Question

If Rs.10,000 amounts to Rs.13,310 in 3 years at compound Interest,find the rate of interest.​

Answer 1

Answer:

Step-by-step explanation:

Let the required rate be x% p.a.

Here,A=Rs.13310,P=Rs.10000,t=3 years.

Using formula:

A=p(1+R/100)*n

⟹13310=10000(1+x/100)*3

 13310=10000[  100 +(100+x)]*3

 ⟹  10000/ 13310  =[  100 +(100+x)  ]  *3

 ⟹(  10 /11  )  *3

=[  100 +(100+x)  ]  *3

 ⟹ 10/11= 100 (100+x)*3

⟹11×100=10×(100+x)

⟹1100=1000+10x

⟹1100−1000=10x

⟹10x=100

x=10

Hence ,the required rate =10%p.a.

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Answer 2

Given:

• Principal = Rs. 10000
• Amount = Rs. 13,310
• Time = 3 years

What To Find:

We have to find the rate (R) of the compund interest.

How To Find:

To find the rate, we will use,

$\sf{A = P \bigg(1 + \dfrac{R}{100} \bigg)^T}$

And subtitute the values and solve.

Solution:

Using the formula,

⇒ $\sf{A = P \bigg(1 + \dfrac{R}{100} \bigg)^T}$

Substitute the values,

⇒ $\sf{13310 = 10000 \bigg(1 + \dfrac{R}{100} \bigg)^3}$

Take 10000 to LHS,

⇒ $\sf{ \dfrac{13310}{10000} = \bigg( 1 + \dfrac{R}{100} \bigg)^3$

Take 100 as the LCM in the brackets,

⇒ $\sf{ \dfrac{13310}{10000} = \bigg( \dfrac{100 + R}{100} \bigg)^3$

Take the cube (3) to LHS,

⇒ $\sf{\sqrt[3]{\dfrac{13310}{10000}} = \dfrac{100 + R}{100}}$

Simplify 13310 by 10000,

⇒ $\sf{\sqrt[3]{1.331} = \dfrac{100 + R}{100}}$

Find the cube root of 1.331,

⇒ $\sf{1.1 = \dfrac{100 + R}{100}}$

Take 100 to LHS,

⇒ 1.1 × 100 = 100 + R

Multiply 1.1 by 100,

⇒ 110 = 100 + R

Take 100 to LHS,

⇒ 110 - 100 = R

Subtract 100 from 110,

⇒ 10 = R

∴ Thus, the rate of interest is 10 %.