Question

plase quickly response​

Answer 1

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

Let we assume that

Sum invested at the rate of 12 % per annum be Rs x

Sum invested at the rate of 10 % per annum be Rs y

According to statement,

Annual interest earned = Rs 1145

We know,

Interest received on a sum of Rs p invested at the rate of r % per annum for n years is

$\boxed{\tt{ \: I \: = \: \dfrac{p \times r \times n}{100} \: }}$

So, we get

$\rm :\longmapsto\:\dfrac{x \times 12 \times 1}{100} + \dfrac{y \times 10 \times 1}{100} = 1145$

$\rm :\longmapsto\:\dfrac{12x }{100} + \dfrac{10y}{100} = 1145$

$\rm :\longmapsto\:\boxed{\tt{ 12x + 10y = 114500 }}- - - (1)$

According to second condition

Sum invested at the rate of 12 % per annum be Rs y

Sum invested at the rate of 10 % per annum be Rs x

According to statement,

Annual interest earned = Rs 1055

So,

$\rm :\longmapsto\:\dfrac{x \times 10 \times 1}{100} + \dfrac{y \times 12 \times 1}{100} = 1055$

$\rm :\longmapsto\:\dfrac{10x }{100} + \dfrac{12y}{100} = 1055$

$\rm :\longmapsto\:\boxed{\tt{ 10x + 12y = 105500 }}- - - (2)$

Now, On adding equation (1) and (2), we get

$\rm :\longmapsto\:22x + 22y = 220000$

$\rm :\longmapsto\:22(x + y) = 220000$

$\rm :\longmapsto\:\boxed{\tt{ x + y= 10000}} - - - - (3)$

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

$\rm :\longmapsto\:2x - 2y = 9000$

$\rm :\longmapsto\:2(x - y) = 9000$

$\rm :\longmapsto \:\boxed{\tt{ x - y = 4500}} - - - - (4)$

On adding equation (3) and (4), we get

$\rm :\longmapsto\:2x = 14500$

$\bf\implies \:x = 7250$

On substituting the value of x in equation (3), we get

$\rm :\longmapsto\:7250 + y = 10000$

$\rm :\longmapsto\:y = 10000 - 7250$

$\bf\implies \:y = 2750$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

So,

Sum invested at the rate of 12 % per annum be Rs 7250

Sum invested at the rate of 10 % per annum be Rs 2750