A sum of ₹12,000 is invested for 2 years at a certain rate of interest, compounded annually. Had it been invested at a rate 2% higher, it would have earned ₹ 504 more. The original rate of interest is

Correct option is B

Given:
Principal (P) = ₹12000
Time (T) = 2 years
Difference in Compound Interest = ₹504
Difference in rate = 2%
Formula Used:
CI = $$P(1 + \frac{R}{100})^n - P$$​​
Difference = $$P(1 + \frac{R+2}{100})^2 - P(1 + \frac{R}{100})^2$$​​
Solution:
Let the original rate be R.
​$$12000 [ (1 + \frac{R+2}{100})^2 - (1 + \frac{R}{100})^2 ] = 504$$​​

​$$(1 + \frac{R+2}{100})^2 - (1 + \frac{R}{100})^2 = 0.042$$​​

Use the algebraic identity $$a^2 - b^2 = (a-b)(a+b):$$​

​$$(1 + \frac{R+2}{100} - 1 - \frac{R}{100})(1 + \frac{R+2}{100} + 1 + \frac{R}{100}) = 0.042$$​

​$$(\frac{2}{100})(2 + \frac{2R+2}{100}) = 0.042$$​

​$$2 + \frac{2R+2}{100} = 2.1$$​

​$$\frac{2R+2}{100} = 0.1$$​

2R + 2 = 10

2R = 8

R = 4
The original rate of interest is 4%.
Final Answer
So the correct answer is (b)