Question

At what rate percent per annum, compound Interest will Rs.10,000 , amount to Rs. 13310 in three years?

Answer 1

Answer:

Rate of interest is 10%.    

Step-by-step explanation:

Given : Compound Interest will Rs.10,000 , amount to Rs. 13310 in three years.

To find : At what rate percent per annum?

Solution :

We know the compound formula,

$A=P(1+\frac{r}{100})^t$

Where, A is the amount A=Rs.13310

P is the principal P=Rs.10000

t is the time t=3 years.

r is the rate of interest.

Substituting the values in the formula,

$13310=10000(1+\frac{r}{100})^3$

$\frac{13310}{10000}=(1+\frac{r}{100})^3$

$1.331=(1+\frac{r}{100})^3$

Taking cube roots both side,

$\sqrt[3]{1.331}=\sqrt[3]{(1+\frac{r}{100})^3}$

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

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

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

$0.1\times 100=r$

$10=r$

Therefore, Rate of interest is 10%.

Answer 2

Answer:

$Rate \: of \: interest (r)=10\%$

Step-by-step explanation:

Given principal (P) = Rs10000,

Amount (A)= Rs13310

Time(T) = 3 years

Number of times interest paid (n) = 3

Let the rate of interest = r

$We \: know \: that \\A=P \left(1+\frac{r}{100}\right)^{n}$

$\implies 10000=13310\times \left(1+\frac{r}{100}\right)^{3}$

$\implies \frac{10000}{13310}= \left(1+\frac{r}{100}\right)^{3}$

$\implies \frac{1000}{1331}= \left(1+\frac{r}{100}\right)^{3}$

$\implies \frac{10^{3}}{11^{3}}= \left(1+\frac{r}{100}\right)^{3}$

$\implies \left(\frac{11}{10}\right)^{3}=\left(1+\frac{r}{100}\right)^{3}$

/* On finding cube root ,we get

$\implies \left(\frac{11}{10}\right)=\left(1+\frac{r}{100}\right)$

$\implies \frac{11}{10}-1 =\frac{r}{100}$

$\implies \frac{11-10}{10}=\frac{r}{100}$

$\implies \frac{1}{10}\times 100=r$

$\implies 10 = r$

Therefore,

$Rate \: of \: interest (r)=10\%$

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