£2000 is invested at x% interest over three years. At the end of three years the money has grown to £2963. To the nearest percentage find the rate of interest.
If you could help me out I would greatly appreciate it. I've tried to figure it out numerous ways and have failed. Thanks. :)
If simple interest,
2000(1+3r) = 2963
r = 16%
If annually compounded interest,
2000(1+r)^3 = 2963
r = 14%
Thanks! :)
Of course, I'd be happy to help you solve this problem!
To find the rate of interest, we'll need to use the formula for calculating compound interest:
\[ A = P \times \left(1 + \frac{r}{100}\right)^n \]
where:
- A is the final amount (in this case, £2963)
- P is the principal amount (initial investment) (£2000)
- r is the rate of interest (which we're trying to find)
- n is the number of compounding periods (in this case, 3 years)
We're given the values of A and P, and we know that n is 3 years. We need to solve for r.
Rearranging the formula, we get:
\[ \left(1 + \frac{r}{100}\right)^n = \frac{A}{P} \]
Substituting the given values, we have:
\[ \left(1 + \frac{r}{100}\right)^3 = \frac{2963}{2000} \]
To solve for r, we need to isolate the (1 + r/100) term. Taking the cube root of both sides, we have:
\[ 1 + \frac{r}{100} = \left(\frac{2963}{2000}\right)^{\frac{1}{3}} \]
Now we can solve for r by subtracting 1 and multiplying by 100:
\[ r = \left(\left(\frac{2963}{2000}\right)^{\frac{1}{3}} - 1\right) \times 100 \]
Let's calculate this:
\[ r = \left(\left(\frac{2963}{2000}\right)^{\frac{1}{3}} - 1\right) \times 100 \approx 9.48 \]
So, the approximate rate of interest, to the nearest percentage, is 9.48%.
I hope this helps! Let me know if you have any further questions.