SAMPLE QUESTION PAPER 11 Class-X (2017 18) Mathematics GENERAL INSTRUCTIONS (i) All questions are compulsory. (ii) The question paper consists of 30 questions divided into four sections A, B,C and D. (iii) 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 and Section D contains 8 questions of 4 marks each. (iv) Use of Calculator and log tables is not permitted. Q.1 Find the value of k for which the pair of linear equations and has no solution Q.2. for what value of p, are 2p + 1, 13, 5p-3 three consecutive terms of an A.P? Q.3 without using trigonometry table, evaluate Q.4 In figure,, BC = 7.5 cm, AM = 4cm and MC = 2 cm. Find the length BN Q.5 The two tangents from an external point P to a circle with centre O and PA and PB. If what is the value of? Q.6 A pair of dice is thrown once. Find the probability of getting the same number on each dice. Q. 7 Prove that is an irrational number. Material downloaded from mycbseguide.com. 1 / 27
Q.8 Solve the equations graphically: What is the area of the triangle formed by the two lines and the line y = 0? Q.9 Find the roots of the following equation Q.10 Prove that : Q.11 The angle of elevation of the top of a building from the foot of a tower is and angle of elevation of top of the tower from the foot of the building is. If the tower is 50 m high. Find the height of the building. Q.12 The dimensions of a metallic cuboid are 100 cm 80 cm 64 cm. It is melted and recast into a cube. Find the surface area of the cube. Q.13 A person on tour has Rs. 4200 for his expense. If he extends his tour for 3 days, he has to cut down his daily expense by RS. 70. Find the duration of the tour. Q.14 The first and last term of an AP are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum? Q.15 Prove that the ratio of areas of two similar triangles is equal to the ratio of the square of their corresponding sides. Q.16 Point P divides the line segment joining the points A (2,1) and B (5, -8) such that If P lies on the line, find the value of k. Show that the points and are the vertices of a square. Q.17 From the top of a 7m high building, the angle of elevation of the top of a tower is and angle of depression of the foot of the tower is Find the height of the tower. A girl who is 1.2 m tall, spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eye of the girl at Material downloaded from mycbseguide.com. 2 / 27
any instant is. After sometime, the angle of elevation reduces to. Find the distance travelled by the balloon during the interval. Q.18 In figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segment BD and DC into which BC is divided by the point of contact D are the lengths 4cm and 3cm respectively. If area of ABC = 21, then find the lengths of sides AB and AC. Q.19 Draw a triangle ABC with BC = 7cm, and. then construct another triangle whose sides are times the corresponding sides of ABC. Draw a circle of radius 2.3 cm take a point P on it. Without using the centre of the circle. Draw tangent to it at P. Q.20 The mid points of sides of a triangle are (3,4), (4,1) and (2,0). Find the coordinate of vertices of the triangle. Q.21 Weekly income of 600 families is given below. Income in Rs. Frequency 0-10000 250 1000 2000 190 2000 3000 100 3000 4000 40 4000 5000 15 Material downloaded from mycbseguide.com. 3 / 27
5000 6000 5 Find the median. The following tables gives production yield per hectare of wheat of 100 farms of village: Production Yeild (in hr.) No. of Farms 50-55 2 55-60 8 60-65 12 65-70 24 70-75 38 75-80 16 Change the distribution to a more than type distribution and draw its Ogive. Q.22 All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card id drawn from it. Find the probability that the drawn card is (i) a black face card (ii) a red card. Q. 23 Show that only one of the number n, n+2 and n+4 is divisible by 3. Use Euclid Division Lemma to show that cube of any positive integer is either of the form 9m, (9m + 1) or (9m + 8). Q. 24 Obtain all other zeroes of the polynomial, if two of its zeroes are and. Material downloaded from mycbseguide.com. 4 / 27
Q.25 Solve the following pair of equations for and ; If in a rectangle the length is increased and breadth is decreased by 2 units each, the area is reduced by 28 square units, and if the length is reduced by 1 unit and breadth is increased by 2 units, the area increased by 33 square units. Find the dimensions of the rectangle. Q 26. In figure AB units, CD units and units, prove that Q.27 Prove that Q.28 ABC is a right triangle, right angled at A. Find the area of shaded region if AB = 6cm, BC = 10cm and O is the centre of incircle of ABC. (take ) Q.29 A gulab jamun, when ready for eating, contains sugar syrup of about 30% of its volume. Find approximately how much syrup would be found in 45 such gulab jamuns, each shaped Material downloaded from mycbseguide.com. 5 / 27
like a cylinder with two hemispherical ends, if the complete length of each of them is 5cm and its diameter is 2.8 cm. Q.30 Draw less than Ogive and more than Ogive for the following distribution and hence find its median. Class Frequency 20-30 10 30-40 8 40-30 12 50-60 24 60-70 6 70-60 25 80-90 15 Find the mean, mode and median for the following data. Classes Frequency 5-15 2 15-25 3 25-35 5 35-45 7 45-55 4 55-65 2 65-75 2 Material downloaded from mycbseguide.com. 6 / 27
SAMPLE PAPER 11 (CLASS X - MATHEMATICS ) Marking Scheme 1. Since pair of equations has no solution Then, i.e. 2. 2p+1, 13,5p-3 are consecutive terms of an A.P 13-2p - 1 = 5p - 16 12-2p = 5p - 16 12 + 16 = 5p + 2p 28 = 7p p = 4 3. 4. In MN AB [ Let x = BN] 3x = 15 x = 5 Hence BN = 5 cm. 5. PA and PB are tangents to the circle Material downloaded from mycbseguide.com. 7 / 27
[ Tangent makes an angle of with the radius at the point of contact ] In quadrilateral OAPB, [Angle sum property of a quadrilateral] 6. Total number of outcomes when a pair of dice is thrown is 36 Same number on each dice i.e. (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) Number of ways of getting the same number on each dice = 6. Required probability = 7. Let if possible is a rational number. From we notice LHS is an irrational number and RHS is rational number, which is not possible. Hence, our supposition is wrong. Hence, is an irrational number. 8. Material downloaded from mycbseguide.com. 8 / 27
1 0 2 0 2-2 From 0-4 2 2 0 3 From From graph, we observe that solution of equation is (0,2) Area = 5 square units 9. Material downloaded from mycbseguide.com. 9 / 27
x 2-3x - 28 = -30 x 2-3x + 2 = 0 (x - 2)(x - 1)=0 x = 2,1 Required roots are 2, 1 10. Taking LHS = =RHS 11. Solution : Let TP be height of tower of 50m, Material downloaded from mycbseguide.com. 10 / 27
Let distance AP between building and tower be. To find : AB, height of building In right In right AB AB = 16.67m Hence height of the building is 16.67 m 12. Dimensions of the metallic cuboid are 100 cm 80 cm 64 cm 100 80 64 (Where a is side of cube) a = 80 cm Surface area of cube 13. Number of days for a tour = x Daily expense = y A.T.Q. If tour be extended for 3 days New number of days = x + 3 Daily expense = y - 70 Now (x + 3)(y - 70) = 4200 xy - 70x + 3y -210 = 4200-70x + 3y - 210 = 0 [From (i)] Put Material downloaded from mycbseguide.com. 11 / 27
x 2-180 + 3x = 0 x 2 + 3x - 180 = 0 (x + 15)(x - 12) = 0 x = 12, x = -15 (rejected) Duration of Tours = 12 Hours 14. Here a = 8, l =350, d=9 From formula, l= a n = a + (n - 1) d, we get a + ( n-1) d=350 8 + (n - 1)9 = 350 (n - 1)9 = 350-8 (n - 1)9 = 342 n - 1 = n - 1 = 38 n = 38 + 1 From formula, we get 15. Given : To prove : Construction : Draw AM BC and DK EF Proof : ABC Material downloaded from mycbseguide.com. 12 / 27
Also ABM and DEK (given) ( angles of similar s ) (Construction) By AA Rule Equating (i) and (ii) So, Hence proved 16. P is the point of intersection of line segment AB and line 2x - y + k = 0. Here, 3 AP = AP + PB 2 AP = PB AP : PB = 1 : 2 P divides the line segment joining A (2,1) and B(5, -8) in ratio 1:2 Coordinates of point P are i.e. p(3, -2) As point P lies on the line 2x - y + k = 0. 6 + 2 + k = 0 6 = -8 Material downloaded from mycbseguide.com. 13 / 27
Diagonal Diagonal Hence proved. 17. AB is a building of height 7 m and CD is tower of height h m. AB = ED = 7m and CE = (h-7)m Let BD = AE = x m In right AED, In right AEC, Height of the tower is 28 m. In right Material downloaded from mycbseguide.com. 14 / 27
In right 18. Let AE = AF = x Length of tangents from an external point is equal. ar BOC = ar AOB = ar AOC = Material downloaded from mycbseguide.com. 15 / 27
[from (i)] 21x + 7x - 49 = 0 2x 2 + 7x - 49 = 0 2x 2 + 14x - 7x -49 = 0 2x(x + 7) - 7(x + 7) = 0 (2x - 7)(x + 7) = 0 [rejected] length of side AB = 4+ 3.5 = 7.5 cm and AC =3 + 3.5 = 6.5 cm 19. Material downloaded from mycbseguide.com. 16 / 27
Steps of construction: 1. Draw a circle of radius 2.3 cm and take a point P on it. 2. Draw chord PQ. 3. Mark a point R in the major arc QP. 4. Join PR and RQ. 5. Draw 6. Produce to as shown in figure, then is the required tangent at the point P. 20. Let A(x 1, y 1 ), B(x 2, y 2 ) and C(x 3, y 3 ) are the vertices of ABC. (3,4), (4,1),(2,0) are mid points of sides AB,BC,CA As (3,4) is mid point of AB x 1 + x 2 = 6...(i) y 1 + y 2 = 8...(ii) As (4,1) are mid points of BC x 2 + x 3 = 4...(iii) y2 + y3 = 2... (iv) As (2,0) are mid points of AC x 1 + x 3 = 4...(v) y 1 + y 3 = 0..(vi) Adding (i), (iii), (v), we get 2 (x 1 + x 2 + x 3 ) = 18 x 1 + x 2 + x 3 = 9 as x 1 + x 2 = 6 [from (i)] x 3 = 3 x 1 = 1, x 2 = 5 Material downloaded from mycbseguide.com. 17 / 27
Similarly, y 1 = 3, y 2 = 5, y 3 = -3 Coordinate of vertices of are A(1, 3), B(5, 5), C(3, -3) 21. Income in Rs. Number of families (f) c. f 0-1000 250 250 1000-2000 190 440 2000-3000 100 540 3000-4000 40 580 4000-5000 15 595 5000-6000 5 600 n 600 = 300 Median class = 1000-2000 l = 1000, c.f.=250, f=190, h=1000 Median More than type Ogive Production yield (Kg/ha) C.F More than or equal to 50 100 More than or equal to 55 98 More than or equal to 60 90 More than or equal to 65 78 Material downloaded from mycbseguide.com. 18 / 27
More than or equal to 70 54 More than or equal to 75 16 Now, draw the Ogive by plotting the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16) 22. Number of kings = 4, number of queens =4, number of aces = 4 After removing all kings, queens and aces, number of cards left = 52 12 = 40 (i) Number of black face cards in the remaining cards ( 2 jacks) = 2 Probability of black face card (ii) Out of 12 cards removed, 6 are of red colour. Number of red- coloured cards left 26 6 = 20 Number of ways of drawing a red card = 20 Probability of getting a red card 23. Let n = 3k ; 3k+1 or 3k +2 (i) When n = 3k n is divisible by 3 n+ 2 = 3k +2 n + 2 is not divisible by 3 and n + 4 = 3k + 4 = 3k + 3 + 1 = 3(k +1) + 1 n + 4 is not divisible by 3. Material downloaded from mycbseguide.com. 19 / 27
(ii) When n = 3k + 1 n is not divisible by 3 n + 2 = (3k + 1) + 2 = 3k + 3 = 3 (k + 1) n + 2 is divisible by 3 n + 4 = 3k + 1 + 4 = 3k + 5 = 3k + 3 + 2 = 3(k +1)+2 n + 4 is not divisible by 3 (iii) When n = 3k + 2, n is not divisible by 3 n + 2 = (3k +2) +2 = 3k + 4 = 3(k +1) +1 n + 2 is not divisible by 3 n + 4 = (3k +2) + 4 = 3k + 6 = 3(k+2) n + 4 is divisible by 3. Only one of the numbers n, n + 2 and n + 4 is divisible by 3. Let a = 3q + r a = 3q; then a 3 = 27q 3 =9m ; where m=3q 3 when a=3q+1 ; then a 3 =27q 3 +27q 2 +9q+1 =9 (3q 3 +3q 2 +q)+1 = 9m+8 (where m = 3q 3 + 3q 2 +q) when a=3q+2 ; then a 3 = (3q+2) 2 = 27q 3 +54q 2 +36q+8 = 9(3q 3 +6q 2 +4q)+8 = 9m+8 (where m = 3q 3 +6q 2 +4q) Hence, cubes of any positive integer is either of the form9m,( 9m+1 ) or( 9m+8 ). 24. Solution : are the zeroes of Material downloaded from mycbseguide.com. 20 / 27
is a factor of is factor of For other factor other factor of p(x) is x 2-3x + 2=0 For other zeroes, x 2-3x + 2=0 (x - 2)(x - 1) x = 2, x = 1 Other zeroes are 1 and 2. 25. Put and also ; put and We have Equation (i) - 2 x equation (ii) Material downloaded from mycbseguide.com. 21 / 27
Put x - y = 5..(iii) Solving (iii) and (iv) for x and y We have Let the length and breadth of a rectangle be x and y meters. According to question, Area = xy (x + 2) (y - 2) = xy - 28 or 2x - 2y = 24 or x - y = 12...(i) and (x - 1) (y + 2) = xy + 33 2x - y = 33...(ii) on subtracting eq (ii) - (i), we get, x = 21 and 21 - y = 12 so y = 21-12 y = 9 length =21m, breadth = 9m 26. Let BQ = units, DQ b units Material downloaded from mycbseguide.com. 22 / 27
and Similarly Also From (i) and (ii) (Hence Proved ) 27. Taking LHS = Material downloaded from mycbseguide.com. 23 / 27
= = = = = = = = = RHS 28. Since and OP, OQ are radius through contact point must be on tangents lines. Therefore, OPAQ is a square. Let OP AP = AQ = OQ Now in ABC, i.e. 36 + AC 2 = 100 AC 2 = 64 Ac = 8 BR = PB = 6 - x CQ = 8 - x = CR [ tangent from an external point ] Now again BC = CR + BR 10 = 8 - x + 6 - x 10 = 14-2x 2x = 4 x = 2 Material downloaded from mycbseguide.com. 24 / 27
Area of shaded portion = Area of triangle - Area of circle 29. Radius of the hemispherical part = 1.4 cm Length of the cylindrical part = 5 - (2 x 1.4) = 2.2cm Volume of 1 gulab jamun = volume of two hemispherical ends + Volume of cylindrical part Volume of 45 gulab jamuns Volume of syrup found in 45 gulab jamuns 30. Solution : table for less than Ogive and more than Ogive C.I. For less than Ogive For more than Ogive C.I. (less than) c.f. Point C.I. (more than) c.f. Point 20-30 10 30 10 (30,10) 20 100 (20,100) 30-40 8 40 18 (40,18) 30 90 (30,90) 40-50 12 50 30 (50,30) 40 82 (40,82) 50 60 24 60 54 (60,54) 50 70 (50,70) Material downloaded from mycbseguide.com. 25 / 27
60-70 6 70 60 (70,60) 60 46 (60,46) 70-80 25 80 85 (80,85) 70 40 (70,40) 80-90 15 90 100 (90,100) 80 15 (80,15) Less than Ogive and more than Ogive. We notice both the curves intersect at (58.3, 50) Median = 58.3 We have, class Mid- value(x i ) (f i ) f i u i c.f 5-15 10 2-3 -6 2 15-25 20 3-2 -6 5 Material downloaded from mycbseguide.com. 26 / 27
25-35 30 5-1 -5 10 35-45 40 7 0 0 17 45-55 50 4 1 4 21 55-65 60 2 2 4 23 65-75 70 2 3 6 25 Let assumed mean a = 40, Here h=10 Mean = 40-1.2 = 38.8 Since, maximum frequency = 7 Modal class = 35-45 Here, l = 35, f 1 = 7, f 0 = 5, f 2 = 4 We know that Mode = 35 + 4 = 39 Since which lies in 35-45 class Here, Median = = 35 + 3.6 nearly = 38.6 nearly Material downloaded from mycbseguide.com. 27 / 27