RS Aggarwal Solutions for Class 7 Maths Chapter 3–Decimals

Class 7: Maths Chapter 3 solutions. Complete Class 7 Maths Chapter 3 Notes.

RS Aggarwal Solutions for Class 7 Maths Chapter 3–Decimals

RS Aggarwal 7th Maths Chapter 3, Class 7 Maths Chapter 3 solutions

Ex 3A Solutions

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
(i) 6.5, 16.03, 0.274, 119.4
In these decimals, the greatest places of decimal is 3
6.5 = 6.500
16.03 = 16.030
0. 274 = 0.274
119.4 = 119.400 are like decimals.
(ii) 3.5, 0.67, 15.6, 4
In these decimal, the greatest place of decimal is 2
3.5 = 3.50
0.67 = 0.67
15.6 = 15.60
4 = 4.00 are the like decimals

Question 5.
Solution:
(i) Among 78.23 and 69.85,
78.23 is greater than 69.85 (78 > 69)
78.23 > 69.85
(ii) Among 3.406 and 3.46,
3.406 is less than 3.46 (40 < 46)
3.406 < 3.46
(iii) Among 5.68 and 5.86,
5.68 is less than 5.86 (68 < 86)
5.68 < 5.86
(iv) Among 14.05 and 14.005
14.5 is greater than 14.005 (05 > 00)
14.5 >14.005
(v) Among 1.85 and 1.805,
1.85 is greater than 1.805 (85 > 80)
1.85 > 1.805
(vi) Among 0.98 and 1.07,
0.98 is less than 1.07 (0 < 1)
0.98 < 1.07

Question 6.
Solution:
(i) 4.6, 7.4, 4.58, 7.32, 4.06
Converting the given decimals into like decimals, we get:
4.60, 7.40, 4.58, 7.32, 4.06.
We see that 4.06 < 4.58 < 4.60 < 7.32 < 7.40.
Writing in ascending order, 4.06, 4.58, 4.6, 7.32, 7.4
(ii) 0.5, 5.5, 5.05, 0.05, 5.55
Converting the given decimals into like decimals, we get:
0. 50, 5.50, 5.05, 0.05, 5.55
We see that 0.05 < 0.50 < 5.05 < 5.50 < 5.55.
Writing in ascending order, 0.05, 0.50, 5.05, 5.5, 5.55
(iii) 6.84, 6.84, 6.8, 6.4, 6.08
Converting the given decimals into like decimals
6.84, 6.48, 6.80, 6.40, 6.08
We see that 6.08 < 6.40 < 6.48 < 6.80 < 6.84
Writing in ascending order,
6.08, 6.4, 6.48, 6.8, 6.84
(iv) 2.2, 2.202, 2.02, 22.2, 2.002
Converting them into like decimals
2.200, 2.202, 2.020, 22.200, 2.002 we see that
2.002 < 2.020 < 2.200 < 2.202 < 22.200
Now writing in ascending order,
2.002, 2.020, 2.2, 2.202, 22.2

Question 7.
Solution:
(i) 7.4, 8.34, 74.4, 7.44, 0.74
Converting them into like decimals,
7.40, 8.34, 74.40, 7.44, 0.74
we see that
74.40 > 8.34 > 7.44 > 7.40 > 0.74
Writing in descending order,
74.4, 8.34, 7.44, 7.4, 0.74
(ii) 2.6, 2.26, 2.06, 2.007, 2.3
Converting them into like decimals,
2.600, 2.260, 2.060, 2.007, 2.300
We see that
2.600 > 2.300 > 2.260 > 2.060 > 2.007
Writing in descending order,
2.6, 2.3, 2.26, 2.06, 2.007

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Ex 3B Solutions

Question 1.
Solution:
Converting them into like decimals 16.00, 8.70, 0.94, 6.80 and 7.77
Now, adding them,
16.0 + 8.70 + 0.94 + 6.80 + 7.77 = 40.21

Question 2.
Solution:
Converting them into like decimals 18.600, 206.370, 8.008, 26.400, 6.900
Adding we get
18.600 + 206.370 + 8.008 + 26.400 + 6.900 = 266.278

Question 3.
Solution:
Converting them into like decimals, 63.50, 9.70, 0.80, 26.66, 12.17
Adding we get:
63.50 + 9.70 + 0.80 + 26.66 + 12.17 = 112.83

Question 4.
Solution:
Converting them into like decimals 17.400, 86.390, 9.435, 8.800, 0.060
Adding we get:
17.400 + 86.390 + 9.435 + 8.800 + 0.060 = 122.085

Question 5.
Solution:
Converting them into like decimals 26.900, 19.740, 231.769, 0.048
Now adding we get:
26.900 + 19.740 + 231.769 + 0.048 = 278.457

Question 6.
Solution:
Converting them into like decimals 23.800, 8.940, 0.078 and 214.600
Now adding we get:
23.800 + 8.940 + 0.078 + 214.600 = 247.418

Question 7.
Solution:
Converting them into like decimals.
6.606, 66.600, 666.000,0.066, 0.660
Now adding we get:
6.606 + 66.600 + 666.000 + 0.066 + 0,660 = 739.932

Question 8.
Solution:
9.090, 0.909, 99.900, 9.990, 0.099
Now adding we get:
9.090 + 0.909 + 99.900 + 9.990 + 0.099 = 119.988

Subtract:
Question 9.
Solution:
14.79 from 72.43
72.43 – 14.79 = 57.64

Question 10.
Solution:
Converting them into like decimals, We get
36.74 and 52.60
Now 52.60 – 36.74 = 15.86

Question 11.
Solution:
Converting them into like decimals, We get
13.876 and 22.000
22.000 – 13.876 = 8.124

Question 12.
Solution:
Converting them into like decimals, We get
15.079 and 24.160
24.160 – 15.079 = 9.081

Question 13.
Solution:
Converting them into like decimals We get
0.680 and 1.007
1.007 – 0.680 = 0.327

Question 14.
Solution:
Converting them into like decimals,
We get 0.4678 and 5.0500
5.0500 – 0.4678 = 4.5822

Question 15.
Solution:
Converting them into like decimals,
We get 2.5307 and 8.0000
8.0 – 2.5307 = 5.4693

Question 16.
Solution:
There are like decimals
9.1 – 6.732 = 2.269

Question 17.
Solution:
Converting them into like decimals,
We get 5.746 and 9.100
9.100 – 5.746 = 3.354

Question 18.
Solution:
Converting into like decimals, we get,
63.59 and 92.00
Required number = 92.00 – 63.58 = 28.42

Question 19.
Solution:
Converting into like decimals, we get:
8.100 and 0.813
Required number = 8.100 – 0.813 = 7.287

Question 20.
Solution:
Converting them into like decimals, we get: 32.67 and 60.10
Required number = 60.10 – 32.67 = 27.43

Question 21.
Solution:
Converting into like decimals, we get 74.3 and 26.87
Required number = 74.30 – 26.87 = 47.43

Question 22.
Solution:
Cost of notebook = Rs. 23.75
Cost ofpencil = Rs. 2.85
Costofpen =Rs. 15.90
Total cost = Rs. 42.50
Amount gave to the shop keeper = 50 rupees
Balance amount got = Rs 50.00 – Rs 42.50 = 7.50

Ex 3C Solutions

Question 1.
Solution:
We know that by multiplying by 10, the decimal point is shifted one place to its right side.
(i) 73.92 x 10 = 739.2
(ii) 7.54 x 10 = 75.4
(iii) 84.003 x 10 = 840.03
(iv) 0.83 x 10 = 8.3
(v) 0.7 x 10 = 7.0
(vi) 0.032 x 10 = 0.32

Question 2.
Solution:
We know that by multiplying a decimal by 100, two decimal points are shifted to it right side
(i) 2.397 x 100 = 239.7
(ii) 6.83 x 100 = 683.0
(iii) 2.9 x 100 = 290
(iv) 0.08 x 100 = 8
(v) 0.6 x 100 = 60
(vi) 0.003 x 100 = 0.3

Question 3.
Solution:
We know that by multiplying a decimal by 1000, three places of decimal are shifted to its right.
(i) 6.7314 x 1000 = 6731.4
(ii) 0.182 x 1000 = 182
(iii) 0.076 x 1000 = 76
(iv) 6.25 x 1000 = 6250
(v) 4.8 x 1000=4800
(vi) 0.06 x 1000 = 60

Question 4.
Solution:
(i) 5.4 x 16 = 86.4 (One place of decimal)
(ii) 3.65 x 19 = 69.35 (Two place of decimal)
(iii) 0.854 x 12 = 10.2468 (Three place of decimal)
(iv) 36.73 x 48 = 1763.04 (Two places of decimal)

(v) 4.125 x 86=354.750 (Three places of decimal)
= 354.75

Question 5.
Solution:
(i) 7.6 x 2.4= 18.24
{Sum of decimal places = 1 + 1 = 2}

Question 6.
Solution:
(i) 13 x 1.3 x 0.13 = 2.197
{Sum of decimal places = 1 + 2 = 3}

(ii) 2.4 x 1.5 x 2.5 = 9.000 = 9
{Sum of decimal places = 1 + 1 + 1 = 3}
(iii) 0.8 x 3.5 x 0.05 = 0.1400 = 0.14
{Sum of decimal places = 1 + 1 + 2 = 4}
(iv) 0.2 x 0.02 x 0.002 = 0.000008
{Sum of decimal places = 1 + 2 + 3 = 6}
(v) 11.1 x 1.1 x 0.11 = 1.3431
{Sum of decimal places = 1 + 1 + 2 = 4}

(vi) 2.1 x 0.21 x 0.021 = 0.00926
21 x 21 = 441
441 x 21 = 9261
{Sum of decimal places = 1 + 2 + 3 = 6}

Question 7.
Solution:
(i) (1.2)²= 1.2 x 1.2 = 1.44
{Sum of decimal places = 1 + 1 = 2}
(ii) (0.7)² = 0.7 x 0.7 = 0.49
{Sum of decimal places = 1 + 1 = 2}
(iii) (0.04)² = 0.04 x 0.04 = 0.0016
{Sum of decimal places = 2 + 2 = 4}
(iv) (0.11)² = 0.11 x 0.11 =0.0121
{Sum of decimal places = 2 + 2 = 4}

Question 8.
Solution:
(i) (0.3)³ = 0.3 x 0.3 x 0.3 = 0.027
{Sum of decimal places = 1 + 1 + 1 = 3}
(ii) (0.05)³= 0.05 x 0.05 x 0.05 = 0.000125
{Sum of decimal places = 2 + 2 + 2 = 6}
(iii) (1.5)³ = 1.5 x 1.5 x 1.5 = 3.375
{Sum of decimal places = 1 + 1 + 1 = 3}

Question 9.
Solution:
Distance covered in one hour = 62.5 km
Distance covered in 18 hours = 62.5 x 18 km = 1125.0 km

Question 10.
Solution:
Weight of one tin of oil = 16.8 kg
Weight of 45 tins = 16.8 x 45 kg = 756.0 kg = 756 kg

Question 11.
Solution:
Weight of wheat in one bag = 97.8 kg
weight of wheat in 500 bags = 97.8 x 500 kg = 48900.0 kg = 48900 kg

Question 12.
Solution:
Weight of one bag = 48.450 kg
Weight of 16 bags = 48.450 x 16 = 775.200 kg

Question 13.
Solution:
Quantity of sauce in one bottle = 0.845 kg
quantity of sauce in 72 bottles = 0.845 x 72 kg = 60.840 kg

Question 14.
Solution:
Quantity of jam in one bottle = 925 .
Quantity of jam in 25 bottles = 925 x 25 g = 23135 g = 23.125 kg

Question 15.
Solution:
Oil in one drum = 16.850 litres
Oil in 48 drums = 16.850 x 48 = 808.800 = 808.800 litres

Question 16.
Solution:
Cost of 1 kg rice = Rs 56.80
Cost of 16.25 kg of rice = Rs 56.80 x 16.25 = Rs 923.0000 = Rs 923

Question 17.
Solution:
Cost of one metre of cloth = Rs 108.5 0
Costof 18.5 metres of cloth = Rs 108.50 x 18.5 = Rs 2007.250 = Rs 2007.25

Question 18.
Solution:
Distance covered in one litre = 8.6 km
Distance covered in 36.5 litres = 8.6 x 36.5 km = 313.90 km = 313.9 km

Question 19.
Solution:
Charges for 1 km = Rs 9.80
Charges for 106.5 km = Rs 9.80 x 106.5 = Rs 1043.700 = Rs 1043.70

Ex 3D Solutions

Question 1.
Solution:
We know that a decimal divided by 10, the decimal point is shifted to the left by one place. Therefore
(i) 131.6 ÷ 10 = 13.16
(ii) 32.56 ÷ 10 = 3.256
(iii) 4.38 ÷ 10 = 0.438
(iv) 0.34 ÷ 10 = 0.034
(v) 0.08 ÷ 10 = 0.008
(vi) 0.062 ÷ 10 = 0.0062

Question 2.
Solution:
We know that decimal divided by 100, the decimal point is shifted to the left by two place. Therefore
(i) 137.2 ÷ 100 = 1.372
(ii) 23.4 ÷ 100= 0.234
(iii) 4.1 ÷ 100 = 0.047
(iv) 0.3 ÷ 100 = 0.003
(v) 0.58 ÷ 100 = 0.0058
(vi) 0.02 ÷ 100 = 0.0002

Question 3.
Solution:
We know that a decimal divided by 1000, the decimal point is shifted to the left by three places. Therefore:
(i) 1286.5 ÷ 1000= 1.2865
(ii) 354.16 ÷ 1000 = 0.35416
(iii) 38.9 ÷ 1000 = 0.0389
(iv) 4.6 ÷ 1000 = 0.0046
(v) 0.8 ÷ 1000 = 0.0008
(vi) 2 ÷ 1000 = 0.002

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:
Cost of 24 chairs = Rs. 9255.60
Cost of one chair = Rs. 9255.6024 = Rs. 385.65

Question 9.
Solution:
Length of cloth for one shirt = 1.8 m
Total length of piece of cloth = 45 m
Number of shirts will be = 45 ÷ 1.8

Question 10.
Solution:
A car covers in 2.4 litre = 22.8 km
It will cover in 1 litre = 22.82.4 km = 9.5 km

Question 11.
Solution:
Oil in one tin = 16.5 l
Total oil = 478.5 l

Question 12.
Solution:
Weight of 37 bags of sugar=3644.5 kg
Weight of one bag of sugar = 3644.5 ÷ 37

Question 13.
Solution:
Capacity of 69 buckets = 586.5 litres
Capacity of 1 bucket = 586.5 ÷ 69

Question 14.
Solution:
Number of pieces in 1.15 m = 1

Question 15.
Solution:
Total weight of cement = 1792.8 kg
Cement in one bag = 49.8 kg
Number of bags = 1792.8 ÷ 49.8

Question 16.
Solution:
Total thickness = 1.89 m = 189 cm
Thickness of one piece = 0.3 5 cm
Number of pieces = 189 ÷ 0.35

Question 17.
Solution:
Product of two decimals = 261.36
One decimal = 17.6
Second decimal = 261.36 ÷ 17.6

Ex 3E Solutions

OBJECTIVE QUESTIONS
Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(b)

Question 2.
Solution:
(c)

Question 3.
Solution:
(b)

= 208100 = 2.08

Question 4.
Solution:

Question 5.
Solution:
(b) 70g = 701000 = 0.07 kg

Question 6.
Solution:
(c) 5 kg 6 g = 561000 kg = 5.006 kg

Question 7.
Solution:
(c) 2 km 5 m = 251000 km = 2.005 km

Question 8.
Solution:
(c)
1.007 – 0.7 = 1.007 – 0.700 = 0.307

Question 9.
Solution:
(b)
0.1 – 0.03 = 0.10 – 0.03 = 0.07

Question 10.
Solution:
(c)
3.5 – 3.07 = 3.50 – 3.07 = 0.43

Question 11.
Solution:
(c)
0.23 x 0.3 = 0.069

Question 12.
Solution:
(b)
0.02 x 30 = .60 = .6

Question 13.
Solution:
(b)
0.25 x 0.8 = 0.200 = 0.2

Question 14.
Solution:
(c)
0.4 x 0.4 x 0.4 = 0.064

Question 15.
Solution:
(b)
1.1 x .1 x .01 = .0011

Question 16.
Solution:
(a)

Question 17.
Solution:
(b)
1.02 ÷ 6 = 1.026 = 0.17

Question 18.
Solution:

Question 19.
Solution:
(b)

Question 20.
Solution:
(a)

Question 21.
Solution:
(c)

RS Aggarwal Solutions for Class 7 Maths Chapter 3: Download PDF

RS Aggarwal Solutions for Class 7 Maths Chapter 3–Decimals

Download PDF: RS Aggarwal Solutions for Class 7 Maths Chapter 3–Decimals PDF

Chapterwise RS Aggarwal Solutions for Class 7 Maths :

• Chapter 1–Integers
• Chapter 2–Fractions
• Chapter 3–Decimals
• Chapter 4–Rational Numbers
• Chapter 5–Exponents
• Chapter 6–Algebraic Expressions
• Chapter 7–Linear Equations in One Variable
• Chapter 8–Ratio and Proportion
• Chapter 9–Unitary Method
• Chapter 10–Percentage
• Chapter 11–Profit and Loss
• Chapter 12–Simple Interest
• Chapter 13–Lines and Angles
• Chapter 14–Properties of Parallel Lines
• Chapter 15–Properties of Triangles
• Chapter 16–Congruence
• Chapter 17–Constructions
• Chapter 18–Reflection and Rotational Symmetry
• Chapter 19–Three-Dimensional Shapes
• Chapter 20–Mensuration
• Chapter 21–Collection and Organisation of Data (Mean, Median and Mode)
• Chapter 22–Bar Graphs
• Chapter 23–Probability

About RS Aggarwal Class 7 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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