Question

What annual payment will discharge a debt of Rs 7620 due in 3 years at $16\dfrac{2}{3}%$ per annum interest?
(a) 5430
(b) 4430
(c) 3430
(d) 2430

Answer 1

Hint: To solve this problem, we should know the basics of compound interest (such that it is compounded annually). This is given by the formula of total amount (A) as –
A = $P{{(1+r)}^{t}}$
Here, P = principal amount, r = interest rate (in decimals), t = time period (in years). We will use this to solve the problem.

Complete step-by-step answer:

Before solving the problem, let us introduce a few basic notions from the question in hand, which would help us to use the formula (A = $P{{(1+r)}^{t}}$) of compound interest effectively. Let the annual payment in consideration be x (thus, with reference to formula, x = principal amount). Since, the debt to be discharged is Rs 7620 in 3 years, we have,
$\dfrac{x}{(1+r)}+\dfrac{x}{{{(1+r)}^{2}}}+\dfrac{x}{{{(1+r)}^{3}}}=7620$ -- (1)
Here, each term represents the total amount paid for each year starting from the first year.
Since, r = $16\dfrac{2}{3}%=\dfrac{50}{3}%=\dfrac{50}{300}=\dfrac{1}{6}$ (here, perform the conversion from percent to decimals).
Thus, in (1), we have,
$\dfrac{x}{\left( 1+\dfrac{1}{6} \right)}+\dfrac{x}{{{\left( 1+\dfrac{1}{6} \right)}^{2}}}+\dfrac{x}{{{\left( 1+\dfrac{1}{6} \right)}^{3}}}=7620$
$\dfrac{6x}{7}+\dfrac{36x}{49}+\dfrac{216x}{343}=7620$
$\dfrac{294x+252x+216x}{343}=7620$
$\dfrac{762x}{343}=7620$
x = 3430
Thus, the annual payment that will discharge a debt of Rs 7620 due in 3 years at $16\dfrac{2}{3}%$ per annum interest is Rs 3430. Hence, the correct option is (c) 3430.

Note: While solving this problem, one should know that interest gets compounded annually. Basically, the amount of interest increases every year. Thus, if the interest is x for the first year, the interest would be higher than x for the second year. Thus, the problem cannot be solved by directly multiplying 3 to the interest amount of the first year (however, this can be done for simple interest cases).