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class 9 maths sample papers 2018-19, Exercises of Mathematics

The Central Board of Secondary Education (CBSE) is a national level board of education in India for public and private schools, controlled and managed by Union Government of India. CBSE has asked all schools affiliated to follow only NCERT curriculum.[2] There are approximately 20,299 schools in India and 220 schools in 28 foreign countries affiliated to the CBSE

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Uploaded on 04/19/2020

riya-vinoo
riya-vinoo 🇮🇳

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Download class 9 maths sample papers 2018-19 and more Exercises Mathematics in PDF only on Docsity! PRACTICE QUESTION PAPER-! CLASS-IX SUBJECT : MATHEMATICS Time : 3 Hrs. M.M. 80 General Instruction: 1. 2. All questions are compulsory. The paper consists of 30 questions divided into four section A, B, C, D. Section A comprises of 6 questions of 1 mark each. Section B compmrises of 6 questions of 2 marks each. Section C comprises of 10 questions of 3 marks each. Section D comprises of 8 questions of 4 marks each. There is no over all choice in this question paper. Although internal choices has been provided in some questions. SECTION-A Write the formula used to calculate the total surface area of a hemispherical solid of radius 'r’. If each side of a trianagle is doubled then how many times the area of triangle increased? A oO B If 2x = y then find the value of y? -7 Represent 5 on the number line. In which quadrants y co-ordinates are negative? How many solutions are there for equation y = x + 2? 10. 11. 12. SECTION-B Write the coefficients of x? in each of following. (i) 2-x2 +x (ii) J2x -1 P Ss Q R T In figure ZPQR = ZPRQ then prove that ZPQS = ZPRT. Q\ 60° Oo Find the value of x°. The angles of a quadrilateral are in the ratio 3: 5:9: 13. Find the greatest angle of the quadrilateral. PQRS is a rhombus with ZQPS = 50° find ZRQS. A AD and BC are equal perpendiculars to a line segment AB as given in figure, show that CD bisects AB. 26. 27. 28. 29. 30. Given below is the data of students who participated in different activities. Activity Sports Meditation Yoga Walking Number of Girls 40 35 100 120 Draw the bar graph. For the given data. Sides of a triangle are in ratio 12 : 17 : 25 and its perimeter is 540 cm. Find its area. OR A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m, the non parallel sides are 14 m and 13 m find the area of the field. If x + y +z =0 show that, x3 + y3 + 29 = 3xyz Draw the graph of following linear equation in two variables. x+y=4. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of 7.50 per m?. OR The volume of a right circular cone is 9856 cm’. If the diameter of the base is 28 cm find. (i) Height of the cone. (ii) Curved surface area of the cone. Pn SOLUTION 3nr? 3 times 2x=y x+y = 180° x + 2x = 180° x=60 y= 120° -7/5 1 1 1 n t t t t t S322 32 5 5 5 5 5 Ill and IV Quadrants Infinitely many solutions. ()-1, (iO ZPQR + ZPQS = ZPRQ+ ZPRT ZPQR + ZPQS = ZPQR+ ZPRT ZPQS = ZPRT x= 2x 60° = 120 (Linear Pair) ( ZPQR= ZPRQ) (angle subtended by an arc at the centre is double the angle subtended by the same arc on the remaining part of the circle). Let angles be 3x, 5x, 9x, 13x 3x + 5x + 9x + 13x = 360° 30x = 360° x =12° Greatest angle = 13x = 13 x 2 = 156. ZRQP + ZQPS = 180° ZRQP = 180°— 50° ZRQP = 130° ZRQS = 65° 12. INAOBC and AOAD 2B = ZA=90° (given) BC =AD (given) ZBOC = ZAOD (V.0.A) AOBC = AOAD (By AAS congruency rule) OB =OA (CPCT) CD bisects AB. 13. (5+ V7) (2+ V5)=10 + 5V5 + 2V7 + V35 14. i) P (2 He ay= 24 . (i)PC lead) = 799 ii) P (One Hi a= 11 (ii) P (One Heat = 30 (iii) P (Not Head) = = 15. No. of terms (n) = 10 (even) S the (5 ttn term Median = 44 7 __ jedian 2 _ 5'"term +6" term 2 X+x+2 63=—>—_ => x=62 2 OR M = 130 =13 jean = 10 = Mode = 12 16. (998) = (1000-2) (A-B)? = A3 — B? - 3AB (A-B) (998) = (1000)° — (2) - 3 x 1000 (1000-2 ) = 1000012000 — 600008 = 994011992 17. Construction of triangle. = 3, 4 2 23. InAAODand ACOD > OA= OC (diagonal of || gm) OD =OD A c (Common) AD =CD (Sides of rhombus) AAOD = ACOD AY (By SSS ) B ZAQD = ZCOD (CPCT) ZAOD + ZCOD = 180° (Linear Pair) <=. ZAOD + ZAOD = 180° (-- ZAOD = ZCOD) ZAOD = 90° Diagonals of rhombus are | to each other. 1 81 74 V2 7-2 24 742 742° 7-J2 49-4 = 72 47 AT 25. Correct proof of theorem. 26. 120 100 2 6 80 6 cs 60 Zz 40 20 o+ & 2 Gg Activity ——> g € “8 g Moip 27. _ Calcualtion of sides 120, 170 and 250 cm Area = ./S(S—a)(S— b)(S-c) Area = 9000 cm? A_10m_B 14m CM D 10m E¢———+C 15m OR ar ABEC = 84 cm? (Hero’s) Area x2 n= = 1.2m 1 Area of trapezium = 3h [a+b] = 196 m2 28. Using identity x3 + y® + z9 — 3xyz = (x + y +2) (x2 +y? +z? — xy — yz—- zx) if x+y+z=0 RHS => Ox [x?+y*+z*-xy-yz-zx]=0 LHS => x°+y3+z3-3xyz=0 Hence x? + y3 + z? = 3xyz 29. Correct graph forx +y=4 30. Area of 4 walls = 2h [| + b] = 54 m? Area of ceiling = L x b = 20 m2 Area to be white washed = 74 m? Cost = Area x Rate = 7 555 OR 1 Volume = = rer? h = 9856 =12x 1x 14% 14h ( h=48m (ii) CSA for cone = ml =50cm CSA = 2200 cm?
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