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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)2 – 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)n = (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 :
👉 Chapter 10 👈
