Question

A Sum of 64000 produces an interest of 4921 when compounded annually for 3 years, find rate of interest.​

Answer 1

Step-by-step explanation:

Solution :-

Given that

Principle (P) = Rs. 64000

Compound Interest (C.I) = Rs. 4921

Time (T) = 3 years

The interest is calculated compounded annually then number of times the interest is calculated for 3 years (n ) = 3

Let the rate of interest be R %

We know that

Amount = Principle + Interest

=> A = 64000+4921

=> A = 68921

Amount after 3 years = Rs. 68921

A = P[1+(R/100)]^n

On substituting these values in the above formula then

=> 68921 = 64000 [ 1+ (R/100) ]³

=> 68921/64000 = [(100+R)/100]³

=> (41×41×41)/(40×40×40) = [(100+R)/100]³

=> (41)³/(40)³ = [(100+R)/100]³

=> (41/40)³ = [(100+R)/100]³

On comparing both sides then

=> 41/40 = (100+R)/100

On applying cross multiplication then

=> 40(100+R) = 41×100

=> 4000+40R = 4100

=> 40R = 4100-4000

=> 40R = 100

=> R = 100/40

=> R = 5/2 % or

=> R = 2.5% or

=> R = 2 1/2 %

Answer:-

The required rate of interest is 5/2% or 2 1/2% or 2.5%

Check:-

A = P[1+(R/100)]^n

On substituting these values in the above formula then

=> A = 64000 [ 1+ (5/2×100) ]³

=> A = 64000[1+(1/40)]³

=> A = 64000[(40+1)/40]³

=> A = 64000(41/40)³

=> A = 64000×[(41×41×41)/(40×40×40)]

=> A = 64000×(68921/64000)

=> A = 68921

Amount = Rs. 68921

Compound Interest = A-P

=> C.I. = 68921-64000

Therefore, C I. = Rs. 4921

Verified the given relations in the given problem.

Used formulae:-

→ A = P[1+(R/100)]^n

→ Amount = Principle + Interest

• A = Amount
• P = Principle
• R = Rate of Interest
• n = Number of times the interest is calculated compounded per annum

Answer 2

Answer:

Given :-

• A sum of Rs 64000 produces an interest of Rs 4921 when compounded annually for 3 years.

To Find :-

• What is the rate of interest.

Formula Used :-

$\clubsuit$ Amount Formula :

$\dashrightarrow \sf\boxed{\bold{\pink{Amount =\: Principal + Interest}}}$

$\bigstar$ Amount Formula when the interest is compounded annually Formula :

$\longrightarrow \sf\boxed{\bold{\pink{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n}}}$

where,

• A = Amount
• P = Principal
• r = Rate of Interest
• n = Time Period

Solution :-

First, we have to find the amount :

Given :

• Principal = Rs 64000
• Interest = Rs 4921

According to the question by using the formula we get,

$\implies \sf Amount =\: Rs\: 64000 + Rs\: 4921$

$\implies \sf\bold{\green{Amount =\: Rs\: 68921}}$

Now, we have to find the rate of interest :

Given :

• Amount = Rs 68921
• Principal = Rs 64000
• Time Period = 3 years

According to the question by using the formula we get,

$\implies \sf 68921 =\: 64000\bigg(1 + \dfrac{r}{100}\bigg)^3$

$\implies \sf 68921 =\: 64000\bigg(\dfrac{100 + r}{100}\bigg)^3$

$\implies \sf \dfrac{68921}{64000} =\: \bigg(\dfrac{100 + r}{100}\bigg)^3$

$\implies \sf \bigg(\dfrac{41}{40}\bigg)^3 =\: \bigg(\dfrac{100 + r}{100}\bigg)^3$

$\implies \sf \dfrac{41}{40} =\: \dfrac{100 + r}{100}$

By doing cross multiplication we get,

$\implies \sf 40(100 + r) =\: 41(100)$

$\implies \sf 4000 + 40r =\: 4100$

$\implies \sf 40r =\: 4100 - 4000$

$\implies \sf 40r =\: 100$

$\implies \sf r =\: \dfrac{10\cancel{0}}{4\cancel{0}}$

$\implies \sf r =\: \dfrac{10}{4}$

$\implies \sf\bold{\red{r =\: 2.5\%}}$

$\therefore$ The rate of interest is 2.5% per annum.