1. ## Problemo

When a child is born you put a sum of money to the account so that the child can lift 17000 when she is 18 How much have you put in the first time if the intrest is 6%???

2. Originally Posted by Rambo
When a child is born you put a sum of money to the account so that the child can lift 17000 when she is 18 How much have you put in the first time if the intrest is 6%???
Let the original sum be x

The sum after $1$ year is $1.06 \times x$

After $2$ years $1.06 \times (1.06 \times x)=1.06^2 ~x$

After $3$ years $1.06 \times (1.06^2 ~ x)=1.06^3 ~x$

After $N$ years $1.06 \times (1.06^{n-1} ~ x)=1.06^N ~x$

So the fund has $1.06^{18} ~x=17000$ on the girls 18th birthday, now solve for $x$.

(This is an initial sum of $x$ compounded annulaly for 18 years at a rate of 6%)

RonL

3. i dont know how to solve help anyone

4. Originally Posted by CaptainBlack
Let the original sum be x

The sum after $1$ year is $1.06 \times x$

After $2$ years $1.06 \times (1.06 \times x)=1.06^2 ~x$

After $3$ years $1.06 \times (1.06^2 ~ x)=1.06^3 ~x$

After $N$ years $1.06 \times (1.06^{n-1} ~ x)=1.06^N ~x$

So the fund has $1.06^{18} ~x=17000$ on the girls 18th birthday, now solve for $x$.

(This is an initial sum of $x$ compounded annulaly for 18 years at a rate of 6%)

RonL
Originally Posted by Rambo
i dont know how to solve help anyone
If you can't do basic algebra you are in trouble, in this case I will show you

$x=\frac{17000}{1.06^{18}}$

Now its just a job for your calculator.

RonL

5. i solved it already its 17000*(1.06)^-18=

thats much easier than your suggestions

6. Originally Posted by Rambo
i solved it already its 17000*(1.06)^-18=

thats much easier than your suggestions
No its not, its the same thing, except more key stokes

RonL

7. Originally Posted by Rambo
i solved it already its 17000*(1.06)^-18=

thats much easier than your suggestions
This post is sailing close to the wind, this is verging on the insulting.

If someone attempts to help you, gratitude is the correct response not distain.

RonL