Problems and Solutions

mathsyperson · 2006-02-18 01:48:50

Jai Ganesh · 2006-02-18 02:33:13

Well done, mathsyperson!

Problem # k + 94

If 1/a + 1/b + 1/c = 1/(a+b+c); a+b+c≠0 and abc≠0, what is the value of (a+b)(b+c)(c+a)?

irspow · 2006-02-18 03:06:10

Nice job, Mathsyperson! I spent like twenty minutes to come up with the same answer using the disk method of integration.

irspow · 2006-02-18 05:00:24

This is not my type of problem, but here goes nothing.

*edit

Duh! I forgot to subtract abc from both sides of the equation.

Last edited by irspow (2006-02-19 01:14:16)

Jai Ganesh · 2006-02-18 19:13:54

irspow, please check the solution you posted again

Jai Ganesh · 2006-02-18 19:36:18

Problem # k + 95

1 man or 2 women or three boys can do a work in 55 days. In how many days can 1 man, 1 woman and 1 boy do the same work?

krassi_holmz · 2006-02-18 22:07:26

So a=-c or a=-b and

Last edited by krassi_holmz (2006-02-18 22:26:54)

krassi_holmz · 2006-02-18 22:31:02

k+95:
1m-1/55 per day
1w-1/110 per day
1b-1/165 per day
t-the numbers of days:
t(1/55+1/110+1/165) = 1;
t(1/30)=1;
t=30 days.

irspow · 2006-02-19 01:22:20

krassi_holmz · 2006-02-19 02:24:03

Yeah

Jai Ganesh · 2006-02-19 02:25:03

24805BP~The-Simpsons-Mr-Burns-Excellent.jpg

Jai Ganesh · 2006-02-19 03:48:19

Problem # k + 96

Can a majority of the numbers from 1 to a million be represented as the sum of a square and a (non-negative) cube?

mathsyperson · 2006-02-19 06:01:50

I don't know if it was appropriate to say QED in that, but it's just really cool to say. ^_^

Jai Ganesh · 2006-02-19 16:39:00

mathsyperson, you are correct!

Ricky · 2006-02-19 17:52:50

By mathimatical law, you can only say QED when you start out with "Pf" or "Proof". But we'll let you go with a warning this time.

Jai Ganesh · 2006-02-19 18:37:08

Problem # k + 97

Two men set out at the same time to walk towards
each other from points A and B, 72 km apart. The first man
walks at the constant rate of 4 km/hr. The second man walks
1 km the first hour, 2 km the second hour, 3 km the third hour
and so on. When will the two men meet?

mathsyperson · 2006-02-20 02:49:11

John E. Franklin · 2006-02-20 11:56:23

On k + 89, I meant to multiply by 3, but then I still would be wrong.
I wish there was a theorum to get the number that is close to this 3.
I'll have to work on that. The 3 multiplier would be like having 4 + 4 + 4 in the numerator.
I know it is wrong, but I feel the need to research this further.

John E. Franklin · 2006-02-20 13:07:51

On k + 89, I was below with this 3 times correction still by factor of 1.062801932.
So with the new correction factor, it's kind of dumb, but, you'll laugh, but...
percentage that it was from machine B knowing first that the piece is
defective = 35% ((4 x 3)/(5 + 4 + 2)) * correction factor, where correction factor = (5 + 4 + 2)/(3*(.35 x 4 + .25 x 5 + .40 x 2))

irspow · 2006-02-20 14:32:25

Last edited by irspow (2006-02-20 16:00:28)

Jai Ganesh · 2006-02-20 17:20:59

Solution to Problem # k + 97

After 8 hours, the first man has walked (8x4) kilometers, i.e. 32 kilometers.
After 8 hours, the second man has walked (1+2+3+4+5+6+7+8) kilometers, i.e. 36 kilometers. In the ninth hour, the first man is walking at 4 km/hr and the second is walking at 9 km/hr. Let them meet after time t hours.
Hence, 4t+9t=4, 13t=4, t=4/13 hours = 18.46 minutes approximately.
Hence, they would meet after 8 hours and 18.46 minutes approximately.

mathsyperson is correct

Jai Ganesh · 2006-02-20 18:13:40

Problem # k + 98

A and B are two stations 300 km apart. Two trains T1
and T2 start from A and B respectively, towards each other at
the same time. T1 reaches B nine hours and T2 reaches A four
hours after they meet. Find the difference between the speeds
of T1 and T2.

irspow · 2006-02-21 03:37:38

lol. You love this instantaneous velocity. I stand by my k+97 solution(s). There was nothing in the question that required the solution agreed upon. Distance travelled does not prove constant velocity during a time period. I have yet to see an instance of infinite acceleration either. My equations describe the motion of both people as correctly as your assumptions.

I am not trying to dispute your solution, just pointing out that the wording of a problem is just as vitally important as the mathematics that lead to a solution. Your solution and mine are equally valid given the way in which it was worded. More details would be needed to invalidate either one.

My answers are more correct though! Just kidding. Why quibble over less than 30 seconds?

Last edited by irspow (2006-02-21 03:42:33)

Jai Ganesh · 2006-02-21 03:46:47

I concede you're correct too, irspow! The time given to solve the problem was about a minute! I thought the best way to do that was the way I solved

irspow · 2006-02-21 03:49:19

Thanks for the concession. lol. Interesting how both are correct and yet different with a problem stated as such, no?

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