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Case Study Questions for Class 7 Maths Chapter 1 Integers

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Here in this article, we are providing case study questions for Class 7 Maths Chapter 1 Integers.

Maths Class 7 Chapter List

Latest chapter list (2023-24).

There is total 13 chapters.

Chapter 1 Integers Case Study Questions Chapter 2 Fractions and Decimals Case Study Questions Chapter 3 Data Handling Case Study Questions Chapter 4 Simple Equations Case Study Questions Chapter 5 Lines and Angles Case Study Questions Chapter 6 The Triangles and its Properties Case Study Questions Chapter 7 Comparing Quantities Case Study Questions Chapter 8 Rational Numbers Case Study Questions Chapter 9 Perimeter and Area Case Study Questions Chapter 10 Algebraic Expressions Case Study Questions Chapter 11 Exponents and Powers Case Study Questions Chapter 12 Symmetry Case Study Questions Chapter 13 Visualising Solid Shapes Case Study Questions

Old Chapter List

Chapter 1 Integers Chapter 2 Fractions and Decimals Chapter 3 Data Handling Chapter 4 Simple Equations Chapter 5 Lines and Angles Chapter 6 The Triangles and its Properties Chapter 7 Congruence of Triangles Chapter 8 Comparing Quantities Chapter 9 Rational Numbers Chapter 10 Practical Geometry Chapter 11 Perimeter and Area Chapter 12 Algebraic Expressions Chapter 13 Exponents and Powers Chapter 14 Symmetry Chapter 15 Visualising Solid Shapes

Deleted Chapter:

Chapter 7 Congruence of Triangles
Chapter 10 Practical Geometry

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RD Sharma Solutions
Chapter 1 Integers

RD Sharma Solutions for Class 7 Maths Chapter - 1 Integers

RD Sharma Solutions for Class 7 Chapter 1 Integers PDF is provided here. Students can download the PDF of these solutions from the given links. Class 7 is a stage where several important topics are introduced. These crucial topics are discussed here in a particular way. The solutions are provided in accordance with the latest syllabus of CBSE , which, in turn, helps the students to build a strong foundation and secure excellent marks in the final exam. Therefore, we, at BYJU’S, provide answers to all questions uniquely and briefly. The solutions to all questions in RD Sharma books are given here in a step-by-step way to help the students understand effectively. In this chapter, students will learn about the multiplication and division of integers and the various properties of these operations on integers.

RD Sharma Solutions Class 7 Maths Chapter 1 Integers
RD Sharma Solutions Class 7 Maths Chapter 2 Fractions
RD Sharma Solutions Class 7 Maths Chapter 3 Decimals
RD Sharma Solutions Class 7 Maths Chapter 4 Rational Numbers
RD Sharma Solutions Class 7 Maths Chapter 5 Operations on Rational Numbers
RD Sharma Solutions Class 7 Maths Chapter 6 Exponents
RD Sharma Solutions Class 7 Maths Chapter 7 Algebraic Expressions
RD Sharma Solutions Class 7 Maths Chapter 8 Linear Equations in One Variable
RD Sharma Solutions Class 7 Maths Chapter 9 Ratio and Proportion
RD Sharma Solutions Class 7 Maths Chapter 10 Unitary Method
RD Sharma Solutions Class 7 Maths Chapter 11 Percentage
RD Sharma Solutions Class 7 Maths Chapter 12 Profit and Loss
RD Sharma Solutions Class 7 Maths Chapter 13 Simple Interest
RD Sharma Solutions Class 7 Maths Chapter 14 Lines and Angles
RD Sharma Solutions Class 7 Maths Chapter 15 Properties of Triangles
RD Sharma Solutions Class 7 Maths Chapter 16 Congruence
RD Sharma Solutions Class 7 Maths Chapter 17 Constructions
RD Sharma Solutions Class 7 Maths Chapter 18 Symmetry
RD Sharma Solutions Class 7 Maths Chapter 19 Visualising Solid Shapes
RD Sharma Solutions Class 7 Maths Chapter 20 Mensuration I (Perimeter and Area of Rectilinear Figures)
RD Sharma Solutions Class 7 Maths Chapter 21 Mensuration II (Area of Circle)
RD Sharma Solutions Class 7 Maths Chapter 22 Data Handling I (Collection and Organisation of Data)
RD Sharma Solutions Class 7 Maths Chapter 23 Data Handling II (Central Values)
RD Sharma Solutions Class 7 Maths Chapter 24 Data Handling III (Constructions of Bar Graphs)
RD Sharma Solutions Class 7 Maths Chapter 25 Data Handling IV (Probability)
Exercise 1.1 Chapter 1 Integers
Exercise 1.2 Chapter 1 Integers
Exercise 1.3 Chapter 1 Integers
Exercise 1.4 Chapter 1 Integers

RD Sharma for Class 7 Maths Chapter 1 Integers

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Access answers to Maths RD Sharma Solutions for Class 7 Chapter 1 – Integers

Exercise 1.1 Page No: 1.5

1. Determine each of the following products:

(i) 12 × 7

(ii) (-15) × 8

(iii) (-25) × (-9)

(iv) 125 × (-8)

(i) Given 12 × 7

Here we have to find the products of given numbers

12 ×7 = 84

Because the product of two integers of like signs is equal to the product of their absolute values.

(ii) Given (-15) × 8

(-15) ×8 = -120

Because the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.

(iii) Given (-25) × (-9)

(-25) × (-9) = + (25 ×9) = +225

(iv) Given 125 × (-8)

125 × (-8) = -1000

2. Find each of the following products:

(i) 3 × (-8) × 5

(ii) 9 × (-3) × (-6)

(iii) (-2) × 36 × (-5)

(iv) (-2) × (-4) × (-6) × (-8)

(i) Given 3 × (-8) ×5

Here we have to find the product of given number.

3 × (-8) × 5 = 3 × (-8 × 5)

=3 × -40 = -120

Since the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.

(ii) Given 9 × (-3) × (-6)

9 × (-3) × (-6) = 9 × (-3 × -6) [∵ the product of two integers of like signs is equal to

the product of their absolute values.]

=9 × +18 = +162

(iii) Given (-2) × 36 × (-5)

(-2) × 36 × (-5) = (-2 × 36) × -5 [∵ the product of two integers of like signs is equal to

=-72 × -5 = +360

(iv) Given (-2) × (-4) × (-6) × (-8)

(-2) × (-4) × (-6) × (-8) = (-2 × -4) × (-6 × -8) [∵ the product of two integers of like signs is

equal to the product of their absolute values.]

=-8 × -48 = +384

3. Find the value of:

(i) 1487 × 327 + (-487) × 327

(ii) 28945 × 99 – (-28945)

(i) Given 1487 × 327 + (-487) × 327

By using the rule of multiplication of integers, we have

1487 × 327 + (-487) × 327 = 486249 – 159249

(ii) Given 28945 × 99 – (-28945)

28945 × 99 – (-28945) = 2865555 + 28945

Since the product of two integers of like signs is equal to the product of their absolute values.

4. Complete the following multiplication table:

Second number

Is the multiplication table symmetrical about the diagonal joining the upper left corner to the lower right corner?

From the table it is clear that, the table is symmetrical about the diagonal joining the upper left corner to the lower right corner.

5. Determine the integer whose product with ‘-1’ is

(i) Given 58

Here we have to find the integer which is multiplied by -1

We get, 58 × -1 = -58

(ii) Given 0

We get, 0 × -1 = 0 [because anything multiplied with 0 we get 0 as their result]

(iii) Given -225

We get, -225 × -1 = 225

Exercise 1.2 Page No: 1.8

(i) 102 by 17

(ii) -85 by 5

(iii) -161 by -23

(iv) 76 by -19

(v) 17654 by -17654

(vi) (-729) by (-27)

(vii) 21590 by -10

(viii) 0 by -135

(i) Given 102 by 17

We can write given question as 102 ÷ 17

102 ÷ 17 = |102/17| = |102|/|17| [by applying the mod]

= 102/17 = 6

(ii) Given -85 by 5

We can write given question as -85 ÷ 5

-85 ÷ 5 = |-85/5| = |-85|/|5| [by applying the mod]

= -85/5 = -17

(iii) Given -161 by -23

We can write given question as -161 ÷ -23

-161 ÷ -23 = |-161/-23| = |-161|/|-23| [by applying the mod]

= 161/23 = 7

(iv) Given 76 by -19

We can write given question as 76 ÷ -19

76 ÷ -19 = |76/-19| = |76|/|-19| [by applying the mod]

= 76/-19 = -4

(v) Given 17654 by -17654

We can write given question as 17654 ÷ -17654

17654 ÷ -17654 = |17654/-17654| = |17654|/|-17654| [by applying the mod]

= 17654/-17654 = -1

(vi) Given (-729) by (-27)

We can write given question as (-729) ÷ (-27)

(-729) ÷ (-27) = |-729/-27| = |-729|/|-27| [by applying the mod]

= 729/27 = 27

(vii) Given 21590 by -10

We can write given question as 21590 ÷ -10

21590 ÷ -10 = |21590/-10| = |21590|/|-10| [by applying the mod]

= 21590/-10 = -2159

(viii) Given 0 by -135

We can write given question as 0 ÷ -135

0 ÷ -135 = 0 [because anything divided by 0 we get the result as 0]

Exercise 1.3 Page No: 1.9

Find the value of

1. 36 ÷ 6 + 3

Given 36 ÷ 6 + 3

According to BODMAS rule we have to operate division first then we have to do addition

Therefore 36 ÷ 6 + 3 = 6 + 3 = 9

2. 24 + 15 ÷ 3

Given 24 + 15 ÷ 3

Therefore 24 + 15 ÷ 3 = 24 + 5 = 29

3. 120 – 20 ÷ 4

Given 120 – 20 ÷ 4

According to BODMAS rule we have to operate division first then we have to do subtraction

Therefore 120 – 20 ÷ 4 = 120 – 5 = 115

4. 32 – (3 × 5) + 4

Given 32 – (3 × 5) + 4

According to BODMAS rule we have to operate in brackets first then move to addition and subtraction.

Therefore 32 – (3 × 5) + 4 = 32 – 15 + 4

= 32 – 11 = 21

5. 3 – (5 – 6 ÷ 3)

Given 3 – (5 – 6 ÷ 3)

According to BODMAS rule we have to operate in brackets first then we have move to subtraction.

Therefore 3 – (5 – 6 ÷ 3) = 3 – (5 – 2)

= 3 –3 = 0

6. 21 – 12 ÷ 3 × 2

Given 21 – 12 ÷ 3 × 2

According to BODMAS rule we have to perform division first then move to multiplication and subtraction.

Therefore, 21 – 12 ÷ 3 × 2 = 21 – 4 × 2

= 21 – 8 = 13

7. 16 + 8 ÷ 4 – 2 × 3

Given 16 + 8 ÷ 4 – 2 × 3

According to BODMAS rule we have to perform division first followed by multiplication, addition and subtraction.

Therefore, 16 + 8 ÷ 4 – 2 × 3 = 16 + 2 – 2 × 3

= 16 + 2 – 6

8. 28 – 5 × 6 + 2

Given 28 – 5 × 6 + 2

According to BODMAS rule we have to perform multiplication first followed by addition and subtraction.

Therefore, 28 – 5 × 6 + 2 = 28 – 30 +2

= 28 – 28 = 0

9. (-20) × (-1) + (-28) ÷ 7

Given (-20) × (-1) + (-28) ÷ 7

Therefore, (-20) × (-1) + (-28) ÷ 7 = (-20) × (-1) – 4

= 20 – 4 = 16

10. (-2) + (-8) ÷ (-4)

Given (-2) + (-8) ÷ (-4)

According to BODMAS rule we have to perform division first followed by addition and subtraction.

Therefore, (-2) + (-8) ÷ (-4) = (-2) + 2

11. (-15) + 4 ÷ (5 – 3)

Given (-15) + 4 ÷ (5 – 3)

Therefore, (-15) + 4 ÷ (5 – 3) = (-15) + 4 ÷ 2

12. (-40) × (-1) + (-28) ÷ 7

Given (-40) × (-1) + (-28) ÷ 7

(-40) × (-1) + (-28) ÷ 7 = (-40) × (-1) – 4

= 40 – 4

13. (-3) + (-8) ÷ (-4) -2 × (-2)

Given (-3) + (-8) ÷ (-4) -2 × (-2)

(-3) + (-8) ÷ (-4) -2 × (-2) = -3 + 2 -2 × (-2)

= -3 + 2 + 4

= 6 – 3

14. (-3) × (-4) ÷ (-2) + (-1)

Given (-3) × (-4) ÷ (-2) + (-1)

(-3) × (-4) ÷ (-2) + (-1) = -3 × 2 -1

= – 6 – 1

Exercise 1.4 Page No: 1.12

Simplify each of the following:

1. 3 – (5 – 6 ÷ 3)

According to removal of bracket rule firstly remove inner most bracket

We get 3 – (5 – 6 ÷ 3) = 3 – (5 – 2)

= 3 – 3

2. -25 + 14 ÷ (5 – 3)

Given -25 + 14 ÷ (5 – 3)

We get -25 + 14 ÷ (5 – 3) = -25 + 14 ÷ 2

RD Sharma Solutions for class 7 Chapter 1 Integers Exercise 1.4 image 1

According to removal of bracket rule first we have to remove vinculum we get

= 25 – ½ {5 + 4 – (5 – 4)}

Now by removing the innermost bracket we get

= 25 – ½ {5 + 4 – 1}

By removing the parentheses we get

= 25 – ½ (8)

Now simplifying we get

= 25 – 4

RD Sharma Solutions for class 7 Chapter 1 Integers Exercise 1.4 image 3

= 27 – [38 – {46 – (15 – 11)}]

Now by removing inner most bracket we get

= 27 – [38 – {46 – 4}]

= 27 – [38 – 42]

Now by removing braces we get

= 27 – (-4)

RD Sharma Solutions for class 7 Chapter 1 Integers Exercise 1.4 image 5

By removing innermost bracket we get

= 36 – [18 – {14 – (11 ÷ 2 × 2)}]

= 36 – [18 – {14 – 11}]

Now by removing the parentheses we get

= 36 – [18 – 3]

Now remove the braces we get

= 36 – 15

6. 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]

Given 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]

First remove the inner most brackets

= 45 – [38 – {20 – (6 – 3) ÷ 3}]

= 45 – [38 – {20 – 3 ÷ 3}]

Now remove the parentheses we get

= 45 – [38 – 19]

= 45 – 19

RD Sharma Solutions for class 7 Chapter 1 Integers Exercise 1.4 image 7

Now first remove the vinculum we get

= 23 – [23 – {23 – (23 – 0)}]

Now remove the innermost bracket we get,

= 23 – [23 – {23 – 23}]

By removing the parentheses we get,

= 23 – [23 -0]

Now we have to remove the braces and on simplifying we get,

= 23 – 23

RD Sharma Solutions for class 7 Chapter 1 Integers Exercise 1.4 image 9

First we have to remove the vinculum from the given equation we get,

= 2550 – [510 – {270 – (90 – 150)}]

= 2550 – [510 – {270 – (-60)}]

= 2550 – [510 – {270 + 60}]

Now remove the parentheses we get,

= 2550 – [510 – 330]

Now we have to remove braces

= 2550 – 180

RD Sharma Solutions for class 7 Chapter 1 Integers Exercise 1.4 image 11

First we have to remove vinculum from the given equation,

= 4 + 1/5 [{-10 × (25 – 10)} ÷ (-5)]

Now remove the innermost bracket, we get

= 4 + 1/5 [{-10 × 15} ÷ -5]

Now by removing the parentheses we get,

= 4 + 1/5 [-150 ÷ -5]

By removing the braces we get,

= 4 + 1/5 (30)

On simplifying we get,

RD Sharma Solutions for class 7 Chapter 1 Integers Exercise 1.4 image 13

Now we have to remove innermost bracket

= 22 – ¼ {-5 – (- 48 ÷ – 16)}

After removing innermost bracket

= 22 – ¼ {-5 – 3}

= 22 – ¼ (-8)

RD Sharma Solutions for class 7 Chapter 1 Integers Exercise 1.4 image 15

First we have to remove vinculum from the given equation then we get,

= 63 – [(-3) {-2 – 5}] ÷ [3 {5 + 2}]

Now remove the parentheses from the above equation

= 63 – [(-3) (-7)] ÷ [3 (7)]

= 63 – [21] ÷ [21]

= 63 – 1

RD Sharma Solutions for class 7 Chapter 1 Integers Exercise 1.4 image 17

First we have to remove the innermost brackets then we get,

= [29 – (-2) {6 – 4}] ÷ [3 × {5 + 6}]

Now remove the parentheses in the above equation,

= [29 + 2 (2)] ÷ [3 × 11]

Now remove all braces present in the above equation,

= 33 ÷ 33

13. Using brackets, write a mathematical expression for each of the following:

(i) Nine multiplied by the sum of two and five.

(ii) Twelve divided by the sum of one and three.

(iii) Twenty divided by the difference of seven and two.

(iv) Eight subtracted from the product of two and three.

(v) Forty divided by one more than the sum of nine and ten.

(vi) Two multiplied by one less than the difference of nineteen and six.

(i) 9 (2 + 5)

(ii) 12 ÷ (1 + 3)

(iii) 20 ÷ (7 – 2)

(iv) 2 × 3 -8

(v) 40 ÷ [1 + (9 + 10)]

(vi) 2 × [(19 -6) -1]

RD Sharma Solutions for Class 7 Maths Chapter 1 – Integers

Chapter 1 Integers contains four exercises. RD Sharma Solutions are given here, which include the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.

Multiplication of integers
Properties of multiplication
Division of integers
Properties of division
Operator precedence
Use of brackets
Removal of brackets

Chapter Brief of RD Sharma Solutions for Class 7 Maths Chapter 1 – Integers

Chapter 1 – Integers primarily covers three topics, which are multiplication and division of integers and operator precedence. It also explains how to use and remove the brackets. Integers can be defined as a “number that can be written without a fractional component”. Here, the students will be thorough with the precedence of fundamental operations of addition, subtraction, multiplication and division.

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NCERT Solutions Class 7 Maths Chapter 1 Integers

NCERT solutions for class 7 maths chapter 1 integers comprise exercises based on differentiating integers from the higher sets, their properties and operations. Integers are non-fractional numbers that can hold a positive, negative, or zero value. These numbers are significantly important in performing arithmetic operations due to their prefixed signs. NCERT solutions class 7 maths chapter 1 integer covers all the important concepts based on the properties of these numbers in detail with examples.

The first few exercises in these class 7 maths NCERT solutions chapter 1 integers explain the representation of integers on a number line along with the revision of integers concepts studied in the previous classes with suitable examples. Sample problems covered in this chapter are sufficient for students to gain an in-depth understanding of integers and their properties. You can find some of these in the exercises given below and also find some of these in the exercises given below.

NCERT Solutions Class 7 Maths Chapter 1 Ex 1.1
NCERT Solutions Class 7 Maths Chapter 1 Ex 1.2
NCERT Solutions Class 7 Maths Chapter 1 Ex 1.3
NCERT Solutions Class 7 Maths Chapter 1 Ex 1.4

NCERT Solutions for Class 7 Maths Chapter 1 PDF

NCERT solutions for class 7 maths are well-structured to provide the optimal math learning of complete math syllabus. To get the best results in exams, one can easily plan preparation and revision with these competent resources. These solutions are also available for free pdf download as given below.

☛ Download Class 7 Maths NCERT Solutions Chapter 1 Integers

NCERT Class 7 Maths Chapter 1 Download PDF

$NCERT Solutions Class 7 Maths Chapter 1 Integers$

NCERT Solutions for Class 7 Maths Chapter 1 Integers

With a thorough practice of the above exercises, students will attain the right approach required for solving arithmetic operations on integers. The regular practice of these exercises will also guide the students through the types of questions and the stepwise solution to each of them. The chapter-wise detailed analysis of NCERT Solutions Class 7 Maths Chapter 1 Integers is given below.

Class 7 Maths Chapter 1 Ex 1.1 - 10 Questions
Class 7 Maths Chapter 1 Ex 1.2 - 4 Questions
Class 7 Maths Chapter 1 Ex 1.3 - 9 Questions
Class 7 Maths Chapter 1 Ex 1.4 - 7 Questions

☛ Download Class 7 Maths Chapter 1 NCERT Book

Topics Covered: Properties of integers and their arithmetic operations ( addition , subtraction , multiplication, and division ) are the most important topics. Other topics included in these class 7 maths NCERT solutions chapter 1 are the properties of commutativity, associativity under addition and multiplication , and the distributive property of multiplication of integers.

Total Questions: Class 7 maths chapter 1 Integers Chapter 1 consists of a total of 30 questions of which 9 are easy, 10 are moderate and11 are long answer type questions.

List of Formulas in NCERT Solutions Class 7 Maths Chapter 1

NCERT solutions class 7 maths chapter 1 covers lots of important concepts based on integers that are crucial for strengthening the math foundation. Some important concepts in NCERT solutions for class 7 maths chapter 1 are given below:

Commutative Property: a + b = b + a
Associative Property: a + (b + c) = (a + b) + c
Distributive Property: a × (b – c) = a × b – a × c

Important Questions for Class 7 Maths NCERT Solutions Chapter 1

Ncert solutions for class 7 maths video chapter 1, faqs on ncert solutions class 7 maths chapter 1, what is the importance of ncert solutions class 7 maths chapter 1 integers.

NCERT Solutions Class 7 Maths Chapter 1 deals with integers and their concepts that strengthen the student’s theoretical and practical knowledge. This enables students to score well in their examinations as NCERT solutions lay a strong mathematical foundation.

Do I Need to Practice all Questions Provided in NCERT Solutions for Class 7 Maths Integers?

Most questions in the NCERT Solutions Class 7 Maths Integers are important. Questions based on addition, subtraction, multiplication, and division of integers, should be regularly practiced while studying. The associative and distributive property of multiplication is one topic that needs regular practice.

What are the Important Topics Covered in NCERT Solutions Class 7 Maths Chapter 1?

The important sub-topics covered in NCERT Solutions Class 7 Maths Chapter 1 are properties of arithmetic operations of integers. Integers Chapter 1 is very well structured to include all the important sub-topics.

How Many Questions are there in Class 7 Maths NCERT Solutions Chapter 1 Integers?

There are a total of 30 questions in NCERT class 7 maths chapter 1 integers. Out of these 30 questions, 20 are carefully paced to provide a gradual understanding of the context of the theory. These 20 questions can be further categorized into long answers, moderate level, and easy ones. Students can plan and set a practice time for each of these subcategories.

What are the Important Formulas in NCERT Solutions Class 7 Maths Chapter 1?

The important formulas covered in the NCERT Solutions Class 7 Maths Chapter 1 are based on integer’s properties of addition and multiplication like commutative, associative, or distributive, etc. These formulas are also important for students to acquire a clear understanding of applying arithmetic operators to integers.

Why Should I Practice NCERT Solutions Class 7 Maths Integers Chapter 1?

Practicing NCERT Solutions Class 7 Maths Integers Chapter 1 can help develop excellent mathematical skills and accuracy. This will help in enhancing the student’s math skills. While practicing the various arithmetic operations on integers, make sure to use all the tips that have been mentioned in NCERT solutions.

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CBSE Class 7 Math

Class 7 Maths Chapter 1 Integers Practice Questions And Answers

by Sanjusha · October 18, 2019

CLASS 7 MATHEMATICS | INTEGERS – Chapter 1 – Practice Questions

Answer the following:

1. Write down a pair of integers whose: a) Sum is -5 b) Difference is -8 c) Sum is 0.

2. Write a pair of negative integers whose difference gives 6.

3. Fill in the blanks to make the following statements true:

a) (-2) + (-4) = (-4) + ———– b) -24 + ——– = -24 c) 12 + ——– = 0

4. Find each of the following products:

a) (-6) x (4) x 8 b) (-1) x (-4) x (-5) c)(-10) x (-4) x 0

5. Replace the blank with an integer to make it a true statement.

a) (-2) x —– = 16 b) 4 x ——- = -24 c) (-5) x (-4) = ———

6. Evaluate each of the following:

a) (-40)/10 b) 20/ (-4) c) 0/ (-13)

7. Fill in the blanks:

a) 65/—- = 65 b) (-21)/ —- = -1 c) 10/ —- = -2

8. In a test containing 10 questions, (+5) marks are given for every correct answer and (-2) marks are given for every incorrect answer.

a) Lalitha gets five correct and five incorrect answers. What is her score? b) Nalini gets six correct and four incorrect answers. What is her score?

9. The sum of two integers is 125. If one of them is -28, find the other integers.

10. If a = -6, b= 4 and c = -2, verify that:

a) a+ (b +c) = (a +b) +c b) a x (b +c) = a x b + a x c

1. a) (-10) and 5.

b) (-10) and (-2) c) (-5) and +5

2. (-10) and (-16)

3. a) (-2) + (-4) = (-4) + (-2)

b) (-24) + 0 = – 24 c) 12 + (-12) = 0

4. a)(-6) x 4 x 8 = -192

b) (-1) x (-4) x (-5) = -20 c) (-10) x (-4) x 0 = 0

5. a) (-2) x (-8) = 16

b) 4 x (-6) = -24 c) (-5) x (-4) = 20

6. a)-40/10 = -4

b) 20/ -4 = -5 c) 0/ -13 = 0

7. a) 65/1 = 65

b) (-21)/21 = -1 c) 10 / -5 = -2

8. a) Score for five correct answers = 5 x 5 = 25

Score for 5 incorrect answers = 5 x -2 = -10 Total score = 25 + (-10) = 15

b) Score for six correct answers = 6 x 5 = 30 Score for four incorrect answers = 4 x (-2) = -8 Total score = 30 + (-8) = 22

9. Sum of two integers = 125 (given)

One number = -28 (given) Other number = 125 – (-28) = 125 + 28 = 153.

10. a = -6, b= 4, c = -2 (given)

a) LHS = a + (b +c) = -6 + (4 + -2) = (-6) +2 = -4 RHS = (a + b) + c = (-6 + 4) + -2 = -2 + -2 = -4

So verified.

b) LHS = (-6) x (4 + -2) = (-6) x 2 = -12 RHS = (-6 x 4) + (-6 x -2) = -24 + 12 = -12 So verified.

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CBSE Class 7 Mathematics/Integers – Chapter 1 Extra Questions for Practice/Model Questions

October 10, 2023

by Sanjusha · Published October 10, 2023

CBSE Class 7 Mathematics/ Model Question Paper/ Yearly Examination.

March 14, 2021

by Sanjusha · Published March 14, 2021

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Integers class 7 extra questions with answers and PDF Download

Integers class 7

What are Integers ?

Properties of integers:

Negative integers are on the left side of 0
Positive integers are on the right side of the zero
0 is neither +ve nor -ve.
Integers are closed under addition. It means, for any two integers a and b, a + b is an integer.
Integers are also closed under subtraction. Thus, if a and b are two integers then a – b is also an integer.
Addition is commutative for integers. In general, for any two integers a and b, we can say a + b = b + a

$\neq$

Addition is associative for integers. In general for any integers a, b and c, we can say a + (b + c) = (a + b) + c
For any two positive integers a and b, we can say a × (– b) = (– a) × b = – (a × b)
Integers are closed under multiplication. In general, a × b is an integer, for all integers a and b
Multiplication is commutative for integers. For any two integers a and b, a × b = b × a
Distributivity of multiplication over addition is true for integers. In general, for any integers a, b and c, a × (b + c) = a × b + a × c
Integers are not closed under division.
Division is not commutative for integers.

Questions on integers for class 7 pdf

Sums on integers for class 7(Solved Examples )

Example 1: Arrange the following integers in ascending order -102, -39, -51, -5 , 0 , -6, 35 and 7 .

Sol. Ascending order of the given integers are : -102 < -51< -39 < -6< -5 < 0< 7< 35

Example 2: Find the sum of -72 , 237 , 84 , 72, -184 , -37 .

Sol. (-72) +237+84+72+(-184)+(-37)=(-72) +393 +(-184)+(-37)

= -72+393-184-37

=393-293 =100

Example 3: Write down a pair of integers whose (a) sum is –3 (b) difference is –5 (c) difference is 3 (d) sum is 0 Sol. (a) (–1) + (–2) = –3 or (–5) + 2 = –3 or (-7)+(4) =-3

(b) (–9) – (– 4) = –5 or (–2) – 3 = –5 or (-7) -(-2)= -5

(d) (–10) + 10 = 0 or 4 + (–4) = 0 or 3 +(-3) =0 (You can make more pairs.)

Example 4: Solve the following: (i) (-10) × (-5) + (-6) (ii) (-10) × [(-13) + (-10)] (iii) (-5) × [(-6) + 5] Sol. (i) (-10) × (-5) + (-6) = 50 – 6 = 44

(ii) (-20) × [(-13) + (-10)] = (-20) × (-23) = 460

(iii) (-5) × [(-6) + 5]= (-5) x (-1) =5

Example 5: In a class test containing 15 questions, 4 marks are given for every correct answer and (–2) marks are given for every incorrect answer. Anil attempts all questions but only 10 of her answers are correct. What is his total score? Sol. (i) Marks given for one correct answer = 4 So, marks given for 10 correct answers = 4 × 10 = 40 Marks given for one incorrect answer = – 2 So, marks given for 5 (= 15 – 10) incorrect answers = (–2) × 5 = –10 Therefore, Gurpreet’s total score = 40 + ( –10) = 30

Example 6: Is (–15) × [(–7) + (–1)] = (–15) × (–7) + (–15) × (–1)?

Sol. (–15) × [(–7) + (–1)]= (-15) x (-8) = 120

(–15) × (–7) + (–15) × (–1)= 105 + 15 = 120

Hence (–15) × [(–7) + (–1)] = (–15) × (–7) + (–15) × (–1)

Example 7: Find each of the following products: (a) (–20) × (–2) × (–5) × 7 (b) (–1) × (–5) × (– 4) × (– 6)

Sol. (a) (–20) × (–2) × (–5) × 7 =– 20 × (–2 × –5) × 7 = [–20 × 10] × 7 = – 1400

(b) (–1) × (–5) × (– 4) × (– 6) = [(–1) × (–5)] × [(– 4) × (– 6)] = 5 × 24 = 120

Example 8: A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?

$10 \times 5^{\circ} C$

Example 9: Find: (a) 125 ÷ (–25) (b) 80 ÷ (–5)

Sol. (a) 125 ÷ (-25) = (-125) ÷ 25 = =5

(b) 80 ÷ (-5) =(-80) ÷ 5 = -16

Example 10: Find: (a) (–36) ÷ (– 4) (b) (–201) ÷ (–3)

Sol. (a) (-36) ÷ (-4)= 36 ÷ 4 = 9

(b) (-201) ÷ (-3) = 201÷ 3 = 67

Example 11: Solve the following :

(a) 3+2-1 x 4 ÷ 2

Sol. 3+2-1 x 4 ÷ 2 = 3+2-1 x 2 = 3+2-2 = 3

(b) 53 x 2 -1 x 6

Sol. 53 x 2 -1 x 6 = 106-6 =100

Sol. 7 x 3 +8-2= 21 +8 – 2 = 29-2 = 27

(d) 12+(-3) +5 – (-2)

Sol. 12+(-3) +5 – (-2) = 12-3+5+2 =9+7 = 16

Unsolved Examples with answers

(i) Complete the following pattern: (a) 7, 3, – 1, – 5, _____, _____, _____. (b) – 2, – 4, – 6, – 8, _____, _____, _____. (c) 15, 10, 5, 0, _____, _____, _____. (d) – 11, – 8, – 5, – 2, _____, _____, _____.

$\square$

(iii) Solve the following: (a) (-15) × 8 + (-15) × 4 (b) [32 + 2 × 17 + (-6)] ÷ 15

(iv) a × (b – c) = a × b – …………

(v) (a) For any integer a, what is (–1) × a equal to? (b) Determine the integer whose product with (–1) is 24

(vi) Replace the blank with an integer to make it a true statement. (a) (–3) × _____ = 24 (b) 5 × _____ = –40 (c) _____ × (– 9) = –63 (d) _____ × (–12) = 132

(vii) For any integer a, a ÷ 1 = ……………..

(viii) For any integer a, a ÷ (- 1) =………….

(ix) Evaluate the following : (a) (-526)-(-217) (b) (-31) +31 (c) [(-6) x (-8) ] x 5 (d) -13 x (7-8) (e) (-3) x 8 x (-5)

(x) Match the following Column I Column II

(a) a × 1 (i) Additive inverse of a

(b) 1 (ii) Additive identity

(d) a × ( – 1) (iv) a ÷ ( – b)

(e) a × 0 (v) a ÷ b

(f) ( –a) ÷ b (vi) a

(g) 0 (vii) – a

(h) a ÷ (–a) (viii) 0

(i) –a (ix) –1

Ans. (i) (a) -9, -13, -17, -21 (b) -10, -12, -14, -16 (c) -5, -10, -15, -20 (d) 1, 4, 7, 10

(ii) (a) < (b) < (c) > (d) < (e) > (iii) (a) -180 (b) 4 (iv) a x c

(v) (a) -a (b) -24 (vi) (a) -8 (b) -8 (c) 7 (d) -11 (vii) a (viii) -a

(ix) (a) -309 (b) 0 (c) 240 (d) 13 (e) 120

(x) (a) → (vi), (b) → (iii), c → (v), d → (vii), e → (viii), f → (iv) g → (ii), h → (ix), i → (i)

DOWNLOAD PDF Integers class 7 worksheet

MCQ questions for class 7 maths integers

(a) < (b) > (c) = (d) none of these

(ii) Solve 40-(-39) + (-5)

(a) 74 (b) 64 (c) 60 (d) 0

(iii) When the integers 12, 0, 5, – 5, – 8 are arranged in descending or ascending order, then find out which of the following integers always remains in the middle of the arrangement.

(a) 0 (b) 5 (c) – 8 (d) – 5

(iv) Next three consecutive numbers in the pattern 11, 8, 5, 2, –, –, –are (a) 0, – 3, – 6 (b) – 1, – 5, – 8 (c) – 2, – 5, – 8 (d) – 1, – 4, – 7

(v)The …………… is an additive identity for integers.

(a)1 (b) 0 (c)-1 (d) 2

(vi) (–1) × (–1) × (–1) × (–1) × (–1) = ………….

(a) 1 (b) -1 (c) 0 (d) 2

(vii) …………………… is the multiplicative identity for integers.

(a) 0 (b) 1 (c) -1 (d) 2

(viii) 0 ÷ a = …………… for a ≠ 0

(a) 0 (b) 1 (c) -1 (d) not defined

(ix) any integer divided by zero is ……………………………

(a) 0 (b) meaningless (c) 1 (d) -1

(x) (– 85) × 43 + 43 × ( – 15) is equal to ?

(a) -4300 (b) 4300 (c) 430 (d) -430

Ans. (i) (a) (ii) (a) (iii) 0 (iv) (d) (v) (b) (vi)( b) (vii) (b) (viii) (a) (ix) (b) (x) (a)

Integers Class 7 Maths quiz

Maths quiz for class 7 integers

Find the absolute value of -5 .

2. Add the integers +15, -4 , + 8 and -6 .

3. Write a negative integer and a positive integer whose sum is -5 .

$-7^{\circ} C$

5. In a quiz team A scored -40 , 10 and 0 and team B scored 10 , 0 and -30 in three successive round . Which team scored more ?

6. Subtract the sum of -1878 and 878 from 2000 .

7. Subtract -134 from the sum of 37 and -87 .

8. If the sum of two integers is -1700 and one of them is 560 . Find the other number.

9. Determine the integer whose product with -17 is -153 .

10. Find the value (-23){(-5) +25}

11. Anil is standing in the middle of a bridge which is 30 m above the water level of a river. If a 35 m deep river is flowing under the bridge , then the vertical distance between the foot of Anil and bottom level of the river is?

12. [(– 10) × (+ 9)] + ( – 10) is equal to

13. –16 ÷ [8 ÷ (–2)] is equal to

14. (– 35) × 30 = – 30 × _______.

15. .Write a pair of integers whose product is – 36 and whose difference is 15.

Your score is

The average score is 67%

Restart quiz

Integers class 7 (extra questions with answers)

A. State whether the following statements are correct or incorrect.

(i) When two positive integers are added we get a positive integer.

(ii) When two negative integers are added we get a positive integer.

(iii) When a positive integer and a negative integer are added, we always get a negative integer.

(iv) Additive inverse of an integer 8 is (– 8) and additive inverse of (– 8) is 8.

(v) For subtraction, we add the additive inverse of the integer that is being subtracted, to the other integer.

(vi) (–10) + 3 = 10 – 3

(vii) 8 + (–7) – (– 4) = 8 + 7 – 4

(viii) 25 × (–21) = (–25) × 21

(ix) –1 is a multiplicative identity of integers?

(x) The distributivity of multiplication over addition is true for integers.

(xi) Is division associative for integers?

(xii) When we change the order of integers, their sum remains the same.

(xiii) When we change the order of integers their difference remains the same.

(xiv) a ÷ b = b ÷ a

(xv) a – b = b – a

B . Answer the following questions:

(i) Write a pair of integers whose sum is zero (0) but difference is 8.

(ii) Write a pair of integers whose product is – 15 and whose difference is 8.

(iii) On multiplying or dividing two integers, If the signs of both the integers are same, the sign of the answer is ………………..

(iv) On multiplying or dividing two integers, If the signs of both the integers are different, the sign of the answer is ………………

(v) If a, b and c are integers then a x (b+c)= ………….+…………..

(vi) The product of three negative integers is a …………….. integer.

(vii) The product of four negative integers is a …………………. integer.

(viii) If the number of negative integers in a product is even, then the product is a ……………….. integer; if the number of negative

integers in a product is odd, then the product is a …………………. integer.

(ix) What will be the sign of the product if we multiply together: (a) 8 negative integers and 3 positive integers?

(b) 5 negative integers and 4 positive integers?

(x) (-5) x (-10) =………… x (-5)

(xi) The sum of two integers is 116. If one of them is -79, find the other integers.

(xii) The product of three integers does not depend upon the grouping of integers and this is called the …………………….. for multiplication of integers

Ans. A. (i) True (ii) False (iii) False (iv) True (v) True (vi) False (viii) False (viii) True (ix) False (x) True (xi) False (xii) True (xiii) False (xiv) False (v) False

B. (i) 4 and -4 (ii)5, -3 and 3 ,-5. (iii) positive (iv) negative (v) (a x b)+(b x c) (vi) negative (vii) positive (viii) positive, negative (ix) (a) positive (b) negative (x) -10 (xi) 195 (xii) associative property

Integers class 7 extra questions word problems

(i) A spacecraft is at 4525km above the earth’s surface.if it descends at the rate of 5km per minute. What will be its position after 9 hours.

(ii)A fruit merchant earns a profit of Rs. 66 per bag of orange sold and a loss of Rs. 44 per bag of grapes sold.

(a)Merchant sells 1800 bags of orange and 2500 bags of grapes. What is the profit or loss?

(b) What is the number of bags of oranges to be sold to have neither profit nor loss if the number of grapes bags are sold is 900 bags?

(iii) Nathan ended round one of a quiz with 200 points. In round two, he scored -300 points and in the third round, he gained 200 points. What was his total score at the end of the third round?

(iv) A submarine submerges from the surface of the sea at the rate of 10 m/min. How long will it take to reach 300 m below the sea level?

(v)Roman civilization began in 509 B.C. and came to an end in 476 A.D. How long did Roman civilization last?

(vi)Metal mercury is a liquid at room temperature. Its melting point is -39°C. The freezing point of alcohol is -114°C. How much warmer is the melting point of mercury than the freezing point of alcohol?

(vii)The temperature in city-X was 10°C in the morning which dropped by 15°C in the evening. What is the temperature in the evening?

(viii)Everest, the highest peak in Asia, is 29,028 feet above sea level. The Dead Sea is 1,312 feet below sea level. What is the level difference between these two places? 7.

(ix)The temperature recorded in a city on Monday is 50°. If the temperature increases by 10° on Tuesday and falls by 11° on Wednesday. Find the temperature recorded on Wednesday

(x)In an examination,3 marks are awarded for every correct answer and1mark is deducted for every incorrect answer. If a student gets 15questions correct and7questions incorrect, find the marks scored by him.

(xi)Andrew has deposited $5500 and has withdrawn $4800 two times consecutively. Find the amount in his account.

(xii) A bird is flying 1200 feet above the sea level. At a particular point, it is directly above a fish floating at 400 feet below the sea level. By how much the bird must descend its flight to be as far as from the sea level, as the fish is?

(xiii)Ryan was playing a game where catching a fish in the basket adds 10 points to his score. If he catches an octopus instead, he loses 5 points. What is his total if he catches 3 fish and 5 octopus?

(xiv)Jason is moving from his hometown to Alaska. He is expecting a big temperature change, and he is getting ready. The hottest it gets in his hometown is 32 degrees celsius. The coldest it gets in Alaska is approximately – 45 degrees celsius. What temperature change will Jason experience?

(xv)The temperature of a hot iron rod drops by 8°𝐶 every hour. If the temperature of the iron rod is 118°𝐶, find the temperature after 4 hours.

(xvi)A submarine submerges at the rate of 5 m/min. If it descends from 20 m above the sea level, how long will it take to reach 250 m below sea level?

(xvii)At Srinagar temperature was − 5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?

(xviii)A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?

(xix)Mohan deposits Rs 2,000 in his bank account and withdraws Rs 1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal.

(xx)Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance travelled towards west? By which integer will you represent her final position from A?

(xxi) A cement company earns a profit of Rs 8 per bag of white cement sold and a loss of Rs 5 per bag of grey cement sold.

(a) The company sells 3, 000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?

(b) What is the number of white cement bags it must sell to have neither profit nor loss, if the number of grey bags sold is 6,400 bags.

(xxii) An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach − 350 m.

Ans. (i) 1825 km (ii) (a) profit amount is rupees 8800 . (b) 600 bags of oranges

(iii) 300 (iv) 30 minutes (v) 985 degrees (vi) -75 (vii) -5 (viii) 30340 feet

(ix) 49 degrees (x) 38 (xi) $ -4100 (xii) 800 feet (xiii) 5 points (xiv)77

degrees (xv) 86 degrees celcius (xvi) 54 minutes (xvii) Monday = −5°C, on

tuesday −7°C , on Wednesday −3ºC (xviii) 6200 m (xix) rs. 358 (xx) −10 km (i.e.,

Rita is now in west direction) (xxi) (a) loss of Rs 1000 (b) 4000 bags of white

cement (xxii) 1 hour

NCERT Solutions for Class 7 Maths Chapter 1 Integers PDF Download

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NCERT Solutions for Class 7 Maths Chapter 1 Integers, NCERT Solutions for Class 7 Maths Chapter 1 Integers PDF, NCERT Class 7 Maths Solutions Chapter 1 Integers, NCERT Solutions for Class 7 Maths Chapter 1 Integers Revision, NCERT Solutions for Class 7 Maths Chapter 1 Integers Theory

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Important Questions for CBSE Class 7 Maths (2023-24)

CBSE Class 7 Maths Chapter wise Important Questions - Free PDF Download

Class 7 Maths syllabus is set to create a foundation of knowledge among the students so that they can learn the advanced concepts in the higher classes easily. To make this conceptual base stronger, Vedantu has prepared a list of Class 7 Maths Important Questions for all chapters. Students will be able to find new patterns of questions in this list and can practice them to enhance their knowledge. They will learn new ways to use the formulas and concepts by practising solving these questions.

All these questions are formulated by following the standard CBSE rules . These questions focus on the conceptual development of students of Class 7. By solving these Important Questions for Class 7 Maths , you will strengthen your problem-solving skills and develop an idea of how to approach critical questions easily. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Register Online for NCERT Solutions Class 7 Science tuition on Vedantu.com to score more marks in CBSE board examination.

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Also, check CBSE Class 7 Maths Important Questions for other chapters

Chapterwise Important Questions for CBSE Class 7 Maths

NCERT Class 7 Maths Important Questions Summary

Students need to organize their study schedules properly to give attention to all the subjects. The overall development will ensure a good score in the final exam. Mathematics is a crucial subject to study and move forward to learn more concepts in the next classes. For this, you should pay attention to your learning process. Gather the ideal study material necessary for learning the new mathematical concepts and practice. The ideal way to practice is by downloading the NCERT Class 7 Maths Important Questions.

These questions have been composed by expert mathematics teachers. They have the best idea of how to challenge the students intellectually by asking conceptual questions. Apart from solving exercise questions, you will need a better platform to analyze your problem-solving skills. By attempting to solve something new and different from the exercise questions, you will be able to develop new methodologies and skills. These skills will then be used to develop more knowledge of mathematics in the following years.

14 chapters are covered in this section. You will find excellent questions framed by the teachers for all these chapters. Once you have completed the textbook exercises, you can proceed to solve these questions and check the solutions. Compare your answers with that of the solution to assess your skills. Learn how the teachers have approached and solved these problems using specific techniques. Practice these techniques of solving the Class 7 Important Questions for Maths to improve your preparation level and stay ahead of the class.

Why Should You Download Class 7 CBSE Maths Important Questions?

Many students follow one or two books apart from the mathematics textbook prescribed by NCERT . By following these books, they avail of an advanced platform for learning new skills to solve maths problems. These skills are ideal to save time, to avoid mistakes, and to score better during an exam. For those students, these important questions for every chapter will be a perfect tool to judge their skills.

These files can be easily downloaded in PDF format and stored on the computers. You can access them offline to practice as per your convenience. Let us check why you should get these Class 7 Maths Important Questions as a part of your study material.

Quick Review of Your Problem-Solving Skills

There are multiple chapters in Class 7 Maths NCERT syllabus . It is not possible to have the same efficiency of solving questions related to all the chapters. How can you determine the sections you need to focus on more? By solving these important questions for each chapter, you will get a clear idea of how efficient you are. Review your skills and make sure that the syllabus has been covered perfectly.

Important Questions for Exams

Most schools do not focus on the exercises to frame questions for an exam. The teachers like to pay attention to different types of questioning formats to assess the intellect and problem-solving skills of the students of Class 7 . It is beneficial to learn different types of question formats for all chapters beforehand. You can do this by downloading the Important Questions for Class 7 Maths . Check the different formats of mathematical questions and focus on solving techniques by referring to their solutions.

Clarification of Doubts in Different Chapters

The prime reason for checking and solving the new questions suggested by the experienced teachers is to clarify your doubts in different chapters. Most of the time, students are unaware of the doubts unless they are revealed while solving a problem. This is the reason why teachers suggest solving questions from different books after completing the exercises in the textbook. By attempting to solve these important questions and checking their solutions, your underlying doubts will also get resolved.

Download the important questions chapter-wise and solve them at your convenience. Complete the respective chapters first and then indulge in practising these questions.

Class 7 Maths Important Related Links

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FAQs on Important Questions for CBSE Class 7 Maths (2023-24)

1. Where Can You Find the Best NCERT Class 7 Maths Important Questions?

Ans: Vedantu is preferred to be the best platform for finding important questions related to the Class 7 Maths syllabus. These questions are ideal in terms of quality and skill development among students.

2. How Solving Important Questions can Help Your Preparation?

Ans: The underlying benefits of solving and practising Class 7 Important Questions for Maths are enhancing problem-solving skills, increasing efficiency, and helping to manage time during an exam. Learn new techniques from the solutions provided for each set.

3. How many chapters does the Math NCERT book of Class 7 consist of?

Ans: The Maths NCERT book of Class 7 consists of 15 chapters. These are:

Chapter 1 - Integers

Chapter 2–Fractions and Decimals

Chapter 3–Data Handling

Chapter 4–Simple Equations

Chapter 5–Lines and Angles

Chapter 6–The Triangles and its Properties

Chapter 7 – Congruence of Triangles

Chapter 8 – Comparing Quantities

Chapter 9 – Rational Numbers

Chapter 10 – Practical Geometry

Chapter 11 – Perimeter and Area

Chapter 12 – Algebraic Equation

Chapter 13 – Exponents and Powers

Chapter 14 - Symmetry

Chapter 15 – Visualizing Solid Shapes

4. How many exercises does each chapter of Class 7 Maths contain?

Ans: The number of exercises in each chapter of Class 7 Maths is given below:

Chapter 1 – four exercises

Chapter 2– seven exercises

Chapter 3 – four exercises

Chapter 4 – four exercises

Chapter 5 – two exercises

Chapter 6 – five exercises

Chapter 7 – two exercises

Chapter 8 – three exercises

Chapter 9 – two exercises

Chapter 10 – five exercises

Chapter 11 – four exercises

Chapter 12 – four exercises

Chapter 13 – three exercises

Chapter 14 – three exercises

Chapter 15 – four exercises

5. What should I do to get the important questions of all the chapters of Class 7 Maths?

Ans: Here is the technique which can help students to get important questions of all the chapters of Class 7 Maths in one place:

Visit the link of Important Questions for CBSE Class 7 Maths, Chapter wise Solutions.

After visiting the link, you will reach the official website of Vedantu.

You will discover the index of chapters of Class 7 Maths on the Vedantu's official page.

Here from the list, pick out the chapter you want to study.

After choosing the lesson, the important questions of that chapter will open.

There when you will hit the option of “Download PDF”, the PDF file of important questions will be downloaded. These solutions are available at free of cost on Vedantu(vedantu.com) and mobile app.

6. What are the perks of studying Class 7 Maths from the NCERT book?

Ans: There are many benefits of studying Class 7 Maths from the NCERT book. Some of them are:

The Maths NCERT book is prescribed by the CBSE for Class 7.

The content of this book is designed by well-qualified teachers.

The dialect used in this book is very easy to understand.

Every concept of the chapter is explained in detail in this book.

Solving the questions given in this book can help students to score well in their exams.

Most of the questions asked in the exam are taken from the NCERT book.

7. How can students prepare an effective study plan for Class 7 Maths?

Ans: For creating an effective study plan for Class 7 Maths do follow the given steps:

First of all, build a timetable so that you can get time to study every chapter.

Be thorough with your syllabus and the blueprint of the exam paper.

Solve each question of the NCERT book.

Practice the worksheets given by your teachers.

Do not miss any school lectures.

Prepare notes so that you can easily remember all the formulas.

Avoid learning new topics before the day of examination.

Important Questions for CBSE Class 7

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RS Aggarwal Solutions Class 7 Chapter 1 Integers

RS Aggarwal Solutions for Class 7 Maths Chapter 1 Integers are available here. This chapter has a total of five exercises. Study path has prepared solutions to these exercises by our expert math teachers to help you to get good marks in exams. RS Aggarwal Solutions for Class 7 Chapter 1 has a ton of questions. We at Study Path solved each questions step by step with detailed explanations. Students must practice from practice these problems to score high marks in Maths.

Class 7 RS Aggarwal Solutions Chapter 1 Integers

Class 7 rs aggarwal solutions chapter 1 ex 1a.

RS Aggarwal Solutions Class 7 Chapter 1 Integers Exercise 1A 0001

Class 7 RS Aggarwal Solutions Chapter 1 Ex 1B

RS Aggarwal Solutions Class 7 Chapter 1 Integers Exercise 1B 00001

Class 7 RS Aggarwal Solutions Chapter 1 Ex 1C

RS Aggarwal Solutions Class 7 Chapter 1 Integers Exercise 1C 001

Class 7 RS Aggarwal Solutions Chapter 1 Ex 1D

RS Aggarwal Solutions Class 7 Chapter 1 Integers Exercise 1D MCQs 0001

Class 7 RS Aggarwal Solutions Chapter 1 Test Paper

RS Aggarwal Solutions Class 7 Chapter 1 Integers Test Paper 00001

Chapter Brief of Chapter 1 Integers Class 7 RS Aggarwal Solutions

Below we have summarised the topics that has been discussed in this chapter.

Addition of Integers
Properties of Addition of Integers
Subtraction of Integers
Properties of Subtraction of Integers
Multiplication of Integers
Properties of Multiplication of Integers
Division of Integers
Properties of Division of Integers

We at Study Path work hard to provide you with the best solutions and study materials for free. We hope these solutions help you in your studies. Now, if have any doubts on RS Aggarwal Class 7 Solutions, please Comment below. We will definitely try to help you with this.

Unit 4: Integers

Negative numbers.

Negative symbol as opposite (Opens a modal)
Intro to negative numbers (Opens a modal)
Number opposites challenge Get 3 of 4 questions to level up!
Interpreting negative numbers (temperature and elevation) Get 3 of 4 questions to level up!

Comparing integers

Ordering negative numbers (Opens a modal)
Negative numbers, variables, number line (Opens a modal)
Ordering rational numbers Get 3 of 4 questions to level up!
Compare rational numbers using a number line Get 3 of 4 questions to level up!

Adding and subtracting

Adding numbers with different signs (Opens a modal)
Adding & subtracting negative numbers (Opens a modal)
Adding & subtracting negative numbers Get 5 of 7 questions to level up!
Addition & subtraction: find the missing value Get 3 of 4 questions to level up!

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CBSE Important Questions Class 7 Maths Chapter 1

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Important Questions Class 7 Mathematics Chapter 1 – Integers

Mathematics is an important subject that we need in our daily life too. Students must solve questions to clear their concepts and boost their confidence. The first chapter of Class 7 Mathematics under CBSE curriculum is integers.

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Students have learned integers in their previous class. In this chapter, they will learn how to put the integers on the number line, their properties, and the addition and multiplication of integers. It is a very important chapter. Students must practice the textbook exercise and questions from other sources to build their concepts.

Extramarks is a leading company that provides a wide range of study materials related to CBSE and NCERT. Our experts have made the Important Questions Class 7 Mathematics Chapter 1 to help students in regular practice. They collected the questions from different sources such as the textbook exercises, CBSE sample papers, CBSE past years’ question papers and important reference books. They have solved the questions too. Hence, the question series will help students increase their exam marks.

Extramarks is a leading company that helps students by providing all the important study materials related to CBSE and NCERT. You may register on our official website and download these study materials. You will find the CBSE syllabus, NCERT textbooks, CBSE past years’ question papers, CBSE sample papers, CBSE revision notes, CBSE extra questions, NCERT solutions, NCERT important questions, vital formulas and many more.

Important Questions Class 7 Mathematics Chapter 1 – With Solutions

The experts of Extramarks have made this question series so that students can solve the questions daily. They collected the questions from the textbook exercises, CBSE sample papers and important reference books. They have included a few questions from the past years’ question papers so that students may have an idea regarding questions in exams. Experienced professionals have further checked the answers to ensure the best quality of the content. Thus, the Important Questions Class 7 Mathematics Chapter 1 will help students to score better in exams. The questions are-

Question 1. Following number line given below shows the temperature present in degree celsius at different places on a particular day.

Image Source: Internet / NCERT Textbook

(i) Observe the number line and write down the temperature of the places marked on it.

By observing the above number line, we can find out the temperature of the cities as follows,

The temperature in the city of Lahulspiti is -8°C.

The temperature in the city of Srinagar is -2°C

The temperature in the city of Shimla is 5°C.

The temperature in the city of Ooty is 14°C.

The temperature in the city of Bengaluru is 22°C.

(ii) What is the temperature difference between the hottest and the coldest places among the cities stated above?

From the above number line, we can observe that,

The temperature at the given hottest place, that is, Bengaluru, is 22°C.

The temperature at the given coldest place, that is, Lahulspiti, is -8°C

The temperature difference between the hottest and the coldest place is given as = 22°C – (-8°C)

= 22°C + 8°C

= 30° Celsius

Hence, the total temperature difference between the hottest and the coldest place is 30oC.

(iii) What is the temperature difference between the cities of Lahulspiti and Srinagar?

From the above-given number line,

∴The temperature difference between the cities Lahulspiti and Srinagar is = -2oC – (8oC)

= – 2°C + 8°C

(iv) Can we say that the temperature of Srinagar and Shimla taken together is less than the temperature present at Shimla? Is it also less than the temperature present at Srinagar?

The temperature in the city of Srinagar =-2°C

The temperature in the city of Shimla = 5°C

The temperature of the cities Srinagar and Shimla taken together becomes = – 2°C + 5°C

= 3° degree C

5°C > 3°C

Hence, the temperature of the cities Srinagar and Shimla taken together is indeed less than the temperature present at Shimla.

3° > -2°

And No, the temperature of the cities Srinagar and Shimla taken together is not less than the temperature of the city Srinagar.

Question 2. Mohan deposits ₹ 2,000 in his bank account and then withdraws ₹ 1,642 from it the following day. Now, if the withdrawal of the amount from the account is represented by a negative integer, then how will you represent the total amount deposited? Also, Find the balance in Mohan’s account after the withdrawal.

Withdrawal of these amounts from the account is represented by a negative integer.

Then, the deposit of the amount to the account is represented by a positive integer.

From the above question,

The total amount that is deposited in the bank account by the Mohan = ₹ 2000

The total amount that is withdrawn from the bank account by the Mohan is = – ₹ 1642

Final Balance in Mohan’s account after the withdrawal = amount deposited + amount is withdrawn

= ₹ 2000 + (-₹ 1642)

= ₹ 2000 – ₹ 1642

Hence, the total balance in Mohan’s account after the withdrawal is ₹ 358

Question 3. In the following quiz, positive marks are given for every correct answer and negative marks are given for each incorrect answer. If Jack’s scores in the quiz for five successive rounds were 25, – 5, – 10, 10, and 15 so, what was his total at the end?

Jack’s scores in the five successive rounds are 25, -5, -10, 15 and 10

Hence, Their total score of Jack at the end will be = 25 + (-5) + (-10) + 15 + 10

= 25 – 5 – 10 + 15 + 10

∴ Now, Jack’s total score at the end is 35.

Question 4. In the city of Srinagar, temperature was – 5°C on Monday, and then it dropped by two °C on Tuesday. What was the temperature of the city of Srinagar on Tuesday? On Wednesday, the temperature rose by 4°C. What was the temperature on this day?

The temperature on Monday at Srinagar is = -5C

The temperature on Tuesday at the city of Srinagar is dropped by 2C = Temperature on Monday – 2C

= -7 celsius

The temperature on Wednesday at the city Srinagar rose by 4C = Temperature on Tuesday + 4C.

= -3 celsius

Thus, the temperature on days Tuesday and Wednesday was found to be -7C and -3C, respectively.

Question 5. In a magic square, every row, column and diagonal has the same sum. Check which of these following is a magic square.

Firstly we consider the square (i)

Now By adding these numbers in each of the rows, we get,

= 5 + (- 1) + (- 4) equals to 5 – 1 – 4 = 5 – 5 = 0

= -5 + (-2) + 7 equals to – 5 – 2 + 7 = -7 + 7 = 0

= 0 + 3 + (-3) = 3 – 3 = 0

By adding these numbers in every column we receive,

= 5 + (- 5) + 0 is equal to 5 – 5 = 0

= (-1) + (-2) + 3 equals to -1 – 2 + 3 = -3 + 3 = 0

= -4 + 7 + (-3) equals to -4 + 7 – 3 = -7 + 7 = 0

By adding these numbers in diagonals, we receive,

= 5 + (-2) + (-3) is equal to 5 – 2 – 3 = 5 – 5 = 0

= -4 + (-2) + 0 is equal to – 4 – 2 = -6

Because the sum of one diagonal is not always equal to zero,

Hence, (i) is not a magic square.

Now, we should consider the square (ii)

By adding these numbers to each rows we receive,

= 1 + (-10) + 0 is equal to 1 – 10 + 0 = -9

= (-4) + (-3) + (-2) equal to -4 – 3 – 2 = -9

= (-6) + 4 + (-7) becomes equal to -6 + 4 – 7 = -13 + 4 = -9

By adding these numbers in each column we receive,

= 1 + (-4) + (-6) equals to 1 – 4 – 6 = 1 – 10 = -9

= (-10) + (-3) + 4 equals to -10 – 3 + 4 = -13 + 4

= 0 + (-2) + (-7) equals to 0 – 2 – 7 = -9

= 1 + (-3) + (-7) equals to 1 – 3 – 7 = 1 – 10 = -9

= 0 + (-3) + (-6) equal to 0 – 3 – 6 = -9

Hence This (ii) square is a magic square because the sum of each row, each column and the diagonal becomes equal to -9 (negative).

Question 6. Verify a – (– b) is equal to a + b for the following values of alphabets a and b.

(i) a = 21, b = 18

a = 21 and b = 18

So To verify a – (- b) is equal to a + b

Let us take the Left Hand Side (LHS) = a – (- b)

= 21 – (- 18)

Now, lets take Right Hand Side (RHS) = a + b

By comparing both the LHS and the RHS.

Hence, the value of a and b are verified.

(ii) a = 118, b = 125

a = 118 and b = 125

To verify this a – (- b) = a + b

= 118 – (- 125)

= 118 + 125

Now, take the Right Hand Side (RHS) = a + b

By comparing both the LHS and the RHS

Hence, the values of a and b are verified.

(iii) a = 75, b = 84

a = 75 and b = 84

To verify that the a – (- b) = a + b

= 75 – (- 84)

Now, the Right Hand Side (RHS) = a + b

By comparing both LHS and RHS, we find that,

Hence, the value of a and b is verified as.

(iv) a = 28, b = 11

a = 28 and b = 11

To verify that a – (- b) = a + b

Let us now take Left Hand Side (LHS) = a – (- b)

= 28 – (- 11)

Now, Right Hand Side (RHS) = a + b

Question 7 . A water tank has stepped inside it. A monkey is sitting on the utter topmost step (which is the first step). The water level is present at the ninth step.

(i) He jumps three steps down the stairs and then successively jumps back two steps upwards. In how many jumps will the Monkey reach the following water level?

Let us consider the steps moved down are represented by a positive integer, and then the steps moved up are represented by a negative integer.

Initially, the Monkey is sitting on the topmost step, which is the first step.

In the 1st jump monkey will be at the step = 1 + 3 = 4 steps

In the 2nd jump monkey will be at the step = 4 + (-2) = 4 – 2 = 2 steps

In the 3rd jump monkey will be at the step = 2 + 3 = 5 steps

In the 4th jump monkey will be at the step = 5 + (-2) = 5 – 2 = 3 steps

In the 5th jump monkey will be at the step = 3 + 3 = 6 steps

In the 6th jump monkey will be at the step = 6 + (-2) = 6 – 2 = 4 steps

In the 7th jump monkey will be at the step = 4 + 3 = 7 steps

In the 8th jump monkey will be at the step = 7 + (-2) = 7 – 2 = 5 steps

In the 9th jump monkey will be at the step = 5 + 3 = 8 steps

In the 10th jump monkey will be at the step = 8 + (-2) = 8 – 2 = 6 steps

In the 11th jump monkey will be at the step = 6 + 3 = 9 steps

∴Monkey took a total of 11 jumps (i.e., 9th step) to reach the water level.

(ii) After drinking water, the Monkey wants to go back. For this, the Monkey jumps four steps up and then successively jumps back two steps down in his every move. In how many total jumps will he reach back to the top step?

Let us consider the steps moved down are represented by the positive integers, and then the steps moved up are represented by the negative integers.

Initially, the Monkey is sitting on the ninth step, i.e., at the water level.

In the 1st jump monkey will be at the step = 9 + (-4) = 9 – 4 = 5 steps

In the 2nd jump monkey will be at the step = 5 + 2 = 7 steps

In the 3rd jump monkey will be at the step = 7 + (-4) = 7 – 4 = 3 steps

In the 4th jump monkey will be at the step = 3 + 2 = 5 steps

In the 5th jump monkey will be at the step = 5 + (-4) = 5 – 4 = 1 step

∴ Hence the Monkey took five jumps to reach back to the top step, i.e., the first step.

Question 8. Fill in the blanks to make the following statements true:

(i) (–5) + (– 8) = (– 8) + (…………)

Let us assume that the missing integer is x,

= (–5) + (– 8) which equals to (– 8) + (x)

= – 5 – 8 = – 8 + x

= – 13 = – 8 + x

By sending – 8 from the RHS to the LHS, it becomes 8,

= – 13 + 8 = x

Now substitute the x value in the place of the blank place present,

(–5) + (– 8) = (– 8) + (- 5) … [This following equation is present in the form of the Commutative law of Addition]

(ii) –53 + ………… = –53

= –53 + x = –53

By sending – 53 from the LHS to the RHS, it becomes 53,

= x = -53 + 53

Now substitute the following x value in the blank place,

= –53 + 0 = –53 … [This equation is present in the form of Closure property of Addition]

(iii) 17 + ………… = 0

= 17 + x = 0

By sending 17 from the LHS to the RHS, it becomes -17,

= x = 0 – 17

Now substitute this x value in the blank place,

= 17 + (-17) = 0 … [This equation is present in the form of Closure property of Addition]

= 17 – 17 = 0

(iv) [13 + (– 12)] + (…………) = 13 + [(–12) + (–7)]

= [13 + (– 12)] + (x) = 13 + [(–12) + (–7)]

= [13 – 12] + (x) = 13 + [–12 –7]

= [1] + (x) = 13 + [-19]

= 1 + (x) = 13 – 19

= 1 + (x) = -6

By sending one from the LHS to the RHS, it becomes -1,

= x = -6 – 1

Now substitute the following x value in the blank place value,

= [13 + (– 12)] + (-7) equals to 13 + [(–12) + (–7)] … [This equation is present in the form of the Associative Property of Addition]

(v) (– 4) + [15 + (–3)] equals to [– 4 + 15] +…………

= (– 4) + [15 + (–3)] is equal to [– 4 + 15] + x

= (– 4) + [15 – 3)] equals to [– 4 + 15] + x

= (-4) + [12] = [11] + x

= 8 = 11 + x

Now, By sending 11 from the RHS to the LHS, it becomes -11,

= 8 – 11 = x

Now substitute the x value in the place of the blank place,

= (– 4) + [15 + (–3)] equals to [– 4 + 15] + -3 … [The following equation is in the form of the Associative property of the Addition]

Question 9. Find the product using the suitable properties:

(i) 26 × (– 48) + (– 48) × (–36)

This given equation is in the form of the Distributive law of the Multiplication property over Addition.

= a × (b + c) becomes equal to (a × b) + (a × c)

Let, a = -48, b = 26, c = -36

= 26 × (– 48) + (– 48) × (–36)

= -48 × (26 + (-36)

= -48 × (26 – 36)

= -48 × (-10)

= 480 … [∵ (- × – = +)

(ii) 8 × 53 × (–125)

The given equation is present in the form of the Commutative law of Multiplication.

= a × b = b × a

= 8 × [53 × (-125)]

= 8 × [(-125) × 53]

= [8 × (-125)] × 53

= [-1000] × 53

(iii) 15 × (–25) × (– 4) × (–10)

This given equation is in the form of the Commutative law of the Multiplication property.

= 15 × [(–25) × (– 4)] × (–10)

= 15 × [100] × (–10)

= 15 × [-1000]

(iv) (– 41) × 102

This given equation is in the form of a Distributive law of the Multiplication property over Addition.

= a × (b + c) = (a × b) + (a × c)

= (-41) × (100 + 2)

= (-41) × 100 + (-41) × 2

= – 4100 – 82

(v) 625 × (–35) + (– 625) × 65

This given equation is in the form of the Distributive law of Multiplication over Addition.

= 625 × [(-35) + (-65)]

= 625 × [-100]

Question 10. A certain freezing process requires that the room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the final room temperature 10 hours after the actual process begins?

Answer 10:-

From the above question, it is given that

Let us take the lowered temperature as a negative integer,

Initial temperature will be= 40oC

Change in temperature per hour is = -5oC

Change in temperature after 10 hours will be = (-5) × 10 = -50oC

∴The final room temperature after the 10 hours of freezing process = 40oC + (-50oC)

Question 11. In a class test containing about ten questions, five marks are awarded for each correct answer and (–2) marks are awarded for every incorrect answer and 0 for questions which are not attempted.

(i) Mohan gets four correct answers and six incorrect answers on his test. What is his total score?

Marks awarded for one correct answer is = 5

The total marks awarded for his four correct answers are = four × 5 = 20 marks.

Marks awarded for 1 wrong answer = -2 (negative)

Total marks awarded for 6 wrong answers is = 6 × -2 = -12

∴Total score obtained by Mohan = 20 + (-12)

(ii) Reshma gets five correct answers and similarly five incorrect answers; what is her total score?

Total marks awarded for 5 correct answer becomes = 5 × 5 = 25

Marks awarded for one wrong answer is = -2

Total marks awarded for 5 wrong answer becomes = 5 × -2 = -10

∴Total score obtained by Reshma is = 25 + (-10)

(iii) Heena gets two correct answers and five incorrect answers out of the seven questions she attempts. What is her final score?

Total marks awarded for 2 correct answer is = 2 × 5 = 10

Marks awarded for the questions which are not attempted is = 0

∴Total score obtained by Heena is = 10 + (-10)

Question 12. A cement company earns a profit of around ₹ 8 per bag of white cement that is sold and simultaneously a loss of ₹ 5 per bag of grey cement that is sold.

(i) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?

We denote profit by a positive integer and loss by a negative integer,

So From the above question,

The Cement company earns a profit on selling one bag of white cement = ₹ 8 per bag.

The cement company earns a total profit on selling 3000 bags of white cement = 3000 × ₹ 8

And also the,

Loss on selling one bag of grey cement is = – ₹ 5 per bag.

Loss on selling the 5000 bags of the grey cement = 5000 × – ₹ 5

= – ₹ 25000

Total loss or profit earned by these cement companies is = profit + loss.

= 24000 + (-25000)

Hence, a loss of ₹ 1000 will be incurred by the company.

(ii) What is the number of white cement bags that must sell to have neither a profit nor loss if the total number of grey bags sold is 6,400 bags?

We denote the profit as a positive integer and the loss as a negative integer,

The cement company earns the profit on selling one bag of white cement as = ₹ 8 per bag.

Now Let the number of white cement bags present be x.

The cement company earns a profit on selling these x bags of white cement as = (x) × ₹ 8

Loss on selling one bag of grey cement becomes = – ₹ 5 per bag.

Loss on selling 6400 bags of grey cement becomes = 6400 × – ₹ 5

= – ₹ 32000

According to the above question,

Company to have neither profit nor loss, must sell,

= Profit + loss = 0

= 8x + (-32000) =0

By sending -32000 from the LHS to the RHS, it becomes 32000

= 8x = 32000

= x = 32000/8

Hence, the 4000 bags of white cement should sell to have neither profit nor loss.

Question 13. Evaluate each of the following:

(i) (–30) ÷ 10

= (–30) ÷ 10

When we divide the negative integer by a positive integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.

(ii) 50 ÷ (–5)

= (50) ÷ (-5)

When we divide the positive integer by a negative integer, we first divide them as whole numbers and then apply the minus sign (-) before the quotient.

(iii) (–36) ÷ (–9)

= (-36) ÷ (-9)

When we divide the negative integer by a similar negative integer, we first divide these as whole numbers and then put the positive sign (+) before the quotient.

(iv) (– 49) ÷ (49)

= (–49) ÷ 49

When we divide the negative integer by a positive integer, we first divide these as whole numbers and then put the minus sign (-) before the quotient.

(e) 13 ÷ [(–2) + 1]

= 13 ÷ [(–2) + 1]

= 13 ÷ (-1)

When we divide the positive integer by a negative integer, we first divide these as whole numbers and then put the minus sign (-) before the quotient.

(f) 0 ÷ (–12)

= 0 ÷ (-12)

When we divide zero by a negative integer, it gives zero.

(g) (–31) ÷ [(–30) + (–1)]

= (–31) ÷ [(–30) + (–1)]

= (-31) ÷ [-30 – 1]

= (-31) ÷ (-31)

When we divide the negative integer by a negative integer, we first divide these as whole numbers and then put the positive sign (+) before the quotient.

(h) [(–36) ÷ 12] ÷ 3

First, we have to solve these integers within the bracket,

= [(–36) ÷ 12]

= (–36) ÷ 12

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.

(i) [(– 6) + 5)] ÷ [(–2) + 1]

The given question can be written as,

= [-1] ÷ [-1]

Question 14. Verify that a ÷ (b + c) is not equal to (a ÷ b) + (a ÷ c) for each of the following symbols of a, b and c.

(i) a = 12, b = – 4, c = 2

From the above question, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)

Given, a = 12, b = – 4 (negative), c = 2

Now, consider that the LHS = a ÷ (b + c)

= 12 ÷ (-4 + 2)

= 12 ÷ (-2)

When we divide a following positive integer by any of the negative integers, we first divide them as a whole number and then put the minus sign (-) before their quotient.

Then, consider that the RHS is equal to = (a ÷ b) + (a ÷ c)

= (12 ÷ (-4)) + (12 ÷ 2)

= (-3) + (6)

By comparing the LHS and RHS, we get,

= LHS ≠ RHS

Hence, the given values have been verified.

(ii) a = (–10), b = 1, c = 1

Given, a = (-10), b = 1, c = 1

= (-10) ÷ (1 + 1)

= (-10) ÷ (2)

When we divide a negative integer by any other positive integer, we first divide them as a whole number and then put the minus sign (-) before the quotient.

Then, consider RHS = (a ÷ b) + (a ÷ c)

= ((-10) ÷ (1)) + ((-10) ÷ 1)

= (-10) + (-10)

By comparing LHS and RHS

Hence, the given values are verified.

Question. Fill in the following blanks:

(a) 369 ÷ _____ = 369

= 369 ÷ x = 369

= x = (369/369)

Hence, put the valve of x in the blank place.

= 369 ÷ 1 = 369

(b) (–75) ÷ _____ = –1

= (-75) ÷ x = -1

= x = (-75/-1)

Now, put the above valve of x in the blank place.

= (-75) ÷ 75 = -1

= (-206) ÷ x = 1

= x = (-206/1)

= (-206) ÷ (-206) = 1

(d) – 87 ÷ _____ = 87

= (-87) ÷ x = 87

= x = (-87)/87

= (-87) ÷ (-1) = 87

(e) _____ ÷ 1 = – 87

= (x) ÷ 1 = -87

= x = (-87) × 1

So, put the valve of x in the blank.

= (-87) ÷ 1 = -87

(f) _____ ÷ 48 = –1

= (x) ÷ 48 = -1

= x = (-1) × 48

Now, put the above valve of x in the following blank.

= (-48) ÷ 48 = -1

Question 15. The temperature at 12 noon was 10 degrees C above zero. If it decreases at the rate of 2C per hour until midnight, at what time would the temperature be eight °C below zero? Also, What would be the temperature at midnight?

From the above question, it is given that,

The temperature at the beginning, which is, at 12 noon, is = 10C

The rate of change of temperature becomes = – 2C per hour.

Temperature present at 1 PM = 10 + (-2) = 10 – 2 = 8° C

Temperature present at 2 PM = 8 + (-2) = 8 – 2 = 6° C

Temperature present at 3 PM = 6 + (-2) = 6 – 2 = 4°C

Temperature present at 4 PM = 4 + (-2) = 4 – 2 = 2°C

Temperature present at 5 PM = 2 + (-2) = 2 – 2 = 0°C

Temperature present at 6 PM = 0 + (-2) = 0 – 2 = -2°C

Temperature present at 7 PM = -2 + (-2) = -2 -2 = -4°C

Temperature present at 8 PM = -4 + (-2) = -4 – 2 = -6°C

Temperature present at 9 PM = -6 + (-2) = -6 – 2 = -8°C

∴At 9 PM, the temperature will be 8° C below zero.

The temperature at mid-night which is at 12 AM

Change in the temperature in every 12 hours = -2°C × 12 = – 24°C

So, at midnight the temperature will be = 10 + (-24)

At midnight the temperature will be 14°C below 0.

Question 16. In the following class test, (+ 3) marks are given for every correct answer, (–2) marks are given for every the incorrect answer and no marks are given for not attempting any question.

(i) Radhika scored 20 marks. If she has got around 12 correct answers, then how many questions has she attempted that are incorrect?

(ii) Mohini scores –5 (negative) marks on this test, and though she has got seven correct answers. How many questions has she attempted incorrectly?

Marks awarded for 1 correct answer is = + 3

(i) Radhika, in the test, scored 20 marks

Total marks awarded for every 12 correct answers is = 12 × 3 = 36

Marks awarded for every incorrect answer = Total score – Total marks awarded for 12 correct questions.

So, the number of incorrect answers done by Radhika = (-16) ÷ (-2)

(ii) Mohini scored a total of -5 marks

Total marks awarded for her 7 correct answers is = 7 × 3 = 21

Marks awarded for her incorrect answers = Total score – Total marks awarded for the 12 correct answers.

Hence, the number of incorrect answers made by Mohini = (-26) ÷ (-2)

Question 17. An elevator descends down into a mine shaft at the rate of 6 m per min. If the descent starts from 10 meters above the ground level, how much time will it take to reach – 350 m?

The initial height of the elevator becomes = 10 m

Final depth of elevator is = – 350 m … [the distance descended is denoted by a negative integer]

The total distance to descend by the elevator becomes = (-350) – (10)

Time taken by the elevator to descend (negative) -6 m is = 1 min

So, the total time taken by the elevator to descend – 360 m becomes = (-360) ÷ (-60)

= 60 minutes

= 1 hour Benefits of Solving Important Questions Class 7 Mathematics Chapter 1

Practice is the key to success. The practice habit is very important for students because it will help them in many ways. It will help them to score better in exams. Apart from this, practice will clear doubts, generate interest in the subject matter, and strengthen the concepts. Thus, students must practice sums regularly to improve their exam preparation. The Important Questions Class 7 Mathematics Chapter 1 will help students in many ways. These are-

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The subject matter experts of Extramarks understand the student’s needs. They have built the question series to help students with their exam preparation. They have collected all the vital questions so students can find them in a single article. Sometimes, students need more than the textbook. Hence, they can follow the Class 7 Mathematics Chapter 1 Important Questions because they will find chapter-wise questions for each subject. Regular practice will strengthen their ideas, and they can solve any question that comes in exams. Thus, the question series will help them to score better in exams.

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Q.1 Which one of the following statements is false?

1. For any two positive integers a and b, a ÷ (–b) = – a ÷ b, where b ≠ 0.

2. The commutativity, associativity and distributivity of integers help to make calculations simpler.

3. The product of three integers does not depend upon the grouping of integers.

4. Division is closed for integers.

Option 4. Explanation

Division is not closed for integers. For example: 2 ÷ 6 =

is not an integer.

Q.2 Which one of the following is false?

Marks: 1 1. Sum of integers a and b is an integer.

2. a + b = b + a, for all integers a and b

3. a – b = b – a, for all integers a and b

4. a + (b + c) = (a + b) + c, for all integers a, b and c

Ans Option3 Explanation

a – b = b – a, for all integers a and b is false. For example, 2 – 4 = – 2 and 4 – 2 = 2 Thus, 2 – 4 ≠ 4 – 2

Q.3 What is the difference between a temperature of 7º C above zero and a temperature of 3º C below zero?

Ans Option 1. Explanation

Difference between a temperature of 7º C above zero and a temperature of 3º C below zero = 7º C – (– 3º C) = 7º C + 3º C = 10º C

Q.4 A plane is flying at the height of 8750 m above sea level. At a particular point, it is exactly above a submarine floating 1340 m below sea level. What is the vertical distance between them?

Marks: 2 Ans

Height of the plane above sea level = 8750 m Distance of submarine below sea level = – 1340 m Vertical distance = 8750 m – (– 1340 m) = 8750 m + 1340 m = 10,090 m

Q.5 A man walks 22 m towards east and then 17 m towards west. The position of the man with respect to his starting point is ______________.

1.5 m towards west

2.5 m towards east

3.39 m towards east

4.39 m towards west

Ans Option 2. Explanation

Let 22 m towards east be represented by +22, then –17 m represents 17 m towards west. On adding, +22 – 17 = +5 (positive) The position of the man with respect to his starting point = 5 m towards east

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Faqs (frequently asked questions), 1. is class 7 mathematics chapter 1 easy.

Class 7 Mathematics Chapter 1 under CBSE curriculum is about integers. Students will study the properties of integers, how to add and multiply integers and how to put them on the number line. The concepts may be new to them, but they have studied integers in Class 6. They can easily understand the concepts if they follow the textbook seriously. The chapter is relatively easy. Students can take help from the Important Questions Class 7 Mathematics Chapter 1 to solve questions from the chapter.

2. How can the Important Questions Class 7 Mathematics Chapter 1 help students?

The experts of Extramarks have made the question series after taking help from several sources. They have collated the questions from the textbook exercise, CBSE sample papers, important reference books and NCERT exemplar. They have included questions from CBSE past years’ question papers too. Apart from this, they have solved the questions for students, and experienced professionals have further checked the answers. Thus, the Important Questions Class 7 Mathematics Chapter 1 will help the students to practice the sums regularly. It will boost their confidence and increase their marks in exams.

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