Interest math question.?

Question

So I haven't done this type of question in a long time, and left my notes at school. Any help would be appreciated.

Question:
Mrs. Mathsgreat has spent her entire life teaching mathematics to grateful students. She has now saved $1000 and plans to deposit it into a fund that will pay her 10 percent interest each year. Her husband, Mr. Mathsgreat, says that it will take ten years for the money to double. Is this correct? Show how you would determine how long it would take algebraically or graphically.

I think there is a certain equation you're supposed to use?

Answer 1

Use the compound interest formula:
S = P(1+i)^n.
Here S is the amount accumulated
P is the principal
i is the interest rate per year
n is the number of compounding periods.
So in your case
2000 =1000(1+.1)^n = (1.1)^n
So 1.1^n = 2
n log 1.1 = log 2
n = log 2/ log 1.1 = 7.27 years(approximately)
So you will have more than double your
money in 8 years.

Answer 2

P = 1000
r = 10% = .1

Simple Interest

A = P(1 +rt)
A = 1000(1+.1(10))
A = 1000(1 + 1)
A = 1000(2)
A = 2000
It will double

Compound Interest

A = P(1 + r)^n
A = 1000(1 + .1)^10
A = 1000(1.1)^10
A = 1000(2.594)
A = 2593.74
More than doubles

Answer 3

8 years. $1000 times 1.1 = $1100 after 1 year
$1100 x 1.1=$1210 after 2
$1210 x 1.1=$1331 after 3
$1331 x 1.1=$1464.10 after 4
$1464.10 x 1.1=$1610.51 after 5
$1610.51 x 1.1=$1771.56 after 6
$1771.56 x 1.1=$1948.72 after 7
and
$1948.72 x 1.1=$2143.59 after 8 years. You need to take into account the new balance at the end of each year before you calculate the interest at 10%

Answer 4

well, the equation is

Money=1000*1.1^x

So plug in 10 for x and see if it is 2000 dollars

Answer 5

yes
it's A= P (1+r)^t
so just check if 1000 (1.1)^10 = 2000 or more

Answer 6

you need to know what type of interest you are talking about.
simple, compound, or continuous.