Question

11. Samita has a recurring deposit account in a bank of 2000 per month at the rate of
10% p.a. If she gets *83100 at the time of maturity, find the total time for which the
account was held.
Hint
Let the account be held for x months, then
2000 x + 2000 x
x(x + 1) 10
X . = 83100.
212 100​

Answer 1

Answer:

36 months or 3 years

Step-by-step explanation:

Given Data:

P = Rs. 2000

R = 10% p.a

N = ? ( We need to find )

M.V = Rs. 83100

We know,

MV = Pn + I

I = MV - Pn

I = 83100 - (2000 x n)

We knows,

I = $Pn(n+1)r\over2400$

thus,

83100 - 2000n = $Pn(n+1)r\over2400$

83100 - 2000n = $2000n(n+1)10\over2400$

83100 - 2000n = $100n(n+1)\over12$

12 * ( 83100 - 2000n ) = 100n² + 100n

997200 - 24000n = 100n² + 100n

100n² + 100n + 24000n - 997200 = 0

...Dividing the whole equation by 100...

n² + 241n - 9972 = 0

n² - 36n + 277n - 9972 = 0

n( n - 36 ) + 277( n - 36 ) = 0

Thus,

Either n - 36 = 0

or, n + 277 = 0

This means

n = 36

or

n = -277 (not possible)

Thus n = 36 months or 3 years