Compound Interest Practice Test Set 1

The amount after 3 years is ₹8,028 and after 6 years is ₹12,042 . The growth over the additional 3 years is 12,042/8,028 = 3/2.

Therefore, the amount after 3 years is 1.5 times the original principal: 8,028 = P $\times 3/2$.

The amount increases from ₹6,600 to ₹7,920 during the 3rd year. Interest for that year = 7,920 - 6,600 = ₹1,320 .

The 3rd-year rate is 1,320/6,600 $\times 100 = 20%$. Hence, the annual growth factor is 1.20.

The amount grows from year 3 to year 5, a period of 2 years. The growth factor is 19,326.12/15,972 = 1.21.

The annual growth factor is √1.21 = 1.10, so the annual rate is 10%.

Simple interest for 3 years is ₹7,200 . Therefore, yearly simple interest = 7,200/3 = ₹2,400 .

$Rate = (2,400/24,000) \times 100 = 10% per annum.$

For 3 years, the difference between compound interest and simple interest is P $\times (R/100)² \times (3 + R/100)$.

$Substitute R = 10%: 186 = P \times 0.1² \times 3.1 = P \times 0.031.$