MATHS MCQ SOLVED AND UNSOLVED

As per CBSE Sample Paper Issued On 02 Sep 2021... Sample Papers Mathematics (Standard) CBSE Class 10 (Term I) Authors Chitra Dhingra Vishal Kumar Mehta Visit https://telegram.me/booksforcbse for more books. ARIHANT PRAKASHAN (School Division Series)

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Contents þ One Day Revision 1-10 þ The Qualifiers 11-28 þ CBSE Question Bank 29-41 þ Latest CBSE Sample Paper 42-55 Sample Paper 1 59-69 Sample Paper 2 70-82 Sample Paper 3 83-96 Sample Paper 4 97-110 Sample Paper 5 111-123 Sample Paper 6 124-136 Sample Paper 7 137-148 Sample Paper 8 149-161 Sample Paper 9 162-172 Sample Paper 10 173-185 Watch Free Learning Videos Video Solutions of CBSE Sample Papers Chapterwise Important MCQs CBSE Case Based MCQs CBSE Updates Much more valuable content will be available regularly

Syllabus Mathematics (Standard) CBSE Class 10 (Term I) Marks No. Unit Name I Number Systems II Algebra III Coordinate Geometry IV Geometry V Trigonometry VI Mensuration VII Statistics Probability Total Internal Assessment Total Internal Assessment Marks Total Marks Periodic Tests Multiple Assessments marks for Portfolio the term Student Enrichment Activities-practical work UNIT I NUMBER SYSTEMS . Pair of Linear Equations in Two Variables . Real Number Pair of linear equations in two variables and Fundamental Theorem of Arithmetic - graphical method of their solution, statements after reviewing work done consistency inconsistency. Algebraic conditions earlier and after illustrating and motivating for number of solutions. Solution of a pair of through examples. Decimal representation linear equations in two variables algebraically - of rational numbers in terms of by substitution and by elimination. Simple terminating non-terminating recurring situational problems. Simple problems on decimals. equations reducible to linear equations. UNIT II ALGEBRA UNIT III COORDINATE GEOMETRY . Polynomials . Coordinate Geometry Zeroes of a polynomial. Relationship between zeroes and coefficients of LINES In two-dimensions Review: Concepts of quadratic polynomials only. coordinate geometry, graphs of linear equations. Distance formula. Section formula internal division

UNIT IV GEOMETRY . Motivate In a triangle, if the square on one side is equal to sum of the squares . Triangles on the other two sides, the angle opposite to the first side is a right angle. Definitions, examples, counter examples of similar triangles. UNIT V TRIGONOMETRY . Prove If a line is drawn parallel to one . Introduction to Trigonometry side of a triangle to intersect the other Trigonometric ratios of an acute angle of a two sides in distinct points, the other two right-angled triangle. Proof of their sides are divided in the same ratio. existence well defined . Values of the trigonometric ratios of , and . . Motivate If a line divides two sides of a Relationships between the ratios. triangle in the same ratio, the line is Trigonometric Identities parallel to the third side. Proof and applications of the identity sin A + cos A = . Only simple identities to be . Motivate If in two triangles, the given corresponding angles are equal, their corresponding sides are proportional and UNIT VI MENSURATION the triangles are similar. . Areas Related to Circles . Motivate If the corresponding sides of Motivate the area of a circle; area of sectors two triangles are proportional, their and segments of a circle. Problems based corresponding angles are equal and the on areas and perimeter circumference of two triangles are similar. the above said plane figures. In calculating area of segment of a circle, problems should . Motivate If one angle of a triangle be restricted to central angle of and is equal to one angle of another triangle only. Plane figures involving triangles, and the sides including these angles are simple quadrilaterals and circle should be proportional, the two triangles are taken. similar. UNIT VII STATISTICS PROBABILITY . Motivate If a perpendicular is drawn from the vertex of the right angle of a . Probability right triangle to the hypotenuse, the Classical definition of probability. Simple triangles on each side of the problems on finding the probability perpendicular are similar to the whole of an event. triangle and to each other. . Motivate The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. . Prove In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

MCQs Preparation Tips Focus on Theory Learn to Identify Wrong Answers MCQs can be formed from any part or The simplest trick is, observe the options line of the chapter. So, strong command first and take out the least possible one on theory will increase your chances to and repeat the process until you reach solve objective questions correctly and the correct option. quickly. Analyse your Performance Practice of Solving MCQs During the practice of MCQs, you can Cracking an MCQ-based examination identify your weak & strong requires you to be familiar with the topics/chapter by analysing of incorrect question format, so continuous practice answers, in this way you will get an will make you more efficient in solving awareness about your weaker topics. MCQs. Practice through Sample Papers Speed & Accuracy Solving more & more papers will make In MCQ-based examination, you need you more efficient and smarter for both speed and accuracy, if your exams. Solve lots of Sample Papers accuracy is good but speed is slow then given in a good Sample Papers book. you might attempt less questions resulting in low score. Attempting MCQs in Exams 1. Read the paper from beginning to Step by step solution is not required in MCQ type questions, it is a waste end & attempt those questions first in of time, you will not get extra marks which you are confident. Now move for this. on to those questions which requires thinking and in last attempt those 4. Most of the time, you need not to questions for which you need more attention. solve the MCQ completely to get the correct option. You can start thinking 2. Read instructions of objective in reverse order and choose the best fit option. questions carefully and find out what is being asked, a bit carelessness can 5. As there is no negative marking for lead you to incorrect answer. incorrect answers, so don't leave any 3. Tick/Write down the correct option question unanswered. Use your guess if you have not exact idea only while filling the OMR sheet. about the correct answer.

CBSE Sample Paper Mathematics Standard Class X (Term I) 1 ONE DAY REVISION Revise All the Concepts in a Day Just Before the Examination... G Real Number (ii ) For any three positive integers a, b and c, the relation between these numbers and their HCF and Prime, Coprime and Composite numbers LCM is Prime numbers are those numbers, which have no HCF (a, b,c ) factors other than 1 and the number itself. = a ´ b ´ c ´ LCM (a, b, c ) (c, a) e.g. 2, 3, 5, 7, 11, … LCM (a, b) ´ LCM (b, c ) ´ LCM Coprime numbers are those numbers, which do not or LCM (a, b, c ) have any common factor other than 1. = a ´ b ´ c ´ HCF(a, b, c ) a) e.g. 2 and 9 are coprime numbers. HCF(a, b) ´ HCF(b, c ) ´ HCF (c, Composite numbers are those numbers, which have Real Numbers atleast 1 factor other 1 and the number itself. A number, which is either rational or irrational, is called e.g. 4, 6, 24, … a real number. Factor Tree Rational Numbers A chain of factors which is represented in the form of a A number that can be expressed as p, where p, q are tree, is called factor tree. q Fundamental Theorem of Arithmetic integers and q ¹ 0, is called a rational number. Fundamental theorem of arithmetic states that every Irrational Numbers ONE DAY REVISION composite number can be written (factorised) as the product of primes and this factorisation is unique, apart A number that cannot be expressed in the form p , from the order in which the prime factors occur. It is q also called unique factorisation theorem. \\ Composite number = Product of prime numbers where p, q are integers and q ¹ 0, is called an irrational Relation between Numbers and their HCF number. and LCM Useful Theorems (i ) For any two positive integers a and b, the relation between these numbers and their HCF and LCM is Theorem 1 Let p be a prime number and a be a HCF (a, b) ´ LCM (a, b) = a ´ b positive integer. If p divides a2, then p divides a. Þ HCF (a, b) = a ´ b LCM (a, b) Theorem 2 2 is irrational number, then 2 2 is or LCM (a, b) = a ´ b irrational number. HCF (a, b) Decimal Expansions of Rational Numbers 1. Terminating Decimal Expansion The number which terminates (i.e. ends completely) after a finite

02 CBSE Sample Paper Mathematics Standard Class X (Term I) number of steps in the process of division, is said non-repeating decimal expansion. These numbers are to be terminating decimal expansion. e.g. 1.25, called irrational numbers. 3.14, etc. e.g. 1. 030030003..., 3, etc. 2. Non-terminating Decimal Expansion The number which does not terminate in the Important Theorems on Decimal process of division, is said to be non-terminating Expansion of Rational Numbers decimal expansion. Theorem 3 Let x be a rational number whose decimal There are following two types of non-terminating expansion terminates. Then, x can be expressed in the decimal expansions form p/q, where p, q are coprimes and the prime factorisation of q is of the form 2n5m, where n and m are (i ) Non-terminating Repeating Expansion The number, which does not terminate but non-negative integers. repeats the particular number again and again in the process of division, is said to be Theorem 4 (Converse of Theorem 3) Let x = p/q non-terminating repeating decimal or recurring be a rational number, such that the prime factorisation decimal expansion. The repeated digit is of q is of the form 2n5m, where n and m are denoted by bar ‘-’ e.g. 1 = 0.333 K = 0.3 non-negative integers. Then, x has a decimal 3 expansion, which terminates. (ii ) Non-terminating Non-repeating Decimal Theorem 5 Let x = p/q be a rational number, Expansion The number, which neither terminates such that the prime factorisation of q is not of the form nor repeats the particular number in the process 2n5m, where n and m are non-negative integers. Then, x of division, is said to be a non-terminating has a decimal expansion, which is non-terminating repeating (recurring). G Polynomials A polynomial in one variable x, is an algebraic Geometrical Meaning of the Zeroes of a Polynomial expression of the form The geometrical meaning of the zeroes of a polynomial p( x) = an xn + an- 1xn- 1 + an- 2 xn- 2 means that the curve intersect the X-axis, the + ... + a2 x2 + a1x + a0 intersection point is said to be zeroes of the curve. where n is a positive integer and constants Relationship between Zeroes and a0, a1, a2,..., an are known as coefficients of Coefficients of a Polynomial polynomial. The zeroes of a polynomial are related to its Degree of a Polynomial coefficients. The highest power (exponent) of x in a polynomial f (x), (i) For a Linear Polynomial The zero of the linear is called the degree of the polynomial f(x). polynomial ax + b is - b = - Constant term . Types of Polynomials a Coefficient of x (i) Linear Polynomial A polynomial of degree one, is (ii) For a Quadratic Polynomial Let a and b be the called linear polynomial. zeroes of quadratic polynomial p(x) = ax2 + bx + c, (ii) Quadratic Polynomial A polynomial of degree two, is called quadratic polynomial. a ¹ 0, then (iii) Cubic Polynomial A polynomial of degree three, is \\Sum of zeroes, a + b called cubic polynomial. = – Coefficient of x = - b (iv) Biquadratic Polynomial A polynomial Coefficient of x2 a ONE DAY REVISION of degree four, is called biquadratic polynomial. and product of zeroes, ab Value of a Polynomial at Given Point = Constant term = c If p(x) is a polynomial and a is a real value, then the Coefficient of x2 a value obtained by putting x = a in p(x) , is called the value of p(x) at x = a and it is denoted by p(a). Formation of Quadratic and Cubic Polynomials Zeroes of a Polynomial If a and b are the zeroes of a quadratic polynomial, A real number k is said to be a zero of a polynomial f(x), if f(k ) = 0. then quadratic polynomial will be k [x2- (sum of zeroes) x + product of zeroes] i.e. k [x2 - (a + b)x + ab], where k is some constant.

CBSE Sample Paper Mathematics Standard Class X (Term I) 03 G Pair of Linear Equations in Two Variables Two linear equations in the same two variables, say x a1 = b1 Coincident Infinitely System is and y, are called pair of linear equations a2 b2 lines many consistent (or system of pair equations) in two variables. = c1 Parallel solutions (dependent) The general form of pair of linear equations in two c2 lines No solution System is variables x and y is a1 = b1 ¹ c1 inconsistent a1 x + b1 y + c1 = 0 a2 b2 c2 and a2 x + b2 y + c2 = 0, Algebraic Methods for Solving a Pair of Linear Equations where a1, b1,c1 and a2, b2,c2 are all real numbers and a12 + b12 ¹ 0, a22 + b22 ¹ 0. Solution of a Pair of Linear Equations There are three methods for solving a pair of linear in Two Variables equations Any pair of values of x and y which satisfies both the 1. Substitution Method equations, a1x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0, is called a solution of a given pair of linear equations. In this method, value of one variable can be found out in terms of other variable from one of the given Solution of a Pair of Linear equation and this value is substituted in other equation, Equations by Graphical Method then we get an equation in one variable, which can be solved easily. Let us consider a pair of linear equations in two variables, a1x + b1y +c1 = 0 and a2 x + b2 y + c2 = 0. 2. Elimination Method To find the solution graphically, there are three cases In this method, one variable out of the two variables is arise eliminated by making the coefficients of that variable equal in both the equations. Case I When the graph of system of linear equations will represent two intersecting lines, then coordinates of After eliminating that variable, the left equation is an point of intersection say equation in another variable, which can be solved (a, b) is the solution of the pair of linear equations. This easily. is called consistent pair of linear equations. Case II When the graph of system of linear equations Value of one variable obtained in this way can be will represent two parallel lines, then there is no point of substituted in any one of the two equations to find the intersection and consequently there is no pair of values value of other variable. of x and y which satisfy both equations. Thus, given system of equations have no solution. This is called Equations Reducible to a Pair of inconsistent pair of linear equations. Linear Equations Case III When the graph of system of linear equations Sometimes, equations are not linear but they can be will represent coincident or overlapping lines, there are infinitely many common points. Thus, the given system reduced to a pair of linear equations by making some of equations have infinitely many solutions. suitable substitutions. Such pair of linear equations is called dependent pair (i) If the given equations involve 1 and 1, then put of linear equations and it is always consistent. xy Nature of Lines and Consistency 1 = p and 1 = q to convert into linear form. xy The nature of lines and consistency corresponding to (ii) If the given equations involve x 1 a and y 1 , linear equations a1x + b1y + c1 = 0 and ± ± b a2 x + b2 y + c2 = 0, is shown in the table given below ONE DAY REVISION then put x 1 a = p and y 1 b = q to convert into ± ± Compare Graphical Algebraic Consis- linear form. the ratios representation interpre- tency (iii) If the given equations involve 1 and 1 , tation x + y x- y a1 ¹ b1 Intersecting Exactly System is then put 1 y = p and 1 =q to convert into a2 b2 lines one consistent x+ x-y solution linear form. (unique)

04 CBSE Sample Paper Mathematics Standard Class X (Term I) G Coordinate Geometry Cartesian System Section Formulae The system used to describe the position of a point in a In section formula, we find the coordinates of a point which divides the given line segment internally (or plane, is called cartesian system. In cartesian system, externally) in a given ratio. there are two mutually perpendicular straight lines XX¢ and YY ¢, which intersect each other at origin point O. Internal Division of a Line Segment Y Let A (x1, y1) and B(x2, y2) are two points and P (x, y) is 3 a point on the line segment joining A and B such that AP : BP = m1 : m2, then point P is said to divide line 2 P (x, y) segment AB internally in the ratio m1 : m2. 1 90° X' –3 –2 –1 O X m1 m2 B 123 (x2 , y2) –1 (x1, y1) A –2 P –3 The coordinates of point P are given by Y' æèç m1x2 + m2 x1 m1y2 + m2 y1 ÷öø. m1 + m2 , m1 + m2 The horizontal line XOX¢ is called X-axis (or abscissa) and the vertical line YOY ¢ is called Y-axis (or ordinate). Generally, for finding internal division ratio, we consider Distance between Two Points P divides AB in the ratio k : 1, then the coordinates of in a Cartesian Plane the point P will be The distance between any two points P(x1, y1)and æçè kx2 + x1 , ky2 + y1 öø÷. Q( x2, y2 ) is given by k + 1 k + 1 PQ = ( x2 - x1)2 + ( y2 - y1)2 Coordinates of Mid-point of Line Segment or PQ = ( x1 - x2 )2 + ( y1 - y2 )2 If the point P divides the line segment equally i.e. 1 : 1, then the coordinates of P will be + + Collinear Points çæè x1 2 y1 , y1 2 y2 ø÷ö. This is also called mid-point When three or more than three points lie on a same formula. line, then they are called collinear points. Suppose A, B and C are three points, then the Note Trisection of the line segment means, a line is condition for collinearity of three points is divided into three equal line segment AB + BC = AC AP Q B or AC + CB = AB or BA + AC = BC i.e. AP = PQ = QB. ONE DAY REVISION

CBSE Sample Paper Mathematics Standard Class X (Term I) 05 G Triangles Similar Polygons (i) AAA Similarity Criterion Two polygons of the same number of sides are similar, if In two triangles, if corresponding angles are equal, then their corresponding sides are proportional and (i) all the corresponding angles are equal and hence the two triangles are similar. (ii) all the corresponding sides are in the same ratio (or Note If two angles of one triangle are respectively equal to two angles of another triangle, then the two proportion). triangles are similar. AAA similarity criterion can be consider as AA similarity criterion. DC H G (ii) SSS Similarity Criterion 6 and 9 If in two triangles, three sides of one triangle are A8B E 12 F proportional (i.e., in the same ratio) to the three sides of the other triangle, then their corresponding angles are If only one condition from (i) and (ii) is true for two equal and hence the two triangles are similar. polygons, then they cannot be similar. (iii) SAS Similarity Criterion Similar Triangles If one angle of a triangle is equal to one angle of the Two triangles are said to be similar, if other triangle and the sides including these angles are proportional, then the two triangles are similar. (i) their corresponding angles are equal and Theorem 1 If a perpendicular is drawn from the vertex (ii) their corresponding sides are proportional. of the right angle of a right angled triangle to the hypotenuse, then triangles on both sides of the Symbolically it can be represented by the perpendicular are similar to the whole triangle and to symbol ‘~’. each other. A Theorem 2 (Pythagoras Theorem) In a right angled P triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. B CQ R Theorem 3 (Converse of Pythagoras Theorem) In a e.g. In DABC and DPQR, if right angled triangle, the square of the hypotenuse is ÐA = ÐP, ÐB = ÐQ , ÐC = ÐR equal to the sum of the squares of the other two sides. and AB = BC = AC. AC2 = AB2 + BC2 PQ QR PR A Then, DABC is similar to DPQR. Conversely If DABC is similar to DPQR, then ÐA = ÐP, ÐB = ÐQ, ÐC = ÐR and AB = BC = AC PQ QR PR Basic Proportionality Theorem (BPT) BC Theorem 1 (Thales Theorem) If a line is drawn parallel Area of Similar Triangles ONE DAY REVISION to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided Theorem 1 The ratio of the areas of two similar in the same ratio. triangles is equal to the square of the ratio of their corresponding sides. Theorem 2 (Converse of Basic Proportionality Theorem) If a line divides any two sides of a triangle in AP the same ratio, then the line must be parallel to the third side. Criteria for Similarity of Triangles B CQ R We have some criteria for congruency of two triangles ar (DABC) ( AB)2 æçè AC ÷öø 2 æçè BC ÷øö 2 involving only three pairs of corresponding parts ar (DPQR) (PQ ) PR QR (elements) of two triangles. Similarly, we have some = = = criteria for similarity of two triangles, which are given below:

06 CBSE Sample Paper Mathematics Standard Class X (Term I) G Introduction to Trigonometry Trigonometric Ratios A Popular Technique to Remember T-ratios i.e. PBP The ratios of the sides of a right angled triangle with respect to its acute angles, are called trigonometric HHB ratios. Pandit (P ) Badari (B ) Prasad (P ) Trigonometric ratios are also called T-ratio. Trigonometric ratios of ÐA in right angled D ABC are Har (H ) Har (H ) Bholay (B ) defined below. X C Hypotenuse (H) HP Side opposite to Ð A [i.e. Perpendicular (P)] ZB Y ASide adjacent B Then, sin q = P , cos q = B, tan q = P to ÐA [i.e. Base (B)] H HB Þ cosec q = H, sec q = H, cot q = B P BP (i) sin A = Side opposite to ÐA çæèi.e. P ö÷ø where, P is perpendicular, B is base and H is Hypotenuse H hypotenuse. = BC Important Points AC (i) In an isosceles right DABC, right angled (ii) cos A = Side adjacent to ÐA çèæi.e. B øö÷ = AB at B, the trigonometric ratios obtained by taking Hypotenuse H AC either ÐA or ÐC, both give the same value. (iii) tan A = Side opposite to ÐA æèçi.e. P ÷øö = BC (ii) The value of each of the trigonometric ratios of an Side adjacent to ÐA B AB angle does not depend on the size of the triangle. It only depends on the angle. (iv) cosec A = Hypotenuse ÐA èæçi.e. H ÷öø = AC Side opposite to P BC (iii) It is clear that the values of the trigonometric ratios (v) sec A = Hypotenuse ÐA æçèi.e. H öø÷ = AC of an angle do not vary with the lengths of the sides Side adjacent to B AB of the triangle, if the angle remains the same. (vi) cot A = Side adjacent to ÐA èæçi.e. B ÷øö = AB (iv) As, the hypotenuse is the longest side in a right Side opposite to ÐA P BC angled triangle, the value of sin A or cos A is always less than 1 (or in particular equal to 1) whereas the Similarly trigonometric ratios of ÐC are value of sec A or cosec A is always greater than or equal to 1. (a) sin C = AB (b) cos C = BC AC AC (c) tan C = AB (d) cosecC = AC Relation Between Trigonometric Ratios BC AB (i) sin A = 1 ,cosec A = 1 (e) sec C = AC (f ) cot C = BC cosec A sin A BC AB (ii) cos A = 1 , sec A = 1 C ONE DAY REVISION Hypotenuse (H) sec A cos A Side adjacent to Ð A [i.e. Base (B)] (iii) tan A = 1 , cot A = 1 cot A tan A AB (iv) tan A = sin A Side opposite to ÐC cos A [i.e. Perpendicular (P)] (v) cot A = cos A sin A

CBSE Sample Paper Mathematics Standard Class X (Term I) 07 Values of Trigonometric Ratios for Important Points Some Specific Angles (i) The value of sin q increase from 0 to 1 and cos q Angles 0° 30° 45° 60° 90° decrease from 1 to 0, where 0 £ q £ 90°. sin q 0 1 13 1 (ii) In the case of tan q, the values increase from 0 to ¥, 2 where 0 £ q £ 90°. 22 (iii) In the case of cot q, the values decrease from ¥ to cos q 31 1 0, where 0 £ q £ 90°. 1 2 22 0 (iv) In the case of cosec q, the values decrease from ¥ to 1, where 0 £ q £ 90°. tan q 0 1 3¥ 31 (v) In the case of sec q, the values increase from 1 to ¥, where 0 £ q £ 90° . cosec q ¥ 2 2 (vi) Division by 0 is not allowed, since 1/0 is 2 31 indeterminate (not defined). 2 Trigonometric Identity sec q 1 3 22 ¥ For any acute angle q, we have 1 (i) sin2 q + cos2 q = 1 (ii) sec2 q - tan2 q = 1 30 cot q ¥ 31 (iii) 1 + cot2 q = cosec2 q Here, ¥ = undefined Note sin2 q = (sin q)2 but sin q2 ¹ (sin q)2. The same is true for all other trigonometric ratios. Representation of a Trigonometric Ratio in Terms of Any Other Trigonometric Ratio sin q sin q cos q tan q cot q sec q cosec q cos q sin q (1- cos2 q) tan q 1 (1+ tan2 q) (sec2 q - 1) 1 (1- sin2 q) cos q (1+ cot2 q) sec q cosec q 1 (1+ tan2 q) cot q 1 (cosec 2 q - 1) (1+ cot2 q) sec q cosec q tan q sin q (1- cos2 q) tan q 1 (sec2 q - 1) 1 cot q (1- sin2 q) cos q cot q 1 (cosec2 q - 1) sec q cos q 1 cosec q (1- sin2 q) tan q cot q (sec2 q - 1) cosec2 q - 1 sin q (1- cos2 q) cosec q 1 (1+ tan2 q) (1+ cot2 q) sec q (cosec2 q - 1) 1 cot q sec q (1- sin2 q) cos q (1+ tan2 q) (sec2 q - 1) cosec q tan q (1+ cot2 q) 1 1 sin q (1- cos2 q) ONE DAY REVISION

08 CBSE Sample Paper Mathematics Standard Class X (Term I) G Areas Related to Circles Circular Ring Circle It is a plane figure bounded by the circumference of two concentric circles of two different radii. A circle is the locus of a point which moves in a plane such a way that its distance from a fixed point remains RO the same. The fixed point is called the centre and the r given constant distance is known as radius of the circle. ONE DAY REVISION O Area of ring = p(R2 - r 2) sq units. Radius A Diameter B Sector of a Circle Circumference (Perimeter) of a Circle The region enclosed by two radii and the The distance covered by travelling once around a circle corresponding arc of a circle is called the sector of is called the circumference or the length of boundary circle. of a circle. \\ Circumference = p ´ Diameter = p ´ 2r units In the figure, unshaded region OACBO is called the where, r is the radius of circle. major sector and shaded region OAPBO is called the minor sector of a circle. Area of a Circle Length of an Arc of a Sector The space occupied in a circular region is called area of a circle. The arc corresponding to a sector is called the arc of \\ Area of a circle = pr 2 sq units the sector. Semi-circle C A diameter divides the circle into two parts, each part is Major called semi-circle. sector r O AB rq (i) Perimeter of semi-circle = 2pr + 2r AB 2 P l sector = (pr + 2r ) units Minor (ii) Area of semi-circle = 1 (pr 2) sq units Length of an arc of a sector, l = q ´ 2pr 2 360° Quadrant of a Circle Area of Sector of a Circle If a circle is divided into four equal parts, then each part (i) Area of the sector = q ´ pr 2 of a circle is said to be quadrant of a circle. 360° (ii) Area of sector in terms of length of arc = 1 lr 2 (iii) Area of the major sector = pr 2 - Area of minor sector (iv) Area of minor sector = pr 2 - Area of major sector A rB Note (i) Perimeter of a quadrant = 2pr + 2r (i) If q = 180°, then sector becomes a semi-circular 4 region and its area = 1 pr 2. pr 2 = çèæ 2 + 2r ÷öø units (ii) If q = 90°, then sector becomes a quadrant of a (ii) Area of a quadrant = æç pr 2 ÷ö sq units circle and its area = 1 pr 2. è 4ø 4

CBSE Sample Paper Mathematics Standard Class X (Term I) 09 Segment of a Circle The segment containing the minor arc is called the minor segment and the segment containing the major The region bounded by a chord and the corresponding arc is called the major segment. arc of the circle is called the segment of the circle. (i) Area of a segment D = Area of corresponding sector Major - Area of triangle formed by chord segment and the radii of the circle. O (ii) Area of major segment = pr 2 - Area of minor segment q (iii) Area of minor segment AB = pr 2 - Area of major segment P Minor segment G Probability Probability of an Event ONE DAY REVISION (or Probability of occurrence of an Event) Probability is the study of the chances (or likelihood) of events happening. By means of probability, the If E is an event associated with a random experiment, chance (or likelihood) of events is measured by a then probability of E, denoted by P(E), represents the number lying from 0 to 1. chance of occurrence of event E. Experiment e.g. If E denotes the event of getting an even number in a single throw of a die, then P(E) represents the chance An operation which produces some well defined of occurrence of event E, i.e. the chance of getting 2, 4 outcomes, is called an experiment. or 6. e.g. Tossing a coin, throwing a dice, etc. Compound Event (i) Random experiment If an experiment is repeated A collection of two or more elementary events under identical conditions and they do not produce associated with an experiment is called a compound the same outcomes every time, then it is said to be event. e.g. In the random experiment of tossing of two random coins simultaneously, if we define the event of getting (or probabilistic) experiment. exactly one head, then it is a collection of elementary events (or outcomes) HT and TH. So, it is a compound (ii) Deterministic experiment If an experiment is event. repeated under identical conditions and they produce the same outcomes every time, then it is is Equally Likely Outcomes said to be deterministic experiment. The outcomes of a random experiment are said to be An event for an experiment is the collection of some equally likely, when each outcome is as likely to occur outcomes of the experiment. We generally denote it by as the other, i.e. when we have no reason to believe capital letter E. that one is more likely to occur than the other. e.g. Getting an even number in a single throw of a die e.g. When a die is thrown, all the six outcomes, i.e. 1, is an event. This event would consist of three 2, 3, 4, 5 and 6 are equally likely to appear. So, the outcomes, namely 2, 4 and 6. outcomes 1, 2, 3, 4, 5 and 6 are equally likely outcomes. Elementary Event Favourable Outcomes An event having only one outcome of the random The outcomes which ensure the occurrence of an experiment is called an elementary event. e.g. In event are called favourable outcomes to the event. e.g. tossing of a coin, the possible outcomes are head (H ) The favourable outcomes to the event of getting an and tail (T ). Getting H or T are known as elementary even number when a die is thrown are 2, 4 and 6. events. Complement of an Event/Negation of Occurrence of an Event an Event An event E associated to a random experiment is Let E be an event associated with a random said to be occur (or happen) in a trial, if the outcome experiment. Then, we can define the complement of of trial is one of the outcomes that favours E. event E or negation of event E, denoted by E, e.g. If a die is rolled and the outcome of a trial is 4, then as an event which occurs if and only if E does we say that event getting an even number has not occur. happened (or occurred).

10 CBSE Sample Paper Mathematics Standard Class X (Term I) e.g. Let E be the event of getting an even number in a Impossible Event single throw of a die. Then, its complement can be define as event E of getting an odd number, as E is consisting 2, An event which is impossible to occur, is called an 4 and 6. Therefore, E would consist 1, 3 and 5. impossible event and probability of impossible event is always zero. Note E and E are called complementary events. e.g. In throwing a die, there are only six possible Theoretical (Classical) Definition of Probability outcomes 1, 2, 3, 4, 5 and 6. Let we are interested in getting a number 7 on throwing a die. Since, no Let us assume all the outcomes of an experiment are face of the die is marked with 7. So, 7 cannot come equally likely and E is an event associated with the in any throw. Hence, getting 7 is an impossible experiment, then the theoretical probability event. (or classical probability) of the event E is given by Then, P (getting a number 7) = 0 = 0 P(E) = Number of outcomes favourable to E 6 Total number of outcomes Sure Event or Certain Event = n (E) n (S ) An event which is sure to occur, is called a sure event or certain event and probability of sure event (i) Probability of an event can never be negative. is always 1. e.g. Suppose we want to find the probability of getting a number less than 7 in a (ii) The sum of the probabilities of complementary events single throw of a die having numbers 1 to 6 on its of an experiment is 1. six faces. i.e. If E and E are complementry events. We are sure that, we shall always get a number Then, P( E) + P( E) = 1or P( E) = 1- P( E) less than 7, whenever we throw a die. So, getting a or P( E) = 1- P( E) number less than 7 is a sure event. Then, P (getting a number less than 7) = 6 = 1 where, P(E) represents the probability of occurrence of an event E and P( E) represents the 6 probability of non-occurrence of an event E . ONE DAY REVISION

CBSE Sample Paper Mathematics Standard Class X (Term I) 11 THE QUALIFIERS Chapterwise Set of MCQs to Check Preparation Level of Each Chapter 1. Real Numbers Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct. Select the correct option as your answer. 1. If two positive integers a and b are written as a = x 4 y2 and b = x 2 y3 , where x and y are prime numbers, then LCM (a, b) is (b) x 6y 5 (d) None of these (a) x8y6 (c) x 4y 3 2. The HCF and LCM of two numbers are 23 and 828. When the first number is completely divided by 2 the quotient is 23. The other number is (a) 420 (b) 410 (c) 414 (d) 425 3. Two natural numbers whose difference is 66 and the LCM is 360, are (a) 180 and 114 (b) 90 and 24 (c) 120 and 54 (d) 130 and 64 4. The HCF and LCM of the smallest composite number and the smallest prime number are respectively. (a) 2 and 2 (b) 2 and 4 (c) 4 and 4 (d) 8 and 4 5. If HCF of two numbers is 4 and their product is 160, then their LCM is (a) 40 (b) 60 (c) 80 (d) 120 6. If LCM = (32, a) = 64 and HCF (32, a) = 4, then a is equal to THE QUALIFIER (a) 16 (b) 8 (c) 20 (d) 10 7. The sum of two numbers is 528 and their HCF is 33, then the number of pairs satisfying the given condition is (a) 5 (b) 3 (c) 4 (d) 2

12 CBSE Sample Paper Mathematics Standard Class X (Term I) 8. The ratio of LCM and HCF of second smallest prime number and second smallest composite number is (a) 2 : 5 (b) 2 : 1 (c) 1 : 2 (d) 5 : 2 9. If the LCM of two prime number a and b (a > b) is 253, then the value of 8b - 3a is (a) 16 (b) 17 (c) 19 (d) 18 10. If x 2 = 1 + 2 + 5, then x is (b) rational (d) integer 36 6 (a) irrational (c) whole number 11. Prime factors of the denominator of a rational number with the decimal expansion 62.47 are (a) 2 and 35 (b) 2 and 5 (c) 3 and 5 (d) 4 and 5 12. The decimal expansion of the rational number 53 , will terminate after how many 23 ´ 5 places of decimal? (a) 1 (b) 3 (c) 4 (d) 2 13. The smallest number by which 1 should be multiplied, so that its decimal expansion 17 terminator after one decimal place is (a) 17 (b) 17 100 10 (c) 100 (d) 10 17 17 14. The rational form of 0.325 is in the form of p, then q - p is q (a) 640 (b) 650 (c) 668 (d) 670 15. What smallest number must be multiplied in the denominator, so that the decimal number 14588 will be terminated? 625 (a) 4 (b) 18 (c) 16 (d) 20 THE QUALIFIER Answers 1. (c) 2. (c) 3. (b) 4. (b) 5. (a) For Detailed Solutions 6. (b) 7. (c) 8. (b) 9. (c) 10. (a) Scan the code 11. (b) 12. (b) 13. (b) 14. (c) 15. (c)

CBSE Sample Paper Mathematics Standard Class X (Term I) 13 2. Polynomials Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct. Select the correct option as your answer. 1. If 2 is a zero of polynomial p(x) = 5x 2 + 3x - 4k, then the value of k is (a) 15 (b) 13 2 2 (c) 11 (d) None of these 2 2. The graph of the polynomial p(x) is given as below, Y X¢ X –4 –3 –2 –1 0 1 2 3 Y¢ The zeroes of p(x) are (b) - 2 , 0, 2 and 3 (a) - 3, - 2 , 0 and 0 (d) - 3, 0, 2 and 3 (c) - 3, - 2 , 0 and 2 3. The number of zeroes of given polynomial graph are Y X¢ X Y¢ (a) 4 (b) 3 (c) 2 (d) 1 4. Which of the following is not the graph of a quadratic polynomial? YY (a) X¢ X (b) X¢ X Y¢ Y¢ Y Y (c) X¢ X (d) X¢ X THE QUALIFIER Y¢ Y¢ 5. If one zero of the polynomial x 2 - 4x + 1 is 2 - 3, then the other zero is (a) 2 - 3 (b) 2 + 3 (c) 3 - 2 (d) None of these

14 CBSE Sample Paper Mathematics Standard Class X (Term I) 6. The number of polynomials having zeroes as 3 and - 7 is (a) 11 (b) 2 (c) 3 (d) more than 2 7. If p(x) = ax 2 + bx + c and a + b + c = 0, then one zero is (a) c (b) - b a a (c) b (d) Cannot determined a 8. Find a quadratic polynomial, whose zeroes are 3 and 1. 4 (a) 2 x 2 + 13x - 3 (b) 4x 2 + 13x - 3 (c) 4x 2 - 13x - 3 (d) 4x 2 - 13x + 3 9. If 2 and 3 are zeroes of polynomial 3x 2 - 2kx + 2m, then the values of k and m are respectively (b) 15 and - 9 (c) 15 and 9 (d) None of these (a) 15 and 9 2 2 2 10. If one zero of the polynomial (a 2 + 4)x 2 + 9x + 4a is the reciprocal of the other, then the value of a is (b) - 2 (c) 2 (d) - 2 and - 3 (a) 2 and 3 11. If zeroes of the polynomial p(x) = - 8x 2 + (k + 5)x + 36 are negative to each other, then the value of k is (b) 5 (c) 4 (d) 3 (a) - 5 12. If a and b are zeroes of the polynomial p(x) = x 2 - p(x + 1) + c such that (a + 1) (b + 1) = 0, then the value of c is (c) - 1 (d) 1 2 2 (a) - 1 (b) 1 13. Suppose a and b are zeroes of the quadratic polynomial p(x) = x 2 - (k + 5)x + 3(2k - 3) such that a + b = ab. Then, the value of k is 2 (a) 19 (b) - 19 (c) 19 (d) None of these 3 4 4 14. If one zero of the polynomial p(x) = 2x 2 - 5x - (2k + 1) is twice the other zero, then the value of k is (b) - 17 (c) 9 (d) None of these (a) 17 9 17 9 15. If the square of difference of the zeroes of the quadratic polynomial p(x) = x 2 + px + 45 THE QUALIFIER is equal to 144, then the value of p are (a) ± 9 (b) ± 12 (c) ± 15 (d) ± 18 Answers 1. (b) 2. (c) 3. (b) 4. (c) 5. (b) For Detailed Solutions 6. (d) 7. (b) 8. (d) 9. (c) 10. (c) Scan the code 11. (a) 12. (a) 13. (c) 14. (b) 15. (d)

CBSE Sample Paper Mathematics Standard Class X (Term I) 15 3. Pair of Linear Equations in Two Variables Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct. Select the correct option as your answer. 1. The line x = h and y = k, where h ¹ 0 and k ¹ 0, are (a) parallel (b) intersecting (c) overlapping (d) None of these 2. The pair of equations x = 0 and x = 4 has (a) no solution (b) one solution (c) two solutions (d) infinitely many solutions 3. If the system of equations 2x + ky = 8 and x + y = 6 has no solution, then the value of k is (a) 2 (b) - 3 (c) - 2 2 (d) 3 2 4. One equation of a pair of dependent linear equation is 3x + 5y = 4. Then, second equation can be (b) 15x + 25y = 20 (a) 15x + 25y = 15 (d) 15x - 25y = 20 (c) - 15x - 25y = 20 5. If the lines given by 2x + ky = 1 and 3x - 5y = 7 has unique solution, then the value of k is (a) all real values of k (b) all real values of k except k = - 10 3 (c) all real value of k except k = 10 3 (d) None of these 6. If the system of equations (k + 2)x + 21y - 3k = 0 and 4x + 7y - 10 = 0 has infinitely many solutions, then value of k is (a) 11 (b) 10 (c) 12 (d) - 12 7. The pair of linear equations 2x + 3y = 5 and 4x + 6y = 10 is (a) consistent (b) inconsistent (c) dependent consistent (d) None of these 8. The nature of the lines representing the linear equations 2x - y = 3 and 4x - y = 5 is (a) intersecting (b) parallel (c) coincident (d) None of these 9. The solution of the pair of systems 7x - 4y = 3 and x + 2y = 3 is THE QUALIFIER (a) 10 and 8 (b) 8 and 13 (c) 1 and 1 (d) None of these 10. The solution of the system of equations x + y = 2 and ax - by = a 2 - b 2 is ab (a) a and b (b) b and a (c) - a and - b (d) - b and - a

16 CBSE Sample Paper Mathematics Standard Class X (Term I) 11. The area of figure formed by the lines x = - 3 and y = 2 and along with coordinate axes is (a) 5 sq units (b) 6 sq units (c) 3 sq units (d) None of these 12. The graph of linear equations 2x + y = 6 and 4x - 2y = 4 is shown below. Find the area of triangle formed by lines and X-axis. Y 6 5 4 4x – 2y – 4 = 0 3 2C 1 X¢ –4 –3 –2 –1 O AD B X –5 12 34 5 –1 –2 2x + y = 6 –3 –4 Y¢ (a) 3 sq units (b) 2 sq units (c) 1 sq units (d) None of these 13. The solution of 7 x + y = 343 and 343x - y = 7 is (a) x = 4 and y = 5 (b) x = 5 and y = 4 33 33 (c) x = - 5 and y = 4 (d) None of these 33 14. The solution of the pair of equations 2 + 3 = 2 and 4 - 9 = - 1 is xy xy (a) x = 4 and y = 9 (b) x = - 4 and y = - 9 (c) x = 2 and y = 3 (d) None of these 15. The sum of the digits of a two-digit number is 9. Also nine times this number is twice the number obtained by reversing the order of the digits. The original number is (a) 17 (b) 16 (c) 18 (d) 15 THE QUALIFIER Answers 1. (b) 2. (a) 3. (a) 4. (b) 5. (b) For Detailed Solutions 6. (b) 7. (c) 8. (a) 9. (c) 10. (a) Scan the code 11. (b) 12. (b) 13. (b) 14. (a) 15. (c)

CBSE Sample Paper Mathematics Standard Class X (Term I) 17 4. Coordinate Geometry Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct. Select the correct option as your answer. 1. The distance between the points A(- 1, - 5) and B(- 6, 7) is (a) 12 units (b) 13 units (c) 14 units (d) 15 units 2. If the distance between the points A(x, 2) and B(3, - 6) is 10 units, then the positive value of x is (a) 6 (b) 9 (c) - 3 (d) 5 3. A triangle with vertices A(4, 0), B(- 1, - 1) and C(3, 5) is/an (a) equilateral triangle (b) isosceles triangle (c) right triangle (d) right isosceles triangle 4. In the given figure, the area of DABC (in sq. units) is Y 4C 3 2 X¢ A1 D 5B X –4 –3 –2 –1 O 1 23 4 –1 –2 –3 Y¢ (a) 18 sq units (b) 12 sq units (c) 14 sq units (d) 16 sq units 5. If three points A(0, 0), B(3, 3) and C(3, l) form an equilateral triangle, then l is equal to (a) 2 (b) - 3 (c) - 4 (d) ± 3 6. Find the points on Y-axis which is equidistant from two points A(- 3, 4) and B(3, 6) on the same plane. (b) P(0, 5) THE QUALIFIER (a) P(0, - 5) (d) None of these (c) P(0, ± 5) 7. If the points A(1, - 1), B(5, y) and C(9, 5) are collinear, then the value of y is (a) 2 (b) 3 (c) - 2 (d) 4

18 CBSE Sample Paper Mathematics Standard Class X (Term I) 8. Find the equation of the perpendicular bisector of line segment joining points A(- 2, 3) and B(4, 7). (b) 3x - 2 y = 13 (a) 3x + 2 y = 13 (d) 3x + 2 y = - 13 (c) 2 x - 3y = 13 9. Find the ratio, in which the line segment joining points A(- 3, 10) and B(6, - 8) is divided by C(- 1, 6). (a) 3 : 4 (b) 2 : 7 (c) 7 : 2 (d) None of these 10. The vertices of a parallelogram in order are A(2, 3), B(3, y), C(x, 6) and D(4, 5). Then, (x, y) is equal to (b) (4, 5) (a) (- 5, 4) (d) (4, - 5) (c) (5, 4) 11. Find the coordinates of the centroid of a triangle, whose vertices are A(0, 6),B(8, 12) and C(8, 0). (b) çèæ136 , 6øö÷ (a) çæè136 , - 6øö÷ (c) (16, 2) (d) None of these 12. If the centroid of triangle formed by the points P(a, b), Q(b, c) and R(c, a) is a origin. Then, the value of a + b is equal to (a) 3c (b) c (c) - c (d) c 3 13. The point P which divides the line segment joining the points A(2, - 5) and B(5, 2) in the ratio 2 : 3, lies in the quadrant (a) IV (b) III (c) II (d) I 14. If vertices of a DABC are A(5, 1), B(1, 5) and C(- 3, - 1). Then, find the length of median AD. (a) 6 units (b) 35 units (c) 37 units (d) 41 units 15. Point P divides the line segment joining R(- 6, 10) and S(3, - 8) in the ratio l : 1. If point P lies on the line 2x - y + 4 = 0, then the value of l is (a) - 1 (b) 1 (c) 1 (d) - 1 2 2 THE QUALIFIER Answers 1. (b) 2. (b) 3. (d) 4. (c) 5. (d) For Detailed Solutions 6. (b) 7. (a) 8. (a) 9. (b) 10. (c) Scan the code 11. (b) 12. (c) 13. (a) 14. (c) 15. (b)

CBSE Sample Paper Mathematics Standard Class X (Term I) 19 5. Triangles Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct. Select the correct option as your answer. 1. In the figure, if MN||QR, PM = 6 cm, QM = 8 cm and QR = 28 cm, then MN is equal to P 6 cm M N 8 cm Q 28 cm R (a) 20 cm (b) 32 cm (c) 12 cm (d) 16 cm 2. In the given figure, DE||BC. If AD = x + 2, DB = x - 2, AE = x + 3 and EC = x - 4, then the value of x C E (a) 3 A DB 2 (b) 3 (c) 3 8 5 (d) - 2 3 3. In DPQR is such that PQ = 4 cm, QR = 3 cm PR = 3.5 cm. If DPQR ~ DXYZ and YZ = 5 cm, then perimeter of DXYZ is PX 4 cm THE QUALIFIER Q 3 cm R Y 5 cm Z3.5 cm (a) 17.5 cm (b) 16 cm (c) 18.5 cm (d) 22.5 cm 4. In the given figure ÐM = ÐN and PM = PN, then DPQR is MQ NR P (a) equilateral triangle MN (c) right triangle QR (b) isosceles triangle (d) None of these

20 CBSE Sample Paper Mathematics Standard Class X (Term I) 5. In the given figure ÐP = ÐT, PQ = 6 cm, QR = 15 cm, PR = 12 cm and RT = 4 cm, then the value of RS is P 6 cm 12 cm R S Q 15 cm 4 cm T (a) 6 cm (b) 5 cm (c) 8 cm (d) 9 cm 6. In the given figure ÐMNP = 90° and ON ^ MP. If ON = 12 cm and MO = 6 cm, then the value of OP is M 6 cm O 12 cm NP (a) 16 cm (b) 36 cm (c) 24 cm (d) 18 cm 7. In DABC,ÐA is obtuse angle, PB ^ PC and QC ^ QB, then the value of AB ´ AQ is PQ A (a) AP ´ AB B C (d) AQ ´ AP (b)AC ´ AP (c) AB ´ AC 8. If two coconut trees 15 m and 25 m high are 70 m apart, then the height of the point of intersection of the line joining the top of each tree to the foot of the opposite tree is A P R (a) 9 19 m B h (d) 9 22 m 56 SQ 56 (b) 9 20 m 56 (c) 9 21 m 56 THE QUALIFIER 9. Diagonal of a trapezium ABCD intersect each other at the point O, AB||CD and AB : CD = 2 : 3, then the ratio of the areas of DAOB and DCOD is DC O (a) 4 : 9 AB (c) 16 : 36 (b) 9 : 16 (d) None of these

CBSE Sample Paper Mathematics Standard Class X (Term I) 21 10. Two isosceles triangles have equal vertical angles and their area are in the ratio 49 : 64, then the ratio of their corresponding heights. (a) 8 : 7 (b) 9 : 6 (c) 7 : 9 (d) 7 : 8 11. In DABC, AD is bisector of ÐA, if BD = 6 cm, DC = 8 cm and AB = 6 cm, then AC is A 6 cm B 6 cm D 8 cm C (a) 12 cm (b) 4.5 cm (c) 7 cm (d) 6 cm 12. The hypotenuse of a right triangle is 6 m more than the twice of the shortest side. If the third side is 2 m less than the hypotenuse, then the sides of triangle is (a) 10, 12 and 14 (b) 10, 24 and 26 (c) 12, 13 and 15 (d) None of these 13. In the given figure, PR = 12 cm and PL = 8 cm, then QM is P 12cm 8cm M L Q 6 cm R (a) 6 cm (b) 9 cm (c) 4 cm (d) 5 cm 14. An equilateral triangle is inscribed in a circle of radius 8 cm, then the side of equilateral triangle is (a) 16 cm (b) 4 3 cm (c) 14 cm (d) 12 cm 15. In the given figure, AD ^ BC, then AC 2 + 2BC × BD is A (a) AB2 + BC2 BDC (c) AD2 (b) AB2 + AD2 (d) AB2 - BC2 Answers THE QUALIFIER 1. (c) 2. (d) 3. (a) 4. (b) 5. (b) For Detailed Solutions 6. (c) 7. (b) 8. (c) 9. (a) 10. (d) Scan the code 11. (d) 12. (b) 13. (c) 14. (b) 15. (a)

22 CBSE Sample Paper Mathematics Standard Class X (Term I) 6. Trigonometry Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct. Select the correct option as your answer. 1. If secq = 3, then the value of 2 cos2 q + 2 cot 2 q - 9 2 (a) - 91 (b) 91 (c) 44 (d) 90 15 15 15 44 2. If 3 cot A = 4 cos A, then the relation between sec A and tan A is (a) 4 sec A = 3 tan A (b) 3 sec2 A - 4 tan2 A = 0 (c) 4 sec2 A - 3 tan2 A = 0 (d) 3 sec A - 4 tan A = 0 3. In the given figure, ÐP = q and ÐR = f, R x +2 f x q Q P then the value of x 3 + x 2 tan q is (c) 2 (d) x 2 (a) - x2 (b) - x 2 x2 2 2 4. The positive minimum value of cosec q is (a) 0 (b) 1 (c) 2 (d) 1 2 5. If - x tan 45° sin 60° + cos 60°× cot 45° = 4, then the value of x is 5 (a) 3 - 3 (b) 10 5 - (c) 3 (d) 3 5 10 6. The value of (sin 30°× cos 60° + cos 30° sin × 60°) is (a) 2 (b) 1 (c) 1 2 THE QUALIFIER (d) 1 4 7. If DABC is right angled at C, then the value of sin(A + B) is (a) 1 (b) 0 (c) - 1 (d) 1 2

CBSE Sample Paper Mathematics Standard Class X (Term I) 23 8. In an acute angled DABC, if cot(A + B - C) = 1 and cosec(B + C - A) = 2, then the angles A, B and C is (b) 75° , 75° and 135° (a) 105° , 135° and 75° 22 22 (c) 75° , 75° and 135° (d) 105° , 135° and 75° 22 2 22 2 9. If cot q = 4, then the value of é 1 - sin 2 q ù 3 ê + cos 2 ú êë1 qûú (a) 16 (b) 16 41 35 (c) 25 (d) None of these 16 10. If cot q = 1, then the value of (1 + sin 2 q) - æèçç 1 + tan 2 q ÷ø÷ö cot 2 q (a) 1 (b) - 1 2 2 (c) 1 (d) 0 11. The value of + 1 + - 1 - 2 is cos q cos q 1 1 (a) 2 sec2 q (b) 2 cot 2 q (c) 2 secq (d) secq × tanq 12. If m = cosq - sin q and n = cosq + sin q, then the value of sec2 q is (m + n)2 (m + n)2 (a) (b) 2 4 4 (c) (m + n)2 (d) None of these 13. The value of 1 + tan 2 a is 1 + seca (a) seca (b) cosec a (c) cos a (d)cot a 14. If çèæ tan q - 1 ÷öø = x, then the value of èçæ tan 3 q - 1 q öø÷ is tan q tan 3 (a) x2(x + 2) (b) x(x 2 - 2 ) (c) x(x 2 + 3) (d) x(x 2 - 3) 15. The value of 2(sin 6 q + cos6 q) - 3(sin 4 q + cos4 q) + 1 is (a) 0 (b) 1 THE QUALIFIER (c) - 1 (d) 2 Answers 1. (a) 2. (d) 3. (d) 4. (b) 5. (b) For Detailed Solutions 6. (c) 7. (a) 8. (b) 9. (a) 10. (b) Scan the code 11. (b) 12. (c) 13. (a) 14. (c) 15. (a)

24 CBSE Sample Paper Mathematics Standard Class X (Term I) 7. Area Related to Circles Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct. Select the correct option as your answer. 1. If the area of circular field is 30800 m 2 , then the perimeter of the field is (a) 720 m (b) 360 2 m (c) 360 m (d) None of these 2. The radii of two circles are 14 cm and 7 cm, respectively, then the radius of the circle which has circumference equal to sum of the circumference of the two circles. (a) 21 cm (b) 19 cm (c) 18 cm (d) 20 cm 3. The number of revolution made by circular wheel of radius 1.4 m in rolling a distance 176 m is (a) 40 (b) 20 (c) 30 (d) 15 4. The ratio of the outer and inner circumference of a circular path is 24 : 23. If the path is 5 m wide, then the diameter of the inner circle is (a) 210 m (b) 220 m (c) 200 m (d) 230 m 5. The circumference of circle is 110 cm. The side of a square inscribed in the circle is (a) 35 cm (b) 35 2 cm 2 (c) 35 2 cm (d) 70 cm 2 6. The short and long hands of a clock are 10 cm and 12 cm along, respectively. The sum of the distances travelled by their tips in one days is (a) 1282 cm (b) 1789 cm (c) 1936 cm (d) 1848 cm 7. In a circle of radius 14 cm, an arc subtends an angle of 60° at the centre, then the length of the arc and area of sector are THE QUALIFIER O 14 cm 60° (a) 44 cm and 305 cm2 (b) 22 cm and 154 cm2 33 33 (c) 88 cm and 616 cm2 (d) None of these 33 8. The length of minute hand of a clock is 16 cm, then the area swept by the minute hand in 1 min is (a) 15.75 cm2 (b) 17.65 cm2 (c) 10.27 cm2 (d) 19.85 cm2

CBSE Sample Paper Mathematics Standard Class X (Term I) 25 9. In the given figure, sectors of two concentric circle of radii 14 cm and 7 cm are shown, then the area of the shaded region is 7 cm 30° 14 cm (a) 35.5 cm2 (b) 36.5 cm2 (c) 38.5 cm2 (d) 42.5 cm2 10. The minute hand of a clock is 12 cm long, then the area of the face of the clock described by the minute hand in 30 min is (a) 220 cm2 (b) 226.28 cm2 (c) 246 cm2 (d) 315 cm2 11. An chord 16 cm long is drawn in a circle whose radius is 16 cm. Then, the area of segment is O qq A M B 16 cm (b) 60 cm2 (a) 80 cm2 (c) 23.24 cm2 (d) 86.14 cm2 12. In the adjoining figure OACBO respresents a quadrants of a circle of radius 3.5 cm with centre O, then the area of shaded portion A C 3 cm (a) 4.37 cm2 OB (c) 10. 5 cm2 3.5 cm (b) 6.25 cm2 (d) None of these 13. In the given figure, ABCD is a square of side 8 cm with E, F, G and H as the mid-points of sides AB, BC, CD and DA respectively. Then, area of the shaded portion is E AB HF THE QUALIFIER (a) 32 cm2 DC (c) 98 cm2 G (b) 64 cm2 (d) 84 cm2

26 CBSE Sample Paper Mathematics Standard Class X (Term I) 14. In the given figure PQRS is a square of length 20 2 cm. If DPEQ is an isosceles triangle inscribed in the semi-circle with diameter PQ, then the area of the shaded region is PQ SE R (a) 328 cm 2 (b) 428 cm 2 (c) 600 cm 2 (d) 628 cm 2 15. In the given figure, ABCD is a square of side 8 cm and A, B, C and D are centres of equal circle touching externally in pairs, then the area of the shaded region. DC (a) 88 cm 2 AB 7 (b) 96 cm 2 (c) 78 cm 2 7 7 (d) 92 cm 2 7 Answers 1. (d) 2. (a) 3. (b) 4. (d) 5. (c) For Detailed Solutions 6. (c) 7. (d) 8. (c) 9. (c) 10. (b) Scan the code 11. (c) 12. (a) 13. (a) 14. (b) 15. (b) 8. Probability Directions (Q. Nos. 1-15) Each of the question has four options out of which only one is correct. Select the correct option as your answer. 1. Two fair coins are tossed, then the probability of getting at the least one head is (a) 3 (b) 1 4 4 (c) 1 (d) 3 2 8 THE QUALIFIER 2. In a simultaneous tossing of three coins, the probability of getting at most one head is (a) 3 (b) 1 4 2 (c) 3 (d) 2 8 3

CBSE Sample Paper Mathematics Standard Class X (Term I) 27 3. A card is drawn from a well shuffled deck of 52 cards, then the probability of getting the jack of heart is (a) 1 (b) 1 13 26 (c) 1 (d) None of these 52 4. Cards marked with numbers 5 to 75 are placed in a box and mixed thoroughly. One card is drawn from the box. Then, the probability that the number on the card is even is (a) 35 (b) 35 71 70 (c) 36 (d) 36 71 70 5. A die is thrown once, then the probability of getting a number lying between 3 and 6 is (a) 1 (b) 5 3 6 (c) 2 (d) 1 3 6 6. A die is thrown once, then the probability of getting a number which is not a factor of 30 is (a) 1 (b) 1 6 3 (c) 2 (d) None of these 3 7. A letter of English alphabets is chosen at random. Then, the probability that it is a letter of the word ‘ARIHANT’ is (a) 4 (b) 9 13 26 (c) 7 (d) 5 26 26 8. A child has a die whose six faces show the letters as given below ABCDE ZO The die is thrown once, then the probability of getting Z is (a) 2 (b) 1 3 3 (c) 1 (d) 1 6 2 9. An integer is chosen at random between 1 and 100, then the probability that it is THE QUALIFIER divisible by 8 is (a) 6 (b) 5 49 49 (c) 3 (d) 2 49 49

28 CBSE Sample Paper Mathematics Standard Class X (Term I) 10. The probability of guessing the correct answer to certain question is p . If the probability 10 of not guessing the correct answer to same questions is 1, then the value of p is 4 (a) 15 (b) 8 23 (c) 15 4 (d) None of these 11. A number x is chosen at random from the numbers -3, - 2, - 1, 0, 1, 2, 3, then the probability that x 2 £ 4 is (a) 5 (b) 4 7 7 (c) 5 (d) 6 7 7 12. Suppose, you drop a die at random on the square region shown in figure. What is the probability that it will land inside the circle of diameter 1 m ? 5m 5m (a) p (b) p 24 100 (c) p (d) None of these 40 13. 10 defective pens are accidentally mixed with 140 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot, the probability that the pen taken out is a good one is (a) 7 (b) 6 15 15 (c) 14 (d) 15 15 14 14. A box contains 80 discs, which are numbered from 1 to 80. If one disc is drawn at random from the box, then the probability that it bears a number divisible by 5 is (a) 2 (b) 8 5 45 (c) 3 (d) None of these 5 15. In a family of 3 children, then the probability of having atmost one boy is THE QUALIFIER (a) 5 (b) 1 8 8 (c) 7 (d) 3 88 Answers 1. (a) 2. (b) 3. (c) 4. (a) 5. (a) For Detailed Solutions 6. (a) 7. (c) 8. (c) 9. (a) 10. (a) Scan the code 11. (c) 12. (b) 13. (c) 14. (d) 15. (d)

CBSE Sample Paper Mathematics Standard Class X (Term I) 29 CBSE QUESTION BANK Case Study Based Questions Real Numbers (iv) 7 ´ 11 ´ 13 ´ 15 + 15 is a (a) Prime number 1. To enhance the reading skills of grade X (b) Composite number (c) Neither prime nor composite students, the school nominates you and two (d) None of the above of your friends to set up a class library. There are two sections- Section A and (v) If p and q are positive integers such that Section B of grade X. There are 32 students in p = ab2 and q = a2b, where a and b are section A and 36 students in section B. prime numbers, then the LCM (p, q) is (a) ab (b) a2b2 (c) a3b2 (d) a3b3 Sol. (i) (c) Given, number of students in Section A = 32 Number of students in Section B = 36 The minimum number of books acquire for the class library = LCM of (32, 36) =2 ´2 ´2 ´2 ´2 ´3 ´3 = 25 ´ 32 (i) What is the minimum number of books = 32 ´ 9 = 288 (ii) (b) Given, product of the two numbers you will acquire for the class library, so = LCM ´ HCF that they can be distributed equally \\ 32 ´ 36 = LCM (32, 36) ´ HCF (32, 36) Þ 32 ´ 36 = 288 ´ HCF (32, 36) among students of Section A or Section B? Þ HCF (32, 36) = 32 ´ 36 = 4 (a) 144 (b) 128 288 (iii) (a) The prime factors of 36 are (c) 288 (d) 272 36 = 2 ´ 2 ´ 3 ´ 3 = 22 ´ 32 (ii) If the product of two positive integers is equal to the product of their HCF and (iv) (b) 7 ´ 11 ´ 13 ´ 15 + 15 = 15 ´ (7 ´ 11 ´ 13 + 1) CBSE QUESTION BANK LCM is true, then the HCF (32, 36) is Since, the number is divisible by a number other than (a) 2 (b) 4 itself and 1. (c) 6 (d) 8 Hence, it is a composite number. (v) (b) Given, p = ab2 and q = a2b (iii) 36 can be expressed as a product of its primes as LCM ( p, q ) = Product of the greatest power of each (a) 22 ´ 32 (b) 21 ´ 33 prime factor involved in the numbers, with highest power = a2 ´ b2 (c) 23 ´ 31 (d) 20 ´ 30

30 CBSE Sample Paper Mathematics Standard Class X (Term I) 2. A seminar is being conducted by an (iii) (a) LCM of (60, 84, 108) = Product of the greatest power of each prime factor involved in the numbers Educational Organisation, where the with highest power participants will be educators of different = 22 ´ 33 ´ 5 ´ 7 = 4 ´ 27 ´ 35 = 3780 subjects. The number of participants in Hindi, English and Mathematics are 60, 84 (iv) (d) Now, HCF (60, 84, 108) ´ LCM (60, 84, 108) and 108 respectively. = 12 ´ 3780 = 45360 (v) (d) Number 108 can be expressed as a product of its prime as 22 ´ 33. (i) In each room the same number of 3. A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and ask for your help in completing a quiz for the audience. Observe the following factor tree and answer the following : x participants are to be seated and all of them being in the same subject, hence 5 2783 maximum number participants that can accommodated in each room are (a) 14 (b) 12 (c) 16 (d) 18 y 253 (ii) What is the minimum number of rooms 11 z required during the event? (a) 11 (b) 31 (c) 41 (d) 21 (i) What will be the value of x ? (iii) The LCM of 60, 84 and 108 is (a) 15005 (b) 13915 (a) 3780 (b) 3680 (c) 4780 (d) 4680 (c) 56920 (d) 17429 (iv) The product of HCF and LCM of 60, 84 (ii) What will be the value of y ? and 108 is (a) 23 (b) 22 (c) 11 (d) 19 (a) 55360 (b) 35360 (iii) What will be the value of z ? (c) 45500 (d) 45360 (a) 22 (b) 23 (c) 17 (d) 19 (v) 108 can be expressed as a product of its (iv) According to Fundamental Theorem of Arithmetic 13915 is a primes as (b) 23 ´ 33 (a) Composite number (a) 23 ´ 32 (b) Prime number (c) Neither prime nor composite (c) 22 ´ 32 (d) 22 ´ 33 (d) Even number Sol. (i) (b) Given, number of students in each subject are (v) The prime factorisation of 13915 is Hindi = 60, English = 84 and Mathematics = 108. The prime factors of each subject students are (a) 5 ´ 113 ´ 132 (b) 5 ´ 113 ´ 232 60 = 2 ´ 2 ´ 3 ´ 5 = 22 ´ 3 ´ 5 84 = 2 ´ 2 ´ 3 ´ 7 = 22 ´ 3 ´ 7 (c) 5 ´ 112 ´ 23 (d) 5 ´ 112 ´ 132 108 = 2 ´ 2 ´ 3 ´ 3 ´ 3 = 22 ´ 33 CBSE QUESTION BANK Sol. The maximum number of participants that can (i) (b) x = 5 ´ 2783 = 13915 accommodated in each room (ii) (c) We have, 2783 = y ´ 253 = HCF (60, 84, 108) Þ y = 2783 = 11 253 = Product of the smallest power of each common prime factor involved in the numbers (iii) (b) We have, 253 = 11 ´ z Þ z = 253 = 23 = 22 ´ 3 = 12 11 (ii) (d) The minimum number of rooms required during (iv) (a) Here, 13915 = 5 ´ 2783 Since, 13915 has factor other than 1 and the number the event is itself. It is a composite number. = Total number of participants Maximum participants in one room (v) (c) 13915 = 5 ´ 11 ´ 11 ´ 23 = 5 ´ 112 ´ 23 = 252 = 21rooms. 12

CBSE Sample Paper Mathematics Standard Class X (Term I) 31 Polynomials Sol. 1. The below picture are few natural examples (i) (c) In the standard form of quadratic polynomial ax2 + bx + c; ‘a’ is a non-zero real number, and b of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an and c are any real number. arch in the shape of a parabola. (ii) (d) In a quadratic polynomial, if roots are equal, then discriminant, D = 0. (iii) (b) Given, a and 1 are the zeroes of quadratic a polynomial 2 x2 - x + 8k. Now, product of zeroes, 1 Constant term a´ = a Coefficient of x2 Þ 1 = 8k Þ k = 2 = 1 2 84 (iv) (c) Given equation is x2 + 1 = 0. x2 = - 1 In structures, their curve represents an y= x2 +1 efficient method of load, and so can be (0, 1) found in bridges and in architecture in a variety of forms. Hence, it is neither touches nor intersects X-axis. (i) In the standard form of quadratic (v) (c) Given, sum of roots = - p polynomial, ax2 + bx + c, a, b and c are and product of roots = - 1 p (a) All are real numbers \\ Required quadratic polynomial (b) All are rational numbers. = k[ x2 - (Sum of roots) x + Product of roots] (c) a is a non-zero real number, b and c are = k é x2 - (- p)x + æ 1 öù = k æ x2 + px - 1ö any real numbers. ê ç- ÷ú ç ÷ (d) All are integers. è p øû ë è pø (ii) If the roots of the quadratic polynomial 2. An asana is a body posture, originally and are equal, where the discriminant D = b2 - 4ac, then still a general term for a sitting meditation pose, and later extended in hatha yoga and (a) D > 0 (b) D < 0 modern yoga as exercise, to any type of pose or position, adding reclining, standing, (c) D ³ 0 (d) D = 0 inverted, twisting, and balancing poses. In the figure, one can observe that poses (iii) If a and 1 are the zeroes of the quadratic can be related to representation of quadratic a polynomial. polynomial 2x2 - x + 8k, then k is (a) 4 (b) 1 (c) -1 (d) 2 4 4 (iv) The graph of x2 + 1 = 0 (a) Intersects X-axis at two distinct points. CHAKRASANA CBSE QUESTION BANK (b) Touches X-axis at a point. TRIKONASANA (c) Neither touches nor intersects X-axis. (d) Either touches or intersects X-axis. (v) If the sum of the roots is -p and product of the roots is - 1 , then the quadratic p polynomial is (a) k æ -px 2 + x + ö (b) k æ px 2 - x ö ç p 1÷÷ ç p - 1÷÷ ç ø ç è è ø (c) k æ x 2 + px - 1 ö (d) k æ x 2 - px + 1 ö çç p ÷÷ çç p ÷÷ è ø è ø

32 CBSE Sample Paper Mathematics Standard Class X (Term I) (i) The shape of the poses shown is (v) (b) Let p( x) = 4 3x2 + 5x - 2 3 (a) Spiral = 4 3x2 + (8 - 3)x - 2 3 (b) Ellipse [by splitting middle term] (c) Linear = 4 3 x2 + 8x - 3x - 2 3 (d) Parabola = 4x( 3x + 2) - 3( 3x + 2) (ii) The graph of parabola opens = (4x - 3)( 3x + 2) downwards, if ………… . For finding the zeroes, put p( x) = 0 (a) a ³ 0 (b) a = 0 \\ (4x - 3)( 3x + 2) = 0 (c) a < 0 (d) a > 0 Þ 4x - 3 = 0 and 3x + 2 = 0 (iii) In the graph, how many zeroes are there Þ x = 3 and x = - 2 for the polynomial? 43 –2 4 3. Basketball and soccer are played with a 1 spherical ball. Even though an athlete (a) 0 (b) 1 (c) 2 (d) 3 dribbles the ball in both sports, a basketball player uses his hands and a soccer player (iv) The two zeroes in the below shown graph uses his feet. Usually, soccer is played are outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial. Y X¢ X –2 –1 1 2 3 4 Y¢ (a) 2, 4 (b) - 2, 4 (c) - 8, 4 (d) 2, - 8 (v) The zeroes of the quadratic polynomial 4 3x2 + 5x - 2 3 are CBSE QUESTION BANK (a) 2 , 3 (i) The shape of the path traced shown is 34 (b) - 2 , 3 34 (c) 2 , - 3 34 (d) - 2 , - 3 34 Sol. (i) (d) The shape of given poses are parabolic. (a) Spiral (ii) (c) The graph of parabola opens downwards, if a < 0. (b) Ellipse (c) Linear (iii) (c) Number of zeroes is equal to number of times (d) Parabola intersects parabola on the X-axis. (ii) The graph of parabola opens \\ Number of zeroes = 2. downwards, if ………… . (iv) (b) The curve intersect X-axis at points x = - 2 and x = 4. (a) a = 0 (b) a < 0 Hence, two zeroes in the given graph are - 2 and 4. (c) a > 0 (d) a ³ 0

CBSE Sample Paper Mathematics Standard Class X (Term I) 33 (iii) Observe the following graph and answer. Pair of Linear Equations in Two Variables 6 1. A test consists of ‘True’ or ‘False’ questions. –4 –3 –2 –1 2 2 34 One mark is awarded for every correct 1 answer while 1 mark is deducted for every –2 4 wrong answer. A student knew answers to –6 some of the questions. Rest of the questions he attempted by guessing. He answered 120 questions and got 90 marks. Type of Marks given for Marks deducted question correct answer for wrong answer In the above graph, how many zeroes are True/False 1 0.25 there for the polynomial? (a) 0 (b) 1 (i) If answer to all questions he attempted by (c) 2 (d) 3 guessing were wrong, then how many questions did he answer correctly? (iv) The three zeroes in the above shown graph are (ii) How many questions did he guess? (a) 2, 3, - 1 (b) - 2, 3, 1 (iii) If answer to all questions he attempted by guessing were wrong and answered 80 (c) - 3, - 1, 2 (d) -2, - 3, - 1 correctly, then how many marks he got? (v) What will be the expression of the (iv) If answer to all questions he attempted by polynomial? guessing were wrong, then how many (a) x3 + 2x2 - 5x - 6 (b) x3 + 2x2 - 5x + 6 questions answered correctly to score 95 (c) x3 + 2x2 + 5x - 6 marks? (d) x3 + 2x2 + 5x + 6 Sol. Let the number of questions whose answer is known to the student be x and questions attempted by Sol. guessing be y. (i) (d) The shape of the path traced shown in the given Then, x + y = 120 … (i) figure is the form of parabola. and x - 1 y = 90 Þ 4x - y = 360 … (ii) (ii) (b) The graph of parabola opens downwards, if a < 0. 4 (iii) (d) In the given graph, we see that curve intersect the On adding Eqs. (i) and (ii), we get X-axis at three points. Hence, number of zeroes in the 5x = 480Þ x = 480 = 96 given polynomial are 3. 5 (iv) (c) The given curve intersect the X-axis at points Put x = 96 in Eq. (i), we get x = - 3, -1and 2. Hence, three zeroes in the given graph are -3, - 1, 2. 96 + y = 120 Þ y = 120 - 96 = 24 (v) (a) Since, given polynomial has three zeroes. (i) He answered 96 questions correctly. So, it will be a cubic polynomial. Now, sum of zeroes = - 3 - 1 + 2 = - 2 (ii) He guesses only 24 questions. Sum of product of two zeroes at a time = - 3 ´ (-1) + (-1) ´ 2 + 2 ´ (-3) (iii) In out of 120 questions attempted 80 answered are CBSE QUESTION BANK =3-2 -6=-5 correct and 40 guessing answered are wrong. and product of all zeroes = - 3 ´ - 1 ´ 2 Then, he got the marks = 80 - 1 of 40 =6 4 \\ Required cubic polynomial = 80 - 1 ´ 40 = 80 - 10 = 70 = x3 - (Sum of zeroes) x2 4 (iv) According to the given condition, x - 1 of (120 - x) = 95Þ x - 1 ´ (120 - x) = 95 44 + (Sum of product of two zeroes at a time) x Þ 4x - 120 + x = 380 - (Product of three zeroes) Þ 5x = 500 = x3 - (-2) x2 + (-5)x - (6) Þ x = 100 = x3 + 2 x2 - 5x - 6 Hence, he answered correctly 100 questions to score 95 marks.

34 CBSE Sample Paper Mathematics Standard Class X (Term I) 2. Amit is planning to buy a house and the But, it is also given, the cost of laying tiles in kitchen at the rate of `50 per m2. layout is given below. The design and the measurement has been made such that \\Total cost of laying tiles in the kitchen = 35 ´ 50 areas of two bedrooms and kitchen together is 95 sq m. = ` 1750 x2 y 3. It is common that Governments revise travel 5m Bedroom 1 Bath Kitchen fares from time to time based on various room factors such as inflation ( a general increase in prices and fall in the purchasing value of 2m Living room money) on different types of vehicles like 5m auto, Rickshaws, taxis, Radio cab etc. The Bedroom 2 auto charges in a city comprise of a fixed charge together with the charge for the 15 m distance covered. Study the following situations Based on the above information, answer the following questions: (i) Form the pair of linear equations in two variables from this situation. (ii) Find the length of the outer boundary of the layout. (iii) Find the area of each bedroom and Name of Distance travelled Amount kitchen in the layout. the city (km) paid (in `) (iv) Find the area of living room in the layout. City A 10 75 15 110 (v) Find the cost of laying tiles in kitchen at City B 8 91 the rate of ` 50 per sq m. 14 145 Sol. (i) From the given figure we see that area of two Situation 1 In city A, for a journey of 10 km, bedrooms = 2(5x) = 10x m2 the charge paid is ` 75 and for a journey of 15 Area of kitchen = 5 ´ y = 5y m2 km, the charge paid is ` 110. According to the question, Situation 2 In a city B, for a journey of 8 km, the charge paid is ` 91 and for a journey of 14 Area of the two bedrooms and area of kitchen km, the charge paid is ` 145. = 95 Refer situation 1 \\ 10x + 5y = 95 (i) If the fixed charges of auto rickshaw be `x and the running charges be ` y km/h, Þ 2 x + y = 19 [divide both sides by 5] … (i) the pair of linear equations representing the situation is Also, length of the home = 15 cm (a) x + 10y = 110, x + 15y = 75 (b) x + 10y = 75, x + 15y = 110 \\ x + 2 + y = 15 Þ x + y = 13 … (ii) (c) 10x + y = 110, 15x + y = 75 (d) 10x + y = 75, 15x + y = 110 Hence, pair of linear equations is (ii) A person travels a distance of 50 km. 2 x + y = 19 and x + y = 13 The amount he has to pay is (a) ` 155 (ii) The length of the outer boundary of the layout (b) ` 255 (c) ` 355 = 2(l + b) =2(15 + 12) (d) ` 455 CBSE QUESTION BANK = 2(27) = 54 m (iii) On solving Eqs. (i) and (ii), we get x = 6 and y = 7 \\Area of each bedroom = 5 ´ x = 5 ´ 6 = 30 m2 and area of kitchen = 5 ´ y = 5 ´ 7 = 35 m2 (iv) Area of living room = 15 ´(5 + 2) - Area of bedroom 2 = 15 ´ 7 - 5 ´ 6 = 105 - 30 = 75 m2 (v) Since, area of kitchen = 5 ´ y = 5 ´ 7 = 35 m2

CBSE Sample Paper Mathematics Standard Class X (Term I) 35 Refer situation 2 (ii) (c) On solving the above equations, we get (iii) What will a person have to pay for x + 10y = 75 travelling a distance of 30km ? x + 15y = 110 -- - (a) ` 185 (b) ` 289 - 5y = - 35Þ y = 7 (c) ` 275 (d) ` 305 \\ x + 10 ´ 7 = 75 (iv) The graph of lines representing the Þ x = 75 - 70 = 5 conditions are: (situation 2) To travel a distance of 50 km, a person has to pay Y amount = x + 50y = 5 + 50 ´ 7 25 (20, 25) = 5 + 350 = ` 355 20 (iii) (b) As per the situation 2, the pair of linear is (a) 15 x + 8y = 91 … (i) 10 and x + 14y= 145 … (ii) 5 (0, 5) (30, 5) On solving Eqs. (i) and (ii), we get X¢ X x + 8y = 91 5 10 15 20 25 30 35 0 x + 14y = 145 -- - –5 –10 - 6y = - 54 Y¢ Þ y=9 Put y = 9 in Eq. (i), we get Y 25 x + 8 ´ 9 = 91 Þ x = 91 - 72 = 19 20 To travel a distance of 30 km, a person has to pay amount = x + 30 ´ y (b) 15 X¢ = 19 + 30 ´ 9 = 19 + 270 = ` 289 10 (0, 10) (20, 10) (iv) (c) In situation 2, the intersection point of two lines is 5 X (19, 9), which is shown in figure (c). (12.5, 0) Similar Triangles 0 5 10 15 20 25 30 35 –5 1. Vijay is trying to find the average height of a –10 (5, –10) (25, –10) tower near his house. He is using the properties of similar triangles.The height of Y¢ Vijay’s house, if 20 m when Vijay’s house casts a shadow 10m long on the ground. (c) Y X X¢ 25 At the same time, the tower casts a shadow 50 m long on the ground and the house of 20 Ajay casts 20 m shadow on the ground. 15 10 (11, 10) (19, 9) (5, 10) 5 0 5 10 15 20 25 30 35 –5 Y¢ Y 25 20 CBSE QUESTION BANK (d) 15 (15, 15) X¢ (0, 10) (35, 10) 10 Vijay’s Tower house 5 Ajay’s house 0 5 10 15 20 25 30 35 X (i) The height of the tower is –5 (15, 5) (a) 20 m (b) 50 m Y¢ (c) 100 m (d) 200 m Sol. (ii) What will be the length of the shadow of (i) (b) As per the situation 1, the pair of linear equation the tower when Vijay’s house casts a representing the situation is x + 10y = 75 shadow of 12 m? and x + 15y = 110 (a) 75 m (b) 50 m (c) 45 m (d) 60 m

36 CBSE Sample Paper Mathematics Standard Class X (Term I) (iii) What is the height of Ajay’s house? 2. Rohan wants to measure the distance of a (a) 30 m (b) 40 m pond during the visit to his native. He marks points A and B on the opposite edges of a (c) 50 m (d) 20 m pond as shown in the figure below. To find the distance between the points, he makes a (iv) When the tower casts a shadow of 40 m, right-angled triangle using rope connecting Bwith another point C are a distance of 12 m, same time what will be the length of the connecting C to point D at a distance of 40 m from point C and the connecting D to the shadow of Ajay’s house? point A which at distance of 30 m from D such the Ð ADC = 90°. (a) 16 m (b) 32 m A B 12 m C (c) 20 m (d) 8 m (v) When the tower casts a shadow of 40 m, same time what will be the length of the shadow of Vijay’s house? (a) 15 m (b) 32 m (c) 16 m (d) 8 m Sol. (i) (c) Let CD = h m be the height of the tower. Let 30 m BE = 20 m be the height of Vijay’s house and GF be 40 m the height of Ajay’s house. D D E G (i) Which property of geometry will be used 20 m to find the distance AC ? h (a) Similarity of triangles (b) Thales Theorem A 10 m B C F 20 m H (c) Pythagoras Theorem (d) Area of similar triangles 50 m 50 m (ii) What is the distance AC ? DACD ~ ABE (a) 50 m (b) 12 m AC CD (c) 100 m (d) 70 m \\= (iii) Which is the following does not form a AB EB Þ 50 h Þ h = 100 m Pythagoras triplet? = 10 20 (a) (7,24,25) (b) (15,8,17) (ii) (d) Given AB = 12 m, let AC = h (c) (5,12,13) (d) (21,20,28) In similar DABE and DACD, (iv) Find the length AB? (b) 38 m (a) 12 m (d) 100 m AB BE 12 20 (c) 50 m = Þ= (v) Find the length of the rope used. AC CD h 100 Þ h = 12 ´ 100 = 12 ´ 5 = 60 m 20 (iii) (b) Let height of Ajay’s house be GF = h1 (a) 120 m (b) 70 m Since, DHFG ~ DHCD (c) 82 m (d) 22 m HF FG Sol. \\= (i) (c) To find the distance AC in the given figure, we HC CD use Pythagoras theorem. Þ 20 h1 (ii) (a) In right DADC, use Pythagoras theorem, = 50 100 Þ h1 = 20 ´ 100 = 40 m AC = ( AD)2 + (CD)2 = (30)2 + (40)2 50 = 900 + 1600 = 2500 = 50 m CBSE QUESTION BANK (iv) (a) Given, HC = 40 cm (iii) (d) (a) Now, 242 + 72 = 576 + 49 = 625 = (25)2, Let length of the shadow of Ajay’s hour be HF = l m which forms a Pythagoras triplet (b) (15)2 + (8)2 = 225 + 64 = 289 = (17)2, Since, DHFG ~ DHCD \\ HF = FG which forms a Pythagoras triplet Þ HC CD (c) (12)2 + (5)2 = 144 + 25 = 169 = (13)2, l = 40 Þ l = 40 ´ 40 = 16 m 40 100 100 which form a Pythagoras triplet. (d) (20)2 + (21)2 = 400 + 441= 881¹ (28)2, (v) (d) Given, AC = 40 cm Let length of the shadow of Vijay’s house be AB = l m which does not form a Pythagoras triplet. Since, DABE ~ ACD (iv) (b) Since, AC = 50 m \\ \\ AB = AC - BC = 50 - 12 = 38 m AB = EB Þ AC CD (v) (c) The length of the rope used = BC + CD + DA = 12 + 40 + 30 = 82 m l = 20 Þ h = 20 ´ 40 = 8 m 40 100 100

CBSE Sample Paper Mathematics Standard Class X (Term I) 37 3. A scale drawing of an object is the same (v) The length of AB in the given figure shape at the object but a different size. The A scale of a drawing is a comparison of the length used on a drawing to the length it x cm 3 cm C represents. The scale is written as a ratio. B The ratio of two corresponding sides in 6 cm E similar figures is called the scale factor. 4 cm D Scale factor Length in range = Corresponding length in object If one shape can become another using (a) 8 cm (b) 6 cm revising, then the shapes are similar. Hence, two shapes are similar when one can (c) 4 cm (d) 10 cm become the other after a resize, flip, slide or turn. In the photograph below showing Sol. Given, scale factors = 1: 200 the side view of a train engine. Scale factor is 1 : 200. It means that length of 1 cm on the photograph above corresponds to a length of 200 cm (or 2 m) of the actual engine. (i) (d) Since, length of the model is 11 cm. Therefore, the overall length of the engine = 11 ´ 200 = 2200 cm = 22 m (ii) (d) The similarity of any two polygons will affect that they are not the mirror image of one another. This means that a length of 1 cm on the (iii) (a) The actual width of the door = 0.35 ´ 200 cm photograph above corresponds to a length of 200 cm or 2 m, of the actual engine. The = 70 cm scale can also be written as the ratio of two lengths. = 0.7 m (i) If the length of the model is 11cm, then (iv) (b) If two similar triangles have a scale factor 5 : 3, then their altitudes have a ratio 25 : 15. the overall length of the engine in the (v) (c) In the given BC|| DE. photograph above, including the \\ DABC ~ ADE, couplings(mechanism used to connect) is AB BC (a) 22 cm (b) 220 cm Þ= AD DE (c) 220 m (d) 22 m x3 (ii) What will affect the similarity of any two Þ= polygons? x+ 4 6 (a) They are flipped horizontally Þ x =1 x+ 4 2 (b) They are dilated by a scale factor Þ 2x = x + 4 (c) They are translated down Þ x = 4 cm (d) They are not the mirror image of one Coordinate Geometry another. (iii) What is the actual width of the door, if the 1. In order to conduct Sports Day activities in CBSE QUESTION BANK width of the door in photograph is 0.35 your School, lines have been drawn with chalk powder at a distance of 1 m each, in a cm? rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of (a) 0.7 m (b) 0.7 cm 1 m from each other along AD, as shown in given figure below. Niharika runs 1/4 th the (c) 0.07 cm (d) 0.07 m distance AD on the 2nd line and posts a green flag. Preet runs 1 th distance AD on (iv) If two similar triangles have a scale factor 5 : 3 which statement regarding the two 5 triangles is true? the eighth line and posts a red flag. (a) The ratio of their perimeters is 15 : 1 (b) Their altitudes have a ratio 25 :15 (c) Their medians have a ratio 10 : 4 (d) Their angle bisectors have a ratio 11 : 5

38 CBSE Sample Paper Mathematics Standard Class X (Term I) D C (iii) (c) Distance between these flags = GR = (8 - 2)2 + (25 - 20)2 = (6)2 + (5)2 = 36 + 25 = 61 m (iv) (a) The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. G Let this point be A( x, y). R x = 2 + 8, y = 25 + 20 22 x = 10 = 5, y = 45 = 22.5 22 Hence, A( x, y) =(5, 22.5) (v) (a) Let the point at which Joy post his flag be B( x, y). 2 1:3 1 G(2, 25) B R(8, 20) A B Then, x = 1 ´ 8 + 3 ´ 2, y = 1 ´ 20 + 3 ´ 25 1 2 3 4 5 6 7 8 9 10 (i) Find the position of green flag 1+ 3 1+ 3 (a) (2, 25) (b) (2, 0.25) x = 14 = 3.5, y = 95 = 23.75 » 24 44 (c) (25, 2) (d) (0, - 25) Hence, B( x, y) = (3.5, 24) (ii) Find the position of red flag (a) (8, 0) (b) (20, 8) Areas Related to Circles (c) (8, 20) (d) (8, 0.2) 1. Pookalam is the flower bed or flower (iii) What is the distance between both the pattern designed during Onam in Kerala. It is similar as Rangoli in North India flags? and Kolam in Tamil Nadu. During the festival of Onam , your school is planning to (a) 41 m (b) 11 m conduct a Pookalam competition. Your friend who is a partner in competition, (c) 61 m (d) 51 m suggests two designs given below. Observe these carefully. (iv) If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? (a) (5, 22.5) (b) (10, 22) A A (c) (2, 8.5) (d) (2.5, 20) B (v) If Joy has to post a flag at one-fourth distance from green flag ,in the line segment joining the green and red flags, then where should he post his flag? BC D (a) (3.5, 24) (b) (0.5, 12.5) C (c) (2.25, 8.5) (d) (25, 20) Design I This design is made with a circle of radius 32 cm leaving equilateral triangle Sol. ABC in the middle as shown in the given figure. CBSE QUESTION BANK (i) (a) It can be observed that Niharika posted the green flag at 1th distance of AD Design II This Pookalam is made with 4 9 circular design each of radius 7 cm. i.e. 1 ´ 100 = 25 m from the starting point 4 of 2nd line. Refer Design I Therefore, the coordinates of this point G is (i) The side of equilateral triangle is (2, 25). (ii) (c) Preet posted red flag at 1th distance of AD, (a) 12 3 cm (b) 32 3 cm 5 (c) 48 cm (d) 64 cm i.e. 1 ´ 100 = 20 m from the starting point of 8th line. (ii) The altitude of the equilateral triangle is 5 (a) 8 cm (b) 12 cm Therefore the coordinates of this point R is (8, 20). (c) 48 cm (d) 52 cm

CBSE Sample Paper Mathematics Standard Class X (Term I) 39 Refer Design II Now, area of one circle = pr2 = p (7)2 = 22 ´ (7)2 = 154 cm2 (iii) The area of square is (b) 1764 cm2 7 (a) 1264 cm2 (d) 1944 cm2 (c) 1830 cm2 \\Area of nine circles = 9 ´ 154 = 1386 cm2 Area of square ABCD = (Side)2 = (42)2 = 1764 cm2 (iv) Area of each circular design is Hence, area of the remaining portion of the handkerchief (a) 124 cm2 (b) 132 cm2 = Area of square - Area of nine circles = 1764 - 1386 = 378 cm2 (c) 144 cm2 (d) 154 cm2 (v) Area of the remaining portion of the (iii) (b) 1764 cm 2 (iv) (d) 154 cm2 square ABCD is (b) 260 cm2 (v) (a) 378 cm2 (a) 378 cm2 (d) 278 cm2 (c) 340 cm2 2. A brooch is a small piece of jewellery which Sol. has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some Directions (i-ii) brooch are shown below. Observe them carefully. Given, radius of the circle = 32 cm Let the side of the equilateral DABC be a cm. Let h be the height of the triangle. A a a O h AB B a/2 D a/2 C We know that in an equilateral triangle, centroid and circumcentre coincide. \\ AO = 2 h cm 3 [Q centroid divides the median in the ratio 2 : 1] which is equal to the radius of circle. …(i) Design A Brooch A is made with silver wire \\ 2 h = 32 Þ h = 48 cm in the form of a circle with diameter 28 mm. The wire used for making 4 diameters which 3 divide the circle into 8 equal parts. Now, we draw a perpendicular from vertex A to side BC which bisects BC at D. In right angled DADB, Design B Brooch b is made two colours_ Gold and silver. Outer part is made with AB2 = BD2 + AD2 [by Pythagoras theorem] Gold. The circumference of silver part is 44 mm and the gold part is 3mm wide Þ a2 = æ a ö2 + h2 Þ h2 = a2 - a2 = 3a2 everywhere. ç ÷ è2 ø 4 4 Þ (48)2 = 3a2 [from Eq. (i)] CBSE QUESTION BANK 4 Refer to Design A Þ a2 = 3072 (i) The total length of silver wire required is Þ a = 3072 [taking positive square root] = 32 3 cm (a) 180 mm (b) 200 mm (i) (b) 32 3 cm (c) 250 mm (d) 280 mm (ii) (c) 48 cm (ii) The area of each sector of the brooch is Directions (iii-v) (a) 44 mm2 (b) 52 mm2 (c) 77 mm2 (d) 68 mm2 Given, radius of each circle, r = 7 cm \\Diameter of circle, d = 14 cm [Q diameter = 2 ´ radius] Refer to Design B In the given figure, horizontal three circles touch each other. \\Length of a side of square = 3 ´ Diameter of one circle (iii) The circumference of outer part (golden) is = 3 ´ 14 = 42 cm (a) 48.49 mm (b) 82.2 mm (c) 72.50 mm (d) 62.86 mm

40 CBSE Sample Paper Mathematics Standard Class X (Term I) (iv) The difference of areas of golden and Probability silver parts is 1. On a weekend Rani was playing cards with (a) 18p mm2 (b) 44p mm2 her family. The deck has 52 cards.If her brother drew one card. (c) 51p mm2 (d) 64p mm2 (v) A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 p mm ? (a) 2 (b) 3 (c) 4 (d) 5 Sol. (i) (b) Given, diameter of circle, d = 28 mm \\Circumference of circle = pd [Qd = 2r] (i) Find the probability of getting a king of = 22 ´ 28 = 88 mm red colour. 7 (a) 1 (b) 1 (c) 1 (d) 1 Now, length of 4 diameters = 4 ´ 28 = 112 mm 26 13 52 4 Total length of the silver wire = pd + 4d (ii) Find the probability of getting a face card. = 88 + 112 = 200 mm (a) 1 (b) 1 (c) 2 (d) 3 26 13 13 13 (ii) (c) Here, we see that total circle is divided into 8 parts. (iii) Find the probability of getting a jack of \\ Area of each sector = 1 ´ Area of circle 8 hearts. = 1 ´ pr2 8 (a) 1 (b) 1 (c) 3 (d) 3 = 1 ´ 22 ´ 14 ´ 14 26 52 52 26 87 (iv) Find the probability of getting a red face = 77 mm2 card. Directions (iii-iv) (a) 3 (b) 1 (c) 1 (d) 1 13 13 52 4 (v) Find the probability of getting a spade. r (a) 1 (b) 1 (c) 1 (d) 1 26 13 52 4 Silver 3 mm Sol. Total number of cards in one deck of cards is 52. R Gold \\ Total number of outcomes = 52 We have, circumference of silver part = 44 mm (i) (a) Let E1 = Event of getting a king of red colour \\ Number of outcomes favourable to E1 = 2 \\ 2pr = 44 [Q there are four kings in a deck of playing cards out of which two are red and two are black] Þ r = 44 = 7 mm Hence, probability of getting a king of red colour, 22 2 P(E1) = 2 = 1 ´ 52 26 7 \\ R = r + 3=7 + 3 (ii) (d) Let E2 = Event of getting a face card \\ Number of outcomes favourable to E2 = 12 CBSE QUESTION BANK = 10 mm [Q in a deck of cards, there are 12 face cards, namely 4 kings, (iii) (d) Circumference of golden part = 2pR 4 jacks, 4 queens] = 2 ´ 22 ´ 10 7 = 62.86 mm Hence, probability of getting a face card, (iv) (c) Difference of areas = pR2 - pr2 = p(R2 - r2 ) P(E2 ) = 12 = 3 = (102 - 72 ) p = 51p mm2 52 13 (v) (c) Required number of revolutions (iii) (b) Let E3 = Event of getting a jack of heart \\ Number of outcomes favourable to E3 = 1 Distance covered [Q there are four jack cards in a deck, = namely 1 of heart, 1 of club, 1 of spade and 1 of diamond] Circumference 80p 80p =4 == Hence, probability of getting a jack of heart, 2pR 2p ´ 10

CBSE Sample Paper Mathematics Standard Class X (Term I) 41 P(E3 ) = 1 the top face of the dice is less than or 52 equal to 12 ? (b) 5 (c) 1 (iv) (*) Let E4 = Event of getting a red face card. (a) 1 36 18 (d) 0 \\Number of outcomes favourable to E4 = 6 [Q in a deck of cards, there are 12 face cards (iv) Rahul got next chance. What is the out of which 6 are red cards] probability that he got the sum of the two Hence, probability of getting a red face card, numbers appearing on the top face of the 6 3 P(E4 ) = 52 = 26 dice is equal to 7 ? (v) (d) Let E5 = Event of getting a spade (a) 5 (b) 5 (c) 1 (d) 0 \\ Number of outcomes favourable to E5 = 13 9 36 6 [Q in a deck of cards, there are 13 spades, 13 clubs, 13 hearts and 13 diamonds] (v) Now it was Ravi’s turn. He rolled the dice. What is the probability that he got Hence, probability of getting a spade, the sum of the two numbers appearing on P(E5 ) = 13 = 1 the top face of the dice is greater than 8 ? 52 4 (a) 1 (b) 5 (c) 1 (d) 5 36 18 18 2. Rahul and Ravi planned to play Business Sol. Total number of out comes = 6 ´ 6 = 36 (board game) in which they were supposed to use two dice. (i) (b) Let E1 = Event of getting sum 8 \\Number of favourable outcomes to E1 = 5 i.e. (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) \\ P(E1) = 5 36 (ii) (d) Let E2 = Event of getting sum 13. Since, we can’t get sum more than 12. \\ P(E2 ) = 36 =0 0 (iii) (a) Let E3 = Event of getting sum less than or equal to 12. \\Number of favourable outcomes = 36 (i) Ravi got first chance to roll the dice. What As sum of all the out comes is less than or equal to 12. is the probability that he got the sum of \\ P(E3 ) = 36 = 1 36 the two numbers appearing on the top face of the dice is 8? (iv) (c) Let E4 = Event of getting sum 7. (a) 1 (b) 5 (c) 1 (d) 0 \\Number of favourable outcomes to E2 = 6 26 36 18 (ii) Rahul got next chance. What is the i.e. (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) (6, 1) probability that he got the sum of the two \\ P(E4 ) = 6 = 1 36 6 numbers appearing on the top face of the dice is 13? (v) (d) Let E5 = Event of getting sum greater than 8 i.e. getting sum equal to 8, 9, 10, 11, 12 (a) 1 (b) 5 (c) 1 (d) 0 CBSE QUESTION BANK 36 18 \\Favourable outcomes = 10 i.e. (3, 6), (4, 5), (5, 4), (iii) Now it was Ravi’s turn. He rolled the (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6). dice. What is the probability that he got the sum of the two numbers appearing on \\ P(E5 ) = 10 = 5 36 18

42 CBSE Sample Paper Mathematics Standard Class X (Term I) Latest CBSE SAMPLE PAPER Latest Sample Question Paper for Class XII (Term I) Issued by CBSE on 2 Sept 2021 Mathematics Class 10 (Term I) Time : 90 Minutes Instructions 1. The question paper contains three parts A, B and C. 2. Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 3. Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 4. Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions. 5. There is no negative marking. Maximum Marks : 40 Section A Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted. 1. The ratio of LCM and HCF of the least composite and the least prime numbers is (a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) 1 : 3 2. The value of k for which the lines 5x + 7y = 3 and 15x + 21y = k coincide is (a) 9 (b) 5 (c) 7 (d) 18 3. A girl walks 200m towards East and then 150m towards North. The distance of the girl from the starting point is Latest CBSE SAMPLE PAPER (a) 350 m (b) 250 m (c) 300 m (d) 225 m 4. The lengths of the diagonals of a rhombus are 24cm and 32cm, then the length of the altitude of the rhombus is (a) 12 cm (b) 12.8 cm (c) 19 cm (d) 19.2 cm 5. Two fair coins are tossed. What is the probability of getting at the most one head? (a) 3 (b) 1 (c) 1 (d) 3/8 4 4 2 6. DABC ~ DPQR. If AM and PN are altitudes of DABC and DPQR respectively and AB 2 : PQ2 = 4 : 9, then AM : PN = (a) 16 : 81 (b) 4 : 9 (c) 3 : 2 (d) 2 : 3 7. If 2 sin 2 b - cos2 b = 2, then b is (a) 0° (b) 90° (c) 45° (d) 30°

CBSE Sample Paper Mathematics Standard Class X (Term I) 43 8. Prime factors of the denominator of a rational number with the decimal expansion 44.123 are (a) 2, 3 (b) 2, 3, 5 (c) 2, 5 (d) 3, 5 9. The lines x = a and y = b are (b) parallel (d) None of these (a) intersecting (c) overlapping 10. The distance of point A(- 5, 6) from the origin is (a) 11 units (b) 61 units (c) 11 units (d) 61 units 11. If a 2 = 23 / 25, then a is (a) rational (b) irrational (c) whole number (d) integer 12. If LCM (x, 18) = 36 and HCF (x, 18) = 2, then x is (a) 2 (b) 3 (c) 4 (d) 5 13. In DABC right angled at B, if tan A = 3, then cos A cosC - sin A sin C is equal to (a) - 1 (b) 0 (c) 1 (d) 3 / 2 14. If the angles of DABC are in ratio 1 : 1 : 2, respectively (the largest angle being angle C), then the value of sec A - tan A is cosec B cot B (a) 0 (b) 1/2 (c) 1 (d) 3 / 2 15. The number of revolutions made by a circular wheel of radius 0.7 m in rolling a distance of 176 m is (a) 22 (b) 24 (c) 75 (d) 40 16. DABC is such that AB = 3 cm, BC = 2 cm, CA = 2.5 cm. If DABC ~ DDEF and EF = 4 cm, then perimeter of DDEF is (a) 7.5 cm (b) 15 cm (c) 22.5 cm (d) 30 cm 17. In the figure, if DE||BC, AD = 3 cm, BD = 4 cm and BC = 14 cm, then DE equals Latest CBSE SAMPLE PAPER A DE B C (a) 7 cm (b) 6 cm (c) 4 cm (d) 3 cm (d) 3 18. If 4 tan b = 3, then 4 sin b - 3 cosb = (c) 2/3 4 4 sin b + 3 cosb (a) 0 (b) 1/3