Posted by **Corleone** on Friday, June 5, 2009 at 10:04pm.

suppose the population of a town increases by 3% each year. the population of the town today is 25,000. use compound interest table to find its population ten years from now. (must give the answer to the nearest thousand)

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**Reiny**, Friday, June 5, 2009 at 10:14pm
the calculation is

25000(1.03)^10

which on a calculator is a piece of cake

If you have to use "compound interest tables ??" you probably have a column for 4%, go down that table to the row n=10 to find 1.34392

multiply that by 25000

you should get $33597.91

- math -
**Reiny**, Friday, June 5, 2009 at 10:15pm
the above contains a typo, I obviously meant 3% in the third last line

- math -
**MathMate**, Friday, June 5, 2009 at 10:17pm
Compound interest table for 3% p.a., compounded each year, obtained by multiplying the factor of the previous year by 1.03. The value is also equal to 1.03^{n} where n is the number of years compounded.

Year growth factor

0 1.0000

1 1.0300

2 1.0609

3 1.0927

4 1.1255

5 1.1593

6 1.1941

7 1.2299

8 1.2668

9 1.3048

10 1.3439

After 10 years, the population is

25000*1.3439

=33598

=34000 (to the nearest thousand).

- math -
**MathMate**, Friday, June 5, 2009 at 10:20pm
Oops, it does not seem to like pasting from Excel!

Here is the compound interest table:

year growth factor

0 1.0000

1 1.0300

2 1.0609

3 1.0927

4 1.1255

5 1.1593

6 1.1941

7 1.2299

8 1.2668

9 1.3048

10 1.3439

After the tenth year, the population is

25000*1.3439

=33598

=34000 (to the nearest thousand)

- math -
**MathMate**, Friday, June 5, 2009 at 10:24pm
OK, I give up pasting!

The growth factor for 10 years is 1.3439.

Multiply this with 25000 gives 33598, and rounding to the nearest thousand gives 34000.

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