SOLUTION: PLEASE HELP ME IF ANY ONE CAN....... AND TELL ME TOO ACTUALLY WHAT ARE THEY ASKING ABOUT MAY BE I AM NOT GETTING THE QUESTION.....
THE DIFFERENCE BETWEEN COMPOUND INTEREST AND SI
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Question 935622: PLEASE HELP ME IF ANY ONE CAN....... AND TELL ME TOO ACTUALLY WHAT ARE THEY ASKING ABOUT MAY BE I AM NOT GETTING THE QUESTION.....
THE DIFFERENCE BETWEEN COMPOUND INTEREST AND SIMPLE INTEREST AT THE RATE FOR $5000 FOR 2 YEARS IS $72.FIND THE RATE OF INTEREST p.a.?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
THE DIFFERENCE BETWEEN COMPOUND INTEREST AND SIMPLE INTEREST AT THE RATE FOR $5000 FOR 2 YEARS IS $72. FIND THE RATE OF INTEREST p.a.?
equation for simple interest is i = n * r * p
equation for compound interest is i = p * (1+ r)^n - p
you are being told that the difference between them after 2 years is equal to 72.
replace what you know in both equations.
i = n * r * p becomes i = 2 * r * 5000
i = p * (1 + r)^n - p becomes i = 5000 * (1 + r)^2 - 5000
the difference between them is 72.
compound interest will be greater than simple interest.
the equation you now want to solve is:
5000 * (1 + r)^2 - 5000 - (2 * r * 5000) = 72
you need to use this equation to solve for r.
simplify the equation to get:
5000 * (1 + r)^2 - 5000 - 10000 * r = 72
since (1 + r)^2 = r^2 + 2r + 1, then your equation becomes:
5000 * (r^2 + 2r + 1) - 5000 - 10000 * r = 72
distribute the multiplication to get:
5000 * r^2 + 10000 * r + 5000 - 5000 - 10000 * r = 72
combine like terms to get:
5000 * r^2 = 72
divide both sides of this equation by 5000 to get:
r^2 = 72 / 5000
take the square root of both sides of this equation to get:
r = plus or minus sqrt(72/5000) = .12
your interest rate is 12%.
with simple interest, i = n * r * p becomes i = 2 * .12 * 5000 which becomes i = 1200
with compound interest, i = p * (1 + r)^n - p becomes i = 5000 * (1.12)^2 - 5000 which becomes i = 6272 - 5000 which becomes i = 1272.
the difference between them is 1272 - 1200 = 72.
the solution is that r = .12 and it is confirmed to be good.
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