Let principal = P, rate of interest = r%
CI earned at the end of three years = P(1 + r)^{3} - P = 1456
=> P(3r^{2} + 3r + r^{3}) = 1456
CI earned at the end of two years = P(1 + r)^{2} - P = 880
=> P(r^{2} + 2r) = 880
Dividing one by the other we get:
We can cancel â€˜râ€™ and solve the resulting quadratic. However, let us see if we can spot something in the numbers.
Remember that the amounts at the end of three years and two years are linked to (1 + r)^{3}and (1 + r)^{2} respectively.
Now observe that 1728 = 12^{3} and 144 is 12^{2}.
So,
=>r = 20%
CI at the end of 2 years = (1.2^{2} â€“ 1)P => 880 = 0.44P => P = Rs. 2000.
Alternatively:
(r^{2} + 3r + 3) x 440 = (r + 2) x 728
(r^{2} + 3r + 3) x 55 = (r + 2) x 91
55r^{2} + 165r + 165 = 91r + 182
55r^{2} + 74r -17 = 0
55r^{2} + 85r - 11r - 17 = 0
5r (11r + 17) -1 (11r + 17) = 0
r = 0.2 or a negative number. Or, r has to be 20%.
Correct Answer: Rs. 2,000