Question

Find the ratio of CI to S.I on a certain sum at 5% per annum for 2 years?​

Answer 1

GIVEN :-

• Time ( T ) = 2 years.
• Rate ( R ) = 5 % P.A.

TO FIND :-

• The ratio of CI to SI.

SOLUTION :-

Let the sum be "x".

$\\ : \implies \displaystyle \sf \: \dfrac{CI}{SI} = \frac{P \bigg(1 + \dfrac{R}{100} \bigg) ^{n} - P}{\bigg(\dfrac{PRT}{100}\bigg)}$

$: \implies \displaystyle \sf \: \dfrac{CI}{SI} = \frac{x \bigg(1 + \dfrac{5}{100} \bigg) ^{2} - x }{\bigg(\dfrac{x \times 5 \times 2}{100}\bigg)}$

$: \implies \displaystyle \sf \: \dfrac{CI}{SI} = \frac{ x\bigg( \dfrac{100 + 5}{100} \bigg) ^{2} - x}{\bigg(\dfrac{10x}{100}\bigg)}$

$: \implies \displaystyle \sf \: \dfrac{CI}{SI} = \frac{ x\bigg( \dfrac{105}{100} \bigg) ^{2} - x }{\bigg(\dfrac{10x}{100}\bigg)}$

$: \implies \displaystyle \sf \: \dfrac{CI}{SI} = \frac{x \bigg( \dfrac{11025}{10000} \bigg) - x}{\bigg(\dfrac{10x}{100}\bigg)}$

$: \implies \displaystyle \sf \: \dfrac{CI}{SI} = \frac{ \bigg(\dfrac{11025x}{10000} \bigg)-x}{\bigg(\dfrac{10x}{100}\bigg)}$

$: \implies \displaystyle \sf \: \dfrac{CI}{SI} = \frac{1025x}{10000} \times \frac{100}{10x}$

$: \implies \displaystyle \sf \: \dfrac{CI}{SI} = \frac{1025}{1000}$

$: \implies \displaystyle \sf \: \dfrac{CI}{SI} = \frac{41}{40}$

$: \implies \underline{ \boxed{ \displaystyle \sf \: {CI} :{SI} =41 : 40}}$

Answer 2

Step-by-step explanation:

$\frak {Given} \begin{cases} &\sf{Time\ =\ \bf 2\ \sf years.} \\ &\sf{Rate\ =\ \bf 5\ \%\ \sf P.A.} \end{cases}$

To find:- Ratio of CI to SI ?

❒ By taking the sum be x.

__________________

$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf : \implies {\dfrac{CI}{SI}\ =\ \dfrac{P \bigg (1\ +\ \dfrac{R}{100} \bigg)^n\ -\ P}{\bigg (\dfrac{P\ R\ T}{100} \bigg)}}$

__________________

$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}} \\ \\ \\ \sf : \implies {\dfrac{CI}{SI}\ =\ \dfrac{x \bigg (1\ +\ \dfrac{5}{100} \bigg)^2 \ -\ x}{\bigg (\dfrac{x\ \times\ 5\ \times\ 2}{100} \bigg)}} \\ \\ \\ \sf : \implies {\dfrac{CI}{SI}\ =\ \dfrac{x \bigg (\dfrac{100\ +\ 5}{100} \bigg)^2\ -\ x}{\bigg (\dfrac{10x}{100} \bigg)}} \\ \\ \\ \sf : \implies {\dfrac{CI}{SI}\ =\ \dfrac{x \bigg (\dfrac{105}{100} \bigg)^2\ -\ x}{\bigg (\dfrac{10x}{100} \bigg)}} \\ \\ \\ \sf : \implies {\dfrac{CI}{SI}\ =\ \dfrac{x \bigg (\dfrac{11025}{10000} \bigg)\ -\ x}{\bigg (\dfrac{10x}{100} \bigg)}} \\ \\ \\ \sf : \implies {\dfrac{CI}{SI}\ =\ \dfrac{\bigg (\dfrac{11025x}{10000} \bigg)\ -\ x}{\bigg (\dfrac{10x}{100} \bigg)}} \\ \\ \\ \sf : \implies {\dfrac{CI}{SI}\ =\ \dfrac{1025x}{10000}\ \times\ \dfrac{100}{10x}} \\ \\ \\ \sf : \implies {\dfrac{CI}{SI}\ =\ \dfrac{1025}{1000}} \\ \\ \\ \sf : \implies {\dfrac{CI}{SI}\ =\ \dfrac{41}{40}} \\ \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf CI\ :\ SI\ =\ 41\ :\ 40.}}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ required\ ratio\ is\ \bf 41\ :\ 40.}}$