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Jim Brier
11-16-2002, 05:49 AM
Hal gives Tom as many raffle tickets as Tom first had and Gary as many as Gary first had. Then Tom gives Hal and Gary as many tickets as they then had. Finally, Gary gives Tom and Hal as many tickets as they currently had. Hal, Tom, and Gary end up with 40 tickets each. How many tickets did Hal, Tom, and Gary each have to start with?

pudley4
11-16-2002, 11:43 AM
Work it backwards:

At the end, G=40, T=40, H=40

If Gary gave Tom and Hal enough tickets to double their holdings, then before this transaction, T=40/2=20, H=40/2=20, G=40+20+20=80

If Tom gave tickets to Hal and Gary, then before he did, H=10, G=40, T=70

If Hal gave tickets to Gary and Tom, then before he did, G=20, T=35, H=65

SunTzu68
11-16-2002, 04:30 PM
Hal=70, Tom=40, Gary=10. I'm not sure how the previous player got 65+35+20 to equal 120 (3 x 40), but his methodology was essentially correct.

Jim Brier
11-17-2002, 05:09 AM
The answer is Hal originally had 65 tickets, Tom originally had 35 tickets, and Gary originally had 20 tickets. Pudley's approach is better than mine. I worked the problem from the "front end" which is far more cumbersome.

Jim Brier
11-17-2002, 05:10 AM
(n/t)

11-17-2002, 10:05 AM
65+35+20 does indeed work out to 120, I apologize for my previous post. However, it is still not the right answer. Let me work it through, and maybe someone can show me where my mistake is.

Hal Tom Gary
70 40 10
Hal than doubles Tom's and Gary's tickets making it:
20 80 20
Tom than doubles Hal's and Gary's tickets making it:
40 40 40- The right ending point.

If we were to use the other set of starting numbers it works out this way:
Hal Tom Gary
65 35 20
10 70 40
20 20 80-which is not the 40, 40, 40 that was what they ended with. So the correct answer has to be 70, 40, and 10.

Let me say that I also used the long approach to this, creating an algebra equation to solve it. Pudley's approach was much better. I gave this problem to my 10 yo step-son in the car and he solved it in less than 5 minutes (it took me longer to explain the problem to him than it did for him to solve it.) I think they teach math in a much better way than they did when we were in school, using common sense and estimating.

SunTzu68
11-17-2002, 10:09 AM
65+35+20 does indeed work out to 120, I apologize for my previous post. However, it is still not the right answer. Let me work it through, and maybe someone can show me where my mistake is.

Hal Tom Gary
70 40 10
Hal than doubles Tom's and Gary's tickets making it:
20 80 20
Tom than doubles Hal's and Gary's tickets making it:
40 40 40- The right ending point.

If we were to use the other set of starting numbers it works out this way:
Hal Tom Gary
65 35 20
10 70 40
20 20 80-which is not the 40, 40, 40 that was what they ended with. So the correct answer has to be 70, 40, and 10.

Let me say that I also used the long approach to this, creating an algebra equation to solve it. Pudley's approach was much better. I gave this problem to my 10 yo step-son in the car and he solved it in less than 5 minutes (it took me longer to explain the problem to him than it did for him to solve it.) I think they teach math in a much better way than they did when we were in school, using common sense and estimating.

SunTzu68
11-17-2002, 10:46 AM
OOOOOOOOOOOps, my mistake........I only had the first two trades...........

Jim Brier
11-17-2002, 06:53 PM
(n/t)

Jim Brier
11-17-2002, 06:54 PM
(n/t)

11-18-2002, 01:37 AM
Problem type: Word definition reducing to a set of three linear algebraic equations in three unknowns. This is not a probability problem.
Let: x =Hals original amount of raffle tickets
y = Toms original amount of raffle tickets
z = Garys original amount of raffle tickets

Performing the three transfers of tickets per the word problem and simplifying the equations gives:
x  y  z =10
-x + 3y  z = 20
-x  y + 7z = 40

Just solve these equations if you know algebra, or use Mathcad or whatever.

11-18-2002, 01:41 AM
I suggest you take a class in algebra.

11-18-2002, 09:05 AM
throw the last guys post into a matrix, and punch it out

Homer
11-18-2002, 03:29 PM
That's a pain in the ass (at least for me). I would just multiply both sides of equation 1 by 2 to get z = 20. Then I'd multiply equation 3 by 2 to solve for x = 65. Then solve any of the three with x and z plugged in to get y = 35. Different strokes for different folks I guess...

-- Homer