Question

Samita has a recurring deposit account in a bank of Rs.2000, per month at a rate of 10% p.a. If she gets Rs.83100 at the time of maturity, find the total time for which the account was held.

Answer 1

Hint: According to given in the question we have to find the time for which the account was held as Samita has a recurring deposit account in a bank of 2000rs, per month at a rate of 10% p.a. If she gets 83100rs at the time of maturity, first of all we have to find the simple interest (S.I.) for the amount 2000rs at the rate of 10% p.a. with the help of the formula as given below:
Formula used:
Simple Interest (S.I.)$ = P \times \dfrac{{n(n + 1)}}{2} \times \dfrac{1}{{12}} \times \dfrac{R}{{100}}$ ..............(a)
Where, n is the time, P is the recurring deposit amount, and R is the rate of interest.
After finding the simple interest now, we will obtain quadratic equation in form of n and now we have to find the value of maturity (M.V.) with the help of the formula as given below:
$MV = P \times n + S.I$ ..................(b)
Where, P is the amount at the time of maturity, n is the time, and S.I. is the simple interest.
So, on placing all the values in the formula (b) we will obtain a quadratic expression in form of n now, we have to solve the obtained quadratic expression to obtain the value of n, which is the total time for which the account was held.

Complete step by step answer:
Given,
Recurring deposit amount (P) = 2000rs
Rate of interest (R) = 10% per annum
Matured value (MV) = 83100rs
Step 1: As given in the question, to find the total time for which the account was held for Samita who has a recurring deposit account in a bank of 2000rs, per month at a rate of 10% p.a. If she gets 83100rs at the time of maturity, first of all we have to find the simple interest with the help of the formula (a) as mentioned in the solution hint. On substituting all the values in formula (a)
Simple Interest (S.I.)$ = 2000 \times \dfrac{{n(n + 1)}}{2} \times \dfrac{1}{{12}} \times \dfrac{{10}}{{100}}$
On solving the obtained expression just above,
S.I.$ = 100 \times \dfrac{{n(n + 1)}}{{12}}$
S.I.$ = 25 \times \dfrac{{{n^2} + n}}{3}$
Step 2: Now, as we have obtained the simple interest so we can use the formula (b) as mentioned in the solution hint to find the total time for which the account was held. On substituting the value of S.I. in formula (b)
$
   \Rightarrow 83100 = (2000 \times n) + \dfrac{{25}}{3}{n^2} + \dfrac{{25}}{3}n \\
   \Rightarrow 83100 = \dfrac{{6000n + 25{n^2} + 25n}}{3} \\
 $
Applying cross-multiplication in the expression as obtained just above,
$
   \Rightarrow 83100 \times 3 = 6000n + 25{n^2} + 25n \\
   \Rightarrow 25{n^2} + 6025n - 249300 = 0................(1) \\
 $
Step 3: As from the step 2 we have obtained a quadratic expression in form of n so to find the value of n which is the total time for which the account was held we have to solve this quadratic expression (1) by finding the roots but first of all we have to divide the whole expression (1) by 25.
Dividing with 25 on the both sides of the expression (1),
 $
   \Rightarrow \dfrac{{25{n^2} + 6025n - 249300}}{{25}} = 0 \\
   \Rightarrow {n^2} + 241n - 9972 = 0 \\
 $
On solving the obtained quadratic expression,
$
   \Rightarrow {n^2} + 277n - 36n - 9972 = 0 \\
   \Rightarrow n(n + 277) - 36(n + 277) = 0 \\
   \Rightarrow (n + 277)(n - 36) = 0 \\
 $
Step 4: Now, from the step 3 we can see that we have obtained to roots of the obtained quadratic expression so to find the value of n we have to solve both of the roots hence,
$
   \Rightarrow (n + 277) = 0 \\
   \Rightarrow n = - 277 \\
 $
Which is not possible because, number of days can’t be negative.
$
   \Rightarrow (n - 36) = 0 \\
   \Rightarrow n = 36 \\
 $
Hence, number of days $n = 36$ months

Hence with the help of the formulas (a) and (b) we have obtained the total time for which the account was held is 36 months or 3 years.

Note:
To find the total time for which the account was held it is necessary to find the simple interest on the amount which was deposited at a particular rate of interest for a particular time.
If the interest is given for per annum but the amount deposited per month then it is necessary to find the interest that should be applied for per month.
On solving the quadratic equation we know that only two roots would be obtained.