how many possible solutions?

gyanshrestha · 2007-11-16 16:41:45

tow equations

m + w + c =100

2m+w+(1/2)c=100

for these two equations how many solutions are possible.
assuming that m for man, w for woman and c for children.

luca-deltodesco · 2007-11-16 20:09:58

ive never done a question like this before of fiding the number of natural solutions, only to solve in the reals for unique solutions or to find if there is no solution, but ill give it a go.

I assume m,w,c ∈ {N, 0} (m,w,c must be more than or equal to 0 and a whole number)

m + w + c = 100
2m + w + 0.5c = 100

c = 100 - m - w = 200 - 4m - 2w

100 - m - w = 200 - 4m - 2w

3m + w = 100

the max value of m is therefore 33, giving w = 1, and therefore c = 66 (if m was more than 33, w would be negative)
the min value of m is 0, giving w = 100, and c = 0 (m = 0 is an existant boundary condition)

w = 100-3m
c = 100-w-m = 2m

so there are 34 solutions of the form: 0 <= m <= 33, w = 100 - 3m, c = 2m

Last edited by luca-deltodesco (2007-11-16 20:12:51)

Khushboo · 2007-11-18 23:37:10

m + w + c =100.......I

2m+w+(1/2)c=100........II

substitute 2m+w+(1/2)c in equation I, after substitution we get m+w+c=2m+w+(1/2)c
on solving the equation we get c=2m, this clearly shows that c>m as c is two times m

Now substitute c with 2m, on subsitution we get the following equation 3m+w=100
Inorder to make both the lefthandside equal to 100, one can subsitute following values for m=(0,1,2......33) but as we need the count we will need the values which are whole numbers. The maximum value for m can be 33. On substituting 33 for m we get w=1and c=66.
Possible solutions for (m,w,c)
Max (33,1,66)

No. of possible solutions can be 34.

Hope the approach is correct.

Regards

Last edited by Khushboo (2007-11-18 23:44:25)

gyanshrestha · 2007-11-18 23:51:07

thanks for yours(both of you) great responses
one doubt is here.

both of you conclude that
c=2m and
3m+w=100
the second one suggest that possible value of m are 0,1,2....33
and both of u suggest that there are 34 solutions

i think it is not. because c=2m suggest that m is only even then possible value of m are 0, 2 , 4 , 6, ........32 then there are only 17 only solutions. am i right?

sorry in advance if any mistake.

Last edited by gyanshrestha (2007-11-18 23:51:50)

mathsyperson · 2007-11-19 00:39:32

c=2m means that c can only take even values. m can be whatever it wants.

gyanshrestha · 2007-11-19 22:41:57

oh sorry
i got wrong conclusion

Math Is Fun Forum

#1 2007-11-16 16:41:45

how many possible solutions?

#2 2007-11-16 20:09:58

Re: how many possible solutions?

#3 2007-11-18 23:37:10

Re: how many possible solutions?

#4 2007-11-18 23:51:07

Re: how many possible solutions?

#5 2007-11-19 00:39:32

Re: how many possible solutions?

#6 2007-11-19 22:41:57

Re: how many possible solutions?

Board footer