Maths Standard Class X Past Paper 2023-24 PDF

Marking Scheme Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) TIME: 3 hours MAX.MARKS: 80...

Marking Scheme Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) TIME: 3 hours MAX.MARKS: 80 SECTION A Section A consists of 20 questions of 1 mark each. 1. (b) xy2 1 2. (b) 1 zero and the zero is ‘3’ 1 3. 𝑎1 𝑏1 𝑐1 1 (b) = ≠ 𝑐2 𝑎2 𝑏2 4. (c) 2 distinct real roots 1 5. (c) 7 1 6. (a) 1:2 1 7. (d) infinitely many 1 8. 𝑎𝑐 1 (b) 𝑏+𝑐 9. (b) 100° 1 10. (d) 11 cm 1 11. √𝑏 2 −𝑎2 1 (c) 𝑏 12. (d) cos A 1 13. (d) 60° 1 14. (a) 2 units 1 15. (a) 10m 1 16. 4−𝜋 1 (b) 4 17. 22 1 (b) 46 18. (d) 150 1 19. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of 1 assertion (A) 20. (c) Assertion (A) is true but reason (R) is false. 1 SECTION B Section B consists of 5 questions of 2 marks each. 21. Let us assume, to the contrary, that √2 is rational. 𝑎 ½ So, we can find integers a and b such that √2 = 𝑏 where a and b are coprime. So, b √2 = a. Squaring both sides, we get 2b2 = a2. ½ Therefore, 2 divides a2 and so 2 divides a. So, we can write a = 2c for some integer c. Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2. ½ This means that 2 divides b2, and so 2 divides b Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factors other than 1. ½ This contradiction has arisen because of our incorrect assumption that √2 is rational. So, we conclude that √2 is irrational. 1 22. ABCD is a parallelogram. ½ AB = DC = a Point P divides AB in the ratio 2:3 2 3 AP = a , BP = a 5 5 point Q divides DC in the ratio 4:1. ½ 4 1 DQ = a , CQ = a 5 5 ∆ APO ~ ∆ CQO [AA similarity] 𝐴𝑃 𝑃𝑂 𝐴𝑂 ½ = = 𝐶𝑄 𝑄𝑂 𝐶𝑂 2 ½ AO a 2 CO = 5 1 = 1 ⇒ OC = ½ OA a 5 23. PA = PB; CA = CE; DE = DB [Tangents to a circle] ½ Perimeter of ∆PCD = PC + CD + PD = PC + CE + ED + PD = PC + CA + BD + PD = PA + PB 1 Perimeter of ∆PCD = PA + PA = 2PA = 2(10) = 20 ½ cm 24. ∵ tan(𝐴 + 𝐵) = √3 ∴ 𝐴 + 𝐵 = 600 …(1) ½ ∵ tan(𝐴 − 𝐵) = 1 ∴ 𝐴 − 𝐵 = 30 0 …(2) ½ √3 ½ Adding (1) & (2), we get 2A=900 ⟹ 𝐴 = 450 ½ Also (1) –(2), we get 2𝐵 = 300 ⟹ 𝐵 = 450 [or] 3 2 cosec230 + x sin260 – tan230 = 10 4 2 2 √3 3 1 ⇒ 2(2)2 + x ( ) − 4 ( ) = 10 1 2 √3 3 3 1 ⇒ 2(4) + x (4) − 4 (3) = 10 ½ 3 1 ⇒ 8 + x (4) − 4 = 10 ⇒ 32 + x (3) −1 = 40 ½ ⇒ 3x = 9 ⇒ x = 3 25. ∠𝐴 ∠𝐵 ∠𝐶 Total area removed = 360 π𝑟 2 + 360 π𝑟 2 + 360 π𝑟 2 ½ ∠𝐴+ ∠𝐵+ ∠𝐶 = π𝑟 2 360 180 = π𝑟 2 ½ 360 180 22 = X X (14)2 ½ ½ 360 7 = 308 cm2 [or] The side of a square = Diameter of the semi-circle = a Area of the unshaded region ½ = Area of a square of side ‘a’ + 4(Area of a semi-circle of diameter ‘a’) The horizontal/vertical extent of the white region = 14-3-3 = 8 cm ½ Radius of the semi-circle + side of a square + Radius of the semi-circle = 8 cm 2  2 (radius of the semi-circle) + side of a square = 8 cm 2a = 8 cm ⇒ a = 4 cm ½ Area of the unshaded region = Area of a square of side 4 cm + 4 (Area of a semi-circle of diameter 4 cm) 1 = (4)2 + 4 X 2 π(2)2 = 16 + 8π cm2 ½ SECTION C Section C consists of 6 questions of 3 marks each 26. Number of students in each group subject to the given condition = HCF (60,84,108) ½ HCF (60,84,108) = 12 ½ 60 Number of groups in Music = =5 12 ½ 84 Number of groups in Dance = =7 ½ 12 108 ½ Number of groups in Handicrafts = =9 ½ 12 Total number of rooms required = 21 27. P(x) = 5x2 + 5x + 1 ½ −𝑏 −5 α+β = = = -1 𝑎 5 ½ 𝑐 1 αβ = = ½ 𝑎 5 𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 - 2𝛼𝛽 1 ½ = (-1)2 – 2 ( ) 5 2 3 ½ =1- = 5 5 1 1 ½ 𝛼 +𝛽 = + −1 −1 𝛼 𝛽 (𝛼+𝛽) (−1) = = 1 = -5 𝛼𝛽 5 28. Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the original number = 10x + y When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s Digit. ½ So the obtain by reversing the digits= 10y + x According to the given condition. (10x + y) + (10y + x) = 66 i.e., 11(x + y) = 66 ½ i.e., x + y = 6 ---- (1) We are also given that the digits differ by 2, ½ therefore, either x – y = 2 ---- (2) ½ or y – x = 2 ---- (3) If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2. ½ In this case, we get the number 42. If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4. ½ In this case, we get the number 24. Thus, there are two such numbers 42 and 24. [or] 1 1 Let √𝑥 be ‘m’ and √𝑦 be ‘n’, ½ Then the given equations become 2m + 3n = 2 ½ 4m - 9n = -1 3  ( 2m + 3n = 2) X- 2 ⇒ −4𝑚 − 6𝑛 = −4 …(1) 4m - 9n = -1 4𝑚 − 9𝑛 = −1 …(2) Adding (1) and (2) 1 We get −15𝑛 = −5 ⇒ 𝑛 = 3 ½ 1 Substituting n = 3 in 2m + 3n = 2, we get ½ 2m + 1 = 2 2m = 1 1 m= 2 1 1 1 m= ⇒ √𝑥 = 2 ⇒ x = 4 and n = ⇒ √𝑦 = 3 ⇒ y = 9 2 3 29. ∠OAB = 30° ∠OAP = 90° [Angle between the tangent and the radius at the point of contact] ∠PAB = 90° - 30° =60° ½ AP = BP [Tangents to a circle from an external point] ∠PAB = ∠PBA [Angles opposite to equal sides of a triangle] ½ In ΔABP, ∠PAB + ∠PBA + ∠APB = 180° [Angle Sum Property] 60° + 60° + ∠APB = 180° ∠APB = 60° ½ ∴ ΔABP is an equilateral triangle, where AP = BP = AB. PA = 6 cm ½ In Right ΔOAP, ∠OPA = 30° 𝑂𝐴 tan 30° = 𝑃𝐴 1 𝑂𝐴 ½ = √3 6 6 OA = = 2√3𝑐𝑚 ½ √3 [or] Let ∠ TPQ = 𝜃 ∠ TPO = 90° [Angle between the tangent and the radius at the point of contact] ½ ∠ OPQ = 90° - 𝜃 TP = TQ [Tangents to a circle from an external point] ½ ∠ TPQ = ∠ TQP = 𝜃 [Angles opposite to equal sides of a triangle] ½ In ΔPQT, ∠PQT + ∠QPT + ∠PTQ = 180° [Angle Sum Property] ½ 𝜃 + 𝜃 + ∠PTQ = 180° ∠PTQ = 180° - 2 𝜃 ½ ∠PTQ = 2 (90° - 𝜃) ½ ∠PTQ = 2 ∠ OPQ [using (1)] 30. Given, 1 + sin2θ = 3 sin θ cos θ Dividing both sides by cos2θ, 1 + tan2θ = 3 tan 𝜃 cos2 𝜃 ½ sec2θ + tan2θ = 3 tan 𝜃 ½ 1 + tan2θ + tan2θ = 3 tan 𝜃 ½ 1 + 2 tan2θ = 3 tan 𝜃 ½ 2 tan θ - 3 tan 𝜃 +1 = 0 2 If tan θ = x, then the equation becomes 2x2 -3x + 1 = 0 4  1 ⇒ (𝑥 − 1)(2𝑥 − 1) = 0 x = 1 or 2 1 tan θ = 1 or 1 2 31. Length Number of CI Mid x d fd [in mm] leaves (f) 118 – 126 3 117.5- 126.5 122 -27 -81 127 – 135 5 126.5– 135.5 131 -18 -90 136 – 144 9 135.5– 144.5 140 -9 -81 145 – 153 12 144.5 – 153.5 a = 149 0 0 153.5 – 162.5 158 9 45 2 154 – 162 5 163 – 171 4 162.5 – 171.5 167 18 72 ½ ½ 172 – 180 2 171.5 – 180.5 176 27 54 ∑ 𝑓𝑑 −81 Mean = a + = 149 + ∑𝑓 40 = 149 – 2.025 = 146.975 Average length of the leaves = 146.975 SECTION D Section D consists of 4 questions of 5 marks each 32. Let the speed of the stream be x km/h. The speed of the boat upstream = (18 – x) km/h and the speed of the boat downstream = (18 + x) km/h. 1 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 24 The time taken to go upstream = = hours 𝑠𝑝𝑒𝑒𝑑 18−𝑥 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 24 the time taken to go downstream = = hours 1 𝑠𝑝𝑒𝑒𝑑 18+𝑥 According to the question, 24 24 - =1 1 18−𝑥 18+𝑥 24(18 + x) – 24(18 – x) = (18 – x) (18 + x) x2 + 48x – 324 = 0 x = 6 or – 54 1 Since x is the speed of the stream, it cannot be negative. Therefore, x = 6 gives the speed of the stream = 6 km/h. 1 [or] Let the time taken by the smaller pipe to fill the tank = x hr. Time taken by the larger pipe = (x – 10) hr ½ 1 Part of the tank filled by smaller pipe in 1 hour = 𝑥 1 1 Part of the tank filled by larger pipe in 1 hour = 𝑥−10 3 75 ½ The tank can be filled in 9 8 = hours by both the pipes together. 8 8 ½ Part of the tank filled by both the pipes in 1 hour = 75 5  1 1 8 Therefore, + = ½ 𝑥 𝑥−10 75 8x2 – 230x +750 = 0 30 1 x = 25, 8 Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by ½ the larger pipe will become negative, which is logically not possible. Therefore, the time taken individually by the smaller pipe and ½ the larger pipe will be 25 and 25 – 10 =15 hours, respectively. 33. (a) Statement – ½ Given and To Prove – ½ Figure and Construction ½ 3 Proof – 1 ½ G [b] Draw DG ‖ BE 𝐴𝐵 𝐴𝐸 ½ In Δ ABE, = [BPT] 𝐵𝐷 𝐺𝐸 CF = FD [F is the midpoint of DC] ---(i) ½ 𝐷𝐹 𝐺𝐸 In Δ CDG, = = 1 [Mid point theorem] 𝐶𝐹 𝐶𝐸 ½ GE = CE ---(ii) ∠CEF = ∠CFE [Given] CF = CE [Sides opposite to equal angles] ---(iii) ½ From (ii) & (iii) CF = GE ---(iv) From (i) & (iv) GE = FD 𝐴𝐵 𝐴𝐸 𝐴𝐵 𝐴𝐸 ∴ = ⇒ = 𝐵𝐷 𝐺𝐸 𝐵𝐷 𝐹𝐷 34. Length of the pond, l= 50m, width of the pond, b = 44m 21 Water level is to rise by, h = 21 cm = m 100 21 Volume of water in the pond = lbh = 50 x 44 x m3 =462 m3 1 100 Diameter of the pipe = 14 cm 7 Radius of the pipe, r = 7cm = m 100 Area of cross-section of pipe = πr² 22 7 7 154 1 = x x = m2 ½ 7 100 100 10000 Rate at which the water is flowing through the pipe, h = 15km/h = 15000 m/h Volume of water flowing in 1 hour = Area of cross-section of pipe x height of water ½ coming out of pipe 154 = (10000 × 15000) 𝑚3 1 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑛𝑑 Time required to fill the pond = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑓𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑛 1 ℎ𝑜𝑢𝑟 1 462 ×10000 = = 2 hours 154 × 15000 Speed of water if the rise in water level is to be attained in 1 hour = 30km/h [or] 6  Radius of the cylindrical tent (r) = 14 m Total height of the tent = 13.5 m Height of the cylinder = 3 m Height of the Conical part = 10.5 m ½ Slant height of the cone (𝑙) = √ℎ2 + 𝑟 2 = √(10.5)2 + (14)2 = √110.25 + 196 1 = √306.25 = 17.5 m Curved surface area of cylindrical portion = 2πrh 22 = 2x ×14×3 1 7 = 264 m2 Curved surface area of conical portion =πrl 22 = ×14×17.5 1 7 =770 m2 ½ Total curved surface area = 264 m2 + 770 m2 = 1034 m2 Provision for stitching and wastage = 26 m2 ½ Area of canvas to be purchased = 1060 m2 Cost of canvas = Rate × Surface area ½ = 500 x 1060 = ₹ 5,30,000/- 35. Number of Cumulative Marks obtained students frequency 20 – 30 p p 30 – 40 15 p + 15 40 – 50 25 p + 40 1 50 – 60 20 p + 60 60 – 70 q p + q + 60 70 – 80 8 p + q + 68 ½ 80 - 90 10 p + q + 78 ½ 90 p + q + 78 = 90 p + q = 12 𝑛 −𝑐𝑓 Median =(𝑙) + 2.h 𝑓 45−(𝑝+40) ½ 50 = 50 +. 10 20 45−(𝑝+40) ½. 10 = 0 20 45 − (𝑝 + 40) = 0 ½ P=5 ½ 5 + q = 12 q=7 𝑓1−𝑓0 1 Mode = 𝑙 +.h 2𝑓1−𝑓0−𝑓2 7  25−15 = 40 +. 10 2(25)−15−20 100 = 40 + = 40 + 6.67 = 46.67 15 SECTION E 36. (i) Number of throws during camp. a = 40; d = 12 1 𝑡11 = 𝑎 + 10𝑑 = 40 + 10 × 12 = 160 𝑡ℎ𝑟𝑜𝑤𝑠 (ii) a = 7.56 m; d = 9cm = 0.09 m ½ n = 6 weeks ½ tn = a + (n-1) d ½ = 7.56 + 6(0.09) = 7.56 + 0.54 ½ Sanjitha’s throw distance at the end of 6 weeks = 8.1 m (or) a = 7.56 m; d = 9cm = 0.09 m ½ tn =11.16 m ½ tn = a + (n-1) d 11.16 = 7.56 + (n-1) (0.09) ½ 3.6 = (n-1) (0.09) 3.6 n-1 = = 40 0.09 n = 41 ½ Sanjitha’s will be able to throw 11.16 m in 41 weeks. (iii) a = 40; d = 12; n = 15 𝑛 ½ Sn = [2a + (n-1) d] 2 15 Sn = [2(40) + (15-1) (12)] 2 15 = [80 + 168] 2 15 ½ = =1860 throws 2 37. (i) Let D be (a,b), then Mid point of AC = Midpoint of BD 1+6 2+6 4+𝑎 3+𝑏 ½ ( , )=( , ) 2 2 2 2 4+a=7 3+ b = 8 a=3 b=5 Central midfielder is at (3,5) ½ 8  (ii) GH = √(−3 − 3)2 + (5 − 1)2 = √36 + 16 = √52 = 2√13 ½ ½ GK = √(0 + 3)2 + (3 − 5)2 = √9 + 4 = √13 ½ HK = √(3 − 0)2 + (1 − 3)2 = √9 + 4 = √13 ½ GK +HK = GH ⇒G,H & K lie on a same straight line [or] CJ = √(0 − 5) + (1 + 3) = √25 + 16 = √41 2 2 ½ ½ CI =√(0 + 4)2 + (1 − 6)2 = √16 + 25 = √41 Full-back J(5,-3) and centre-back I(-4,6) are equidistant from forward C(0,1) 5−4 −3+6 1 3 ½ Mid-point of IJ = ( , )=( , ) ½ 2 2 2 2 C is NOT the mid-point of IJ (iii) A,B and E lie on the same straight line and B is equidistant from A and E ⇒ B is the mid-point of AE 1+𝑎 4+𝑏 ½ ( , ) = (2, −3) ½ 2 2 1 + 𝑎 = 4 ; a = 3. 4+b = -6; b = -10 E is (3,-10) 38. 80 (i) tan 45° = 𝐶𝐵 ⇒ CB = 80m 1 80 ½ (ii) tan 30° = 𝐶𝐸 ½ 1 80 ½ ⇒ = ½ √3 𝐶𝐸 ⇒ CE = 80√3 Distance the bird flew = AD = BE = CE-CB = 80√3 – 80 = 80(√3 -1) m ½ (or) ½ 80 tan 60° = 𝐶𝐺 80 ½ ⇒ √3 = ½ 𝐶𝐺 80 ⇒ CG = √3 Distance the ball travelled after hitting the tree =FA=GB = CB -CG 80 1 GB = 80 - = 80 (1 - )m √3 √3 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 20(√3 + 1) ½ (iii) Speed of the bird = = m/sec 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 2 20(√3 + 1) ½ = x 60 m/min = 600(√3 + 1) m/min 2 9