Respuesta :
Solution :
Given :
Principal amount, P = Rs. 1000
Time period = 12 months
The maturity value = Rs. 12,715
We know that,
[tex]$ SI = \frac{PTR}{100}$[/tex]
[tex]$SI = 1000 \times \frac{n(n+1)}{2 \times 12} \times \frac{R}{100}$[/tex]
[tex]$SI = 1000 \times \frac{12(12+1)}{2 \times 12} \times \frac{R}{100}$[/tex]
SI = 65 R
So we know,
maturity value = principal amount + SI
12715 = 1000 + 65 R
65 R = 12715 - 1000
65 R = 11715
R = 18%
So the rate is 18%
Answer:
[tex]R=11\%[/tex] p.a.
Step-by-step explanation:
Given:
the principal amount deposited each month, [tex]P=Rs. 1000[/tex]
amount after maturity of one year, [tex]A=Rs. 12715[/tex]
We have the formula as:
[tex]I=\frac{PR}{100}\times\frac{T(T+1)}{2\times 12}[/tex]
where:
R = rate of interest per annum
T = time in months
[tex]A-12P=\frac{PR}{100}\times\frac{T(T+1)}{2\times 12}[/tex] [since the principal is deposited each month]
[tex]715=\frac{1000\times R}{100}\times \frac{12\times 13}{24}[/tex]
[tex]R=11\%[/tex] p.a.
