Algebra problem - old but new

UdarM

New member
Joined
May 28, 2016
Messages
3
There are 100 people and £100. The Men got £10, Women = £0.5 and the Children £0.125
How many Men received £10, how many Women received £0.5 and how many Children received £0.125?
I know the final result but don't know HOW (step by step) to get there.
Answer: Men = 7, Women = 49 and Children = 44

I have tried the following as a start:

x + y + z = 100
10x+0.5y+0.125z = 100

Thanks in advance for your help!
 
There are 100 people and £100. The Men got £10, Women = £0.5 and the Children £0.125
How many Men received £10, how many Women received £0.5 and how many Children received £0.125?
I know the final result but don't know HOW (step by step) to get there.
Answer: Men = 7, Women = 49 and Children = 44

I have tried the following as a start:

x + y + z = 100
10x+0.5y+0.125z = 100

Thanks in advance for your help!

First I'll convert it to an one equation - two variables - one constraint (x,y,z are non-negative integers)

z = 100 - y - x

80x + 4y + z = 800 → 79x + 3y = 700 → y = (700 -79x)/3

Now try different values 'x' → 0 ≤ x ≤ 10 and find for which value/s of 'x' you get integer 'y'.
 
There are 100 people and £100. The Men got £10, Women = £0.5 and the Children £0.125
How many Men received £10, how many Women received £0.5 and how many Children received £0.125?

I know the final result but don't know HOW (step by step) to get there.
This exercise has enough information for two equations, but it requires three unknowns. As such, it is not solvable, by algebraic methods. Some guess-and-check will be required, as well as making assumptions (such as that people can occur only in non-negative whole numbers). ;)
 
Thank you!

First I'll convert it to an one equation - two variables - one constraint (x,y,z are non-negative integers)

z = 100 - y - x

80x + 4y + z = 800 → 79x + 3y = 700 → y = (700 -79x)/3

Now try different values 'x' → 0 ≤ x ≤ 10 and find for which value/s of 'x' you get integer 'y'.


Thanks for response.
I got to this point previously but i thought that it can be also a different way of solving it without trying different numbers.
 
!!!

This exercise has enough information for two equations, but it requires three unknowns. As such, it is not solvable, by algebraic methods. Some guess-and-check will be required, as well as making assumptions (such as that people can occur only in non-negative whole numbers). ;)

Thanks for prompt response and clarification!
 
Thanks for response.
I got to this point previously but i thought that it can be also a different way of solving it without trying different numbers.

Why didn't you share that in your initial response?!

y = (700 - 79x)/3 = 233 1/3 - (79x)/3

x has to be ≤ 10. 700, 3 and 79 are relatively prime. So x has to be > 1.

Since y ≤ 100 → x > 5

Anyway, since 'y' has to be integer, (700-79x) must be divisible by 3. x itself must not be divisible by 3.

79*9 = 711 → x ≤ 8 → x can be either 7 or 8

Now what....
 
Last edited by a moderator:
Top