# Thread: Proportion

1. ## Proportion

An amount of money was divided between some people in such a way that if there had been 4 more people, everyone would have got $16 less. But if there had been 4 less people, everyone would got$24 more. How many people were there in the group?

How could I solve this?

2. Originally Posted by geton
An amount of money was divided between some people in such a way that if there had been 4 more people, everyone would have got $16 less. But if there had been 4 less people, everyone would got$24 more. How many people were there in the group?

How could I solve this?
1. Let
x denote the number of people in the group;
y denote the amount of money each person will get.

2. "Translate" the text into equations:

$\left| \begin{array}{rcl}(x+4)(y-16)&=&xy \\ (x-4)(y+24)&=& xy\end{array}\right.$

3. Expand the brackets and subtract the 1st equation columnwise from the 2nd equation. After moving some stuff around you'll get:

$y = 5x-4$ ----- [A]

4. Substitute in the 1st equation the variable y by 5x - 4. Expand the brackets and solve for x.

5. Plug in the value of x into the equato [A] to calculate y.

6.
Spoiler:
I've got x = 20 and y = 96

3. Originally Posted by earboth
1. Let
x denote the number of people in the group;
y denote the amount of money each person will get.

2. "Translate" the text into equations:

$\left| \begin{array}{rcl}(x+4)(y-16)&=&xy \\ (x-4)(y+24)&=& xy\end{array}\right.$

3. Expand the brackets and subtract the 1st equation columnwise from the 2nd equation. After moving some stuff around you'll get:

$y = 5x-4$ ----- [A]

4. Substitute in the 1st equation the variable y by 5x - 4. Expand the brackets and solve for x.

5. Plug in the value of x into the equato [A] to calculate y.

6.
Spoiler:
I've got x = 20 and y = 96
Thank you so much

4. Originally Posted by geton
An amount of money was divided between some people in such a way that if there had been 4 more people, everyone would have got $16 less. But if there had been 4 less people, everyone would got$24 more. How many people were there in the group?

How could I solve this?
Alternatively, the sum of money is constant, while the number of people and each one's share varies.

The number of people multiplied by each one's share always equals the sum of money.

$p=$ number of people; $s=$ each one's share.

$ps=(p+4)(s-16)=(p-4)(s+24)$

$\Rightarrow\ ps=ps-16p+4s-64=ps+24p-4s-96$

We require $p$ only, so subtract $ps$ from all three expressions...

$4s-16p-64=24p-4s-96=0$

Add these to eliminate $s$ as both are zero...

$8p-160=0$