NCERT Solutions for Class 7 Maths Chapter 1 Integers (EX 1.3) Exercise 1.3

It might be difficult to decide between several books and notes at times, especially while preparing for exams. It is crucial to select quality books so that all exam-related subjects are covered and presented in an understandable manner. NCERT Books are among the top books for CBSE-affiliated schools, according to a majority of teachers. NCERT books are also essential for a variety of exams, including IAS, UPSC, and others. To understand these books better, students are advised to make use of learning tools such as the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and more.

The significance of NCERT books :

  • Easy to read

All people can understand NCERT books because they are written in a simple manner. These books are easy to understand because many students use them to prepare for different tests. To easily understand and organise the texts from NCERT books, students can refer to the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and more tools like it.

  • Explanation of topics in detail

All the topics included in the syllabus are covered in the basics and fundamentals of NCERT Books. Additionally, it covers a number of crucial themes in great detail. As a result, students will be fully prepared for their exams once they have thoroughly covered all the topics from the NCERT Books. One way to thoroughly cover the NCERT texts is to practice with the help of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, and other similar tools available on the Extramarks website. Exams require students to retain all the material covered in the syllabus. Students are advised to refer to the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, and other helpful resources that are available on the Extramarks website.

  • Written by Professionals

The authors of NCERT Books are experts in their fields. Each topic has been thoroughly investigated by subject-matter experts before compiling these solutions. The experts at Extramarks created tools like NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 which make understanding these concepts even easier.

  • Ease of Access

Students may read and download any NCERT book online by visiting the official website of NCERT. After going through the NCERT books, students can make use of NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and many such tools on the Extramarks website.

  • Good for Different Exams

The government suggests NCERT books for CBSE exams. Students taking the UPSC and IAS Exams can also use NCERT books. Numerous additional exams can be passed with ease if NCERT books are well-read. Students can be thorough with the NCERT books with the help of resources like NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and more.

  • Most of the questions in exams are from the NCERT books

The fact that the majority of exam questions are either drawn directly from NCERT books or are strikingly similar to the ones already included in the book is one of the critical advantages of reading NCERT books. Students should be thorough with the NCERT books and they can do so with the help of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and other resources.

Mathematicians work with a variety of numbers. Even though each number is unique, they have certain things in common. These numbers are divided into many groups based on their qualities in order to be understood. The following are a few of the systems groups:

  • Natural numbers
  • Whole numbers
  • Integers
  • Rationals
  • Real numbers

The NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 are the solutions to Chapter 1 of Class 7 Mathematics and this chapter deals with Integers.

What are Integers? 

The collection of whole numbers and negative numbers is known as an Integer in Mathematics. Integers, like whole numbers, do not include the fractional portion. Integers can therefore be defined as numbers that can be positive, negative, or zero, but not as fractions. On Integers, students can carry out all arithmetic operations, including addition, subtraction, multiplication, and division. Integer examples include 1, 2, 5, 8, -9, -12, etc. “Z” stands for an Integer.

What is the meaning of Integers –

The Latin term “Integer,” which implies entire or intact, is where the word “Integer” first appeared. Zero, positive numbers, and negative numbers make up the particular set of numbers known as Integers.

Integer examples include – 1, -12, and 6.

Symbol – The letter “Z” stands for an Integer.

Unknown or unidentified Integers are denoted in mathematical equations by lowercase, italicised letters from the “late middle” of the alphabet. P, q, r, and s are the most prevalent.

Denumerable sets include the set Z. Denumerability is the property that even if a set may include an endless number of members, any element in the set may be represented by a list that implies its identity.

Students can sometimes get overwhelmed by the amount of new information provided in this chapter and topic. Students are advised by the experts at Extramarks to practice with the help of NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and other such tools to help them understand this new topic. The NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 provides detailed step-by-step solutions to the NCERT questions that can be very helpful for the students trying to complete their homework or preparing for their examinations.

There are three different types of Integers –

  • Zero
  • Positive Numbers
  • Negative Numbers

Students can understand all these numbers with the help of tools like the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and more tools that are available on the Extramarks website.

  • Positive Numbers

Positive numbers are ones that have a plus symbol (+) before them. Positive numbers are typically shown as plain numbers without the addition sign (+). Every positive number is greater than 0, as are the numbers to its left as well as the negative numbers. Positive numbers are shown to the right of the origin on a number line ( zero).

Examples are 1, 5, 500, 66, 89, etc.

  • Negative Numbers

The numbers that are preceded by a minus sign are considered negative numbers (-). Mentioning the sign of negative Integers is required. On a number line, negative numbers are shown to the left of the origin (zero). Examples are -8, -10, -1000, and -1919.

  • Zero

Zero cannot be positive or negative, i.e., it lacks the +ve or -ve sign, making it a neutral number. If both of the Integers have the same sign, adding them results in a result that has the same sign as both of the values.

Rules:

  • (+) + (+) = +
  •  (-) + (-) = –

To add two Integers with different signs: Subtraction will occur if one of the numbers has a different sign, and the result will reflect the sign of the larger number. Students can calculate the difference between the absolute values and assign it the same sign as the absolute value of the Integer with the highest difference.

NCERT Solutions for Class 7 Maths Chapter 1 Integers (EX 1.3) Exercise 1.3 (include PDF) 

The easy accessibility of resources such as the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, can make it extremely simple for students to prepare for exams. The use of resources like the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 can help students comprehend various types of problems with ease and effectively prepare for all the questions found in the exercise covered in the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3.

There are numerous resources accessible for this particular chapter, one of which is the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3. On the Extramarks website, students can find a variety of resources for Chapter 1 of Class 7 Mathematics, including the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3.

Access NCERT Solutions for Class 7 Maths Chapter 1 – Integers 

Students ought to utilise the tools offered by Extramarks professionals to avoid circumstances where they can lose interest in a subject. One of the many resources that students may easily access at Extramarks is the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3. With the aid of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, students can boost their exam preparation. The NCERT Class 7 Maths Chapter 1 Exercise 1.3 available on the Extramarks website can help the students in effectively completing the Class 7 Maths Chapter 1 Exercise 1.3. The NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, can be practised by students in order to help them comprehend each question in detail.

The availability of tools like the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 can make the process of exam preparations to be very easy for the students. It is tools like the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 that can efficiently assist the students in understanding different types of questions with ease and prepare efficiently all the types of questions asked in Class 7 Maths Exercise 1.3.

The Class 7 Maths Chapter 1 Exercise 1.3 Solutions are just one of the many tools available for this particular chapter.

Exercise 1.3 

Qualified mathematicians developed the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3. Students can access materials for lower elementary and middle school that were created using the same methodology as the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, on the Extramarks website. To make sure they thoroughly understand all that is taught in class and what is required of them to write about in their exams, experts urge students to use resources like NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3. Without a solid grounding in the teaching strategies used in elementary school classrooms, students cannot understand the middle and senior secondary school curricula. If they lack a solid foundation in a subject, students may lose interest in it.

The likelihood that a student will comprehend the content is higher when they use resources like the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and many other tools available on the Extramarks website. The NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 give them the chance to practice the questions and acquire the proper methodology for reaching a conclusion. The step-by-step solutions included in materials like the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 are developed while keeping in mind the NCERT guidelines and the correct method for solving the problems.

Chapter 1 Integers 

In a subject like Mathematics, where even the marks for each individual step are given, the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 can play a key role in supporting students in efficiently studying for their exams. Finding the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 is made easier by the Extramarks website. They are also easy to comprehend. Extramarks’ Mathematics experts created the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, making them reliable sources as well. Mathematics specialists work to provide materials like the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 that are easy for everyone to grasp and benefit from.

Introduction 

Anytime, regardless of their location or the time, students can get the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, and other materials of similar nature on the Extramarks website. Students can now arrange their planning in a very practical way. In order to help students work through and practice Chapter 1 questions, the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 are created to help with this. The answers to every question in the exercises are detailed and include step-by-step instructions for the students. The NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, and all the other NCERT solutions that are accessible on the Extramarks’ website can help students considerably improve their skills and gain great confidence for their exam preparations.

Properties of Addition and Subtraction of Integers 

Students will study how to add and subtract numbers with the same sign and various signs when learning about the addition and subtraction of Integers. The number line can be used to add and subtract signed Integers. When performing operations on numbers, specific guidelines must be followed.

In contrast to adding two negative Integers, which produces a sum with a negative sign, adding two positive Integers produces positive Integers. However, adding two different signed Integers will only produce subtraction, with the sign of the result matching that of the larger number. Here are a few examples:

  • 2+2 = 4
  • 2 + (-2) = 0
  • -2 + (-2) = -4
  • -2 – (-2) = 0

The students can better understand this by practising. They can do so with the help of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and other similar tools available on the Extramarks website.

Subtraction and Addition –

The two main arithmetic operations in Mathematics are addition and subtraction. In addition to these two operations, Multiplication and Division are the other two fundamental operations we learn in elementary Mathematics.

The “addition” stands for the values that have been added to the original value. For instance, if students add more than two balls to a basket, it will contain a total of four balls. Similar to this, subtracting two balls from a basket containing four balls results in the basket having just two balls, which demonstrates subtraction.

In addition to Integers, rational and irrational numbers can also be added to and subtracted from. Both operations are therefore applicable to all real and complex numbers. Additionally, when conducting algebraic operations, the addition and subtraction of algebraic expressions are done according to the same laws.

To practice this theme, students are advised to make use of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 easily available on the Extramarks website.

Multiplication of Integers 

Multiplication is the repeated addition of numbers, according to its definition. However, the guidelines for multiplying Integers differ from those for addition. There are three options available. As follows:

  • Multiplication of a positive number by a negative number,
  • between two positive numbers,
  • and between a positive and a negative number.

When two Integers with comparable sign numbers are multiplied, the result is always positive. So, whether two positive numbers or two negative numbers are added together, the result will always be positive. While an Integer with different signs that is both positive and negative will always result in a negative number.

To understand this concept students must consult the Extramarks’ website as many students may find this to be slightly challenging, they can make the most of their exam preparation with the help of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 to practice different questions that require this.

Division of Integers 

The division is the distribution of numbers, whereas the multiplication is the adding up of numbers. The reverse of multiplication is the operation of dividing numbers. Integer division rules, however, are identical to multiplication rules. However, it need not always be the case that the quotient is an Integer.

Rule 1: A pair of positive Integers will always have a positive quotient.

Rule 2: A pair of negative Integers will always have a positive quotient.

Rule 3: A positive Integer and a negative Integer will never have a positive quotient.

Similar to multiplying, students should divide the Integers without the sign, and then add the sign in accordance with the table’s rules. Positive quotient results from the division of two Integers with similar signs, while negative quotient results from the division of two Integers with opposite signs.

All of these subtopics might seem confusing and overwhelming to many students and they may change that with the help of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and other tools that are targeted towards this chapter.

Secret Tips to Understand the Chapter 

The best way to understand any chapter in Mathematics is to practice as many questions as possible. With the help of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, students can very conveniently do that. The more the students practice with the help of tools like the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, the higher are their chances of mastering the topics that are generally considered complex and difficult. Students are advised to thoroughly go through the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, and practice with their help frequently. Students should not wait until their exams to use the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3. They should use them for their homework as well as class tests, etc.

NCERT Solutions for Class 7 Maths 

In computer programming and coding, if objects must be on or off, i.e., 1 or 0, Integers are used. Everything that a computer performs can be reduced to a series of ones and zeros; this is referred to as being “binary.” In Mathematics, Integers are crucial. They are useful not just in Mathematics but also in daily life.

In practically every field, Integers are useful for computing the effectiveness of positive or negative numbers.

  • Students can determine their position using Integers.
  • Calculating how many steps should be taken—more or fewer—to get better results is also helpful.
  • The intensity of actual life and intense emotions, however, cannot be quantified.

The use of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 can further help students practice and understand all that is taught in this chapter.

NCERT Solution Class 7 Maths of Chapter 1 All Exercises 

Integers are employed in a variety of everyday applications, including clocks, addresses, thermometers, and money. Maps, altitude measurements, and hockey scores are other applications for Integers.

Whole numbers, often known as positive Integers, are frequently used in everyday life. Roadway speed limits are posted along with highway numbers. To determine the speed to travel on a certain road, one uses Integers. Clock numbers are used to read the time and set alarms. Integers are used in buildings and homes have numerals on them that serve as both address numbers and floor numbers within buildings. Integers are used in maps to show direction and information. Students can understand this concept better when they practice with the help of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 and other similar tools made for this particular chapter that are easily available on the Extramarks website.

Negative Integers may be more difficult to employ in everyday situations. The use of positive Integers can be seen, for instance, in bank calculations, hockey game scorekeeping, altitude, and thermometer readings. All of these examples use positive Integers, although they also make use of negative Integers. For temperatures below zero, a thermometer utilises negative Integers. In hockey, when the first team scores, it is a plus; however, when the other team scores on their line, it is a minus. Negative numbers are used to indicate elevation below sea level. In banks and credit unions, debits are represented by negative Integers, and credits are represented by positive Integers.

Students are advised by the experts at Extramarks that they practice this topic and its NCERT questions as much as possible with the help of tools such as the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 to understand the real-life use of Integers. The NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 can help not only with the practice part but also with understanding the application part of the topic.

NCERT Solutions for Class 7 

The ease of accessibility of the internet has made it much easier than before for students to refer to different types of study material apart from their course books. The importance of making use of trustworthy sources of information and practice tools is great. The tools like the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 can really make it easy for the students to practice at their own convenience.

The availability of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 on the Extramarks website means that the students can easily practice and feel confident in their examination preparations.

The specialists at Extramarks develop resources like the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 to help students feel less stressed during the exam process. The purpose of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3 is to aid students in becoming more exam-ready. Students who have done their homework are more assured of their preparation. Students can develop their confidence by using the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3.

Q.1

Find each of the following products: (a) 3×(–1) (b) (–1)×225 (c) (–21)×(–30) (d) (–316)×(–1) (e) (–15)×0×(–18) (f) (–12)×(–11)×(10)(g) 9×(–3)×(–6) (h) (–18)×(–5)×(–4)(i) (–1)×(–2)×(–3)×4(j) (–3)×(–6)×(–2)×(–1)

Ans.

( a ) 3 × ( −1 )= −3 ( b ) ( −1 ) × 225= −225 ( c ) ( −21 ) × ( −30 ) = 630 ( d ) ( −316 ) × ( −1 )= 316 ( e ) ( −15 ) × 0 × ( −18 )= 0 ( f ) ( −12 ) × ( −11 ) × ( 10 )= 1320 ( g ) 9 × ( −3 ) × ( − 6 )= 162 ( h ) ( −18 ) × ( −5 ) × ( − 4 )= −360 ( i ) ( −1 ) × ( −2 ) × ( −3 ) × 4= −24 ( j ) ( −3 ) × ( −6 ) × ( −2 ) × ( −1 )= 36 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa caqGGaGaae4maiaabccacqGHxdaTcaqGGaWaaeWaaeaacqGHsislca qGXaaacaGLOaGaayzkaaGaeyypa0ZaauIhaeaacqGHsislcaaIZaaa aaqaamaabmaabaGaaeOyaaGaayjkaiaawMcaaiaabccadaqadaqaai abgkHiTiaabgdaaiaawIcacaGLPaaacaqGGaGaey41aqRaaeiiaiaa bkdacaqGYaGaaeynaiabg2da9iaabccadaqjEaqaaiabgkHiTiaaik dacaaIYaGaaGynaaaaaeaadaqadaqaaiaabogaaiaawIcacaGLPaaa caqGGaWaaeWaaeaacqGHsislcaqGYaGaaeymaaGaayjkaiaawMcaai aabccacqGHxdaTcaqGGaWaaeWaaeaacqGHsislcaqGZaGaaGimaaGa ayjkaiaawMcaaiaacckacqGH9aqpdaqjEaqaaiaaiAdacaaIZaGaaG imaaaacaGGGcaabaWaaeWaaeaacaqGKbaacaGLOaGaayzkaaGaaeii amaabmaabaGaeyOeI0Iaae4maiaabgdacaqG2aaacaGLOaGaayzkaa GaaeiiaiabgEna0kaabccadaqadaqaaiabgkHiTiaabgdaaiaawIca caGLPaaacqGH9aqpdaqjEaqaaiaaiodacaaIXaGaaGOnaaaaaeaada qadaqaaiaabwgaaiaawIcacaGLPaaacaqGGaWaaeWaaeaacqGHsisl caqGXaGaaeynaaGaayjkaiaawMcaaiaabccacqGHxdaTcaqGGaGaaG imaiaabccacqGHxdaTcaqGGaWaaeWaaeaacqGHsislcaqGXaGaaeio aaGaayjkaiaawMcaaiabg2da9maaL4babaGaaGimaaaacaGGGcaaba WaaeWaaeaacaqGMbaacaGLOaGaayzkaaGaaeiiamaabmaabaGaeyOe I0IaaeymaiaabkdaaiaawIcacaGLPaaacaqGGaGaey41aqRaaeiiam aabmaabaGaeyOeI0IaaeymaiaabgdaaiaawIcacaGLPaaacaqGGaGa ey41aqRaaeiiamaabmaabaGaaeymaiaaicdaaiaawIcacaGLPaaacq GH9aqpdaqjEaqaaiaaigdacaaIZaGaaGOmaiaaicdaaaaabaWaaeWa aeaacaqGNbaacaGLOaGaayzkaaGaaeiiaiaabMdacaqGGaGaey41aq RaaeiiamaabmaabaGaeyOeI0Iaae4maaGaayjkaiaawMcaaiaabcca cqGHxdaTcaqGGaWaaeWaaeaacqGHsislcaqGGaGaaeOnaaGaayjkai aawMcaaiabg2da9maaL4babaGaaGymaiaaiAdacaaIYaaaaaqaamaa bmaabaGaaeiAaaGaayjkaiaawMcaaiaabccadaqadaqaaiabgkHiTi aabgdacaqG4aaacaGLOaGaayzkaaGaaeiiaiabgEna0kaabccadaqa daqaaiabgkHiTiaabwdaaiaawIcacaGLPaaacaqGGaGaey41aqRaae iiamaabmaabaGaeyOeI0IaaeiiaiaabsdaaiaawIcacaGLPaaacqGH 9aqpdaqjEaqaaiabgkHiTiaaiodacaaI2aGaaGimaaaaaeaacaGGGc WaaeWaaeaacaqGPbaacaGLOaGaayzkaaGaaeiiamaabmaabaGaeyOe I0IaaeymaaGaayjkaiaawMcaaiaabccacqGHxdaTcaqGGaWaaeWaae aacqGHsislcaqGYaaacaGLOaGaayzkaaGaaeiiaiabgEna0kaabcca daqadaqaaiabgkHiTiaabodaaiaawIcacaGLPaaacaqGGaGaey41aq RaaeiiaiaabsdacqGH9aqpdaqjEaqaaiabgkHiTiaaikdacaaI0aaa aaqaamaabmaabaGaaeOAaaGaayjkaiaawMcaaiaabccadaqadaqaai abgkHiTiaabodaaiaawIcacaGLPaaacaqGGaGaey41aqRaaeiiamaa bmaabaGaeyOeI0IaaeOnaaGaayjkaiaawMcaaiaabccacqGHxdaTca qGGaWaaeWaaeaacqGHsislcaqGYaaacaGLOaGaayzkaaGaaeiiaiab gEna0kaabccadaqadaqaaiabgkHiTiaabgdaaiaawIcacaGLPaaacq GH9aqpdaqjEaqaaiaaiodacaaI2aaaaaaaaa@166C@

Q.2

Verify the following:(a)18×[7+(–3)]=[18×7]+[18×(–3)](b)(–21)×[(–4)+(–6)]=[(–21)×(–4)]+[(–21)×(–6)]

Ans.

( a ) 18 × [ 7 + ( –3 ) ]=18×[ 4 ]=72 and [ 18 × 7 ] + [ 18 × ( –3 ) ]=[ 126 ]+[ −56 ]=70 So, 18 × [ 7 + ( –3 ) ]≠[ 18 × 7 ] + [ 18 × ( –3 ) ] ( b ) ( −21 ) × [ ( −4 ) + ( −6 ) ]=( −21 )×[ −10 ]=210 and [ ( −21 ) × ( −4 ) ] + [ ( −21 ) × ( −6 ) ]=[ 84 ]+[ 126 ]=210 So, ( –21 ) × [ ( – 4 ) + ( – 6 ) ]=[ ( –21 ) × ( – 4 ) ] + [ ( –21 ) × ( – 6 ) ] MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa aeaacaqGXaGaaeioaiaabccacqGHxdaTcaqGGaWaamWaaeaacaqG3a GaaeiiaiabgUcaRiaabccadaqadaqaaiaacobicaqGZaaacaGLOaGa ayzkaaaacaGLBbGaayzxaaGaeyypa0JaaGymaiaaiIdacqGHxdaTda WadaqaaiaaisdaaiaawUfacaGLDbaacqGH9aqpcaaI3aGaaGOmaaqa aiaabggacaqGUbGaaeizaaqaamaadmaabaGaaeymaiaabIdacaqGGa Gaey41aqRaaeiiaiaabEdaaiaawUfacaGLDbaacaqGGaGaey4kaSIa aeiiamaadmaabaGaaeymaiaabIdacaqGGaGaey41aqRaaeiiamaabm aabaGaai4eGiaabodaaiaawIcacaGLPaaaaiaawUfacaGLDbaacqGH 9aqpdaWadaqaaiaaigdacaaIYaGaaGOnaaGaay5waiaaw2faaiabgU caRmaadmaabaGaeyOeI0IaaGynaiaaiAdaaiaawUfacaGLDbaacqGH 9aqpcaaI3aGaaGimaaqaaiaabofacaqGVbGaaiilamaaL4babaGaae ymaiaabIdacaqGGaGaey41aqRaaeiiamaadmaabaGaae4naiaabcca cqGHRaWkcaqGGaWaaeWaaeaacaGGtaIaae4maaGaayjkaiaawMcaaa Gaay5waiaaw2faaiabgcMi5oaadmaabaGaaeymaiaabIdacaqGGaGa ey41aqRaaeiiaiaabEdaaiaawUfacaGLDbaacaqGGaGaey4kaSIaae iiamaadmaabaGaaeymaiaabIdacaqGGaGaey41aqRaaeiiamaabmaa baGaai4eGiaabodaaiaawIcacaGLPaaaaiaawUfacaGLDbaaaaaaba WaaeWaaeaacaqGIbaacaGLOaGaayzkaaaabaWaaeWaaeaacqGHsisl caqGYaGaaeymaaGaayjkaiaawMcaaiaabccacqGHxdaTcaqGGaWaam WaaeaadaqadaqaaiabgkHiTiaabsdaaiaawIcacaGLPaaacaqGGaGa ey4kaSIaaeiiamaabmaabaGaeyOeI0IaaeOnaaGaayjkaiaawMcaaa Gaay5waiaaw2faaiabg2da9maabmaabaGaeyOeI0IaaGOmaiaaigda aiaawIcacaGLPaaacqGHxdaTdaWadaqaaiabgkHiTiaaigdacaaIWa aacaGLBbGaayzxaaGaeyypa0JaaGOmaiaaigdacaaIWaaabaGaaeyy aiaab6gacaqGKbaabaWaamWaaeaadaqadaqaaiabgkHiTiaabkdaca qGXaaacaGLOaGaayzkaaGaaeiiaiabgEna0kaabccadaqadaqaaiab gkHiTiaabsdaaiaawIcacaGLPaaaaiaawUfacaGLDbaacaqGGaGaey 4kaSIaaeiiamaadmaabaWaaeWaaeaacqGHsislcaqGYaGaaeymaaGa ayjkaiaawMcaaiaabccacqGHxdaTcaqGGaWaaeWaaeaacqGHsislca qG2aaacaGLOaGaayzkaaaacaGLBbGaayzxaaGaeyypa0ZaamWaaeaa caaI4aGaaGinaaGaay5waiaaw2faaiabgUcaRmaadmaabaGaaGymai aaikdacaaI2aaacaGLBbGaayzxaaGaeyypa0JaaGOmaiaaigdacaaI WaaabaGaae4uaiaab+gacaGGSaaabaWaauIhaeaadaqadaqaaiaaco bicaqGYaGaaeymaaGaayjkaiaawMcaaiaabccacqGHxdaTcaqGGaWa amWaaeaadaqadaqaaiaacobicaqGGaGaaeinaaGaayjkaiaawMcaai aabccacqGHRaWkcaqGGaWaaeWaaeaacaGGtaIaaeiiaiaabAdaaiaa wIcacaGLPaaaaiaawUfacaGLDbaacqGH9aqpdaWadaqaamaabmaaba Gaai4eGiaabkdacaqGXaaacaGLOaGaayzkaaGaaeiiaiabgEna0kaa bccadaqadaqaaiaacobicaqGGaGaaeinaaGaayjkaiaawMcaaaGaay 5waiaaw2faaiaabccacqGHRaWkcaqGGaWaamWaaeaadaqadaqaaiaa cobicaqGYaGaaeymaaGaayjkaiaawMcaaiaabccacqGHxdaTcaqGGa WaaeWaaeaacaGGtaIaaeiiaiaabAdaaiaawIcacaGLPaaaaiaawUfa caGLDbaaaaaaaaa@1C03@

Q.3

(i) For any integera, what is (–1)×a equal to ?(ii) Determine the integer whose product with (–1) is(a)–22 (b) 37 (c) 0

Ans.

( i ) ( −1 )×a=−a (ii) (a) 22 ×( −1 )=−22 (b) −37 ×( −1 )=37 (c) 0 ×( −1 )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caaMe8+aaeWaaeaacqGHsislcaqGXaaacaGLOaGaayzkaaGaey41aq Raaeyyaiaab2dacqGHsislcaqGHbaabaGaaeikaiaabMgacaqGPbGa aeykaiaabccacaqGOaGaaeyyaiaabMcacaqGGaWaauIhaeaacaaIYa GaaGOmaaaacqGHxdaTdaqadaqaaiabgkHiTiaaigdaaiaawIcacaGL PaaacqGH9aqpcqGHsislcaaIYaGaaGOmaaqaaiaabIcacaqGIbGaae ykaiaabccadaqjEaqaaiabgkHiTiaabodacaqG3aaaaiabgEna0oaa bmaabaGaeyOeI0IaaGymaaGaayjkaiaawMcaaiabg2da9iaaiodaca aI3aaabaGaaeikaiaabogacaqGPaGaaeiiamaaL4babaGaaeimaaaa cqGHxdaTdaqadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH9a qpcaaIWaaaaaa@7353@

Q.4

Starting from (–1)×5, write various products showing somepattern to show (–1)×(–1)=1

Ans.

−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 Thus, −1×( −1 )=0+1=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacqGHsislcaaIXaGaey41aqRaaGynaiab g2da9iabgkHiTiaaiwdaaeaacqGHsislcaaIXaGaey41aqRaaGinai abg2da9iabgkHiTiaaisdacqGH9aqpcqGHsislcaaI1aGaey4kaSIa aGymaaqaaiabgkHiTiaaigdacqGHxdaTcaaIZaGaeyypa0JaeyOeI0 IaaG4maiabg2da9iabgkHiTiaaisdacqGHRaWkcaaIXaaabaGaeyOe I0IaaGymaiabgEna0kaaikdacqGH9aqpcqGHsislcaaIYaGaeyypa0 JaeyOeI0IaaG4maiabgUcaRiaaigdaaeaacqGHsislcaaIXaGaey41 aqRaaGymaiabg2da9iabgkHiTiaaigdacqGH9aqpcqGHsislcaaIYa Gaey4kaSIaaGymaaqaaiabgkHiTiaaigdacqGHxdaTcaaIWaGaeyyp a0JaaGimaiabg2da9iabgkHiTiaaigdacqGHRaWkcaaIXaaabaGaae ivaiaabIgacaqG1bGaae4CaiaabYcadaqjEaqaaiabgkHiTiaabgda cqGHxdaTdaqadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH9a qpcaaIWaGaey4kaSIaaGymaiabg2da9iaaigdaaaaaaaa@8D8E@

Q.5

Find the product, using suitable properties: ( a ) 26 × ( – 48 ) + ( – 48 ) × ( –36 ) ( b ) 8 × 53 × ( –125 ) ( c ) 15 × ( –25 ) × ( – 4 ) × ( –10 ) ( d ) ( – 41 ) × 102 ( e ) 625 × ( –35 ) + ( – 625 ) × 65 ( f ) 7 × ( 50 – 2 ) ( g ) ( –17 ) × ( –29 ) ( h ) ( –57 ) × ( –19 ) + 57 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGa a8hiaiaa=rhacaWFObGaa8xzaiaa=bcacaWFWbGaa8NCaiaa=9gaca WFKbGaa8xDaiaa=ngacaWF0bGaa8hlaiaa=bcacaWF1bGaa83Caiaa =LgacaWFUbGaa83zaiaa=bcacaWFZbGaa8xDaiaa=LgacaWF0bGaa8 xyaiaa=jgacaWFSbGaa8xzaiaa=bcacaWFWbGaa8NCaiaa=9gacaWF WbGaa8xzaiaa=jhacaWF0bGaa8xAaiaa=vgacaWFZbGaa8Noaaqaai aa=bkadaqadaqaaiaa=fgaaiaawIcacaGLPaaacaWFGaGaa8Nmaiaa =zdacaWFGaGaa831aiaa=bcadaqadaqaaiaa=nbicaWFGaGaa8hnai aa=HdaaiaawIcacaGLPaaacaWFGaGaa83kaiaa=bcadaqadaqaaiaa =nbicaWFGaGaa8hnaiaa=HdaaiaawIcacaGLPaaacaWFGaGaa831ai aa=bcadaqadaqaaiaa=nbicaWFZaGaa8NnaaGaayjkaiaawMcaaiaa =bcadaqadaqaaiaa=jgaaiaawIcacaGLPaaacaWFGaGaa8hoaiaa=b cacaWFxdGaa8hiaiaa=vdacaWFZaGaa8hiaiaa=DnacaWFGaWaaeWa aeaacaWFtaIaa8xmaiaa=jdacaWF1aaacaGLOaGaayzkaaaabaGaa8 hOamaabmaabaGaa83yaaGaayjkaiaawMcaaiaa=bcacaWFXaGaa8xn aiaa=bcacaWFxdGaa8hiamaabmaabaGaa83eGiaa=jdacaWF1aaaca GLOaGaayzkaaGaa8hiaiaa=DnacaWFGaWaaeWaaeaacaWFtaIaa8hi aiaa=rdaaiaawIcacaGLPaaacaWFGaGaa831aiaa=bcadaqadaqaai aa=nbicaWFXaGaa8hmaaGaayjkaiaawMcaaiaa=bcacaWLjaWaaeWa aeaacaWFKbaacaGLOaGaayzkaaGaa8hiamaabmaabaGaa83eGiaa=b cacaWF0aGaa8xmaaGaayjkaiaawMcaaiaa=bcacaWFxdGaa8hiaiaa =fdacaWFWaGaa8Nmaaqaaiaa=bkadaqadaqaaiaa=vgaaiaawIcaca GLPaaacaWFGaGaa8Nnaiaa=jdacaWF1aGaa8hiaiaa=DnacaWFGaWa aeWaaeaacaWFtaIaa83maiaa=vdaaiaawIcacaGLPaaacaWFGaGaa8 3kaiaa=bcadaqadaqaaiaa=nbicaWFGaGaa8Nnaiaa=jdacaWF1aaa caGLOaGaayzkaaGaa8hiaiaa=DnacaWFGaGaa8Nnaiaa=vdacaWFGa WaaeWaaeaacaWFMbaacaGLOaGaayzkaaGaa8hiaiaa=DdacaWFGaGa a831aiaa=bcadaqadaqaaiaa=vdacaWFWaGaa8hiaiaa=nbicaWFGa Gaa8NmaaGaayjkaiaawMcaaaqaaiaa=bkadaqadaqaaiaa=Dgaaiaa wIcacaGLPaaacaWFGaWaaeWaaeaacaWFtaIaa8xmaiaa=DdaaiaawI cacaGLPaaacaWFGaGaa831aiaa=bcadaqadaqaaiaa=nbicaWFYaGa a8xoaaGaayjkaiaawMcaaiaa=bcacaWLjaGaaCzcaiaaxMaacaWLja WaaeWaaeaacaWFObaacaGLOaGaayzkaaGaa8hiamaabmaabaGaa83e Giaa=vdacaWF3aaacaGLOaGaayzkaaGaa8hiaiaa=DnacaWFGaWaae WaaeaacaWFtaIaa8xmaiaa=LdaaiaawIcacaGLPaaacaWFGaGaa83k aiaa=bcacaWF1aGaa83naaaaaa@F7B5@

Ans.

( a ) 26 × ( − 48 ) + ( − 48 ) × ( −36 ) =( −48 )×26+( −48 )×( −36 ) =( −48 )[ 26−36 ] =( −48 )[ −10 ] = 480 ( b ) 8 × 53 × ( −125 ) =424×( −125 ) = 53000 ( c ) 15 × ( −25 ) × ( − 4 ) × ( −10 ) =( 15×−25 )×( −4×−10 ) =( −375 )×40 = −15000 ( d ) ( − 41 ) × 102 = −4182 ( e ) 625 × ( −35 ) + ( − 625 ) × 65 =−21875+( −40625 ) = −62500 ( f ) 7 × ( 50 − 2 ) =7×( 48 ) = 336 ( g ) ( −17 ) × ( −29 ) = 493 ( h ) ( −57 ) × ( −19 ) + 57 =1083+57 = 1140 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa caqGGaGaaeOmaiaabAdacaqGGaGaey41aqRaaeiiamaabmaabaGaey OeI0IaaeiiaiaabsdacaqG4aaacaGLOaGaayzkaaGaaeiiaiabgUca RiaabccadaqadaqaaiabgkHiTiaabccacaqG0aGaaeioaaGaayjkai aawMcaaiaabccacqGHxdaTcaqGGaWaaeWaaeaacqGHsislcaqGZaGa aeOnaaGaayjkaiaawMcaaaqaaiabg2da9maabmaabaGaeyOeI0IaaG inaiaaiIdaaiaawIcacaGLPaaacqGHxdaTcaaIYaGaaGOnaiabgUca RmaabmaabaGaeyOeI0IaaGinaiaaiIdaaiaawIcacaGLPaaacqGHxd aTdaqadaqaaiabgkHiTiaaiodacaaI2aaacaGLOaGaayzkaaaabaGa eyypa0ZaaeWaaeaacqGHsislcaaI0aGaaGioaaGaayjkaiaawMcaam aadmaabaGaaGOmaiaaiAdacqGHsislcaaIZaGaaGOnaaGaay5waiaa w2faaaqaaiabg2da9maabmaabaGaeyOeI0IaaGinaiaaiIdaaiaawI cacaGLPaaadaWadaqaaiabgkHiTiaaigdacaaIWaaacaGLBbGaayzx aaaabaGaeyypa0ZaauIhaeaacaaI0aGaaGioaiaaicdaaaaabaWaae WaaeaacaqGIbaacaGLOaGaayzkaaGaaeiiaiaabIdacaqGGaGaey41 aqRaaeiiaiaabwdacaqGZaGaaeiiaiabgEna0kaabccadaqadaqaai abgkHiTiaabgdacaqGYaGaaeynaaGaayjkaiaawMcaaaqaaiabg2da 9iaaisdacaaIYaGaaGinaiabgEna0oaabmaabaGaeyOeI0IaaGymai aaikdacaaI1aaacaGLOaGaayzkaaaabaGaeyypa0ZaauIhaeaacaaI 1aGaaG4maiaaicdacaaIWaGaaGimaaaaaeaacaGGGcWaaeWaaeaaca qGJbaacaGLOaGaayzkaaGaaeiiaiaabgdacaqG1aGaaeiiaiabgEna 0kaabccadaqadaqaaiabgkHiTiaabkdacaqG1aaacaGLOaGaayzkaa GaaeiiaiabgEna0kaabccadaqadaqaaiabgkHiTiaabccacaqG0aaa caGLOaGaayzkaaGaaeiiaiabgEna0kaabccadaqadaqaaiabgkHiTi aabgdacaaIWaaacaGLOaGaayzkaaaabaGaeyypa0ZaaeWaaeaacaaI XaGaaGynaiabgEna0kabgkHiTiaaikdacaaI1aaacaGLOaGaayzkaa Gaey41aq7aaeWaaeaacqGHsislcaaI0aGaey41aqRaeyOeI0IaaGym aiaaicdaaiaawIcacaGLPaaaaeaacqGH9aqpdaqadaqaaiabgkHiTi aaiodacaaI3aGaaGynaaGaayjkaiaawMcaaiabgEna0kaaisdacaaI WaaabaGaeyypa0ZaauIhaeaacqGHsislcaaIXaGaaGynaiaaicdaca aIWaGaaGimaaaaaeaadaqadaqaaiaabsgaaiaawIcacaGLPaaacaqG GaWaaeWaaeaacqGHsislcaqGGaGaaeinaiaabgdaaiaawIcacaGLPa aacaqGGaGaey41aqRaaeiiaiaabgdacaaIWaGaaeOmaaqaaiabg2da 9maaL4babaGaeyOeI0IaaeinaiaabgdacaqG4aGaaeOmaaaaaeaada qadaqaaiaabwgaaiaawIcacaGLPaaacaqGGaGaaeOnaiaabkdacaqG 1aGaaeiiaiabgEna0kaabccadaqadaqaaiabgkHiTiaabodacaqG1a aacaGLOaGaayzkaaGaaeiiaiabgUcaRiaabccadaqadaqaaiabgkHi TiaabccacaqG2aGaaeOmaiaabwdaaiaawIcacaGLPaaacaqGGaGaey 41aqRaaeiiaiaabAdacaqG1aaabaGaeyypa0JaeyOeI0IaaGOmaiaa igdacaaI4aGaaG4naiaaiwdacqGHRaWkdaqadaqaaiabgkHiTiaais dacaaIWaGaaGOnaiaaikdacaaI1aaacaGLOaGaayzkaaaabaGaeyyp a0ZaauIhaeaacqGHsislcaaI2aGaaGOmaiaaiwdacaaIWaGaaGimaa aaaeaaaeaadaqadaqaaiaabAgaaiaawIcacaGLPaaacaqGGaGaae4n aiaabccacqGHxdaTcaqGGaWaaeWaaeaacaqG1aGaaGimaiaabccacq GHsislcaqGGaGaaeOmaaGaayjkaiaawMcaaaqaaiabg2da9iaaiEda cqGHxdaTdaqadaqaaiaaisdacaaI4aaacaGLOaGaayzkaaaabaGaey ypa0ZaauIhaeaacaaIZaGaaG4maiaaiAdaaaaabaGaaiiOamaabmaa baGaae4zaaGaayjkaiaawMcaaiaabccadaqadaqaaiabgkHiTiaabg dacaqG3aaacaGLOaGaayzkaaGaaeiiaiabgEna0kaabccadaqadaqa aiabgkHiTiaabkdacaqG5aaacaGLOaGaayzkaaaabaGaeyypa0Zaau IhaeaacaaI0aGaaGyoaiaaiodaaaaabaWaaeWaaeaacaqGObaacaGL OaGaayzkaaGaaeiiamaabmaabaGaeyOeI0IaaeynaiaabEdaaiaawI cacaGLPaaacaqGGaGaey41aqRaaeiiamaabmaabaGaeyOeI0Iaaeym aiaabMdaaiaawIcacaGLPaaacaqGGaGaey4kaSIaaeiiaiaabwdaca qG3aaabaGaeyypa0JaaGymaiaaicdacaaI4aGaaG4maiabgUcaRiaa iwdacaaI3aaabaGaeyypa0ZaauIhaeaacaaIXaGaaGymaiaaisdaca aIWaaaaaaaaa@6377@

Q.6

i For any integer a, what is –1 × a equal to?ii Determine the integer whose product with –1 isa –22 b 37 c 0

Ans.

( i ) ( −1 )×a=−a (ii) (a) 22 ×( −1 )=−22 (b) −37 ×( −1 )=37 (c) 0 ×( −1 )=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caaMe8+aaeWaaeaacqGHsislcaqGXaaacaGLOaGaayzkaaGaey41aq Raaeyyaiaab2dacqGHsislcaqGHbaabaGaaeikaiaabMgacaqGPbGa aeykaiaabccacaqGOaGaaeyyaiaabMcacaqGGaWaauIhaeaacaaIYa GaaGOmaaaacqGHxdaTdaqadaqaaiabgkHiTiaaigdaaiaawIcacaGL PaaacqGH9aqpcqGHsislcaaIYaGaaGOmaaqaaiaabIcacaqGIbGaae ykaiaabccadaqjEaqaaiabgkHiTiaabodacaqG3aaaaiabgEna0oaa bmaabaGaeyOeI0IaaGymaaGaayjkaiaawMcaaiabg2da9iaaiodaca aI3aaabaGaaeikaiaabogacaqGPaGaaeiiamaaL4babaGaaeimaaaa cqGHxdaTdaqadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH9a qpcaaIWaaaaaa@7353@

Q.7

Starting from -1×5, write various products showing some pattern to show -1×-1=1

Ans.

−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 Thus , −1×( −1 )=0+1=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacqGHsislcaaIXaGaey41aqRaaGynaiab g2da9iabgkHiTiaaiwdaaeaacqGHsislcaaIXaGaey41aqRaaGinai abg2da9iabgkHiTiaaisdacqGH9aqpcqGHsislcaaI1aGaey4kaSIa aGymaaqaaiabgkHiTiaaigdacqGHxdaTcaaIZaGaeyypa0JaeyOeI0 IaaG4maiabg2da9iabgkHiTiaaisdacqGHRaWkcaaIXaaabaGaeyOe I0IaaGymaiabgEna0kaaikdacqGH9aqpcqGHsislcaaIYaGaeyypa0 JaeyOeI0IaaG4maiabgUcaRiaaigdaaeaacqGHsislcaaIXaGaey41 aqRaaGymaiabg2da9iabgkHiTiaaigdacqGH9aqpcqGHsislcaaIYa Gaey4kaSIaaGymaaqaaiabgkHiTiaaigdacqGHxdaTcaaIWaGaeyyp a0JaaGimaiabg2da9iabgkHiTiaaigdacqGHRaWkcaaIXaaabaGaae ivaiaabIgacaqG1bGaae4CaiaabYcadaqjEaqaaiabgkHiTiaabgda cqGHxdaTdaqadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH9a qpcaaIWaGaey4kaSIaaGymaiabg2da9iaaigdaaaaaaaa@8D8E@

Q.8

Find the product, using suitable properties: a 26 × – 48 + – 48 × –36 b 8 × 53 × –125 c 15 × –25 × – 4 × –10 d – 41 × 102 e 625 × –35 + – 625 × 65 f 7 × 50 – 2 g –17 × –29 h –57 × –19 + 57

Ans.

( a ) 26 × ( − 48 ) + ( − 48 ) × ( −36 ) =( −48 )×26+( −48 )×( −36 ) =( −48 )[ 26−36 ] =( −48 )[ −10 ] = 480 ( b ) 8 × 53 × ( −125 ) =424×( −125 ) = 53000 ( c ) 15 × ( −25 ) × ( − 4 ) × ( −10 ) =( 15×−25 )×( −4×−10 ) =( −375 )×40 = −15000 ( d ) ( − 41 ) × 102 = −4182 ( e ) 625 × ( −35 ) + ( − 625 ) × 65 =−21875+( −40625 ) = −62500 ( f ) 7 × ( 50 − 2 ) =7×( 48 ) = 336 ( g ) ( −17 ) × ( −29 ) = 493 ( h ) ( −57 ) × ( −19 ) + 57 =1083+57 = 1140 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa caqGGaGaaeOmaiaabAdacaqGGaGaey41aqRaaeiiamaabmaabaGaey OeI0IaaeiiaiaabsdacaqG4aaacaGLOaGaayzkaaGaaeiiaiabgUca RiaabccadaqadaqaaiabgkHiTiaabccacaqG0aGaaeioaaGaayjkai aawMcaaiaabccacqGHxdaTcaqGGaWaaeWaaeaacqGHsislcaqGZaGa aeOnaaGaayjkaiaawMcaaaqaaiabg2da9maabmaabaGaeyOeI0IaaG inaiaaiIdaaiaawIcacaGLPaaacqGHxdaTcaaIYaGaaGOnaiabgUca RmaabmaabaGaeyOeI0IaaGinaiaaiIdaaiaawIcacaGLPaaacqGHxd aTdaqadaqaaiabgkHiTiaaiodacaaI2aaacaGLOaGaayzkaaaabaGa eyypa0ZaaeWaaeaacqGHsislcaaI0aGaaGioaaGaayjkaiaawMcaam aadmaabaGaaGOmaiaaiAdacqGHsislcaaIZaGaaGOnaaGaay5waiaa w2faaaqaaiabg2da9maabmaabaGaeyOeI0IaaGinaiaaiIdaaiaawI cacaGLPaaadaWadaqaaiabgkHiTiaaigdacaaIWaaacaGLBbGaayzx aaaabaGaeyypa0ZaauIhaeaacaaI0aGaaGioaiaaicdaaaaabaWaae WaaeaacaqGIbaacaGLOaGaayzkaaGaaeiiaiaabIdacaqGGaGaey41 aqRaaeiiaiaabwdacaqGZaGaaeiiaiabgEna0kaabccadaqadaqaai abgkHiTiaabgdacaqGYaGaaeynaaGaayjkaiaawMcaaaqaaiabg2da 9iaaisdacaaIYaGaaGinaiabgEna0oaabmaabaGaeyOeI0IaaGymai aaikdacaaI1aaacaGLOaGaayzkaaaabaGaeyypa0ZaauIhaeaacaaI 1aGaaG4maiaaicdacaaIWaGaaGimaaaaaeaacaGGGcWaaeWaaeaaca qGJbaacaGLOaGaayzkaaGaaeiiaiaabgdacaqG1aGaaeiiaiabgEna 0kaabccadaqadaqaaiabgkHiTiaabkdacaqG1aaacaGLOaGaayzkaa GaaeiiaiabgEna0kaabccadaqadaqaaiabgkHiTiaabccacaqG0aaa caGLOaGaayzkaaGaaeiiaiabgEna0kaabccadaqadaqaaiabgkHiTi aabgdacaaIWaaacaGLOaGaayzkaaaabaGaeyypa0ZaaeWaaeaacaaI XaGaaGynaiabgEna0kabgkHiTiaaikdacaaI1aaacaGLOaGaayzkaa Gaey41aq7aaeWaaeaacqGHsislcaaI0aGaey41aqRaeyOeI0IaaGym aiaaicdaaiaawIcacaGLPaaaaeaacqGH9aqpdaqadaqaaiabgkHiTi aaiodacaaI3aGaaGynaaGaayjkaiaawMcaaiabgEna0kaaisdacaaI WaaabaGaeyypa0ZaauIhaeaacqGHsislcaaIXaGaaGynaiaaicdaca aIWaGaaGimaaaaaeaadaqadaqaaiaabsgaaiaawIcacaGLPaaacaqG GaWaaeWaaeaacqGHsislcaqGGaGaaeinaiaabgdaaiaawIcacaGLPa aacaqGGaGaey41aqRaaeiiaiaabgdacaaIWaGaaeOmaaqaaiabg2da 9maaL4babaGaeyOeI0IaaeinaiaabgdacaqG4aGaaeOmaaaaaeaada qadaqaaiaabwgaaiaawIcacaGLPaaacaqGGaGaaeOnaiaabkdacaqG 1aGaaeiiaiabgEna0kaabccadaqadaqaaiabgkHiTiaabodacaqG1a aacaGLOaGaayzkaaGaaeiiaiabgUcaRiaabccadaqadaqaaiabgkHi TiaabccacaqG2aGaaeOmaiaabwdaaiaawIcacaGLPaaacaqGGaGaey 41aqRaaeiiaiaabAdacaqG1aaabaGaeyypa0JaeyOeI0IaaGOmaiaa igdacaaI4aGaaG4naiaaiwdacqGHRaWkdaqadaqaaiabgkHiTiaais dacaaIWaGaaGOnaiaaikdacaaI1aaacaGLOaGaayzkaaaabaGaeyyp a0ZaauIhaeaacqGHsislcaaI2aGaaGOmaiaaiwdacaaIWaGaaGimaa aaaeaadaqadaqaaiaabAgaaiaawIcacaGLPaaacaqGGaGaae4naiaa bccacqGHxdaTcaqGGaWaaeWaaeaacaqG1aGaaGimaiaabccacqGHsi slcaqGGaGaaeOmaaGaayjkaiaawMcaaaqaaiabg2da9iaaiEdacqGH xdaTdaqadaqaaiaaisdacaaI4aaacaGLOaGaayzkaaaabaGaeyypa0 ZaauIhaeaacaaIZaGaaG4maiaaiAdaaaaabaGaaiiOamaabmaabaGa ae4zaaGaayjkaiaawMcaaiaabccadaqadaqaaiabgkHiTiaabgdaca qG3aaacaGLOaGaayzkaaGaaeiiaiabgEna0kaabccadaqadaqaaiab gkHiTiaabkdacaqG5aaacaGLOaGaayzkaaaabaGaeyypa0ZaauIhae aacaaI0aGaaGyoaiaaiodaaaaabaWaaeWaaeaacaqGObaacaGLOaGa ayzkaaGaaeiiamaabmaabaGaeyOeI0IaaeynaiaabEdaaiaawIcaca GLPaaacaqGGaGaey41aqRaaeiiamaabmaabaGaeyOeI0Iaaeymaiaa bMdaaiaawIcacaGLPaaacaqGGaGaey4kaSIaaeiiaiaabwdacaqG3a aabaGaeyypa0JaaGymaiaaicdacaaI4aGaaG4maiabgUcaRiaaiwda caaI3aaabaGaeyypa0ZaauIhaeaacaaIXaGaaGymaiaaisdacaaIWa aaaaaaaa@6376@

Q.9

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

Ans.

Given initial temprature=40°C Change in temprature per hour =−5°C Change in temprature after 10 hours=−5°C×10=−50°C Final temprature=40°C−50°C= −10°C MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGhbGaaeyAaiaabAhacaqGLbGaaeOB aiaabccacaqGPbGaaeOBaiaabMgacaqG0bGaaeyAaiaabggacaqGSb GaaeiiaiaabshacaqGLbGaaeyBaiaabchacaqGYbGaaeyyaiaabsha caqG1bGaaeOCaiaabwgacqGH9aqpcaaI0aGaaGimaiabgclaWkaado eaaeaacaqGdbGaaeiAaiaabggacaqGUbGaae4zaiaabwgacaqGGaGa aeyAaiaab6gacaqGGaGaaeiDaiaabwgacaqGTbGaaeiCaiaabkhaca qGHbGaaeiDaiaabwhacaqGYbGaaeyzaiaabccacaqGWbGaaeyzaiaa bkhacaqGGaGaaeiAaiaab+gacaqG1bGaaeOCaiaabccacqGH9aqpcq GHsislcaaI1aGaeyiSaaRaam4qaaqaaiaaboeacaqGObGaaeyyaiaa b6gacaqGNbGaaeyzaiaabccacaqGPbGaaeOBaiaabccacaqG0bGaae yzaiaab2gacaqGWbGaaeOCaiaabggacaqG0bGaaeyDaiaabkhacaqG LbGaaeiiaiaabggacaqGMbGaaeiDaiaabwgacaqGYbGaaeiiaiaabg dacaqGWaGaaeiiaiaabIgacaqGVbGaaeyDaiaabkhacaqGZbGaeyyp a0JaeyOeI0IaaGynaiabgclaWkaadoeacqGHxdaTcaaIXaGaaGimai abg2da9iabgkHiTiaaiwdacaaIWaGaeyiSaaRaam4qaaqaaiaabAea caqGPbGaaeOBaiaabggacaqGSbGaaeiiaiaabshacaqGLbGaaeyBai aabchacaqGYbGaaeyyaiaabshacaqG1bGaaeOCaiaabwgacqGH9aqp caaI0aGaaGimaiabgclaWkaadoeacqGHsislcaaI1aGaaGimaiabgc laWkaadoeacqGH9aqpdaqjEaqaaiabgkHiTiaaigdacaaIWaGaeyiS aaRaam4qaaaaaaaa@C4D0@

Please register to view this section

FAQs (Frequently Asked Questions)

1. Are Integers a part of Algebra?

Algebra has Integers and whole numbers etc but the Integers are not exclusively of use in Algebra Mathematics. With the use of the NCERT Solutions For Class 7 Maths Chapter 1 Exercise 1.3, students can understand this better and practice better too.

2. Is it possible to download Extramarks’ study materials in PDF format?

Yes, students can download the study materials available on the Extramarks’ website in PDF format. Students can download the study material in PDF for offline access.