Question

10) Madhavrao gets 6445 at the end of one year when he deposits 500
per month in a recurring deposii scheme. Find rate of interest

Answer 1

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

Given that,

Sum deposited every month, P = Rs 500

Time = 1 year

Maturity Value = Rs 6445

Let assume that, rate of interest be r % per annum.

We know that,

Maturity Value ( M. V. ) on a certain sum of money Rs P per month invested at the rate of r % per annum for n months is

$\bold{ \red {\boxed{\text{M. V.} =nP \: + \: \text{P} \times \dfrac{ \text{n(n + 1)}}{2 \times 12} \times \dfrac{ \text{r}}{100} }}}$

So, on substituting the values, we get

$\rm :\longmapsto\:6445=12 \times 500 + 500 \times \dfrac{ \text{12(12+ 1)}}{2 \times 12} \times \dfrac{ \text{r}}{100}$

$\rm :\longmapsto\:6445=6000 + 500 \times \dfrac{ \text{13}}{2} \times \dfrac{ \text{r}}{100}$

$\rm :\longmapsto\:6445 - 6000 = \dfrac{65r}{2}$

$\rm :\longmapsto\:445 = \dfrac{65r}{2}$

$\rm :\longmapsto\:r = \dfrac{890}{65}$

$\rm :\longmapsto\:r = \dfrac{178}{13}$

$\rm :\longmapsto\:r = 13.69 \: \%$

Answer 2

Answer:

14 %

Step-by-step explanation:

Monthly Deposit = 500Rs (P)

Time = 1*12=12 (n)

Rate = ? (r)

Maturity value = 6455 Rs (m.v)

n(n+1)p / 2

12*13*500/2

12*13*500/2*r/100*1/12

13*5/2

=65/2

amount = pn+1

6455=(12*500)+65/2

6455-6000

455=65/2r

r = 910/65

r = 14 %