CBSE Sample Paper-05 (Solved) SUMMATIVE ASSESSMENT I MATHEMATICS Class IX. Time allowed: 3 hours Maximum Marks: 90. Section A.

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1 CBSE Sample Paper-05 (Solved) SUMMATIVE ASSESSMENT I MATHEMATICS Class IX Time allowed: hours Maximum Marks: 90 General Instructions: a) All questions are compulsory. b) The question paper consists of 1 questions divided into four sections A, B, C and D. c) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of marks each, Section C contains 10 questions of marks each and Section D contains 11 questions of 4 marks each. d) Use of calculator is not permitted. 1. The p form of the number 0.8 is q. In figure the measure of a is Section A. The distance of the point (-6, -) from y-axis is Two angles of triangles are 65 and 45 respectively. Find third angles. Section B Write the following numbers in ascending order: 6, 7, 8 p x x 5x Find the zeroes of the polynomial ( ) 4 7. Find the remainder when x 6x x x x is divided by ( ) 8. In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that: ROS 1 ( QOS POS) 9. In a ABC, 0A + 6B 5C. Determine A, B and C.

2 10. Draw a triangle ABC where vertices A, B and C are (0, ), (, ) and (,) Section C respectively. 11. Find five rational numbers between 5 and Simplify: Simplify: a b + a a + b b a + b + b a a b f y y 8y y 5y 5 by y. 1. Divide ( ) If the polynomials px x x and each case is the same. Find the value of p. What must be added to ( x x 4x 1) ( x )? 4 + are divided by x, then the remainder in x x p 15. Factorize: a px + a qx apy 4aqy + pz + qz + to obtain a polynomial which is exactly divisible by 16. If a point C lies between two points A and B such that AC BC, then point C is called the midpoint of line segment AB. Prove that every line segment has one and only one mid-point. 17. In the figure, if AOC + BOD 66, then find all the four angles. If the figure, if AOC + BOC BOD 8, then find the all four angles. 18. If a line is perpendicular to one of the two given parallel lines then prove that it is also perpendicular to the other line.

3 19. In a triangle ABC, A + B 84 and B + C 146. Find the measure of each of the angles of the triangle. 0. In the figure, find x and y, if AB DF and AD FG. 1. Represent 5 on number line.. Rationalize the denominator of Section D Simplify: Ram has two rectangles in which their areas are given: (a) 5a 5a + 1 (b) 5y + 1y 1 (i) Give possible expressions for the length and breadth of each of the rectangles. (ii) Which mathematical concept is used in this problem? (iii) Which value is depicted in this problem? 4. Factorize , if x 1 is a factor of it. x x x Factorize by using factor theorem: y 7y Factorize : x + x 6. If lines AB, AC, AD and AE are parallel to a line l, then points A, B, C. D and E are collinear. 7. In the figure, AB CD and PQ RS, find the angles marked.

4 8. Two plane mirrors are placed perpendicular to each other, as shown in the figure. An incident ray AB to the first mirror is first reflected in the direction of BC and then reflected by the second mirror in the direction of CD. Prove that AB CD. 9. In the figure, it is given that A C and AB BC. Prove that ABD CBE. 0. Draw the graph of linear equation: 8x y The side of a square exceeds the side of another square by 4 cm and the sum of the areas of the two squares is 400 sq. cm. Find the dimensions of the squares.

5 CBSE Sample Paper-05 (Solved) SUMMATIVE ASSESSMENT I MATHEMATICS Class IX units L.C.M. of 6, and 4 is (Solutions) SECTION-A and 6 < 51 < 401 x 6 < 8 < x x x x x( x ) ( x ) 0 ( x )( x ) Zeroes are and. 7. By remainder theorem, 4 ( ) ( ) ( ) ( ) ( ) f ( ) f QOS POS ( QOR + ROS) POS 90 + ROS POS ( 90 POS) + ROS ( ROP POS) + ROS ROS < 51 < 401 Hence, ROS 1 ( QOS POS) 9. Given 0A 6B 5C A B C [Dividing by 0] A : B : C 1 : 5 : 6 Let A x, B 5x and C 6x x + 5x + 6x 180 1x 180 x 15 Hence A 15, B 75 and C 90

6 10. r + s 11. A rational number between r and s is Therefore a rational number between 5 and A rational number between and Hence five rational numbers between 5 and a b + a a a b a + b + b a + b b b a a ( a b ) ( a + b ) b are,,,,

7 A x px + 4x + x Let ( ) 15. ( ) ( ) x B x x 4x + p g x According to question, A( ) B ( ) p( ) 4( ) ( ) 4 ( ) 4( ) 7 p p 7 p p p 1 f x x x x p Let ( ) and g ( x) x Let k be added to f ( x ) so that it may be exactly divisible by ( x ). ( ) ( 4 1) p x x x + x + k p ( ) ( ) ( ) ( ) k k 0 1+ k 0 k a px a qx apy 4aqy pz qz ( a px + a qx) + ( apy 4aqy) + ( pz + qz) a x( p + q) ay( p + q) + z ( p + q) ( p + q)( a x ay + z) 16. Given AC BC.(i) If possible let D be another mid-point of AB AD DB.(ii) Subtracting eq. (i) from eq. (ii), we get AD AC DB CB CD CD CD 0 CD 0 C and D coincide. Hence every line segment has one and only one mid-point. 17. AOC + BOD 66.(i) But BOD AOC [Vertically opposite] AOC + AOC 66

8 Now AOC + BOC BOC 180 BOC 47 AOD BOC AOD 47 [Linear pair] AOC + BOC + BOD 8.(i) AOC + BOC + BOD + AOD 60.(ii) From eq. (i) and eq. (ii), we get, 8 + AOD 60 AOD BOC, BOD 158 and AOC Given : l, m, n are three lines such that m n and l m. To prove: l n Proof : Since l m 1 90.(i) Now, m n and transversal intersects them. 1..(ii) [Corresponding angles] From eq. (i) and (ii), we get, l n 19. Given A + B 84 (i) And B + C 146 (ii) Adding eq. (i) and (ii), we get, A + B + B + C 0 ( A + B + C) + B B 0 B 50 Putting the value of B in eq. (i), we get, A A 4 Putting the value of B in eq. (ii), we get, 50 + C 146 C y [Straight angle] y 55 (i) Now AB is parallel to FD and transversal AD cuts them. D A [Alternate angles] D 65

1. Again ADFG, transversal FD cuts them. F D F 65 In triangle EFG, x + 65 + 55 18

9 1. Again ADFG, transversal FD cuts them. F D F 65 In triangle EFG, x x 60.(ii) x + F + y ( + ) ( 10 ) ( ) ( ) ( )( ) ( 6 ) ( 5) ( ) + ( ) 1. (i) (a) Area 5a 5a + 1 a a a a( 5a ) 4( 5a ) ( 5a )( 5a 4) So possible length and breadth are ( 5a ) and ( 5 4) a units respectively.

10 (b) Area 5y + 1y 1 5y + 8y 15y 1 7y ( 5y + 4) ( 5y + 4) ( 7y )( 5y + 4) So possible length and breadth are ( 7y ) and ( y + ) 5 4. (ii) Factorization of Polynomials. (iii) Expression of one s desires and news is very necessary. 4. Let us divide x x + 14x 10 by x 1to get the other factors ( x 1)( x x + 10) x x x f y y 7y + 6 Let ( ) ( x 1)( x 1x 10x + 10) ( x 1) x( x 1) 10( x 1) ( x 1)( x 1)( x 10) The constant term in f ( y ) is 6 and its factors are ± 1, ±, ±, ± 6. On putting y 1in given expression, we get, ( ) ( ) ( ) f ( ) ( ) ( ) So ( 1) f y is a factor of f ( y). Now we divide ( ) f y y 7y + 6 by y 1 to get other factors.

11 5. y 7 y + 6 ( y 1)( y + y 6) ( y 1)( y + y y 6) x 1 + x x x x ( ) ( y 1) y ( y + ) ( y + ) ( y 1)( y + )( y ) x 1 x x x x x 1 1 x x x x x x x x x x x x x + x x x x 6. Given : Lines AB, AC, AD and AE are parallel to line l. To prove: A, B, C, D and E are collinear. Proof : Since AB, AC, AD and AE are all parallel to line l. Therefore point A is outside l and lines AB, AC, AE are drawn through A and each line is parallel to l. But by parallel lines axiom, one and only one line can be drawn through A outside it and parallel to l. This is possible only when A, B, C, D and E all lie on the same line. Hence A, B, C, D and E are collinear. 7. PQRS 1+ EFS EFS BEQ EFD 54 [consecutive interior angles are supplementary when lines are parallel] 7 90 [Linear pair] [Vertically opposite angles] [Straight angles] EFD [Vertically opposite angles]

12 8. Let BO and CO be the normals to the mirrors. As mirrors are perpendicular to each other. SO their normals BO and CO are perpendicular. BOC 90 In right angled triangle OBC, + 90.(i) 1 [Angle of incident Angle of reflection] 4 [Angle of incident Angle of reflection] On adding, (ii) On adding eq. (i) and (ii), we get, ABC + BCD 180 But ABC and BCD are consecutive interior angles formed when the transversal BC intersect AB and CD. ABCD 9. In s AOE and COD, A C and AOE COD [Vertically opposite angles] A + AOE C + COD 180 AEO 180 CDO [ A + AEO 180 and C + COD + CDO 180 ] AEO CDO (i) Now, AEO + OEB 180 [ Angles of a linear pair] And CDO + ODB 180 [ Angles of a linear pair] AEO + OEB CDO + ODB OEB ODB CEB ADB (ii) [ OEB CEB and ODB ADB] In ADB and CBE, A C [Given] ADB CEB [From eq. (ii)] And AB BC [Given] ADB CBE [By AAS]

13 0. We have, 8x y y 8x + 4 8x 4 y + Table of coordinates: x y points A B C Join the points A, B, C. The straight line AC is the graph of the linear equation 8x y Let S1 and S be the two squares. Let the side of the square S be x cm in length. Then the side of square S1 is ( x + 4) cm. Area of square S1 ( x + 4) And Area of square S x We are given that, Area of square S1 + Area of square S 400 cm x x 400 ( ) x x x x + 8x x 400 x + x x + 4x x( x + 16) 1( x + 16) 0 ( x )( x ) x 16,1 As the length of the side of a square cannot be negative, therefore x 1 Side of square S1 x cm and side of square S 1 cm.

Q4. In ABC, AC = AB and B = 50. Find the value of C. SECTION B. Q5. Find two rational numbers between 1 2 and.

Q4. In ABC, AC = AB and B = 50. Find the value of C. SECTION B. Q5. Find two rational numbers between 1 2 and. SUMMATIVE ASSESSMENT 1 (2013 2014) CLASS IX (SET I) SUBJECT : MATHEMATICS Time: 3 hours M.M. : 90 General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 31 questions