- 1. SUMMATIVE ASSESSMENT - II MATHS CBSE Samle Papers For Class 10 Maths - SAII - Solved - Paper 1 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 .LearnC BSE.in
- 2. SECTION – A 1. If - and 1 are the sum and product of roots of a quadratic equation respectively, then the equation is _____. a) = 0 b) = 0 c) = 0 d) = 0 = 0 = 0 = 0 2. The tops of two poles of height 25 m and 15 m are connected by a wire. If the wire makes an angle of 30∘ with the horizontal, then find the length of wire. a) 30 m b) 20 m c) 5 m d) 10 m According to the given figure, AB = 25 – 15 = 10 m The length of the string = AC sin = ⇒ = AC = 20 m 3. As shown in the given image , from an external point S, tangents SA and SB are drawn to circle O. CD is tangent to the circle at E. If AS = 12 cm, find the perimeter of ∆SCD. a) 20 cm b) 24 cm c) 25 cm d) 12 cm CA = CE and DE = DB (tangents from external points C and D) ------ (1) Perimeter of ∆SCD = SC + CD + SD = SC + CE + ED + SD = SC + CA + DB + SD (from (1)) = SA + SB = 12 + 12 = 24 cm Perimeter of ∆SCD = 24 cm www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 3. 4. TA and TB are tangents to circle O from point T. TA = 4x - 7 and TB = 3x + 5, find TA. a) 12 cm b) 24 cm c) 41 cm d) 36 cm Since tangents drawn from an external point to a circle are equal, TA = TB -----(1) 4x - 7 = 3x + 5 x = 12 cm TA = 4(12) – 7 = 48 – 7 = 41 cm TA = 41 cm 5. If AB = 12 cm and CD = 9 cm are two chords of a circle, then which of the following is true? a) AB is nearer to the centre of the circle b) CD is nearer to the centre of the circle c) AB and CD are equidistant from the centre of the circle d) Can’t say Of any two chords of a circle, the one which is larger is nearer to the centre of the circle 6. Which constant must be added or subtracted to solve the equation 5 - √ x + 3 = 0 by the method of completing the square? a) b) c) √ d) 5 - √ x + 3 = 0 ⇒ - √ x + = 0 ⇒ - 2( √ + √ ( √ ) = 0 ∴ √ should be added and subtracted √ = = www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 4. ∴ the number of red balls in the bag is 6 7. Given below are two similar triangles, ∆ABC and ∆AB'C' .Which of the following statements are true? a) AB' = AB b) AB = AB' c) AB = AB d) None of the above AB' is 5 parts: AA1 = AA2 = AA3 = AA4 = AA5 AB is 7 parts: AA1 = AA2 = AA3 = AA4 = AA5 = AA6 = AA7 The ratio of the sides of ABC to that of corresponding sides of the triangle AB'C' is 7:5 8. A rope that attaches a swing to a tree is 2.6 m long and the maximum difference between trajectories is an angle of 150⁰. The maximum distance travelled by the swing when the swing angle is maximum is a) 4.2 m b) 6.8 m c) 13.2 m d) 8.6 m length of the arc = 2 r = × 2 × × 2.6 = 6.8 m SECTION – B 9. If O ( , 2 ) is the midpoint of the line segment joining the points Q ( -7 , 3 ) and R (-5 , 1) , then find the value of m. Applying the Mid-point Formula for the line joining Q (-7,3) and R(-5,1),we get = = = -6 m = -6 x 5 = -30 10. A bag contains 12 green balls and some red balls. If the probability of drawing a red ball is half that of a green ball, then the number of red balls in the bag is? No. of green balls = 12 Let no. of red balls = x ∴ Total no. of balls = 12 + x Given P (red ball) = P(green ball) = ⇒ x = 6 www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 5. 11. Find the numerical difference of the roots of - 7x - 9 = 0. Roots of the equation - 7x - 9 = 0 are x = √ ⇒ √ ⇒x = √ , √ ⇒ Numerical difference of the roots = √ √ = √ = √ 12. Two unbiased dice are thrown. Find the probability that i) the total score is more than 5 ii) the total score is less than 5 i) Total outcomes are 36 (i) The favorable outcomes are: (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6). ∴ P = = ii) The favourable outcomes are: (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1) = 6 ∴ P = = 13. Given below is the group-wise heights of children in cms in a sports academy: Group A - 131, 141, 115, 161 Group B - 119, 125, 131, 137 Group C - 102, 130, 142, 150 Group D - 105, 104, 103, 101 Which of the following groups forms an AP? Consider the group A 131,141,115,161 a2 – a1 = 10 a3 – a2 = -26 a4 – a3 = 46 as the difference is not same they do not form AP. Consider the group B 119,125,131,137 a2 – a1 = 6 a3 – a2 = 6 a4 – a3 = 6 www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 6. as the difference is same they form an AP. Consider the group C 102,130,142,150 a2 – a1 = 28 a3 – a2 = 12 a4 – a3 = 8 as the difference is not same it does not form an AP. Consider the group D 105,104,103,101 a2 – a1 = -1 a3 – a2 = -1 a4 – a3 = -2 as the difference is not same they do not form A 14. A horse is tied to a pole fixed at the middle of the side of a square field of edge 65 m with a rope of 18 m long. Find the area of the field in which the horse can graze? As the pole is on side of square, θ = 180 Area of the sector = × × r² = × × 18 ×18 = 509.14 m² Area of the field in which the horse can graze is 509.14 m² SECTION – C 15. The surface area of a metallic sphere is 616 cm2 . It is melted and recast into a cone of height 28 cm. Find diameter of the base of cone so formed. TSA of sphere = 4 r2 4 r2 = 616 r2 = 616 r2 = 49 r = 7 cm Volume of sphere = Volume of cone r3 = r2 h 7 7 7 = r2 28 r2 = 49 R = 7 cm Diameter of cone = 7 2 = 14 cm www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 7. 16. Find the length of wire of diameter cm that can be drawn from a solid sphere of radius 12 cm (Write the answer in m) Radius of wire (r) = = Volume of sphere = × πr3 = × 3.14 × 12 × 12 × 12 Volume of wire = Volume of sphere 3.14 × × × h = × 3.14 × 12 × 12 × 12 h = = 25600 cm = 256 m Diameter of well = d = 6 m Radius of well = r = =3 m Height of well = h = 30 m Length of platform = l = 24 m Breadth of platform = b = 12 m Let height of platform =H According to given condition we have: Volume of Well = Volume of Earth Dug out ⇒ = l × b × H ⇒ ×3×3×30=24×12×H ⇒H = × 3 × 3 × 30 × × ⇒H = 2.946 m 17. If tangents PA and PB from a point P to a c ircle with centre O are inclined to each other at angle of 70⁰, then find POA. ∠OAP = 90 ∠OBP = 90 ∠APB = 70 Sum of the angles of a Quadrilateral = 360⁰ ∠AOB = 360 - (90 + 90 +70 ) ∠AOB = 110 ∠POA = (∠AOB) ∠POA = 55 18. Determine the ratio in which the line segment joining the points A (1, -1) and B (4, 9) is divided by the line . Let the line divide the line segment in the ratio k:1. Then the co-ordinates of the point of division are : ( ) www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 8. Since this point also lies on the given line , 3( ) – 2( ) + 4 = 0 = 0 -2k + 9 = 0 => k = Therefore the ratio is 9: 2 19. Sum of the areas of two squares is 640 m2 . If the difference of their perimeters is 64 m, find the sides of the two squares. Let the sides of two squares be a and b meters. Given a2 +b2 = 640 and 4a−4b = 64 ⇒4(a−b) = 64 ⇒a−b = 16 →(1) (a−b2 )=162 ⇒a2 + b2 −2ab = 256 640 − 2ab = 256 ⇒−2ab = 256 − 640 ⇒ 2ab = 384 (a + b)2 = a2 + b2 + 2ab = 640 + 384 = 1024 ⇒a + b = √ ⇒a + b = 32 →(2) Solving equation (1) and (2), a = 24 ⇒ a – b = 16 ⇒ 24 – b = 16 ⇒ −b = 16 – 24 = -8 b = 8 m, a = 24 m 20. A lot consists of 52 laptops of which 46 are good, 4 have only minor defects and 2 have major defects. Smith will buy a phone if it is good but the trader will only buy it only if it has no major defect. One phone is selected at random, from the lot. Find the probability that i) it is acceptable to Smith ii) it is acceptable to the trader iii) it is acceptable neither to Smith nor to the trader Total number of laptops = 52 No. of good laptops = 46 Minor defect laptops = 4 , Major defect laptops = 2 i) P(good laptops) = = ii) No. of good laptops + Laptops with minor defect = 46 + 4 = 50 ∴ P(acceptable to the trader ) = = iii) P(acceptable neither to Smith nor to the trader) = = www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 9. 21. In a nursery, 37 plants have been arranged in the first row, 35 in the second, 33 in the third and so on. If there are 5 plants in last row, how many plants are there in the nursery? The given sequence is: 37, 35, 33……, 5 Here a = 37 ; d= -2 and an = 5 an = a + (n-1)d 5 = 37 + (n-1)-2 5 = 37 +2 -2n 5 = 39 -2n 2n = 34 n = 17 Sn = (a + an) (37 + 5) × 42 =17 x 21 = 357 There are 357 roses on the flower bed.. 22. An electrician needs to reach a point 1.6 m below the top of the pole of height 8 m to undertake the repair work. If he has to incline the ladder at 60 degrees angle to the horizontal, what should be the length of the ladder to enable him to reach the required position? Let BP be the required pole of height 8 m. Let A be the point below the top P such that AP = 1.6 m BA = 8 - 1.6 = 6.4 m In ABC triangle, Sin 60 = √ = AC = √ = √ 4.26√ = 7.38 m The length of the ladder should be 7.38 m 23. A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical portion is 53 m, find the area of canvas needed to make the tent. Area of Canvas Required = CSA of Cylindrical part + CSA of Conical part = 2 × (r) (h) + × (r) (l) = (r) [2 (h) + (l)] = (52.5) [2 (3) + 53] = 22 × 7.5 × 59 = 9735 m2 www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 10. 24. Draw a line segment of 8 cm and then divide it internally in the ratio 3:2. To divide a line segment of 8 cm internally in the ratio 3:2 a) Draw a line segment AB = 8 cm b) Draw another ray Making an acute angle BAX with AB c) Along AX mark off 5(2 +3 ) points A1,A2,A3,A4 and A5 such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 d) Join BA5 e) Through A3 draw a line A3P parallel to BA5 AP : PB = 3 : 2 Section - D 25. If the co-ordinates of the mid points of sides AB, BC and CA of ∆ABC are (1, 1), (2, -3) and (3, 4) respectively, then find the co-ordinates of the centroid. Let A = (x1, y1) B = (x2, y2) and C = (x3, y3) Let P be the mid-point of AB. ⇒( ) = (1, 1) ⇒x1 + x2 = 2, y1 + y2 = 2 →(1) Q is the mid-point of BC ⇒( ) =(2,−3) ⇒x2 + x3 = 4, y2 + y3 = −6 →(2) R is the mid-point of CA ⇒ ( ) = (3, 4) ⇒x3 + x1 = 6, y3 + y1 = 8 →(3) From (1), (2) and (3), x1 + x2 + x2 + x3 + x3 + x1 = 12 y1 + y2 + y2 + y3 + y3 + y1 = 2 – 6 + 8 ⇒x1 + x2 + x3 = 6 y1 + y2 + y3 = 2→(4) ∴ Co-ordinates of the centroid of D ABC are ( ) = ( ) = ( ) www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 11. 26. A park’s jogging track has its left and right ends as semi-circles. The distance between the two inner parallel line segments is 60 m and the length of the straight part of the track is 118 m. If the track is 12 m wide, find i) the length of the track along its inner edge. ii) the area of the track. i) Radius of inner semi-circle = = 30 m Circumference of inner semicircle = πr = × 30 = m The length of the track along its inner edge = length of two inner parallel lines + circumference of two inner semicircles. = 118 + 118 + + = 236 + 188.57 = 424.57 m ii) Area of the track = area of two rectangles ABCD and EFGH + 2( area enclosed between outer and inner semi-circles) ---- (1) Area of rectangle ABCD = length x width of track = 118 x 12=1416 m Area of two rectangles=2 x 1080=2832 sq m ----(2) Diameter of outer semi-circle= 60 + width of track + width of track = 60 + 12 + 12=84 m Radius of outer semi-circle = = 42 m Area enclosed between semi-circles = Area of bigger semi-circle - area of smaller semi-circle = [(r1 2 - r2 2 ] = [(422 - 302 ] = (1764 - 900) = (864) = 1357.7 m2 ⇒Area enclosed by left and right semi-circles=2 x 1357.7 =2715.4 m2 -------- (3) Putting (2) and (3) in (1), we get Area of the track = area of two rectangles formed in the track + area enclosed by the semi-circles = 2832 + 2715.4 = 4988.56 m2 27. If TA and TB are two tangents drawn to a circle with centre O from an external point T, prove that ATB = 2OAB. OAT = OBT = 90 (radius is perpendicular to tangent) In quadrilateral OATB OAT + ATB + TBO + BOA = 360 90 +ATB + 90 +BOA = 360 ATB +BOA = 180 -----(1) www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 12. If two tan gents are drawn from an external point to a circle then the tangents are equal in length. AF = AE , FB = BD , EC = CD Perimeter of triangle ABC = AB + BC+ CA = AF + FB + BD + DC + CE +EA = AF + BD + BD + CE + CE + A F (Since AF = AE, FB = BD, EC = CD) = AF + AF + BD + BD + CE + CE = 2(AF + BD+ CE) Perimeter of triangle ABC = 2(AF + BD+ CE) In OAB, OA = OB OAB = OBA (angles opp. to equal sides are equal) OAB + OBA + BOA = 180 2(OAB) + BOA = 180 BOA = 180 - 2(OAB) ----(2) from (1) and (2) ATB = 2(OAB) Hence proved. 28. In the given figure, D, E, and F are the points where the incircle of the triangle ABC touches the sides BC, CA, and AB respectively. Show that AF + BD + CE = AE + BF + CD = (perimeter of triangle ABC) AF + BD + CE = (perimeter of ABC) ……..(1) Perimeter of ABC = AB + BC + AC = AF + BF + BD + CD + AE + EC = AE + BF + BF + CD + AE + CD (Since AF = AE, BF = BD, EC = CD) = AE + AE + BF + BF + CD + CD = 2(AE + BF + CD) Perimeter of ABC = 2(AE + BF + CD) AE + BF + CD = (perimeter of ABC) ……..(2) From (1) and (2) ,we get AF + BD + CE = AE + BF + CD = (perimeter of ABC) Hence proved. www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 13. 29. The sum of two numbers is 20 and the sum of their reciprocals is . Find the numbers. Let one number be x ∴ other number = 20 − x Given ( ) + ( ) = ⇒ = ⇒ 480 = 100x − 5x2 ⇒ 5x2 − 100x + 480 = 0 ⇒ x2 − 20x + 96 = 0 ⇒x2 − 12x − 8x + 96 = 0 ⇒x(x − 12) − 8(x −1 2) = 0 ⇒(x − 12) (x − 8) = 0 ⇒x = 12 or x = 8 ∴ Numbers are 8 and 12 30. From a solid cylinder of height h = 16 cm and diameter d = 8 cm, a conical cavity of same height and same diameter is carved out. Find the volume and total surface area of the remaining solid. Height of cylinder (H) = 16 cm Diameter of cylinder (D) = 8 cm Radius of cylinder(R) = = 4 cm Height of cone (h) = 16 cm Radius of cone = 4 cm Vol. of remaining solid = vol. of cylinder - vol. of cone = × 4 × 4 × 16 – × × 4 × 4 × 16 = × 4 × 4 × 16 [ ] = 536.38 cm3 l = √ = 16.5 cm TSA of remaining solid = C.S.A. of cylinder + C.S.A. of cone + Area of the base = 2 × × 4 × 16 + × 4 × 16.5 + × 4 × 4 = + + = = 660 cm2 Vol. of remaining solid = 536.38 cm3 TSA of remaining solid = 660 cm2 www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 14. =(TP + PB) (TP – PB) (from (1)) 31. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? i) The amount of air present in a balloon when th of the air remaining in the ballon is removed by a vacuum pump at a time. ii) The cost of digging a well after every meter of digging, if it costs Rs 125 for the first meter and increases by Rs. 65 for every subsequent meter. i) Let amount of air initially present in the balloon be u vacuum pump removes 1/5th of the air remaining in the cylinder So the amount of air left when vacuum pump pumps out air = u, u-(1/5)u=4u/5,4/5u-1/5u=3/5u,.... = v,(4/5)u,(3/5)u,... So, the sequence we get is u,(4/5)u,(3/5)u,... The difference between consecutive terms is, u − u = − , − = - Difference between consecutive terms is constant. Therefore, it is an arithmetic progression. ii) Cost of digging 1 meter of well = Rs 125 Cost of digging the well = 125,125+65= Rs 190,190+65 = Rs. 255,..... we get a sequence of the form 125, 190, 255 .... The difference between consecutive terms is, 255-190 = 190-125 = 65 Difference between consecutive terms is constant. Therefore, it is an arithmetic progression. 32. TBA is a secant to the circle with center O intersecting the circle at A and B. TC is a tangent segment, prove that TA × TB = TC2 Given: A secant TBA to a circle C with center O intersecting the circle at A and B and TC is a tangent segment. To prove : TA × TB = TC2 Construction: Draw OP ⊥ AB. Join OP, OT and OA Proof: Since OP ⊥ AB ⇒ PA = PB … (1) (Perpendicular from the centre to the chord bisects the chord) TA × TB = (TP + PA) (TP – PB) www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 15. = TP2 – PB2 In right ΔOPT, OT2 = OP2 + PT2 ⇒ TP2 = OT2 – OP2 ∴ TA × TB = OT2 – OP2 – PB2 = OT2 – (OP2 + PB2 ) In right ΔOBP OB2 = OP2 + PB2 ∴ TA × TB = OT2 – OB2 = OT2 – OC2 ------- (2) (OB = OC = r) In right ∆OCT, OT2 = TC2 + OC2 ⇒ TC2 = OT2 – OC2 ------ (3) From (2) and (3) ∴ TA × TB = PT2 33. Stancy's house has an overhead tank in the shape of a cylinder which is filled by pumping water from a cuboidal sump. The sump has dimensions 1.57 m 1.44 m 95cm. the overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the sump after the overhead tank has been completely filled with water from the sump which had been full. Compare the capacity of the tank with that of the sump. (Use pi as 3.14 ) Radius of Overhead tank = 60 cm = 0.6 m Height of Overhead tank = 95 cm = 0.95 m Volume of water in overhead tank = πr2 h = 3.14 0.6 0.6 0.95 = 3.14 0.36 0.95 m3 Dimensions of the sump = 1.57 m 1.44 m 95 cm Volume of water in the sump when full = lbh =1.57 1.44 0.95 = 2 3.14 0.36 0.95 m3 Volume of water in the sump after filling the tank = 2 3.14 0.36 0.95 - 3.14 0.36 0.95 = 3.14 0.36 0.95 (2 - 1) = 3.14 0.36 0.95 m3 Area of the base of the sump = 1.57 1.44 = 1.57 4 0.36 = 2 3.14 0.36 m3 Height of water in sump = = = 47.5 cm = = Height of water in sump = 47.5 cm Capacity of the tank is half the capacity of sump www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in
- 16. 34. From a point on the ground 60 m away from the foot of a tower, the angle of elevation of the top of the tower is 30 . The angle of elevation to the top of a water tank (on the top of the tower) is 45 . Find the (i) height of the tower (ii) the depth of the tank. i) Let h be the height of the tower BM In right triangle CBM, we have = tan 30° = √ h = √ = √ = 20√ = 34.34 m Thus, the height of the tower is 20√ = 34.64 m approx. ii) Let the depth AM of the water tank be x m In right triangle ABC, = tan 45° = h + x = 60 m (Given) x = 60 – 34.64 = 25.36 m Hence, the depth of the water tank = 25.36 m Approx. www.learnCBSE.in www.learnCBSE.in w w w .LearnC BSE.in