Step-by-step explanation:
[tex]\large\underline{\sf{Solution-}}[/tex]
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
[tex]\boxed{\tt{ \: I \: = \: \dfrac{p \times r \times n}{100} \: }}[/tex]
So, we get
[tex]\rm :\longmapsto\:\dfrac{x \times 12 \times 1}{100} + \dfrac{y \times 10 \times 1}{100} = 1145[/tex]
[tex]\rm :\longmapsto\:\dfrac{12x }{100} + \dfrac{10y}{100} = 1145[/tex]
[tex]\rm :\longmapsto\:\boxed{\tt{ 12x + 10y = 114500 }}- - - (1)[/tex]
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,
[tex]\rm :\longmapsto\:\dfrac{x \times 10 \times 1}{100} + \dfrac{y \times 12 \times 1}{100} = 1055[/tex]
[tex]\rm :\longmapsto\:\dfrac{10x }{100} + \dfrac{12y}{100} = 1055[/tex]
[tex]\rm :\longmapsto\:\boxed{\tt{ 10x + 12y = 105500 }}- - - (2)[/tex]
Now, On adding equation (1) and (2), we get
[tex]\rm :\longmapsto\:22x + 22y = 220000[/tex]
[tex]\rm :\longmapsto\:22(x + y) = 220000[/tex]
[tex]\rm :\longmapsto\:\boxed{\tt{ x + y= 10000}} - - - - (3)[/tex]
On Subtracting equation (2) from equation (1), we get
[tex]\rm :\longmapsto\:2x - 2y = 9000[/tex]
[tex]\rm :\longmapsto\:2(x - y) = 9000[/tex]
[tex]\rm :\longmapsto \:\boxed{\tt{ x - y = 4500}} - - - - (4)[/tex]
On adding equation (3) and (4), we get
[tex]\rm :\longmapsto\:2x = 14500[/tex]
[tex]\bf\implies \:x = 7250[/tex]
On substituting the value of x in equation (3), we get
[tex]\rm :\longmapsto\:7250 + y = 10000[/tex]
[tex]\rm :\longmapsto\:y = 10000 - 7250[/tex]
[tex]\bf\implies \:y = 2750[/tex]
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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
