find the value of r ?

Question

if a man invests an amount of R10,000 for 1 year at an interest rate of r % and then re-invests his capital plus interest at double the initial interest rate for the second year, he receives a final amount of R12,862.

Answer 1

Let u = 100r.
Then:
(1 + u)(1 + 2u) = 1.2862
2u^2 + 3u - 0.2862 = 0
u = ( -3 +/- sqrt(9 + 8*0.2862) ) / 4
= ( -3 +/- 3.36 ) / 4
Discarding the negative root:
u = 0.36 / 4
= 0.09
r = 9%.

Answer 2

He starts with 10000.
After one year, he has 10000(1 + r/100), or after distributing:
10000 + 100r.
This amount is invested at 2r%, so after the second year he has:
(10000 + 100r)(1 + 2r/100). Set this equal to 12862 and solve for r.

(10000 + 100r)(1 + r/50) = 12862. Multiplying,
10000 + 300r + 2r² = 12862. Dividing by 2,
r² + 150r + 5000 = 6431. Subtracting 6431,
r² + 150r - 1431 = 0. Using the quadratic formula,
r = {-(150) ± √[(150)² - 4(1)(-1431)] } / [2(1)]
= {-150 ± √[22500 + 5724] } / [2]
= {-150 ± √[28224] } / [2]
= {-150 ± 168 } / [2]
= -75 ± 84.

Since a negative interest rate doesn't make sense,
r = -75 + 84 = 9.

Answer 3

What is your question? Also we need to know:
Is the interest compounded yearly?, monthly? weekly?, daily?, or continuously?