# By using formula,find compound interest and amount Principle RS 6400Rate of interest 12%Time period 2 year formula A = p(1+r*100) N

Answers (1)

Here, P = Rs 6400

R = 12%

T = 2 years

then compound interest ,

$P (1 + \frac{R}{100})^{T} - P \\ C.I = P[(1 + \frac{R}{100})^{T} - 1] \\ C.I = 6400 [ (1 + \frac{12}{100})^{2} - 1] \\ C.I = 6400 [(1.12)^{2} - 1]\\ C.I = 6400 [1.2544 -1]\\ C.I = 6400 \times .2544 \\ C.I = Rs. 1628.16$

## Most Viewed Questions

### Preparation Products

##### Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
Buy Now
##### Knockout NEET Aug 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
Buy Now
##### Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
Buy Now
##### Knockout NEET Aug 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
Buy Now
##### Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course

Exams
Articles
Questions