Question

Please solve question no. 28.........

Answer 1

solution is given in the picture above

Answer 2

Answer-

He had invested 67500 at 10% and 65250 at 8% interest rate.

Solution-

Let's assume x amount was invested at the rate of 10% and y amount was invested at the rate of 8%

i₁ = interest gained at 10%,

i₂ = interest gained at 8%

Applying the formula for interest,

$i=\frac{P\times R\times t}{100}$

$i_1=\frac{x\times 10\times 1}{100} =\frac{x}{10}$

$i_2=\frac{y\times 8\times 1}{100} =\frac{8y}{100}=\frac{2y}{25}$

The total interest he got,

$i=i_1+i_2=1350$

$\Rightarrow \frac{x}{10}+\frac{2y}{25}=1350$  ---------------------1

Had he interchanged the amount, interest he could have gained,

$i_3=\frac{y\times 10\times 1}{100} =\frac{y}{10}$

$i_4=\frac{x\times 8\times 1}{100} =\frac{8x}{100}=\frac{2x}{25}$

The total interest he could have gotten,

$i=i_3+i_4=1350-45=1305$

$\Rightarrow \frac{y}{10}+\frac{2x}{25}=1305$  -----------------------2

Solving eq 1 and 2 we can get the amounts,

$\Rightarrow x=67500,\ y=65250$