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CBSE Class 9 Maths Sample Paper - Studymate

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CBSE Class 9 Maths Sample Paper - Studymate

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  1. Sample Paper (2018-19) Mathematics (Set-1) Date : 04-02-19 Duration : 3 Hrs. Max. Marks : 80 Class IX Instructions: 4 All questions are compulsory. 4 The question paper consists of 30 questions divided into four sections A, B, C, and D. 4 Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each. 4 There is no overall choice. However, an internal choice has been provided in 2 questions of 1 mark each, 2 questions of 2 marks each, 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions. 4 Use of calculators is not permitted. Section – A 1. 2. Find the total surface area of a hemisphere of radius 7 cm. ABCD is a parallelogram having an area of 60cm2. P is any point on CD. Find the area of triangle APB. ( )− 1 ( ). 3. Simplify : 3 + − 3 2 3 4. Is (2, 0) a solution of x = 2y + 4? OR 5. The graph of the linear equation 2x – y = 4 cuts x -axis at ? PQRS is a parallelogram in which ∠PSR = 125°. What is the measure of ∠RQT? S R 125° P Q T 6. Find the class size, if the class marks of a frequency distribution are 6, 10, 14, 18, 22, 26 and 30. OR What is the class mark of the class interval 45 – 52? Section – B 7. Find the area of a triangle when two sides are 24 cm and 10 cm and the perimeter of triangle is 62 cm. If mean of seven observations 28, 32, x, x + 2, x + 5, 43, 45 is 38, find x. OR The following observations are arranged in ascending order; 26, 29, 42, 53, x, x + 2, 70, 75, 82, 93. If median is 65, find the value of x. 8. [1]

  2. 9. A bag contains cards numbered from 1 to 28. A card is drawn at random from the bag. Find the probability that the card bears a number (a) Which is a multiple of 6 10. Show that diagonals of a rhombus are perpendicular to each other. OR In a parallelogram, if diagonals are equal, show that it is a rectangle. (b) Which is greater than or equal to 27 =1 11. If a point C lies between two points A and B such that AC = BC, then prove that AC Explain using figure. 12.In the given figure, PQ||MN, find the measure of x. M AB . 2 N 110° 120° P x Q Section – C in the formp q ¹ 13. Express 0 123 . , 0 , p and q are integers. OR q 14.Factorise : 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz. 15. In the figure, it is given that LM = MN and LP = QN. Prove that DLMQ ≅ NMP. If 5x–3. 32x – 8 = 225, find the value of x. M L P Q N OR 16. Show that line segments joining the mid points of opposite sides of a quadrilateral bisect each other. OR Points M and N are taken on diagonal AC of a parallelogram ABCD such that AM = CN. Prove that BMDN is a parallelogram. D Prove that sum of any two sides of a triangle is greater than twice the median on the third side. C N M A B [2]

  3. 17. In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are the diagonals, if ∠DBC = 55° and ∠BAC = 45°, find ∠BCD. D A C B 18. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 revolutions to move once     22 7 . over to level a playground. Find the area of playground in m2. π =   OR A room is 22 m long, 15 m wide and 6 m high. Find the area of its four walls and cost of painting it at the rate of `12 per m2. 19. If mean of the following data is 20.2, find the missing frequency p : x f 10 6 15 8 20 p 25 10 30 6 20. Find the coordinates of the point: (a) which lies on x and y axes both. (b) whose ordinate is –4 and which lies on y-axis. (c) whose abscissa is 5 and which lies on x-axis. 21. If figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays 1 2 R ( ) ROS QOS POS . OP and OR. Prove that : ∠ = ∠ −∠ S P O Q 22. A die is rolled 120 times and the outcomes are recorded as follows : Outcomes Frequency Find the probability of getting (i) An even number (iii) An odd number greater than 1 1 even no. < 6 40 odd no. > 1 35 6 25 20 (ii) One Section – D 23. Solve the equation 2x + 1 = x – 3 and represent the solution(s) on (i) The number line. (ii) The Cartesian plane. OR Draw the graphs of two lines whose equations are x + y – 6 = 0 and x – y – 2 = 0 on the same graph paper. Also, find the coordinates of the point of intersection of the lines. 3 3 2 2 3 3 2 2 + − − + 24. If x y and , then find the value of x2 + y2. = = [3]

  4. 25. The polynomials ax3 + 3x2 – 3 and 2x3 – 5x + a when divided by (x – 4) leave the remainders R1 and R2 respectively. Find a, if R1 + R2 = 0. 26. D, E and F are respectively the mid-points of the sides BC, CA and AB of DABC. Show that 1 4 27. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of common chord. OR Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. ∆ ( )= ∆ ( ) ar ar DEF ABC . 28. Construct a triangle ABC, in which ∠B = 60°, ∠C = 45° and AB + BC + CA = 11 cm. 29. Draw the histogram and frequency polygon for the following data : Cost of living index 440 – 460 460 – 480 480 – 500 500 – 520 520 – 540 540 – 560 560 – 580 580 – 600 Total 30. Twenty seven solid iron spheres, each of radius ‘r’ and surface area ‘S’ are melted to form a sphere with surface area S′ and radius r′. Find the (i) Radius r′ of the new sphere OR At a fair, a stall keeper in one of the food stalls has a large cylinder of radius 15 cm filled upto a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses of radius 3 cm upto a height of 8 cm and sold for ` 3 each. How much money does the stall keeper receive by selling the juice completely? vvvvv Number of months 2 4 3 5 3 2 1 4 24 Ratio of S and S′. (ii) [4]

  5. Hints/Solutions to Sample Paper (2018-19) Mathematics (Set-1) Date : 04-02-19 Duration : 3 Hrs. Max. Marks : 80 Class IX TSA of a hemisphere = 3pr2 22 7 1. 2 3 × × 7 7 462 cm . = × = 2. ar(DAPB) = 30 cm2 3 3 + . 2 3. 4. 5. 6. x = 2y + 4, if y = 0 then x = 4 \ (2, 0) is not a solution of the given equation. OR (2, 0) is the required point. ∠RQT = 55° Class size = 4 OR Class mark = 45 2 2 a = 24 cm, b = 10 cm, c = 28 cm Semiperimeter(s) = 31 cm 52 97 + 48 5 . = = 7. \ ( ) ( ) Area of ∆ = s s ( a s ) b s c − − − 31 7 21 3 = × × × 2 = 21 31 cm sum of observations No. of observations 155 7 3x + 155 = 266 3x = 111 \ x = 37 OR th 8. Mean = 3 + x 38 = th       2   n 2 n 2       1 + + Median = th th 5 6 + [ ] n 65  10 = = 2 2 2 + x 65 65 1 = ⇒ = + x 2 \ x = 64 4 28 2 28 1 7 1 14 9. (a) = = (b) 10. Rhombus is also a parallelogram \ AO = CO Now, in DAOB and DCOB AO = CO D C O A B [5]

  6. 11. 12. 13. \ Also, ∠AOB + ∠COB = 180° (Linear pair) ⇒ 2∠AOB = 180° [From (i)] ⇒ ∠AOB = 90° Hence, AC⊥BD. OR Refer to NCERT, Ex. 8.1, Q. 2 Refer to NCERT, Ex. 5.1, Q. 4 Draw EF||MN Also, PQ||MN (given) \ EF||PQ ∠ 1 + ∠ 110° = 180° ⇒∠ 1 = 70° ∠ 2 + ∠ 120° = 180° ⇒∠ 2 = 60° \ x = ∠ 1 + ∠2 = 70° + 60° = 130° Let x = 0. 123 ⇒ 100 x = 12.3 ⇒ 1000x = 123.3 ⇒ 1000x – 100x = 123.3 – 12.3 900x = 111 111 900 300 BO = BO (common) AB = CB (sides of rhombus) DAOB ≅DCOB (By SSS) ∠AOB = ∠COB (cpct) ...(i) N M 110° 120° P 1 2 E F Q ...(i) ..(ii) [Multiply eq. (i) by 100] ...(iii) [Multiply eq. (ii) by 10] [Subtracting (ii) from (iii)] 37 x = = 37 300 \ 0 123 . = 14. 15. 16. 17. 5x – 3 . 32x – 8 = 225 5x – 3 . 32x – 8 52 . 32 ⇒ x – 3 = 2 and 2x – 8 =2 [Comparing powers of equal bases] \ x = 5 (4x)2 + (–2y)2 + (3z)2 + 2(4x)(–2y) + 2(–2y)(3z)+ 2(3z)(4x) =(4x – 2y + 3z)2 [Qa2 + b2 + c2 + 2ab + 2bc +2ca = (a + b + c)2] = (4x – 2y + 3z) (4x – 2y + 3z) LM = NM (given) ⇒ ∠1 = ∠2 (isosceles D theorem) LP = QN (given) ⇒ LQ = NP [LP + PQ = NQ + PQ] So, by SAS congruency, DLMQ ≅DNMP OR Produce median AD to point E such that AD = DE In DADB and DEDC BD = CD (AD is median on BC) AD= ED (By construction) ∠ADB = ∠EDC (vertically opp. angles) \ DADB ≅ DEDC (by SAS) ⇒ AB = EC (CPCT) Now, in DACE AC + EC > AE AC + AB > 2AD [Q AB = EC as in equation (i) and AD = ED] Refer to NCERT, Ex. 8.2, Q. 6. OR Refer to NCERT, Ex. 8.1, Q. 9 ∠DAC = ∠DBC (Angles in same segment) \ ∠DAC = 55° ∠BAD = ∠DAC + ∠BAC = 55° + 45° = 100° Now, ∠BAD + ∠BCD = 180° [ABCD is cyclic] OR M 1 2 L P Q N A ...(i) D B C E [6]

  7. \ ∠BCD = 80° 42 100 120 100 21 50 = Radius (r) = 42 cm = m m = 18. 6 5 m m Length (h) = 120 cm = \ Area of playground = No. of revolutions × CSA of roller = 500 × 2prh 22 7 21 50 6 5m2 = 1584 m2 500 × × 2 = × × 19. Area of four walls = 2 (l + b) × h = 2 (22 + 15) × 6 = 444 m2 Cost of painting = Area of four walls × rate = 444 × 12 = `5328 OR x 10 15 20 25 30 Mean = 20.2 f 6 8 p 10 6 Σf = 30 + p fx 60 120 20p 250 180 Σfx = 610 + 20p fx f Now, Mean =Σ Σ 610 30 20 p + + p ⇒ ⇒ 606 + 20.2p = 610 + 20p ⇒ 0.2p = 4 20 2 . = 4 . p = ⇒ \p = 20 0 2 20. 21. (a) ∠POR = ∠QOR (each 90°) ⇒ ∠POS + ∠ROS = ∠QOR ⇒ ∠POS + ∠ROS + ∠ROS = ∠QOR + ∠ROS [Add ∠ROS both sides] ⇒ ∠POS + 2∠ROS = ∠QOS 1 2 (i) 120 120 2 35 120 24 Refer to NCERT, Chapter 4, Example No. 9 (0, 0) (b) (0, –4) (c) (5, 0) S R ( ) QOS POS ∠ROS = 40 + ∠ 60 −∠ 1 \ P O Q 20 25 120 5 24 22. (ii) = = = 7 = (iii) 23. y-axis (0, 6) x + y – 6 = 0 OR 6 5 4 3 2 1 x + y – 6 = 0 x – y – 2 = 0 (4, 2) x 0 6 4 y 6 0 2 x – y – 2 = 0 (3, 1) (6, 0) x 3 2 0 (2, 0) x-axis –2 –1 y 1 0 –2 1 2 3 4 5 6 –1 –2 Coordinates of point of intersection is (4, 2). (0, –2) 2 2 =( ) +( ) + ( ) −( ) − ( ) −( 2 2 6 ) 24 20 6 )( ) 3 2 2 2 3 2 3 3 2 2 3 3 2 2 + − + + 5 2 6 x = × = + 24. 2 ( ) 3 2 2 2 =( ) +( )( ) 3 2 2 2 3 2 3 3 2 2 3 3 2 2 − + − − 5 2 6 y = × = − 2 ( ) 3 2 2 ( ) + )+ ( 2 2 \ x y 5 2 6 5 + = + − ( ( )= 25 24 20 6 25 98 = + + + − 25. R1 = a(4)3 + 3(4)2 – 3 = 64 a + 45 R2 = 2(4)3 – 5(4) + a = 108 + a Now, R1 + R2 = 0 [Given] \ 64a + 45 + 108 + a = 0 ⇒ 65a + 153 = 0 a = −153 65 \ [7]

  8. 26. 27. 28. 29. Refer to NCERT, Ex, 9.3, Q. 5 Refer to NCERT, Ex. 10.3, Q. 3 OR Refer to NCERT, Chapter 10 Theorem 10.8 Refer to NCERT, Page No.195 Example No. 1 No. of Months 6 5 4 3 2 1 5 4 E 4 3 3 I C F 2 D 2 G 1 B H J A 420 440 460 480 500 520 540 560 580 600 620 Cost of Living Index 30. Refer to NCERT, Ex. 13.8, Q. 9 OR Volume of orange juice = pr2h = p(15)2 × 32 cm3 Volume of orange juice in one glass = p(3)2 × 8 cm3 2 3 ( π ) × 3 15 32 8 cm cm π \ No of glasses filled with orange juice = 2 3 ( ) × \ Cost of orange juice = No. of glasses filled with juice × Rate Stall keeper receives `300 by selling the juice completely. = 100 glasses = 100 × 3 = ` 300 vvvvvvvvvv [8]

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