A father in his will left all his money to his children in the following manner:
$1000 to the first born and 1/10 of what then remains, then
$2000 to the second born and 1/10 of what remains, then
$3000 to the third born and 1/10 of what remains, and so on.
When this was done, each child had the same amount. How many children were there?
How would you solve this problem? This would be easy if it stated the total amount of money -_-
Answer: 9
F=1000+ 1/10 (A-1000)
S=2000 + 1/10 (9/10(a-1000) -2000)
T=3000 + 1/10 (9/10(a-1000-2000-3000)
subtract the first from the second equation.
S-F=2000-1000 + 9/100 (a-1000)-1/10(a-1000) - 200
0= 1000+a(.09 -.1) -90 +100 -200
0= 820 -.01a
a=82,000 dollars.
Now the first kid got 1000+ 1/10 (81000) or 9,100 dollars.
Since they all got the same, the number of kids has to be 82000/9,1000 or nine kids. Of course, there was some change left over at the end.
Could you show me a step by step on how to distribute the second child?
"S=2000 + 1/10 (9/10(a-1000) -2000) "
I don't know how to properly distribute "9/10(a-1000) -2000)"
I know F = (9000 + x) / 10
were did you get 9/10 from
To solve this problem, we can use a reverse approach starting from the amount of money each child had at the end.
Let's assume there are n children.
We know that each child receives an equal amount at the end. Let's call this final amount x.
According to the given information, the first child receives $1000 and 1/10 of what remains. So, their total amount is $1000 + (1/10) * (x - $1000).
Similarly, the second child receives $2000 and 1/10 of what remains. So, their total amount is $2000 + (1/10) * [(x - $1000) - $2000] = $2000 + (1/10) * (x - $3000).
Following this pattern, the third child's total amount is $3000 + (1/10) * [(x - $1000) - $3000] = $3000 + (1/10) * (x - $4000).
We can continue this for all other children. The nth child's total amount would be $n * 1000 + (1/10) * (x - $n * 1000) = $n * 1000 + (1/10) * (x - $n * 1000).
Since all children have the same amount, we can set up an equation:
$x = $n * 1000 + (1/10) * (x - $n * 1000)
To simplify, we can multiply both sides by 10:
10x = 10 * $n * 1000 + x - $n * 1000
10x - x = 10 * $n * 1000 - $n * 1000
9x = 9 * $n * 1000
Dividing both sides by 9, we get:
x = $n * 1000
This means that the final amount x must be a multiple of 1000.
Now, let's consider the options for the final amount x. We'll start with the smallest possible multiple of 1000, which is $1000.
If x = $1000, then the equation becomes:
$1000 = $n * 1000
Dividing both sides by $1000, we find:
1 = $n
Since we can't have a fractional number of children, this option is not valid.
If x = $2000, the equation becomes:
$2000 = $n * 1000
Dividing both sides by $1000, we find:
2 = $n
Again, this is not a valid option, as we can't have a non-integer number of children.
Continuing this process, we can test x = $3000, $4000, $5000, and so on.
Eventually, when x = $9000, the equation becomes:
$9000 = $n * 1000
Dividing both sides by $1000, we find:
9 = $n
Now, we have a valid solution. When x = $9000, we have 9 children.
Therefore, there are 9 children in this scenario.