WITH COMPLETE SOLUTIONS

QUESTION BANKOSWAAL

Mathematics

Class

7

OSWAAL BOOKSPublished by :

0562-2857671, 25277811/11, Sahitya Kunj, M.G. Road, Agra -282002 (UP) India

0562-2854582 contact@oswaalbooks.com www.OswaalBooks.com

CONTENTS

l Syllabus v - vii

1. Integers 1 - 9

2. Fractions and Decimals 10 - 21

3. Data Handling 22 - 33

4. Simple Equations 34 - 40

5. Lines and Angles 41 - 46

6. The Triangle and Its Properties 47 - 52

7. Congruency of Triangles 53 - 58

8. Comparing Quantities 59 - 64

9. Rational Numbers 65 - 71

10. Practical Geometry 72 - 77

11. Perimeter and Area 78 - 85

12. Algebraic Expressions 86 - 90

13. Exponents and Powers 91 - 96

14. Symmetry 97 - 100

15. Visualising Solid Shapes 101 - 103

qq

( iii )�

PREFACE

Year after year CBSE has been introducing changes in the curriculum of

various classes. We, at Oswaal Books, closely follow every change made by the

Board and endeavor to equip students with the latest study material to prepare

for the Examinations.

The latest offering from us are these Question Banks. These will provide

comprehensive practice material for every chapter. These are prepared by

experienced teachers who have translated their expertise into making

important questions from every chapter in order to facilitate wholesome

learning of every concept.

Highlights of our Question banks:

• Question Bank strictly as per the NCERT Curriculum

• Variety of Questions from NCERT Textbooks

• A synopsis of the important points from every chapter

• Value Based Questions as specified by CBSE Board

• Answers follow the marking scheme and the prescribed word limit

We feel extremely happy to offer our Question Banks and hope that with them,

every student will discover a more thorough way of preparing and thereby

excelling in their examinations. Though we have taken enough care to ensure

our products to be error free, yet we welcome any feedback or suggestions that

come our way for improvisation.

We wish you good luck for the forthcoming academic year!!

–Publisher

( iv )�

Number System (50 hrs)

(i) Knowing our Numbers : Integers

• Multiplication and division of integers (through patterns). Division by zero is meaningless

• Properties of integers (including identities for addition & multiplication, commutative, associative,

distributive) (through patterns). These would include examples from whole numbers as well. Involve

expressing commutative and associative properties in a general form. Construction of counter

examples, including some by children. Counter examples like subtraction is not commutative.

• Word problems including integers (all operations)

(ii) Fractions and rational numbers:

• Multiplication of fractions

• Fraction as an operator

• Reciprocal of a fraction

• Division of fractions

• Word problems involving mixed fractions

• Introduction to rational numbers (with representation on number line)

• Operations on rational numbers (all operations)

• Representation of rational number as a decimal.

• Word problems on rational numbers (all operations)

• Multiplication and division of decimal fractions

• Conversion of units (length & mass)

• Word problems (including all operations)

(iii) Powers:

• Exponents only natural numbers.

• Laws of exponents (through observing patterns to arrive at generalisation.)

(i) am an am+n (ii) (am)n =amn

(iii)am

an = am-n, where m-n N

Algebra (20 hrs)

Algebraic Expressions

• Generate algebraic expressions (simple) involving one or two variables

• Identifying constants, coefficient, powers

• Like and unlike terms, degree of expressions e.g., x2y etc. (exponent 3, number of variables)

Course StructureMathematics Class VII

( v )

• Addition, subtraction of algebraic expressions (coefficients should be integers).

• Simple linear equations in one variable (in contextual problems) with two operations (avoid

complicated coefficients)

Ratio and Proportion (20 hrs)

• Ratio and proportion (revision)

• Unitary method continued, consolidation, general expression.

• Percentage- an introduction.

• Understanding percentage as a fraction with denominator 100

• Converting fractions and decimals into percentage and vice-versa.

• Application to profit and loss (single transaction only)

• Application to simple interest (time period in complete years).

Geometry (60 hrs)

(i) Understanding shapes:

• Pairs of angles (linear, supplementary, complementary, adjacent, vertically opposite) (verification and

simple proof of vertically

opposite angles)

• Properties of parallel lines with transversal (alternate, corresponding, interior, exterior angles)

(ii)Properties of triangles:

• Angle sum property (with notions of proof & verification through paper folding, proofs using property

of parallel lines, difference between proof and verification.)

• Exterior angle property

• Sum of two sides of a triange is it’s third side

• Pythagoras Theorem (Verification only)

(iii) Symmetry

• Recalling reflection symmetry

• Idea of rotational symmetry, observations of rotational symmetry of 2-D objects. (900, 1200, 1800)

• Operation of rotation through 900 and 1800 of simple figures.

• Examples of figures with both rotation and reflection symmetry (both operations)

• Examples of figures that have reflection and rotation symmetry and vice-versa

(iv) Representing 3-D in 2-D:

• Drawing 3-D figures in 2-D showing hidden faces.

• Identification and counting of vertices, edges, faces, nets (for cubes cuboids, and cylinders, cones).

• Matching pictures with objects (Identifying names)

( vi )

• Mapping the space around approximately through visual estimation.

(v) Congruence

• Congruence through superposition (examples blades, stamps, etc.)

• Extend congruence to simple geometrical shapes e.g. triangles, circles.

• Criteria of congruence (by verification) SSS, SAS, ASA, RHS

(vi) Construction

(Using scale, protractor, compass)

• Construction of a line parallel to a given line from a point outside it.(Simple proof as remark with the

reasoning of alternate angles)

• Construction of simple triangles. Like given three sides, given a side and two angles on it, given two

sides and the angle between them.

Mensuration (15 hrs)

• Revision of perimeter, Idea of, Circumference of Circle

Area

Concept of measurement using a basic unit area of a square, rectangle, triangle, parallelogram and

circle, area between two rectangles and two concentric circles.

Data handling (15 hrs)

(i) Collection and organisation of data – choosing the data to collect for a hypothesis testing.

(ii) Mean, median and mode of ungrouped data – understanding what they represent.

(iii)Constructing bargraphs

(iv) Feel of probability using data through experiments. Notion of chance in events like tossing coins,

dice etc. Tabulating and counting occurrences of 1 through 6 in a number of throws. Comparing the

observation with that for a coin.Observing strings of throws, notion of randomness.

( vii )

EARN WHILE YOU LEARNGive us Feedback and make money for it!

We at Oswaal Books try our best to make sure that our publications are error free. At the same time we also acknowledge that it is humane to make errors. It is this understanding that makes us strive to improve our publications on an on going basis.

So in case if you have any suggestions/comments or ideas, we will be excited to hear from you. You can either email us at contact@oswaalbooks.com or fill out the form below.

For each minor error, we will pay you Rs. 5 and for every major error we will pay you Rs. 10. These errors will be approved by our panel of authors and errors which have already been brought to our notice by some other reader will not be valid.

IMPORTANT NOTE : This is not a competition. This is an effort to make our books better for many more readers to come.

FEEDBACK - FORM

Date :

QUESTION BANK

Mathematics, Class-VII

Your name with complete address & telephone number :__________________________________________

____________________________________________________________________________________________

First Name ______________________________ Last Name________________________________________

Date of Birth___________________________________________________________ Sex M/F______________

Address__________________________________________________________Pincode ___________________

Tel : Mobile E-mail _________________________________________________________________________

Name of your School / College _________________________________________________________________

___________________________________________________________________________________________

Name of the teacher or coaching class with address where you are studying :____________________

___________________________________________________________________________________________

Name and Address of the book-seller from where you have purchased this book :_________________

___________________________________________________________________________________________

Who recommended you this book :____________________________________________________________

___________________________________________________________________________________________

What else needs to be incorporated in the book according to your considered view :_______________

___________________________________________________________________________________________

Any error (s)/irrelevent matter : ______________________________________________________________

___________________________________________________________________________________________

Any suggestions you may be pleased to offer :___________________________________________________

___________________________________________________________________________________________

Fill in Hindi or English and send this Feedback Form to

SWAATI JAIN, Oswaal Books 'Oswaal House' 1/11 Sahitya Kunj, M.G. Road, AGRA-282 002.

Receiving Dt.:___________(Office use only)

( viii )�

INTEGERS [ 1

LET’S REVISE1. Counting numbers (natural numbers) are 1, 2, 3, 4, ........2. Whole numbers are 0, 1, 2, 3, ........3. ........., – 4, – 3, – 2, – 1, 0, 1, 2, 3, ......., this sequence is called the set of integers.4. 0 is called the additive Identity, as 0 + a = a for any number a.5. When two positive integers are added, we get a positive integer. e.g. 18 + 2 = 20.6. When two negative integers are added, we get a negative integer. e.g. – 6 + (– 2) = – 87. The additive inverse of any integer x is – x, and the additive inverse of (– x) is x.8. If a and b are integers, then (a + b), (a – b) and (a × b) are also integers.9. If a and b are integers and a + b = b + a, then addition is commutative for integers.

10. In the same way multiplication is commutative for integers , i.e., a × b = b × a.11. If there are three integers a, b and c, then

(i) addition is associative, i.e., (a + b) + c = a + (b + c)(ii) multiplication is associative, i.e., (a × b) × c = a × (b × c)(iii) multiplication is distributed over addition, i.e., a × (b + c) = (a × b) + (a × c) anda × (b – c) = (a × b) – (a × c)

12. 1 is the multiplicative identity. a.1 = a = 1.a for any number a.13. For integers, multiplication of same signs is positive and multiplication of opposite signs is

negative.14. If 0 is multiplied to any integer, product is always 0. a.0 = 0 for any number a.

15.Integer

0 = not defined (), but 0Integer = 0, for non-zero integers.

EXERCISE

Objective Type Questions

(Each question carries 1 mark)Q. 1. Team A scored – 40, 10, 0 and team B scored

10, 0, – 40 in there successive rounds. Whichteam scored more. Can we say that we canadd integers in any order ? (NCERT)

Q. 2. Find the pattern and complete the following :(a) 7, 3, – 1, – 5, ................ .(b) 15, 10, 5, 0, .................. .

Q. 3. Fill in the blanks to make the followingstatements true :(a) (– 5) + (– 8) = (– 8) + (........)(b) – 53 + ........ = – 53.

Integers

1CHAPTER

OSWAAL CBSE Question Bank, Mathematics Class - VII2 ]

Q. 4. Match the column :Integer Additive inverse 10 76– 76 – 10.

Q. 5. True and false :(a) If 0 is multiplied to any integer, product will be zero.(b) 1 × a = a × 1 = 0.

Q. 6. Fill in the blanks :(a) (– 5) + (– 7) = ......... .(b) (– 11) × 0 = ........ .

Q. 7. Which of the following is the additiveinverse of – 27 ?(a) – 27 (b) 27(c) 0 (d) 1

Q. 8. Which of the following is the value of (– 12)× (– 2) × (– 5) ?(a) – 120 (b) 120(c) 0 (d) 1

Q. 9. Which of the following is the value of (– 4)× [(– 5) + (– 3)] ?(a) – 32 (b) 120(c) 32 (d) – 23

Q. 10. A shop keeper makes a profit of ` 5 on eachpen and incurs a loss of ` 2 on each pencilbox. What will be his net profit if he sellsten pens and ten pencil boxes ?(a) 20 (b) 30(c) 50 (d) 100

Q. 11. Which of the following is the simplest formof [(– 5) + (– 7)]/[(– 2) + (– 1)] ?(a) – 12 (b) 12(c) 4 (iv) – 4

Answers1. Total runs scored by team A in three successive

records= – 40 + 10 + 0 ½= – 30

The total runs scored by team B in threesuccessive records

= 0 + 10 + (– 40) ½= – 30

Yes, we can add integers in any order2. (a) – 9, – 13 (b) – 5 – 10 ½+½3. (a) – 5 (b) 0 ½+½4. Integer Additive inverse

10 – 10 ½– 76 76 ½

5. (a) True ½(b) False ½

6. (a) – 12 ½(b) 0 ½

7. (b) (– 27) – (– 27) 27 18. (a) (– 12) × (– 2) × ( – 5)

– 12 × 2 × 5 – 120 1

9. (c) (– 4) × [(– 5) + (– 3)]– 4 × (– 8) 32 1

10. (b) Profit on 10 pens = 5 × 10 = ` 50Loss on 10 pencil boxes = 2 × 10 = ` 20Total Profit = 50 – 20 = ` 30 1

11. (c) [(– 5) + (– 7)]/[(–2) + (– 1)][–12]/[–3]4 1

Very Short Answer Type Questions(Each question carries 2 marks)

Q. 1. Write a pair of integer whose difference is(–10).

Sol. The first integer (50, 60)= 50 ½

The second integer = 60 ½(50 – 60) = – 10 1

Q. 2. Write a pair of integers whose sum is 0.Sol. (– 10, 10), (10, – 10), (2, – 2), (– 2, 2)

The first intetger = – 10 ½The second integer = 10 ½

(– 10 + 10) = 0 1Q. 3. If the total of two integers is (– 53) and one

of them is 9, find other integer.Sol. Let other integer be x, then according to the

question9 + x = – 53 1

x = – 53 – 9= – 62. 1

Q. 4. Write a pair of integers whose sum is(– 7). (NCERT)

Sol. We can have (– 2) + (– 5) = – 7 1 – 2 and – 5 may be the required pair. 1

Q. 5. Evaluate the following :(a) (– 30) 10(b) (– 36) (– 9). (NCERT)

Sol. (a) (– 30) 10 = – 3 1(b) (– 36) (– 9) = 4. 1

INTEGERS [ 3Q. 6. Result of the following is an integer or not :

(a) (– 8) (– 2)(b) 3 (– 8).

Sol. (a) (– 8) (– 2) = 4 (integer) 1

(b) 3 (– 8) = 38 , which is not an integer..

1Q. 7. Evaluate :

(a) 13 [(– 2) + 1](b) 0 (– 12).

Sol. (a) 13 [(– 2) + 1]= 13 [– 2 + 1]= 13 (– 1) = – 12 1

(b) 0 ( – 12) = 0. 1Q. 8. Find the values of a and b in following

algebraic expressions :(a) 1 + a = 3(b) b – 2 = 1.

Sol. (a) Since,1 + a = 3

a = 3 – 1 a = 2 1(b) Since,

b – 2 = 1 b = 1 + 2 b = 3. 1

Q. 9. Find the product :(– 1) × (– 1) × (– 1) ............ 5 times.

Sol. (– 1) × (– 1) × (– 1) ...... 5 timesIf odd number of (–ve) integers are multiplied,then the sign of product will be negative. 1Since 5 is odd number.So, result = – 1. 1

Q. 10. Write the multiplicative identity for integers.Sol. Multiplicative identity for integers is 1.

as 1.a = a = a.1 2Q. 11. Find each of the following products :

(i) (– 115) × 8(ii) 9 × ( – 3) × (– 6)(iii) (– 12) × (– 13) × (– 5)

Sol. (i) we have,(– 115) × 8 = – (115 × 8) = – 920 ½

(ii) We have,9 × (– 3) × (– 6)= {9 × (– 3)} × (– 6)= – (9 × 3) × (– 6)= – 27 × (– 6) = 27 × 6 = 162 ½(iii) we have,(– 12) × (– 13) × (– 5)

= {(– 12) × (– 13)} × (– 5) ½= (12 × 13) × (– 5)= 156 × (– 5) = – (156 × 5) = – 780 ½

Q. 12. Evaluate each of the following products :(i) (– 1) × (– 2) × ( – 3) × ( – 4) × (– 5)(ii) (– 3) × (– 6) × (– 9) × (– 12)

Sol. (i) Since the number of negative integers inthe product is odd. Therefore, their productis negative.Thus, we have(– 1) × (– 2) × (– 3) × (– 4) × (– 5)= – (1 × 2 × 3 × 4 × 5)= – (2 × 3 × 4 × 5)= – (6 × 4 × 5)= – (24 × 5) = – 120 1(ii) Since the number of negative integers inthe given product is even. Therefore, theirproduct is positive. Thus, we have.(– 3) × ( – 6) × (– 9) × (– 12)= (3 × 6 × 9 × 12)= (18 × 9 × 12) [? 3 × 6 = 18]= (162 × 12) [? 18 × 9 = 162]= 1944 1

Q. 13. Find the value of(i) 15625 × (– 2) + ( – 15625) × 98(ii) 18946 × 99 – (– 18946)(iii) 1569 × 887 – 569 × 887

Sol. (i) 15625 × ( – 2) + (– 15625) × 98= (– 15625) × 2 + (– 15625) × 98= (– 15625) × (2 + 98)= (– 15625) × 100= – (15625 × 100)= – 1562500 ½(ii) 18946 × 99 – ( – 18946)= 18946 × 99 + 18946= 18946 × 99 + 18946 × 1

[? 18946 = 18946 × 1]= 18946 × (99 + 1)

[Using a × b + a × c = a × (b + c)]= 18946 × 100= 1894600 1(iii) 1569 × 887 – 569 × 887= (1569 – 569) × 887

[? b × a – c × a = (b – c) × a]= 1000 × 887= 887000 ½

OSWAAL CBSE Question Bank, Mathematics Class - VII4 ]

Short Answer Type Questions(Each question carries 3 marks)

Q. 1. Verify a – (– b) = a + b for the followingvalues of a and b.(a) a = 42, b = 36(b) a = 236, b = 250(c) a = 150, b = 168(d) a = 50, b = 22

Sol. (i) a = 42, b= 36To prove,

a – (– b) = (a + b)Putting the values of a and b we have,42 – (– 36) 42 + 36 78a + b= 42 + 36 78 1

(ii) a = 236, b = 250To prove,

a – (– b) = (a + b)Putting the values of a and b we have,

236 – (– 250) = 236 +250 = 486a + b = 236 + 250 = 486 1

(iii) a = 150, b = 168To prove,

a – (– b) = (a + b)Putting the values of a and b we have,

150 – (– 168) = 150 + 168 = 318a + b = 150 + 168 = 318 ½

(iv) a = 50, b = 22To prove,

a – (– b) = (a + b)Putting the values of a and b we have,

50 – (– 22) = 50 + 22 = 72a + b = 50 + 22 = 72 ½

Q. 2. Find each of the following products :

(a) (– 20) × (– a) × 9

(b) (– 18) × (– 5) × (– 1) × 7

(c) (– 5) × (– 4) × (– 8)Sol. (a) (– 20) × (– a) × 9

20 × a × 9 180a 1(b) (– 18) × (– 5) × (– 1) × 7 – 18 × 5 × 1 × 7 – 630 1(c) (– 5) × (– 4) × (– 8) – 5 × 4 × 8 – 160 1

Q. 3. Find the product using suitable properties :(a) 26 × (– 48) + (– 36) × (– 48)(b) 15 × (– 25) + (– 4) × (– 25)(c) 7 × (50 – 2).

Sol. (a) 26 × (– 48) + (– 36) × (– 48)[a × (– b) = – (a × b)

(– a) × (– b) = (a × b)]= [– (26 × 48)] + (36 × 48)= (– 1248) + 1728 = 480 1

OR26 × (– 48) + (– 36) × (– 48)= [26 + (– 36)] × (– 48)= (– 10) × (– 48), by distributive property

= 480 1(b) 15 × (– 25) + (– 4) × (– 25)= [15 + (– 4)] × (– 25), by distributive property= (11) × (– 25)

= – 275 1(c) 7 × (50 – 2)

= 7 × 50 – 7 × 2= 350 – 14 = 336. 1

(distributivity of multiplicationover subtraction)

Q. 4. Verify : (– 30) × [13 + (– 3)]= [(– 30) × 13] + [(– 30) × (– 3)].

(NCERT) 3Sol. L.H.S. = (– 30) × [13 + (– 3)]

= (– 30) × (13 – 3)= (– 30) × 10= – 300 1

R.H.S. = [(– 30) × 13] + [(– 30)× (– 3)]

= (– 390) + 90= – 300 1

L.H.S. = R.H.S., which is verified.1

Q. 5. What is the :(a) sign of the product of 15 negative integers ?(b) product of 108 negative integers and 0 ?(c) sign of product of 22 negative integers ?

Sol. (a) Since 15 is odd, so the sign of product of15 negative integer is negative. 1(b) Since product of integer is an integer andproduct of integer and zero is zero.Product = 0. 1

(c) Sign of product of 22 negative integers ispositive, because 22 is even. 1

Q. 6. Starting from (– 1) × 5 write various product,showing some pattern to show(– 1) × (– 1) = 1. (NCERT) 3

Sol. (– 1) × 5 = – 5(– 1) × 4 = – 4 [ = (– 5) – (– 1)]

INTEGERS [ 5(– 1) × 3 = – 3 [= (– 4) – (– 1)] 1(– 1) × 2 = – 2 [= (– 3) – (– 1)](– 1) × 1 = – 1 [(– 2) – (– 1)] 1(– 1) × 0 = 0 [= (– 1) – (– 1)]

(– 1) × (– 1) = 1 [= 0 – (– 1)] 1Q. 7. Subtract : 3

(a) 63 from (– 100 + 7)(b) – 222 from (– 666).

Sol. (a)We have to subtract 63 from (– 100 + 7)Since (– 100 + 7) = – 93 – 93 – 63 = – 156. 1½(b) We have, (– 666) – (– 222)

= (– 666) + 222= – 444. 1½

Q. 8. Solve the following : 3(a) 738 + (– 99) + 100 – (– 400)(b) 76 × (– 42) + 76 × 50.

Sol. (a) 738 + (– 99) + 100 – (– 400)= 738 – 99 + 100 + 400= 738 + 1 + 400= 1139 1½

(b) By distributive property, 76 × (– 42) + 76 × 5076 × [(– 42) + 50] = 76 × 8

= 608. 1½Q. 9. Write the additive inverse of the following : 3

(a) – 100 (b) 72(c) – 3.

Sol. Since we know that the total of any integerand its additive inverse is zero.(a) Additive inverse of – 100 = 100 as 100 + (– 100) = 0 1(b) Additive inverse of 72 = – 72 as 72 + (– 72) = 0 1(c) Additive inverse of – 3 = 3 as 3 + (– 3) = 0 1

Q. 10 Find the value of 44875 × 99 – (– 44875) usingthe property. 3

Sol. 44875 × 99 – (– 44875)= 44875 × 99 – 44875 × – 1= 44875 × [99 – (– 1)](using distributivity over addition) 1

= 44875 × (99 + 1) 1= 44875 × 100= 4487500. 1

Q. 11. Simplify : 24 – 4 ÷ 2 × 3Sol. we have

24 – 4 ÷ 2 × 3= 24 – 2 × 3 2

[Performing division – 4 ÷ 2 = – 2]

= 24 – 6 1[Performing multiplication 2 × 3 = 6]

= 18 [Performing subtraction]Q. 12. Simplify : (– 20) + ( – 8) ÷ (– 2) × 3Sol. we have ,

(– 20) + ( – 8) ÷ ( – 2) × 3= (– 20) + 4 × 3 2= ( – 20) + 12 [Performing multiplication] 1= – 8 [Performing subtraction]

Q. 13. Simplify : (– 5) – (– 48) ÷ (– 16) + ( – 2) × 6Sol. we have ,

(– 5) – (– 48) ÷ (– 16) + ( – 2) × 6= (– 5) – 3 + (– 2) × 6[Performing division]1= (– 5) – 3 + (– 12) 2

[Performing multiplication]= – 5 – 3 – 12= – 20 [Performing addition]

Q. 14. Divide :(i) 84 by 7 (ii) – 91 by 13(iii) – 98 by – 14 (iv) 324 by – 27

Sol. (i) we have ,

84 ÷ 7 = |84| 84|7| 7

= 12 ½

(ii) we have,

– 91 ÷ 13 = |–91| 91|13| 13

= – 7 ½

(iii) we have,

– 98 ÷ (– 14) = |–98| 98|–14| 14 = 7 1

(iv) we have,

324 ÷ (– 27) = |324| 324|–27| 27 = – 12 1

Q. 15. Find the quotient in each of the following :(i) (– 1728) ÷ 12(ii) (– 15625) ÷ (– 125)(iii) 3000 ÷ (– 100)

Sol. (i) we have ,

(–1728) ÷ 12= |–1728| 1728

|12| 12 = – 144 1

(ii) we have ,

(– 15625) ÷ (– 125) = |–15625| 15625| 125| 125

= 125 1

(iii) we have ,

30000 ÷ (– 100) = |30000|| 100|

OSWAAL CBSE Question Bank, Mathematics Class - VII6 ]

= – 30000100

= – 300 1Q. 16. Find the value of

(i) [32 + 2 × 17 + (– 6)] ÷ 15(ii) ||– 17| + 17 | ÷ || – 25 | – 42|

Sol. (i) we have,[32 + 2 × 17 + ( – 6) ÷ 15]= [32 + 34 + (– 6)] ÷ 15 ½= (66 – 6) ÷ 15 ½= 60 ÷ 15

= 6015

½

= 4(ii) we have,||– 17| + 17 | ÷ || – 25 | – 42|= | 17 + 17|÷ |25 – 42| ½= |34| ÷ |–17| ½= 34 ÷ 17

= 3417

½

= 2Q. 17. Simplify :

{36 ÷ (– 9)] ÷ {(– 24) ÷ 6}Sol. we have ,

{36 ÷ (– 9)} ÷ {(– 24) ÷ 6} 1

= |36| |24| 36 24| 9| |6| 9 6

1

= (– 4) ÷ (–4) | 4| 4| 4| 4

= 1 1

Long Answer Type Questions(Each question carries 4 marks)

Q. 1. A plane is flying at the height of 5000 mabove the sea level. At a particular point, itis exactly above a submarine floating 1200m below the sea level. What is the verticaldistance between them. (NCERT)

Sol. Height of plane above sea level= 5000 m (given) 1

Height of submarine below sea level= 1200 m (given) 1

Total vertical distance between planeand submarine = (5000 + 1200) m

= 6200 m 2Q. 2. A water tank has steps inside it. A monkey

is sitting on the top most step (i.e., the firststep). The water level is at the ninth step.(i) He jumps 3 steps down and then jumps

back 2 steps up. In how many jumpswill he reach the water level ?

(ii) After drinking water, he wants to goback. For this, he jumps 4 steps up andthen jumps 2 steps down in every move.In how many jumps will he reach backthe top step.

Sol. Let the steps moved down be represented bypositive integers and the steps moved up berepresented by negative integers. 1(i) Initially, the monkey was at step = 1

After Ist jump,the monkey will be at step

= 1 + 3 = 4After 2nd jump,

the monkey will be at step= 4 + (– 2) = 2

After 3rd jump,the monkey will be at step

= 2 + 3 = 5After 4th jump,

the monkey will be at step= 5 + (– 2) = 3

After 5th jump,the moneky will be at step

= 3 + 3 = 6After 6th jump,

the moneky will be at step= 6 + (– 2) = 4

After 7th jump,the moneky will be at step

= 4 + 3 = 7After 8th jump,

the moneky will be at step= 7 + (– 2) = 5

After 9th jump,the moneky will be at step

= 5 + 3 = 8After 10th jump,the moneky will be at step

= 8 + (– 2) = 6After 11th jump,the moneky will be at step

= 6 + 3 = 9Clearly, the monkey will be at water level (i.e.,9th step) after 11 jumps. 3(ii) The monkey will reach back at the topstep after 5 jumps. 1

Q. 3. In a class test containing 10 questions, 5marks are awarded for every correct answerand (–2) marks are awarded for every

Oswaal CBSE Question BanksMathemaitcs For Class 7

Publisher : Oswaal Books ISBN : 9789386681102 Author : Panel Of Experts

Type the URL : http://www.kopykitab.com/product/11507

Get this eBook

25%OFF