Question

(b)
A person invests 2500 per month in
a recurring deposit account and gets
31625 as maturity amount. If the rate
of interest is 10%, find the time in years
till the maturity of the account. [3]​

Answer 1

Answer:

1 year or 12 months is the answer

Step-by-step explanation:

Use formula

M.V. = Pxn + P x n(n+1)/2*12 * r/100

Equation will be n^2 + 241n - 3036 = 0

(n-12)(n+253)

thus n = 12 months or 1 year

Answer 2

Hey there,

Here's the answer to your question:

According to the question, a recurring deposit account is used.

Therefore, it is a case of compound interest.

We know, interest compounded annually is given by this formula -

$A = P(1+\frac{r}{n})^n^t$

where, A is interest, P is principal invested amount, r is rate of interest, n is number of times the principal amount is compounded and, t is the total frame of time the principal amount has been compounded.

From the question,

A = 31625 Rs

P = 2500 Rs

r = 10%

n = 12 (invested every month)

t = ?

Therefore,

$31625 = 2500(1 + \frac{10}{12} )^1^2^t\\\\\frac{31625}{2500} = (1+\frac{5}{6} )^1^2^t\\\\12.65 = (\frac{11}{6} )^1^2^t\\\\12.65 = 1.83^1^2^t\\\\log(1.83)^1^2^.^6^5 = 12t\\\\\frac{log(1.83)^1^2^.^6^5}{12} =t\\\\ \frac{\frac{log1.83}{log12.65}}{12} = t\\\\\frac{\frac{0.262}{1.1}}{12} = t\\\\\frac{0.24}{12} = t\\\\t = 0.02$

Therefore, time taken = 0.02 years

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