Posted: Tue Mar 01, 2005 11:10 am Post subject: Need some help to solve the puzzle.
Hi,
It will be great if some one helps me to solve the following question.
Question :
At a certain festivety a rich man decided to distribute free rice to deserving people. He had altogether 100 Kg of rice and wanted to distribute the grain to 100 people in such a manner that each old person received 3 Kg , each young person 2 Kg and each child 1/2 Kg. How many old persons , young persons and children were there ?
you formed a couple of nice equations
in the second one you have simply said that the sum of the old , men and youth would be equal to 100.
Only question is when you solve the equation you seemed to have not evaluated the left side of the rows of equations
Is there a reason for that
Posted: Wed May 04, 2005 11:12 am Post subject: RAW
3x + 2y + (z/2) =100 => 6x + 4y + z = 200 .............eqn 1
x + y + z = 100 => 6x + 6y +6z = 600 .............eqn 2
Subtract 1 from 2 ... 2y + 5z = 400 => z = (400 -2y)/5 => z = 80 -2y/5
Substitute values for y such that z is positive .( Remember x,y,z > 0)
The sets of solutions are
z 78 , 76 , 74 , 72 , 70 , 68
y 5 , 10 , 15 , 20 , 25 , 30
x 17 , 14 , 11 , 8 , 5 , 2
Do u follow the pattern ??????? x->old y->youth z->children
Points for Review :
1. I have said x + y + z = 100 because the problem statement says that there are 100 people . The people could be old(x), youth(y) or children(z).
Hence the equation .
2. I have actually multiplied the 2nd eqn by 6 so as to eliminate one variable . I've infact used both the equations .
6x + 4y + z = 200 .............eqn 1
x + y + z = 100 => 6x + 6y +6z = 600 .............eqn 2
Subtract 1 from 2 , we get
2y + 5z = 400 ---eqn 3 . Equation 3 is got from the first 2 eqns . If you're not clear plz seek for clarity . I'll do my best to reply .
Hi Aswathy,
Mr.Kandhi has given exact answer in a concise way. Mr.Srikanth has given possible six set of answers. But out of the possible six answers, we need to think a while that there needs to be much older people, minimum young people so that we can accomodate maximum possible children. Because the importance lies in the order: Older , children and Young ones in descendending order.
Now coming to the problem, we have two equations:3O+2Y+(1/2)C = 100 and O+Y+C = 100 where O=old, Y=young and C= children.
Eliminating C, we get 5O+3Y=100 which gives Y<100/3 i.e., 33.33 say 33
so Y<33 .
Eliminating C, we get (3/2)C-O = 100 which means two conditions here.
C should be even and C>66.66 say 67.
Eliminating O, we have -Y-(5/2)C = -200 which gives Y<200.
So, summing the following derived conditions, we have
C=even,
C=67 to 79
Y=1 to 33
So with these trying for different combinations, we get five possible combinations as Mr.Srikanth has stated.
C,Y,O
68,30,2
70,25,5
72,20,8
74,15,11
76,10,14
78,5,17
So as I first said, we need to accomodate maximum older persons and minimum young ones.
So 78,5,17 is the best combination and is the answer as Mr. Kandhi already stated.
If you have any problem, feel free to contact me at tgbharati@yahoo.com
Have a great day!
Bharathi.
