Question
The compound interest on a certain sum of money at a certain rate for 2 years is Rs.40.80 and the simple interest on the same sum is Rs.40 at the same rate and for the same time. The rate of interest is
Answer: Option C
Answer: (c)Let the principal be P and rate of interest be r per cent per annum. Then,C. I = P$[(1 + r/100)^2 - 1]$40.80 = P$[(1 + r/100)^2 - 1]$...(i)S.I. = ${P.r.t}/100 ⇒ 40 = {Pr × 2}/100$ ...(ii)${40.80}/40 = P[(1 + r/100)^2 - 1]/{{2Pr}/100}$ ⇒ 1.02= $100/{2r}[1 + r^2/10000 + {2r}/100 - 1]$1.02 = $r/200$ +1$r/200$ = 1.02 - 1r = 0.02 × 200 = 4% per annum.Using Rule 10,Here, C.I. = Rs.40.80, S.I. = Rs.40, R = ?C.I.= S.I.$(1 + R/200)$40.80 = 40$(1 + R/200)$$4080/4000 = 1 + R/200$$408/400 = {200 + R}/200$408 = 400 + 2R2R = 8 ⇒ R = 4%
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Answer: (c)Let the principal be P and rate of interest be r per cent per annum. Then,C. I = P$[(1 + r/100)^2 - 1]$40.80 = P$[(1 + r/100)^2 - 1]$...(i)S.I. = ${P.r.t}/100 ⇒ 40 = {Pr × 2}/100$ ...(ii)${40.80}/40 = P[(1 + r/100)^2 - 1]/{{2Pr}/100}$ ⇒ 1.02= $100/{2r}[1 + r^2/10000 + {2r}/100 - 1]$1.02 = $r/200$ +1$r/200$ = 1.02 - 1r = 0.02 × 200 = 4% per annum.Using Rule 10,Here, C.I. = Rs.40.80, S.I. = Rs.40, R = ?C.I.= S.I.$(1 + R/200)$40.80 = 40$(1 + R/200)$$4080/4000 = 1 + R/200$$408/400 = {200 + R}/200$408 = 400 + 2R2R = 8 ⇒ R = 4%
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