Dark Mode

608 views

1 vote

Best answer

Principle Amount = Rs.$10$

Total Amount given (in 11 monthly installments) = $Rs. 11 \times 1= Rs. 11$

We know, if the rate is $r\%$ per annum,then Simple Interst per year = $r\%$ of P

& $SI = \dfrac{P \times r \times t}{100}$

& Amount Returnable = Principle + SI

Here, in $1^{st}$ month $Rs. 1$ is returned back, So, in next month interest has to be given upon $Rs.(10-1) = Rs. 9$

$1^{st}$ months interest = $\dfrac{10 \times r \times \dfrac{1}{12}}{100}$

$2^{nd}$ months interst = $\dfrac{9 \times r \times \dfrac{1}{12}}{100}$

$3^{rd}$ months interest = $\dfrac{8 \times r \times \dfrac{1}{12}}{100}$

$4^{th}$ months interest = $\dfrac{7 \times r \times \dfrac{1}{12}}{100}$

$5^{th}$ months interest = $\dfrac{6 \times r \times \dfrac{1}{12}}{100}$

.......................................

....................

..................

.............

We know, Total Interest = Rs. 1

∴ $1 = \dfrac{10 \times r \times \dfrac{1}{12}}{100} + \dfrac{9 \times r \times \dfrac{1}{12}}{100} + \dfrac{8 \times r \times \dfrac{1}{12}}{100} + \dfrac{7 \times r \times \dfrac{1}{12}}{100} + .....$

Or, $1 = \dfrac{r}{12 \times 100} \times [10 + 9 + 8 + 7 + 6 + ... + 1]$

Or, $1 = \dfrac{r}{12 \times 100} \times \big[10 \times \dfrac{(10 + 1)}{2} \big] $ $\qquad \big[ \text{∵ Sum of The numbers from 1 to n = }n \times \dfrac{(n+1)}{2}\big]$

Or, $1= \dfrac{r}{12 \times 100} \times 10 \times \dfrac{11}{2}$

Or, $r = \dfrac{1 \times 12 \times 100 \times 2}{10 \times 11}$

Or, $r = \dfrac{12 \times 10 \times 2}{11}$

Or, $r = \dfrac{240}{11}$

Or, $r = 21\dfrac{9}{11}\%$

∴ $\color{maroon}{\text{Rate of Interest}}$ = $\color{green}{21\dfrac{9}{11}\%}$

1 vote

Let the rate be $R$% per annum

Amount paid( if paid at the end of 11 months ) = $P + S.I$

= $P + \dfrac{P*R*T}{100}$

= $10 + \dfrac{10*R*(11/12)}{100}$ $\because$ $T = \dfrac{11}{12} \ years$

= $10 + \dfrac{11*R}{120}$

Now it is given in the question that it has to be returned in $11 \ monthly \ installments$ of $Rs. \ 1 \ each$

Total effective payment = $(Rs.1 + interest \ on \ Rs.1 \ for \ 10 \ months) + (Rs.1 + interest \ on \ Rs.1 \ for \ 9 \ months) + ... + (Rs.1 + interest \ on \ Rs.1 \ for \ 1 \ month) + Rs.1$

= $(1 + \dfrac{1*R*(\dfrac{10}{12})}{100} ) + (1 + \dfrac{1*R*(\dfrac{9}{12})}{100}) + ......... + 1$

= $(1 + \dfrac{10*R}{1200} ) + (1 + \dfrac{9*R}{1200} ) + ......... + 1$

= $11 + \dfrac{R*[1+2+3+4+......10]}{1200}$

= $11 + \dfrac{\dfrac{11*10}{2}}{1200}$

= $11 + \dfrac{11*R}{240}$

$10 + \dfrac{11*R}{120}$ = $11 + 11*\dfrac{11*R}{240}$

$11*\dfrac{R}{120} - 11*\dfrac{R}{240} = 1$

$11*\dfrac{R}{240} = 1$

$R = \dfrac{240}{11}$

$R = 21\dfrac{9}{11}\%$

Amount paid( if paid at the end of 11 months ) = $P + S.I$

= $P + \dfrac{P*R*T}{100}$

= $10 + \dfrac{10*R*(11/12)}{100}$ $\because$ $T = \dfrac{11}{12} \ years$

= $10 + \dfrac{11*R}{120}$

Now it is given in the question that it has to be returned in $11 \ monthly \ installments$ of $Rs. \ 1 \ each$

Total effective payment = $(Rs.1 + interest \ on \ Rs.1 \ for \ 10 \ months) + (Rs.1 + interest \ on \ Rs.1 \ for \ 9 \ months) + ... + (Rs.1 + interest \ on \ Rs.1 \ for \ 1 \ month) + Rs.1$

= $(1 + \dfrac{1*R*(\dfrac{10}{12})}{100} ) + (1 + \dfrac{1*R*(\dfrac{9}{12})}{100}) + ......... + 1$

= $(1 + \dfrac{10*R}{1200} ) + (1 + \dfrac{9*R}{1200} ) + ......... + 1$

= $11 + \dfrac{R*[1+2+3+4+......10]}{1200}$

= $11 + \dfrac{\dfrac{11*10}{2}}{1200}$

= $11 + \dfrac{11*R}{240}$

$10 + \dfrac{11*R}{120}$ = $11 + 11*\dfrac{11*R}{240}$

$11*\dfrac{R}{120} - 11*\dfrac{R}{240} = 1$

$11*\dfrac{R}{240} = 1$

$R = \dfrac{240}{11}$

$R = 21\dfrac{9}{11}\%$