Marking Scheme Class- X Session- 2021-22 TERM 1 Subject- Mathematics (Standard) SECTION A QN Correct HINTS/SOLUTION MAR Option KS 1 (b) Least composite number is 4 and the least prime number is 2. LCM(4,2) : 1 HCF(4,2) = 4:2 = 2:1 2 (a) 𝑎₁ 𝑏₁ 𝑐₁ 1 For lines to coincide: 𝑎₂ = 𝑏₂ = 𝑐₂ 5 7 −3 so, 15 = 21 = −k i.e. k= 9 3 (b) By Pythagoras theorem 1 The required distance =√(200² + 150²) = √(40000+ 22500) = √(62500) = 250m. So the distance of the girl from the starting point is 250m. 4 (d) 1 1 1 Area of the Rhombus = 2 d₁d₂ = 2 x 24 x 32= 384 cm². Using Pythagoras theorem 1 1 side² = (2d₁)² + (2d₂)² = 12² +16² = 144 +256 =400 Side = 20cm Area of the Rhombus = base x altitude 384 = 20 x altitude So altitude = 384/20 = 19.2cm 5 (a) Possible outcomes are (HH), (HT), (TH), (TT) 1 Favorable outcomes(at the most one head) are (HT), (TH), (TT) So probability of getting at the most one head =3/4 6 (d) Ratio of altitudes = Ratio of sides for similar triangles 1 So AM:PN = AB:PQ = 2:3 7 (b) 2sin2β – cos2β = 2 1 Then 2 sin2β – (1- sin2β) = 2 3 sin2β =3 or sin2β =1 β is 90ᵒ 8 (c) Since it has a terminating decimal expansion, 1 so prime factors of the denominator will be 2,5 9 (a) Lines x=a is a line parallel to y axis and y=b is a line parallel to x axis. So they 1 will intersect. 10 (d) Distance of point A(-5,6) from the origin(0,0) is 1 √(0 + 5)2 + (0 − 6)2 = √25 + 36 = √61 units 11 (b) a²=23/25, then a = √23/5, which is irrational 1 12 (c) LCM X HCF = Product of two numbers 1 36 X 2 = 18 X x x=4 13 (b) tan A= √3 = tan 60° so ∠A=60°, Hence ∠C = 30°. 1 So cos A cos C- sin A sin C = (1/2)x (√3/2) - (√3/2)x (1/2) =0 14 (a) 1x +1x +2x =180°, x = 45°. 1 ∠A , ∠B and ∠C are 45°, 45° and 90°resp. sec A tan A sec 45 tan 45 √2 1 – = – = – = 1-1= 0 cosec B cot B cosec 45 cot 45 √2 1

15 (d) total distance 176 1 Number of revolutions= = 22 circumference 2 X 7 X 0.7 = 40 16 (b) perimeter of ∆ABC BC 1 = perimeter of ∆DEF EF 7.5 2 = . So perimeter of ∆DEF = 15cm perimeter of ∆DEF 4 17 (b) Since DE∥ BC, ∆ABC ~ ∆ADE ( By AA rule of similarity) 1 AD DE 3 DE So = i.e. = . So DE = 6cm AB BC 7 14 18 (a) Dividing both numerator and denominator by cosβ, 1 4 𝑠𝑖𝑛𝛽−3 cos 𝛽 4 𝑡𝑎𝑛𝛽−3 3−3 = = =0 4 sin 𝛽+3 cos 𝛽 4 tan 𝛽+3 3+3 19 (d) 𝑎₁ 𝑏₁ 𝑐₁ 1 -2(–5x + 7y = 2) gives 10x – 14y = –4. Now = 𝑏₂ = 𝑐₂ = -2 𝑎₂ 20 (a) Number of Possible outcomes are 26 1 Favorable outcomes are M, A, T, H, E, I, C, S probability = 8/26 = 4/13 SECTION B 21 (c) Since HCF = 81, two numbers can be taken as 81x and 81y, 1 ATQ 81x + 81y = 1215 Or x+y = 15 which gives four co prime pairs- 1,14 2,13 4,11 7, 8 22 (c) Required Area is area of triangle ACD = ½(6)2 1 = 6 sq units 23 (b) tan α + cot α = 2 gives α=45°. So tan α = cot α = 1 1 tan20α + cot 20α = 120 + 120 = 1+1 = 2 24 (a) Adding the two given equations we get: 348x + 348y = 1740. 1 So x +y =5 25 (c) LCM of two prime numbers = product of the numbers 1 221= 13 x 17. So p= 17 & q= 13 ⸫3p - q= 51-13 =38 26 (a) Probability that the card drawn is neither a king nor a queen 1 52−8 = 52 = 44/52 = 11/13 27 (b) Outcomes when 5 will come up at least once are- 1 (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 28 (c) 1+ sin2α = 3 sinα cos α 1 sin2α + cos2α + sin2α = 3 sinα cos α 2 sin2α - 3sinα cos α + cos2α = 0 (2sinα -cos α)( sinα- cosα) =0 ⸫cotα = 2 or cotα = 1 29 (a) Since ABCD is a parallelogram, diagonals AC and BD bisect each other, ∴ mid 1 point of AC= mid point of BD

𝑥+1 6+2 3+4 5+𝑦 ( 2 , 2 )=( 2 , 2 ) Comparing the co-ordinates, we get, 𝑥+1 3+4 = 2 . So, x= 6 2 6+2 5+𝑦 Similarly, 2 = . So, y= 3 2 ∴(x, y) = (6,3) 30 (c) ∆ACD ~∆ ABC( AA ) 1 AC AD ∴ = (CPST) AB AC 8/AB = 3/8 This gives AB = 64/3 cm. So BD = AB – AD = 64/3 -3 = 55/3cm. 31 (d) Any point (x, y) of perpendicular bisector will be equidistant from A & B. 1 ∴ √(𝑥 − 4)2 + (𝑦 − 5)2 = √(𝑥 + 2)2 + (𝑦 − 3)2 Solving we get -12x – 4y + 28=0 or 3x + y – 7=0 32 (b) cot 𝑦 ° AC/ BC 1 = = CD/ BC = CD/ 2CD = ½ cot 𝑥 ° 𝐴𝐶/𝐶𝐷 33 (a) The smallest number by which 1/13 should be multiplied so that its decimal 1 1 13 1 expansion terminates after two decimal points is 13/100 as 13 x 100 = 100 = 0.01 Ans: 13/100 34 (b) 1 ∆ABE is a right triangle & FDGB is a square of side x cm ∆AFD ~∆ DGE( AA ) AF FD ∴ = (CPST) DG GE 16 − 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 ∴ 1 9k −1 8k+3 coordinates of P are ( k+1 , ) k+1 9k −1 8k+3 Since P lies on the line x – y +2=0, then - +2 =0 k+1 k+1 9k -1 -8k-3 +2k+2 =0 which gives k=2/3 36 (c) 1 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 = ½ x 14 x14 = 98cm²

37 (d) 1 .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, ѳ= 60° 60 √3 = 8x(360 x π x 1²- 4 x 1²) = 8(π/6 - √3/4)cm² 38 (b) Sum of zeroes = 2 + ½ = -5/p 1 i.e. 5/2 = -5/p . So p= -2 Product of zeroes = 2x ½ = r/p i.e. r/p = 1 or r = p = -2 39 (c) 2πr =100. So Diameter = 2r =100/π = diagonal of the square. 1 side√2 = diagonal of square = 100/ π ∴ side = 100/√2π = 50√2/π 40 (b) 3x+y = 243 = 35 1 So x+y =5-----------------------------------(1) 243x-y = 3 (35) x-y = 31 So 5x -5y =1--------------------------------(2) 𝑎₁ 𝑏₁ Since : 𝑎₂ ≠ 𝑏₂ , so unique solution SECTION C 41 (c) Initially, at t=0, Annie’s height is 48ft 1 So, at t =0, h should be equal to 48 h(0) = -16(0)² + 8(0) + k = 48 So k = 48 42 (b) When Annie touches the pool, her height =0 feet 1 i.e. -16t² + 8t + 48 =0 above water level 2t² - t -6 =0 2t² - 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 43 (d) t= -1 & t=2 are the two zeroes of the polynomial p(t) 1 Then p(t)=k (t- -1)(t-2) = k(t +1)(t-2) When t = 0 (initially) h₁ = 48ft p(0)=k(0²- 0 -2)= 48 i.e. -2k = 48 So the polynomial is -24(t²- t -2) = -24t² + 24t + 48. 44 (c) A polynomial q(t) with sum of zeroes as 1 and the product as -6 is given by 1 q(t) = k(t² - (sum of zeroes)t + product of zeroes) = k(t² -1t + -6) ………..(1) When t=0 (initially) q(0)= 48ft

q(0)=k(0²- 1(0) -6)= 48 i.e. -6k = 48 or k= -8 Putting k = -8 in equation (1), reqd. polynomial is -8(t² -1t + -6) = -8t² + 8t + 48 45 (a) When the zeroes are negative of each other, 1 sum of the zeroes = 0 So, -b/a = 0 (k-3) - −12 = 0 k−3 + 12 = 0 k-3 = 0, i.e. k = 3. 46 (a) Centroid of ΔEHJ with E(2,1), H(-2,4) & J(-2,-2) is 1 2+−2+ −2 1+4+ −2 ( , ) = (-2/3, 1) 3 3 47 (c) If P needs to be at equal distance from A(3,6) and G(1,-3), such that A,P and G 1 are collinear, then P will be the mid-point of AG. 3+1 6+ −3 So coordinates of P will be ( 2 , 2 ) = (2, 3/2) 48 (a) Let the point on x axis equidistant from I(-1,1) and E(2,1) be (x,0) 1 then √(𝑥 + 1)2 + (0 − 1)2 = √(𝑥 − 2)2 + (0 − 1)2 x2 + 1 + 2x +1 = x2 + 4 - 4x +1 6x = 3 So x = ½ . ∴ the required point is (½, 0) 49 (b) Let the coordinates of the position of a player Q such that his distance from 1 K(-4,1) is twice his distance from E(2,1) be Q(x, y) Then KQ : QE = 2: 1 2 X 2+1 X−4 2 X 1+1 X 1 Q(x, y) = ( , ) 3 3 = (0,1) 50 (d) Let the point on y axis equidistant from B(4,3) and C(4,-1) be (0,y) 1 then √(4 − 0)2 + (3 − 𝑦)2 = √(4 − 0)2 + (𝑦 + 1)2 16 + y2 + 9 - 6y = 16 + y2 + 1 + 2y -8y = -8 So y = 1 . ∴ the required point is (0, 1)