
Mr. Nirav borrowed rupees Rs. $50000$ from the bank for $5$ years. The rate of interest is $9\% $ per annum compounded monthly. Find the payment he makes monthly if he pays back at the beginning of each month.
Answer
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Hint: When the payments are made at the beginning of each time period then it is known as the Annuity Due. They are also known as the Recurring Payments. The most common example of the recurring payment is the monthly rent on an apartment.
Complete step-by-step answer:
The money borrowed by Mr. Nirav $P = 50000$
The time period for the money borrowed $n = 5{\rm{ years}}$
And the rate of interest
$
r = 9\% {\rm{ per annum}}\\
r = \dfrac{9}{{100}}\\
r = 0.09
$
Let us assume the amount paid by Mr. Nirav at the beginning of each month is $a$.
Then, the present value of an annuity due when payments are made at the beginning of each month is given by this formula,
$P = \dfrac{{12a}}{r}\left( {1 + \dfrac{r}{{12}}} \right)\left[ {1 - {{\left( {1 + \dfrac{r}{{12}}} \right)}^{ - 12n}}} \right]$
Substituting the values in the formula we get,
$
50000 = \dfrac{{12a}}{{0.09}}\left( {1 + \dfrac{{0.09}}{{12}}} \right)\left[ {1 - {{\left( {1 + \dfrac{{0.09}}{{12}}} \right)}^{ - 12 \times 5}}} \right]\\
50000 = \dfrac{a}{{0.0075}}\left( {1 + 0.0075} \right)\left[ {1 - {{\left( {1 + 0.0075} \right)}^{ - 60}}} \right]\\
375 = a\left( {1.0075} \right)\left[ {1 - {{\left( {1.0075} \right)}^{ - 60}}} \right]\\
375 = a\left[ {1.0075 - {{\left( {1.0075} \right)}^{ - 59}}} \right]
$
We can assume
${\left( {1.0075} \right)^{ - 59}} = x$
Then, taking logarithm of both sides we get,
\[
\log x = - 59\log \left( {1.0075} \right)\\
\log x = - 59\left( {0.0032} \right)\\
\log x = - 0.1914
\]
Solving this we get,
$x = 0.6434$
Substituting this value in the expression we get,
$
375 = a\left[ {1.0075 - 0.6434} \right]\\
375 = a\left[ {0.3641} \right]\\
a = \dfrac{{375}}{{0.3641}}
$
Solving this we get,
$a = 1029.94$
Therefore, the payments made by Mr. Nirav at the beginning of each month is \[{\rm{Rs}}{\rm{.\;}}1029.94\].
So, the correct answer is “Option C”.
Note: It should be noted that in the formula used to calculate the annuity due we have multiplied the value of payment at the beginning of each month $a$ with 12 because in a one year of time period there are 12 months and the payment is made at the beginning of each of these months so the total payment in one year would be $12a$. Similarly, when calculating the rate of interest we have divided the rate $r = 0.09$ with 12 because the rate of interest given is for one year and we have to put the value of the rate for a month, that is why the rate of interest per month is written as $\dfrac{{0.09}}{{12}}$ in the formula.
Complete step-by-step answer:
The money borrowed by Mr. Nirav $P = 50000$
The time period for the money borrowed $n = 5{\rm{ years}}$
And the rate of interest
$
r = 9\% {\rm{ per annum}}\\
r = \dfrac{9}{{100}}\\
r = 0.09
$
Let us assume the amount paid by Mr. Nirav at the beginning of each month is $a$.
Then, the present value of an annuity due when payments are made at the beginning of each month is given by this formula,
$P = \dfrac{{12a}}{r}\left( {1 + \dfrac{r}{{12}}} \right)\left[ {1 - {{\left( {1 + \dfrac{r}{{12}}} \right)}^{ - 12n}}} \right]$
Substituting the values in the formula we get,
$
50000 = \dfrac{{12a}}{{0.09}}\left( {1 + \dfrac{{0.09}}{{12}}} \right)\left[ {1 - {{\left( {1 + \dfrac{{0.09}}{{12}}} \right)}^{ - 12 \times 5}}} \right]\\
50000 = \dfrac{a}{{0.0075}}\left( {1 + 0.0075} \right)\left[ {1 - {{\left( {1 + 0.0075} \right)}^{ - 60}}} \right]\\
375 = a\left( {1.0075} \right)\left[ {1 - {{\left( {1.0075} \right)}^{ - 60}}} \right]\\
375 = a\left[ {1.0075 - {{\left( {1.0075} \right)}^{ - 59}}} \right]
$
We can assume
${\left( {1.0075} \right)^{ - 59}} = x$
Then, taking logarithm of both sides we get,
\[
\log x = - 59\log \left( {1.0075} \right)\\
\log x = - 59\left( {0.0032} \right)\\
\log x = - 0.1914
\]
Solving this we get,
$x = 0.6434$
Substituting this value in the expression we get,
$
375 = a\left[ {1.0075 - 0.6434} \right]\\
375 = a\left[ {0.3641} \right]\\
a = \dfrac{{375}}{{0.3641}}
$
Solving this we get,
$a = 1029.94$
Therefore, the payments made by Mr. Nirav at the beginning of each month is \[{\rm{Rs}}{\rm{.\;}}1029.94\].
So, the correct answer is “Option C”.
Note: It should be noted that in the formula used to calculate the annuity due we have multiplied the value of payment at the beginning of each month $a$ with 12 because in a one year of time period there are 12 months and the payment is made at the beginning of each of these months so the total payment in one year would be $12a$. Similarly, when calculating the rate of interest we have divided the rate $r = 0.09$ with 12 because the rate of interest given is for one year and we have to put the value of the rate for a month, that is why the rate of interest per month is written as $\dfrac{{0.09}}{{12}}$ in the formula.
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