8. Find the compound interest on Rs. 14000
in 2 years when rate of interest for
successive years are 6% and 10%
respectively. *
(3 Points)
Answers
Answer:
Let principal (p) = Rs. 100
For 1st year
P = Rs. 100
R = 10%
T = 1 year
straight I equals fraction numerator 100 cross times 10 cross times 1 over denominator 100 end fraction equals 10
A = 100 + 10 = Rs. 110
For 2nd year
P = Rs. 110
R = 11%
T = 1 year
A = 110 + 12.10 = Rs. 122.10
If Amount is Rs. 122.10 on a sum of Rs. = 100
If amount is Rs. 1, sum =
If amount is Rs. 6593.40, sum =
= Rs. 5400
Answer:
Given that
Principal, P = Rs 14, 000
Rate of interest = 6 % per annum for 1 first year
Rate of interest = 10 % per annum for second year.
So,
Amount is given by
\tt \: Amount, A =14000 \bigg(1 + \dfrac{6}{100} \bigg) \bigg(1 + \dfrac{10}{100} \bigg)Amount,A=14000(1+1006)(1+10010)
\tt \: Amount, A =14000 \times \dfrac{53}{50} \times \dfrac{11}{10}Amount,A=14000×5053×1011
\tt \: Amount, A = \: Rs \: 16324Amount,A=Rs16324
And
\tt \:Compound \: Interest \: = Amount \: - \: PrincipalCompoundInterest=Amount−Principal
\tt \therefore \: Compound \: Interest \: = 16324 - 14000∴CompoundInterest=16324−14000
\tt \therefore \: Compound \: Interest \: = \: Rs \: 2324∴CompoundInterest=Rs2324