| View previous topic :: View next topic |
| Author |
Message |
Panda
Joined: 25 Oct 2008
|
 Posted: Thu Oct 08, 2009 10:49 pm Post subject: A math problem for 4th grade, can you solve it? 3 more added |
 |
|
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 am adding more interesting math problems here. Those are not all for 4th grade, some applies to the game theory.
1) ABCDEF X C = CDEFAB
what is this 6-digits number?
2) There is a bridge, it takes A 1min, B 3min, C 6min, D 8min, E 12min to cross. In an evening, all the 5 people want to cross the bridge, but only 2 people are allowed each time, and they have a lamp that will go out in 30 min. How can all these 5 manage to cross before the lamp go out?
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 (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 prisoner(s) take, the last prisoner doesnt have to take all left). 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.
Last edited by Panda on Sun Oct 11, 2009 6:01 pm; edited 14 times in total |
|
| Back to top |
|
 |
andrewchon
Joined: 16 Nov 2008
Location: Back in Oz. Living in ISIS Aust.
|
| Posted: Thu Oct 08, 2009 11:02 pm Post subject: |
|
|
| I think it's 700 + 400 = 1100m |
|
| Back to top |
|
|
Panda
Joined: 25 Oct 2008
|
| Posted: Thu Oct 08, 2009 11:07 pm Post subject: |
|
|
To andrewchon:
What is the reasoning? |
|
| Back to top |
|
|
TheUrbanMyth
Joined: 28 Jan 2003
Location: Retired
|
| Posted: Thu Oct 08, 2009 11:26 pm Post subject: Re: A math problem for 4th grade, can you solve it? |
|
|
| Panda wrote: |
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?
|
Impossible to tell. We don't know the speed of the swimmers. |
|
| Back to top |
|
|
Goku
Joined: 10 Dec 2008
|
| Posted: Thu Oct 08, 2009 11:27 pm Post subject: Re: A math problem for 4th grade, can you solve it? |
|
|
| TheUrbanMyth wrote: |
| Panda wrote: |
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?
|
Impossible to tell. We don't know the speed of the swimmers. |
I was thinking the same thing.
Maybe one of them arthritis or scurvy |
|
| Back to top |
|
|
Panda
Joined: 25 Oct 2008
|
| Posted: Thu Oct 08, 2009 11:29 pm Post subject: |
|
|
To Guko and TheUrbanMyth
 it took me 20 min to solve. |
|
| Back to top |
|
|
andrewchon
Joined: 16 Nov 2008
Location: Back in Oz. Living in ISIS Aust.
|
| Posted: Thu Oct 08, 2009 11:38 pm Post subject: |
|
|
Panda:
Have mercy sensei, for I have sinned.
I've not thought of Newtonian motion for over twenty years. |
|
| Back to top |
|
|
Underwaterbob
Joined: 08 Jan 2005
Location: In Cognito
|
| Posted: Fri Oct 09, 2009 12:32 am Post subject: |
|
|
| The answer is 1700m. And if fourth grade kids use the same kind of math I used to solve the problem then wow... |
|
| Back to top |
|
|
Underwaterbob
Joined: 08 Jan 2005
Location: In Cognito
|
| Posted: Fri Oct 09, 2009 12:44 am Post subject: |
|
|
My method:
The two swimmers leave each Island at velocities V1 and V2.
V1 = 700/t1 = (x+400)/t2
V2 = (x-700)/t1 = (2x-400)/t2
Where t1 and t2 are the times they meet and x is the distance between the two islands. By cross multiplying we can then simplify:
700/t1 = (x+400)/t2
700/(x+400) = t1/t2
and
(x-700)/t1 = (2x-400)/t2
(x-700)/(2x-400) = t1/t2
We now have:
700/(x+400) = (x-700)/(2x-400)
Which simplifies to:
x^2-1700x = 0
Plug A = 1, B= 1700 and C=0 into good ol' (-B+,-sqr(B^2-4AC))/2A (or just look at it closely) and the two solutions are 0 (impossible) and 1700 (the answer).
They really do this in fourth grade? I know some of the middle school math gives me trouble, but I can usually work it out and it's not much more complex than this question.
Last edited by Underwaterbob on Fri Oct 09, 2009 12:52 am; edited 1 time in total |
|
| Back to top |
|
|
Panda
Joined: 25 Oct 2008
|
| Posted: Fri Oct 09, 2009 12:49 am Post subject: |
|
|
Underwaterbob, You got it~!
actually you only need the simplified equation:
700/(X-700)=(X+400)/(2X-400)
X is the distance between A and B.
Both sides of the equation is the speed ratio of the two people.
Left side is the ratio they meet first time.
Right side is the ratio they meet second time.
Its actually quite easy if you think of it.
Last edited by Panda on Fri Oct 09, 2009 2:18 am; edited 2 times in total |
|
| Back to top |
|
|
Hightop
Joined: 11 Jun 2003
|
| Posted: Fri Oct 09, 2009 1:20 am Post subject: |
|
|
| So in 4th grade they are teaching math like this yet in English class they are learning to count to 10 and say 'I like apple'. Something must be wrong. |
|
| Back to top |
|
|
Konglishman
Joined: 14 Sep 2007
Location: Nanjing
|
| Posted: Fri Oct 09, 2009 1:24 am Post subject: Re: A math problem for 4th grade, can you solve it? |
|
|
| Panda wrote: |
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?
|
Clearly, we cannot assume that the two swimmers are swimming at the same speed. So, we must assign them with different variables for their velocities.
Now, lets label the islet A to be at the origin and islet B will be a distance b away.
In the first situation when they meet:
700 = V1*t
700 = b - V2*t
Consequently,
t = 700/V1
700 = b - 700*V2/V1
Hence,
b = 700(1 + V2/V1)
Now, for the sake of argument, I am going to assume that V2 > V1. So, the second swimmer will reach islet A before the first swimmer reaches islet B.
When the second swimmer reaches islet A:
0 = b - V2*ta
ta = b/V2 = 700(1/V1 + 1/V2)
At the same time, the first swimmer will have swam:
V1*ta = 700(1 + V1/V2)
Hence, the first swimmer still needs to swim:
b - V1*ta = 700(V2/V1 - V1/V2) = 700(V2^2 - V1^2)/V1V2
Obviously, the first swimmer will reach islet B after this much more time has passed:
tb = (b - V1*ta)/V1 = 700(V2^2 - V1^2)/(V1^2 V2)
At this point, the second swimmer has turned around and swam:
b2 = V2*tb = 700(V2^2 - V1^2)/V1^2 = 700([V2/V1]^2 - 1)
Note that the last algebraic expression will now function as the second swimmer's new y-intercept.
So, when they meet again:
b - 400 = V2*t2 + b2
b - 400 = b - V1*t2
-400 = -V1*t2
t2 = 400/V1
b - 400 = 400V2/V1 + b2
700(1 + V2/V1) - 400 = 400V2/V1 + 700([V2/V1]^2 - 1)
700 + 700V2/V1 - 400 = 400V2/V1 + 700[V2/V1]^2 - 700
300 + 700V2/V1 = 400V2/V1 + 700[V2/V1]^2 - 700
700[V2/V1]^2 - 300V2/V1 - 1000 = 0
7[V2/V1]^2 - 3V2/V1 - 10 = 0
V2/V1 = (3 +/- sqrt[9 - 4(7)(-10)])/14
V2/V1 = (3 +/- sqrt[289])/14
V2/V1 = (3 +/- 17)/14
V2/V1 = 20/14 = 10/7 since the each velocity was defined as positive.
Finally,
b = 700(1 + V2/V1) = 700(1 + 10/7) = 700*17/7 = 1700 meters
There you have it. |
|
| Back to top |
|
|
Joe666
Joined: 19 Nov 2008
Location: Jesus it's hot down here!
|
| Posted: Fri Oct 09, 2009 1:56 am Post subject: |
|
|
| Konglishman - You need a chick something fierce!!! Your gonna POP any day now!!! |
|
| Back to top |
|
|
Panda
Joined: 25 Oct 2008
|
| Posted: Fri Oct 09, 2009 2:13 am Post subject: |
|
|
Joe666
Its 10 times fun than scoring any chicks.......
Konglishman rocks.....................(I was like......wow) |
|
| Back to top |
|
|
benji
Joined: 21 Jul 2009
|
| Posted: Fri Oct 09, 2009 3:49 am Post subject: |
|
|
| Korean kids are not doing algebra word problems in the fourth grade. At least not in the public schools. |
|
| Back to top |
|
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
