| Let the rate of interest be R % . 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.
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At what rate per cent will a sum of 640 be compounded to 774.40 in two years?
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