Compound Interest Quiz Set 014

Let P, A, r and n have their usual meanings. For the first deposit n = 2, and for the second deposit n = 1. So total amount is P × $((1 + r/100)^2 + (1 + r/100))$ = $P/10000$ × $((100 + r)^2 + 100(100 + r))$ = $P/10000 × (100 + r)$ × $(100 + r + 100)$ which equals ${P × (100 + r) × (200 + r)}/10000.$ Putting r = 7 and P = 60000 and cancelling 10000, we get 6 × 107 × 207 = Rs. 132894. Please note that the rate of interest will be 1/2 because the compounding is half yearly.

Let P, A, r and n have their usual meanings. For the first deposit n = 2, and for the second deposit n = 1. So total amount is P × $((1 + r/100)^2 + (1 + r/100))$ = $P/10000$ × $((100 + r)^2 + 100(100 + r))$ = $P/10000 × (100 + r)$ × $(100 + r + 100)$ which equals ${P × (100 + r) × (200 + r)}/10000.$ Putting r = 7 and P = 30000 and cancelling 10000, we get 3 × 107 × 207 = Rs. 66447. Please note that the rate of interest will be 1/2 because the compounding is half yearly.

The amount is 7000000 + 1103375. So 8103375 = 7000000 × $(105/100)^n$. So $8103375/7000000$ = $(105/100)^n$, which can be put in the form $(105/100)^3$ = $(105/100)^3$, so n = 3 years.

If d is the difference, r is the rate and P is the principal, then the shortcut formula for the difference between compound and simple interest over a period of 2 years is d = P × $(r/100)^2$. So rate = 100 × $√{d/P}$ = 100 × $√{8/20000}$ = 2%.

The amount is 90000 + 3636. So 93636 = 90000 × $(102/100)^n$. So $93636/90000$ = $(102/100)^n$, which can be put in the form $(102/100)^2$ = $(102/100)^n$, so n = 2 years.