Question

Mr Prabhakar opens a recurring deposit account of rupees 300 per month at 8% simple interest per annum. on maturity he gets Rs 9930. find the period for which he continued with the account. ​

Answer 1

Answer:

Time period = 4.5 yrs

Step-by-step explanation:

SI = 300per month

= 3600 per annum

P = 9930

R = 8% p.a.

T = ?

SI = (P×R×T)/100

=>3600 = (9930×8×T)/100

=>3600 = 79440×T /100

=>3600 = 7944×T/10

=> T= 3600×10/7944

=> T= 4.5

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Answer 2

Given:

Mr Prabhakar opens a recurring deposit account of rupees 300 per month at 8% simple interest per annum. on maturity he gets Rs 9930.

To find:

The period for which he continued with the account. ​

Solution:

Let the number of months Mr. Prabhakar deposited be "n" months

Mr Prabhakar opens a recurring deposit account of rupees 300 per month ⇒ 300n

Number of months of interest  = n(n + 1)/2

Principle per month = 300 × n(n + 1)/2 = 150n(n + 1)

Interest per month at 8%

SI = PTR / 100

= 150n(n + 1) × 1 × 8 / 100

= n(n + 1)

Since, the total amount of maturity is Rs 9930

300n + SI = 9930

300n + n(n + 1) = 9930

300n + n² + n = 9300

n² + 301n - 9930 = 0

(n + 331) (n - 30) = 0

n = -331, 30

∴ n = 30

The period for which Mr Prabhakar continued with the account

= 30 months