SAMPLE QUESTION PAPER 09 Class-X (2017 18) Mathematics Time allowed: 3 Hours Max. Marks: 80 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. Section D contains 8 questions of 4 marks each. (iv) There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions. (v) Use of calculators is not permitted. SECTION A 1. Find the zeroes of the quadratic polynomial. 2. A ladder of length H metre makes an angle of 60 o with the ground when placed against a wall. If the distance between the feet of the ladder and the wall be 5 m, then determine H. 3. Captain of a team tosses two different coins, one golden colour and other of silver colour simultaneously. What is the probability that he gets atleast one head? 4. Write the sum of the first 100 natural numbers. 5. Find the coordinate of the mid-point of the line segment joining the points whose coordinates are and. 6. From an outside point A, two tangents AB and AC are drawn to touch the circle with centre at O (in fig-1). Given BAC = 30 o. Find AOB. Material downloaded from mycbseguide.com. 1 / 21
SECTION B 7. Find the values of y for which the distance between the points P(2, 3) and Q(10, y) is 10 units. 8. Find the discriminant of the equation and hence find the nature of its roots. 9. The sum of first 30 and 40 terms of an AP are respectively 2265 and 4020, then find the common difference of the AP. 10. The height and base diameter of a solid cylinder are 16 m and 2r m respectively. The cylinder is melted and recast to 12 solid spheres of same base diameter. Find r. 11. Two concentric circles are of radius 7 cm and cm. Find the length of the chord of the bigger circle, which touches the smaller circle. 12. If ABC and DEF are two similar triangles and their areas are 81 cm 2 and 144 cm 2 respectively. The bases of the triangles are respectively BC and EF. If EF = 15 cm, find BC. SECTION C 13. The denominator of a fraction is one more than thrice of the numerator. If the sum of the fraction and its reciprocal is, then find the fraction. 14. Use Euclid s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. 15. An army contingent of 616 members is to march behind an army band of 32 members in Material downloaded from mycbseguide.com. 2 / 21
a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? 16. The ratio of the sum of n-terms of two APs is (2n + 4) : (5n +2), find the ratio of their 12 th terms. Or Find the sum of all 3 digit numbers which leave remainder 3 when divided by 5. 17. The vertices of a ABC are A(4, 9), B and C(9, 4). A line is drawn to intersect sides AB and AC at P and Q respectively, such that. Find the area of APQ. Or The vertices of a ABC is (1, 2), (3, 1) and (2, 5). Point D divides AB in the ratio 2:1 and P is the mid-point of CD. Find the coordinates of the point P. 18. In fig-2, ABC and ABD are on the same base AB and on opposite sides of AB. If CD intersects AB at O, then show that,. 19. In fig-3, PQ and RS are two parallel tangents to a circle with centre O and another tangent EF with point of contact C intersecting PQ at E and RS at F. Prove that EOF = 90. Material downloaded from mycbseguide.com. 3 / 21
Or Prove that, the lengths of tangents drawn from an external point to a circle are equal. 20. Prove that:. 21. A bag contains 5 red ball and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag. Or A bag contains 8 red balls and x blue balls. the odd against drawing a blue ball are 2:5. what is the value of x? 22. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting : a) a king of red colour (b) a spade (c) a face card SECTION D 23. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction? Or The sum of a 2 digit number and number obtained by reversing the order of the digits is 99. If the digits of the number differ by 3. Find the number. 24. Find the common difference of an A.P. whose 1 st term is 100 and the sum of whose first six terms is 5 times the sum of the next six terms. Material downloaded from mycbseguide.com. 4 / 21
25. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle. Or Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and B = 90. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle. 26. Prove that, 27. Find the area of the segment APB shown in fig-4, if radius of the circle is 14 cm, and (use ) 28. There are two poles either on each bank of a river, just opposite to each other. One pole is 60 metre high. From the top of this pole, the angles of depression of the top and the foot of the other pole are 30 o and 45 o respectively. Find the width of the river and the height of the other pole. 29. In fig-5, a toy is in the form of a cone mounted on a hemisphere of common base of radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy. [ ] Material downloaded from mycbseguide.com. 5 / 21
30. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median and mode of the data. Monthly Consumption Number of Consumer 65-85 85-105 105-125 125-145 145-165 165-185 185-205 4 5 13 20 14 8 4 Or If the mean of the following distribution is 27, find the value of p. Also find the median and mode. Class-interval 0-10 10-20 20-30 30-40 40-50 No of workers 8 p 12 13 10 Material downloaded from mycbseguide.com. 6 / 21
CBSE SAMPLE PAPER 09 CLASS X - Mathematics Solutions SECTION- A 1. 3x 2 + 7x + 2 So the zeroes are. 2. Therefore, H = 10 m. 3. Total outcomes = 2 2 = 4 4. Sum of the first 100 natural numbers = 5. Coordinates of the required mid-point = 6. Since tangents drawn from an external point to a circle subtend equal angles at the centre of the circle. Material downloaded from mycbseguide.com. 7 / 21
= SECTION- B 7. By question, Hence values of y are 3, 9 8. Discriminant of is = 80-80 = 0 Therefore roots of the given quadratic equation are real and equal. 9. We know that the sum of first n-terms of an AP (whose first term is and common difference is d) is By question, Again by question, Subtracting (i) from (ii), we get, 10d = 50 d = 5 Hence the common difference of the AP is 5. 10. The volume of the solid cylinder = Volume of each sphere = [since its base diameter is equal to that of the cylinder] By question,. Hence r = 1 m Material downloaded from mycbseguide.com. 8 / 21
11. Let C is the centre of two concentric circles of radii 5cm and 3 cm. Let AB, a chord of the bigger circle which touches the smaller circle at M. CM = cm, CA = 7 cm [radius through point of contact is to tangent] [by Pythagoras Theorem ] Therefore, AB = 2AM = 2 6 = 12 cm. [Line segment, drawn from centre of a circle perpendicular to any chord, bisects the chord ] Hence, the length of the chord is 12 cm. 12. We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore for the given problem, Hence, BC = 11.25 cm SECTION C 13. Let the fraction be According to first condition, y = 3x + 1. So the fraction becomes Again by the 2 nd condition, Material downloaded from mycbseguide.com. 9 / 21
Therefore. x = 5 (since, x >0) Hence the required fraction = 14. According to Euclid s division lemma, Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 r < b. Let a be any positive integer and b = 3. Then, let a = 3q + r, where q > 0 and 0< r < 3 a = 3q or a = 3q + 1 or a = 3q +2 We have three cases : i) when a = 3q, then, a3 = (3q) 3 = 9(3q 3 ) = 9k [where k = 3q 3 is an integer] ii) when a = 3q + 1, then, a 3 = (3q + 1) 3 = 9(3q 3 ) + 9(3q 2 ) + 9q + 1 = 9(3q 3 +3q 2 +3q) + 1 = 9k1 + 1 [where k 1 = 3q 3 + 3q 2 + 3q is an integer] iii) When a = 3q + 2 then, a 3 = (3q + 2) 3 = 9(3q 3 ) + 9(6q 2 ) + 9(4q) + 8 = 9(3q 3 + 6q 2 + 4q) + 8 = 9k 2 + 8 [where k 2 = 3q 3 + 6q 2 + 4q is an integer] Therefore, the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. 15. To find the maximum number of columns in which members can march, we have to calculate the HCF of 32 and 616. Using Euclid s algorithm, 616 = 32 19 + 8, 32 = 8 4 + 0 Therefore the HCF of 32 and 616 is 8 Hence they can march in 8 columns each. 16. Let, be the first terms of the two APs also let be their common differences respectively. According to question, This is an identity. Material downloaded from mycbseguide.com. 10 / 21
Putting n = 23, we get, Hence ratio of their 12 th terms is 50 : 117. 17. Since P divides AB internally in the ratio 2 : 3, So, abscissa x-coordinate of P = and ordinate y-coordinate of P = Therefore coordinates of P are (3, 5) Since, Q divides AC internally in the ratio 2 : 3 Again, abscissa x-coordinate of Q = and ordinate y-coordinate of Q = Therefore coordinates of Q are (6, 7) Using, formula :, area of APQ = 18. Given that, ABC and ABD are on the same base AB but on opposite sides of AB. Material downloaded from mycbseguide.com. 11 / 21
To prove that, Construction :Let CM AB and DN AB are drawn. Proof: In CMO and DNO, CMO = DNO = 90 o COM = DON [ vertically opposite ] CMO ~ DNO [by A-A property ] So, (ii) and (iii) From (ii) and (iii), Hence proved. 19. Given, two tangents PQ and RS, which touch the circle, with centre O, at points A and B. Also, PQ RS. Another tangent EF touches the circle at C and meets the lines PQ and RS at E and F respectively. To prove,. Material downloaded from mycbseguide.com. 12 / 21
Construction : O, C joined. Proof :. [Since, radius through point of contact is perpendicular to the tangent] In OA = OC [radius of same circle] EA = EC [Tangents drawn from outside point to a circle are equal] Similarly, it can be shown that, Since PQ RC and EF is a transversal, So,. Hence proved. 20. Material downloaded from mycbseguide.com. 13 / 21
= RHS Hence proved. 21. Let the number of blue balls in the bag be x. Therefore total number of balls in the bag = 5 + x Let probability of drawing a blue ball be P(blue). Let probability of drawing a red ball be P(red). Then, and According to the question, P(blue) = 2 P(red) x=10 [since x can not be negative ] Hence number of blue balls in the bag is 10. Material downloaded from mycbseguide.com. 14 / 21
22. Total number of favourable outcomes = 52. a) We know that, cards of diamond and heart are of red colour. There exist one king of Diamond and one king of heart. Number of outcomes favourable to getting a king of red colour = 2 P(getting a spade) b) A well-shuffled deck of 52 cards contains 13 cards of spade. So, number of outcomes favourable to getting a spade = 13. P(getting a king of red colour) c) A well-shuffled deck of 52 cards contains 12 face cards. So, number of outcomes favourable to getting a face card = 12. P(getting a face card) SECTION - D 23. Let the numerator and the denominator be x and y respectively. the fraction is According to question we get, and Material downloaded from mycbseguide.com. 15 / 21
Now, and Substituting y in (ii) with the help of (i), 2x = x + 2 +1 x = 3 1 Now from (i), y = 3 +2 =5. Hence the fraction is. 24. Let the 1 st term and common difference of the AP be a and d respectively. Here a = 100. We know that sum of first n-terms of an AP,. Now sum of first 6-terms of the AP is. and sum of next 6-terms of the AP = S 12 - S 6 = 6(200 + 11d) - 3(200 + 5d) = 1200 + 66d - 600-15d = 600 + 51d According to the question, 600+15d = 5(600+51d) 600 + 15d = 3000 + 255d 600 3000= 255d 15d 240d = 2400 d = 10 Hence the common difference of the AP is ( 10). Material downloaded from mycbseguide.com. 16 / 21
25. Steps of construction : Draw a circle with help of a bangle. Take two non parallel chords AB and CD Draw perpendicular bisectors of these chords. Mark point O where two bisectors intersect. Point O is center of the circle. Take a point P outside the circle and join PO. Bisect PO. Let M be mid-point of PO. Taking M as centre and MO as radius, draw a circle. Let it intersect the given circle at the points Q and R. Join PQ and PR. 26. Hence proved. 27. Area of the segment APB = Area of the sector OAPBO area of OAB. Material downloaded from mycbseguide.com. 17 / 21
Here area of the sector OAPBO = Let, OM AB is drawn. Given, Since, perpendicular drawn from centre of a circle to a chord bisects the chord, So, here, Now Therefore area of OAB = Hence, Area of the segment APB = Area of the sector OAPBO area of OAB = 28. Let two poles AB(= 60 m) and PQ be either on each bank of a river. Material downloaded from mycbseguide.com. 18 / 21
Let PQ = h m. and width of the river BQ = x m. In figure, PR AB. AR = AB BR = AB PQ = (60 h) m Acc to question, Angle of depression of P from A = XAP = APR (alternate angle) = and angle of depression of Q from A = XAQ = AQB (alternate angle) = Now in AQB, x =60 Now in APR, [since x = 60] Material downloaded from mycbseguide.com. 19 / 21
Hence the width of the river is 60 m and height of the other pole is 25.36 m. 29. Total height of the toy is 31 cm in fig-5, AD = 31 cm. Radius of the common base is 7 cm. OC = OB = OD = 7 cm. AO = AB OD = 31 7 = 24 cm. Surface area of the toy = slant surface area of the cone + curved surface area of the hemisphere Here, slant surface area of the cone = sq. unit Curved surface area of the hemisphere = Hence the total surface area of the toy =( 550 + 308) cm 2 = 858 cm 2. 30. For table = 1 Monthly consumption (in units) Number of consumer (fi) Cumulative frequency 65-85 4 4 85-105 5 9 105-125 13 22 Material downloaded from mycbseguide.com. 20 / 21
125-145 20 42 145-165 14 56 165-185 8 64 185-205 4 68 Total N=68 For Median. Here,, then, which lies in the interval 125-145. Median class is 125-145. Here, l = 125, n = 68, f = 20, cf = 22 and h = 20. For Mode : In the given data, maximum frequency is 20 and it corresponds to the class interval 125-145 Modal class = 125-145 and = 125, f 1 = 20, f 0 = 13, f 2 = 14 and h = 20 Now, Hence Median= 137 units, Mode = 135.77 units. Material downloaded from mycbseguide.com. 21 / 21