Question

Find the amount and compound interest on Rs.14,000 in 2 years when the
rates of interest for successive years are 6% and 10% respectively.


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Answer 1

Given :-

$\sf{ \: \: \bull \: \: Sum\:_{(money)}=14000}$

$\sf{ \: \: \bull \: \: Time\:_{(annually)}=2\: years}$

$\sf{ \: \: \bull \: \: Rate\:_{(interest)}=r_1=6\%}$

$\sf{ \: \: \bull \: \: Rate\:_{(interest)}=r_1=10\%}$

To Find :-

$\sf{ \: \: \bull \: \: Amount\:_{(after\:2\: years)}=?}$

$\sf{ \: \: \bull \: \: Interest\:_{(after\:2\: years)}=?}$

Formula Used :-

Let us consider a sum of Rs P invested at the successive rate of interest x % per annum for 'a' years and at the rate of y % for 'b' years, then

$\tt \: Amount, A =P { \bigg(1 + \dfrac{x}{100} \bigg) }^{a} { \bigg(1 + \dfrac{y}{100} \bigg) }^{b}$

$\large\underline{\bold{Solution-}}$

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

$\tt \: Amount, A =14000 \bigg(1 + \dfrac{6}{100} \bigg) \bigg(1 + \dfrac{10}{100} \bigg)$

$\tt \: Amount, A =14000 \times \dfrac{53}{50} \times \dfrac{11}{10}$

$\tt \: Amount, A = \: Rs \: 16324$

And

$\tt \:Compound \: Interest \: = Amount \: - \: Principal$

$\tt \therefore \: Compound \: Interest \: = 16324 - 14000$

$\tt \therefore \: Compound \: Interest \: = \: Rs \: 2324$

Additional Information :-

1. If rate of interest is compounded Half yearly or semi-annual or bi-annual, we have to change time and rate by multiply time by 2 and for changing rate we have to divide by 2.

2. If the rate of interest is compoundedquarterly, then we have to multiply by 4 in time and dividing by 4 in rate of interest.

Answer 2

Answer:

your ans is in the attachment above⬆️