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November 29, 2014

November 29, 2014

Posted by **Anonymous** on Wednesday, November 28, 2007 at 8:51pm.

$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

- Math - Algebra -
**bobpursley**, Wednesday, November 28, 2007 at 9:37pmF=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.

- Math - Algebra -
**Anonymous**, Wednesday, November 28, 2007 at 9:40pmCould 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)"

- Addition to this -
**Anonymous**, Wednesday, November 28, 2007 at 9:40pmI know F = (9000 + x) / 10

- Addition to this -
- Math - Algebra -
**anonymous**, Tuesday, February 21, 2012 at 8:09pmwere did you get 9/10 from

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