Compound Interest Quiz Set 002

Let SI, P, r, t have usual meanings. Then, for 2 years, SI = (P × r × 2)/100. So P = $(50 × SI)/r$. The compound interest for 2 years by shortcut formula is ${P × r × (200 + r)}/10000$. Putting P here, it becomes, ${{(50 × SI)/r} × r × (200 + r)}/10000$ = ${SI × (r + 200)}/200$ = ${7000 × (6 + 200)}/200$ = Rs. 7210.

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 = 9 and P = 60000 and cancelling 10000, we get 6 × 109 × 209 = Rs. 136686. 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 = 9 and P = 20000 and cancelling 10000, we get 2 × 109 × 209 = Rs. 45562.

The amount is 2000000 + 185454. So 2185454 = 2000000 × $(103/100)^n$. So $2185454/2000000$ = $(103/100)^n$, which can be put in the form $(103/100)^3$ = $(103/100)^3$, so n = 3 years.

In this case r = $16/4$% and n = 3 because compounding is quarterly, and in 9 months there are three quarters. So A = 3000000 × $(1 + 4/100)^3$, which equals 3 × 104 × 104 × 104, i.e., Rs. 3374592.