Simple Interest Question?

♥♦♣♠

of course you can use that formula to solve this problem.

I = P • R • N / 100

...............where I = interest

........................P = principal amount

........................R = rate of interest

........................N = number of years

I = $4,500 - $3,000

..= $1,500

P = $3,000

N = 4 years

.............. substitute the given values to the equation;

1,500 = 3,000 • R • 4 / 100

...............in the right expression, 3,000 / 100 would result to 30;

1,500 = 30 • R • 4

1,500 = 120 • R

...............divide both sides of the equation by 120;

.........................remember that performing the same mathematical operation

.........................on both sides of the equation would still maintain its equality.

1,500 / 120 = 120 • R/ 120

12.5 = R

...or

R = 12.5 % per annum.

░░░░░░░░░░░░░░░

☺☺☺☺☺☺☺☺☺☺

let's check:

Every year, the initial deposit would have the following interest:

.......= $ 3,000 x 12.5 %

.......= $ 3,000 x 0.125

.......= $ 375

At the end of 4 years, the total interest would be:

.......= $ 375 • 4

.......= $ 1,500

So, Dela would now have:

.......= $ 3,000 + $ 1,500

.......= $ 4,500

☺☺☺☺☺☺☺☺☺☺

☻☻☻☻☻☻☻☻☻☻

Immunitas' answer is what you'll get if the interest is COMPOUNDED annually,

where the interest for the year is added to the principal amount.......

But you specified it as a Simple Interest Question, right ?!?!?!?!

if it is compounded, this would be the scenario :

1st year:

............3,000 • 10.67 %

.........= 3,000 • 0.1067

.........= 320.05

.....................this would now be added to the original principal amount;

2nd year:

............( 3,000 + 320.05) • 10.67 %

.........= 3,320.05 • 0.1067

.........= 354.19

.....................this would now be added to the principal amount of the second year;

3rd year:

............( 3,000 + 320.05 + 354.19 ) • 10.67 %

.........= 3,674.24 • 0.1067

.........= 391.97

.....................this would now be added to the principal amount of the third year;

4th year:

............( 3,000 + 320.05 + 354.19 + 391.97 ) • 10.67 %

.........= 4,066.21 • 0.1067

.........= 433.79

the total interest would be

.....= 320.05 + 354.19 + 391.97 + 433.79

.....= $ 1,500

And, Dela would now have:

.......= $ 3,000 + $ 1,500

.......= $ 4,500

remember, this is for the compounded annually scenario.....

☻☻☻☻☻☻☻☻☻☻

A = P(1+i)^n

A = total after maturarity

P = principle

i = interest

n = number of years

to solve for i

4500 = 3000(1+i)^4

(1+i)^4 = 1.5

1+i = 1.1067

i = 0.1067

to answer you question, interest is 10.67%