MathsWiz–7

A course in Mathematics

S.K. GuptaPrincipal (Retd.)

Birla Vidya Mandir, NainitalFormer Chairman

Indian Public Schools’ Conference

Anubhuti GangalM.A. (Gold Medalist), M.Ed.

Formerly, Senior Faculty MemberThe Daly College, Indore

Birla Vidya Mandir, Nainital

Book

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MathsWiz–7

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© 2016, S.K. Gupta & Anubhuti Gangal

All rights reserved. No part of this publication may be reproduced or copied in any material form (including photocopying or storing it in any medium in form of graphics, electronic or mechanical means and whether or not transient or incidental to some other use of this publication) without written permission of the publisher. Any breach of this will entail legal action and prosecution without further notice.Jurisdiction : All disputes with respect to this publication shall be subject to the jurisdiction of the Courts, Tribunals and Forums of New Delhi, India only.

First Published in 2016First Impression 2016

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MathsWiz–7

Disclaimer : While the authors of this book have made every effort to avoid any mistakes or omissions and have used their skill, expertise and knowledge to the best of their capacity to provide accurate and updated information, the authors and S. Chand do not give any representation or warranty with respect to the accuracy or completeness of the contents of this publication and are selling this publication on the condition and understanding that they shall not be made liable in any manner whatsoever. S.Chand and the authors expressly disclaim all and any liability/responsibility to any person, whether a purchaser or reader of this publication or not, in respect of anything and everything forming part of the contents of this publication. S. Chand shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in this publication.Further, the appearance of the personal name, location, place and incidence, if any; in the illustrations used herein is purely coincidental and work of imagination. Thus the same should in no manner be termed as defamatory to any individual.

PrefacePrefaceMathsWiz, a series of nine textbooks for KG to 8 is a course based on the National Curriculum Framework and the guidelines provided therein.

The content of the series is student-centred and activity-based with emphasis on developing problem-solving skills, encouraging the child to think creatively and work independently. The methodology facilitates the teacher, the student and the parent to ensure full involvement of the child in the classroom and at home.

All the mathematical concepts are presented in a very simple and easy-to-understand form. The concepts are explained by taking examples from our daily life situations. The examples and problems also make use of modern tools, gadgets and technology commonly used. An abundant use of visual tools such as diagrams, illustrations, cartoons, tables and charts makes learning fun and helps in greater retention. The approach helps create passion for mathematics in children rather than fear for the subject. It encourages them to enquire, explore and discover rather than only learn by rote.

Each book of the series is accompanied by an interactive student CD to help concept-building by showing its application in daily life. The CD also supplements the book content through visuals, interactive practice and additional information.

Teacher’s Manuals with extensive teaching ideas and solutions are also available separately.

Web Support includes interactive tests, worksheets, term-wise updated papers, unit-wise test papers to support classroom teaching.

The salient features of the series are Test Your Understanding exercises for immediate practice have been given almost after every topic. Exercises have been designed to include all types of questions, especially the Multiple Choice Questions (MCQs)

and Higher Order Thinking Skills (HOTS). Chapter Assessment provided at the end of each chapter to assess the child’s understanding of the concepts given

therein. Maths Lab Activities as per CBSE guidelines have been incorporated. Formative Assessment is a part of the on-going scholastic evaluation of the child. It is a kind of Unit Test – A

Self-assessment of the concepts taught in the class. Mental Maths exercises have been given to develop skills in rapid calculations. Maths Alert to warn against likely mistakes and misconceptions. Summative Assessment at the end of each term has been skillfully prepared incorporating all types of questions

such as:

(i) Very Short Answer Type Questions (Concept Review, True-False, Matching Questions) (ii) Short Answer Type Questions (iii) Multiple Choice Questions (MCQs) (iv) Higher Order Thinking Skills (HOTS)

We are thankful to the management and the editorial team of S. Chand And Company Pvt. Ltd., New Delhi, for their help and support in the publication of the books of this series.

Suggestions and feedback for the improvement of this newly introduced series from the principals, teachers, students and parents would be most welcome. You may write in at anubhutigangal@hotmail.com

Authors

MathsWiz–7

UNIT I: NUMBER SYSTEM (Periods 50)

(i) Knowing Our Numbers: Integers  • Multiplicationanddivisionofintegers(throughpatterns).Divisionbyzeroismeaningless. • Propertiesofintegers(includingidentitiesforadditionandmultiplication, commutative, associative, distributive)

(throughpatterns).Thesewouldincludeexamplesfromwholenumbersaswell.Involveexpressingcommutativeandassociativepropertiesinageneralform.Constructionofcounter-examples,includingsomebychildren.Counterexamplelikesubtractionisnotcommutative.

• Wordproblemsincludingintegers(alloperations)(ii) Fractions and Rational Numbers:

• Multiplicationoffractions • Fractionasanoperator • Reciprocalofafraction • Divisionoffractions • Wordproblemsinvolvingmixedfractions • Introductiontorationalnumbers(withrepresentationonnumberline) • Operationsonrationalnumbers(alloperations) • Representationofrationalnumberasadecimal • Wordproblemsonrationalnumbers(alloperations) • Multiplicationanddivisionofdecimalfractions • Conversionofunits(lengthsandmass) • Wordproblems(includingalloperations)(iii) Powers: • Exponentsonlynaturalnumbers. • Lawsofexponents(throughobservingpatternstoarriveatgeneralization.)

(i) +=.m n m na a a (ii) =( )m n mna a (iii) −= ,m

m nn

aa

a wherem – n ∈ N (iv) =. ( )m m ma b ab

UNIT II: ALGEBRA (Periods 20)Algebraic Expressions: • Generatealgebraicexpression(simple)involvingoneortwovariables • Identifyingconstants,coefficient,powers • Likeandunliketerms,degreeofexpression,e.g.,x2yetc.(exponent≤ 3numberofvariables≤2) • Addition,subtractionofalgebraicexpressions(coefficientsshouldbeintegers). • Simple linearequations inonevariable (in contexualproblems)with twooperations (avoid complicated

coefficients).

UNIT III: COMMERCIAL ARITHMETIC (Periods 20)

• Ratioandproportion(revision) • Unitarymethod,continuedconsolidation,generalexpression • Percentage—anintroduction • Understandingpercentageasafractionwithdenominator100. • Convertingfractionsanddecimalsintopercentageandviceversa. • Applicationtoprofitandloss(singletransactiononly) • Applicationtosimpleinterest(timeperiodincompleteyears)

SyllabusSyllabus

MathsWiz–7

UNIT IV: GEOMETRY (Periods 60)

(i) Understanding Shapes: • Pairsofangles(linear,supplementary,complementary,adjacentverticallyopposite)(verificationandsimple

proofofverticallyoppositeangles) • Propertiesofparallellineswithtransversal(alternate,corresponding,interior,exteriorangles). (ii) Properties of Triangles: • Anglessumproperty(withnotionsofproofandverificationthroughpaperfolding,proofsusingpropertyof

parallellines,differencebetweenproofandverification.) • Exteriorangleproperty • Sumoftwosidesofa∆>it’sthirdside • Pythagoras’Theorem(Verificationonly)(iii) Symmetry: • Recallingreflectionsymmetry • Ideaofrotationalsymmetry,observationsofrotationalsymmetryof2Dobjects,(90°,120°,180°) • Operationofrotationthrough90°and180°ofsimplefigures • Examplesoffigureswithbothrotationandreflectionsymmetry(bothoperations) • Examplesoffiguresthathavereflectionandrotationsymmetryandvice-versa(iv) Representing 3D in 2D: • Drawing3Dfiguresin2Dshowinghiddenfaces • Identificationandcountingofveritiesedges,faces,nets(forcubescuboids,andcylinders,cones) • Matchingpictureswithobjects(identifyingnames) • Mappingthespacearoundapproximatelythroughvisualestimation(v) Congruence: • Congruencethroughsuperposition(examples—blades,stamps,etc.) • Extendcongruencetosimplegeometricalshapes,e.g.,triangles,circles. • Criteriaofcongruence(byverification)SSS,SAS,ASA,RHS.(vi) Construction (Using Scale, Protractor, Compasses): • Constructionofalineparalleltoagivenlinefromapointoutsideit.(Simpleproofasremarkwiththereasoning

ofalternateangles) • Constructionofsimpletriangles.Likegiventhreesides,givenasideandtwoanglesonit,giventwosidesand

theanglebetweenthem.

UNIT V: MENSURATION (Periods 15)

• Revisionofperimeter,Ideaofπ,CircumferenceofcircleArea:

Conceptofmeasurementusingabasicunitareaofasquare,rectangle,triangle,parallelogramandcircle,areabetweentworectanglesandtwoconcentriccircles

UNIT VI: STATISTICS (Periods 15)Data Handling: (i) Collectionandorganisationofdata–choosingthedatatocollectforahypothesistesting (ii) Mean,medianandmodeofungroupeddata,understandingwhattheyrepresent (iii) Constructingbargraphs (iv) Feelofprobabilityusingdata throughexperiments.Notionof chance inevents like tossingcoins,diceetc.

Tabulatingandcountingoccurrencesof1through6inanumberofthrows.Comparingtheobservationwiththatforacoin.Observingstringsofthrows,notionofrandomness.

MathsWiz–7

UNIT 1: NUMBER SYSTEM

1. Integers 1 – 19

2. Fractions 20 – 37

3. Decimals 38 – 52

4. Rational Numbers 53 – 77

5. Exponents and Powers 78 – 92

Formative Assessment–1

UNIT 2: ALGEBRA

6. Algebraic Expressions 93 – 104 7. Linear Equations in One Variable 105 – 116

UNIT 3: COMMERCIAL ARITHMETIC

8. Ratio and Proportion and Unitary Method 117 – 131

9. Percentage and Its Applications 132 – 157

Formative Assessment–2

Summative Assessment–1

UNIT 4: GEOMETRY

10. Lines and Angles 158 – 178

11. The Triangle and Its Properties 179 – 202

12. Practical Geometry 203 – 207

13. Congruence of Triangles 208 – 226

Formative Assessment–3

UNIT 5: MENSURATION

14. Perimeter and Area 227 – 251

UNIT 6: STATISTICS

15. Data Handling 252 – 267

16. Chance and Probability 268 – 277

UNIT 4: GEOMETRY (CONTD.)

17. Symmetry 278 – 295

18. Visualising Solid Shapes (Representing 3D in 2D) 296 – 313

Summative Assessment–2

Maths Lab Activities 314 – 323

Vedic Maths 324 – 325 Answers 326 – 338

ContentsContents

First we will revise quickly all that you have studied about integers in Class VI and then proceed further.

QUICK RECALLWhat are Integers? The sum of two whole numbers is always a whole number, so the set of whole numbers is closed under addition, but is this true for subtraction also? The answer is no. For example when 4 is subtracted from 9, the answer is 5, which is a whole number. But when we subtract 9 from 4 what do we get? We get –5 which is not a whole number. Thus, this operation cannot be performed if we have the set of whole numbers to work with. So we extend the set of whole numbers and include negative natural numbers also. This new set comprising whole numbers and negative of natural numbers is called the set of integers. Thus, I or Z = {....–3, –2, –1, 0, 1, 2, 3,.....} is the set of integers.

The numbers +1, +2, +3, .........., etc. are called positive integers. The numbers –1, –2, –3, .........., etc. are called negative integers.

Zero is neither negative nor positive.

Both the positive and negative integers are called directed numbers since they indicate direction. Another name given to them is signed numbers because of the + or – sign which is a part of them.

1. For every positive natural number ‘+ a’ there exists an opposite number called the negative of ‘a’ represented by ‘– a’.

2. The sum of any integer and its negative is always zero, i.e., a + (–a) = 0. 3. For the sake of simplicity we omit the + sign and write a instead of + a, where a is a positive integer.

Representing Integers on a Number Line We draw a line and mark a point O in the middle of that line. This point denotes the number 0. Since negative numbers are opposites of positive numbers, therefore if positive numbers +1, +2, +3, +4, ......... are marked at 1 unit, 2 units, 3 units, 4 units and so on from the 0-mark to the right of it, then the points at distances 1 unit, 2 units, 3 units, 4 units and so on from the 0-mark to the left of it shall represent negative integers –1, –2, –3, –4, ........... and so on.

Note that

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1 Integers

UNIT 1: NUMBER SYSTEM

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Predecessor and Successor Predecessor: One less than a given integer is called its predecessor. Thus, on the number line each integer

is the predecessor of the integer just on the right of it. Examples 0 is the predecessor of 1, –1 that of 0, –2 that of –1, and so on. Successor: One more than a given integer is called its successor. Thus, on the number line each integer is

the successor of the integer just on the left of it. Examples –2 is the successor of –3, –1 that of –2, 0 that of –1, 1 that of 0 and so on.

Comparison of Integers For any two integers represented on the number line the one indicated on the right is greater. Examples

–4 > –5, –3 > –4, –2 > –3, –1 > –2, 0 > –1, 1 > 0, 2 > 1, etc.

1. Every positive integer is greater than the negative integer. 2. Zero is less than every positive integer. 3. Zero is greater than every negative integer. 4. The greater the number, the lesser is its opposite, i.e., if a and b are two integers such that

a > b, then –a < –b. Likewise if a < b then –a > –b, e.g., if 5 > 3, then –5 < –3 and if 2 < 5 then –2 > –5.

Absolute Value of an Integer

The absolute value of an integer is the numerical value of the integer regardless of its sign. The symbol ‘| |’ is used to represent the absolute value of an integer. Thus |+8| = 8, |–8| = 8. This shows that the absolute value of an integer is either equal to or greater than the integer, but never less than the integer.

If a represents an integer, then |a | = a if a is + ve or 0 = – a if a is –ve.

Addition of Integers Using the concept of absolute value, we state the rules for addition of integers as under:

Rule 1. To add two integers of like signs, find the sum of their absolute values and place the common sign before the sum.

Rule 2. To add two integers of unlike signs, find the difference of their absolute values and place the sign of the integer which has the larger absolute value before the difference.

Note that

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Examples (i) (+ 6) + (+ 3) = 9 Absolute values of addends are 6 and 3. Sum = 9. Both addends are +ve . ∴ Sum is also +ve. (ii) (– 9) + (– 3) = –12 Absolute values of addends are 9 and 3. Sum = 12. Both addends are –ve. ∴ Sum is also –ve. (iii) (+ 8) + (–11) = –3 Difference between 11 and 8 is 3. The greater absolute value is that of 11 and it has –ve sign. ∴ Sum is –ve.

Subtraction of Integers

If a and b are two integers, then a – b is equal to a + (–b), i.e., to subtract b from a, change the sign of b and add it to a. a – b = a + (–b) Examples:

(a) (b)

(c) (d)

EXERCISE 1 (A)

1. Fill in the blanks: (i) The greatest negative integer is ............. . (ii) 0 is greater than every ................. integer. (iii) The absolute value of an integer is either ................. or ................. than the integer, but never

................. than the integer. (iv) The successor of –91 is ................. . (v) The predecessor of –380 is ................. . 2. Replace each blank by < or > to make the given statement true:

(i) –8 ......... –18 (ii) –4 ........ 0 (iii) –357 ......... –537 (iv) –71 .......... 17.

3. (i) Arrange the following integers in ascending order.

–8, –4 , 0, –11, 9, 4, 6, 13, –27, 19

(ii) Arrange the following integers in descending order.

–6, –11, 12, –32, –23, 14, 0, 32, 16, –19, –18

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4. Add: (i) –18, 14 (ii) –26, –3 (iii) 46, –46 (iv) 400, –31, –71 (v) –613, –421, –700 5.Subtractthefirstintegerfromthesecondineachcase: (i) 6, 18 (ii) –3, 6 (iii) –27, –54 (iv) 12, –7

6. Find the value of: (i) –37 –(–15) –2 (ii) 16 – [14 –(–2) – (–6)] 7. Using>,<or=,fillintheboxestomakeeachstatementtrue.

(i) (–1) + (–5) (–7) – (–5) (ii) 42 – 35 + 12 42 – 35 – 12

(iii) 6 – (–2) + 5 –1 – (–2) – 3 (iv) –135 + 64 + 41 –159 – (–81) + 160

Multiple Choice Questions (MCQs)

8. Which of the following on simplification is not negative?

(a) –8 – (–4) + 5 (b) 7 + (–11) – 5

(c) –4 – 2 + (–8) (d) –10 – (–2) + 4

9. In a quiz, positive marks were given for correct answers and negative marks for incorrect

answers. If Rishi’s scores in six successive rounds were 30, –10, –12, –5, –15, 20, what is his total score at the end?

(a) –12 (b) 8 (c) – 8 (d) 12 Higher Order Thinking Skills (HOTS) 10. What is the value of 0 – 1 + 2 – 3 + 4 – 5 + 6

– 7 . . . . . . . + 16 – 17 + 18 – 19 + 20?

MULTIPLICATION AND DIVISION OF INTEGERSMultiplication of Integers Study the following illustrations:

A. You know that the multiplicaton of whole numbers is repeated addition.

For example: 3 × 4 = Three ‘fours’ = 4 + 4 + 4 = 12 Likewise, we have 3 × (– 4) = Three ‘minus fours’ = (– 4) + (– 4) + (– 4) = –12 ... (1) Further, by commutative property, you know that 3 × 4 = 4 × 3, and 4 × 3 = Four ‘threes’ = 3 + 3 + 3 + 3

Thus, you can also write, 3 × 4 = Three ‘fours’ = 4 + 4 + 4 Likewise, 4 × (–3) = Four ‘minus threes’ = (– 3 ) + (– 3 ) + (– 3 ) + (– 3 ) = –12 ...(2) Hence, from (1) and (2),

3 × (– 4) = 4 × (–3) = – 12 = – (3 × 4)

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Note that as the multiplier goes on decreasing by 1 (3 → 2, 2 → 1, 1 → 0) so the product goes on increasing by 8 (e.g., – 24 is 8 more is than – 32, –16 is 8 more than –24 and so on. If the multiplier decreases further by 1, i.e., 0 becomes –1, –1 becomes –2 and so on then as the pattern of products obtained suggests the product will go on increasing by 8, i.e., 0 will become 8, 8 will become 16 and so on.

B. Above, we have found out the product of two integers with unlike signs, i.e., one positive and the other negative. Now let us investigate what would be the product when both the integers are negative. We have seen in (A) that

4 × – 8 = – 32

Similarly, 3 × – 8 = – 24 2 × – 8 = – 16

1 × – 8 = – 8

0 × – 8 = 0

The above pattern suggests that – 1 × – 8 = 8 – 2 × – 8 = 16 – 3 × – 8 = 24, and so on i.e., the product of two negative integers is positive. The discussions done in A and B lead us to the following two rules of multiplication.

Rule 1. To determine the product of two integers with unlike signs, find the product of the numbers regardless of their signs (i.e., the absolute values) and give minus sign to the product.

e.g. (i) 3 × (– 6) = – 18 (ii) 7 × (– 4) = –28 6 × (– 3) = – 18 4 × (– 7) = –28 Rule 2. To determine the product of two integers with like signs, i.e., with the same sign, find the product of their absolute values and give plus sign to the product.

e.g. (i) 3 × 6 = 18 (ii) 7 × 4 = 28 (– 3) × (– 6) = 18 (– 7) × (– 4) = 28

First Number × Second Number =

+ × + + + × _ _

_ × + _

_ × _ +

Hence, (–) × (+) = (–) and (–) × (–) = (+).

Remarks: Observe that (i) –1 × –1 = 1. (ii) –1 × –1 × –1 = (–1 × –1) × –1 = 1 × –1 = –1. (iii) –1 × –1 × –1 × –1 = (–1 × –1) × (– 1 × –1) = 1 × 1 = 1.

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(iv) –1 × –1 × – 1 × –1 × –1 = (–1 × – 1) × (–1 × –1) × –1 = 1 × 1 × –1 = 1 × – 1 = –1. (v) –1 × –1 × –1 × –1 × –1 × –1 = (–1 × –1) × (–1 × –1) × (–1 × –1) = 1 × 1 × 1 = 1. Therefore it is easy to conclude that:

Ifminussignappearsanoddnumberoftimesinaproductasin(ii)and(iv),thefinalproductisnegative.

Ifminussignappearsanevennumberoftimesinaproductasin(i),(iii)and(v)thefinalproductis positive.

Ex. 1. Find each of the products: (i) 40 × (– 7) (ii) – 3 × 25 (iii) – 5 × – 8 (iv) – 896 × 0 (v) (– 10) × (– 10) × (– 10) (vi) (– 3) × (– 6) × (– 9) × (– 12) (vii) (– 8) × (– 43) × 0 (viii) (– 5) × (6) × (– 7) × (– 20) Sol. (i) 40 × – 7 = – (40 × 7) = – 280. (ii) – 3 × 25 = – (3 × 25) = – 75. (iii) – 5 × – 8 = 40, since (–) × (–) = + (iv) – 896 × 0 = 0 (v) (– 10) × (– 10) × (– 10) = (– 10 × – 10) × – 10 = (100) × – 10 = – 1000. (vi) (– 3) × (– 6) × (– 9) × (– 12) = [( – 3) × (– 6)] × [(– 9) × (– 12)] = 18 × 108 = 1944. (vii) (– 8) × (– 43) × 0 = – 8 × 0 = 0. (viii) – 5 × 6 × – 7 × – 20 = (– 5 × 6) × ( – 7 × – 20) = – 30 × 140 = – (30 × 140) = – 4200. Ex. 2. Compare: (– 2 – 5) × (– 6) and (– 2) + (– 5) × (– 6). Sol. (– 2 – 5) × (– 6) = – 7 × – 6 = 42 – 2 + (– 5) × (– 6) = – 2 + 30 = 28. Since 42 > 28, so (– 2 – 5) × (– 6) is greater. Ex. 3. Use the number line to work out – 7 × 2. Sol. – 7 × 2 = – (7 × 2) = – (2 × 7) so on the number

line – 7 × 2 means two jumps of 7 to the left. By making two jumps of 7 we move 14 spaces to

the left and reach point – 14. Hence, – 7 × 2 = – 14. Ex. 4. Answer True or False: For all non-zero integers a and b, a × b is always greater than either a or b. Sol. False, because if a and b are both +ve integers, then the product will be greater than both a

and b but if either of a and b is a negative integer then the product will be smaller than a and b.

EXERCISE 1 (B)

1. Find each of the following products: (i) 7 × 3 (ii) –7 × 3 (iii) 7 × – 3 (iv) –7 × – 3 (v) –4 × 4 (vi) 8 × –9 (vii) –17 × 0 (viii) –24 × – 5 (ix) –845 × 12 (x) 500 × – 19

2. Simplify: (i) –4 × – 2 × 1 (ii) –5 × – 6 × –1 (iii) –15 × – 8 × – 283 × 0

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(iv) –14 × –5 × – 8 (v) – 17 × – 15 × – 12 (vi) –7 × 29 × – 8 × 0

(vii) –48 × 20 × –1 × –5 (viii) 1 9 4 1

12× − × × − (ix) –

2 25 2385× ×

3. Fill in the blanks:

(i) The product of ............. and (–1) is –35.

(ii) ............. × (–8) = 96.

(iii) The sign of the product of 8 negative integers and 1 positive integer is ............. .

(iv) –7 × 29 × 0 × –8 = ............. .

4. State which is greater:

(i) (7 + 8) × 9 or 7 + (8 × 9) (ii) (7 – 8) × 9 or 7 – (8 × 9) (iii) (– 7 – 8) × 9 or 7 – (– 8 × – 9)

5. Use the number line to work out:

(i) – 4 × 3 (ii) 5 × – 2

6. The temperature of a certain place fell by 2°C (a change of – 2°C) each day for a week. What is the change of temperature over the whole week?

7. For each of the following statements write a true (T) or false (F):

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

(ii) Of the two integers, if one is negative, then their product must be negative.

(iii) The product of a negative and a positive integer may be zero.

(iv) If a > 1, then there is no integer b such that a × b = b × a = b.

Multiple Choice Questions (MCQs) 8. Which of the following expressions are equal

to –20?

I –4 × 5 II –32 + 10 – (–2) III –6 × 2 – [–2 × –4] IV 5 × (–2) + (–3) × 4

(a) I only (b) I and II

(c) I, II and III (d) I, II, III and IV

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

(a) –10°C (b) 10°C (c) 20°C (d) –5°C

Higher Order Thinking Skills (HOTS)

10. Use < or > to compare –2 × (–1) × 4 × 2 × (–3) and (–2) + (–1) + 4 + 2 – (–3).

Division of Integers You know that division facts come from multiplication facts. This is shown in the following examples: 3 × 4 = 12, so 12 ÷ 4 = 3; 8 × 7 = 56, so 56 ÷ 7 = 8

In general, for any whole numbers a and b ( )0 , ab cb

≠ = means a = b × c. The same definition of quotient

is extended to integers. Thus, for any integers a and b (b ≠ 0), ab

= c means a = b × c.

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MathsWiz–7

Now study the following examples:A. (+ 3) × (+ 2) = + 6, so (+ 6) ÷ (+ 2) = + 3; (+ 3) × (– 2) = – 6, so (– 6) ÷ (– 2) = + 3

Ex. 1. Find:

(i) – 20 ÷ 4 (ii) 18 ÷ (– 9) (iii) – 15 ÷ (– 3) (iv) 0 ÷ (– 832) (v) – 56 ÷ (– 1) (vi) 7049 ÷ (– 7049) (vii) – 400000 ÷ (– 800). Sol. (i) – 20 ÷ 4 = – (20 ÷ 4) = – 5. (ii) 18 ÷ (– 9) = – (18 ÷ 9) = – 2. (iii) – 15 ÷ (– 3) = + (15 ÷ 3) = 5. (iv) 0 ÷ (– 832) = – (0 ÷ 832) = 0. (v) – 56 ÷ (– 1) = + (56 ÷ 1) = 56. (vi) 7049 ÷ (– 7049) = – (7049 ÷ 7049) = – 1. (vii) – 400000 ÷ (– 800) = + (400000 ÷ 800) = 500. Ex. 2. Use a number line to work out: (i) – 6 ÷ 3 (ii) – 8 ÷ – 4 Sol. (i) Think if we divide – 6 into 3 equal parts, how big is each part?

66 33

− ÷ = −

The single jump of – 6 can be split into 3 equal jumps of – 2. So – 6 ÷ 3 = – 2. (ii) Think how many lots of – 4 go into – 8?

So – 8 ÷ (–4) = 2.

First number ÷ Second number =

+ ÷ + + + ÷ – – – ÷ + – – ÷ – +

Since the dividend is positive and the divisor is negative, therefore,the quotient is negative. And since 72 ÷ 8 = 9, the answer is –9.

i.e., if the dividend and the divisor are both positive or both negative, the quotient is positive.

B. (–3) × (+ 2) = – 6, so (– 6) ÷ (+2) = –3 (–3) × (– 2) = + 6, so (+ 6) ÷ (– 2) = – 3

i.e., if either the dividend or the divisor is positive and the other is negative, the quotient is negative.

Consider the following example: Geeta was asked to find + 72 ÷ – 8. She reasoned and divided as under :

Maths Wiz Book 7

Publisher : SChand Publications ISBN : 9789385401237Author : S K Gupta AndAnubhuti Gangal

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