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Page 1 Sample Paper 25 Solutions Sample Paper 25 Solutions Class- X Exam - 2023-24 Mathematics - Standard Time Allowed: 3 Hours Maximum Marks : 80 General Instructions : 1. This Question Paper has 5 Sections A-E. 2. Section A has 20 MCQs carrying 1 mark each 3. Section B has 5 questions carrying 02 marks each. 4. Section C has 6 questions carrying 03 marks each. 5. Section D has 4 questions carrying 05 marks each. 6. Section E has 3 case based integrated units of assessment (04 marks each) with sub-parts. 7. All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. 8. Draw neat figures wherever required. Take 7 22 π = wherever required if not stated. Section - A Section A consists of 20 questions of 1 mark each. 1. If the angle of depression of an object from a 75 m high tower is 30c, then what is the distance of the object from the tower? (a) 75 2 m (b) m75 3 (c) 75 m (d) 75 2 5. m Ans : [Board Term-2 OD 2017 We have tan 30c OB AB = 3 1 OB 75 = OB = m75 3 Thus (b) is correct option. 2. If one zero of a quadratic polynomial ( ) kx 3x k 2 + + is 2, then the value of k is (a) 6 5 (b) 6 5 - (c) 5 6 (d) 5 6 - Ans : [Board 2020 Delhi Basic] We have p x( ) kx 3x k 2 = + + Since, 2 is a zero of the quadratic polynomial p( )2 = 0 k k () () 2 3 2 2 + + = 0 4 6 k k + + = 0 5k =− 6 & k 5 6 =− Thus (d) is correct option. 3. Two concentric circles are of radii 10 cm and 8 cm, then the length of the chord of the larger circle which touches the smaller circle is (a) 6 cm (b) 12 cm (c) 18 cm (d) 9 cm Ans : Let O be the centre of the concentric circles of radii cm10 and 8 cm, respectively. Let AB be a chord of the larger circle touching the smaller circles at P. Then, AP = PB and OP = AB
Page 2 Sample Paper 25 Solutions Applying Pythagoras theorem in TOPA, we have OA2 OP AP 2 2 = + 100 64 AP2 = + AP 2 = 100 64 36 − = & AP = 6 cm AB = 2AP = 2 6 # = 12 cm Thus (b) is correct option. 4. If 1⁄2 is a root of the equation x kx 0 2 4 + − 5 = , then the value of k is (a) 2 (b) -2 (c) 4 1 (d) 2 1 Ans : We have x kx 4 2 5 + − = 0 Since, 1⁄2 is a root of the given quadratic equation, it must satisfy it. Thus k 2 1 2 1 4 5 2 b l + − b l = 0 k 4 1 2 4 5 + − = 0 k 4 1 2 + − 5 = 0 2 4 k - = 0 & k = 2 Thus (a) is correct option. 5. Consider the data: Class 65- 85 85- 105 105- 125 125- 145 145- 165 165- 185 185- 205 Frequency 4 5 13 20 14 7 4 The difference of the upper limit of the median class and the lower limit of the modal class is (a) 0 (b) 19 (c) 20 (d) 38 Ans : [Board Term-1 Foreign 2013] Class Frequency Cumulative frequency 65-85 4 7 85-105 5 9 105-125 13 22 125-145 20 42 145-165 14 56 165-185 7 63 185-205 4 67 Here, .33 5 N 2 2 = = 67 , which lies in the interval 125 145 - . Hence, upper limit of median class is 145. Here, we see that the highest frequency is 20 which lies in 125-145. Hence, the lower limit of modal class is 125. Required difference = Upper limit of median class - Lower limit of modal class = 145 125 20 − = Thus (c) is correct option. 6. An AP starts with a positive fraction and every alternate term is an integer. If the sum of the first 11 terms is 33, then the fourth term is (a) 2 (b) 3 (c) 5 (d) 6 Ans : (a) 2 We have S11 = 33 a d 2 116 @ 2 1 + 0 = 33 a d + 5 = 3 i.e. a6 = 3 & a4 = 2 Since, alternate terms are integers and the given sum is possible, a4 = 2. Thus (a) is correct option. 7. If the sum of the zeroes of the quadratic polynomial kx 2 3 x k 2 + + is equal to their product, then k equals (a) 3 1 (b) 3 1 - (c) 3 2 (d) 3 2 - Ans : [Board 2020 OD Basic] We have p x( ) kx 2 3 x k 2 = + + Comparing it byax bx c 2 + + , we get a = k , b = 2 and c = 3k .
Page 3 Sample Paper 25 Solutions Sum of zeroes, α β + a b =− k 2 =− Product of zeroes, αβ a c = k 3k = = 3 According to question, we have α β + = αβ k 2 - = 3 & k 3 2 =− Thus (d) is correct option. 8. TABC and TBDE are two equilateral triangle such that D is the mid-point of BC . Ratio of the areas of triangles ABC and BDE is ................. . (a) 1 : 4 (b) 4 : 1 (c) 1 : 3 (d) 3 : 1 Ans : [Board 2020 Delhi Standard] From the given information we have drawn the figure as below. ( ) ( ) BDE ABC area area T T ( ) ( ) BD BC 4 3 2 4 3 2 = ( ) BC BC 2 1 2 2 = ^ h BC 4BC2 2 = 1 4 = = 4 1: Thus (b) is correct option. 9. If sin b a θ = , then cos θ is equal to (a) b a b 2 2 - (b) a b (c) b b a 2 2 - (d) b a a 2 2 - Ans : We have sin θ b a = Hypotenuse Perpendicular = Base b a 2 2 = − So, cos θ Hypotenuse Base = b b a 2 2 = − Thus (c) is correct option. 10. An observer, 1.5 m tall is 20.5 away from a tower m22 high, then the angle of elevation of the top of the tower from the eye of observer is (a) 30c (b) 45c (c) 60c (d) 90c Ans : Let BE = 22 m be the height of the tower and AD = 1 5. m be the height of the observer. The point D be the observer’s eye. We draw DC < AB as shown below. Then, AB = 20 5. m = DC and EC = BE BC − = BE − AD = − 1 5.22 = 20 5. m [BC = AD] Let θ be the angle of elevation make by observer’s eye to the top of the tower i.e. +DCE , tan θ B P = DC CE = . . 20 5 20 5 = tan θ = 1 tan θ = tan 45c & θ = 45c Thus (b) is correct option. 11. HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, then the other number is (a) 36 (b) 35 (c) 9 (d) 81 Ans : [Board 2020 OD Basic] Let y be the second number. Since, product of two numbers is equal to product of LCM and HCM, 54 # y = LCM H # CF 54 # y = 162 27 # y 54 = 162 27 # = 81 Thus (b) is correct option.
Page 4 Sample Paper 25 Solutions 12. Volumes of two spheres are in the ratio 64 : 27. The the ratio of their surface areas is (a) 3 : 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9 Ans : [Board Term-2 OD 2011] Let the radii of the two spheres are r1 and r2 , respectively. Given, ratio of their volumes, V V1 2 : = :64 27 V V 2 1 27 64 = r r 3 4 2 3 3 4 1 3 p p 27 64 = r r 2 3 1 3 27 64 = r r 2 1 3 4 = Now, ratio of their surface area, S S 2 1 r r 4 4 2 2 1 2 p p = r r 2 2 1 2 = r r 2 1 2 = a k 3 4 2 = b l 9 16 = Hence, the required ratio of their surface area is 16 :9. Thus (d) is correct option. 13. Volume of a spherical shell is given by (a) 4 ( ) R r 2 2 π − (b) ( ) R r 3 3 π − (c) 4 ( ) R r 3 3 π − (d) ( ) R r 3 4 3 3 π − Ans : Volume of spherical shell R r 3 4 3 3 3 4 = π π − ( ) R r 3 4 3 3 π = − Thus (d) is correct option. 14. While computing mean of grouped data, we assume that the frequencies are (a) evenly distributed over all the classes (b) centred at the class marks of the classes (c) centred at the upper limits of the classes (d) centred at the lower limits of the classes Ans : (b) centred at the class marks of the classes Frequencies are centred at the class-marks of the classes. So, the option (b) is correct, which is the required answer. Thus (b) is correct option. 15. If an event cannot occur, then its probability is (a) 1 (b) 4 3 (c) 2 1 (d) 0 Ans : [Board Term-2 OD 2012] The event which cannot occur is said to be impossible event and probability of impossible event is zero. Thus (d) is correct option. 16. In Figure, in TABC , DE z BC such that AD = 2 4. cm, AB = 3 2. cm and AC = 8 cm, then what is the length of AE ? (a) 4 cm (b) 6 cm (c) 8 cm (d) 3 cm Ans : [Board 2020 Delhi Basic] We have DE z BC By BPT, DB AD EC AE = AD DB AD + AE EC AE = + AB AD AC AE = . . 3 2 2 4 AE 8 = & AE = cm6 Thus (b) is correct option. 17. The probability of getting a number greater then 3 in throwing a die is (a) 3 1 (b) 4 1 (c) 4 3 (d) 3 2
Page 5 Sample Paper 25 Solutions Ans : [Board Term-2 OD 2017] n S( ) = 6 n E( ) = 2 Required probability, P E( ) ( ) ( ) n S n E = 6 4 3 2 = = Thus (d) is correct option. 18. The distance of the point ( , - )12 5 from the origin is (a) 12 (b) 5 (c) 13 (d) 169 Ans : The distance between the origin and the point ( , x y) is x y 2 2 + . Therefore, the distance between the origin and point ( , - )12 5 d ( ) 12 0 5( )0 2 2 = − − + − = 144 25 + = 169 = 13 units Thus (c) is correct option. 19. Assertion : x y + − 4 0 = and 2 3 x k + −y = 0 has no solution if k = 2. Reason : a x1 1 + + b y c1 = 0 and a x2 2 + + b y c2 = 0 are consistent if a a k k 2 1 2 ! 1 . (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. Ans : For assertion, given equation has no solution if 2 1 k 1 3 ! 4 = − − i.e. 3 4 k 2 holds 2 1 3 ! 4 = : D Assertion is true. Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). Thus (b) is correct option. 20. Assertion : If the outer and inner diameter of a circular path is 10 m and 6 m then area of the path is 16 m2 π . Reason : If R and r be the radius of outer and inner circular path, then area of path is R r 2 2 π ^ h − . (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. Ans : Area of the path 2 10 2 6 2 2 π = ; E b b l l − π = ^ h 25 9 − 16 π = Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). Thus (a) is correct option. Section - B Section B consists of 5 questions of 2 marks each. 21. In the given figure, + + A B = and AD = BE. Show that DE AB || . Ans : [Board Term-1 2012] In TCAB, we have +A = +B (1) By isosceles triangle property we have AC = CB But, we have been given AD = BE (2) Dividing equation (2) by (1) we get, AD CD BE CE = By converse of BPT, DE AB || . Hence Proved