# math

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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)

• math -

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 -

the above contains a typo, I obviously meant 3% in the third last line

• math -

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 1.03n 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 -

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 -

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.