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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 = 14th January 2015 to 21st 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 = 8th April 1999 to 7th 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 = 3rd January 2013 to 17th 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 :
1st invests
R = 4%
T = 1 years
Let, principal = P
∴ SI1 = P × 4 × 1/100
= 4P/100
2nd invests
P = 156500
R = 4 1/2 = 9/2%
T = 1year
SI = PRT/100
=> SI2 = 156500 × 9 × 1/2 × 100
= 9 × 1565/2
According to the question,
SI1 + SI2 = 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,
SI1 = 5400 × r × 8/5 × 100
= 432r/5
For Priya,
SI2 = 6000 × r × 8/5 × 100
= 96r
According to the question,
SI2 – SI1 = 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
=> 16r2 = 100
=> r2 = 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
T1 = 3 1/3 = 13/4 years
R1 = 4 1/2% = 9/2%
Now,
T2 = 2 years
R2 = 5%
Now,
SI1 = x × 9/2 × 13/4/100
= x × 9 × 13/2 × 4 × 100
= 117x/800
and SI2 = x × 5 × 2/100
= x × 10/100
= x/10
According to the question,
SI1 – SI2 = 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 1st part,
r = 5%, T = 4years
P = x
∴ SI = x × 5 × 4/100
= 20x/100
and,
2nd 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
∴ 2nd 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
1st year interest (R1) = 5%, T = 1 year
A1 = P (1 + R1/100)T
= 20,000 (1 + 5/100)1
= 20,000 × 105/100
= 200 × 105
= Rs 21000
2nd year interest (R2) = 7%, Time = 1 year
∴ A2 = P (1 + R2/100)T
= 20,000 (1 + 7/100)1
= 20,000 × 107/100
= Rs 21400
3rd year interest (R3) = 10%, Time = 1 year
A3 = P (1 + R3/100)T
= 20,000 (1 + 10/100)1
= 20,000 (11/10)
= 22000
∴ Total amount = A1 + A2 + A3
= 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
R1 = 10%
R2 = 10%
T = 2 years
∴ A = P (1 + R1/100) (1 – R1/100)
= 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.
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