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Mathematics

th

8 EDITION “978-93-5463-499-4”

ISB N SYLLABUS COVERED

KARNATAKA TEXTBOOK SOCIETY

PUBLISHED BY

C O P YR IG H T

RESERVED BY THE PUBLISHERS

All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without written permission from the publishers. The author and publisher will gladly receive information enabling them to rectify any error or omission in subsequent editions.

OSWAAL BOOKS & LEARNING PVT. LTD. 1/11, Sahitya Kunj, M.G. Road, Agra - 282002, (UP) India

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D I S C L A IM ER

Oswaal Books has exercised due care and caution in collecting all the data before publishing this book. In spite of this, if any omission, inaccuracy or printing error occurs with regard to the data contained in this book, Oswaal Books will not be held responsible or liable. Oswaal Books will be grateful if you could point out any such error or offer your suggestions which will be of great help for other readers. ) Printed at Upkar( 2Printing Unit, Agra

LET THE ADVENTURE BEGIN

The new way of learning; Blended Learning The pandemic introduced us all to a phenomenon which now seems to be the way forward for learners & teachers alike, it is Blended Learning. In just a span of a year, we have witnessed a rapid advancement in e-learning. Many researchers say that, in no time e-learning will become mainstream. Oswaal Books identified this as an opportunity and thus we decided to prepare students for this turbulent yet a very useful change, hence this book is a hybrid edition. Through this hybrid edition, we aim to help the students learn at their own pace. Hence, we wish to make learning adaptive in order to simplify it for every student.

India is currently one of the youngest economies in the world. Hence, it’s imperative for our education system to churn out more learners than ever before. It is this ‘learner’s mindset’ which will set us apart from the rest of the world. Through Education, we must sow the seeds of Curiosity today to reap the benefits of Intellect tomorrow.

Why this book? This book prepares students for the new world & its new challengesit focuses on nurturing problem solving and thinking skills. • It is strictly based on the latest KSEEB blueprint and design for 2021-2022 SSLC exams. • It contains fully solved questions from NCERT textbook issued by Karnataka Textbook Society. • Students get ample practice with all Typologies of Questions: Like VSA Questions (Very Short Answer), SA Questions (Short Answer), MCQs (Multiple Choice Questions) and LA Questions (Long Answer). • The book helps speed up last-minute preparations with synopsis given at the beginning of every chapter which covers all the essential points. • Equipped with innovative learning tools like ‘Mind Maps’ and high-quality figures, this book is a great tool for qualitative learning and developing a deeper understanding of concepts.

Our Heartfelt Gratitude! At last, we would like to thank our authors, editors and reviewers. Special thanks to students who send us suggestions and constantly help improve our books. We promise to always strive towards ‘Making Learning Simple’ for all of you. Wish you all Happy Learning and a Successful 2021-22!!

(3)

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(4)

CONTENTS • Latest Question Paper Design • SSLC Solved Paper, June 2020 with Model Answers (Issued by KSEEB)

9 - 9 11 - 28

• SSLC Solved Paper, April 2019 with Model Answers (Issued by KSEEB)

29 - 50

• SSLC Model Question Paper, July 2018-19 (Issued by KSEEB)

51 - 54

• Toppers' Answers, March 2018 (Issued by KSEEB)

55 - 72

• Mind Maps

73 - 88

1. Arithmetic Progressions

1 - 21

Topic-1 : Arithmetic Progression

8. Real Numbers

193 - 206

Topic-1 : Euclid’s Division Algorithm.

2. Triangles

22 - 72

Topic-2 : Prime Factorization (H.C.F., L.C.M.).

Topic-1 : Basic Proportionality Theorem

Topic-3 : LCM. and formula fLCM.(a, b)

Topic-2 : Similar Triangles [AA, SSS, SAS]



Topic-3 : Areas of Similar Triangles.

Topic-4 : Irrational Numbers/Rational Numbers

Topic-4 : Pythagorean Triplets

9. Polynomials

Topic-5 : Pythagoras Theorem

Topic-1 : Degree, Value and

3. Pair of Linear Equations



in Two Variables

73 - 113





× H.C.F.(a,b) = a × b}

Zero of a Polynomial.

Topic-2 : Division Algorithm

Topic-1 : Graphical Solution of Linear







Equations in Two Variables

Topic-3 : Remainder Theorem.





Consistency / Inconsistency

Topic-4 : Factor Theorem









of Linear Equations and Equations

10. Quadratic Equations





reducible to Linear Equations

Topic-1 : Roots of the equations. 114 - 129

Topic-1 : Chord & Tangents of a Circle.



for Polynomials.

Topic-2 : Algebraic Methods to Solve Pair

4. Circles

and Factorization.





and its coefficients and



formation of equations.

130 - 149



6. Constructions

150 - 171

Topic-3 : Nature of the roots









and discriminant.

11. Introduction to Trigonometry

in a given ratio

Topic-2 : Construction.

Topic-1 : Trigonometric Ratios:

Topic-3 : Circle, chord, tangents and





Topic-2 : Trigonometric Identities.



their properties.



12. Some Application of



Trigonometry

similar to given triangle

7. Co-ordinate Geometry

172 - 192

256 - 278

Complementary angles & Values.

Topic-4 : Construction of a Triangle

228 - 255

Topic-2 : Relation between roots

5. Areas Related to Circles Topic-1 : Division of a line segment

207 - 227

279 - 289

Topic-1 : Height and Distance

Topic-1 : Co-ordinates and Quadrants.

13. Statistics

Topic-2 : Distance Between Two Points.

Topic-1 : Mean, Median and Mode

Topic-3 : Section Formula and Mid-Point.

Topic-2 : Cumulative Frequency Graph

(5)

290 - 351

...Contd. Contents 14. Probability

352 - 365

Topic-2 : Area and Volume of Cone.

Topic-1 : Empirical Probability.

Topic-3 : Area and Volume of Frustum.

Topic-2 : Theoretical Probability.

Topic-4 : Area and Volume of Sphere.

15. Surface Area and Volumes Topic-1 : Area and Volume of Cylinder



366 - 391

Topic-5 : Area and Volume of Hemisphere. Topic-6 : Area and Perimeter of 2D figures. 

and Hollow Cylinder.

Scan the QR Code to download the fully Board Solved Paper 2020 Science

(6)

30 DAYS OF ONLY GRATITUDE! Take it as a challenge; practice gratitude every day. When you’ll look around yourself, you’ll find umpteen number of things to be grateful for. Practicing gratitude everyday will only multiply those things in your life & will ignite positive emotions in you. Here are a few things you could be grateful for.

#1 About your body.

#6 A Smell you love.

#11 A food you love.

#16 A person you look up to.

#2 What you find beautiful.

#7 Something that makes you smile.

#12 An ability of yours.

#17 A personality trait of yours.

#3

So, get started today!

#4

A song you love.

#8

An accomplishment of yours.

#9

A happy memory.

#13

Something you like about where you live.

#14

A person.

You’re looking forward to.

#18

#19

An item you use every day.

A freedom you are grateful for.

#24

#5 A friend.

#10 A person in your family.

#15 A life lesson.

#20 A holiday you love.

#21

#22

#23

A technology.

Something made you laugh.

Something nice.

#26

#27

#28

#29

#30

Something that brings hope.

A compliment you have received.

Something you are passionate about.

Something in nature.

A gift you received.

(7)

A book magazine or podcast.

#25 Another person.

POSITIVE AFFIRMATIONS “Affirmations are like a seed planted in soil. Poor soil, poor growth. Rich soil, abundant growth. The more you choose to think thoughts that make you feel good, the quicker the affirmations work.”

- Louise Hay

I am confident.

I am strong.

I am joyful.

I love who I am.

I can achieve my goals.

I care about others.

I am compassionate.

I make good decisions.

I am important.

I am responsible.

I am diligent.

I am a leader.

I am thoughtful.

I like myself. It’s going to be a great day.

I believe in my dreams.

I am talented.

I learn from my mistakes.

I am brave.

I am loved.

I make friends easily.

I choose a positive attitude.

I am generous.

I am enough.

I accept and love myself.

I am worthy. I am open to new experiences.

I am great just the way I am.

I am unique.

I am beautiful.

I work hard.

I am wonderfully made.

I am deserving of good things.

I radiate joy and love.

I am creative.

I am grateful.

I am honest.

Good things happen to me.

I believe in me.

I am patient.

I am loving.

I help my family.

I am kind.

I respect myself and I respect others.

Our mind starts believing what we repeatedly think or say. We, at Oswaal Books, resonate with this belief. So, we want all our readers to create their own positive affirmations! A positive affirmation is something spoken aloud that you want to believe or want to be true. Repeating positive affirmations daily can help shift your internal dialogue from negative to positive. So let’s get started!

(8)

LATEST QUESTION PAPER DESIGN FOR SSLC EXAMINATION MATHMATICS WEIGHTAGE TO CONTENTS - DIMENSION-I Sl. No.

Chapter Name

Marks allotted

1*

Arithmetic progressions

6

2*

Triangles

8

3*

Pair of Linear equations in two variables

8

4*

Circles

4



5

Areas related to circles

3



6

Constructions

5

7*

Coordinate Geometry

5

8*

Real numbers

4

9*

Polynomials

6

10*

Quadratic equations

6

11*

Introduction to Trigonometry

5

12

Some Applications of Trigonometry

4

13*

Statistics

6

14

Probability

3

15*

Surface areas and volumes

7

A1

Proofs in mathematics



A2

Mathematical modelling





Total

80

* Indicates internal choice question unit

(9)

( 10 )

SSLC Karnataka June 2020 Class-X

SOLVED PAPER

Mathematics

Time : 3 Hours 15 Minutes

Max. Marks : 80

General Instructions to the Candidate :

1. This question paper consists of 38 objective and subjective types of questions.



2. This question paper has been sealed by reverse jacket. You have to cut on the right side to open the paper at the time of commencement of the examination. Check whether all the pages of the question paper are intact.



3. Follow the instructions given against both the objective and subjective types of questions.



4. Figures in the right hand margin indicate maximum marks for the questions.



5. The maximum time to answer the paper is given at the top of the question paper. It includes 15 minutes for reading the question paper.

Code No : 81 - E

CCERF CCERR Revised

I. Four alternatives are given for each of the following questions / incomplete statements. Choose the correct alternative and write the complete answer along with its letter of alphabet: (8 × 1 = 8) 1. In the pair of linear equations a1x + b1y + c1 = 0 a1 b1 ≠ and a2x + b2y + c2 = 0, if then the a 2 b2

(A) equations have no solution



(B) equations have unique solution



(C) equations have three solutions



(D) equations have infinitely many solutions.

2.

In an arithmetic progression, if an = 2n + 1, then the common difference of the given progression is



(A) 0

(B) 1



(C) 2

(D) 3

3. The degree of a linear polynomial is

(A) 0

(B) 1



(C) 2

(D) 3

4. If 13 sin q = 12, then the value of cosec q is 12 (A) 5

12 (C) 13

(B)

13 5

13 (D) 12

5. In the figure, if DPOQ ~ DSOR and PQ : RS = 1 : 2, then OP : OS is

6. 7.

8.



(A) 1 : 2 (B) 2 : 1 (C) 3 : 1 (D) 1 : 3 A straight line passing through a point on a circle is (A) a tangent (B) a secant (C) a radius (D) a transversal Lenght of an arc of a sector of a circle of radius r and angle q is    r 2  2 r 2 (A) (B) 360  360     2 r (D)  2 r 180 360 If the area of the circular base of a cylinder is 22 cm2 and its height is 10 cm, then the volume of the cylinder is (A) 2200 cm2 (B) 2200 cm3 (C) 220 cm3 (D) 220 cm2 (C)

II. Answer the following questions : (8 × 1 = 8) 23 9. Express the denominator of in the form of 20

2n × 5m and state whether the given fraction is terminating or non-terminating repeating decimal.

12 ]

Oswaal SSLC Karnataka Chapterwise & Topicwise Question Bank, MATHEMATICS, Class – 10

10. The following graph represents the polynomial y = p (x). Write the number of zeroes that p(x) has. y

y=p(x)

x'

x



Find the polynomial of least degree that should 3 2 be subtracted from p(x) = x – 2x + 3x + 4 so that it is exactly divisible by 2 g(x) = x – 3x + 1. 22. Find the distance between the points (– 5, 7) and (– 1, 3). OR Find the coordinates of the point which divides the line joining the points (1, 6) and (4, 3) in the ratio 1 : 2. 23. The points A(1, 1), B(3, 2) and C(5, 3) cannot be the vertices of the triangle ABC. Justify. 24. Draw a pair of tangents to a circle of radius 3 cm which are inclined to each other at an angle of 60°. IV. Answer the following questions : 25. Prove that

y'

130°

A

C

15. Write,

x 1 1  2 x

OR Find the HCF of 24 and 40 by using Euclid’s division algorithm. Hence find the LCM of HCF (24, 40) and 20. 26. To save fuel, to avoid air pollution and for good health two persons A and B ride bicycle for a distance of 12 km to reach their office. As the cycling speed of B is 2 km/h more than that of A, B takes 30 minutes less than that of A to reach the office. Find the time taken by A and B to reach the office. 27. If x = p tan q + q sec q and y = p sec q + q tan q then prove that 2 2 2 2 x – y = q – p . OR Prove that

11. Find the value of tan 45° + cot 45°. 12. Find the coordinates of the mid-point of the line joining the points (x1, y1) and (x2, y2). 13. State “Basic proportionality theorem”. 14. In the figure AB and AC are the two tangents drawn from the point A to the circle with centre O. If ∠BOC = 130° then find ∠BAC. B

O

(9 × 3 = 27)

5 is a irrational number.

in the standard form of a

quadratic equation. 16. Write the formula to find the total surface area of the cone whose radius is ‘r’ units and slant height is ‘l’ units. III. Answer the following questions : (8 × 2 = 16) 17. Solve : 2x + y = 11 x + y = 8 18. Find the sum of 5 + 8 + 11 + ... upto 10 terms using the formula. 19. Find the value of k, if the pair of linear equations 2x – 3y = 8 and 2(k – 4) x – ky = k + 3 are inconsistent. 20. Find the discriminant of the equation 2x2 – 5x + 3 = 0 and hence write the nature of the roots. 21. If one zero of the polynomial p(x) = x2 – 6x + k is twice the other then find the value of k. OR

cot 2 ( 90  ) cosec 2   2 2 sec   cosec 2  tan   1 1  2   cos2  sin 28. Find the median of the following data :



Class-Interval

Frequency

20 – 40

7

40 – 60

15

60 – 80

20

80 – 100

8

OR Find the mode of the following data : Class-Interval

Frequency

1–3

6

3–5

9

5–7

15

7–9

9

9 – 11

1

[ 13

SOLVED PAPER - 2020

29. The following table give the information of daily income of 50 workers of a factory. Draw a ‘less than type ogive’ for the given data. 0

Less than 120

8

Less than 140

20

Less than 160

34

Less than 180

44

Less than 200

50

30. A bag contains 3 red balls, 5 white balls and 8 blue balls. One ball is taken out of the bag at random. Find the probability that the ball taken out is (a) a red ball, (b) not a white ball. 31. Prove that the “lengths of tangents drawn from an external point to a circle are equal”. 32. Construct a triangle ABC with sides BC = 3 cm, AB = 6 cm and AC = 4.5 cm. Then construct a 4 of the corresponding triangle whose sides are 3 sides of the triangle ABC. 33. ABCD is a rectangle of length 20 cm and breadth 10 cm. OAPB is a sector of a circle of radius 10 2 cm. Calculate the area of the shaded region. [Take p = 3.14]

120°

V. Answer the following questions : (4 × 4 = 16) 34. Find the solution of the pair of linear equations by graphical method. x + y = 7 3x – y = 1 35. There are five terms in an Arithmetic Progression. The sum of these terms is 55, and the fourth term is five more than the sum of the first two terms. Find the terms of the Arithmetic progression. OR In an Arithmetic Progression sixth term is one more than twice the third term. The sum of the fourth and fifth terms is five times the second term. Find the tenth term of the Arithmetic Progression. 36. A tower and a pole stand vertically on the same level ground. It is observed that the angles of depression of top and foot of the pole from the top of the tower of height 60 m is 30° and 60° respectively. Find the height of the pole.

Less than 100

cm

Number of workers

21

Daily Income

A 30° 60°

60 m

E

C



h

OR

A hand fan is made up of cloth fixed in between the metallic wires. It is in the shape of a sector of a circle of radius 21 cm and of angle 120° as shown in the figure. Calculate the area of the cloth used and also find the total length of the metallic wire required to make such a fan.

B

D

37. A container opened from the top is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of the milk which can completely fill the container at the rate of Rs. 20 per litre. [Take p = 3.14] VI. Answer the following question : (1 × 5 = 5) 38. State and prove Pythagoras theorem.

qqq

14 ]

Oswaal SSLC Karnataka Chapterwise & Topicwise Question Bank, MATHEMATICS, Class – 10

Model Answers - June 2020

Issued by KSEEB for Set CCERR-RF REVISED 1. (B) Equations have unique solution. Detailed Answer a b If 1 ≠ 1 then the equations have unique solution. a 2 b2 Ans. (B) 2. (C) 2 Detailed Answer an = 2n + 1 Common difference (d) = an – an – 1 = 2n + 1 – 2(n – 1) – 1 = 2n + 1 – 2n + 2 – 1 d = 2 Ans. (C) 3. (B) 1 Detailed Answer The degree of linear polynomial ax + b where a ≠ 0 is one. Ans. (B) 13 4. (D) 12 Detailed Answer ⇒

13 sin q = 12 12 sin q = 13

\

cosec q 

1 sin  13 cosec q = 12



Ans. (D)

PQ PO OQ = = SO OR SR 1 OP = 2 OS

Detailed Answer length of arc 

Detailed Answer 20 = 2 × 2 × 5 2 = 2 × 5 23 Given fraction is terminating after two decimals 20 i.e. 1.15 10. p(x) have three zeroes. Detailed Answer Polynomial y = p(x) intersect the X-axis at 3 points. So number of zeroes that p(x) has 3 (three). 11. 2 Ans. Detailed Answer tan 45° + cot 45° = 1 + 1 = 2 Ans. 12.

 x  x 2 y1  y 2  , P(x, y) =  1 2   2



⇒ Ans. (A) 6. (A) A tangent Detailed Answer A straight line passing through a point on circle is a tangent. Ans. (A)   2 r 7. (D) 360

Ans. (C)

Detailed Answer

5. (A) 1:2 Detailed Answer DPOQ ~ DSOQ

3

= 220 cm 23 23 9. = 2 20 2 ×5

  2 r Ans. (D) 360

8. (C) 220 cm3 Detailed Answer Volume of cylinder = Area of base × height = 22 × 10

 x  x 2 y1  y 2  , Mid-point of line joining AB is  1 2   2 13. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. Detailed Answer Basic proportionality theorem : If a line is drawn parallel to one side of a triangle to intersect the other two sides in a distinct points, then the other two sides are divided in the same ratio. A

D

B In DABC, line l||BC

\ 14.

l

E

C

AD AE = DB CE ∠BAC = 50°

[ 15

SOLVED PAPER - 2020

Detailed Answer B

O

130°

A

C





In quadrilateral OBAC, ∠BOC + ∠OBA + ∠BAC + ∠ACO = 360° 130° + 90° + ∠BAC + 90° = 360° ∠BAC = 360° – 310° ∠BAC = 50° [OB ⊥ AB, OC ⊥ AC] 15. x2 + x – 2 = 0 Detailed Answer x +1 1 = by cross multiplication 2 x

y = 8 – 3 y = 5 \ x = 3, y = 5 Alternate method : Cross multiplication method : x y

x2 + x = 2 2 x + x – 2 = 0 16. TSA of cone = pr (r + l) sq. units Detailed Answer Total surface area of the cone = pr2 + prl 17. 2x + y = 11 x + y = 8 Substitution method : 2x + y = 11 x + y = 8 From equation (ii) y = 8 – x Substitute y = 8 – x in equation (i) We get, 2x + 8 – x = 11 x + 8 = 11 x = 3 Substitute x = 3 in equation (iii) y = 8 – 3 y = 5 \ x = 3, y = 5. Alternate method : Elimination method : 2x + y = 11 x + y = 8 Subtract equation (ii) from equation (i) 2x + y = 11 x + y = 8 x = 3 Substitute x = 3 is equation (ii) 3 + y = 8

...(i) ...(ii) ...(iii)



1

– 11

2

1

1

–8

1

1

x y = = 1 −8 − ( −11) −11 − ( −16 ) 2 −1 x y 1 = = 8  11 11  16 1 x y 1 = = 3 5 1 x 1 = 3 1 x = 3 y 1 = 5 1

y = 5 \ x = 3, y = 5. 18. S10 = 185 \ The sum of the first ten terms is 185. Detailed Answer 5 + 8 + 11 + ... to 10 terms Given sequence is an arithmetic progression \ first term (a) = 5, common difference (d) = 3 n = 10 n \ Sn  [ 2 a  ( n  1)d ] 2

...(i) ...(ii)

1

S10 

10 [ 2  5  (10  1)3] 2

S10 = 5 × (10 + 27) = 185 Hence sum of 10 terms is 185. 19. k = 6 Detailed Answer 2x – 3y – 8 = 0 2(k – 4) x – ky – (k + 3) = 0 Pair of linear equation a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are inconsistent then a1 b c  1  1 a2 b c2 2 2 3 8   \ 2( k − 4 )  k ( k  3) 1 3 8 when   k − 4  k k 3

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