Answer

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Hint: Use compound interest formula which is given as $A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}-P$, where ‘r’ is interest in percentage, ‘n’ is time period in years, ‘P’ is principal amount and ‘A’ is amount after ‘n’ years.

Complete step-by-step answer:

We know that the amount ‘A’ at the end of ‘n’ years at the rate of r% per annum when the interest is compounded annually is given by,

$A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}...........\left( i \right)$

Where P = Initial amount or principal amount that increases by r % annually.

Here, we have 12000 as the principal amount which is increasing at a rate of 12% for the time period of 10 years.

Hence, from the given equation, we have

P = 12000 Rs

r = 12%

n = 10 years

Now, using the equation, we can get amount ‘A’ after 10 years as,

$\begin{align}

& A=12000{{\left( 1+\dfrac{12}{100} \right)}^{10}} \\

& A=12000{{\left( 1+\dfrac{3}{25} \right)}^{10}} \\

\end{align}$

Now, taking LCM inside the bracket, we get

$A=12000{{\left( \dfrac{28}{25} \right)}^{10}}...........\left( ii \right)$

Now, we can solve equation (ii) by multiplying $\dfrac{28}{25}$ to 10 times. But this process would be very lengthy for calculating the value of ‘A’.

So, we can use logarithm and antilogarithm to get value of A as follows

Taking log to both sides to equation (ii), we get.

$\log A=\log \left( \left( 12000 \right){{\left( \dfrac{28}{25} \right)}^{10}} \right)..........\left( iii \right)$

We can use identity of logarithm as

log ab = log a + log b

Applying above identity with equation (iii) we get,

$\log A=\log 12000+\log {{\left( \dfrac{28}{25} \right)}^{10}}$

Now, we can use another identity of ‘log’ as

$\log {{m}^{n}}=n\log m$

So, we get

$\log A=\log 12000+10\log \dfrac{28}{25}........\left( iv \right)$

We have another identity of logarithm

$\log \left( \dfrac{a}{b} \right)=\log a-\log b$

Hence, equation (iv) can be given as

$\log A=\log 12+\log 1000+10\left( \log 28-log25 \right)$

Now, we get values of logarithms of above by using logarithm table as

log 28 = 1.4472

log 25 = 1.3979

log 12 = 1.0792

log 1000 = $\log {{10}^{3}}$ = 3log 10 = 3

Hence, on putting values in above equation, we get

log A = 4.0792 + 0.493

log A = 4.5722

Taking antilog to both sides, we get

A = antilog (4.5722) = 37350

So, the amount after 10 years is Rs 37350.

As we know compound interest can be given by relation.

Compound interest = Total amount after interest – principal value/amount.

So, compound interest after 10 years is,

37350 – 12000 = Rs.25350

Note: One may apply the formula of simple interest as $\dfrac{P\times R\times T}{100}$ where ‘P’ is principal amount, ‘R’ is interest and ‘T’ is time period but that has become wrong. So, be clear with both terms i.e simple and compound interest. One can answer compound interest as ‘A’ i.e. total amount after 10 years. So, don’t confuse compound interest and total amount.

Complete step-by-step answer:

We know that the amount ‘A’ at the end of ‘n’ years at the rate of r% per annum when the interest is compounded annually is given by,

$A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}...........\left( i \right)$

Where P = Initial amount or principal amount that increases by r % annually.

Here, we have 12000 as the principal amount which is increasing at a rate of 12% for the time period of 10 years.

Hence, from the given equation, we have

P = 12000 Rs

r = 12%

n = 10 years

Now, using the equation, we can get amount ‘A’ after 10 years as,

$\begin{align}

& A=12000{{\left( 1+\dfrac{12}{100} \right)}^{10}} \\

& A=12000{{\left( 1+\dfrac{3}{25} \right)}^{10}} \\

\end{align}$

Now, taking LCM inside the bracket, we get

$A=12000{{\left( \dfrac{28}{25} \right)}^{10}}...........\left( ii \right)$

Now, we can solve equation (ii) by multiplying $\dfrac{28}{25}$ to 10 times. But this process would be very lengthy for calculating the value of ‘A’.

So, we can use logarithm and antilogarithm to get value of A as follows

Taking log to both sides to equation (ii), we get.

$\log A=\log \left( \left( 12000 \right){{\left( \dfrac{28}{25} \right)}^{10}} \right)..........\left( iii \right)$

We can use identity of logarithm as

log ab = log a + log b

Applying above identity with equation (iii) we get,

$\log A=\log 12000+\log {{\left( \dfrac{28}{25} \right)}^{10}}$

Now, we can use another identity of ‘log’ as

$\log {{m}^{n}}=n\log m$

So, we get

$\log A=\log 12000+10\log \dfrac{28}{25}........\left( iv \right)$

We have another identity of logarithm

$\log \left( \dfrac{a}{b} \right)=\log a-\log b$

Hence, equation (iv) can be given as

$\log A=\log 12+\log 1000+10\left( \log 28-log25 \right)$

Now, we get values of logarithms of above by using logarithm table as

log 28 = 1.4472

log 25 = 1.3979

log 12 = 1.0792

log 1000 = $\log {{10}^{3}}$ = 3log 10 = 3

Hence, on putting values in above equation, we get

log A = 4.0792 + 0.493

log A = 4.5722

Taking antilog to both sides, we get

A = antilog (4.5722) = 37350

So, the amount after 10 years is Rs 37350.

As we know compound interest can be given by relation.

Compound interest = Total amount after interest – principal value/amount.

So, compound interest after 10 years is,

37350 – 12000 = Rs.25350

Note: One may apply the formula of simple interest as $\dfrac{P\times R\times T}{100}$ where ‘P’ is principal amount, ‘R’ is interest and ‘T’ is time period but that has become wrong. So, be clear with both terms i.e simple and compound interest. One can answer compound interest as ‘A’ i.e. total amount after 10 years. So, don’t confuse compound interest and total amount.

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