Question
On what sum does the difference between the compound interest and the simple interest for 3 years at 10% is 31 ?
Answer: Option A
Answer: (a)Let the sum be xr = 10%, n = 3 yearsS.I. = ${x × r × n}/100$S.I.= ${x × 10 × 3}/100 = 3/10x$C.I.= $[(1 + r/100)^n - 1]x$= $[(1 + 10/100)^3 - 1]x$= $[(11/10)^3 - 1]x$$(1331/1000 - 1)x = 331/1000x$$331/1000x - 3/10x$ = 31or $({331 - 300})/1000x = 31$or $31/1000x$ = 31or x = 1000Sum = Rs.1000Using Rule 6,Here, C.I. - S.I. = Rs.31, R = 10%, T = 3 years, P = ?C.I. - S.I. = P × $(R/100)^2 × (3 + R/100)$31 = P × $(10/100)^2(3 + 10/100)$31 = P × $1/100 × 31/10$ ⇒ P = Rs.1000
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Answer: (a)Let the sum be xr = 10%, n = 3 yearsS.I. = ${x × r × n}/100$S.I.= ${x × 10 × 3}/100 = 3/10x$C.I.= $[(1 + r/100)^n - 1]x$= $[(1 + 10/100)^3 - 1]x$= $[(11/10)^3 - 1]x$$(1331/1000 - 1)x = 331/1000x$$331/1000x - 3/10x$ = 31or $({331 - 300})/1000x = 31$or $31/1000x$ = 31or x = 1000Sum = Rs.1000Using Rule 6,Here, C.I. - S.I. = Rs.31, R = 10%, T = 3 years, P = ?C.I. - S.I. = P × $(R/100)^2 × (3 + R/100)$31 = P × $(10/100)^2(3 + 10/100)$31 = P × $1/100 × 31/10$ ⇒ P = Rs.1000
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