Sample paper for class 10 ncert , cbse, Exams of Mathematics

Download Sample paper for class 10 ncert , cbse and more Exams Mathematics in PDF only on Docsity! CBSE Board Paper Solution-2020 Class : X Subject : Mathematics (Standard) - Theory Set : 1 Code No : 30/5/1 Time allowed : 3 Hours Maximum Marks : 80 Marks General instructions: Read the following instructions very carefully and strictly follow them: (i) This question paper comprises four sections – A, B, C and D. This question paper carries 40 question All questions are compulsory (ii) Section A: Question Numbers 1 to 20 comprises of 20 question of one mark each. (iii) Section B: Question Numbers 21 to 26 comprises of 6 question of two marks each. (iv) Section C: Question Numbers 27 to 34 comprises of 8 question of three marks each. (v) Section D: Question Numbers 35 to 40 comprises of 6 question of four marks each. (vi) There is no overall choice in the question paper. However, an internal choice has been provided in 2 question of the mark, 2 question of one mark, 2 questions of two marks. 3 question of three marks and 3 question of four marks. You have to attempt only one of the choices in such questions. (vii) In addition to this. Separate instructions are given with each section and question, wherever necessary. (viii) Use of calculations is not permitted. Section A Question numbers 1 to 20 carry 1 mark each. Question numbers 1 to 10 are multiple choice questions. Choose the correct option. 1. On dividing a polynomial p(x) by x2 – 4, quotient and remainder are found to be x and 3 respectively. The polynomial p(x) is (A) 3x2 + x - 12 (B) x3 -4x + 3 (C) x2 + 3x - 4 (D) x2 - 4x - 3 Answer: Correct Answer: (B) x3 –4x+3 Explanation: P(x) = (divisor)×(quotient) + Remainder =(x2 – 4)x+3 = x3 – 4x+3 2) In Figure-1, ABC is an isosceles triangle, right- angled at C. Therefore OR Correct Answer: (C) (0, 7/2) Explanation: The centre of a circle is the mid-point of its diameter. End points of the diameter are: (–6, 3) and (6, 4) Coordinates of the centre = ((– 6+6)/2, (3+4)/2) = (0, 7/2) 4) The value(s) of k for which the quadratic equation 2x2 + kx + 2 = 0 has equal roots, is (A) 4 (B) 4 (C) ─ 4 (D) 0 Answer: Correct Answer: (B) ±4 Explanation: The given equation is: 2x2 + kx + 2 =0 Discriminant = b2–4ac Here, b =k, a =2, and c =2 So, Discriminant = k2–4×2×2 = k2–16 A quadratic equation has equal roots if its discriminant is zero. k2–16 = 0 k2 =16 k = ±4 5) Which of the following is not an A.P.? (A) ─ 1.2, 0.8, 2.8 …. (B) 3, 3 + 2, 3 + 2 2, 3 + 3 2, .... (C) 4 7 9 12 , , , ,... 3 3 3 3 (D) -1 -2 -3 , , ,... 5 5 5 Answer: Correct Answer: (C) 4 7 9 12 , , , ,... 3 3 3 3 Explanation: 4 7 9 12 , , , ,... 3 3 3 3 7 4 7 4 3 3 3 3 3 1 9 7 9 7 3 3 3 2 3 3 2 3 3 Difference between consecutive terms is not same. So, this is not an A.P. 6) The pair of linear equations 3x 5y + = 7 and 9x + 10y = 14 is 2 3 (A) consistent (B) inconsistent (C) consistent with one solution (D) consistent with many solutions Answer: Correct Answer: (B) Inconsistent Explanation: Answer: Correct Answer: (C)32/3 Explanation: 3 3 3 2 2/3 4 Volume of sphere = r 3 4 12 r 3 r 3 r 3 9) The distance between the points (m,–n) and (–m, n) is (A) 2 2m + n (B) m + n (C) 2 2m + n2 (D) 2 22m + 2n Answer: Correct Answer: (C) 2 22 m n Explanation: 2 2 2 2 2 2 Distance = m ( m) (–n–n) (m m) (–2n) 2 m n 10) In Figure-3. From an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O. If QPR = 90°, then length of PQ is (A) 3cm (B) 4cm (C) 2cm (D) 22 cm Answer: Correct Answer: (B) 4 cm Explanation: Tangents are drawn from an external point P. So, line joining centre O and point P bisects ∠PQR. OP bisects ∠QPR = 90°. In ∆ OQP, ∠Q = 90° (radius meets tangent at 90°) ∠QPO = 45° = ∠QOP Thus, OQ = PQ = 4 cm Fill in the blanks in question number 11 to 15 11) The probability of an event that is sure to happen is __. Answer: 1 12) Simplest form of 2 2 1 tan A is ______. 1 cot A   Answer: cot2A      2 2 2 2 2 2 2 1 tan A sec A sin A cot A 1 cot A cosec A cos A 13) AOBC is a rectangle whose three vertices are A(0, –3), O(0, 0) and B (4, 0). The length of its diagonal is ____. Answer:      2 2 2 2 In right-angled triangle AOB, AB OA OB 3 4 25 5 19) Form a quadratic polynomial, the sum and product of whose zeroes are (-3) and 2 respectively. OR Can (x2 – 1) be a remainder while dividing x4 – 3x2 + 5x – 9 by (x2 +3)? Justify your answer with reasons. Answer: x2 – (sum of zeroes)x + product of zeroes = x2 – (–3)x+2 = x2 + 3x+2 So, the required polynomial is x2 +3x+2. OR When a polynomial p(x) is divided by another polynomial g(x), then the degree of remainder r(x) < degree of g(x) Therefore, for the given question x2 – 1 cannot be a remainder while dividing x4 – 3x2 + 5x – 9 by x2 + 3 because deg (x2 – 1) = deg (x2 + 3). 20) Evaluate: 0 0 0 2 tan 45 cos 60 sin 30  Answer:      2 tan45° ×cos60° sin30 1 2 1 2 1 2 2 SECTION B Question number 21 to 26 carry 2 marks each. 21) In the given Figure-5, DE ||AC and DF||AE. Prove that BF BE = . FE EC Answer:     In ABC, DE AC So, using basic proportionality theorem, we get BD BE ...(1) DA EC In BAE, DF AE So, using basic proportionality theorem, we get BD BF DA FE  ...(2) From (1) and (2), we get BE BF EC FE 22) Show that 5+2 7 is an irrational number, where 7 is given to be an irrational number. OR Check whether 12n can end with the digit 0 for any natural number n. Answer:                Let us assume, to the contrary, that 5 2 7 is rational. That is, we can find coprime a and b (b 0) such that 5 2 7 2 7 5 1 –5 Rearranging this equation, we get 7 5 2 2 a b a b a a b b b Answer: We have to prove that AB CD BC AD We know that lengths of tangents drawn from a point to a circle are equal. Therefore, from figure, we have DR DS, CR CQ, AS AP, BP BQ Now, LHS AB CD (AP BP) (CR DR) (AS BQ) (CQ                 DS) BQ CQ AS DS BC AD RHS         OR From the given figure, we have AP = 12 cm Since AQ and AB are the tangent to the circle from a common point A, hence AP = AQ = 12 Similarly, PB = BD and CD =CQ Also, AP =AB + PB and AQ = AC + CQ Perimeter of ABC = AB + BD + CD + AC = AB + PB + CQ + AC (since PB = BM and CM = CQ) = (AB + PB) + (CQ + AC) = AP + AQ = 12 +12 = 24 cm Therefore, the perimeter of triangle ABC =24 cm 25) Find the mode of the following distribution: Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Number of Students 4 6 7 12 5 6 Answer: Marks 0-10 10-20 20-30 30-40 40-50 50-60 Number of Students 4 6 7 12 5 6 hence the father's age will be 10 years more than twice the age of the son. Determine their present ages. Answer: Let the numerator of the fraction be and denominator be . Therefore, the fraction is . According to question, 1 1 3 3 1 3 3 ...(1) 1 and 8 4 4 8 4 8 ...(2) 3 3 From equations 1 and 2 ,  we get x y x y x y x y x y x y x y x y x 5 So, the required fraction = . 4 8 4 3 8 3 5 Putting 5 in equation (1), 3 1 5 3 2 2 1 x x x x x y y OR Let the son's present age be . So, father's present age = 3 3 3 years later: Son's age = 3 Father's age = 3 3 3 3 6 But, according to the given condition, 3 years later father's age= 2 3 10 2 6 10 2 x x x x x x x 16 So, we can write 3 6 2 16 3 2 16 6 10 So, son's present age 10 years and father's present age 10 3 3 33 years x x x x x x 28) Use Euclid Division Lemma to show that the square of any positive integer is either of the form 3q or 3q + 1 for some integer q. Answer: 2 2 2 Let be a positive integer and 3. By Euclid's Algorithm, 3 for some integer 0 and 0 3. The possible remainders are 0, 1 and 2. Therefore, can be 3 or 3 1 or 3 2. Thus, 9 or (3 +1) a b a m r m r a m m m a m m          2 2 2 2 2 2 2 1 2 3 1 2 3 or (3 2) 9 or (9 6 1) or (9 12 4) 3 (3 ) or 3(3 2 ) 1 or 3(3 4 1) 1 3 or 3 1 or 3 1 where , and are some positive integers. Hence, square of any positive integer is m m m m m m m m m m m k k k k k k                 either of the form 3 or 3 + 1 for some integer .q q q 29) Find the ratio in which y-axis divides the line segment joining the points (6, - 4) and (-2, -7). Also find the point of intersection. OR Show that the points (7, 100, (-2, 5) and (3, -4) are vertices of an isosceles right triangle. Answer: 1 sinA sec A + tan A 1 sinA Answer:                        2 2 1 sinA LHS 1 sinA 1 sinA 1 sinA 1 sinA 1 sinA 1 1 sinA 1 sin A 1 sinA cos A 1 sinA cosA sinA 1 cosA cosA tanA sec A RHS 31) For an A.P., it is given that the first term (a) = 5, common difference (d) = 3, and the nth term (an) = 50. Find n and sum of first n terms (Sn) of the A.P. Answer: n n n n Here, a 5, d 3, a 50 We need to find S . Firstly, we will find the value of n. We know that a a (n 1)d So, 50 5 (n 1)3 or 50 5 (n 1)3 45 or 1 n 3 or n 16 We know that sum of first n terms of an AP is given by S n 16 16 n a a 2 16 So, S 5 50 2 8 55 or S 440 32) Construct a ΔABC with sides BC = 6 cm, AB = 5 cm and ABC= 60°. Then construct a triangle whose sides are 3 4 of the corresponding sides of ΔABC. OR Draw a circle of radius 3.5 cm. Take a point P outside the circle at a distance of 7 cm from the centre of the circle and construct a pair of tangents to the circle from that point. Answer: Step 1: Draw a ABC with sides AB 5 cm, BC 6 cm and ABC 60 . Step 2: Draw a ray BX making an acute angle with line BC on the opposite side of vertex A. Step Steps of Construction : 1 2 3 4 1 1 2 2 3 3 4 4 3 4 3: Locate 4 points B , B , B , B on BX such that BB B B B B B B . Step 4: Join the points C and B . Step 5: Through the point B , draw a line parallel to CB intersecting line segment BC at point C . Step 6: Draw a line through C parallel to the line AC to intersect line segment AB at A . The required triangle is A BC .       Numbers on spinner 1, 2, 4, 6, 8, 10 Even numbers on spinner 2, 4, 6, 8, 10 Shweta will pick black marble, if spinner stops on even number. Therefore, n Even number 5 n Possible number 6 P Shweta allowed to pici             k a marble P Even number n Even number n Possible number 5 6 Therefore, the probability of allowing Shweta 5 to pick a marble is . 6                    Since, prizes are given, when a black marble is picked. Number of black marbles 6 Total number of marbles 20 Therefore, P getting a prize P a black marble n Black marbles n Total marble 6 20 3 10 Therefore, ii s 3 the probabiltiy of geting prize is . 10 34. In figure – 9, a square OPQR is inscribed in a quadrant OAQB of a circle. If the radius of circle is 6 2 cm, find the area of the shaded region. Answer:           2 2 Given that, OQ 6 2 cm OPQR is a square. Let the side of square a The diagonal of square a 2 Here, OQ is a diagonal of square. a 2 6 2 a 6 cm Area of square OPQR 6 36 cm Radius of the quadrant OAQB Diagonal of                2 2 2 the square OPQR 6 2 cm 90 22 Area of the quadrant OAQB 6 2 360 7 396 cm 7 Area of shaded region Area of the quadrant OAQB Area of square OPQR 396 36 7 144 7 20.6 cm SECTION D 35) 4 3 2 Obtain other zeroes of the polynomial p(x) = 2x - x - 11x + 5x + 5 if two of its zeroes are 5 and - 5. OR              2 3 2 Remainder 10x 33 Therefore, we should add 10x 33 to make it exactly divisible by x 4x 8. Thus, we should add 10x 33 to 2x 3x 6x 7. 36) Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Answer: 2 2 2 Given: ABC DEF Area ΔABC AB BC AC To prove : Area ΔDEF DE EF DE Construction: Draw AL BC and DM EF 1 BC ALArea ΔABC BC AL2Proof: Here 1 1Area ΔDEF EF DM EF DM 2 In ΔALB and ΔDME ALB DME Each 90° and B E Since ΔABC ΔDEF So, ΔALB ΔDME AA similarity criterion AL AB DM DE AB BC AC But Since ΔABC ΔDEF DE EF DF AL Therefore, D 2 BC 2 M EF From 1 and 2 , we have Area ΔABC BC AL BC BC BC Area ΔDEF EF DM EF EF EF 2 2 2 AB BC AC But Since ΔABC ΔDEF DE EF DF This implies that, Area ΔABC AB BC AC Area ΔDEF DE EF DE 37) Sum of the areas of two squares is 544m2. If the difference of their perimeters is 32 m, find the sides of the two squares. OR A motorboat whose speed is 18km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. Answer:    2 2 2 2 Let the sides of first and second square be x any y. Then, Area of first square = x And, Area of second square = y According to the question, x y 544 ... 1 Now, Perimeter of first square = 4x And, Perimeter of second square = 4y                          24 Similarly, the time taken to go downstream = hours 18 x According to the question, 24 24 – 1 18 – x 18 x 24 18 x –24 18 – x or, 1 18 x 18 – x or, 24 18 x –24 18 – x 18 x 18 – x or, 432 + 24x – 432             2 2 2 + 24x = 324 – x or, x 48x –324 = 0 Using the quadratic formula, we get –48 48 – 4 1 –324 x = 2 –48 2304 1296 = 2 –48 3600 = 2 –48 60 = 2 –48 60 –48 – 60 Therefore, x = or x = 2 2  12 –108 x = or x = 2 2 x = 6 or x = –54 Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = –54. Therefore, x = 6 gives the speed of the stream as 6 km/h. 38. A solid toy is in the form of a hemisphere surmounted by a right circular cone of same radius. The height of the cone is 10 cm and the radius of the base is 7 cm. Determine the volume of the toy. Also find the area of the coloured sheet required to cover the toy. 22 (Use = and 149 12.2) 7   Answer: Let ABC be the hemisphere and ADC be the cone standing on the base of the hemisphere. Height of the cone (h1) = 10 cm (Given) Radius of the cone (r1) = 7 cm (Given) Since the hemisphere is surmounted by the right circular cone of same radius, therefore Radius of the hemisphere (r2) = 7 cm                                     2 3 1 1 2 3 3 3 3 So, Volume of the toy = Volume of the cone + Volume of the hemisphere 1 2 = r h r 3 3 1 22 2 22 7 7 10 7 7 7 cm 3 7 3 7 1540 2156 cm 3 3 3696 cm 3 1232 cm Area of the coloured sheet required to cover the toy = CSA of hemisphere + CSA of cone    2 2 2 r r Where  is the slant height of the cone     2 2 1 1 2 2 r h = 7 10 = 49 100 = 149 = 12.2 cm So, 1.6 3 1 Or, x 3 1 3 1 1.6 1.73 1 3 1 1.6 2.73 2 2.184m Therefore, the height of the pedestal is 2.184m. 40) For the following data, draw a 'less than' ogive and hence find the median of the distribution. Age(in years): 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Number of persons: 5 15 20 25 15 11 9 OR The distribution given below show the number of wickets taken by bowlers in one-day cricket matches. Find the mean and the median of the number of wickets taken. Number of wickets 20-60 60-100 100-140 140-180 180-220 220-260 Number of bowlers: 7 5 16 12 2 3 Answer: Age Number of Persons (Cumulative frequency) Less than 10 5 Less than 20 5 + 15 = 20 Less than 30 20 + 20 = 40 Less than 40 40 + 25 = 65 Less than 50 65 + 15 = 80 Less than 60 80 + 11 = 91 Less than 70 91 + 9 = 100 Age No. of Persons (f) Cumulative frequency (cf) 0 – 10 5 5 10 – 20 15 20 20 – 30 20 40 30 – 40 25 65 40 – 50 15 80 50 – 60 11 91 60 – 70 9 100 Plot the points (10, 5), (20, 20), …, (70, 100) on a graph paper. OR 0 10 20 30 40 50 60 70 80 90 100 110 0 10 20 30 40 50 60 70 80 C u m u la ti v e F r e q u e n c y Upper Limits Median (34)