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(i) A=P(1+(........)/(100))^(n)...

(i) `A=P(1+(........)/(100))^(n)`

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To solve the question \( A = P \left( 1 + \frac{...}{100} \right)^n \), we need to identify what the blank space represents in the context of compound interest. ### Step-by-Step Solution: 1. **Understand the formula**: We start with the formula for compound interest, which is given by: \[ A = P \left( 1 + \frac{R}{100} \right)^n \] where: - \( A \) is the amount after time \( n \), - \( P \) is the principal amount (initial investment), - \( R \) is the rate of interest per annum, - \( n \) is the number of years the money is invested or borrowed. 2. **Identify the components**: In the given formula \( A = P \left( 1 + \frac{...}{100} \right)^n \): - The blank space represents the rate of interest \( R \). 3. **Fill in the blank**: Therefore, we can fill in the blank with \( R \): \[ A = P \left( 1 + \frac{R}{100} \right)^n \] 4. **Conclusion**: The complete formula with the filled blank is: \[ A = P \left( 1 + \frac{R}{100} \right)^n \] ### Final Answer: The blank should be filled with \( R \) (the rate of interest per annum). ---
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Formula 1 Let P be the principal and the rate of interest be R% per annum.If the interest is compunded anually then the amount A and the compound interest C.I.at the end of n years are given by nA=P(1+(R)/(100))^(n) and C.I.=A-P=P{(1+(R)/(100))^(n)-1} respectively.

Formula 2 Let P be the principal and the rate of interest be R% per annum.If the interest is compounded annually then the amount A and the compound interest C.I.at the end of n years are given by A=P(1+(R)/(100k))^(nk) and C.I.A-P=P{((1_(R))/(100k))^(nk)-1} respectively

Knowledge Check

  • Directions for questions : Select the correct answer from the given options. The compound interest on a certain sum is given by C.I. = P (1+ (R)/( 100) )^(n)-P . Find C.I. when P= Rs. 1000, R = 10% p.a., and n=2 The following steps are involved in solving the above problem. Arrange them in sequential order. A therefore C.I. = Rs 210 B 1000 ((11)/(10)) ((11)/(10)) - 1000 = 1210 - 100 C Given CI = P (1+ (R)/( 100) )^(n) - P, P = Rs. = 1000, R = 10% p.a., and n=2 . D C.I. = 1000 + (1+ (10)/(100) )^2 - 1000

    A
    BCDA
    B
    DCBA
    C
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    D
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  • The compound interest on a certain sum is given by C.I.=P(1+(R)/(100))^(n)-P . Find C.I. when P=rs1000,R=10% P.a., and n=2. The following steps are involved in solving the above problem. Arrange them in sequential order. (A) therefore C.I.=rs210 (B) 1000((11)/(10))((11)/(10))-1000=1210-100 (C) Given Cl=P(1+(R)/(100))^(n)-P,P=rs1000 , R=10% p.a., and n=2 (D) C.I. = 1000+(1+(10)/(100))^(2)-1000

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    BCDA
    B
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    C
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    D
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    Statement I: If the annual increase in the population be 20% and present population be 10,000 then the population after 4 years is 20,736 .Statement II: the original population of a location be p and the annual growth rate be r%. The population after n years is P(1+(r)/(100))^(n).

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