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CBSE Class 10 Sample / Model Paper 2022 : Mathematics Standard - Term I (with Marking Scheme / Solutions)

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Sample Question Paper Class X Session -2021-22 TERM 1 Subject- Mathematics (Standard) 041 Time Allowed: 90 minutes Maximum Marks: 40 General 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. SECTION A Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted Q No Marks 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 1 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 1 3 A girl walks 200m towards East and then 150m towards North. The distance of the girl from the starting point is (a)350m (b) 250m (c) 300m (d) 225 1 4 The lengths of the diagonals of a rhombus are 24cm and 32cm, then the length of the altitude of the rhombus is (a) 12cm (b) 12.8cm (c) 19 cm` (d) 19.2cm 1 5 Two fair coins are tossed. What is the probability of getting at the most one head? (a) (b) (c) (d) 3/8 1 6 ABC~ PQR. If AM and PN are altitudes of ABC and PQR respectively and AB 2 : PQ2 = 4 : 9, then AM:PN = (a) 16:81 (b) 4:9 (c) 3:2 (d) 2:3 1 7 If 2sin2 cos2 = 2, then is (a) 0 (b) 90 1 (c) 45 (d) 30 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 1 9 The lines x = a and y = b, are (a) intersecting (b) parallel 1 (c) overlapping (d) (None of these) 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 = 23/25, then a is (a) rational (b) irrational (d) integer (c) whole number 1 1 12 13 14 15 If LCM(x, 18) =36 and HCF(x, 18) =2, then x is (a) 2 (b) 3 (c) 4 1 (d) 5 In ABC right angled at B, if tan A= , then cos A cos C- sin A sin C = (a) -1 (b) 0 (c) 1 (d) / If the angles of ABC are in ratio 1:1:2, respectively (the largest angle being ec a C), then the value of is c ec c (a) 0 (b) 1/2 (c) 1 (d) /2 1 1 angle The number of revolutions made by a circular wheel of radius 0.7m in rolling a distance of 176m is (a) 22 (b) 24 (c) 75 (d) 40 1 16 ABC is such that AB=3 cm, BC= 2cm, CA= 2.5 cm. If ABC ~ DEF and 4cm, then perimeter of DEF is (a) 7.5 cm (b) 15 cm (c) 22.5 cm (d) 30 cm 1 17 In the figure, if DE BC, AD = 3cm, BD = 4cm and BC= 14 cm, then DE equals (a) 7cm 18 If 4 tan = 3, then (a) 0 (b) 6cm c i + c (b) 1/3 (c) 4cm EF = 1 (d) 3cm 1 = (c) 2/3 (d) 19 One equation of a pair of dependent linear equations is 5x + 7y = 2. The second equation can be a) 10x+14y +4 = 0 b) 10x 14y+ 4 = 0 c) 10x+14y + 4 = 0 (d) 10x 14y = 4 20 A letter of English alphabets is chosen at random. What is the probability that it is a letter 1 of the word MATHEMATICS ? (a) 4/13 (b) 9/26 (c) 5/13 (d) 11/26 1 SECTION B Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted MARKS QN 21 If sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are (a) 2 (b) 3 (c) 4 (d) 5 1 22 Given below is the graph representing two linear equations by lines AB and CD 1 respectively. What is the area of the triangle formed by these two lines and the line x=0? (a) 3sq. units 23 24 25 (b) 4sq. units (c) 6sq. units (d) 8sq. units If tan + cot = 2, then tan20 + cot20 = (a) 0 (b) 2 (c) 20 (d) 2 If 217x + 131y = 913, 131x + 217y = 827, then x + y is (a) 5 (b) 6 (c) 7 (d) 8 1 20 1 The LCM of two prime numbers p and q (p > q) is 221. Find the value of 3p q. (a) 4 (b) 28 (c) 38 (d) 48 1 26 A card is drawn from a well shuffled deck of cards. What is the probability that the card drawn is neither a king nor a queen? (a) 11/13 (b) 12/13 (c) 11/26 (d) 11/52 1 27 Two fair dice are rolled simultaneously. The probability that 5 will come up at least once is (a) 5/36 (b) 11/36 (c) 12/36 (d) 23/36 1 28 If 1+ sin2 = 3 sin cos , then values of cot are (a) -1, 1 (b) 0,1 (c)1, 2 1 29 30 The vertices of a parallelogram in order are A(1,2), B(4, y), C(x, 6) and D(3,5). Then (x, y) is (a) (6, 3) (b) (3, 6) (c) (5, 6) (d) (1, 4) In the given figure, ACB = CDA, AC = 8cm, AD = 3cm, then BD is (a) 22/3 cm 31 (d) -1,-1 (b) 26/3 cm (c) 55/3 cm 1 1 (d) 64/3 cm The equation of the perpendicular bisector of line segment joining points A(4,5) and B(-2,3) is (a) 2x y +7=0 (b) 3x +2 y 7=0 (c) 3x y 7 =0 (d) 3x + y 7=0 1 32 In the given figure, D is the mid-point of BC, then the value of (a) 2 (b) 1/2 (c) 1/3 c c is 1 (d) 1/4 33 The smallest number by which 1/13 should be multiplied so that its decimal expansion terminates after two decimal places is (a) 13/100 (b) 13/10 (c) 10/13 (d) 100/13 1 34 Sides AB and BE of a right triangle, right angled at B are of lengths 16 cm and 8 cm respectively. The length of the side of largest square FDGB that can be inscribed in the triangle ABE is 1 (a) 32/3cm (b) 16/3cm (c)8/3cm (d) 4/3cm 35 Point P divides the line segment joining R(-1, 3) and S(9,8) in ratio k:1. If P lies on the line x y +2=0, then value of k is (a) 2/3 (b) 1/2 (c) 1/3 (d) 1/4 1 36 In the figure given below, ABCD is a square of side 14 cm with E, F, G and H as the mid points of sides AB, BC, CD and DA respectively. The area of the shaded portion is 1 37 (a) 44cm (b) 49 cm (c) 98 cm (d) 49 /2 cm Given below is the picture of the Olympic rings made by taking five congruent circles of radius 1cm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1cm. Total area of all the dotted regions assuming the thickness of the rings to be negligible is 1 (a) 4( /12- 3/4) cm 38 (b) ( /6 - 3/4) cm (c) 4( /6 - 3/4) cm If 2 and are the zeros of px2+5x+r, then (a) p = r = 2 (b) p = r = - 2 (c) p = 2, r= -2 (d) 8( /6 - 3/4) cm 1 (d) p = -2, r= 2 39 The circumference of a circle is 100 cm. The side of a square inscribed in the circle is (a) 50 2 cm (b) 100/ cm (c) 50 2/ cm (d) 100 2/ cm 1 40 The number of solutions of 3x+y = 243 and 243x-y = 3 is (a) 0 (b) 1 (c) 2 1 (d) infinite SECTION C Case study based questions: Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted. Q41-Q45 are based on Case Study -1 Case Study -1 The figure given alongside shows the path of a diver, when she takes a jump from the diving board. Clearly it is a parabola. Annie was standing on a diving board, 48 feet above the water level. She took a dive into the pool. Her height (in feet) above the water level at any time t in seconds is given by the polynomial h(t) such that h(t) = -16t + 8t + k. 41 42 What is the value of k? (a) 0 (b) - 48 (c) 48 (d) 48/-16 At what time will she touch the water in the pool? (a) 30 seconds (b) 2 seconds (c) 1.5 seconds (d) 0.5 seconds 1 1 43 Rita s height (in feet) above the water level is given by another polynomial p(t) with zeroes -1 and 2. Then p(t) is given by(a) t + t - 2. (b) t + 2t - 1 (c) 24t - 24t + 48. (d) -24t + 24t + 48. 1 44 A polynomial q(t) with sum of zeroes as 1 and the product as -6 is modelling Anu s height in feet above the water at any time t( in seconds). Then q(t) is given by (a) t + t + 6 (b) t + t -6 (c) -8t + 8t + 48 (d) 8t - 8t + 48 The zeroes of the polynomial r(t) = -12t + (k-3)t +48 are negative of each other. Then k is (a) 3 (b) 0 (c) -1.5 (d) -3 1 45 Q46-Q50 are based on Case Study -2 Case Study -2 A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf. It is rectangular in shape - 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet) above the ground. Each team plays with 11 players on the field during the game including the goalie. Positions you might play include Forward: As shown by players A, B, C and D. Midfielders: As shown by players E, F and G. Fullbacks: As shown by players H, I and J. Goalie: As shown by player K Using the picture of a hockey field below, answer the questions that follow: 1 46 47 48 49 50 The coordinates of the centroid of EHJ are (a) (-2/3, 1) (b) (1,-2/3) (c) (2/3,1) (d) ( -2/3,-1) If a player P needs to be at equal distances from A and G, such that A, P and G are in straight line, then position of P will be given by (a) (-3/2, 2) (b) (2,-3/2) (c) (2, 3/2) (d) ( -2,-3) The point on x axis equidistant from I and E is (a) (1/2, 0) (b) (0,-1/2) (c) (-1/2,0) (d) ( 0,1/2) What are the coordinates of the position of a player Q such that his distance from K is twice his distance from E and K, Q and E are collinear? (a) (1, 0) (b) (0,1) (c) (-2,1) (d) ( -1,0) The point on y axis equidistant from B and C is (a) (-1, 0) (b) (0,-1) (c) (1,0) (d) ( 0,1) 1 1 1 1 1 Marking Scheme Class- X Session- 2021-22 TERM 1 Subject- Mathematics (Standard) QN Correct Option 1 (b) SECTION A HINTS/SOLUTION Least composite number is 4 and the least prime number is 2. LCM(4,2) : MAR KS 1 HCF(4,2) = 4:2 = 2:1 2 (a) For lines to coincide: so, = i.e. k= 9 = = 1 = 3 (b) By Pythagoras theorem The required distance = (200 + 150 ) = (40000+ 22500) = (62500) = 250m. So the distance of the girl from the starting point is 250m. 1 4 (d) 1 5 (a) 6 (d) 7 (b) 8 (c) 9 (a) 10 (d) 11 12 (b) (c) Area of the Rhombus = d d = x 24 x 32= 384 cm . Using Pythagoras theorem side = ( d ) + ( d ) = 12 +16 = 144 +256 =400 Side = 20cm Area of the Rhombus = base x altitude 384 = 20 x altitude So altitude = 384/20 = 19.2cm Possible outcomes are (HH), (HT), (TH), (TT) Favorable outcomes(at the most one head) are (HT), (TH), (TT) So probability of getting at the most one head =3/4 Ratio of altitudes = Ratio of sides for similar triangles So AM:PN = AB:PQ = 2:3 2sin2 cos2 = 2 Then 2 sin2 (1- sin2 ) = 2 3 sin2 =3 or sin2 =1 is 90 Since it has a terminating decimal expansion, so prime factors of the denominator will be 2,5 Lines x=a is a line parallel to y axis and y=b is a line parallel to x axis. So they will intersect. Distance of point A(-5,6) from the origin(0,0) is + + = + = units a =23/25, then a = /5, which is irrational LCM X HCF = Product of two numbers 36 X 2 = 18 X x x=4 13 (b) 1 14 (a) tan A= = tan 60 so A=60 , Hence C = 30 . So cos A cos C- sin A sin C = (1/2)x /2) - /2)x (1/2) =0 1x +1x +2x =180 , x = 45 . A , B and C are 45 , 45 and 90 resp. a a = = = 1-1= 0 1 1 1 1 1 1 1 1 1 15 16 (d) Number of revolutions= i (b) i 17 (b) (a) i a 1 = 22 X X 7 . = 40 = = . So perimeter of DEF = 15cm Since DE BC, ABC ~ ADE ( By AA rule of similarity) So 18 . i i a = i.e. = 1 . So DE = 6cm 1 Dividing both numerator and denominator by cos , = = =0 i + 1 a + + 19 (d) -2( 5x + 7y = 2) gives 10x 14y = 4. Now 20 (a) Number of Possible outcomes are 26 Favorable outcomes are M, A, T, H, E, I, C, S = = = -2 1 1 probability = 8/26 = 4/13 SECTION B 21 (c) Since HCF = 81, two numbers can be taken as 81x and 81y, ATQ 81x + 81y = 1215 Or x+y = 15 which gives four co prime pairs1,14 2,13 4,11 7, 8 1 22 (c) 1 23 (b) Required Area is area of triangle ACD = (6)2 = 6 sq units tan + cot = 2 gives =45 . So tan = cot = 1 tan20 + cot 20 = 120 + 120 = 1+1 = 2 24 (a) Adding the two given equations we get: 348x + 348y = 1740. So x +y =5 1 25 (c) 1 26 (a) LCM of two prime numbers = product of the numbers 221= 13 x 17. So p= 17 & q= 13 3p - q= 51-13 =38 Probability that the card drawn is neither a king nor a queen = 27 (b) Outcomes when 5 will come up at least once are(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6) Probability that 5 will come up at least once = 11/36 1 28 (c) 1+ sin2 = 3 sin cos sin2 + cos2 + sin2 = 3 sin cos 2 sin2 - 3sin cos + cos2 = 0 (2sin -cos )( sin - cos ) =0 cot = 2 or cot = 1 1 29 (a) Since ABCD is a parallelogram, diagonals AC and BD bisect each other, mid 1 1 1 = 44/52 = 11/13 point of AC= mid point of BD + + + + ( , )=( , ) Comparing the co-ordinates, we get, + + = . So, x= 6 + 30 (c) 31 (d) 32 (b) 33 (a) 34 (b) = Similarly, (x, y) = (6,3) + . So, y= 3 ACD ~ ABC( AA ) 1 AD AC = CPST AC AB 8/AB = 3/8 This gives AB = 64/3 cm. So BD = AB AD = 64/3 -3 = 55/3cm. Any point (x, y) of perpendicular bisector will be equidistant from A & B. = + + + Solving we get -12x 4y + 28=0 or 3x + y 7=0 = AC/ BC / 1 1 = CD/ BC = CD/ 2CD = The smallest number by which 1/13 should be multiplied so that its decimal expansion terminates after two decimal points is 13/100 as x = = 0.01 Ans: 13/100 1 1 ABE is a right triangle & FDGB is a square of side x cm AFD ~ DGE( AA ) AF FD = CPST DG GE x x = CPST x 8 x 128 = 24x or x = 16/3cm 35 (a) Since P divides the line segment joining R(-1, 3) and S(9,8) in ratio k:1 coordinates of P are ( + , + + ) Since P lies on the line x y +2=0, then 9k -1 -8k-3 +2k+2 =0 which gives k=2/3 36 1 + - + + +2 =0 1 (c) Shaded area = Area of semicircle + (Area of half square Area of two quadrants) = Area of semicircle +(Area of half square Area of semicircle) = Area of half square = x 14 x14 = 98cm 37 1 (d) .o Let O be the center of the circle. OA = OB = AB =1cm. So OAB is an equilateral triangle and AOB =60 Required Area= 8x Area of one segment with r=1cm, = 60 = 8x( x x 1 - x 1 ) = 8( /6 - 3/4)cm 38 (b) Sum of zeroes = 2 + = -5/p i.e. 5/2 = -5/p . So p= -2 Product of zeroes = 2x = r/p i.e. r/p = 1 or r = p = -2 1 39 (c) 1 40 (b) 2 r =100. So Diameter = 2r =100/ = diagonal of the square. side 2 = diagonal of square = 100/ side = 100/ 2 = 50 2/ 3x+y = 243 = 35 So x+y =5-----------------------------------(1) 243x-y = 3 (35) x-y = 31 So 5x -5y =1--------------------------------(2) Since : , so unique solution 1 SECTION C 41 (c) Initially, at t=0, Annie s height is 48ft So, at t =0, h should be equal to 48 h(0) = -16(0) + 8(0) + k = 48 So k = 48 1 42 (b) When Annie touches the pool, her height =0 feet i.e. -16t + 8t + 48 =0 above water level 2t - t -6 =0 2t - 4t +3t -6 =0 2t(t-2) +3(t-2) =0 (2t +3) (t-2) =0 i.e. t= 2 or t= -3/2 Since time cannot be negative , so t= 2seconds 1 43 (d) t= -1 & t=2 are the two zeroes of the polynomial p(t) Then p(t)=k (t- -1)(t-2) = k(t +1)(t-2) When t = 0 (initially) h = 48ft p(0)=k(0 - 0 -2)= 48 i.e. -2k = 48 So the polynomial is -24(t - t -2) = -24t + 24t + 48. 1 44 (c) A polynomial q(t) with sum of zeroes as 1 and the product as -6 is given by q(t) = k(t - (sum of zeroes)t + product of zeroes) = k(t -1t + -6) ..(1) When t=0 (initially) q(0)= 48ft 1 q(0)=k(0 - 1(0) -6)= 48 i.e. -6k = 48 or k= -8 Putting k = -8 in equation (1), reqd. polynomial is -8(t -1t + -6) = -8t + 8t + 48 45 (a) When the zeroes are negative of each other, sum of the zeroes = 0 So, -b/a = 0 (k-3) =0 1 Centroid of EHJ with E(2,1), H(-2,4) & J(-2,-2) is + + + + ( , = (-2/3, 1) 1 Let the point on x axis equidistant from I(-1,1) and E(2,1) be (x,0) then + = + + 2 2 x + 1 + 2x +1 = x + 4 - 4x +1 6x = 3 So x = . the required point is ( , 0) 1 =0 + k-3 = 0, i.e. k = 3. 46 (a) 47 (c) 48 (a) 49 (b) If P needs to be at equal distance from A(3,6) and G(1,-3), such that A,P and G are collinear, then P will be the mid-point of AG. + + So coordinates of P will be , = (2, 3/2) 1 Let the coordinates of the position of a player Q such that his distance from K(-4,1) is twice his distance from E(2,1) be Q(x, y) 1 Then KQ : QE = 2: 1 Q(x, y) = ( X + X = (0,1) 50 (d) , X + X Let the point on y axis equidistant from B(4,3) and C(4,-1) be (0,y) then = + + + 16 + y2 + 9 - 6y = 16 + y2 + 1 + 2y -8y = -8 So y = 1 . the required point is (0, 1) 1

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