Question

a man deposited rs 500 per month for 1 year qnd recived rs 6390 as the maturity value . find the rate of interest​

Answer 1

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

Given that,

Amount deposit per month, P = Rs 500

Time period, n = 1 year = 12 months

Amount received on maturity value, MV = Rs 6390

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

We know,

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

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

So, on substituting the values, we get

$\rm \: 6390 = 500 \times 12 +500 \times \dfrac{12(12 + 1)}{24} \times \dfrac{r}{100}$

$\rm \: 6390 = 6000 + \dfrac{13}{2} \times 5r$

$\rm \: 390 = \dfrac{13}{2} \times 5r$

$\rm \: 30 = \dfrac{1}{2} \times 5r$

$\rm \: 6 = \dfrac{1}{2} \times r$

$\rm\implies \:r \: = \: 12\% \: per \: annum$

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Additional Information

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

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

Answer 2

★ Given :-

• Principal = Rs. 500
• Time = 1 year = 12 months
• Maturity Value = Rs. 6,390

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★ To Find :-

• Rate of Interest = ?

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★ Solution :-

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∓ Here, it is given that, Principal is Rs. 500 . Time is given 1 year [ 12 months ] . Maturity Value is Rs. 6,390 . And, we have to find Rate of Interest .

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• Now, let's solve step by step -----

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◆ Formula Required :-

$\bull \: \sf{M.V = Pn +P \times \frac{n \: \: ( \: n + 1 \: )}{24} \times \frac{R}{100} }$

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◆ Here :-

• M.V denotes to Maturity Value .
• P denotes to Principal .
• N denotes to Time
• R denotes to Rate of Interest .

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∴ By substituting the given values :-

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$\dashrightarrow \: \sf{6390 =500 \times 12 + 500 \times \frac{12 \: ( \: n + 1 \: )}{24} \times \frac{R}{100}} \\ \\ \dashrightarrow \: \sf{6390 = 6000 + 500 \times \frac{12 \: ( \: 12 + 1 \: )}{24} \times \frac{R}{100} } \\ \\ \dashrightarrow \: \sf{6390 = 6000 + 5 \times \frac{13}{2} \times R } \\ \\ \dashrightarrow \: \sf{6390 = 6000 + \frac{13}{2} \times 5 \: R} \\ \\ \dashrightarrow \: \sf{6390 - 6000 = \frac{13}{2} \times 5 \: R } \\ \\ \dashrightarrow \: \sf{390 = \frac{13}{2} \times 5 \: R } \\ \\ \dashrightarrow \: \sf{30 = \frac{1}{2} \times 5 \: R } \\ \\ \dashrightarrow \: \sf{ \frac{30}{5} \times \frac{1}{2} = R} \\ \\ \dashrightarrow \: \sf{6 \times 2 = R} \\ \\ \dashrightarrow \: \sf{R = 12 \: \%}$

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★ Hence :-

• Rate of Interest = 12 %

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