Question

Monika has a cumulative deposit account of Rs.400 per month at 10% per annum simple
interest. If she gets Rs. 30,100 at the time of maturity, then find the total time for which the
account was held. ​

Answer 1

Total time of investment was 5 years.

Step-by-step explanation:

Let the account be held for n months, then by using the formula :

$I=P\times\frac{n(n+1)}{2\times 12}\times\frac{r}{100}$

where P = money deposited per month = 400 Rs.

           r = rate of interest per annum = 10%

Now put the values,

$I=400\times\frac{n(n+1)}{2\times 12}\times \frac{10}{100}$

$I=\frac{5n(n+1)}{3}$

Total money deposited by Monika = 400 × n = 400n

Maturity amount = deposited money + interest

               $30,100=400n+\frac{5n(n+1)}{3}$

               $30,100=\frac{1200n+5n(n+1)}{3}$

               $30,100=\frac{5n^2+1205n}{3}$

5n² + 1205n - 90,300 = 0

n² + 241n -180,60 = 0

(n-60) (n+301) = 0

n = 60, -301    [ n can not be negative ]

n = 60 months or 5 years

Total time of investment was 5 years.

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Answer 2

Given : A man deposits Rs. 400 per month at 10% p.a in a recurring deposit account and gets R.s.30100 at the time of maturity.

To Find : the total time for which the account is held.

Solution:

Man Deposit Rs 400 per month

Account held for n months

Amount deposited = 400 n

Interest Earned on 1st deposit amount =  400 x 10 *(n/12)/100  = (10/3)n

Interest Earned on 2nd deposit amount =  400 x 10 *((n-1)/12)/100  = (10/3)(n-1)

Interest Earned on nth deposit amount =  400 x 10 *(1/12)/100  = (10/3)

Total amount = 400n  + (10/3)n +  (10/3)(n-1) + ... ... ..  + (10/3)   = 30100

= 400n + (10/3)(n)(n+ 1)/2 = 30100

=> 1200n + 5n(n + 1)  = 90300

=>240n + n² + n = 18060

=> n² + 241n - 18060 = 0

=> n² +  301n - 60n - 18060 = 0

=> (n + 301)(n - 60) = 0

=> n = 60

60 months = 5 years

time for which the account is held. is 5 years

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