Question
The difference between compound interest (compounded annually) and simple interest on a certain sum of money at 10% per annum for 2 years is 40. The sum is :
Answer: Option B
Answer: (b)Let the principal be x.Compound interest= P$[(1 + R/100)^t - 1]$= $x[(1 + 10/100)^2 - 1]$= $x[(1.1)^2 - 1]$= x (1.21 - 1) = 0.21xSI = ${x × 2 × 10}/100 = x/5 = 0.2x$According to the question,0.21x - 0.2x = 40$0.01x = 40 ⇒ x = 40/{0.01}$ = Rs.4000Using Rule 6,Here, C.I. - S.I. = Rs.40, R = 10%, T = 2 years, P = ?C.I. - S.I. = P$(R/100)^2$40 = P$(10/100)^2$ ⇒ P = Rs.4000
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Answer: (b)Let the principal be x.Compound interest= P$[(1 + R/100)^t - 1]$= $x[(1 + 10/100)^2 - 1]$= $x[(1.1)^2 - 1]$= x (1.21 - 1) = 0.21xSI = ${x × 2 × 10}/100 = x/5 = 0.2x$According to the question,0.21x - 0.2x = 40$0.01x = 40 ⇒ x = 40/{0.01}$ = Rs.4000Using Rule 6,Here, C.I. - S.I. = Rs.40, R = 10%, T = 2 years, P = ?C.I. - S.I. = P$(R/100)^2$40 = P$(10/100)^2$ ⇒ P = Rs.4000
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