Question

can someone tell the solution​

Answer 1

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

Amount to be deposited per month, P = ₹ x + 240

Rate of interest, r = 100/x % per annum

Time period = x/12 years = x/12 × 12 = x months

Number of instâllments, n = x

Maturity value, MV = ₹ 6312

We know,

Maturity Value (MV) received on maturity on investment of ₹ P per month at the rate of r % per annum for n months is given by

$\bold{ \red {\boxed{\text{MV} = \text{nP} + \text{P} \times \dfrac{ \text{n(n + 1)}}{24} \times \dfrac{ \text{r}}{100} }}}$

So, on substituting the values of MV, n, r, P, we get

$\rm \: 6312 = x(x + 240) + (x + 240) \times \dfrac{x(x + 1)}{24} \times \dfrac{100}{x} \times 100$

$\rm \: 6312 = x(x + 240) + (x + 240) \times \dfrac{(x + 1)}{24}$

$\rm \: 6312 = x(x + 240) + \dfrac{(x + 240)(x + 1)}{24}$

$\rm \: 6312 = (x + 240)\bigg(x + \dfrac{x + 1}{24}\bigg)$

$\rm \: 6312 = (x + 240)\bigg( \dfrac{24x + x + 1}{24}\bigg)$

$\rm \: 6312 = (x + 240)\bigg( \dfrac{25x+ 1}{24}\bigg)$

$\rm \: 25 {x}^{2} + 6000x + x + 240 = 151488$

$\rm \: 25 {x}^{2} + 6001x+ 240 - 151488 = 0$

$\rm \: 25 {x}^{2} + 6001x - 151248 = 0$

Now, its a quadratic equation and to get the values of x, we use Quadratic Formula.

So, using Quadratic formula, we have

$\rm \: x = \dfrac{ - 6001 \: \pm \: \sqrt{ {(6001)}^{2} - 4 \times 25 \times ( - 151248)} }{2 \times 25}$

$\rm \: x = \dfrac{ - 6001 \: \pm \: \sqrt{36012001 + 15124800} }{50}$

$\rm \: x = \dfrac{ - 6001 \: \pm \: \sqrt{51136801} }{50}$

$\rm \: x = \dfrac{ - 6001 \: \pm \: 7151 }{50}$

$\rm \: x = \dfrac{ - 6001 \: + \: 7151 }{50} \: \: \: \: \: \: \{as \: x \: \cancel{ < } \: 0 \}$

$\rm \: x = \dfrac{ 1150 }{50}$

$\bf\implies \:x = 23 \:$

So, Maturity period of RD is 23 months or 1 year 11 months

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ADDITIONAL INFORMATION

Interest (I) received on maturity on investment of ₹ P per month at the rate of r % per annum for n months is given by

$\bold{ \red{\boxed{\text{I} = \text{P} \times \dfrac{ \text{n(n + 1)}}{24} \times \dfrac{ \text{r}}{100} }}}$