Question

A man deposits Rs.900 per month in a RD account at 8%pa. If he gets Rs.1800 as interest then find time of maturity in years. plz answer correctly.​

Answer 1

Question :-

A man deposits Rs.900 per month in a RD account at 8%pa. If he gets Rs.1800 as interest then find time of maturity in years. plz answer correctly.

Solution :-

Given :

• Simple interest (SI) = 1800
• Principal (P) = 900
• Rate of change (R) = 8%

Need to find :

• Number of years (N)

Using formula

${\bigstar\:{\boxed{\rm{SI= \frac{PNR}{100}}}}}$

Now, Substituting all given values

$\sf:\longmapsto{1800 = \frac{900 \times 8 \times N}{100} } \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{1800 \times 100 = 900 \times 8 \times N} \\ \\ \sf:\longmapsto{180000 = 7200 \times N} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{ N= \frac{1800 \cancel{00}}{72 \cancel{00}} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{N = \cancel\frac{180}{72} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf:\longmapsto \red{ N= 2.5} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Number of years is 2.5years.

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Learn more from brainly :

Deepa has a 4-year recurring deposit account in a bank and deposits Rs. 1,800 per month. If she gets Rs. 1,08,450 at the time of maturity, find the rate of interest.

https://brainly.in/question/37751664

Answer 2

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

Given that,

↝ Amount deposited per month, P = Rs 900

↝ Rate of interest, r = 8 % per annum

↝ Interest received, I = Rs 1800.

Let assume that

↝ Rs 900 per month is invested for 'n' months.

We know,

Interest received on a certain sum of money Rs P invested at the rate of r % per annum for n months is

$\bold{ \red{\bf\implies \red{\:{\boxed{\text{I} = \text{P} \times \dfrac{ \text{n(n + 1)}}{2 \times 12} \times \dfrac{ \text{r}}{100} }}}}}$

So, on substituting the values, we get

$\rm :\longmapsto\:1800 =900 \times \dfrac{ \text{n(n + 1)}}{2 \times 12} \times \dfrac{ \text{8}}{100}$

$\rm :\longmapsto\:2 = \dfrac{ \text{n(n + 1)}}{3} \times \dfrac{ \text{1}}{100}$

$\rm :\longmapsto\:n(n + 1) = 600$

$\rm :\longmapsto\: {n}^{2} + n = 600$

$\rm :\longmapsto\: {n}^{2} + n - 600 = 0$

$\rm :\longmapsto\: {n}^{2} +25 n - 24n - 600 = 0$

$\rm :\longmapsto\:n(n + 25) - 24(n + 25) = 0$

$\rm :\longmapsto\:(n + 25)(n - 24) = 0$

$\bf\implies \:n \: = \: 24 \: months$

$\bf\implies \:n \: = \: 2 \: years$