[wolfgang]

Hi dear friends!

Fourty poeple( men,women,and children)entered a resturant.

each man should pay 4$.

each woman 2$.

each 4children 1$.

totally they paid 40$.

howmany men were out of 40?

howmany women?

and howmany children?

have a nice time......

[Guest]

8 men $8*4 =$32

32 children 32/4 = 8 groups of 4 = $8

[Guest]

24 kids=6$, 1man=4$, and 15women=30$ =40$

[Guest]

It has infinite solutions, because we only have 1 equation for 3 variables

4M + 2W + C/4 = 40

[Guest]

1 Man, 15 Women, 24 Kids

8 Men, 0 Women, 32 Kids

These are the only possible solutions, unless you can have half a person...

[Guest]

1 Man, 15 Women, 24 Kids

8 Men, 0 Women, 32 Kids

These are the only possible solutions, unless you can have half a person...

But you can always take away 8 kids and add a woman or vice versa, or take 2 women and add a man and so on, and the numbers will still be integer.

[Guest]

But you can always take away 8 kids and add a woman or vice versa, or take 2 women and add a man and so on, and the numbers will still be integer.

Not if you only have 40 people. Even without that limitation you can't always take away women or men (ie you'll run out eventually) and you can't keep adding children (ie you can't have more than 160 children)

[Guest]

Not if you only have 40 people. Even without that limitation you can't always take away women or men (ie you'll run out eventually) and you can't keep adding children (ie you can't have more than 160 children)

That's true, I think I skipped that line when I read it

And it's also true that the number of solutions wouldn't be infinite if it wasn't for that limitation. It would be quite big though. I'm too lazy to calculate.

[Guest]

As discussed, this is simply a case of a systems of equations: M + W + C = 40 and 4M + 2W + C/4 = 40.

Since there are only two equations and three variables, there are potentially infinite solutions but constraints limit the possibilities.

As noted, each variable must be an integer (no half people) - and nonnegative.

Also, since the final amount paid is an integer, the number of children must be a multiple of four. (10 possibilities)

Furthermore, since both the men's and women's costs are even, the number of children must be a multiple of eight. (5 possibilities)

Finally, the remaining amount of money after taking out the costs for children must be no more than $4 per remaining person (all men) or less than $2 per remaining person (all women).

That leaves only the two remaining possibilities: 8 men, 0 women, and 32 children or 1 man, 15 women, and 24 children.

Interestingly, one could argue that there are 2, 1 or 0 solutions based on different interprtations of the problem:

if "men, women, and children" implies there must be at least one of each, only 1 solution remains;

if it implies there must be at least two of each, no solutions remain.

BTW, without the restriction on the total number of people, there are 110 valid solutions if any variable can be either 0 or 1 and

only 56 solutions if there must be at least two men, two women, and two children.

[wolfgang]

8 men $8*4 =$32

32 children 32/4 = 8 groups of 4 = $8

and where are the women?

[wolfgang]

As discussed, this is simply a case of a systems of equations: M + W + C = 40 and 4M + 2W + C/4 = 40.

Since there are only two equations and three variables, there are potentially infinite solutions but constraints limit the possibilities.

As noted, each variable must be an integer (no half people) - and nonnegative.

Also, since the final amount paid is an integer, the number of children must be a multiple of four. (10 possibilities)

Furthermore, since both the men's and women's costs are even, the number of children must be a multiple of eight. (5 possibilities)

Finally, the remaining amount of money after taking out the costs for children must be no more than $4 per remaining person (all men) or less than $2 per remaining person (all women).

That leaves only the two remaining possibilities: 8 men, 0 women, and 32 children or 1 man, 15 women, and 24 children.

Interestingly, one could argue that there are 2, 1 or 0 solutions based on different interprtations of the problem:

if "men, women, and children" implies there must be at least one of each, only 1 solution remains;

if it implies there must be at least two of each, no solutions remain.

BTW, without the restriction on the total number of people, there are 110 valid solutions if any variable can be either 0 or 1 and

only 56 solutions if there must be at least two men, two women, and two children.

the solution....1 man...15 women....and 24 children is correct!!Bravooo

[Guest]

As discussed, this is simply a case of a systems of equations: M + W + C = 40 and 4M + 2W + C/4 = 40.

Since there are only two equations and three variables, there are potentially infinite solutions but constraints limit the possibilities.

As noted, each variable must be an integer (no half people) - and nonnegative.

Also, since the final amount paid is an integer, the number of children must be a multiple of four. (10 possibilities)

Furthermore, since both the men's and women's costs are even, the number of children must be a multiple of eight. (5 possibilities)

Finally, the remaining amount of money after taking out the costs for children must be no more than $4 per remaining person (all men) or less than $2 per remaining person (all women).

That leaves only the two remaining possibilities: 8 men, 0 women, and 32 children or 1 man, 15 women, and 24 children.

Interestingly, one could argue that there are 2, 1 or 0 solutions based on different interprtations of the problem:

if "men, women, and children" implies there must be at least one of each, only 1 solution remains;

if it implies there must be at least two of each, no solutions remain.

BTW, without the restriction on the total number of people, there are 110 valid solutions if any variable can be either 0 or 1 and

only 56 solutions if there must be at least two men, two women, and two children.

Oops, as for the 110 possible valid solutions, I overlooked solutions with no children so ther eare actually 121 rather than 110.

[Guest]

the solution....1 man...15 women....and 24 children is correct!!Bravooo

and where are the "men"?

I agree with Chuck that you could argue the question could imply multiples of each type. I do not agree though that the question is worded to where at least one of each type is required.

Edited October 11, 2010 by maurice