Question

Dani invested $200 and after 2 years the value of this investment is $224.72.

Calculate the rate of interest when the interest is

i) Simple
ii) Compound

Answer 1

Answer:

Step-by-step explanation:

Principle= 200

Amount=224.72

No.of years=2

i)SIMPLE INTEREST

S.I=Amount-Principle

S.I= pnr/100

Here we have to find r

S.I=Amount-Principle

S.I=224.72-200=24.72

S.I=24.72

r=(S.I×100)÷pn

r=$\frac{24.72 *100}{200*2}}$

r=24.72÷4

rate of interest =6.18

COMPOUND INTEREST

Amount=$P[1+\frac{r}{100} ]^{n}$

Amount=224.72

p=200

n=2years

r=?

224.72=200[1+$\frac{r}{100}$]²

224.72/200=[1+$\frac{r}{100}$]²

1.1236=[1+$\frac{r}{100}$]²            --->(106×106=11236)

(1.06)²=[(100+r)/100]²

Taking square root on both sides

1.06=(100+r)/100

we know that 1.06=106/100

106/100=(100+r)/100

106=100+r

r=106-100

r=6%

Answer 2

Given:

Principle P=200,N=2 years

Amount=$224.72,r=?

Simple Interest(S.I) is used for calculating the interest rate amount of an account.

i) Simple interest

Formula, SI=$\frac{PNR}{100}$

where SI=amount -principle=224.72-200=24.72

determine the interest rate r,

$r=\frac{SI*100}{PN}\\\\r=\frac{24.72*100}{200*2}\\\\r=\frac{24.72}{4}\\\ r=6.18$

Compound interest(C.I) refers to interest on interest.

ii) Compound interest

Amount$=P(1+\frac{r}{100})^n$

$224.72=200(1+\frac{r}{100})^2 \\\frac{224.72}{200}= (1+\frac{r}{100})^2\\1.1236=(1+\frac{r}{100})^2$

Take LCM right side,$1.1236=(\frac{100+r}{100})^2$

Take square on both sides,$(1.11236)^2=[(\frac{100+r}{100})^2]^2$

Then,

$1.06=\frac{100+r}{100}\\\ 1.06*100=100+r \\106=100+r\\r=106-100=6$

Therefore, the rate of interest for simple is 6.18% and the rate of interest for compound is 6%