# Frank ICSE Class 8 Solutions Chapter 9

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## Frank ICSE Mathematics Class 8 Solutions Chapter 9 Simple Interest and Compound Interest

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 9, Simple Interest and Compound Interest. Here students can easily find step by step solutions of all the problems for Simple Interest and Compound Interest, Exercise 9.1 and 9.2 Also our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 9 solutions. Here in this post all the solutions are based on latest Syllabus.

Simple Interest and Compound Interest Exercise 9.1 Solution :

Question no – (1)

Solution :

S.I = 8000 × 1/1 × 4 1/2/100

= 8000 × 5/2 × 9/2/100

= 80 × 5 × 9/2 × 2

= 20 × 5 × 9

= 900 rupees

Therefore, the simple interest will be 900 rupees.

Question no – (2)

Solution :

S.I = PRT/100

or, P = SI × 100/RT

or, 252 × 100/1 1/2 × 3 1/2

= 252 × 100/3/2 × 2/7

= 252 × 100 × 2 × 2/3 × 7

= 4800 rupees

Therefore, sum will be 4800 rupees.

Question no – (3)

Solution :

S.I = PRT/100

or, T SI × 100/PR

Now, S.I = 7650 – 6000

= 1650

T = 1650 × 100/6000 × 51/2

= 165/6 11/2

= 165 × 2/6 × 11

= 5 years

Therefore, it will take 5 years

Question no – (4)

Solution :

Difference on interests pr1t/100 – pr2t/100

= pt/100 (r1 – r2)

= 8400 × 2 1/2/100 × (4 – 3 1/2)

= 84 × 5/2 × 1/2

= 105 rupees

Therefore, the decrease in simple interest will be 105 rupees

Question no – (5)

Solution :

PRT/100 = 8320

or, P = 8320 × 100/RT

or, P = 8320 × 100/6 × 8/12

or, P = 832 × 100/4

or, P = 208000 rupees

Therefore, the principal will be 208000 rupees

Question no – (6)

Solution :

P + SI = 17000

or, P + PRT/100

= 1700

or, P (1 + RT/100)

= 11999

or, P (1 + 3 × 12/100)

= 1700

or, P (1 + 9/25)

= 17000

or, P (34/25)

= 17000

or, P = 17000 × 25/34

= 12500

Therefore, the sum amount Rs 12500.

Question no – (7)

Solution :

P + PRT/100

= 1212

or, P (1 + RT/100)

= 1212

or, (1 + 6 × 3 4/12/100)

= P (1 + 6 × 2/100)

= 1212

or, P 91 + 1/5)

= 1212

or, P (6/5)

= 1212

or, P 1212 × 5/6

= 1010 rupees

Therefore, the sum will be 1010 rupees.

Question no – (8)

Solution :

Suppose x rupees will produce the same interest

x × 6 × 2 1/2/(100)

= 12000 × 5 × 3 1/2/100

or, x × 6 × 5/2

= 12000 × 5 × 7/2

or, x = 12000 × 5 × 7/2/6 × 5/2

or, x = 2000 × 8 × 7/2 × 2/5

or, x = 14000

Therefore, the sum of money Rs 14000.

Question no – (9)

Solution :

= prt/100 = 240

or, 1500 × r2/100 = 2400 [Since, r = t]

or, r² = 240 × 100/100

or, x² = 16

or, r = 4

The time period is 4 years.

Therefore, Rohan return the loan time period is 4 years

Question no – (10)

Solution :

PRT/100 = 12500 × 7 1/2 × (8 + 30 + 31 + 30 + 31 + 16)/365/100

= 125 × 15/2 × 146/365

= 375 rupees

Therefore, the simple interest 375 rupees

Simple Interest and Compound Interest Exercise 9.2 Solution :

Question no – (1)

Solution :

Amount = 5000 (1 + 8/100)2 × (1 + 8/100)2

= 5000 × (108/100)2

= 5000 × 108/100 × 108/100

= 5832 rupees

Therefore, the amount of the compounded 5832 rupees

Question no – (2)

Solution :

= C.I. = Amount – Principal

= 8000 [(1 + 5/100)3 – 1]

= 8000 [{105/100)3 – 1]

= 8000 [(21/20)3 – 1]

= 8000 [9261/8000 – 1]

= 8000 [9261 – 8000/8000]

= 8000 × 1261/8000

= 1261 rupees

Therefore, the compounded 3 years rate 1261 rupees

Question no – (3)

Solution :

Amount = 5000 (1 + 10/100)2

= 5000 (110/100)2

= 5000 (11/10)2

= 5000 × 121/100

= 6050 rupees

Therefore, Anita borrowed compounded calculate the amount 6050 rupees

Question no – (4)

Solution :

S.I = 12,500 × 2 × 8/100

= 2000 rupees

C.I = 12500 [1 + 8/100)2 – 1]

= 12500 [1 + 2/25)2 – 1]

= 12500 [(27/25)2 – 1]

= 12500 [729/625 – 1]

= 12500 [729 – 625/625]

= 12500 [14/625]

= 2080 rupees

Question no – (5)

Solution :

Amount = 80000 (1 + 15/100)3

= 80000 (1 + 3/20)3

= 80000 × 23/20 × 23/20 × 23/20

= 121670 rupees

Therefore, Santosh will get Rs 121670 after 3 years.

Question no – (6)

Solution :

= C.I = 4800 [(1 + 5/100)2 – 1]

= 4800 [(21/20)2– 1]

= 4800 [441/400 – 1]

= 4800 [41/400]

= 492 rupees

Therefore, the compounded interest will be Rs 492.

Question no – (7)

Solution :

Amount = 6250-0 [(1 + 8/100)2

= 62500 (1 + 2/25)2

= 62500 (27/25)2

= 62500 × 729/625

= 72900 rupees

Therefore, the amount will be Rs 72900.

Question no – (8)

Solution :

= C.I = 5120 [(1 + 6 1/4/100)2 – 1]

= 5120 [(1 + 25/4/100)2 – 1

= 5120 (1 + 1/16)2 – 1]

= 5120 [(17/16)2– 1]

= 5120 [289/256 – 1]

= 5120 [917/16)2 – 1]

= 660 rupees

Therefore, compound interest will be Rs 660.

Question no – (9)

Solution :

= C.I = 7290 [91 + 11 1/9/100)3 – 1]

= 7290 [(1 + 100/9)3 – 1]

= 7290 [(1 + 1/9)3 – 1]

= 7290 [(100/9)3 – 1]

= 7290 [1000/729- 1]

= 7290 [271/729]

= 2710 rupees

Therefore, The compound interest will be Rs 2710 at the end of 3 years.

Question no – (10)

Solution :

S.I = 12000 × [(1 + 5/100)2 – 1]

= 12000 [(1 + 1/20)2 – 1]

= 12000 [(21/20)2 – 1]

= 12000 [441/400 – 1]

= 12000 × 41/400

= 1230 rupees

Therefore, the difference between the compound interest and simple interest Rs 1230

Question no – (11)

Solution :

S.I = 7500 × 2 × 5/100 = 750 rupees

C.I = 7500 [91 + 5/100)2 – 1]

= 7500 [(1 + 1/20)2 – 1]

= 7500 (21/20)2 – 1]

= 7500 [441/400 – 1]

= 7500 [41/400]

= 768.75 rupees

C.I – S.I = 768.75 – 750

= 18.75 rupees

Therefore, the difference between compound interest and simple interest Rs 18.75

Question no – (12)

Solution :

Amount = 7500 (1+ 8/2/(100)2

= 7500 (1 + 1/25)2

= 7500 (26/25)2

= 7500 × 676/625

= 8112 rupees

C.I = 8112 – 7500

= 512 rupees

Therefore, the amount will be Rs 8112 and compounded Rs 512.

Question no – (13)

Solution :

Amount = 24000 (1 + 10/2/(100)3

= 24000 (1 + 1/20)3

= 24000 (21/20)3

= 24000 × 21/20 × 21/20 × 21/20

= 27783 rupees

Therefore, the amount will be Rs 27783

Question no – (14)

Solution :

9/12 year = 3/4 year, P = 15000, r = 8%

S.I = 15000 × 8 × 1/4 / 100

= 300 Rs

∴ Amount = 15300 Rs

2nd quarter,

=15300 × 8 × 1/4 / 100 = 306

∴ amount = 15606

3rd quarter,

= 15606 × 8 × 1/4 / 100

= 312.12 Rs

amount = 15918.12 Rs

Question no – (15)

Solution :

Amount when compounded annually

= 60000 (1 + 8/100)1

= 60000 (1 + 2/25)1

= 60000 × 27/25

= 64800 Rs

Difference – 64896 – 64800

= 96 rupees

Therefore, the difference in the amount will be Rs 96.

Question no – (16)

Solution :

Amount = 7500 × (1 + 6/100) × (1 + 4/100)

= 7500 × 106/100 × 104/100

= 8268 rupees

Therefore, the amount will be Rs 8268 after 2 years.

Question no – (17)

Solution :

Amount = 15000 × (1 + 8/100) × (1 + 4/100)

= 15000 × 108/100 × 105/100

= 126 × 105

= 17010 rupees

C.I = 17010 – 15000

= 2010 rupees

Therefore, the amount interest Rs 17010 and the compound interest Rs 2010.

Question no – (18)

Solution :

Let, the sum of money be x rupees

x (1 + 15/100)2 = 31.740

or, x (115/100)2 = 31,740

or, x (23/20)2 = 31740

or, x = 31740 × 20/23

= 24000 rupees

Therefore, the sum of money will be Rs 24000.

Question no – (19)

Solution :

Let, the time by x years

8000 (1 + 5/100)x = 9261

or, 8000 (1 + 1/20)x = 9261

or, 8000 (21/20)x = 9261

or, (21/20)x = (9261/800)

x = 3 years

Therefore, the required time will be 3 years.

Question no – (20)

Solution :

Let, the rate r %

12000 (1 + r/100)3 = 15927

or, (100 + 8/100)3 = 15972/12000

or, (100 + r/100)3 = 1331000/1000000

or, (100 + r/100)3 = (110/100)3

or, r = 10%

Therefore, the rate of interest will be 10%

Question no – (21)

Solution :

Let, the principal be x rupees

x × 5 × 2/100 = 400

or, x/10 = 400

or, x/10 = 400

or, x = 400 × 10

= 4000

The sum will be Rs 4000

(b) Compound interest,

= 4000 [(1 + 5/100)2 – 1]

= 4000 [(1 + 1/20)2 – 1]

= 4000 [(21/20)2 – 1]

or, 4000 [441/400 – 1]

= 4000 × 41/400

410 rupees

Therefore, C.I will be Rs 410.

Revision Exercise Questions Solution :

Question no – (1)

Solution :

Arvind has to give,

= 20,000 × 12 × 2/100

= 4800 rupees

Arvind will get,

= 20000 (1 + 12/100)

= 20,000 (1 + 3/25)2

= 25088 – 20000 = 5088

= 20000 (28/25) (28/25)

= 32 × 784

= 25088

He will given = 5088 – 4800

= 288 rupees

Therefore, he will get Rs 288.

Question no – (2)

Solution :

(a) C.I = 4000 [(1 + 10/100)2 – 1]

= 4000 = [(1 + 1/100)2

= 840 rupees

(b) C.I = 4000 [(1 + 5/100)2 – 1]

= 4000 [(21 + 1/20)– 1]

= 4000 [921/20)2 – 1]

= 4000 [441/400 – 1]

= 400 × 41/400

= 410 rupees

Question no – (3)

Solution :

Suppose, n years are needed

= 3000 (1 + 10/100)n = 3993

or, 3000 (12/10)n = 3993

or, (11/10)n = 3993

or, (11/10)n

= 3993/3000

= 1331/100

or, (11/10)= (11/10)3

n = 3

Therefore, the required time will be 3 years.

Question no – (4)

Solution :

= C.I = 10,000 [(1 + 8/100)2 – 1]

= 1000 [729/625 – 1]

= 10000 × [(27/25)2 – 1]

= 10000 [729/625 – 1]

= 10000 × 104/625

= 1664 rupees

Therefore, the compound interest on Rs 1664.

Question no – (5)

Solution :

= Simple Interest

= 8000 × 5 × 2/100

= 800 rupees

= Compound Interest

= 800 [(1 + 5/100)² – 1]

= 8000 [(1 + 1/20)² – 1]

= 8000 [(21/20)² – 1]

= 8000 [441/400 – 1]

= 8000 × 41/400

= 840 rupees

Difference = 840 – 800

= 40 rupees

Therefore, the difference between compound interest and simple interest Rs 40

Question no – (6)

Solution :

(a) Suppose the sum is x rupees

= x × 5 × 4/100

= 3200 Rs

or, x = 3200 × 5

= 16000 Rs

The sum of will be R 16000.

(b) Compound Interest

= 16000 [(1 + 25/100)2 – 1]

= 16000 [(1 + 1/40)2 – 1]

= 16000 [(41/40)2 – 1]

= 16000 [1681/16000 – 1]

= 16000 × 81/1000

= 810 rupees

Compound interest will be Rs 810

Question no – (7)

Solution :

= Simple interest

= 16000 × 5 1/4/(100)

= 160 × 21/4 × 4

= 3360 Rs

Simple interest Rs 3360

Amount

= 16000 + 3360

= 19360 Rs

The amount will be Rs 29360

Question no – (8)

Solution :

Simple Interest,

= 8000 × 4 1/2 × 24/365/(100)

= 80 × 9/2 × 24/365

= 1728/73 rupees

Therefore, the simple interest Rs 1728/73

Next Chapter Solution :

Updated: June 24, 2023 — 11:34 am