Please Could Someone Show Me The Mathmatical/algebraic Workings For The Below? (The Answer Is Already Known)

[bond]

100 people attend a dinner, made up of a mixture of men, women and children (100 in total). Men are charged £5, women £1 and children 5p. How many of each (men, women and children) attended the event if it cost £100 in total?

[Prudie]

If you work in pence because of the child price and let number of men = m, number of women = w and number of children - c then the money gained from the men for example would be m x 500p, using this method your equation for the total would be:
500m + 100w +5c = 10,000p
This is not enough info for a unique solution though, there are lots of results without knowing more information.

[JJ109]

This is one that cannot be done with simultaneous equations - but can be solved using logic.
I shall do it for you but it may take a bit of time - give me 30 minutes

[JJ109]

One solution is 0 men , 0 children and 100 women
but I am confirming other solutions possible

[Prudie]

The only starting clue is as the total was a whole number of pounds the number of children must have been a multiple of 20.

[zebo]

With 3 variables and only two pieces of information, the total of 100 and the cost of £100, then a unique solution isn't possible. If something like 'there are twice as many men as women' were introduced then it would be possible.

[JJ109]

Yes Prudie and if you go that way you will come to other solutions and rule out possibilities
The total number of people attend = 100
Children attending can only be 0,20,40,60,80,100, otherwise there would be odd pence
0 children leaves M + w = 100 people
5M + w = 100 pounds
Subtract top equation from bottom
4M = 0
M = 0, therefore W = 100
First solution
M = 0, W = 100, C = 0

[angie55]

80 children = £4, 1 woman = £1 and 19 men = £95. Total £100.

[JJ109]

Discount 100 children as that would only bring in 5 pounds
80 children = 4 pounds
20 = M + W
96 = 5M + W
76 = 4M
M = 19 therefore W = 1

Another solution = 80 children 19 men and 1 woman

[JJ109]

60 children leaves
M + W = 40
5M = W = 97
4M = 47
M = 11.75 can't have 0.75 men
This solution does not work
Discount 60 children

Now do the same for 20 and 40 children

[JJ109]

40 children = 2 pounds
M + W = 60
5M + w = 98
4M = 38
M = 9.5
This solution does not work

20 children = 1 pound
M + w = 80
5M + W = 99
4M = 19
M = 4.75 does not work can't have 3/4 of a man

ONLY TWO POSSIBLE SOLUTIONS
0 men 100 women 0 children
19 men 1 woman 80 children

[Mamyalynne]

0 anyone doesn't work...

// made up of a mixture of men, women and children//

Therefore Angie had it up there.

[JJ109]

error in 60 above
60 children = 3 pounds
M + W = 40 people
5M + W = 97
4M = 57
M = 14.25
Can't have 1/4 of a man - so reject 60 children scenario (like before but just corrected my maths)

[aelmpvw]

good luck to the one woman at that dinner

[Mamyalynne]

She has her hands full for sure.

[JJ109]

Anyway despite my long winded series of answers I hope that you can see how to use trial and error to solve these type of questions but using logic to help you to formulate the guesses

[Prudie]

bit of a collection up there but assuming the were no 0 values then 2 solutions are
10 men, 49 women, 20 children
11 men, 44 women, 20 children
12 men, 39 women, 20 children

I could go on

[JJ109]

No Prudie M + W + C have to total 100 people, your examples do not

Reading mine together I have proven that there are only two solutions, and one of them contains no men or children - and isn't allowed.

The unique solution for the question is
19 men 1 woman and 80 children = 100 people
95 pounds + 1 pound + 4 pounds = 100 pounds

[JJ109]

I am now going to try and pull together all my answers to try and help you understand how to do these type of problems.

There are three variables (M,W,C) and only 2 equations possible, therefore you do not have enough information to solve just by using simultaneous equations.....but

You now look at what further information you can glean. Here the only variable denominated in pence is the cost of the children. To get whole numbers of people, it therefore follows that the volume of children can only be in units of 20, to ensure whole pounds. Otherwise the cost cannot be exactly 100 pounds.

As you know there must be some men, some women and some children, you can disregard solutions where there are no children or 100 children.

Therefore the only possible solutions will be where there are 20, 40, 60 or 80 children.

It is now necessary to go through these 4 scenarios in turn and see if they produce a solution. NO OTHER SOLUTIONS apart from these 4 scenarios are possible

Assuming C = 20 (cost 20 x 0.05 = 1 pound)
This leaves 100 - 20 = 80 people costing 100 - 1 = 99 pounds
Now set up simultaneous equations
M + W = 80
5M + W = 99 (men 5 pounds, women 1 pound)
Subtract top equation from bottom)
4M = 19
M = 4.75
As it is impossible to have 3/4 of a man, you can reject the scenario where C = 20

Do the same for C = 40, cost C = 40 x 0.05 = 2 pounds
M + W = 60 (ie 100 - 40)
5M + W = 98 pounds
Subtract top from bottom
4M = 38
M = 9.5
As you can't have half a man, reject the scenario where C = 40

Now C = 60, cost = 60 x 0.05 = 3 pounds
M + W = 40
5M + W = 97
Subtract top from bottom
4M = 57
M = 14.25
As you can't have 1/4 of a man reject this scenario where C = 60

Final possible solution where C = 80, cost = 80 x 0.05 = 4 pounds
M + W = 20 (ie 100 - 80)
5M + W = 96 (ie 100 - 4 pounds)
Subtract top from bottom
4M = 76
M = 19 ...hooray a solution
As M + W = 20 and M = 19 it follows that W = 1

Therefore the ONLY solution is
M = 19, W = 1 and C = 80

Check the values
M = 19 x 5 = 95 pounds
W = 1 x 1 = 1 pound
C = 80 x 0.05 = 4 pounds
Total people = 19 + 1 + 80 =100
Total cost = 95 + 1 + 4 = 100 pounds
Solution correct

and as we went through the only other three possibilities that could yield a solution, and ruled them out, we conclude that this is a UNIQUE solution.

[Prudie]

Just to JJ109 drat of course you're right. When I started I knew they had to sum to 100 and then got carried away and forgot - typical schoolgirl error. So add to my original equation m+w+c = 100
3 unknowns, 2 equations so- solved by you via logic, trial and error