priyanka has a recurring deposit account of rs 1000 per month at 10%per annum if she gets rs 5550 as interest at the time of maturity ,find the total time for which the account was held
Answers
Answer:
Step-by-step explanation:
Recurring deposit interest I is given by,
I = P×[ n(n+1)/(2×12) ] (r/100) .............. (1)
where P is monthly contribution, P = Rs. 1000
n = number of months,
r = rate of interest = 10%
and I = interest = Rs.5550
By substituting the values in eqn.(1), we get
1000×[ n(n+1)/24 ]×(10/100) = 5550 or n(n+1) = (24/100)×5550 = 1332
Therefore, n² + n - 1332 = (n+37) (n-36) = 0
n = -37 or n= 36
Since, n can't be negative.
Hence, the number of months = 36 or 3 yrs
Answer:
Step-by-step explanation:
Recurring deposit interest I is given by,
I = P×[ n(n+1)/(2×12) ] (r/100) .............. (1)
where P is monthly contribution, P = Rs. 1000
n = number of months,
r = rate of interest = 10%
and I = interest = Rs.5550
By substituting the values in eqn.(1), we get
1000×[ n(n+1)/24 ]×(10/100) = 5550 or n(n+1) = (24/100)×5550 = 1332
Therefore, n² + n - 1332 = (n+37) (n-36) = 0
n = -37 or n= 36
Since, n can't be negative.
Hence, the number of months = 36 or 3 yrs