- 1. SUMMATIVE ASSESSMENT MATHS CLASS 10 CBSE Sample Papers - Solutions - SA2 Time: 3 Hrs Max Marks: 90 General Instructions: A) All questions are compulsory. B) The question paper consists of 34 questions divided into four sections A, B, C and D. a. Section A comprises of 8 questions of 1 mark each b. Section B comprises of 6 questions of 2 marks each c. Section C comprises of 10 questions of 3 marks each d. Section D comprises 10 questions of 4 marks each C) Question numbers 1 to 8 in section A are multiple choice questions where you are to select one correct option out of the given four. D) Use of calculator is not permitted. E) An additional 15 minutes time has been allotted to read this question paper only www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 2. SECTION – A 1. A Line cuts X axis at (18, 0) and Y axis at (0, -8). The circum-centre of the triangle formed by the line with the axis is a) (-4, 9) b) (9, - 4) c) (0, 0) d) (18, -8) Midpoint of (18, 0) and (0, -8) is ( ) 2. To construct the triangle similar to a given △ ABC with its sides of the corresponding sides of △ ABC, draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is a) 8 b) 13 c) 5 d) 3 3. If the circumference of a circle is equal to the perimeter of a square, then the ratio of their areas is a) 7 : 11 b) 14 : 11 c) 22 : 7 d) 7 : 22 4. A die is thrown twice. The probability that two will not come up either time is a) b) c) d) None of these The probability that 2 will not come either time: events are (1,1), (1,3), (1,4), (1,5), (1,6), (3,1), (3,3), (3,4), (3,5), (3,6), (4,1), (4,3), (4,4), (4,5), (4,6), (5,1), (5,3), (5,4), (5,5), (5,6), (6,1), (6,3), (6,4), (6,5), (6,6) www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 3. Total events are: 36 P = 5. The number of points on X axis which are at a distance of 2 units from (2,4) is a) 0 b) 1 c) 2 d) 3 The ordinate of point (2, 4) is 4 because 4 > 2 There is no point that is on X axis at a distance of 2 units from (2, 4) 6. For to have equal roots, k can take the values a) 2 and -2 b) 4 and -4 c) 8 and -8 d) 16 and -16 For a quadratic equation to have equal roots D = 0 i.e., 7. The sum of first 11 terms of an AP whose middle term is 30 is a) 320 b) 330 c) 340 d) None of these The middle term [ ] 30 = 330 8. If AB = 4 m and AC = 8 m then angle of observation of point A as observed from C is a) 60 b) 30 c) 45 d) Cannot be determined www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 4. Point on Y axis is (0, 1) SECTION – B 9. Find the values of k for which the given equation has real and distinct roots : _____________________ For a quadratic equation to have real and distinct root D > 0 10. If the 3rd and 9th terms of an AP are4 and -8 respectively, which term of this AP is zero? ------- 1 ----- 2 Subtracting 2 from 1 – – – Substituting in 1 Let 11. A box contains 12 balls out of whichx are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Findx. i) ii) ( ) 12. Find a point on the Y axis which is equidistant from (-3,2) and (1,-2) Let the point on Y axis be (0, y) Distance between (-3, 2) & (0, y) = Distance between (1, -2) and (0, y) √ √ Squaring both sides www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 5. ( ) 13. The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. Find its area. Perimeter of the quadrant = cm Area of the quadrant = cm2 14. Cards marked with numbers 2 to 90 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the card drawn is i) a two digit number________ ii) a number which is a perfect square_________ i) Two digit number from 10, 11…….90 are 81 ii) Perfect squares between 2 to 90 are 4, 9, 16, 25, 36, 49, 64, and 81 SECTION – C 15. In the given figure, O is the centre of the circle. Determine AQB and AMB if PA and PB are tangents. In quadrilateral PAOB, P = 75 , PAO = PBO = 90 AOB = 105 ------ angle sum property of quadrilateral Q = (Angle subtended by the arc AMB at centre = angle subtended at point Q on circle) AMB = 180 – 52.5 = 127.5 (cyclic quadrilateral property) 16. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return down stream to the same spot. Find the speed of stream. Let the speed of stream = x km/h Speed of motor boat upstream = 18 - x km/h and downstream = 18 + x km/h Time = Given: ( ) www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 6. or 6 Since speed cannot be negative is rejected km/h 17. Which term of the AP 3, 10, 17…..will be 84 more than its 13th term? given: Ans: 25th term 18. An airplane when 3000 m high, passes vertically above another plane at an instant, when the angle of elevation of the two airplanes from the same point on the ground are 60 and 45 respectively. Find the vertical distance between airplanes. √ √ √ √ √ m Height = 3000 – 1732 = 1268 m 19. ABCDE is a polygon are A(-1,0), B(4,0), C(4,4), D(0,7) and E(-6,2). Find the area of the polygon. Area whose vertices of polygon = Area of ABC + Area of AEC + Area of EDC [ ] Area of polygon = = 42 square units 20. ABC, an isosceles in which AB = AC, is circumscribed about a circle. Show that BC is bisected at the point of contact. AB = AC ----- given, also AF = AE (tangents from A) AB - AF = AC – AE BF = CE But BF = BD and CE = CD (tangents from external point) BD = CD BC is bisected at the point of contact D. www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 7. 21. The co-efficient of x in the quadratic equation was taken as 17 in place of 13, its roots were found to be -2 and -15. Find the roots of the original equation. Substituting for x = (- 2) in In original equation Original equation: 22. Find the area of the shaded region in figure, where a circular arc of radius 6 cm has been drawn with vertex 'o' of an equilateral OAB of side 12 cm as centre ( ) AOB = 60 angle of major arc = 360 – 60 = 300 Area of major arc COD = cm2 Area of equilateral = √ cm2 Area of the shaded region = 94.286 + 62.352 = 156.638 cm2 23. A sector of circle of radius 12 cm has the angle 120 . It is rolled up so that the two bounding radius are joined together to form a cone. Find the volume of the cone. Radius of the sector = 12 cm Angle of sector = 120 Length of arc = Circumference of the base of the cone = cm Length of the cone = 12 cm Volume of the cone = √ √ √ √ cm3 cm3 www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 8. 24. A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is i) Black and a king _______ ii) Spade or an ace _______ iii) neither a heart nor a king _______ i) ii) iii) SECTION D 25. Solve for x : ( ) ( ) ( ) ( ) or 26. Find the sum of the integers between 100 and 200 that are i) divisible by 9 ii) not divisible by 9 i) Integers divisible by 9 are 108, 117 ….. 198 – ii) Integers not divisible by 9 are: 101, 102 ….. 199 except those in (i) Sum of integers between 100 and 200 that are not divisible by 9 = 14850 – 1683 = 13167 www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 9. 27. From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20m high building are 45 and 60 respectively. Find the height of the tower. Tan m Tan √ √ √ m 28. ABC is a right, right angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of ABC. (Use = 3.14). Let AP = AR = x; BP = BQ = y; CQ = CR = z ----- 1 AB2 + AC2 = BC2 AC2 = BC2 – AB2 102 - 62 = 100 - 36 AC = √ = 8 cm Given AB = 6 cm and BC = 10 cm AB = AP + PB = x + y = 6 (From 1) BC = BQ + QC = y + z = 10 ---- 2 (From 1) AC = CR + RA = x + z = 8 (From 1) Adding : 2(x + y + z) = 24 x + y + z = 12 Substituting --- 2 in the above , we get x = 2 cm AP ⊥ OP, AR ⊥ OR and A = 90 APOR is a square with side x = r = 2 cm Area of shaded region = Area of ABC - Area of circle Area of shaded region = = 24 - 4 = 24 - 4 (3.14) = 24 - (12.56) = 11.44 cm2 www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 10. 29. A metallic bucket is in the shape of a frustum of a cone mounted on a hollow cylindrical base as shown in the figure. If the diameters of two circular ends of the bucket are 45 cm and 25 cm respectively, the total vertical height is 30 cm and that of the cylindrical portion is 6 cm. Find the area of the metallic sheet used to make the bucket. √ √ √ √ √ Lateral Surface Area of frustum of cone = ( ) ( ) SA of base of frustum = SA of hollow cylinder at base = Total SA = cm2 30. If A (5,-1), B (-3,-2) and C (-1,8) are the vertices of ABC, find the length of median through A and the co-ordinates of the centroid. By midpoint formula, co-ordinates of D are ( ) The centroid divides the median in the ration 2 : 1 Co-ordinates of G are ( ) ( ) Length of median AD = √ √ = √ √ units 31. Construct a ABC in which AB = 5 cm, B = 60 and altitude CD = 3 cm. Construct a A1 BC1 and ABC such that each side is 1.5 times that of the corresponding sides of ABC. Steps of construction :Draw AB = 5 cm. Draw ABQ = 60 At A, draw AX ⊥ BA. Mark AY = 3 cm on ray AX At Y, draw a ⊥ that cuts ray BQ at C. Join AC. ABC is the required www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 11. At B draw acute angle ABP Locate 3 points B1 , B2, B3 on BP such that BB1 = B1 B2 = B2 B3 Join AB2 and draw a line parallel to AB2 through B3 intersecting the extended line segment BA at A1 Draw a line through A1 parallel to AC intersecting the extended line BC at C1 A1 BC1 is the required triangle. 32. QR is a tangent at Q. PR ǁ AQ, where AQ is a chord through A and P is a centre, the end point of the diameter AB. Prove that BR is tangent at B. AQ ǁ PR 2 = 3 and 1 = 4 But 1 = 2 (PA = PQ radius) 3 = 4 In PQR and PBR, PR is common 3 = 4 and PQ = PB radius PQR PBR PQR = PBR = 90 ----- cpct. BR is a tangent 33. A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of 35 paise per square cm. (Use √ = 1.7) Area of 6 segments = 6 [Area of sector – Area of OAB] = [ √ ] = [ ] [ ] = [ ] [ ] = cm2 Area of 6 designs = 464.8 cm2 Cost = www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in
- 12. 34. An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar, if one cubic cm of iron weighs 10 g. ( ) Volume of pillar = Vol. of cylinder + Vol. of cone. ( ) ( ) = cm2 Weight of the pillar = 50688 10 g = 506880 506.88 kg www.LearnCBSE.in www.LearnCBSE.in w w w .LearnCBSE.in