Math, asked by asimchaudhary7, 2 months ago

a man borrows 10000 rupees at a compound interest rate of 8 % if he pays 2000 rupees at the end of each year then find the sum outstanding at the end of 3rd year​

Answers

Answered by mathdude500
5

\large\underline{\sf{Solution-}}

Step :- 1

For the first year :-

  • Sum Borrowed, P = Rs 10000

  • Rate of interest, r = 8 percent per annum

  • Time, n = 1 year

We know,

Simple Interest and Compound interest on a certain sum of money Rs P invested at the rate of r % per annum for n years is same and given by

\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}

So,

\rm :\longmapsto\:SI = \dfrac{10000 \times 8\times 1}{100}

\rm :\longmapsto\:SI = 800

Hence,

Amount to be paid at the end of first year,

\rm :\longmapsto\:Amount = P + SI

\rm :\longmapsto\:Amount = 10000 + 800

\rm :\longmapsto\:Amount = Rs \: 10800

Now,

  • Man repay Rs 2000 at the end of the first year,

So,

  • Balance amount = 10800 - 2000 = Rs 8800

Step :- 2

For second year,

  • Sum, P = Rs 8800

  • Rate of interest, r = 8 percent per annum

  • Time, n = 1 year

We know,

Simple Interest and Compound interest on a certain sum of money Rs P invested at the rate of r % per annum for n years is same and given by

\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}

So,

\rm :\longmapsto\:SI = \dfrac{8800 \times 8 \times 1}{100}

\rm :\longmapsto\:SI = 704

Hence,

  • Amount to be paid at the end of second year,

\rm :\longmapsto\:Amount = P + SI

\rm :\longmapsto\:Amount = 8800 + 704

\rm :\longmapsto\:Amount =Rs \:  9504

Now,

  • Man repay Rs 2000 at the end of the second year,

So,

  • Balance amount = 9504 - 2000 = Rs 7504

Step :- 3

For the third year,

  1. Sum, P = Rs 7504

  • Rate of interest, r = 8 percent per annum

  • Time, n = 1 year

We know,

Simple Interest and Compound interest on a certain sum of money Rs P invested at the rate of r % per annum for n years is same and given by

\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}

So,

\rm :\longmapsto\:SI = \dfrac{7504 \times 8 \times 1}{100}

\rm :\longmapsto\:SI =600.32

Hence,

Amount to be paid at the end of third year,

\rm :\longmapsto\:Amount = P + SI

\rm :\longmapsto\:Amount = 7504 + 600.32

\rm :\longmapsto\:Amount = Rs \: 8104.32

Now,

  • Man repay Rs 2000 at the end of the third year,

So,

Balance amount = 8104.32 - 2000 = Rs 6104.32

Additional information :-

1. Amount on certain sum of Rs P invested at the rate of r % per annum compounded annually for n years is

\rm :\longmapsto\:Amount = P {\bigg(1 + \dfrac{r}{100} \bigg) }^{n}

2. Amount on certain sum of Rs P invested at the rate of r % per annum compounded semi - annually for n years is

\rm :\longmapsto\:Amount = P {\bigg(1 + \dfrac{r}{200} \bigg) }^{2n}

3. Amount on certain sum of Rs P invested at the rate of r % per annum compounded quarterly for n years is

\rm :\longmapsto\:Amount = P {\bigg(1 + \dfrac{r}{400} \bigg) }^{4n}

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