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**Joy of Mathematics Class 8 Solutions Chapter 10 Simple and Compound Interest**

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 10 Simple and Compound Interest. Here students can easily find step by step solutions of all the problems for Simple and Compound Interest. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 10.1, 10.2 and 10.3

**Simple and Compound Interest Exercise 10.1 Solution : **

**Question on – (1)**

**Solution :**

Here,

(P) Principal = Rs 5575

(R) Rate of interest = 12%

(T) Time = 5 years

We know,

Simple interest (SI) = PRT/100

= 5575 × 12 × 5/100

= 3345

Therefore, the required simple interest Rs 3345

**Question no – (2)**

**Solution :**

Here,

P = Rs 9750

R = 2.5 per month

T = 2 years

= 2 × 12 month

= 24 month

**∴** SI = PRT/100

= 9750 × 2.5 × 24/100 × 10

= Rs 5850

Therefore, the required simple interest Rs 5850

Now, Amount = Principal + Interest

= 9750 + 5850

= Rs 15600

Therefore, the required amount Rs 15600

**Question no – (3)**

**Solution :**

Here,

Principal (P) = Rs 5600

R = 2 1/8% = 17/8%

Time (T) = 3 1/4 years

= 13/4 years

**∴** Simple interest (SI) = PRT/100

= 5600 × 17 × 13/100 × 8 × 4

= 386.75

Therefore, the interest Rs 386.75

**Question no – (4)**

**Solution :**

Here,

P = 18,000

R = 9%

T = 14^{th }January 2015 to 21^{st} August 2015

Number in days in January = 17

Number in days in February = 28

Number in days in March = 31

Number in days April = 30

Number in days in May = 31

Number in days in June = 30

Number in days in July = 31

Number in days in August = 21

Total number of days = 219

**∴** T = 219/365 = 0.6

Now, SI = 18000 × 9 × 0.6/100 × 10

= Rs 972

Therefore the required interest Rs 972

**Question no – (5)**

**Solution :**

Here,

Principle (P) = 73,000

Rate of interest (R) = 15%

Time = 8^{th} April 1999 to 7^{th} August 1999

Number of days in April = 22

Number of days in May = 31

Number of days in June = 30

Number of days in July = 31

Number of days in August = 7

**∴** Total number of days = 121

**∴** T = 121/365 = 0.33

**∴** SI = 73000 × 15 × 0.33/100 × 100

= Rs 3613.5

**∴** Amount she pay

= Principal + Interest

= 73000 + 3613.5

= Rs 76613.5

**Question no – (6)**

**Solution :**

Here,

Principal (P) = 46,5000

R = 20%

T = 3^{rd} January 2013 to 17^{th} march

Number of days in January = 28

Number of days in February = 28

Number of days in March = 17

**∴** Total number of days = 63

= 0.17

**∴** S.I = 46,5000 × 20 × 0.17/100 × 100

= 158100/10

= Rs 15810

**∴** Amount did he have to pay

= 465000 + 15810

= Rs 480810

**Question no – (7)**

**Solution :**

Here,

Principal = 54,000

Amount = 64,125

**∴** Amount = Principal + interest

=> Interest = Amount – Principal

= 64,125 – 54000

= 10,125

Now,

R = 7 1/2%

= 15/2

**∴** SI = PRT/100

=> T = SI × 100/PR

= 10,125 × 100 × 2/54000 × 15

= 2.5

Therefore, the required time 2.5 years.

**Question no – (8)**

**Solution :**

Here,

Interest = 340

Time = 4 years

Rate (R) = 8 1/2%

= 17/2%

**∴** SI = PRT/100

=> 340 = P × 17 × 4/2 × 100

=> P = 2 × 100 × 340/17 × 4

= Rs 1000

Therefore, required sum deposited in the bank Rs 1000.

**Question no – (9)**

**Solution :**

Here,

P = Rs 1500

R = 3%

T = 3 years

**∴** SI = PRT/100

= 1500 × 3 × 3/100

= 135

**∴** Total SI = 135 + 500

= 635

**∴** He still owner = 500 + 635

= Rs 1135

**Question no – (10)**

**Solution :**

1^{st} invests

R = 4%

T = 1 years

Let, principal = P

**∴** SI_{1} = P × 4 × 1/100

= 4P/100

2^{nd} invests

P = 156500

R = 4 1/2 = 9/2%

T = 1year

SI = PRT/100

=> SI_{2} = 156500 × 9 × 1/2 × 100

= 9 × 1565/2

According to the question,

SI_{1} + SI_{2} = 115750

=> 4P/100 + 9 × 1565/2

= 115750

=> 4P/100 = 115750 – 7042.5

=> 4P/100 = 108707.5

=> 4P = 108707.5 × 100

=> P = 108707.5 × 100/4 × 10

= Rs 2717687.5

**∴** Money at 4% Pa is Rs 2717687.5

**Question no – (11)**

**Solution :**

Here,

Neha borrowed = Rs 5400

Priya borrowed = Rs 6000

Let, the rate of interest = r

Time = 1 3/5 = 8/5 years

For Neha,

SI_{1} = 5400 × r × 8/5 × 100

= 432r/5

For Priya,

SI_{2} = 6000 × r × 8/5 × 100

= 96r

According to the question,

SI_{2} – SI_{1} = 240

=> 96r – 432r/5 = 240

=> 480r – 432r/5 = 240

=> r = 240 × 5/48

= 25%

Therefore, the rate of interest 25%

**Question no – (12)**

**Solution :**

The simple interest on a sum of money is 1/16 of the principal.

Therefore,

Principal = 16x

Simple interest = x

and,

Let, the rate of interest and time same = r

**∴** SI = PRT/100

=> x = 16x × r × r/100

=> 16r^{2} = 100

=> r^{2} = 100/16

=> r = √100/16

= √10 × 10/4 × 4

= 10/4

= 5/2

**∴** Rate of interest 2 1/2%

**Question no – (13)**

**Solution :**

Here,

Principal = 2000

Rate of interest = 5%

Time = 2 years

**∴** SI = 2000 × 5 × 2/100

= 20 × 10

= 200

**∴** After 2 years he should return = Principal + Interest

= 2000 + 200

= Rs 2200

**Question no – (14)**

**Solution :**

Let, the sum be = Rs x

T_{1} = 3 1/3 = 13/4 years

R_{1} = 4 1/2% = 9/2%

Now,

T_{2} = 2 years

R_{2} = 5%

Now,

SI_{1} = x × 9/2 × 13/4/100

= x × 9 × 13/2 × 4 × 100

= 117x/800

and SI_{2} = x × 5 × 2/100

= x × 10/100

= x/10

According to the question,

SI_{1} – SI_{2} = 1221

=> 117x/800 – x/10 = 1221

=> 117x – 80x/800 = 1221

=> 37x/800 = 1221

=> x = 1221 × 800/37

= Rs 26400

Therefore, the required sum Rs 26400

**Question no – (15)**

**Solution :**

Let, the rate of interest be r and sum be P

**∴** Amount = 6P, T = 15years

**∴** SI = Amount – Principal

= 6P – P

= 5P

Now,

S.I = PRT/100

=> 5P = P × r × 15/100

=> r = 100 × 5/15

= 100/3

Therefore, the rate of interest 33 1/3%

**Question no – (16)**

**Solution :**

Let, x and (5400 – x) be invested into two parts.

For 1^{st} part,

r = 5%, T = 4years

P = x

**∴** SI = x × 5 × 4/100

= 20x/100

and,

2^{nd} part

r = 15%, T = 2 years

P = 5400 – x

**∴** SI = (5400 – x) × 2 × 15/100

According to the question, two interest are some,

20x/100 = (5400 – x) × 30/100

=> 2x = 16200 – 3x

=> 5x = 16200

=> x = 3240

**∴** 2^{nd} part = (5400 – 3240) = Rs 2160

**Simple and Compound Interest Exercise 10.2 Solution : **

**Question no – (1)**

**Solution :**

**(a)** Here,

Principal (P) = Rs 300

Rate of interest (R) = 5%

Time (t) = 3 years

**∴** Amount (A) = P (1 + R/100)^{T}

= 300 (1 + 5/100)^{3}

= 300 × (100 + 5/100)^{3}

= 300 × 105/100 × 105/100 × 105/100

= 3472875/10000

= Rs 347.2875

**∴** Compound interest (CI) = A – P

= 347.28 – 300

= Rs 47.28

Therefore the required amount 347.287 and compound interest Rs 47.28

**(b)** Here,

P = 4000

R = 2%

T = 3 years

**∴** A = P (1 + R/100)^{T}

= 4000 (1 + 2/100)^{3}

= 4000 × 102/100 × 102/100 × 102/100

= 4244832/1000

= 4244.832

**∴** CI = A – P

= 4244.832 – 4000

= 244.832

Therefore, the required amount 4244.832 and the compound interest Rs 244.832

**(c)** Here,

P = 5000

R = 5%

T = 3 years

**∴** A = P (1 + R/100)^{T}

= 5000 (1 + 5/100)^{3}

= 5000 (1 + 1/20)^{3}

= 5000 (21/20)^{3}

= 5000 × 21/20 × 21/20 × 21/20

= 105 × 21 × 21/8

= 46305/8

= 5788.125

**∴ **C.I = A – P

= 5788.125 – 5000

= Rs 788.125

Therefore, the required amount Rs 5788.125 and compound interest Rs 788.125

**(d)** Here,

P = 10,000

R = 4%

T = 2 years

Now,

A = P (1 + R/100)^{T}

= 10,000 (1 + 4/100)^{2}

= 10,000 (1 + 1/25)^{2}

= 10,000 × 26/25 × 26/25

= 6760,000/625

= Rs 10816

**∴** CI = A – P

= 10816 – 10,000

= Rs 816

Therefore, the required amount Rs 10816 and the compound interest Rs 816

**Question no – (2)**

**Solution :**

Here,

Principal (P) = Rs 15000

Rate (R) = 12%

Time (T) = 3 years

Now,

Amount (A) = P (1 + R/100)^{T}

= 15000 (1 + 12/100)^{3}

= 15000 (112/100)^{3}

= 15000 × 112 × 112 × 112/100 × 100 × 100

= 21073920/1000

= Rs 21073.92

**∴** Compound interest = A – P

= 21073.92 – 15000

= Rs 6073.92

Therefore the required compound interest Rs 6073.72

**Question no – (3)**

**Solution :**

Here,

Principal (P) = Rs 20,000

1^{st} year interest (R_{1}) = 5%, T = 1 year

A_{1} = P (1 + R1/100)^{T}

= 20,000 (1 + 5/100)^{1}

= 20,000 × 105/100

= 200 × 105

= Rs 21000

2^{nd} year interest (R_{2}) = 7%, Time = 1 year

**∴** A_{2} = P (1 + R_{2}/100)^{T}

= 20,000 (1 + 7/100)^{1}

= 20,000 × 107/100

= Rs 21400

3^{rd} year interest (R_{3}) = 10%, Time = 1 year

A_{3} = P (1 + R_{3}/100)^{T}

= 20,000 (1 + 10/100)^{1}

= 20,000 (11/10)

= 22000

**∴** Total amount = A_{1} + A_{2} + A_{3}

= 21000 + 21400 + 22000

= Rs 64400

Therefore, compound interest = Amount – principal

= 64400 – 20,000

= Rs 24400

**Question no – (4)**

**Solution :**

Here,

Principal (P) = 5000

Rate (R) = 10%

Time (T) = 5years

**∴** SI = PRT/100

= 5000 × 10 × 5/100

= Rs 2500

and

A = P (1 + R/100)^{T}

= 5000 (1 + 10/100)^{5}

= 5000 × 11 × 11 × 11 × 11 × 11/10 × 10 × 10 × 10 × 10

= 805,255/100

= Rs 8052.55

**∴** Compound interest = A – P

= 8052.55 – 5000

= Rs 3052.55

**∴** Difference between compound and simple interest = CI – SI

= (3052.55 – 2500) Rs

= Rs 552.55

Therefore, the required difference Rs 552.55

**Question no – (5)**

**Solution :**

Here,

Principal (P) = 15000

Time (T) = 3 years

Rate of interest (R) = 5%

**∴** Amount = P (1 + R/100)^{T}

= 15000 (1 + 5/100)^{3}

= 15000 × (1 + 1/20)^{3}

= 15000 (21/20)^{3}

= 15000 × 21 × 21 × 21/20 × 20 × 20

= Rs 17364.375

Therefore, Aryan will get on amount Rs 17364.375 of the fixed deposit.

**Question no – (6)**

**Solution :**

Let, the sum be = x

Rate (R) = 8%

Time (T) = 1.5 years

Now,

SI = PRT/100

=> 360 = P × 8 × 1.5/100 × 10

=> P = 3600 × 100/8 × 15

=> P = 3000

Now,

Principal (P) = Rs 3000

Rate (R) = 8%

Time (T) = 1.5 × 2

= 3 year.

**∴** Amount (A) = P (1 + R/100)^{T}

= 3000 (1 + 8/100)^{3}

= 3000 (27/25)^{3}

= 3000 × 27 × 27 × 27/25 × 25 × 25

= Rs 3779.136

**∴** Compound interest (CI) = A – P

= 3779.136 – 3000

= Rs 779.136

Therefore the required compound interest Rs 779.136

**Question no – (7)**

**Solution :**

Here,

P = 15000

R = 8%

T = 1 1/2 = 3/2

Interest payable half yearly

Now, R = 8/2 = 4%

T = 3/2 × 2 = 3 year

Now,

A = P (1 + R/100)^{T}

= 15000 (1 + 4/100)^{3}

= 15000 (1 + 1/25)^{3}

= 15000 × (26/25)^{3}

= 15000 × 26 × 26 × 26/25 × 25 × 25

= Rs 16872.96

**∴** Compound interest = A – P

= 16872.96 – 15000

= Rs 1872.96

Therefore, the required compound interest Rs 1872.96

**Question no – (8)**

**Solution :**

Here,

T = 2 years

R = 4%

CI = 306

Let, principal = x

Now,

C.I = P (1 + R/100)^{T} – P

=> 306 = x (1 + 4/100)^{2} – x

=> 306 = x (25 + 1/25)^{2} – x

=> 306 = x (26/25)^{2} – x

=> 306 = x {(26/25)^{2} – 1}

=> 306 = x {676 – 625/625}

=> 306 = x × 51/625

=> x = 306 × 625/51

=> x = Rs 3750

Therefore, the required principal Rs 3750

**Question no – (9)**

**Solution :**

Here,

Amount (A) = 8820

Time (T) = 2 years

Rate of interest (R) = 5%

Let, Principal (P) = x

Now,

A = P (1 + R/100)^{T}

=> 8820 = x (1 + 5/100)^{2}

=> 8820 = x (1 + 1/20)^{2}

=> 8820 = x (20 + 1/20)^{2}

=> 8820 = x × 21/20 × 21/20

=> x = 8820 × 20 ×20/21 × 21

= 8000

**∴** Principal = Rs 8000

**Question no – (10)**

**Solution :**

**(a)** Here,

Amount (A) = Rs 729

Principal (P) = Rs 625

Time (T) = 2 years

We have,

A = P (1 + R/100)^{T}

=> 729 = 625 (1 + R/100)^{2}

=> 729/625 = (1 + R/100)^{2}

=> (1 + R/100) = √729/625

=> (1 + R/100) = 27/25 – 1

=> R/100 = 27 – 25/25

=> R/100 = 2/25

=> R = 2 × 100/25

=> R = 8%

Therefore, the required rate 8%

**(b)** Here,

A = Rs 1458

P = Rs 1250

T = 2 years

Rate = R%

Now,

A = P (1 + R/100)^{T}

=> 1458 = 1250 (1 + R/100)^{2}

=> 1458/1250 = (1 + R/100)^{2}

=> √1458/1250 = (1 + R/100)

=> 38/35 = 1 + R/100

=> 1 + R/100 = 38/35

=> R/100 = 38/35 – 1

=> R/100 = 38 – 35/35

=> R/100 = 3/35

=> R = 3 × 100/35

= 300/35

= 60/7

Therefore rate 8 4/7%

**Question no – (11)**

**Solution :**

Here,

Amount (A) = Rs 676

Principal = Rs 625

Rate (R) = 4%

Let, Time = T

Now,

A = P (1 + R/100)^{T}

=> 676 = 625 (1 + 4/100)^{T}

=> 676/625 = (1 + 4/100)^{T}

=> 676/625 = (26/25)^{T}

=> (26/25)^{2} = (26/25)^{T}

=> T = 2

Therefore the required time 2 years.

**Question no – (12)**

**Solution :**

Here,

Amount (A) = Rs 12167

Principal (P) = 8000

Rate (R) = 15%

Time = T

Now,

A = P (1 + R/100)^{T}

=> 12167 = 8000 (1 + 15/100)^{T}

=> 12167/8000 = (1 + 3/20)^{T}

=> 12167/8000 = (23/20)^{T}

=> (23/20)3 = (23/20)^{T}

=> T = 3

Therefore, the required time 3 years.

**Question no – (14)**

**Solution :**

A = P (1 + R/100)^{T}

=> 729 = 625 (1 + 8/100)^{T}

=> 729 = 625 (1 + 2/25)^{T}

=> 729 = 625 (27/25)^{T}

=> 729/625 = (27/25)^{T}

=> (27/25)2 = (27/25)^{T}

=> T = 2

Therefore the required time 2 years.

**Question no – (15)**

**Solution :**

Let, the principal be = x

Rate (r) = 5%

Time (T) = 3 years

Now,

S.I = PRT/100

= x × 5 × 3/100

= 3x/20

and,

C.I = P (1 + R/100)^{T} – P

= x (1 + 5/100)^{3} – x

According to the question,

x (1 + 5/100)^{3} – x- 3x/20 = 183

=> x (1 + 1/20)^{3} – x – 3x/20 = 183

=? x (21/20)^{3} – x – 3x/20 = 183

=> x (9261/8000 – 1 – 3/20) = 183

=> x (9261 – 8000 – 1200/8000) = 183

=> x (9261 – 9200/8000) = 183

=> x (61/8000) = 183

=> x = 183 × 8000/61

= 24000

Therefore, the required principal Rs 24000

**Simple and Compound Interest Exercise 10.3 Solution : **

**Question no – (1)**

**Solution :**

Here,

Initial population (P) = 32000

Rate of growth (R) = 5%

Time (T) = 2 years

**∴** Find population (A) = P (1 + R/100)^{T}

= 32000 (1 + 5/100)^{2}

= 32000 (1 + 1/20)^{2}

= 32000 (21/20)^{2}

= 32000 × 21 × 21/20 × 20

= 1680 × 21

= 35280

**∴** After2 years its population will be 35280

**Question no – (2)**

**Solution :**

Here,

Principal = 40,000

Rate of interest (R) = 8%

Time = 1 year

**∴** Interest is calculated half yearly

Rate (R) = 8/2 = 4%

Time (T) = 1 × 2 = 2 years

**∴** Amount (A) = P (1 + R/100)^{T}

= 40,000 (1 + 4/100)^{2}

= 40,000 (1 + 1/25)^{2}

= 40,000 (26/25)^{2}

= 40,000 × 26 × 26/25× 25

= Rs 43264

**∴** She get after one year Rs 43264

**Question no – (3)**

**Solution :**

Here,

P = Rs 99,000

R_{1} = 10%

R_{2} = 10%

T = 2 years

**∴** A = P (1 + R_{1}/100) (1 – R_{1}/100)

= 99000 (1 + 10/100) (1 – 10/100)

= 99000 (11/10) (9/10)

**Question no – (4)**

**Solution :**

Here,

P = 1000000

r = 10%

T = 2 years

**∴** A = P (1 – r/100)^{T}

= 1000000 (1 – 10/100)^{2}

= 1000000 × 90/10 × 9/10

= 810000

**∴** The value of after 2 years 810000

**Question no – (5)**

**Solution :**

Here,

P = 10000

R = 12.5%

T = 20

**∴** The required amount = 10,000 (1 + 12.5/100)^{20}

= 10,000 (112.5/100)^{20}

= 10,000 (1.125)^{20} à given

= 10,000 × 10.545

= 10,5450

**∴** The amount will be R 10,5450.

**Previous Chapter Solution : **