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Solution : Given<br>Principal= `1000`<br>Amount=`1102.50`<br>Time=`2` years<br>`=>A=P(1+R/100)^n`<br>`=>A/P=(1+R/100)^n`<br>=>`1102.50/1000**100=(1+R/100)^2`<br>`=>441/400=(1+R/100)^2`<br>`=> sqrt (441/400)=(1+R/100)`<br>`=>21/20=1+R/100`<br>`=>21/20-1=R/100`<br>`=>1/20=R/100`<br>`=>R=5%`
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RD Sharma Solutions Class 8 Mathematics Solutions for Compound Interest Exercise 14.2 in Chapter 14 - Compound Interest
Question 45 Compound Interest Exercise 14.2
At what rate percent will a sum of Rs. 1000 amount to Rs. 1102.50 in 2 years at compound interest?
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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