# Brilliant’s Composite Mathematics Class 8 Solutions Chapter 7

## Brilliant’s Composite Mathematics Class 8 Solutions Chapter 7 Compound Interest

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 7, Compound Interest. Here students can easily find step by step solutions of all the problems for Compound Interest, Exercise 7.1, 7.2, 7.3, and 7.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

Compound Interest Solution Exercise 7.1

Question no – (1)

Solution :

C.I.  = 1200 [(1 + 5/100)-1 ] = 1200 [(1+1/20)-1 ]

= 1200 [(21/20)3  – 1] = 1200 [9261/8000-1]

= 1200 [ 9261 – 8000/8000] = 1200 [1261/8000]

= 3 × 1261/20 = 3783/20

This compound interest is Rs 189.15  rupees .

Question no – (2)

Solution :

C.I. =  1000 [(1 + 4/100)– 1 ]

= 1000 [(1+1/25)– 1]

= 1000 [(26/25)2 -1]

= 1000  [ 676/625 -1 ]

= 1000 [51/625]

= 8 × 5/25 = Rs 408/5

The compound interest is 81.6 .

Question no – (3)

Solution :

C.I. = 10000 [(1+10/100)-1 ] = 10000 [(11/10)– 1]

= 10000 [1331/1000 -1] = 10000 [331/1000]

= 10 × 331 = Rs 3310.

Question no – (6)

Solution :

A = 1000 (1+5/100)3

= 1000 (1+1/20) 3

= 1000 (21/20)3

= 1000 × 9261/8000

The amount = Rs 1157.6

Question no – (7)

Solution :

Principal = 5000, interest rate 5%,

Interest for first 6 months = 5000 × 6/12 × 5/100 = 125 rupees

Principal for next 6 months = 5000+125 = 5125 rupees. 1

Interest = 5125 × 6/12 × 5/100

= 205 × 1/2 × 5/4 = 1025/8

Total interest,

= 125 + 128.1

= 253.1 rupees

Compound Interest Solution Exercise 7.2

Question no – (1)

(i) P = Rs 10000; R = 10% per annum; n = 3

Solution :

Amount = 10000 (1+10/100)3

= 10000 (11/10)3

= 10000 × 1331/1000

= 13310 rupees .

C.I. = Amount – Principal

= 13310 – 10000 = 3310 rupees

Therefore, amount will be Rs 13310 and Compound interest is Rs 3310.

Question no – (3)

Solution :

Principal = 64000, rate of interest = 2.5 paise per rupee = 2.5%

C.I. = 64000 [(1+2.5/100)– 1]

= 64000 [(1+1/40)3 – 1]

= 64000 [(41/40)-1]

= 64000 [68921/64000 – 1]

= 64000 [68921 – 64000/64000]

= 4921 rupees.

Hence, compound interest payable after 3 years will be Rs. 4921

Question no – (4)

Solution :

C.I. = 1000 [(1 + 3/2×10/100)– 1]

= 1000 [(1 + 15/100)– 1]

= 1000 [(23/20)3 – 1]

= 1000 [ 12167/8000 – 1]

= 1000 × 4167/-8000

= 4167/8

= Rs 157.62

Thus, the compound interest will be Rs 157.62

Question no – (6)

Solution :

Compound Interest = 15625 [( 1 + 1/4 × 16/100)3  – 1]

= 15625 [(1 + 1/25)– 1 ]

= 15625 [(26/25)3 – 1]

= 15625 [17576 – 15625/15625]

= Rs 1951

Therefore, the Compound Interest Rs 1951.

Question no – (10)

Solution :

The first man will pay = 1000 [(1 + 10.5/100)]

= 1000 × 110.5/100 = 1105 rupees

The second man will pay = 1000 [(1 + 1/2 × 10/100)2

= 1000 ( 1 + 5/100)2

= 1000 × 441/400

= 441 × 2.5 = 1102.5 rupees,

1105 b- 1102.5 = 2.5 rupees .

Therefore, the first person will pay 2.5 rupees more

Question no – (12)

Solution :

Compound interest = 12000 [(1 + 9/100 )– 1 ]

= 12000 [(109/100)2 -1 ]

= 12000 × 1881/10000

= 6 × 1881/8 = 1881 × 1.2 = 2257.2

S.I = 12000 × 9 × 2 /100 = 2160 rupees

Difference,

= 2257.2 – 2160

= 97.2 rupees

Hence, the difference will be Rs 97.2.

Compound Interest Solution Exercise 7.3

Question no – (1)

Solution :

As per the question,

Amount = 164, rate = 5%, period = 2 years

P (1 + 5/100)2 =  0164

Or , P (21/20)2 = 164

Or, P = 164 × 400/441

P = 148.75

Question no – (2)

Solution :

P [(1 + 10/100)– 1] = 331

Or, P [(11/10 )– 1] = 331

Or, P [331/1000] = 331

Or, P = 331 × 1000/331

P = Rs 1000

Question no – (3)

Solution :

1200 (1 + 5/100) = 1323

Or, 1200 × (21/20)n = 1323

Or, ( 21/20)n = 1323/1200

Or, (21/20)n = 441/400 = (21/20)2

n = 2

Question no – (4)

Solution :

64000 (1 + 1/2 × 5/100)2 = 68921

Or, 64000 (41/40)n = 68921

Or, (41/40)n = 68921/69000

Or, (41/40)n = (41/40)3

n = 3

Question no – (5)

Solution :

800 (1 + 5/100)= 882

Or, 800 (21/20)n = 882

Or, (21/20)n  = 882/800

Or, (21/20 )n = 441/400

= (21/20)2

n = 2

Question no – (6)

Solution :

2000 (1 + 1/2 × r/100)= 2315.25

Or, (1 + r/200)3 = 2315.25/2000

Or, (200 + r/200)3 = 1157.375/1000

Or, 200 + r/200 = 3√1157.375/1000

Or, 200 + r/200 = 10.5/10

Or, 10r + 2000 = 2100

Or, 10r = 100

Or, r = 10%

Question no – (7)

Solution :

1000 (1 + r × 1/2/100)12 = 2012.2

Or, (1 + r/200)12 = 2012.2/1000

Or, (1+r/200)12 = 2.012

Or, 1 + r/200 = 12√2.012

r = (12√2.012 – 1) × 200

Question no – (8)

Solution :

5000 (1 + r × 1/2/100)10 = 10000

Or, (1 + r/200)10 = 10000/5000 = 2

Or, (1 + r/200) = 2

Or, 1 + r/200 = 10√2 = 1.072

Or, r/200 = 0.072

Or, r = 14.4%

Question no – (9)

Solution :

P (1 + (r/2) /100)= 9826

Or, P (1 + ( 5/4 )/100)= 9826

Or, P (1 +  1/80)3 = 9826

Or , P (81/80)= 9826

Or, P = 9826 × 80 × 80 × 80 / 81 × 81 × 81

Question no – (10)

Solution :

4400 (1 + 8/100)n = 4576

Or, (108/100)n = 4576/4400

Or, (27/20)n = 104/100

Or, (27/25)n = 26/25

Question no – (11)

Solution

64000 (1 + ( 5/2 )/100)n = 68921

Or, 64000 (41/40)n = 68921

Or, (41/40)= 68921/69000

Or, (41/40)n = (41/40)3

n = 3

The required period is = (6 × 3) = 18 months .

Question no – (12)

Solution :

10000 (1 + r/100)3 = 13310

Or, (1 + r/100)3 = 13310/10000 = 1331/1000

Or, (1 + r/100) =3√ 1331/1000 = 11/10

Or, r/100 = 11/10 – 1 = 1/10

r = 10%

Question no – (13)

Solution :

2000 (1 + ( r/2)/100)3 = 2662

Or, (1 + r/200)3 = 2662/2000 = 1331/1000

Or, (1 + r/200 )3 = ( 11/10 )3

Or, 1 + r/200 = 11/10

Or, r/200 = 11/10 – 1 = 1/10

Or, r/200 = 1/10

Or, r = 20%

Question no – (14)

Solution :

Let  the sum be P, then  the amount will be 2p

p ( 1 + r/100 )3 = 2p

Or, ( 1 + r/100 )3 = 2

Or, ( 1 + r/100 ) = 3 √ 2

Or, r = ( √2 – 1 ) × 200

Question no – (15)

Solution :

Let the  sum be P

P ( 1 + ( 12.5/2 )/100 )3 = 4913

Or, P ( 17/16 )= 4913

Or, P = 4913  × 16  × 16  × 16/17  × 17  × 17

Or, P = 16  × 16  × 16 = 4096  rupees

Question no – (16)

Solution :

Let the sum  be P

P [(1 + 10/100)2 – 1] – p × 10 × 2/100 = 500

Or, P [ 21/100 ] – p/5 = 500

Or, 21 p – 20 p/100 = 500

Or, p= 500 × 100 = 50000  rupees

Question no – (17)

Solution :

Interest on 7396 in 1 year

= 7950.70 – 7396

= 554.70 rupees

rate of  interest,

= 554.70/7396 × 100

= 15/2

= 7 1/2%

Compound Interest Solution Exercise 7.4

Question no – (1)

Solution :

Population after two years

= 16000 ( 1 + 5/100 )= 16000 ( 21/20 )2

= 16000 × 441/400

= 40 × 441

= Rs 17640

Question no – (2)

Solution :

Population after 2 years = 80000 ( 1 + 75/1000 )2

= 8000 (43/40 )2 = 8000 × 43 × 43/1000

= 5 × 1849

= Rs 9245

Question no – (3)

Solution :

Value after 2 years = 1,00,000 ( 1 – 10/100 )2 = 10000 ( 9/10 )2

= 100000 × 81/100 = 81000

Deprecation = 100000 – 81000

= Rs 19000

Question no – (4)

Solution :

Value after 3 years = 7812.50 ( 1 – 12/100 )3 = 7812.50 × ( 22/25 )3

= 7812.5 × 22 × 22/25 × 25 × 25 = 5324

Depreciation = 7812.50 – 5324

= 2488.50

Question no – (6)

Solution :

Final income = 32000 × (1 – 5/100) (1 + 10/100) (1 + 12.5/100)

= 32000 × 19/20 × 11/10 × 9/8

= 20 × 19 × 11 × 9 = 37620

Profit = 37620 – 32000

= 5620 rupees

Question no – (8)

Solution :

Population after 2 years = 490000 ( 1 + 4/100 ) ( 1 + 3/100 )

= 490000 × 26/25 × 103/100 = 196 × 103

= 20188 × 26

= Rs 604888

Question no – (9)

Solution :

Let the previous value be x rupees .

x (1 – 10/100)=7290

Or, x (9/10)3 = 7230

Or, x = 7290 × 1000/729

Or, x = 10000 rupees

Question no – (12)

Solution :

Count of bacteria after 3 hours = 1312500 (1 + 10/100) (1 – 8/100) (1 + 12/100)

= 1312500 (11/10) (23/25) (28/25)

= 210 × 11 × 23 × 28

= Rs 1483240

Question no – (14)

Solution :

Population = 72000 (1 + 7/100) (1 – 10/100) = 72000 (107/100) (9/10)

= 72 × 107 × 9

= 648 × 107

= Rs 69336

Question no – (15)

Solution :

Final in come = 160000 (1 + 5/100) (1 + 10/100) (1 + 12.5/100)

= 160000 (21/20 (11/10) (9/8)

= 21 × 11 × 9 × 100 = 207900

Profit = 207900 – 160000

= Rs 47900

Question no – (16)

Solution :

Since, annual birth rate = 3.3% and annual death rate 1.3%

Population is increase by (3.3 – 1.3) = 2%

Population after 3 years = 125000 (1 + 2/100 )3

= 125000 51 × 51 × 51/50 × 50 × 50

= 51 × 51 × 51

= 2601 × 21

= 132651

Therefore, the population after 3 years will be 132651.

Previous Chapter Solution :

Updated: May 29, 2023 — 6:11 am