# NCERT Solutions Class 7 Mathematics Chapter 1

## NCERT Solutions for Class 7 Mathematics Chapter 1 Integers

The two important aspects of improving Mathematics are  understanding concepts and practising problems on a daily basis. To make room in a highly competitive environment, students must lay a strong foundation of the subject in their early years.

NCERT Class 7 Mathematics Chapter 1 deals with integers. We learned about whole digits and integers in previous classes. We'll now move further into integers, their properties, and operations. Similarly, we will also learn about integer addition and subtraction, integer addition and subtraction properties, integer multiplication and division, and integer multiplication and division properties.

## NCERT Solutions for Class 7 Mathematics Chapter 1

Access NCERT Solutions for Mathematics Chapter 1 – Integers

Chapter 1 encourages students in gaining a better understanding of the number system and solving complex problems  with ease in subsequent classes. As a result, students should be familiar with every topic and practise in-text and end-text questions to erase ‘maths phobia’  and develop interest in Mathematics. Meanwhile, let’s review the key topics covered in Class 7 Mathematics Chapter 1:

 Section Number Section Title 1 Introduction to Integers 2 Properties of Addition and Subtraction of Integers 3 Multiplication of Integers 4 Multiplication of a Positive and Negative Integer 5 Multiplication of two negative integers 6 Properties of Multiplication of Integers 7 Division of Integers 8 Properties of Division of Integers

### 1.1Introduction of Integers

Integers are part of a larger collection of numbers that includes both whole and negative numbers. The student will learn more about integers, their properties, and operations in this chapter. They will also learn about number concepts similar to those covered in the previous class, such as the number line, in this section.

### 1.2Properties of Addition and Subtraction of Integers

Children will learn the addition and subtraction of integers, which will make it easier for them to perform simple calculations in day-to-day life.

### 1.3Multiplication of Integers

Multiplying numbers may be a simple concept. It is important to remember the positive or negative  number sign while multiplying the integers. This is most useful when simplifying an equation.

### 1.4Multiplication of a Positive and Negative Integer

This topic demonstrates with examples how we always get a negative integer by multiplying a positive integer and a negative integer.

### 1.5Multiplication of two negative integers

This topic explains how we always get a positive integer by multiplying a negative integer and a negative integer.

### 1.6Properties of Multiplication of Integers

• By multiplying two positive integers and two negative integers, we get a positive integer.
• We get a negative integer by multiplying a positive integer with a negative integer
• We get  zero by multiplying any integer with zero

### 1.7Division of Integers

It means division in which the fractional part(remainder) is discarded is called integer division.

### 1.8Properties of Division of Integers

• We get a positive integer by dividing two positive integers and two negative integers.
• We get a negative integer by dividing a positive integer with a negative integer.
•  Zero divided  by any number is zero.
•  Any number divided by zero is infinite.

## NCERT Solutions for Class 7 Chapter 1 Mathematics Integers –

Integers are numbers that are not fractions and can be positive, negative, or zero. These numbers can be used for addition, subtraction, multiplication, and division. Integers help evaluate efficiency in both positive and negative numbers in every field. For example   temperature, sea level, and other real-life integers.

Exercise 1.1 will refresh your memory on the number line, how to present integers on the number line, how to arrange integers in ascending and descending order, what is a positive and negative integer, how to add and subtract positive and negative integers, and how to represent them on the number line.

The properties  of Addition and Subtraction of Integers is covered in Exercise 1.2. You'll see how integer addition is commutative for integers but not for integer subtraction.

In Exercise 1.3, you will learn about the properties of  multiplication of integers.

Topics related to the division of integers are covered in Exercise 1.4. Multiplication is the inverse operation of division and its properties

Students will explore the negative set of whole number values, as well as how they are represented on a number line. By understanding the number system, students can easily be acquainted with newly introduced number terminology in higher classes, such as rational numbers, irrational numbers, and so on. Keeping this in mind, Extramarks has designed NCERT Solutions Class 7 Mathematics Chapter 1 in such a way that students can quickly revise these basic Mathematical concepts before moving on to the next academic year.

### Properties of Integers

Numbers for addition and multiplication through patterns are  a part of the properties of integers. They also take into account the whole numbers. Integers involve expression of communicative and associative properties in a general form.

Facts:

• Natural numbers are the counting numbers such as 1, 2, 3, 4, 5, and so on, whereas whole numbers are the set of natural numbers plus zero, such as 0, 1, 2, 3, 4, 5, and so on.
• On a number line, negative integers are represented by points to the left of zero, and positive integers are represented by points to the right of zero.
• For negative integers to the left of zero and positive integers to the right of zero, the integer 0 serves as an additive identity.
• 0 is neither a positive nor a negative integer.
• The numerical value of an integer, regardless of its sign, is its absolute value. | a | denotes the absolute value of an integer a.

### Number Line

Natural numbers, negative and positive numbers, and whole numbers are all represented on a number line. To determine numerical operations, the identities are marked at equal intervals on a line. Number lines are significant because they represent numbers that we use every day.

How to Draw a Number Line:

1. Draw  a straight line of any length.
2. To divide the drawn line into the required number, mark points at fixed distances on it.
3. Any of the points marked on the line in step 2 should be marked as 0.
4. Starting at 0, write the positive numbers + 1, + 2, + 3, and so on on the right-hand side of the line. Similarly, starting at 0, mark the negative integers -1, -2, -3, and so on on the left side.
5. The numbers continue to infinity on both sides of the drawn line, as indicated by the arrowheads on both sides of the drawn line.

## NCERT Solutions for Class 7 Mathematics

In CBSE Class 7, integers, algebraic expressions, fractions, and decimals are all part of the solved exercises. As a result, learning this challenging subject and clarifying their doubts  will aid students in their preparation for  higher classes as well. Extramarks offers CBSE Class 7 Mathematics study materials that will help students achieve higher marks in the exam. Sample papers, past years’ question papers, and NCERT Solutions are all part of our study materials.  Students must practice and revise NCERT solutions to build a strong foundation.

You can use our NCERT textbook solutions to bridge the knowledge gap and stay motivated.  Extramarks subject matter experts have created study materials for CBSE Class 7 Mathematics. They are available  on Extramarks official website.

Chapter 1 - Integers

Chapter 2 - Fractions and Decimals

Chapter 3 - Data Handling

Chapter 4 - Simple Equations

Chapter 5 - Lines and Angles

Chapter 6 - The Triangle 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

## NCERT Solutions for Class 7

In CBSE Class 7, learning the fundamentals is crucial  because the fundamentals learned now will come handy later. Students require solid training and study materials that will assist them in achieving high exam scores and simplifying all of their concepts. Extramarks provides CBSE Class 7 study materials to help students prepare for their exams.

Our CBSE Class 7 study materials are created by experienced faculty . We have textbook solutions, especially NCERT Solutions Class 7, which have simplified solutions to the textbook questions for each chapter.Our textbook solutions also assist students in completing home assignments and  mastering all concepts.

Extramarks offers study materials that are updated regularly to reflect the most recent CBSE Syllabus. The systemic and well-laid-out balanced study plan boosts their performance naturally and effortlessly.

Q.1

$\begin{array}{l}\mathrm{Following}\mathrm{number}\mathrm{line}\mathrm{shows}\mathrm{the}\mathrm{temperature}\mathrm{in}\mathrm{degree}\\ \mathrm{celsius}\left(°\mathrm{C}\right)\mathrm{at}\mathrm{different}\mathrm{places}\mathrm{on}\mathrm{a}\mathrm{particular}\mathrm{day}\\ \left(\text{a}\right)\text{Observe this number line and write the}\\ \text{temperature of the places marked on it}.\\ \left(\text{b}\right)\text{What is the temperature difference between the}\\ \text{hottest and the coldest places among the above}?\\ \left(\text{c}\right)\text{What is the temperature difference between Lahulspiti}\\ \text{and Srinagar}?\\ \left(\text{d}\right)\text{Can we say temperature of Srinagar and Shimla taken}\\ \text{together is less than the temperature at Shimla}?\\ \text{Is it also less than the temperature at Srinagar}?\end{array}$ Ans.

$\begin{array}{l}\text{(a)}\text{\hspace{0.17em}}\text{By observing the number line, the temprature of the}\\ \text{places marked are as follows:}\\ \text{Lahulspiti:}-8°C\\ \text{Srinagar:}-2°C\\ \text{Shimla:}5°C\\ \text{Ooty:}14°C\\ \text{Banglore:}22°C\\ \\ \text{(b)}\text{\hspace{0.17em}}\text{The hottest place is Banglore with temprature}22°C\\ \text{and coldest place is Lahulspiti}\text{\hspace{0.17em}}\text{with temprature}-8°C\text{.}\\ \text{So},\text{\hspace{0.17em}}\text{the temperature difference between the hottest}\\ \text{and the coldest places is}\\ \text{=22}°\text{C}-\left(-8°\text{C}\right)\\ =22°\text{C}+8°\text{C}\\ =\overline{)30°\text{C}}\\ \\ \left(c\right)\text{\hspace{0.17em}}\text{The temperature difference between Lahulspiti and}\\ \text{Srinagar}\\ =-2°\text{C}-\left(-8°\text{C}\right)\\ =-2°\text{C}+8°\text{C}\\ =\overline{)6°\text{C}}\\ \\ \left(d\right)\text{\hspace{0.17em}}\text{The temprature of Srinagar and Shimla are}-2°C\\ \text{and}5°C.\\ \text{So, together their temprature would be}\\ -2°C+5°C=3°C\text{, which is less than temprature of Shimla}\\ \text{Thus, temprature of Srinagar and Shimla taken together}\\ \text{is less than the temprature at Shimla}\text{.}\\ \text{But, it is not less than temprature at Srinagar}\text{.}\end{array}$

Q.2

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{quiz},\mathrm{positive}\mathrm{marks}\mathrm{are}\mathrm{given}\mathrm{for}\mathrm{correct}\mathrm{answers}\\ \mathrm{and}\mathrm{negative}\mathrm{marks}\mathrm{are}\mathrm{given}\mathrm{for}\mathrm{in}\mathrm{correct}\mathrm{answers}.\\ \mathrm{If}\mathrm{Jack}’\mathrm{s}\mathrm{scores}\mathrm{in}\mathrm{five}\mathrm{successive}\mathrm{rounds}\mathrm{were}\\ 25,–5,–10,\text{\hspace{0.17em}}15\mathrm{and}10,\mathrm{what}\mathrm{was}\mathrm{his}\mathrm{total}\mathrm{at}\mathrm{the}\mathrm{end}?\end{array}$

Ans.

$\begin{array}{l}\text{Jack’s scores in five successive rounds were 25,}-5,-10,\\ \text{15 and and 10}\text{.}\\ \text{His total at the end would be}\\ \text{=25+}\left(-5\right)+\left(-10\right)+15+10\\ =50-15\\ =\overline{)35}\end{array}$

Q.3

$\begin{array}{l}\text{At Srinagar temperature was-5°C on Monday and then}\\ \text{it dropped by2°C on Tuesday. What was the temperature}\\ \text{of Srinagar on Tuesday? On Wednesday, it rose by 4°C.}\\ \text{What was the temperature on this day?}\end{array}$

Ans.

$\begin{array}{l}\text{Temprature on Monday=}-5°C\\ \text{Since, temperature dropped by}2°C\text{on Tuesday}\text{.}\\ \text{So, the temperature of Srinagar on Tuesday was}\\ =-5°C-2°C\\ =\text{}\overline{)-7°C}\\ \text{On Wednesday, temperature rose by}4°C,\text{\hspace{0.17em}}\text{so temperature of}\\ \text{Srinagar on Wednesday was}\\ =-7°C+4°C\\ =\overline{)-3°C}\\ \text{Thus, Temperature on Tuesday and Wednesday was}\\ \overline{)-7°C\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}-3°C}\text{respectively}\text{.}\end{array}$

Q.4

$\begin{array}{l}\text{A plane is flying at the height of 5000 m above the sea}\\ \text{level. At a particular point, it is exactly above a submarine}\\ \text{floating 1200 m below the sea level. What is the vertical}\\ \text{distance between them?}\end{array}$ Ans.

$\begin{array}{l}\text{Height of the plane = 5000 m}\\ \text{Depth of the submarine}=\text{}-1200\text{m}\\ \text{So, the distance between plane and submarine}\\ \text{=5000 m}-\text{}\left(-1200\text{\hspace{0.17em}}\text{m}\right)\\ =5000\text{m+1200 m}\\ \text{=}\overline{)\text{6200 m}}\\ \text{Thus, the vertical distance between them is 6200 m}\text{.}\end{array}$

Q.5

$\begin{array}{l}\text{Mohan deposits Rs 2,000 in his bank account and}\\ \text{withdraws Rs 1,642 from it, the next day. If withdrawal}\\ \text{of amount from the account is represented by a negative}\\ \text{integer, then how will you represent the amount deposited?}\\ \text{Find the balance in Mohan’s account after the withdraw.}\end{array}$

Ans.

$\begin{array}{l}\text{Since, withdrawal of amount from the account is represented}\\ \text{by a negative integer, so we take amount deposited as a positive}\\ \text{integer}\\ \text{Amount deposited = Rs 2000}\\ \text{Amount withdrawn}=-\text{Rs 1642}\\ \text{Balance left in Mohan’s account}\\ \text{= Rs 2000}-\text{Rs 1642}\\ \text{=}\overline{)\text{Rs 358}}\\ \text{Thus, the balance in Mohan’s account after withdrawal}\\ \text{is Rs 358}\text{.}\end{array}$

Q.6

$\begin{array}{l}\text{Rita goes 20 km towards east from a point A to the point B.}\\ \text{From B, she moves 30 km towards west along the same road.}\\ \text{If the distance towards east is represented by a positive}\\ \text{integer then, how will you represent the distance travelled}\\ \text{towards west? By which integer will you represent her final}\\ \text{position from A?}\end{array}$ Ans.

$\begin{array}{l}\text{Here, the distance towards east is represented by a positive}\\ \text{integer and the distance travelled towards west will be}\\ \text{represented by a negative integer}\text{.}\\ \text{So, distance travelled in east direction = 20 km}\\ \text{distance travelled in west direction =}-\text{30 km}\\ \text{Distance travelled from A = 20 km+}\left(-30\text{km}\right)\\ =-10\text{\hspace{0.17em}}\text{km}\\ \text{Thus, Rita’s distance travelled from point A}\text{​}\text{will be represented}\\ \text{by a negative integer}\left(-10\text{\hspace{0.17em}}\text{km}\right)\text{.}\\ \text{Rita is in west direction}\text{.}\end{array}$

Q.7

$\begin{array}{l}\text{In a magic square each row, column and diagonal have the}\\ \text{same sum. Check which of the following is a magic square.}\end{array}$ Ans.

$\begin{array}{l}\text{In a magic square, each row, column and diagonal have}\\ \text{the same sum}\text{.}\\ \text{So, in square (i), every row and column sum up to 0}\text{.}\\ \text{However sum of one of its diagonal is not zero}\text{.}\\ -4-2=-6\ne \text{0}\\ \text{So, (i) is not a magic square}\text{.}\\ \text{Similarly, in square (ii) every row, column and diagonal}\\ \text{sum up to}-\text{9}\text{.}\\ \text{Thus, (ii) is a magic square}\text{.}\end{array}$

Q.8

$\begin{array}{l}\text{Verifya}-\left(-\text{b}\right)\text{ = a + b for the following values of a and b.}\\ \left(\text{i}\right)\text{\hspace{0.17em} a = 21,b = 18 }\left(\text{ii}\right)\text{ \hspace{0.17em}a = 118, b = 125}\\ \left(\text{iii}\right)\text{ a = 75,b = 84 }\left(\text{iv}\right)\text{ a = 28, b = 11}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}a=21, b=18}\\ \mathrm{a}-\left(-\mathrm{b}\right)=21-\left(-18\right)=21+18=39\\ \mathrm{a}+\mathrm{b}=21+18=39\\ \mathrm{Thus},\mathrm{a}-\left(-\mathrm{b}\right)=\mathrm{a}+\mathrm{b}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}a=118, b=125}\\ \mathrm{a}-\left(-\mathrm{b}\right)=118-\left(-125\right)=118+125=243\\ \mathrm{a}+\mathrm{b}=118+125=243\\ \mathrm{Thus},\mathrm{a}-\left(-\mathrm{b}\right)=\mathrm{a}+\mathrm{b}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}a=75, b=84}\\ \mathrm{a}-\left(-\mathrm{b}\right)=75-\left(-84\right)=75+84=159\\ \mathrm{a}+\mathrm{b}=75+84=39\\ \mathrm{Thus},\mathrm{a}-\left(-\mathrm{b}\right)=\mathrm{a}+\mathrm{b}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}a=28, b=11}\\ \mathrm{a}-\left(-\mathrm{b}\right)=28-\left(-11\right)=28+11=39\\ \mathrm{a}+\mathrm{b}=28+11=39\\ \mathrm{Thus},\mathrm{a}-\left(-\mathrm{b}\right)=\mathrm{a}+\mathrm{b}\end{array}$

Q.9

$\begin{array}{l}\text{Use the sign of >,< or = in the box to make the statement}\\ \text{true.}\\ \left(\text{a}\right)\left(-\text{8}\right)\text{+}\left(-\text{4}\right)\overline{)}\left(-\text{8}\right)-\left(-\text{4}\right)\\ \left(b\right)\left(–3\right)+7–\left(19\right)\overline{)}15–8+\left(–9\right)\\ \left(c\right)23–41+11\overline{)}23–41–11\\ \left(d\right)39+\left(–24\right)–\left(15\right)\overline{)}36+\left(–52\right)–\left(–36\right)\\ \left(e\right)–231+79+51\overline{)}–399+159+81\end{array}$

Ans.

$\begin{array}{l}\text{(a) (}-\text{8) + (}-\text{4)}\text{}\overline{)<}\text{}\text{}\text{}\text{(}-\text{8)}-\text{(}-\text{4)}\\ \text{(b) (}-\text{3) +7}-\text{(19)}\text{}\overline{)<}\text{}\text{}\text{}\text{15}-\text{8+(}-\text{9)}\\ \text{(c) 23}-\text{41+11}\text{}\text{}\overline{)>}\text{}\text{}\text{}23-41-11\\ \text{(d) 39+(}-\text{24)}-\text{(15)}\overline{)<}\text{}\text{}\text{}36+\text{(}-\text{52)}-\text{(}-\text{36)}\\ \text{(e)}-\text{231+79+51}\overline{)>}\text{}\text{}\text{}-\text{399 + 159 + 81}\end{array}$

Q.10

$\begin{array}{l}\text{A water tank has steps inside it. A monkey is sitting on}\\ \text{the topmost step}\left(\text{i.e., the first step}\right)\text{. The water level is}\\ \text{at the ninth step.}\\ \text{}\left(\text{i}\right)\text{He jumps 3 steps down and then jumps back 2 steps}\\ \text{up.In how many jumps will he reach the water level?}\\ \text{}\left(\text{ii}\right)\text{After drinking water, he wants to go back. For this,}\\ \text{he jumps 4 steps up and then jumps back 2}\\ \text{steps down in every move. In how many jumps will he}\\ \text{reach back the top step?}\\ \text{}\left(\text{iii}\right)\text{If the number of steps moved down is represented}\\ \text{by negative integers and the number of steps moved up}\\ \text{by positive integers, represent his moves in part}\left(\text{i}\right)\text{and}\\ \text{}\left(\text{ii}\right)\text{by completing the following;}\\ \text{}\left(\text{a}\right)\text{– 3 + 2 –}\dots \text{=- 8}\\ \text{}\left(\text{b}\right)\text{4 – 2 +}\dots \text{= 8.}\\ \text{In}\left(\text{a}\right)\text{the sum}\left(\text{– 8}\right)\text{represents going down by eight}\\ \text{steps. So, what will the sum 8 in}\left(\text{b}\right)\text{represent?}\end{array}$ Ans.

$\begin{array}{l}\text{Consider the steps moved down be represented by positive}\\ \text{integers and steps moved up be represented by negative}\\ \text{integers}\text{.}\\ \left(i\right)\text{\hspace{0.17em}}\text{Initially, the monkey was at step = 1}\\ \text{After 1st jump, monkey will be at step = 1+3=4}\\ \text{After 2nd jump, monkey will be at step = 4+}\left(-2\right)\text{=2}\\ \text{After 3rd jump, monkey will be at step = 2+3=5}\\ \text{After 4th jump, monkey will be at step = 5+}\left(-2\right)\text{=3}\\ \text{After 5th jump, monkey will be at step = 3+3=6}\\ \text{After 6th jump, monkey will be at step = 6+}\left(-2\right)\text{=4}\\ \text{After 7th jump, monkey will be at step = 4+3=7}\\ \text{After 8th jump, monkey will be at step = 7+}\left(-2\right)\text{=5}\\ \text{After 9th jump, monkey will be at step = 5+3=8}\\ \text{After 10th jump, monkey will be at step = 8+}\left(-2\right)\text{=6}\\ \text{After 11th jump, monkey will be at step = 6+3=9}\\ \text{Clearly, the monkey will be at water level (i}\text{.e}\text{., 9th step)}\\ \text{after 11 jumps}\text{.}\\ \\ \text{(ii)}\\ \text{Initiall, the monkey was at step = 9}\\ \text{After 1st jump, monkey will be at step = 9+}\left(-4\right)\text{=5}\\ \text{After 2nd jump, monkey will be at step = 5+2=7}\\ \text{After 3rd jump, monkey will be at step = 7+}\left(-4\right)\text{=3}\\ \text{After 4th jump, monkey will be at step = 3+2=5}\\ \text{After 5th jump, monkey will be at step = 5+}\left(-4\right)\text{=1}\\ \text{Clearly,}\text{\hspace{0.17em}}\text{the will reach back at the top step after 5 jumps}\text{.}\\ \text{(iii) If steps moved down are represented by a negative}\\ \text{integers and steps moved up are represented bya positive}\\ \text{integers, then his moves will be as follows:}\\ \text{Moves in part (i):}\\ -\text{3+2}-\text{3+2}-\text{3+2}-\text{3+2}-\text{3+2}-\text{3=}-\text{8}\\ \text{Moves in part (ii):}\\ \text{4}-\text{2+4}-\text{2+4=8}\\ \text{Moves in part (ii) represent goin up 8 steps}\text{.}\end{array}$

Q.11

$\begin{array}{l}\text{Write down a pair of integers whose:}\\ \left(\text{a}\right)\text{ sumis -7}\\ \left(\text{b}\right)\text{ differenceis -10}\\ \left(\text{c}\right)\text{ sumis 0}\end{array}$

Ans.

$\begin{array}{l}\text{(a)}-\text{8+(+1) =}-\text{8+1=}-\text{7}\\ \text{So, the pair is (}-\text{8, 1)}\\ \text{(b)}-\text{12}-\text{(}-\text{2)=}-\text{12+2=}-\text{10}\\ \text{So, the pair is (}-\text{12,}-\text{2)}\\ \text{(c) 5+(}-\text{5)=5}-\text{5=0}\\ \text{So, the pair is (}-\text{5,}-\text{5)}\end{array}$

Q.12

$\begin{array}{l}\begin{array}{l}\left(\text{a}\right)\text{ Write a pair of negative integers whose difference}\\ \text{gives 8.}\\ \left(\text{b}\right)\text{ Write an egative integer and a positive integer whose}\end{array}\\ \begin{array}{l}\text{sumis -5.}\\ \left(\text{c}\right)\text{ Write an egative integer and a positive integer whose}\\ \text{differenceis -3.}\end{array}\end{array}$

Ans.

$\begin{array}{l}\text{(a) To write a pair of negative integers whose difference is 8}\\ \text{So,}-\text{2}-\text{(}-\text{10)=}-\text{2+10=8}\\ \text{So, pair of negative}\mathrm{int}\text{egers is}\overline{)\left(-2,-10\right)}.\\ \left(b\right)\text{To write a negative integer and a positive integer whose}\\ \text{sum is}-\text{5}\text{.}\\ -\text{8+3=}-\text{5}\\ \text{So, a negative integer and a positive}\mathrm{int}\text{eger}\text{\hspace{0.17em}}\text{is}\overline{)\left(-8,3\right)}.\\ \left(c\right)\text{\hspace{0.17em}}\text{To write a negative integer and a positive integer whose}\\ \text{difference is}-\text{3}\\ -2-\text{(+1)=}-\text{2}-\text{1=}-\text{3}\\ \text{So, a negative integer and a positive}\mathrm{int}\text{eger}\text{\hspace{0.17em}}\text{is}\overline{)\left(-2,1\right)}.\end{array}$

Q.13

$\begin{array}{l}\text{In a quiz, team A scored – 40,10,0 and team B scored}\\ \text{10, 0, -40 in three successive rounds.Which team scored}\\ \text{more? Can we say that we can add integers in any order?}\end{array}$

Ans.

$\begin{array}{l}\text{Sum of team A scored}=\text{}-40+10+0=-30\\ \text{Sum of team B scored}=\text{}10+0-40=-30\\ \text{Both teams scored equal}\text{.}\\ \text{Yes, we can add integers in any order}\text{.}\end{array}$

Q.14

$\begin{array}{l}\mathrm{Fill}\text{ }\mathrm{in}\text{ }\mathrm{the}\text{ }\mathrm{blanks}\text{ }\mathrm{to}\text{ }\mathrm{make}\text{ }\mathrm{the}\text{ }\mathrm{following}\text{ }\mathrm{statements}\text{ }\mathrm{true}:\\ \left(\mathrm{i}\right)\mathrm{}\left(–5\right)+\left(.\dots \dots \dots ..\right)=\left(–8\right)+\left(.\dots \dots \dots ..\right)\\ \left(\mathrm{ii}\right)–53+.\dots \dots \dots ..=–53\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}17+.\dots \dots \dots ..=0\\ \left(\mathrm{iv}\right)\left[13+\left(–12\right)\right]+\left(.\dots \dots \dots ..\right)=.\dots \dots \dots ..+\left[\left(–12\right)+\left(–7\right)\right]\\ \left(\mathrm{v}\right)\left(–4\right)+\left[.\dots \dots \dots ..+\left(–3\right)\right]=\left[.\dots \dots \dots ..+15\right]+.\dots \dots \dots ..\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{}\left(-\text{5}\right)\text{}+\text{}\left(\overline{)-8}\right)\text{}=\text{}\left(-\text{8}\right)\text{}+\text{}\left(\overline{)-5}\right)\\ \left(\text{ii}\right)\text{}-\text{53}+\text{}\overline{)0}\text{}=\text{}-\text{53}\\ \left(\text{iii}\right)\text{\hspace{0.17em}}\text{17}+\text{}\overline{)-17}\text{}=\text{}0\\ \left(\text{iv}\right)\text{}\left[\text{13}+\text{}\left(-\text{12}\right)\right]\text{}+\text{}\left(\overline{)-7}\right)\text{}=\text{}\overline{)13}\text{}+\text{}\left[\left(-\text{12}\right)\text{}+\text{}\left(-\text{7}\right)\right]\\ \left(\text{v}\right)\text{}\left(-\text{4}\right)\text{}+\text{}\left[\overline{)15}\text{}+\text{}\left(-\text{3}\right)\right]\text{}=\text{}\left[\overline{)-4}+\text{15}\right]\text{}+\overline{)-3}\end{array}$

Q.15

$\begin{array}{l}\mathrm{Find}\text{ }\mathrm{each}\text{ }\mathrm{of}\text{ }\mathrm{the}\text{ }\mathrm{following}\text{ }\mathrm{products}:\\ \left(\mathrm{a}\right)\text{ }3×\left(–1\right)\\ \left(\mathrm{b}\right)\text{ }\left(–1\right)×225\\ \left(\mathrm{c}\right)\text{ }\left(–21\right)×\left(–30\right)\\ \left(\mathrm{d}\right)\text{ }\left(–316\right)×\left(–1\right)\\ \left(\mathrm{e}\right)\text{ }\left(–15\right)×0×\left(–18\right)\\ \left(\mathrm{f}\right)\text{ }\left(–12\right)×\left(–11\right)×\left(10\right)\\ \left(\mathrm{g}\right)9×\left(–3\right)×\left(–6\right)\\ \left(\mathrm{h}\right)\left(–18\right)×\left(–5\right)×\left(–4\right)\\ \left(\mathrm{i}\right)\left(–1\right)×\left(–2\right)×\left(–3\right)×4\\ \left(\mathrm{j}\right)\text{ }\left(–3\right)×\left(–6\right)×\left(–2\right)×\left(–1\right)\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{3}×\text{}\left(-\text{1}\right)=\overline{)-3}\\ \left(\text{b}\right)\text{}\left(-\text{1}\right)\text{}×\text{225}=\text{}\overline{)-225}\\ \left(\text{c}\right)\text{}\left(-\text{21}\right)\text{}×\text{}\left(-\text{3}0\right)=\overline{)630}\\ \left(\text{d}\right)\text{}\left(-\text{316}\right)\text{}×\text{}\left(-\text{1}\right)=\overline{)316}\\ \left(\text{e}\right)\text{}\left(-\text{15}\right)\text{}×\text{}0\text{}×\text{}\left(-\text{18}\right)=\overline{)0}\\ \left(\text{f}\right)\text{}\left(-\text{12}\right)\text{}×\text{}\left(-\text{11}\right)\text{}×\text{}\left(\text{1}0\right)=\overline{)1320}\\ \left(\text{g}\right)\text{9}×\text{}\left(-\text{3}\right)\text{}×\text{}\left(-\text{6}\right)=\overline{)162}\\ \left(\text{h}\right)\text{}\left(-\text{18}\right)\text{}×\text{}\left(-\text{5}\right)\text{}×\text{}\left(-\text{4}\right)=\overline{)-360}\\ \left(\text{i}\right)\text{}\left(-\text{1}\right)\text{}×\text{}\left(-\text{2}\right)\text{}×\text{}\left(-\text{3}\right)\text{}×\text{4}=\overline{)-24}\\ \left(\text{j}\right)\text{}\left(-\text{3}\right)\text{}×\text{}\left(-\text{6}\right)\text{}×\text{}\left(-\text{2}\right)\text{}×\text{}\left(-\text{1}\right)=\overline{)36}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Verify}\text{ }\mathrm{the}\text{ }\mathrm{following}:\\ \left(\mathrm{a}\right)18×\left[7+\left(–3\right)\right]=\left[18×7\right]+\left[18×\left(–3\right)\right]\\ \left(\mathrm{b}\right)\left(–21\right)×\left[\left(–4\right)+\left(–6\right)\right]=\left[\left(–21\right)×\left(–4\right)\right]+\left[\left(–21\right)×\left(–6\right)\right]\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\\ \text{18}×\text{}\left[\text{7}+\text{}\left(–\text{3}\right)\right]=18×\left[4\right]=72\\ \text{and}\\ \left[\text{18}×\text{7}\right]\text{}+\text{}\left[\text{18}×\text{}\left(–\text{3}\right)\right]=\left[126\right]+\left[-56\right]=70\\ \text{So},\overline{)\text{18}×\text{}\left[\text{7}+\text{}\left(–\text{3}\right)\right]\ne \left[\text{18}×\text{7}\right]\text{}+\text{}\left[\text{18}×\text{}\left(–\text{3}\right)\right]}\\ \left(\text{b}\right)\\ \left(-\text{21}\right)\text{}×\text{}\left[\left(-\text{4}\right)\text{}+\text{}\left(-\text{6}\right)\right]=\left(-21\right)×\left[-10\right]=210\\ \text{and}\\ \left[\left(-\text{21}\right)\text{}×\text{}\left(-\text{4}\right)\right]\text{}+\text{}\left[\left(-\text{21}\right)\text{}×\text{}\left(-\text{6}\right)\right]=\left[84\right]+\left[126\right]=210\\ \text{So},\\ \overline{)\left(–\text{21}\right)\text{}×\text{}\left[\left(–\text{4}\right)\text{}+\text{}\left(–\text{6}\right)\right]=\left[\left(–\text{21}\right)\text{}×\text{}\left(–\text{4}\right)\right]\text{}+\text{}\left[\left(–\text{21}\right)\text{}×\text{}\left(–\text{6}\right)\right]}\end{array}$

Q.17

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{For}\mathrm{any}\mathrm{integera},\mathrm{what}\mathrm{is}\left(–1\right)×\mathrm{a}\mathrm{equal}\mathrm{to}?\\ \left(\mathrm{ii}\right)\mathrm{Determine}\mathrm{the}\mathrm{integer}\mathrm{whose}\mathrm{product}\mathrm{with}\left(–1\right)\mathrm{is}\\ \left(\mathrm{a}\right)–22\left(\mathrm{b}\right)\mathrm{}37\left(\mathrm{c}\right)0\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.17em}}\left(-\text{1}\right)×\text{a=}-\text{a}\\ \text{(ii) (a)}\overline{)22}×\left(-1\right)=-22\\ \text{(b)}\overline{)-\text{37}}×\left(-1\right)=37\\ \text{(c)}\overline{)\text{0}}×\left(-1\right)=0\end{array}$

Q.18

$\begin{array}{l}\mathrm{Starting}\mathrm{from}\left(–1\right)×5,\mathrm{write}\mathrm{various}\mathrm{products}\mathrm{showing}\mathrm{some}\\ \mathrm{pattern}\mathrm{to}\mathrm{show}\left(–1\right)×\left(–1\right)=1\end{array}$

Ans.

$\begin{array}{l}-1×5=-5\\ -1×4=-4=-5+1\\ -1×3=-3=-4+1\\ -1×2=-2=-3+1\\ -1×1=-1=-2+1\\ -1×0=0=-1+1\\ \text{Thus,}\overline{)-\text{1}×\left(-1\right)=0+1=1}\end{array}$

Q.19

$\begin{array}{l}Findtheproduct,usingsuitableproperties:\\ \left(a\right)26×\left(–48\right)+\left(–48\right)×\left(–36\right)\left(b\right)8×53×\left(–125\right)\\ \left(c\right)15×\left(–25\right)×\left(–4\right)×\left(–10\right)\text{}\left(d\right)\left(–41\right)×102\\ \left(e\right)625×\left(–35\right)+\left(–625\right)×65\left(f\right)7×\left(50–2\right)\\ \left(g\right)\left(–17\right)×\left(–29\right)\text{}\text{}\text{}\text{}\left(h\right)\left(–57\right)×\left(–19\right)+57\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{26}×\text{}\left(-\text{48}\right)\text{}+\text{}\left(-\text{48}\right)\text{}×\text{}\left(-\text{36}\right)\\ =\left(-48\right)×26+\left(-48\right)×\left(-36\right)\\ =\left(-48\right)\left[26-36\right]\\ =\left(-48\right)\left[-10\right]\\ =\overline{)480}\\ \left(\text{b}\right)\text{8}×\text{53}×\text{}\left(-\text{125}\right)\\ =424×\left(-125\right)\\ =\overline{)53000}\\ \left(\text{c}\right)\text{15}×\text{}\left(-\text{25}\right)\text{}×\text{}\left(-\text{4}\right)\text{}×\text{}\left(-\text{1}0\right)\\ =\left(15×-25\right)×\left(-4×-10\right)\\ =\left(-375\right)×40\\ =\overline{)-15000}\\ \left(\text{d}\right)\text{}\left(-\text{41}\right)\text{}×\text{1}0\text{2}\\ =\overline{)-\text{4182}}\\ \left(\text{e}\right)\text{625}×\text{}\left(-\text{35}\right)\text{}+\text{}\left(-\text{625}\right)\text{}×\text{65}\\ =-21875+\left(-40625\right)\\ =\overline{)-62500}\\ \\ \left(\text{f}\right)\text{7}×\text{}\left(\text{5}0\text{}-\text{2}\right)\\ =7×\left(48\right)\\ =\overline{)336}\\ \left(\text{g}\right)\text{}\left(-\text{17}\right)\text{}×\text{}\left(-\text{29}\right)\\ =\overline{)493}\\ \left(\text{h}\right)\text{}\left(-\text{57}\right)\text{}×\text{}\left(-\text{19}\right)\text{}+\text{57}\\ =1083+57\\ =\overline{)1140}\end{array}$

Q.20

$\begin{array}{l}\left(\text{i}\right)\text{For any integer a, what is}\left(\text{–1}\right)\text{× a equal to?}\\ \left(\text{ii}\right)\text{Determine the integer whose product with}\left(\text{–1}\right)\text{is}\\ \left(\text{a}\right)\text{–22}\left(\text{b}\right)\text{37}\left(\text{c}\right)\text{0}\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.17em}}\left(-\text{1}\right)×\text{a=}-\text{a}\\ \text{(ii) (a)}\overline{)22}×\left(-1\right)=-22\\ \text{(b)}\overline{)-\text{37}}×\left(-1\right)=37\\ \text{(c)}\overline{)\text{0}}×\left(-1\right)=0\end{array}$

Q.21

$\begin{array}{l}\text{Starting from}\left(\text{-1}\right)\text{×5, write various products showing some}\\ \text{pattern to show}\left(\text{-1}\right)\text{×}\left(\text{-1}\right)\text{=1}\end{array}$

Ans.

$\begin{array}{l}-1×5=-5\\ -1×4=-4=-5+1\\ -1×3=-3=-4+1\\ -1×2=-2=-3+1\\ -1×1=-1=-2+1\\ -1×0=0=-1+1\\ \text{Thus ,}\overline{)-\text{1}×\left(-1\right)=0+1=1}\end{array}$

Q.22

$\begin{array}{l}\text{Find the product, using suitable properties:}\\ \text{}\left(\text{a}\right)\text{26 ×}\left(\text{– 48}\right)\text{+}\left(\text{– 48}\right)\text{×}\left(\text{–36}\right)\text{}\left(\text{b}\right)\text{8 × 53 ×}\left(\text{–125}\right)\\ \text{}\left(\text{c}\right)\text{15 ×}\left(\text{–25}\right)\text{×}\left(\text{– 4}\right)\text{×}\left(\text{–10}\right)\text{}\left(\text{d}\right)\text{}\left(\text{– 41}\right)\text{× 102}\\ \text{}\left(\text{e}\right)\text{625 ×}\left(\text{–35}\right)\text{+}\left(\text{– 625}\right)\text{× 65}\left(\text{f}\right)\text{7 ×}\left(\text{50 – 2}\right)\\ \text{}\left(\text{g}\right)\text{}\left(\text{–17}\right)\text{×}\left(\text{–29}\right)\text{}\left(\text{h}\right)\text{}\left(\text{–57}\right)\text{×}\left(\text{–19}\right)\text{+ 57}\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{26}×\text{}\left(-\text{48}\right)\text{}+\text{}\left(-\text{48}\right)\text{}×\text{}\left(-\text{36}\right)\\ =\left(-48\right)×26+\left(-48\right)×\left(-36\right)\\ =\left(-48\right)\left[26-36\right]\\ =\left(-48\right)\left[-10\right]\\ =\overline{)480}\\ \left(\text{b}\right)\text{8}×\text{53}×\text{}\left(-\text{125}\right)\\ =424×\left(-125\right)\\ =\overline{)53000}\\ \left(\text{c}\right)\text{15}×\text{}\left(-\text{25}\right)\text{}×\text{}\left(-\text{4}\right)\text{}×\text{}\left(-\text{1}0\right)\\ =\left(15×-25\right)×\left(-4×-10\right)\\ =\left(-375\right)×40\\ =\overline{)-15000}\\ \left(\text{d}\right)\text{}\left(-\text{41}\right)\text{}×\text{1}0\text{2}\\ =\overline{)-\text{4182}}\\ \left(\text{e}\right)\text{625}×\text{}\left(-\text{35}\right)\text{}+\text{}\left(-\text{625}\right)\text{}×\text{65}\\ =-21875+\left(-40625\right)\\ =\overline{)-62500}\\ \left(\text{f}\right)\text{7}×\text{}\left(\text{5}0\text{}-\text{2}\right)\\ =7×\left(48\right)\\ =\overline{)336}\\ \left(\text{g}\right)\text{}\left(-\text{17}\right)\text{}×\text{}\left(-\text{29}\right)\\ =\overline{)493}\\ \left(\text{h}\right)\text{}\left(-\text{57}\right)\text{}×\text{}\left(-\text{19}\right)\text{}+\text{57}\\ =1083+57\\ =\overline{)1140}\end{array}$

Q.23

$\begin{array}{l}\text{A certain freezing process requires that room temperature}\\ \text{be lowered from 40°C at the rate of 5°C every hour. What will}\\ \text{be the room temperature 10 hours after the process begins?}\end{array}$

Ans.

$\begin{array}{l}\text{Given initial temprature}=40°C\\ \text{Change in temprature per hour}=-5°C\\ \text{Change in temprature after 10 hours}=-5°C×10=-50°C\\ \text{Final temprature}=40°C-50°C=\overline{)-10°C}\end{array}$

Q.24

$\begin{array}{l}\text{In a class test containing 10 questions, 5 marks are}\\ \text{awarded for every correct answer and}\left(\text{-2}\right)\text{marks are}\\ \text{awarded forevery incorrect answer and 0 for questions}\\ \text{not attempted.}\\ \text{}\left(\text{i}\right)\text{Mohan gets four correct and six incorrect answers.}\\ \text{What is his score?}\\ \text{}\left(\text{ii}\right)\text{Reshma gets five correct answers and five incorrect}\\ \text{answers, what is her score?}\\ \text{}\left(\text{iii}\right)\text{Heena gets two correct and five incorrect answers}\\ \text{out of seven questions she attempts. What is her score?}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Marks given for 1 correct answer}=5\\ \text{Marks given for 4 correct answer}=5×4=20\\ \text{Marks given for 1 wrong answer}=-2\\ \text{Marks given for 6 wrong answer}=-2×6=-12\\ \text{Score obtained by Mohan}=20-12=\overline{)8}\\ \text{(ii)}\\ \text{Marks given for 1 correct answer}=5\\ \text{Marks given for 5 correct answer}=5×5=25\\ \text{Marks given for 1 wrong answer}=-2\\ \text{Marks given for 5 wrong answer}=-2×5=-10\\ \text{Scored obtained by Reshma}=25-10=\overline{)15}\\ \text{(iii) Similarly,}\\ \text{Marks given for 2 correct answer}=5×2\\ \text{Marks given for 5 correct answer}=-2×5=-10\\ \text{Score btained by Heena}=10–10=\overline{)0}\end{array}$

Q.25

$\begin{array}{l}\text{A cement company earns a profit of ₹ 8 per bag of white}\\ \text{cement sold and a loss of ₹ 5 per bag of grey cement sold.}\end{array}$

$\begin{array}{l}\text{}\left(\text{a}\right)\text{The company sells 3,000 bags of white cement and 5,000}\\ \text{bags of grey cement in a month. What is its profit or loss?}\\ \text{}\left(\text{b}\right)\text{What is the number of white cement bags it must sell to}\\ \text{have neither profit nor loss, if the number of grey bags sold}\\ \text{is 6,400 bags.}\end{array}$

Ans.

$\begin{array}{l}\left(a\right)\\ \text{Profit earned while selling 1 bag of white cement}=\text{}₹\text{}8\\ \text{Profit earned while selling 3000 bag of white cement}\\ =\text{}₹\text{}8×3000=24000\\ \text{Loss incurred while selling 1 bag of grey cement}=-₹\text{}5\\ \text{Loss incurred while selling 5000 bag of grey cement}\\ =-₹\text{}5×5000=-₹\text{\hspace{0.17em}}25000\\ \text{Total profit/loss earned = Profit+loss}\\ =₹24000-₹25000=-₹1000\\ \text{Thus, there will be a loss of}₹\text{1000 to the company}\text{.}\\ \text{(b)}\\ \text{Loss incurred while selling 1 bag of grey cement}=\text{}-₹\text{}5\\ \text{Loss incurred while selling 6400 bag of grey cement}\\ =-₹\text{}5×6400=-₹\text{\hspace{0.17em}}32000\\ \text{Let the number of white bag to be sold be}x\text{.}\\ \text{Profit earned selling 1 bag of white cement}=\text{}₹\text{}8\\ \text{Profit earned selling x bag of white cement}=\text{}₹\text{}8x\\ \text{When there is no profit or no loss, we have}\\ \text{Profit earned + Loss incurred}=\text{}0\\ 8x–32000=0\\ 8x=32000\\ x=\frac{32000}{8}\\ =\overline{)4000}\\ \text{Thus},4000\text{\hspace{0.17em}}\text{bags of white cement should be sold}\text{.}\end{array}$

Q.26

$\begin{array}{l}\text{Replace the blank with an integer to make it a true}\\ \text{statement.}\\ \left(\text{a}\right)\text{}\left(-\text{3}\right)\text{× _____ = 27}\\ \left(\text{b}\right)\text{5 × _____ = -35}\\ \left(\text{c}\right)\text{_____ ×}\left(-\text{8}\right)\text{=}-\text{56}\\ \left(\text{d}\right)\text{_____ ×}\left(-\text{12}\right)\text{= 132}\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{}\left(\text{–3}\right)\text{×}\underset{_}{\overline{)\text{–9}}}\text{= 27}\\ \left(\text{b}\right)\text{}\text{\hspace{0.17em}}\text{5 ×}\underset{_}{\overline{)\text{–7}}}\text{= –35}\\ \left(\text{c}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{_}{\overline{)\text{7}}}\text{×}\left(\text{–8}\right)\text{= –56}\\ \left(\text{d}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{_}{\overline{)\text{–11}}}\text{×}\left(\text{–12}\right)\text{= 132}\end{array}$

Q.27

$\begin{array}{l}\text{Evaluate each of the following:}\\ \left(\mathrm{a}\right)\text{ }\left(–30\right)÷10\\ \left(\mathrm{b}\right)\text{ }50÷\left(–5\right)\\ \left(\mathrm{c}\right)\text{ }\left(–36\right)÷\left(–9\right)\\ \left(\mathrm{d}\right)\text{ }\left(–49\right)÷\left(49\right)\\ \left(\mathrm{e}\right)\text{ }13÷\left[\left(–2\right)+1\right]\\ \left(\mathrm{f}\right)\text{ }0÷\left(–12\right)\\ \left(\mathrm{g}\right)\text{ }\left(–31\right)÷\left[\left(–30\right)+\left(–1\right)\right]\\ \left(\mathrm{h}\right)\text{ }\left[\left(–36\right)÷12\right]÷3\\ \left(\mathrm{i}\right)\text{ }\left[\left(–6\right)+5\right)\right]÷\left[\left(–2\right)+1\right]\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{}\left(-\text{3}0\right)\text{}÷\text{1}0=\overline{)-3}\\ \left(\text{b}\right)\text{5}0\text{}÷\text{}\left(-\text{5}\right)=\overline{)-10}\\ \left(\text{c}\right)\text{}\left(-\text{36}\right)\text{}÷\text{}\left(-\text{9}\right)=\overline{)4}\\ \left(\text{d}\right)\text{}\left(-\text{49}\right)\text{}÷\text{}\left(\text{49}\right)=\overline{)-1}\\ \left(\text{e}\right)\text{13}÷\text{}\left[\left(-\text{2}\right)\text{}+\text{1}\right]=13÷\left[-1\right]=\overline{)-13}\\ \left(\text{f}\right)\text{}0\text{}÷\text{}\left(-\text{12}\right)=\overline{)0}\\ \left(\text{g}\right)\text{}\left(-\text{31}\right)\text{}÷\text{}\left[\left(-\text{3}0\right)\text{}+\text{}\left(-\text{1}\right)\right]=\left(-31\right)÷\left[-31\right]=\overline{)1}\\ \left(\text{h}\right)\text{}\left[\left(-\text{36}\right)\text{}÷\text{12}\right]\text{}÷\text{3=}\left[-3\right]÷3=\overline{)-1}\\ \left(\text{i}\right)\left[\left(-\text{6}\right)\text{}+\text{5}\right)\right]\text{}÷\text{}\left[\left(-\text{2}\right)\text{}+\text{1}\right]=\left[-1\right]÷\left[-1\right]=\overline{)1}\end{array}$

Q.28

$\begin{array}{l}\text{Verify that a÷}\left(\text{b+c}\right)\text{¹}\left(\text{a÷b}\right)\text{+}\left(\text{a÷c}\right)\text{ for each of the}\\ \text{following.}\\ \text{values of a, b and c.}\\ \text{(a) a=12, b=-4, c=2}\\ \text{(b) a=(-10), b=1, c=1}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}\mathrm{a}=12,\text{\hspace{0.17em}}\mathrm{b}=-4,\text{\hspace{0.17em}}\mathrm{c}=2\\ \mathrm{a}÷\left(\mathrm{b}+\mathrm{c}\right)=12÷\left(-4+2\right)\\ =12÷\left(-2\right)\\ =-6\\ \text{and}\\ \left(\mathrm{a}÷\mathrm{b}\right)+\left(\mathrm{a}÷\mathrm{c}\right)=\left(12÷\left(-4\right)\right)+\left(12÷2\right)\\ =\left(-3\right)+6=3\\ \mathrm{So},\overline{)\mathrm{a}÷\left(\mathrm{b}+\mathrm{c}\right)\ne \left(\mathrm{a}÷\mathrm{b}\right)+\left(\mathrm{a}÷\mathrm{c}\right)}\\ \left(\mathrm{b}\right)\\ \mathrm{a}=-10,\text{\hspace{0.17em}}\mathrm{b}=1,\mathrm{c}=1\\ \mathrm{a}÷\left(\mathrm{b}+\mathrm{c}\right)=-10÷\left(1+1\right)\\ =-10÷\left(2\right)\\ =-5\\ \text{and}\\ \left(\mathrm{a}÷\mathrm{b}\right)+\left(\mathrm{a}÷\mathrm{c}\right)=\left(-10÷1\right)+\left(-10÷1\right)\\ =\left(-10\right)-10=-20\\ \mathrm{So},\overline{)\mathrm{a}÷\left(\mathrm{b}+\mathrm{c}\right)\ne \left(\mathrm{a}÷\mathrm{b}\right)+\left(\mathrm{a}÷\mathrm{c}\right)}\end{array}$

Q.29

$\begin{array}{l}\text{Fill in the blanks:}\\ \text{}\left(\text{a}\right)\text{ 369 ÷ _____ = 369}\\ \left(\text{b}\right)\text{ }\left(\text{–75}\right)\text{÷ _____ = –1}\\ \left(\text{c}\right)\text{ }\left(\text{–206}\right)\text{÷ _____ = 1}\\ \left(\text{d}\right)\text{ – 87 ÷ _____ = 87}\\ \left(\text{e}\right)\text{ _____ ÷ 1 = – 87}\\ \left(\text{f}\right)\text{ _____ ÷ 48 = –1}\\ \left(\text{g}\right)\text{ 20 ÷ _____ = –2}\\ \left(\text{h}\right)\text{ _____ ÷}\left(\text{4}\right)\text{= –3}\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{369}÷\text{}\underset{_}{\overline{)1}}\text{}=\text{369}\\ \left(\text{b}\right)\text{}\left(-\text{75}\right)\text{}÷\text{}\underset{_}{\overline{)75}}\text{}=\text{}-\text{1}\\ \left(\text{c}\right)\text{}\left(-\text{2}0\text{6}\right)\text{}÷\text{}\underset{_}{\overline{)-206}}\text{}=\text{1}\\ \left(\text{d}\right)-\text{87}÷\text{}\underset{_}{\overline{)-1}}\text{}=\text{87}\\ \left(\text{e}\right)\text{}\underset{_}{\overline{)-87}}\text{}÷\text{1}=\text{}-\text{87}\\ \left(\text{f}\right)\text{}\underset{_}{\overline{)-48}}\text{}÷\text{48}=\text{}-\text{1}\\ \left(\text{g}\right)\text{2}0\text{}÷\text{}\underset{_}{\underset{_}{\overline{)-10}}}\text{}=\text{}-\text{2}\\ \left(\text{h}\right)\text{}\underset{_}{\overline{)-12}}\text{}÷\text{}\left(\text{4}\right)\text{}=\text{}-\text{3}\end{array}$

Q.30

$\begin{array}{l}\text{Write five pairs of integers}\left(\text{a,b}\right)\text{such that a÷b =}-\text{3. One}\\ \text{such pair is}\left(\text{6, }-\text{2}\right)\text{ because 6÷}\left(-\text{2}\right)\text{ =}\left(-\text{3}\right)\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{Five pair of integers are:}\\ \overline{)\left(3,-1\right),\text{\hspace{0.17em}}\left(-3,1\right),\text{\hspace{0.17em}}\left(9,-3\right),\text{\hspace{0.17em}}\left(-9,3\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\left(12,-4\right)}\end{array}$

Q.31

$\begin{array}{l}\text{The temperature at 12 noon was 10°C above zero. If it}\\ \text{decreases at the rate of 2°C per hour until midnight, at}\\ \text{what time would the temperature be 8°C below zero?}\\ \text{What would be the temperature at mid-night?}\end{array}$

Ans.

$\begin{array}{l}\text{Initial temprature}=10°C\\ \text{Change in temprature per hour}=-2°C\\ \text{Temprature at 1:00 PM=10}°C+\left(-2°C\right)=8°C\\ \text{Temprature at 2:00 PM=8}°C+\left(-2°C\right)=6°C\\ \text{Temprature at 3:00 PM=6}°C+\left(-2°C\right)=4°C\\ \text{Temprature at 4:00 PM}=4°C+\left(-2°C\right)=2°C\\ \text{Temprature at 5:00 PM}=2°C+\left(-2°C\right)=0°C\\ \text{Temprature at 6:00 PM}=0°C+\left(-2°C\right)=-2°C\\ \text{Temprature at 7:00 PM}=-2°C+\left(-2°C\right)=-4°C\\ \text{Temprature at 8:00 PM}=-4°C+\left(-2°C\right)=-6°C\\ \text{Temprature at 9:00 PM}=-6°C+\left(-2°C\right)=-8°C\\ \text{Thus, the temprature will be}8°C\text{below zero at 9:00PM}\\ \text{It will take 12 hours to be midnight (i}\text{.e}\text{. 12:00 AM) after 12:00}\\ \text{noon}\text{.}\\ \text{Change in temprature in 12 hours}=\text{}-\text{2}°C×12=-24°C\\ \text{At midnight the temprature would be}=\text{10+}\left(-24\right)=-14°C\\ \text{Thus,}\overline{)\text{the temprature at midnight will be 14}°C\text{\hspace{0.17em}}\text{below}\text{\hspace{0.17em}}0}.\end{array}$

Q.32

$\begin{array}{l}\text{In a class test}\left(\text{+ 3}\right)\text{marks are given for every correct}\\ \text{answer and}\left(\text{-2}\right)\text{marks are given for every incorrect}\\ \text{answer and no marks for not attempting any question.}\\ \text{}\left(\text{i}\right)\text{Radhika scored 20 marks. If she has got 12 correct}\\ \text{answers, how many questions has she attempted}\\ \text{incorrectly?}\\ \text{}\left(\text{ii}\right)\text{Mohini scores -5 marks in this test, though she has}\\ \text{got 7 correct answers. How many questions has she}\\ \text{attempted incorrectly?}\\ \text{}\left(\text{iii}\right)\text{Rakesh scores 18 marks by attempting 16 questions.}\\ \text{How many questions has he attempted correctly and how}\\ \text{many has he attempted incorrectly?}\end{array}$

Ans.

$\begin{array}{l}\text{Marks obtained for 1 right answer}=+3\\ \text{Marks obtained for 1 wrong answer}=-2\\ \left(i\right)\\ \text{Marks obtained by Radhika=20}\\ \text{Marks obtained for 12 correct answers=12}×\text{3=36}\\ \text{Marks obtained for incorrect answer=Total score-Marks}\\ \text{obtained for 12 correct answers}\\ \text{=20-36=–16}\\ \text{Marks obtained for 1 wrong answer=-2}\\ \text{Thus, number of incorrect answer=}\left(-16\right)÷\left(-2\right)=\overline{)8}\\ \text{Thus, Radhika attempted 8 questions wrongly}\text{.}\\ \text{(ii)}\\ \text{Marks scored by Mohini=-5}\\ \text{Marks obtained for incorrect answers}\\ \text{=Total answer-Marks obtained for 12 incorrect answer}\\ =-5-21=-26\\ \text{Marks obtained for 1 wrong answer=-2}\\ \text{Thus, number of incorrect answer=}\left(-26\right)÷\left(-2\right)=\overline{)13}\\ \text{Therefore, Mohini attempted 13 questions wrongly}\text{.}\\ \text{(iii)}\\ \text{Total marks scored by Rakesh =18}\\ \text{Number of questions attempted =16}\\ \text{}\left(\text{Number of correct answers}\right)\left(3\right)\\ +\left(\text{Number of incorrect answers}\right)\left(-2\right)=18\\ ⇒\left(\text{Number of correct answers}\right)\left(5\right)+\left(-32\right)=18\\ ⇒\left(\text{Number of correct answers}\right)=\frac{50}{5}=10\\ So,\left(\text{Number of incorrect answers}\right)=16-10=6\\ \text{Thus, Total number of correct and incorrect answers scored}\\ \text{by Rakesh is 10 and 6 respectively}\text{.}\end{array}$

Q.33

$\begin{array}{l}\text{An elevator descends into a mine shaft at the rate}\\ \text{of 6 m/min. If the descent startsfrom 10 m above}\\ \text{the ground level,how long will it take to reach – 350​ m.}\end{array}$

Ans.

$\begin{array}{l}\text{Distance descended is denoted by a negative integer}\text{.}\\ \text{Initial height}=+10\text{m}\\ \text{Final depth}=-350\text{m}\\ \text{Total distance to be descended by the elevator}=\text{}\left(-350\right)-\left(+10\right)=-360\text{\hspace{0.17em}}\text{m}\\ \text{Time}\text{\hspace{0.17em}}\text{taken by the elevator to be descend}-\text{6 m}=\text{1 min}\\ \text{Thus, time taken by the elevator to descend}-\text{360 m =}\left(-360\right)÷\left(-6\right)=\overline{)60\text{\hspace{0.17em}}\text{minutes}\text{\hspace{0.17em}}=1\text{\hspace{0.17em}}\text{hour}}\end{array}$

##### FAQs (Frequently Asked Questions)
1. How many exercises are included in the NCERT Solutions for Class 7 Mathematics Chapter 1?

There are four exercises in Class 7 Mathematics Chapter 1 Integers.

The first exercise consists of ten questions (Ex 1.1).

The second exercise consists of four questions (Ex 1.2). Question 1 is divided into three sections, question 2 is divided into three sections, and question 4 is divided into five sections.

There are 9 questions in the third exercise (Ex 1.3). Question 2 is made up of two parts, question 3 is made up of two parts, and question 5 is made up of eight parts.

There are seven questions in the fourth exercise (Ex 1.4). Question 1 is divided into nine parts, while question 2 is divided into two parts.

As a result, there are a total of 30 questions in Chapter 1 (Integers) of Class 7 Mathematics.

In the CBSE Class 7 Mathematics Integers Chapter, there are a total of seven exercises.

2. Is it necessary to complete all of the problems in Chapter 1 of the NCERT Solution for Class 7 Mathematics?

Definitely! Because these questions are important from the examination point of view. . Subject matter experts have answered these questions to assist students in completing the exercise with ease.  These NCERT Solutions assist students by encouraging them to become better learners and strive to feed their insatiable curiosity.

3. How long will it take to finish NCERT Class 7 Mathematics Chapter 1?

It takes about a week to complete the NCERT Class 7 Mathematics Chapter 1, but we recommend that you practice , again and again, to significantly improve your performance.

4. Is Class 7 Mathematics chapter Integers difficult?

It is neither easy nor difficult to complete 7th Class Mathematics Chapter 1. Because some parts of this chapter are easy and others are difficult, it falls somewhere in between easy and difficult. The difficulty level of each chapter, however, varies from student to student. As a result, whether the chapter is easy or difficult depends on the individual. Some students find it difficult, while others may find it easy.

5. What are the advantages of using NCERT Solutions Class 7 Mathematics?

Because the CBSE syllabus is based on the NCERT textbooks, NCERT solutions Class 7 Mathematics assist CBSE students in preparing for their exams.  Moreover, these solutions instil a deep understanding of the subject and assist students in resolving their doubts while solving problems and motivates them to improve their performance and stay ahead of the competition.