Karim deposited 9000 rupees after2years he withdraw 4000rs.at the end of 5yearshe get8300 RSVP. Find the rate of interest.
Pls solve the question.
On my "time graph" I picked year 5 as the focal-date
so 9000(1+i)^5 - 4000(1+i)^3 = 8300
let 1+i = x and divide each term by 100
90x^5 - 40x^3 = 83
There is no simple way to solve this equation, and you will
a method such as Newton's Method to solve it.
I went to Wolfram and got x = 1.08242
so i = .08242 or 8.242%
www.wolframalpha.com/input/?i=solve+90x%5E5+-+40x%5E3+%3D+83
check:
amount after 2 years = 90000(1.08242)^2 = 105446.975
withdraw 40000 leaves 65446.975
amount of that after another 3 years = 64446.975(1.08242)^3 = 82999.79
not bad eh?
To find the rate of interest, we need to use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount (8300 rupees)
P = the principal amount (9000 rupees)
r = the annual interest rate (unknown)
n = the number of times that interest is compounded per year (this information is not provided)
t = the number of years (5 years)
Since the compounding frequency (n) is not given, we will assume it is annually.
Now we can rewrite the formula as:
8300 = 9000(1 + r/1)^(1*5)
Simplifying the equation, we get:
8300/9000 = (1 + r)^5
Now, let's solve for (1 + r)^5:
(1 + r)^5 = 8300/9000
(1 + r)^5 ≈ 0.922
To find the value of (1 + r), we take the fifth root of 0.922:
(1 + r) ≈ ∛0.922 ≈ 0.971
Now, we can solve for r by subtracting 1 from both sides:
r ≈ 0.971 - 1 ≈ -0.029
The approximate interest rate is -0.029, or -2.9%.
Note: A negative interest rate suggests that the value of the investment would have decreased over time, which is unusual. It is important to consider if there might be any errors in the given information or calculations.