Question

28. Find the rate at which a sum of money will become four times the original
amount in 2 years, if the interest is compounded half-yearly.
ence​

Answer 1

Step-by-step explanation:

in every two year the interest will be taken as compound interest then the rate will be 313 and it will be taken as simple interest then it will be 939 because it compound interest 1 nearby year but simple interest take one time

Answer 2

Given :-

• Time = 2 years
• Compounded half-yearly
• Principal becomes 4 times.

Answer :-

Forumula to use :

$\longrightarrow \sf amount = principal \bigg(1 + \dfrac{rate}{200} \bigg )^{2 \times time}$

Solution :

• Let the principal be ₹x and rate be r%.
• Amount = 4 × x => ₹4x

Substituting the values,

$\implies \sf 4x = x \bigg(1 + \dfrac{r}{200} \bigg)^{2 \times 2}$

Transposing x to the other side,

$\implies \sf \dfrac{4x}{x} = \bigg(1 + \dfrac{r}{200} \bigg ) ^{4}$

Cancelling x,

$\implies \sf \dfrac{4 \not x}{ \not x} = \bigg(1 + \dfrac{r}{200} \bigg)^{4}$

$\implies \sf 4 = \bigg (1 + \dfrac{r}{200} \bigg )^{4}$

Transposing the power,

$\implies \sf \sqrt[4]{4} = 1 + \dfrac{r}{200}$

$\implies \sf \sqrt{2} = 1 + \dfrac{r}{200}$

Let us take √2 as 1.414

$\implies \sf 1.414 = 1 + \dfrac{r}{200}$

Transposing 1 to the other side of the equation,

$\implies \sf 1.414 - 1 = \dfrac{r}{200}$

$\implies \sf 0.414 = \dfrac{r}{200}$

Transposing 200 to the other side,

$\implies \sf 0.414 \times 200 = r$

$\implies \sf 82.8\% = r$

Therefore the rate% = 82.8%

Some more formulas :-

$\longrightarrow \sf compound \: interest = amount - principal$

$\longrightarrow \sf simple \: interest = \dfrac{ rate\times principal \times time }{100}$

• When interest is compounded yearly :

$\longrightarrow \sf amount = principal \bigg(1 + \dfrac{rate}{100} \bigg)^{time}$

• When interest is compounded quarterly :

$\implies \sf amount = principal \bigg(1 + \dfrac{rate}{400} \bigg) ^{4 \times time}$