A person invested a sum of ₹18,600 at x% p.a. and another sum that is twice the former at (x + 2) % p.a., both at simple interest. If the total interest earned on both investments for $3\frac{1}{2}$years is ₹23,110.50, then the rate of interest p.a. on the second investment is:
Answer & explanation
Correct answer: option 4
Interest = \(\frac{Principal ×Rate × Time }{100}\)
Amount = Principal + Interest
Principal = Rs. 18600 , Time = \(\frac{7}{2}\) years
\(\frac{ 18600 × R × \(\frac{7}{2}\) }{100}\) + \(\frac{ 18600 × ( R + 2) × \(\frac{7}{2}\) }{100}\) = 23110.50
⇒ 1953 R = 23110.50 - 2604
⇒ R = \(\frac{20506.5 }{1953}\)
⇒ R = 10.5%