Question

4. (a) Mr Prabhakar opens a recurring deposit account of 300 per month at 8% simple interest per
annum. On maturity, he gets 9,930. Find the period for which he continued with the account.

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Answer 1

Answer:

Given :

formula : FV = PV [ (1+i)ñ-1 / i ]

FV = 9930

PV = 300

i = 8/100 = 0.08/12 = 0.00666

n = ??

Answer :

9930 = 300 [ (1.0066)n - 1 / 0.0066 ]

9930/300 = [ (1.0066)n - 1 / 0.0066 ]

33.1 × 0.0066 = (1.0066)ñ -1

0.2317 + 1 = (1.0066)*n

(1.0066) ñ = 1.2317

n = 30

key = type 1.007 in calculater and multiple again again and when u get 1.23xx notice the step and minese 1

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