Question

at what rate of compound intrest will rs 20000 become rs24200 after 2 years​

Answer 1

Answer:

2.39 (something)

Step-by-step explanation:

we know that r/=I/pn (here I=4200 p=2000 n=2)

Answer 2

Given ,

Principal = ₹20,000

Amount = ₹24200

Time = 2 years, n = 2

To Find :-

The Rate of Interest.

1st Method :-

Formula :-

$A = P ({1 + \dfrac{r}{100})^n}$

On Inserting the values in the formula

We get ,

$24200 = 20000 ({1 + \dfrac{r}{100})^2}$

${\dfrac{24200}{20000}} = ({1 + \dfrac{r}{100})^2}$

${\dfrac{121}{100}} = ({1 + \dfrac{r}{100})^2}$

$({\dfrac{11}{10})^2} = ({1 + \dfrac{r}{100})^2}$

Since, the powers are same on both sides therefore LHS will be equal to RHS.

${\dfrac{11}{10}} = {1 + \dfrac{r}{100}}$

${\dfrac{11}{10} - 1} = {\dfrac{r}{100}}$

${\dfrac{1 \times 100}{10}}$ = r

Therefore, Rate of Interest = 10%

2nd Method :-

Formula :-

$A = P ({1 + \dfrac{r}{100})^n}$

On Inserting the values in the formula

We get ,

$24200 = 20000 ({1 + \dfrac{r}{100})^2}$

${\dfrac{24200}{20000}} = ({1 + \dfrac{r}{100})^2}$

${\dfrac{121}{100}} = ({1 + \dfrac{r}{100})^2}$

$1.21 = ({1 + \dfrac{r}{100})^2}$

${\sqrt{1.21}} = {1 + \dfrac{r}{100}}$

$1.1 = {1 + \dfrac{r}{100}}$

$1.1 - 1 = {\dfrac{r}{100}}$

$0.1 = {\dfrac{r}{100}}$

0.1 × 100 = r

r = 10

Therefore, Rate of Interest = 10%

Short-hands Used :-

P = Principal

R = Rate

T = Time

I = Interest

CI = Compound Interest

$\bf{ \underline{Some Formulae}} \\ \\ \rightarrow \quad \sf P = \dfrac{100 \times S.I}{R \times T} \\ \\ \\ \rightarrow \quad T = \dfrac{100 \times S.I}{P \times R} \\ \\ \\ \rightarrow \quad R = \dfrac{100 \times S.I}{P \times T} \\ \\ \\ \rightarrow \quad A = P + S.I \\ \\ \\ \rightarrow \quad S.I = \dfrac{P \times R \times T}{100}$