Question

find the number of years after which a sum of rupees 3500 will become rupees 4390.40 at the compound interest of 12%

Answer 1

Given:

A sum of rupees 3500 becomes rupees 4390.40 at the compound interest of 12%

To find:

The number of years

Solution:

Let "n" years represent the no. of years.

To solve the above-given problem we will use the following formula:

$\boxed{\bold{A = P[1 + \frac{R}{100} ]^n}}$

Here we have

P = Principal = Rs. 3500

R = rate of interest = 12%

A = amount = Rs. 4390.40

Now, by substituting the given values in the formula, we get

$4390.40 = 3500 [1 + \frac{12}{100} ]^n$

$\implies \frac{4390.40}{3500} = [ \frac{112}{100} ]^n$

$\implies 1.2544 = [ 1.12 ]^n$

taking log on both sides

$\implies log (1.2544) = log [ 1.12 ]^n$

using the formula → log aᵇ = b log a

$\implies log (1.2544) = n \:log [ 1.12 ]$

$\implies 0.098 = n \:[ 0.049 ]$

$\implies n = \frac{0.098}{0.049}$

$\implies \bold{n = 2\:years}$

Thus, the number of years after which a sum of rupees 3500 will become rupees 4390.40 at the compound interest of 12% is 2 years.

--------------------------------------------------------------------------------------

Also View:

Find the compound interest on rupees 2000 for 1 and half years at 10% p.a. ,when the interest is compounded half yearly  

https://brainly.in/question/15370744

find the amount and the compound interest on rupees 2500 for 2 years at 10% per annum compounded annually

https://brainly.in/question/5817765

find the amount and the compound interest on ₹31250 for 3/2 years at 8%per annum , compounded half yearly

https://brainly.in/question/2024065

Answer 2

Step-by-step explanation:

Given:

A sum of rupees 3500 becomes rupees 4390.40 at the compound interest of 12%

To find:

The number of years

Solution:

Let "n" years represent the no. of years.

To solve the above-given problem we will use the following formula:

\boxed{\bold{A = P[1 + \frac{R}{100} ]^n}}

A=P[1+

100

R

]

n

Here we have

P = Principal = Rs. 3500

R = rate of interest = 12%

A = amount = Rs. 4390.40

Now, by substituting the given values in the formula, we get

4390.40 = 3500 [1 + \frac{12}{100} ]^n4390.40=3500[1+

100

12

]

n

\implies \frac{4390.40}{3500} = [ \frac{112}{100} ]^n⟹

3500

4390.40

=[

100

112

]

n

\implies 1.2544 = [ 1.12 ]^n⟹1.2544=[1.12]

n

taking log on both sides

\implies log (1.2544) = log [ 1.12 ]^n⟹log(1.2544)=log[1.12]

n

using the formula → log aᵇ = b log a

\implies log (1.2544) = n \:log [ 1.12 ]⟹log(1.2544)=nlog[1.12]

\implies 0.098 = n \:[ 0.049 ]⟹0.098=n[0.049]

\implies n = \frac{0.098}{0.049}⟹n=

0.049

0.098

\implies \bold{n = 2\:years}⟹n=2years

Thus, the number of years after which a sum of rupees 3500 will become rupees 4390.40 at the compound interest of 12% is 2 years.