## RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B.

Other Exercises

By using the formula, find the amount and compound interest on :

Question 1.
Solution:
Principal (P) = Rs. 6000
Rate (R) = 9% p.a.
Period (n) = 2 years Question 2.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 11% p.a.
Period (n) = 2 years Question 3.
Solution:
Principal (P) = Rs. 31250
Rate (R) = 8% p.a.
Period (n) = 3 years Question 4.
Solution:
Principal (P) = Rs. 10240
Rate (R) = $$12\frac { 1 }{ 2 }$$% = $$\\ \frac { 25 }{ 2 }$$% p.a.
Period (n) = 3 years Question 5.
Solution:
Principal (P) = Rs. 62500
Rate (R) = 12% p.a.
Period (n) = 2 years 6 months Question 6.
Solution:
Principal (P) = Rs. 9000
Rate (R) = 10% p.a.
Period (n) = 2 years 4 months Question 7.
Solution:
Principal (P) = Rs. 8000
Period (n) = 2 years Question 8.
Solution:
Principal (p) = Rs. 1, 25,000
Rate of interest (r) = 8% p.a.
Period (n) = 3 years Question 9.
Solution:
Price of a buffalo (P) = Rs. 11000
Rate of interest (R) = 10% p.a.
Period (n) = 3 years
Price of buffalo at present Question 10.
Solution:
Amount of loan taken (P)
= Rs. 18000 Question 11.
Solution:
Amount borrowed from Bank (P) = Rs. 24000
Rate (R) = 10% p.a.
Period (n) = 2 years 3 months Question 12.
Solution:
In case of Abhay,
Principal (p) = Rs. 16000  Question 13.
Solution:
Simple interest (S.I.) = Rs. 2400
Rate (R) = 8% p.a.
Period (T) = 2 years Question 14.
Solution:
Difference between C.I. and S.I.
= Rs. 90
Rate (R) = 6% p.a.
Period (n) = 2 years
Let principal (P) = Rs. 100  Question 15.
Solution:
Let sum (p) = Rs. 100
Rate (r) 10% p.a.
Period (t) = 3 years.  Question 16.
Solution:
Amount (A) = Rs. 10240
Rate (r) = $$6\frac { 2 }{ 3 }$$% = $$\\ \frac { 20 }{ 3 }$$% p.a.
Period (n) = 2 years
Let sum = P, then Question 17.
Solution:
Amount (A) = Rs. 21296
Rate (r) = 10% p.a.
Period (n) = 3 years.
Let P be the sum, Then Question 18.
Solution:
Principal (P) = 4000
Amount (A) = Rs. 4410
Period (n) = 2 years
Let r be the rate per cent per annum
We know that, Question 19.
Solution:
Principal (P) = Rs. 640
Amount (A) = Rs. 774.40
Period (n) = 2 years
Let r be the rate per cent per annum.
We know that Question 20.
Solution:
Principal (P) = Rs. 1800
Amount (A) = Rs. 2178
Rate (r) = 10% p.a.
Let n be the number of years,
We know that Question 21.
Solution:
Principal (P) = Rs. 6250
Amount (A) = Rs. 7290
Rate (R) = 8% p.a.
Let n be the time, then Question 22.
Solution:
Present population (P) = 125000
Rate of increasing (R) = 2% p.a.
Period (n) = 3 years Question 23.
Solution:
3 years ago, the population was = 50000
Rate of increase successively (r1, r2, r3) = 4%, 5% and 3% p.a.
Period (n) = 3 years
Present Population Question 24.
Solution:
Population of a city in 2013 = 120000
Increase in next year = 6%
and decrease in the following year = 5%
Population in 2015 Question 25.
Solution:
Initially bacteria = 500000
Increase in bacteria = 2% per hour
Period (n) = 2 hours Question 26.
Solution:
Growth of bacteria in a culture (R1) = 10% in first hour
Decrease in next hour (R2) = 10%
Increase in the third hour (R3) = 10%
Bacteria in the beginning = 20000
Bacteria after 3 hours Question 27.
Solution:
Value of machine (P) = Rs. 625000
Rate of depreciation (R) = 8% p.a.
Period (n) = 2 years Question 28.
Solution:
Value of scooter (P) = Rs. 56000
Rate of depreciation (R) = 10% p.a.
Period = 3 years
Value of scooter after 3 years Question 29.
Solution:
Cost of car = Rs. 34800
Rate of depreciation (R1) = 10% p.a. for first year Question 30.
Solution:
Rate of depreciation (R) = 10% p.a.
Period (n) = 3 years
Present value (A) = Rs. 291600
Value of machine 3 years ago Hope given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B are helpful to complete your math homework.

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