Question

The difference between CI and SI on a certain sum at 20% per annum for 4 year is ₹684. Find the sum. ​

Answer 1

Step-by-step explanation:

$answer}$

sum=2500

Answer 2

$\underline{\bf{Question:-}}$

The difference between CI and SI on a certain sum at 20% per annum for 4 year is ₹684. Find the sum. ​

$\underline{\bf{Given\:that:-}}$

1. Principal (p) = ?
2. Rate of interest = 20%
3. Time = 4 years
4. Difference (CI - SI) = ₹ 684

$\underline{\bf{Equation\:used:-}}$

$\sf{:\implies{Compound\:Interest = CI=P(1+\dfrac{r}{100})^{nt}-1}}$

$\sf{:\implies Simple\:interest=(SI) = \dfrac{P\:\times\:I\:\times\:T}{100}}$

$\underline{\bf{Solution:-}}$

Difference = ₹ 684 (Given)

$\sf{:\implies{[P(1+\dfrac{r}{100})^{nt}-1]-[\dfrac{P\:\times\:I\:\times\:T}{100}]=684}}$

$\sf{:\implies{[P(1+\dfrac{20}{100})^{4}-1]-[\dfrac{P\:\times\:20\:\times\:4}{100}]=684}}$

$\sf{:\implies{[P(1+0.2)^{4}-1]-[\dfrac{80p}{100}]=684}}$

$\sf{:\implies{[P(1.2)^{4}-1]-[\dfrac{80p}{100}]=684}}$

$\sf{:\implies{[1.0736p]-[\dfrac{80p}{100}]=684}}$

$\sf{:\implies{\dfrac{(100\:\times\:1.0736p)-80p}{100}=684}}$

$\sf{:\implies{\dfrac{107.36p-80p}{100}=684}}$

$\sf{:\implies{\dfrac{27.36p}{100}=684}}$

$\sf{:\implies{27.36p=684\times100}}$

$\sf{:\implies{27.36p=68400}}$

$\sf{:\implies{p = \dfrac{68400}{27.36} = 2,500}}$

$\boxed{\bf{:\implies\:\star\:P= 2,500\:\star}}$

$\underline{\sf{Additional\:information:-}}$

Compound interest:- It is the interest that accrues when earning for each specified period of time added to principal thus increasing the principal base on which subsequent interest is computed

Formula for compound interest

$\boxed{\sf{Compound\:Interest = CI=P(1+\dfrac{r}{100})^{nt}-1}}$

Where, P is the principal amount

R is the rate of interest

n is the total conversion i.e. t × no: of conversions per year

$\sf{\left[\begin{array}{ccc}\underline{Conversion\:period}&\underline{description}&\underline{no:\:of conversion\:period\:in\:a\:year}\\1\:day&Comp\:daily&365\\1\:month&Comp\:monthly&12\\3\:months&Comp\:quaterly&4\\6\:months&Comp\:semi-annually&2\\12\:months&Comp\:Annually&1&\end{array}\right]}$