At what rate percent will a sum of ₹ 1000 amount to 1102.50 in 2 years at compound interest?

Question 45 Compound Interest Exercise 14.2

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Answer:

Given details are,

Principal = Rs 1000

Amount = Rs 1102.50

Rate = R% per annum

Let time = 2 years

By using the formula,

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

1102.50 = 1000 (1 + R/100)^2

(1 + R/100)^2 = 1102.50/1000

(1 + R/100)^2 = 4410/4000

(1 + R/100)^2 = (21/20)^2

1 + R/100 = 21/20

R/100 = 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

∴ Required Rate is 5%

Video transcript

hello welcome to leader homework today we are going to see give an example show the sum of potential energy and kinetic energy remains constant if the friction is ignored during a vertical wall of a cylinder tube so what happens if the friction due to the air is ignored the total sum of the potential and kinetic energy each point remains same okay due to the vertical wall of the cylindrical tube okay this is a good example during vertical wall of cylindrical let's go to the friction friction due to air is ignored node so total sum remains total sum remains same at any position so this is a good example of it so i hope you understood this video subscribe to this channel for the regular updates and thanks for watching this video

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At what rate percent will a sum of Rs 1000 amount to Rs 1102.50 in 2 years at compound interest?

\[A = P \left( 1 + \frac{R}{100} \right)^n \]\[1102 . 50 = 1000 \left( 1 + \frac{R}{100} \right)^2 \]\[\frac{1102 . 50}{1000} = \left( 1 + 0 . 01R \right)^2 \]\[ \left( 1 + 0 . 01R \right)^2 = 1 . 1025\]\[ \left( 1 + 0 . 01R \right)^2 = \left( 1 . 05 \right)^2 \]On comparing both the sides, we get: 1 + 0.01R = 1.050.01R = 0.05R = 5

Thus, the required rate percent is 5. 

Concept: Applications of Compound Interest Formula

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