Question

Mr. Sawarn Roy gets Rs.6455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly instàllment.​

Answer 1

Given information to us:

• Mr. Sawarn Roy gets RRs.6455.
• Time period = 1 year
• Rate = 14% p.a.

What is the need to calculate:

• Monthly instàllment?

Usage of formula:

• $\star \: \boxed{\mathrm{I \: = \: P \: \times \: \dfrac{n \: (n \: + \: 1)}{2 \: \times \: 12} } \: \times \: \dfrac{ \text{r}}{100} }$

Solving way:

»» First let us assume the P be Rs.100. Now we would apply all the values in the given formula of interest before that we would convert the year in number of months. After that we would calculate and solve them. Here after the interest which had been calculated would be multiplied by 100 that would have us the amount of money which had been deposited in 12 months.

»» Now we would calculate the maturity value by using the formula of calculating maturity value.

→ Sum deposited every month + Interest

After this we would use the unitary method and solve it.

Calculations:

Here using the formula of interest and substituting the required values in it where value of Principal (P) is Rs.100 and number of months n is 12.

➟ I = 100 × 12(12 + 1) / (2×12) × r/100

➟ 100 × 12 × 13 × 14 / 2 × 12 × 100

➟ 1200 × 13 × 14 / 1200 × 2

➟ 13 × 14 / 2

➟ 182 / 2

➟ 91

Now we would be getting the amount of money which had been deposited in 12 months by multiplying the sum deposited with number of months that is 12.

➟ 12 × 100

➟ 1200

After we got the money which had been deposited in 12 months we would calculate the maturity value by,

➟Total sum which had been deposited + Interest calculated.

Now,

➟ Rs.1200 + Rs.91

➟ Rs.1291

Usage of unitary method:

Maturity value (M.V.) is 6455 .

➟ (100 / 1291) × 6455

➟ 100 × 5

➟ 500

Thus,

• Monthly instàllment paid by Mr. Sawarn Roy is Rs.500.

Answer 2

$\large\underline{\sf{Given- }}$

Amount received after one year, A = Rs 6455

Rate of interest per annum, r = 14 %

Time = 1 year = 12 months.

$\large\underline{\sf{To\:Find - }}$

Amount of instâllment.

$\begin{gathered}\large{\rm{{\underline{Formula \: Used - }}}}  \end{gathered}$

We know that

Amount, A on instâllment of Rs P invested for n months at the rate of r % per annum is given by

$\blue\bigstar \: \red{\boxed{\mathbf{A \: = \: nP \: + \: \dfrac{n \: (n \: + \: 1)P}{24 } \times \dfrac{r}{100} }}}$

where,

A = Amount received on maturity or maturity value

P = Amount of monthly instâllment

r = rate of interest per annum

n = number of monthly instâllment.

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

Given that,

Amount received after one year, A = Rs 6455

Rate of interest per annum, r = 14 %

Time = 1 year = 12 months.

Let assume that amount of monthly instâllment be Rs x

So, using the formula, we get

$\rm :\longmapsto\: \: {{{A \: = \:n P \: + \: \dfrac{n \: (n \: + \: 1)P}{24 } \times \dfrac{r}{100} }}}$

On substituting the values, we get

$\rm :\longmapsto\: \: {{{6455 \: = \: 12P \: + \: \dfrac{ \cancel{12} \: (12 \: + \: 1)P}{\cancel{24} \: \: 2 } \times \dfrac{14}{100} }}}$

$\rm :\longmapsto\: \: {{{6455 \: = \: 12P \: + \: \dfrac{ \: 13P}{ \cancel{2} } \times \dfrac{\cancel{14} \: \: 7}{100} }}}$

$\rm :\longmapsto\: \: {{{6455 \: = \:12 P \: + \dfrac{ 91P}{100} }}}$

$\rm :\longmapsto\: \: {{{6455 \: = \: \dfrac{ 1200P \: + \: 91P}{100} }}}$

$\rm :\longmapsto\: \: {{{6455 \: = \: \dfrac{ 1291P}{100} }}}$

$\rm :\longmapsto\:P = \dfrac{5 \: \: \: \cancel{6455 } \: \: \times 100}{\cancel{1291}}$

$\bf\implies \:P \: = \: 500$

Hence,

The amount of monthly instâllment, P = Rs 500

Additional Information

Interest, I on monthly instâllment of Rs P invested for n months at the rate of r % per annum is given by

$\blue\bigstar \: \red{\boxed{\mathbf{I \: = \: \: \dfrac{n \: (n \: + \: 1)P}{24 } \times \dfrac{r}{100} }}}$

where,

I = interest received along maturity or maturity value

P = Amount of monthly instâllment

r = rate of interest per annum

n = number of monthly instâllment.