same standard, so those problems listed by the OP aren't considered to be any big deal by Korean teachers,and Korean parents. |
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ontheway
Joined: 24 Aug 2005
Location: Somewhere under the rainbow...
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| Posted: Sat Oct 10, 2009 11:20 am Post subject: |
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| panda wrote: |
I am adding more interesting math problems here.
1) ABCDEF X C = CDEFAB
what is this 6-digits number? |
This problem must include the limitation that the letters A B C D E F represent positive integers or zero and that each has a different value.
We can then solve the problem using math and logic:
(Ax100,000+Bx10,000+Cx1,000+Dx100+Ex10+F)xC=Cx100,000+Dx10,000+Ex1,000+Fx100+Ax10+B
We know that we must use 6 of the available digits 0,1,2,3,4,5,6,7,8,9
We can see that C cannot be zero.
Since both ABCDEF and CDEFAB have values in the hundred thousands, we know that A=1 since CxA must =C
Since CxB must be less than 10, and 1 has been used, then C and B can only be 2,3,or 4 and if one of them is 4 the other must be 2.
A few quick trials using B=3 or 2 shows this is not possible:
(If B=3 and C=2 then the last digit on the right should be "3" but Fx2 has to be even.)
(If B=2 and C=3 you find that D=6 and then the computations fail.)
So B must be 4 and C=2
Knowing A=1 B=4 C=2 means ...
Now we know that: 142DEF x 2 = 28EF14
Which means that D must be 8.
We can then see that: 1428EF x 2 = 285F14
So, E must = 5.
To get 14 Fx2=14, so F is 7 and E is 5 so that the tens place on the left side of the equation will be 1. This confirms:
Substituting in the remaining digits:
142857 x 2 = 285714 is the only possible solution, where A,B,C,D,E,F are positive integers each having a different value. |
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Konglishman
Joined: 14 Sep 2007
Location: Nanjing
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| Posted: Sat Oct 10, 2009 1:01 pm Post subject: |
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| Shiktang wrote: |
What about the other problems. Anyone do the 6 digit thingy. Haven't got a pencil right now, but it is a typical problem.
Koreans learn the short cut methods, and they are able to apply these methods to more difficult problems when in highschool. Come exam time they have to solve their problems in the quickest way possible, so the mathematics programs in the elementry schools are geared to teaching quick more efficent ways of solving problems. Most lamebrains brought up in the western way of doing things( that includes most of us) will be forever stuck doing it the longwinded and timecosuming algebraic way.
Their approach to teaching mathematics leaves the methodology employed in your basic western school in the dust.
All elementry school students are held to the same standard, so those problems listed by the OP aren't considered to be any big deal by Korean teachers,and Korean parents. |
While it is great to know shortcuts, the major shortcoming in public Korean mathematics education is that people are not encouraged to show their work. People only seem to care about finding the answer as quickly as possible as opposed to taking time to show the logic of the thought process. To me, that goes against the spirit of mathematics.
Not only that, but some people may come to rely too heavily on the shortcuts and thus when they encounter a solvable problem outside of their repertoire of shortcuts, they then find themselves stumped and do not attempt to make any progress by trying to analyze it in a systematic way. That said, I have seen some exceptional students avoid that trap, but I must say that it does have a tendency to be a problem among some Korean students.
Last edited by Konglishman on Sun Oct 11, 2009 10:15 am; edited 1 time in total |
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Panda
Joined: 25 Oct 2008
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| Posted: Sun Oct 11, 2009 2:04 am Post subject: |
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The problem is for 4th grade kids in China.
Those I added later are for SMARTER kids to play with just for fun. |
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ontheway
Joined: 24 Aug 2005
Location: Somewhere under the rainbow...
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| Posted: Sun Oct 11, 2009 9:13 am Post subject: |
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| Panda wrote: |
3) There are 5 prisoners A,B,C,D and E , whose fate will be decided by the number of beans they get from a bag. There are 100 beans in this bag, 5 prisoners get their beans in sequence (while taking the beans, they get to know how many left in the bag). Those with the most and least beans will be killed in the end (it could be = or >2 people). Who would most likely survive and how?
<-- My answer is different from theirs, I would love to know yours to see if I was wrong. |
What are the parameters?
Does each prisoner get only one draw or is there a succession of draws?
If only one draw, does the last prisoner have to take all that are left?
If a succession of draws what are the rules? |
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ulsanchris
Joined: 19 Jun 2003
Location: take a wild guess
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| Posted: Sun Oct 11, 2009 5:41 pm Post subject: |
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| my god i am useless at math. I don't think i could solve any of those problems if my life depended on it. |
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Panda
Joined: 25 Oct 2008
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| Posted: Sun Oct 11, 2009 5:49 pm Post subject: |
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| ontheway wrote: |
| Panda wrote: |
3) There are 5 prisoners A,B,C,D and E , whose fate will be decided by the number of beans they get from a bag. There are 100 beans in this bag, 5 prisoners get their beans in sequence (while taking the beans, they get to know how many left in the bag). Those with the most and least beans will be killed in the end (it could be = or >2 people). Who would most likely survive and how?
<-- My answer is different from theirs, I would love to know yours to see if I was wrong. |
What are the parameters?
Does each prisoner get only one draw or is there a succession of draws? Yes and Yes
If only one draw, does the last prisoner have to take all that are left? No
If a succession of draws what are the rules? It goes from A->B->C->D->E, only one chance, everyone can tell how many beans left but wont be told how many the previous prisoners take. |
Sorry for the ambiguity, I edited my post already. |
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midgic
Joined: 14 Feb 2004
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| Posted: Sun Oct 11, 2009 7:22 pm Post subject: |
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| Quote: |
| There are 5 prisoners A,B,C,D and E , whose fate will be decided by the number of beans they get from a bag. There are 100 beans in this bag, 5 prisoners get their beans in sequence (while taking the beans, they get to know how many left in the bag). Those with the most and least beans will be killed in the end (it could be = or >2 people). Who would most likely survive and how? |
Case 1: If A takes 19 beans or fewer, then B should take one more bean than A.
C knows the average of the first two, knows that the average is less than 20, and so C should take the minimum number of beans that is still not the least number of beans. So C would take the same number as B.
Similarly, D would take the same number as C and B.
Similarly, E would take the same number as D and C and B.
So A would have the least, while B, C, D, and E would have the same number of beans, one more than A.
So they would all die.
Example:
A takes 15 beans. B should take 16 beans. C knows that 31 beans are gone, so an average of 15.5 beans. C knows that someone must have 15 beans or less so C should take 16 beans. Based on similar reasoning, D and E would take 16 beans. So A has 15 beans while B, C, D, and E have 16 beans.
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Case 2: If A takes 21 beans or more, then B should take one less than A. Actually B can survive by taking any number from 20 beans up to one bean less than A.
C then knows the average of A and B is more than 20, and knows that someone has 21 beans or more. So C takes 20.
Similarly D takes 20.
Now there are fewer than 20 beans left so E will have the least.
In this case, A would have the most and E would have the least.
B, C, and D would survive.
* Obviously, if A chooses a large number of beans, like 50 or 80 or 90, then the beans might be gone before C, D, and E choose, which would mean that C, D, and E would all die.
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Case 3: A takes 20 beans. Then B should take 20. Similarly C, D, and E would take 20 beans. They would all die.
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Conclusion: Assuming everyone chooses logically, A has no chance to survive. B, C, and D can survive as long as A chooses 21 beans or more. E has no chance. So it really depends on A. If A wants to save B, C, and D, then A could choose 21 beans or more. However, if A resents B, C, and D for being able to survive, then maybe A takes 20 beans just to spite them.
Anyway, to answer the question, the person with the best chance to survive is B (I think). |
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Panda
Joined: 25 Oct 2008
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| Posted: Sun Oct 11, 2009 9:13 pm Post subject: |
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Thanks for your reply midgic.
First of all, everyone would have to choose logically(no joke with their lives, thus they have to think very carefully, either I wouldnt assume anyone help another or others or there will be any comminication, thats just drama. In other words, people here will try their best to kill others and save themselves), in that case, its unlikely that A will choose >50 beans or a very small number.
If I were A, I would take 20 beans (20 is the result of 100 devided by 5, which means if all others take >21 or all others take <19. A would die. which is very unlikely. or everyone takes 20, all 5 die, but thats fair then)
B then has to take 21 or 19, cuz he knows C will take the number of (A+B)/2 to make sure he is in the middle, thus B can't let C to stay between him and A.
If B take 21, C has to take 21 or 20.
If B take 19, C has to take 19 or 20.
D will take 20 in all cases.
In summary, there are 4 possibilities for A,B,C,D
1) 20-21-21-20
2) 20-21-20-20
3) 20-19-19-20
4) 20-19-20-20
E has no choice!
E and B, C will die in case 1)
E and B will die in case 2)
In case 3) and 4) all people will die, or E wants to kill himself to save others by choosing a very small number or all left,bt I guess he probably prefers die together.
So my answer is A and D will most likely survive, B and E have no chance to live.
Last edited by Panda on Sun Oct 11, 2009 9:55 pm; edited 2 times in total |
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Panda
Joined: 25 Oct 2008
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| Posted: Sun Oct 11, 2009 9:49 pm Post subject: |
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Now I don't think my answer is right, either, cuz if B will die no matter what he choose, he will just choose 20 , so that 5 will die together, A has to let B live.
So there should be other answers. |
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