Question

Find the simple interest earned in a year if the principal is Rs (5y2 -7y+3) and the rate of interest is (y2+y+11)% p.a?
please answer fastt I need it urgently ​

Answer 1

Step-by-step explanation:

I = P x n x r

100

= 5y2 - 7y + 3 x 1 x y2 + y + 11

100

= 5y2(y2 + y + 11) - 7y(y2 + y + 11) + 3(y2 + y + 11)

100

= 5y4 + 5y3 + 55y2 - 7y3 - 7y2 - 77y + 3y2 + 3y + 33

100

= 5y4 - 2y3 + 51y2 + 33

100

Answer 2

Given,

Principal amount $= (5y^{2}-7y+3 )$

Rate of interest $= (y^{2}+y+11)$

Time period $=1$ year

We have to find the simple interest

Formula for simple interest

Simple interest $= \frac{PTR}{100}$

Now, substitute the given values in the formula,

Simple interest $= \frac{(5y^{2}-7y+3)\times(y^{2}+y+11)\times1}{100}$

$= \frac{5y^{2} (y^{2}+y+11)-7y (y^{2}+y+11)+3 (y^{2}+y+11) }{100}$

$= \frac{ (5y^{4}+5y^{3} +55y^{2} )-(7y^{3}+7y^{2} +77y)+(3y^{2}+3y+33) }{100}$

$= \frac{ (5y^{4}+5y^{3} +55y^{2} )-7y^{3}-7y^{2} -77y+(3y^{2}+3y+33) }{100}$

$= \frac{ 5y^{4}-2y^{3} +51y^{2} -74y+33 }{100}$

Therefore, the simple interest be $\frac{ 5y^{4}-2y^{3} +51y^{2} -74y+33 }{100}$.