Question

find the amount to be paid after 4and half years on a sum of ₹650 of a rate of 12.5% p.a​

Answer 1

$\bf{\Huge{\underline{\sf{ANSWER\::}}}}}$

$\bf{Given}\begin{cases}\sf{Principal,[P]\:=\:Rs.650}\\ \sf{Rate,[R]\:=\:12.5\%}\\ \sf{Time,[n]\:=\:4\frac{1}{2} \:years}\end{cases}}$

$\bf{\Large{\underline{\sf{To\:Find\::}}}}$

The amount of the interest.

$\bf{\Large{\underline{\tt{\red{Explanation\::}}}}}$

We know that formula of the C.I. ;

$\bf{\Large{\boxed{\sf{Amount\:(A)\:=\:P\bigg(1+\dfrac{R}{100} \bigg)^{n} }}}}}}$

Therefore,

$\implies\sf{A\:=\:650\:\bigg(1+\dfrac{12.5*1}{100*2} \bigg)^{4} }$

$\implies\sf{A\:=\:650\:\bigg(1+\dfrac{12.5}{200} \bigg)^{4} }$

$\implies\sf{A\:=\:650\:\bigg(1+\dfrac{12.5*10}{200*10} \bigg)^{4} }$

$\implies\sf{A\:=\:650\:\bigg(1+\cancel{\dfrac{125}{2000}} \bigg)^{4} }$

$\implies\sf{A\:=\:650\:\bigg(1+\dfrac{1}{16} \bigg)^{4} }$

$\implies\sf{A\:=\:650\:\bigg(\dfrac{16+1}{16} \bigg)^{4} }$

$\implies\sf{A\:=\:650\:\bigg(\dfrac{17}{16} \bigg)^{4} }$

$\implies\sf{A\:=\:\cancel{650}\:*\:\dfrac{17}{\cancel{16}} *\dfrac{17}{\cancel{16}} *\dfrac{17}{16} *\dfrac{17}{16} }$

$\implies\sf{A\:=\:\dfrac{2.53*17*17*17*17}{16*16} }$

$\implies\sf{A\:=\:\cancel{\dfrac{212065.03}{256} }}$

$\implies\sf{\red{A\:=\:Rs.828.37}}$

Thus,

The amount is Rs.828.37.

Answer 2

$\huge\bold\green{AnsWer}$

To find :-

The amount of the interest.

We know that formula of the C.I. So , simply by using formula we find out answer

$\tt{A\:=\:650\:\bigg(1+\dfrac{12.5*1}{100*2} \bigg)^{4} }$

$\tt{A\:=\:650\:\bigg(1+\dfrac{12.5}{200} \bigg)^{4} }$

$\tt{A\:=\:650\:\bigg(1+\dfrac{12.5*10}{200*10} \bigg)^{4} }$

$\tt{A\:=\:650\:\bigg(1+\cancel{\dfrac{125}{2000}} \bigg)^{4} }$

$\tt{A\:=\:650\:\bigg(1+\dfrac{1}{16} \bigg)^{4} }$

[tex[\tt{A\:=\:650\:\bigg(\dfrac{16+1}{16} \bigg)^{4} }[/tex]

$\tt{A\:=\:650\:\bigg(\dfrac{17}{16} \bigg)^{4} }$

$\tt{A\:=\:\cancel{650}\:*\:\dfrac{17}{\cancel{16}} *\dfrac{17}{\cancel{16}} *\dfrac{17}{16} *\dfrac{17}{16} }$

$\tt{A\:=\:\dfrac{2.53*17*17*17*17}{16*16} }$

$\tt{A\:=\:\cancel{\dfrac{212065.03}{256} }}$

$\bold\green{A\:=\:Rs.828.37apx.}$