Then,
\[A = P \left( 1 + \frac{R}{100} \right)^n \]
\[774 . 40 = 640 \left( 1 + \frac{R}{100} \right)^2 \]
\[ \left( 1 + \frac{R}{100} \right)^2 = \frac{774 . 40}{640}\]
\[ \left( 1 + \frac{R}{100} \right)^2 = 1 . 21\]
\[ \left( 1 + \frac{R}{100} \right)^2 = \left( 1 . 1 \right)^2 \]
\[\left( 1 + \frac{R}{100} \right) = 1 . 1\]
\[\frac{R}{100} = 0 . 1\]
R = 10
Thus, the required rate of interest is 10 % per annum.
Given details are,
Principal = Rs 640
Amount = Rs 774.40
Time = 2 years
Let rate = R%
By using the formula,
A = P (1 + R/100)^n
774.40 = 640 (1 + R/100)^2
(1 + R/100)^2 = 774.40/640
(1 + R/100)^2 = 484/400
(1 + R/100)^2 = (22/20)^2
By cancelling the powers on both sides,
(1 + R/100) = (22/20)
R/100 = 22/20 – 1
= (22-20)/20
= 2/20
= 1/10
R = 100/10
= 10%
∴ Required Rate is 10% per annum
At what rate per cent will a sum of 64000 be compounded to 68921 in three years?
64000`<br>Amount `A=Rs. 68921`<br>Rate `R=5% `per annum or `5/2` per half-yearly<br>`A=P{1+R/(2times100)}^n`<br>`68921=64000(1+5/200)^(2n)`<br>`68921/64000=(41/40)^(2n)`<br>`(41/40)^3=(41/40)^(2n)`<br>On comparing both the sides, we get:<br>`3=2n`<br>`n=3/2years=1 1/2 `years<br>`therefore` The time `=1 1/2` years.
What is the rate of percent per annum?
Rate = [(100 x P)/ (P x 8)]% = 12.5% per annum.