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MATHEMATICSMATHEMATICS

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2017

TERM-II

( )Solved

Sure ShotSure Shot

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MATHEMATICS–X

SECOND TERM

Marks : 90

UNITS MARKS

II. ALGEBRA (Contd.) 23

III. GEOMETRY (Contd.) 17

IV. TRIGONOMETRY (Contd.) 08

V. PROBABILITY 08

VI. COORDINATE GEOMETRY 11

VII. MENSURATION 23

TOTAL 90

UNIT II : ALGEBRA (Contd.)3. QUADRATIC EQUATIONS (15) Periods

Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solution of the quadraticequations (only real roots) by factorization, by completing the square and by using quadraticformula. Relationship between discriminant and nature of roots.Situational problems based on quadratic equations related to day to day activities to beincorporated.

4. ARITHMETIC PROGRESSIONS (8) PeriodsMotivation for studying Arithmetic Progression Derivation of the nth term and sum of thefirst n terms of A.P. and their application in solving daily life problems.

UNIT III : GEOMETRY (Contd.)2. CIRCLES (8) Periods

Tangent to a circle at a point of contact.1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the

point of contact.2. (Prove) The lengths of tangents drawn from an external point to circle are equal.

3. CONSTRUCTIONS (8) Periods1. Division of a line segment in a given ratio (internally).2. Tangent to a circle from a point outside it.3. Construction of a triangle similar to a given triangle.

SYLLABUS

UNIT IV : TRIGONOMETRY (Contd.)3. HEIGHTS AND DISTANCES : Angle of elevation, Angle of Depression. (8) Periods

Simple problems on heights and distances. Problems should not involve more than two righttriangles. Angles of elevation / depression should be only 30°, 45°, 60°.

UNIT V : STATISTICS AND PROBABILITY2. PROBABILITY (10) Periods

Classical definition of probability. Simple problems on single events (not using set notation).

UNIT VI : COORDINATE GEOMETRY1. LINES (In two-dimensions) (14) Periods

Review : Concepts of coordinate geometry, graphs of linear equations. Distance formula.Section formula (internal division). Area of a triangle.

UNIT VII : MENSURATION1. AREAS RELATED TO CIRCLES (12) Periods

Motivate the area of a circle; area of sectors and segments of a circle. Problems based on areasand perimeter/circumference of the above said plane figures. (In calculating area of segmentof a circle, problems should be restricted to central angle of 60°, 90° and 120° only. Planefigures involving triangles, simple quadrilaterals and circle should be taken.)

2. SURFACE AREAS AND VOLUMES (12) Periods(i) Surface areas and volumes of combinations of any two of the following : cubes, cuboids,

spheres, hemispheres and right circular cylinders/cones. Frustum of a cone.(ii) Problems involving converting one type of metallic solid into another and other mixed

problems. (Problems with combination of not more than two different solids be taken.)

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Questions Paper Designs (Math).pmd 9/1/2016, 12:37 PM1

S-1MATHEMATICS

CONTENTS●●●●● CBSE Question Paper 2016 (SA-II) Delhi (Solved) 1–20●●●●● CBSE Question Paper 2016 (SA-II) Outside Delhi (Solved) 21–35

●●●●● Sample Question Papers 1 & 2 (Solved) S-1–S-30(Based on the latest guidelines issued by CBSE)

Chapterwise Expected Questions (Solved)Unit–II : Algebra1. Quadratic Equations 1–232. Arithmetic Progressions 24–41

Unit–III : Geometry3. Circles 42–684. Constructions 69–76

Unit–IV : Trigonometry5. Heights & Distances 77–99

Unit–V : Probability6. Probability 100–117

Unit–VI : Coordinate Geometry7. Coordinate Geometry 118–139

Unit–VII : Mensuration8. Areas Related to Circles 140–1639. Surface Areas and Volumes 164–194

●●●●● Value Based Questions with Answers VBQ-1-VBQ-13Model Question Papers For Practice

Model Question Paper–1 M-1–M-3Model Question Paper–2 M-4–M-7Model Question Paper–3 M-8–M-11Model Question Paper–4 M-12–M-14Model Question Paper–5 M-15–M-18Model Question Paper–6 M-19–M-21Model Question Paper–7 M-22–M-25Model Question Paper–8 M-26–M-29Model Question Paper–9 M-30–M-33Model Question Paper–10 M-34–M-36

Answers Model Question Papers–1 to 10 M-37–M-41

Content-NEW.pmd 9/1/2016, 2:34 PM1

1

C.B.S.E. QUESTION PAPER 2016 SA-II (SOLVED)

CLASS—X (Delhi)

MATHEMATICS

Time Allowed : 3 Hours Maximum Marks : 90

General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into four sections–A, B, C,

and D. (iii) Section A contains 4 questions of 1 mark each, Section B contains 6 questions

of 2 marks each, Section C contains 10 questions of 3 Marks each and Section D contains 11 questions of 4 marks each.

(iv) Use of calculators is not permitted.

SECTION—A

Q. 1. From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB.

Sol. PA and PB are tangents to the given circle. \ ∠PAO = 90° (Radius is perpendicular to the

tangent at the point of contact O. Now, ∠PAB = 50° (Given) \ ∠OAB = ∠PAO – ∠PAB = 90° – 50° = 40° In DOAB OB = OA (Radii of the circle) \ ∠OAB = ∠OBA = 40° (Angles opposite to equal sides are equal) Now, ∠AOB + ∠OAB + ∠OBA = 180° (Angle sum property) ⇒ ∠AOB = 180° – (40° + 40°) = 100° Q. 2. In figure, AB is a 6 m high pole and CD is a ladder inclined at an

angle of 60° to the horizontal and reaches up to a point D of pole. If

AD = 2.54 m, find the length of the ladder. (use 3 =1.73 ) Sol. AB = AD + DB = 6 m (given) ⇒ 2.54 m + DB = 6 m ⇒ DB = 3.46 m

50°

O

B

A

P

2 MBD SURE SHOT C.B.S.E. SAMPLE PAPERS (X)

Now, in the right ∠d ΔBCD.

BDCD

= sin 60°

⇒ 3 46. mCD

= 32

⇒ CD = 2 3 461 73× .

.m [ 3 = 1.73]

⇒ CD = 4 m Thus, the length of the ladder CD is 4 m. Q. 3. Find the 9th term from the end (towards the first term) of the A. P. 5,

9, 13, ....., 185. Sol. Common difference, d, of the A.P. = 9 – 5 = 4 Last term ‘l’ of the A.P. = 185 nth term from the end of an A.P. = l – (n – 1)d. Thus, the 9th term from the end is = 185 – (9 – 1)4 = 185 – 32 = 153 Q. 4. Cards marked with number 3, 4, 5, ....., 50 are placed in a box and

mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.

Sol. It is given that the box contains cards marked with numbers 3, 4, 5, ....., 50. \ Total number of outcomes = 48 There are six perfect squares, i.e., 4, 9, 16, 25, 36 and 49. \Number of favourable outcomes = 6 Probability that a card drawn at random bears a perfect square

= Number of favourable outcomes

Total number of outcomes= =

648

118

SECTION—B

Questions numbers 5 to 10 carry 2 marks each.

Q. 5. If x = 23

and x = –3 are roots of the quadratic equation ax2 + 7x + b = 0,

find the values of a and b.

Sol. Roots of ax2 + 7x + b = 0 are given as –3 and 23

.

Now, Sum of the roots = – ( )Coefficient of

Coefficient of

x

x2

A

D

CB

60°

2.54

6 m

MATHEMATICS – 2016 PAPERS SA-II 3

⇒ – 3 + 23

= −(7)

a

⇒ −73

= −7a

⇒ a = 3

Also, Product of the roots = Constant term

Coefficient of 2x

⇒ –3 23

= ba

⇒ –2 = b3

⇒ b = – 6 Q. 6. Find the ratio in which y–axis divides the line segment joining the

points A(5, –6) and B(–1, –4). Also find the coordinates of the point of division.

Sol. Let (OX) be a point on the y-axis dividing the line segment AB in the ratio k : 1 using the section formula. We get

(0, a) = –,– –k

kk

k+

+ +

51

4 61

A

(5, 6)�P B

( 1, 4)� �

k : 1

⇒ –

,– –k

kk

k+

+=

+5

10

4 61

= a

Now, – kk

++

51

= 0

⇒ –k + 5 = 0 ⇒ k = 5

Also, − −+

4 61

kk

= a

⇒ – –4 5 6

5 1×+

= a

⇒ a = –133

Thus, the y-axis divides the line segment in the ratio 5 : 1

Also, the coordinates of the point of division are 0133

,−

Q. 7. In figure, a circle is inscribed in a DABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF.

MBD Sure Shot CBSE Sample PapersSolved Class 10 Mathematics (Term-II)

2017

Publisher : MBD GroupPublishers

ISBN : 9789384026141 Author : Panel Of Experts

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