Question

sum of money at simple interest amounts to rs 1045 in 5 yrs and to rs 1111 in 6 yrs the sum is​

Answer 1

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

Let assume that the sum of money invested be Rs P and Let I be the simple interest received every year on Rs P at the rate of r % per annum.

Now, we know that

Amount = Principal + Simple Interest

According to statement,

The sum of money P amounts to Rs 1045 in 5 years.

$\red{\rm :\longmapsto\:P + 5I = 1045 - - - (1)}$

According to second condition,

The sum of money Rs P amounts to Rs1111 in 6 years.

$\red{\rm :\longmapsto\:P + 6I = 1111 - - - (2)}$

On Subtracting equation (2) from equation (1), we get

$\rm :\longmapsto\:P + 6I - (P + 5I) = 1111 - 1045$

$\rm :\longmapsto\:P + 6I - P - 5I = 66$

$\bf\implies \:I = 66$

On substituting I = 66 in equation (1), we get

$\rm :\longmapsto\:P + 5 \times 66 = 1045$

$\rm :\longmapsto\:P + 330 = 1045$

$\rm :\longmapsto\:P= 1045 - 330$

$\rm :\longmapsto\:P= 725$

Hence,

• The sum invested is Rs 725.

Additional Information :-

1. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is

$\rm :\longmapsto\:\boxed{ \bf{ \: A = P {\bigg(1 + \dfrac{r}{100} \bigg) }^{n}}}$

2. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi annually for n years is

$\rm :\longmapsto\:\boxed{ \bf{ \: A = P {\bigg(1 + \dfrac{r}{200} \bigg) }^{2n}}}$

3. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is

$\rm :\longmapsto\:\boxed{ \bf{ \: A = P {\bigg(1 + \dfrac{r}{400} \bigg) }^{4n}}}$

4. Interest on a certain sum of money of Rs P invested at the rate of r % per annum simple interest for n years is

$\rm :\longmapsto\: \boxed{ \bf{ \: I = \frac{P \times r \times n}{100}}}$

Answer 2

Answer:

sum invested=p, interest rate=I

1045=p+pi5=p(1+5i)

1111=p+pi6=p(1+6i)

take ratio

1111/1045=(1+6i)/(1+5i)

1111(1+5i)=1045(1+6i)

1111+5555i=1045+6270i

(6270-5555)I=1111-1045

I=(1111-1045)/(6270-5555)=66/715=0.92=9.2%

put this value of I in any of the above equations

u will get the value of p