CBSE Class 7

RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

Question 1.
Solution:
(i) Draw a number line and locate a point O on it. Let it represent 0 Now \(\frac { 1 }{ 3 }\) has been presented on the number line given below.

(ii) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { 2 }{ 7 }\) has been represented on the number line given below:

(iii) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { 7 }{ 3 }\) has been represented on the number line given below:

(iv) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { 7 }{ 3 }\) has been represented on it as given below:

(v) Draw a number line and locate a point O on it. Let it represented 0. The number \(\frac { 37 }{ 8 }\) has been represented on it as given below:
\(\frac { 37 }{ 8 }\) = 4\(\frac { 5 }{ 8 }\)

(vi) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { -1 }{ 3 }\) has been represented on it as given below:

(vii) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { -3 }{ 4 }\) has been represented on it is as given below:

(viii) Draw a number line and locate a point on it. Let it represent 0. The number \(\frac { -12 }{ 7 }\) has been represented on it as given below:

(ix) Draw a number line and locate a point O on it. Let it represent 0. The number \(\frac { 36 }{ -5 }\) has been represented on it as given below:

(x) Draw a number line and locate is point O on it. Let is represent 0. The number \(\frac { -43 }{ 9 }\) has been represented on it as given below:

Question 2.
Solution:
(i) \(\frac { 5 }{ 6 }\) or 0, \(\frac { 5 }{ 6 }\) is greater as any positive number is always greater than 0.

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
(i) True: All negative numbers lie on the left of 0.
(ii) False: All negative numbers lie on the left of 0.
(iii) True: All positive numbers lie on the right of 0 and all negative numbers on the left of 0.
(iv) False: \(\frac { -18 }{ -13 }\) = \(\frac { 18 }{ 13 }\) which is positive and positive number lie on the left of 0.
(v) True: \(\frac { -5 }{ -8 }\) = \(\frac { 5 }{ 8 }\) which is positive and all positive number lie on the right of negative numbers.
(i), (iii) and (iv) are true.

Question 8.
Solution:
5 rational numbers between -3 and -2.

Question 9.
Solution:

Question 10.
Solution:
L.C.M. of 5 and 2 = 10

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4B
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4C
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4D
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4F
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4G
• RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers CCE Test Paper

Question 1.
Solution:
(i) Rational numbers: The numbers of the form \(\frac { p }{ q }\) where p and q are integers and q ≠ 0, are called rational numbers.

(iv) Yes, there is one rational number (0) which is neither positive nor negative.

Question 2.
Solution:

(viii) \(\frac { 0 }{ 1 }\) are all rational number but \(\frac { 1 }{ 0 }\) and \(\frac { 0 }{ 0 }\) are not rational number as their denominator is zero.

Question 3.
Solution:
(i) Numerator = 8, denominator =19
(ii) Numerator = 5, denominator = – 8
(iii) Numerator =-13, denominator =15
(iv) Numerator = – 8, denominator = -11
(v) Numerator = 9, denominator = 1

Question 4.
Solution:

Question 5.
Solution:
According to the definition, a rational number is positive if both of numerator and denominator have same signs. Therefore
(iii), (iv) and (vi) 8 are positive rational numbers.

Question 6.
Solution:
According to the definition, a rational number is negative if numerator and denominator have opposite sign. Therefore.
(iii), (iv), (v), (vi) are all negative rational numbers.

Question 7.
Solution:
Equivalent rational numbers of each are given below:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:
(i) False, as there is no end of smallest and largest rational number,
(ii) True.
(iii) False, as zero is a rational number but the division of zero is meaningless.
(iv) True.
(v) False, every rational is not a fraction
In a fraction, numerator and denominators is a whole number but the denominator can’t be zero

Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 3 Decimals CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Question 1.
Solution:
Cost of 1 pen = ₹ 32.50
Cost of 24 such pens = ₹ (32.50 x 24) = ₹ 80
Hence, the cost of 24 pens is ₹ 780.

Question 2.
Solution:
Distance covered by the bus in 1 hour = 64.5 km
Distance covered in 18h = (64.5 x 18) km = 1161 km
Hence, the bus can cover a distance of 1161 km in 18h.

Question 3.
Solution:
First, we will find the product 68 x 65 x 4
Now, 68 x 65 x 4 = 4420 x 4 = 17680
Sum of decimal places in the given decimals = (2 + 1 + 2) = 5
So, the product have five decimal places.
0.68 x 6.5 x 0.04 = 0.17680 = 0.1768

Question 4.
Solution:
Total weight of all the bags = 2231 kg
Weight of each bag = 48.5 kg

Question 5.
Solution:

Question 6.
Solution:
Product of the given decimals = 1.824
One decimal = 0.64
The other decimal = 1.824 ÷ 0.64

Hence, the other decimal is 2.85.

Question 7.
Solution:
Thickness of the pile of plywoods = 2.43 m = 2.43 x 100 cm = 243 cm
Thickness of one piece of plywood = 0.45 cm
Required no. of pieces of plywood

Hence, the required number of pieces of plywood is 540.

Question 8.
Solution:
Let the number of sides of the polygon be n.
Length of each side of the polygon = 3.8 cm
Perimeter of the polygon = (3.8 x n) cm
But it is given that its perimeter is 22.8 cm.

Hence, the given polygon has six sides.

Mark (✓) against the correct answer in each of the following :
Question 9.
Solution:
(b) 2.04

Question 10.
Solution:
(b) 1\(\frac { 1 }{ 125 }\)

Question 11.
Solution:
(c) 2.005 kg
2 kg 5 g = (2 x 1000) g + 5 g = (2005)g
= \(\frac { 2005 }{ 1000 }\) kg = 2.005 kg

Question 12.
Solution:
(b) 0.08
We have :

Question 13.
Solution:
(c) 0.011
First, we will find the product 11 x 1 x 1
i.e. 11 x 1 x 1 = 11 x 1 = 11
Sum of decimal places in the given decimals = (1 + 1 + 2) = 4
1.1 x 0.1 x 0.01 = 0.0011 [4 places of decimal]

Question 14.
Solution:
(b) 2.03

Question 15.
Solution:
(c) is correct
Let the number added be x We have :
2.06 + x = 3.1
⇒ x = 3.1 – 2.06
Converting the given decimals into like decimals, we get:
2.06 and 3.10
Thus, required number = (3.10 – 2.06) = 1.04
Hence, 1.04 should be added to 2.06 to get 3.1.

Question 16.
Solution:
(b) 0.06 .
We have :
0.1 – x = 0.04
⇒ x = 0.1 – 0.04
Converting the given decimals into like decimals, we get:
0.10 and 0.04
Thus, required number = (0.10 – 0.04) = 0.06
Hence, 0.06 should be subtracted from 0.1 to get 0.04.

Question 17.
Solution:
(i) 1.001 ÷ 14 = 0.0715


47 x 53 = 2491
Sum of decimal places in the given decimals = (2 + 1) = 3
0.47 x 5.3 = 2.491
(iv) 0.023 x 0.03 = 0.69
Explanation first, we will multiply 23 by 3
23 x 3 = 69
Sum of decimal places in the given decimals = (3 + 2) = 5
0.023 x 0.03 =0.00069
(v) (0.7)² = 0.69
Explanation : (0.7)² = 0.7 x 0.7
First, we will find the product 0.7 x 0.7
Now, 7 x 7 = 49
Sum of decimal places in the given decimals = (1 + 1) = 2
So, the product must have two decimal places.
(0.7)² = 0.7 x 0.7 = 0.49
(vi) (0.05)³ = 0.000125
Explanation : First, we will find the
product 0.05 x 0.05 x 0.05
Now, 5 x 5 x 5 = 125
Sum of decimal places in the given decimals = (2 + 2 + 2) = 6
So, the product must have six decimal places.
(0.05)² = 0.05 x 0.05 x 0.05 = 0.000125

Question 18.
Solution:
(i) False
We have :
0.5 x 0.05 Now, 5 x 5 = 25
Sum of decimal places in the given decimals = (1 + 2) = 3
0.5 x 0.5 = 0.025
(ii) True
We have :
0.25 x 0.8
Now, 25 x 8 = 200
Sum of decimal places in the given decimals = (2 + 1) = 3
0.25 x 0.8 = 0.200 = 0.2
(iii) True
We have :
0.35 ÷ 0.7

(iv) False We have :
0.4 x 0.4 x 0.4
Now, 4 x 4 x 4 = 64
Sum of decimal places in the given decimals = (1 + 1 + 1) = 3
0.4 x 0.4 x 0.4 = 0.064
(v) True
6 cm = \(\frac { 6 }{ 100 }\) m = 0.06 m

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

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

Question 2.
Solution:
(c)

Question 3.
Solution:
(b)

= \(\frac { 208 }{ 100 }\) = 2.08

Question 4.
Solution:

Question 5.
Solution:
(b) 70g = \(\frac { 70 }{ 1000 }\) = 0.07 kg

Question 6.
Solution:
(c) 5 kg 6 g = 5\(\frac { 6 }{ 1000 }\) kg = 5.006 kg

Question 7.
Solution:
(c) 2 km 5 m = 2\(\frac { 5 }{ 1000 }\) 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 = \(\frac { 1.02 }{ 6 }\) = 0.17

Question 18.
Solution:

Question 19.
Solution:
(b)

Question 20.
Solution:
(a)

Question 21.
Solution:
(c)

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

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. \(\frac { 9255.60 }{ 24 }\) = 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 = \(\frac { 22.8 }{ 2.4 }\) 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

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.