Question
A sum of money invested at compound interest amounts to Rs.650 at the end of first year and Rs.676 at the end of second year. The sum of money is :
Answer: Option C
Answer: (c)Interest on Rs.650 for 1 year= 676 - 650 = Rs.26So, r = $26/650 × 100$r = 4% per annumP = $A/[1 + r/100]^t = 650/[1 + 4/100]^1$= $650/{26/25} = 650 × 25/26$ = Rs.625Using Rule 7(i),Here, b - a = 1B = Rs.676, A = Rs.650R% = $(B/A - 1)$ × 100%= $[676/650 - 1] × 100%$= $[{676 - 650}/650] × 100%$= $26/650 × 100% = 100/25$ = 4%Amount= P$(1 + R/100)^1$650 = P$(1 + 4/100)$P = ${650 × 100}/104$ = Rs.625Note : A sum at a rate of interest compounded yearly becomes Rs.$A_1$, in n years and Rs. $A_2$ in (n + 1) years, thenP = $A_1(A_1/A_2)^n$
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Answer: (c)Interest on Rs.650 for 1 year= 676 - 650 = Rs.26So, r = $26/650 × 100$r = 4% per annumP = $A/[1 + r/100]^t = 650/[1 + 4/100]^1$= $650/{26/25} = 650 × 25/26$ = Rs.625Using Rule 7(i),Here, b - a = 1B = Rs.676, A = Rs.650R% = $(B/A - 1)$ × 100%= $[676/650 - 1] × 100%$= $[{676 - 650}/650] × 100%$= $26/650 × 100% = 100/25$ = 4%Amount= P$(1 + R/100)^1$650 = P$(1 + 4/100)$P = ${650 × 100}/104$ = Rs.625Note : A sum at a rate of interest compounded yearly becomes Rs.$A_1$, in n years and Rs. $A_2$ in (n + 1) years, thenP = $A_1(A_1/A_2)^n$
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