## 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**

- RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A
- RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B
- RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11C
- RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11D

**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

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**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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