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ICSE Class IX Sample / Model Paper 2021 : Mathematics : IX - Maths _ Annual Examination _ Samples or Models Papers

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Navin Singh
Kanchrapara English Medium School (KEMS), Bagmore

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ICSE Board Class IX Mathematics Sample Paper 1 Time: 2 hrs Total Marks: 80 General Instructions: 1. Answers to this paper must be written on the paper provided separately. 2. You will NOT be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper. 3. The time given at the head of this paper is the time allowed for writing the answers. 4. This question paper is divided into two Sections. Attempt all questions from Section A and any four questions from Section B. 5. Intended marks for questions or parts of questions are given in brackets along the questions. 6. All working, including rough work, must be clearly shown and should be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. 7. Mathematical tables are provided. SECTION A (40 Marks) (Answer all questions from this Section) Q. 1. (a) Calculate the amount and the compound interest on Rs. 6000 at 10% p.a. for 1 1 years , when the interest is compounded half yearly. 2 (b) If 11 7 11 7 x y 77, find the values x and y. [3] [3] (c) Sonu and Monu had adjacent triangular fields with a common boundary of 25 m. The other two sides of Sonu's field were 52 m and 63 m, while Monu's were 114 m and 101 m. If the cost of fertilization is Rs 20 per sq m, then find the total cost of fertilization for both of Sonu and Monu together. [4] Q. 2. (a) Calculate the area of fig., ABCDE. Given DX = 9 cm, DC = 5 cm FC = 4 cm and XB = 6 cm. Also F and X are the mid-points of EC and AB respectively. [3] (b) If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal. (c) Find x, if 3 4 2x 1 2 8 . 3 5 [4] [3] Q. 3. (a) Given log x = a + b and log y = a b, find the value of log 10x in terms of a and b . [3] y2 (b) The bisector of A of a ABC meets BC at D and BC is produced to E. prove that ABC + ACE = 2 ADC. [3] (c) Using ruler and compass only, construct a trapezium ABCD, in which the parallel sides AB and DC are 3.3 cm apart; AB = 4.5 cm, A = 120 , BC = 3.6 cm and B is obtuse. [4] Q. 4. (a) The following figure shows a right-angled triangle ABC with B = 90 , AB = 15 cm and AC = 25 cm. D is a point in side BC and CD = 7cm. If DE AC, find the length of DE. [4] (b) Prove that the interior angle of a regular pentagon is three times the exterior angle of a regular decagon. [3] (c) If tan + cot = 3, find the value of tan2 + cot2 . [3] SECTION B (40 Marks) (Answer any four questions from this Section) Q. 5. (a) Graphically solve the simultaneous equations: x 2y = 1; x + y = 4 (b) A and B together can do a piece of work in 15 days. If A s one day s work is [4] 3 times 2 B s one day s work; in how many days can A and B do the work alone? [3] (c) How many sides does a regular polygon have, each angle of which is of measure 108o? [3] Q. 6. (a) A person invests Rs. 5600 at 14% p.a. compound interest for 2 years. Calculate i. The interest for the first year. ii. The amount at the end of the first year. iii. The interest for the second year corrected to the nearest rupee. (b) A point P lies on the x-axis and another point Q lies on the y-axis. i. Write the ordinate of point P. [3] [3] ii. Write the abscissa of point Q. iii. If the abscissa of point P is -12 and the ordinate of point Q is 16; calculate the length of line segment PQ. (c) ABC is right angled at B. If m A = 30 and BC = 8 cm. Find the remaining angles and sides. [4] Q. 7. (a) 3n 1 9n 1 Simplify: n(n 1) n 1 3 3n 1 [4] (b) The area of an isosceles triangle is 12 cm2 and the base is 8 cm in length. Find the perimeter of the triangle. (c) In the figure, AC = CD. Prove that BC < CD. [3] [3] Q. 8. (a) If cos = 2 mn , find the value of sin ( > given m > n . m n [4] (b) The mean of 5 numbers is 20. If one number is excluded the mean of the remaining numbers becomes 23. Find the excluded number. [3] (c) 3 equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the new cuboid to that of the sum of the surface areas of three cubes. [3] Q. 9. (a) In the given quadrilateral AD = BC, and P, Q, R and S are the midpoints of the sides AB, BD, CD and AC, respectively. Prove that PQRS is a rhombus. [4] (b) The distance (in km) of 40 engineers from their residence to place of work were found as follows: [3] 5 3 10 20 25 11 13 7 12 31 2 19 10 12 17 18 11 32 17 16 3 7 9 7 8 3 5 12 15 18 12 12 14 2 9 6 15 15 7 6 Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0 - 5 (5 not included). What main feature do you observe from this tabular representation? (c) Solve: log x (8x 3) log x 4 2 [3] Q. 10. (a) Prove that 5 is an irrational number. (b) If tan 1 2 [4] 1 tan 1 tan 2 , find the value of ( 1 + 2) given that tan 1 and 2 1 tan 1 tan 2 1 tan 2 . 3 2 x2 1 (c) If 4 , find the value of 2x3 3 . x x [3] [3] Q. 11. (a) Factorise: 4a3b 44a2b + 112b [3] (b) Find the area of following figure: [3] (c) In ABC, AB = AC = x, BC = 20 cm and the area of the triangle is 250 cm2. Find x. [4] Solution SECTION A (40 Marks) Q. 1. 1 3 (a) P = Rs. 6000, R = 10% p.a., n = 1 years years 2 2 R A P 1 2 100 2n Interest is compounded half yearly 3 10 6000 1 2 100 3 5 6000 1 100 6000 1.05 3 Rs. 6945.75 Amount = Rs. 6945.75 C.I. = 6945.75 6000 = Rs. 945.75 (b) We have 7 x y 7 11 7 11 7 11 11 11 77 77 7 x y 77 11 7 18 2 77 x y 77 4 9 1 77 x y 77 2 2 9 1 x ,y 2 2 77 (c) Q. 2. (a) In DFC, DC2 = DF2 + FC2 [Pythagoras Theorem] 5 DF 4 2 2 2 52 42 DF2 DF2 25 16 DF 3 1 1 Area of DEC 4 4 3 8 3 12 cm2 2 2 FX = DX DF = 9 3 = 6 cm 1 1 Area of trapezium CEBA = 4 4 6 6 6 20 6 60 cm2 2 2 Area of figure ABCDE = area of DEC + area of trapezium ECBA = 12 + 60 = 72 cm2 (b) Given that AB and CD are two chords of a circle with centre O, intersecting at a point E. PQ is the diameter through E, such that AEQ = DEQ. To prove that AB = CD. Draw perpendiculars OL and OM on chords AB and CD respectively. Now, m LOE = 180 90 m LEO ... [Angle sum property of a triangle] = 90 m LEO m LOE = 90 m AEQ m LOE = 90 m DEQ m LOE = 90 m MEQ LOE = MOE In OLE and OME, LEO = MEO LOE = MOE EO = EO OLE OME OL = OM Therefore, cords AB and CD are equidistant from the centre. (c) 3 4 2x 1 2 1/3 4 22/3 2 2 3 8 5 3 2x 1 2x 2 1 2x 2 1 2 81/3 81/3 5 5 2 5 2 1 2x 5 3 2 4x 1 15 x 7 2 Q. 3. (a) Given, log x = a + b and log y = a b 10x log 2 = log 10x log y2 [Using quotient law] y = log 10 + log x 2 log y = 1 + (a + b) 2(a b) = 1 + a + b 2a + 2b = 1 a + 3b (b) Given : In ABC, AD is the bisector of BAC and BC is produced to E To Prove: ABC + ACE = 2 ADC Proof: Let BAD = DAC = x and ABC = y Now, ACE = ABC + BAC .[Exterior angle = Sum of interior opposite s] ACE = y + 2x In ABD, ADC = x + y .[Exterior angle = Sum of interior opposite s] ABC + ACE = y + y + 2x = 2(x + y) ABC + ACE = 2 ADC (c) Steps of Construction: 1. Draw AB = 4.5 cm. 2. Draw BAS = 120 and draw EA AB. 3. From A, cut an arc of measure 3.3 cm on EA such that AX = 3.3 cm. 4. Through X, draw a line QP which is parallel to AB which cuts AS at D. 5. Through B draw an arc taking radius 3.6 cm at C on PQ. 6. Join CB. Thus, ABCD is the required trapezium. Q. 4. (a) We can see that ABC is a right-angles triangle. AB2 + BC2 = AC2 ....[By Pythagoras theorem] 152 + BC2 = 252 BC2 = 400 BC = 20 cm Now BC = DB + CD 20 = DB + 7 DB = 13 cm Again ADB is a right angled triangle. AB2 + DB2 = AD2 ....[By Pythagoras theorem] 152 + 132 = 390 BC = 19.8 cm In the right-angled CDE ED2 + CE2 = CD2 ....[By Pythagoras theorem] ED2 CD2 CE2 72 x2 In the right-angled AED ED2 + AE2 = AD2 ....[By Pythagoras theorem] ED2 AD2 AE2 19.82 25 x 2 Since in both the cases length of ED is same and hence ED2 is also same in both the cases. 72 x 2 = 19.82 25 x 2 72 x 2 = 19.82 625 x 2 + 50x 72 19.82 625 = 50x 281.96 = 50x x = 5.63 cm So, ED2 = 72 5.62 = 17.64 ED = 4.2 cm DE (b) Each interior angle of a regular pentagon 2 5 4 90 5 [n = 5] 6 90 5 108 360 Each exterior angle of a regular decagon 36 [n = 10] 10 Each interior angle of a regular pentagon = 3(Exterior angle of a regular decagon) (c) Given tan + cot = 3, Squaring both sides, tan cot 2 32 tan2 cot 2 2tan cot 9 tan2 cot 2 2tan tan2 cot 2 2 9 tan2 cot 2 7 1 cos 9 cot tan sin Section B (40 Marks) Q. 5. (a) Consider equation, x 2y = 1 x 1 y 2 x y 1 0 3 1 .(1) 5 2 Points are (1, 0), (3, 1) and (5, 2). Now consider equation x + y = 4 .(2) x y 0 4 2 2 4 0 Points are (0, 4), (2, 2) and (4, 0). Now plotting these points on the graph paper, we get Since the lines intersect at (3, 1), therefore the solution is x = 3 and y = 1. (b) In 15 days A and B together can do a piece of work. 1 Therefore, in 1 day they do work 15 Let us assume that A takes x days and B takes y days to do the work alone. 1 So A s one day s work = x 1 B s one day s work = y 1 3 1 . x 2 y 3x 2y 0 2y 3x 3x ....(i) 2 1 1 1 Also, x y 15 1 2 1 x 3x 15 3 2 1 3x 15 3x 75 y x 25 3 25 37.5 2 Hence, A will do the work alone in 25 days and B will do it alone 37 and half days. y (c) Let there be n sides of the polygon. Then, each interior angle is of measure 2n 4 n 90 2n 4 90 108 n (2n 4) 90 108n 180n 360 108n 180n 108n 360 72n 360 n 5 Hence the given polygon has 5 sides. Q. 6. 5600 14 1 Rs. 784 100 (ii) Amount at the end of the first year = 5600 + 784 = Rs. 6384 (iii) Interest for the second year = 6384 14 1 Rs. 893.76 Rs. 894 to the nearest rupee 100 (a) (i) Interest for first year = (b) (i) Since, the point P lies on the x-axis, its ordinate is 0. (ii) Since, the point Q lies on the y-axis, its abscissa is 0. (iii) The co-ordinates of P and Q are ( 12, 0) and (0, 16) respectively. PQ ( 12 0)2 (0 16)2 144 256 400 20 (c) Here m A +m C = 90 as m B = 90 30 + m C = 90 m C = 60 In right-angled ABC, BC tan30 AB 1 8 3 AB AB 8 3 cm sin30 BC AC 1 8 2 AC AC 16 cm Q. 7. (a) 3n 1 9n 1 3n(n 1) 3n 1 n 1 n 1 3 3n(n 1) 3 n 1 n 1 9n 1 3n 1 3 n(n 1) n 1 3 3 3 n 1 n 1 3n 1 n2 n 3 n2 1 3 3 2 n 1 n2 1 3n 1 3 2 2n 2 3n n 3 2 1 (n2 n) (2n 2) 3n 1 n 2 1 n2 n 2n 2 3n 1 n 3 2 1 32 1 9 (b) 1 b 4a2 b2 4 (where b is the base and a is the length of equal sides) Given, b = 8 cm and area =12 cm2 1 8 4a2 82 12 4 Area of an isosceles = 4a2 82 6 4a2 82 36 4a2 100 a2 25 a 5 cm Perimeter = 2a + b = 2 5 + 8 = 18 cm (c) Given AC = CD To prove: BC < CD Proof: In ACD, m ACD = 180 70 = 110 [Linear pair] 70 CAD = ADC = [Angles opposite to equal sides are equal] 35 2 In ABC, m BAC = 70 35 = 35 [ BAC = BAD CAD] m ABC = 180 (70 + 35 ) [Sum of all s of a is 180 ] = 75 BAC < ABC BC < AC So, BC < CD [Since AC = CD] Q. 8. (a) cos = 2 mn m n Now, sin2 cos2 1 sin2 1 cos2 2 mn 1 m n 2 4mn 1 m n 2 m n 4mn 2 m n 2 m2 n2 2mn 4mn m n 2 m2 n2 2mn m n 2 m n 2 m n m n m n m n sin m n 2 2 (b) Mean = 20 Number of terms = 5 Total sum = 20 5 = 100 Let the excluded number be x. Then, 100 x 23 4 100 x = 23 4 = 92 x=8 Hence, the excluded number is 8. (c) Let the side of each of the three equal cubes be 'a' cm. Surface area of one cube = 6a2 cm2 Therefore, sum of surface areas of the three cubes = 3 6a2 = 18a2 cm2 Now, Length of the new cuboid = 3a cm Breadth of the new cuboid = a cm Height of the new cuboid = a cm Total surface area of the new cuboid = 2[(3a a) + (a a) + (a 3a)] = 2[3a2 + a2 + 3a2] = 2[7a2] = 14a2 cm2 Thus, the required ratio of T.S.A. of the new cuboid to that of the sum of the S.A. of the 3 cubes = 14a2 : 18a2 = 7 : 9. Q. 9. (a) Given: In quadrilateral ABCD; AD = BC. P, Q, R, S are the mid-points of AB, BD, CD and AC respectively. To Prove: PQRS is a rhombus. 1 Proof: In ACD, RS||AD and RS AD ....(i) 2 [Line joining the mid-points of the two sides of triangle is parallel and half of the third side.] Similarly, 1 In ABD, PQ||AD and PQ AD ....(ii) 2 1 In BCD, QR||BC and QR BC ....(iii) 2 1 In ABC, SP||BC and SP BC ....(iv) 2 As AD = BC [Given] RS = PQ = QR = SP and RS||PQ and QR||SP [From (i), (ii), (iii) and (iv)] Hence PQRS is a rhombus. (b) Given that we have to construct a grouped frequency distribution table of class size 5. So, the class intervals will be as 0 5, 5 10, 10 15, 15 20, and so on. Required grouped frequency distribution table is as follows: Distance (in km) Tally marks Number of engineers 0 5 5 5 10 11 10 15 11 15 20 9 20 25 1 25 30 1 30 35 2 Total 40 Only 4 engineers have homes at a distance of more than or equal to 20 km from their work place. Most of the engineers have their workplace at a distance of upto 15 km from their homes. (c) log x (8x 3) log x 4 2 8x 3 log x 2 4 8x 3 2 x 4 8x 3 4x 2 4x 2 8x 3 0 4x 2 6x 2x 3 0 2x(2x 3) 1(2x 3) 0 (2x 3)(2x 1) 0 2x 3 0 or 2x 1 0 x 3 1 or x 2 2 Q. 10. (a) Let us assume, on the contrary that 5 is a rational number. Therefore, we can find two integers a, b (b 0) such that 5 a b Where a and b are co-prime integers. 5 a b a 5b a2 5b2 Therefore, a2 is divisible by 5 then a is also divisible by 5. So a = 5k, for some integer k. Now,a2 (5k)2 5(5k 2 ) 5b2 b2 5k 2 This means that b2 is divisible by 5 and hence, b is divisible by 5. This implies that a and b have 5 as a common factor. And this is a contradiction to the fact that a and b are co-prime. So our assumption that 5 is rational is wrong. Hence, 5 cannot be a rational number. Therefore, (b) 1 1 tan 1 tan 2 2 3 tan 1 2 1 tan 1 tan 2 1 1 1 2 3 3 2 5 5 6 tan 1 2 6 6 1 1 6 1 6 5 1 6 6 tan 1 2 1 tan45 1 2 45 5 is irrational. (c) x2 1 4 x x2 1 4x x2 4x 1 0 ....(i) On dividing equation (i) by x, we have 1 x 4 0 x 1 x 4 ....(ii) x On cubing equation (ii) both sides, we have 3 1 3 x + x = 4 1 1 1 x3 + 3 + 3 x x+ = 64 x x x 1 x3 + 3 + 3 4 = 64 x 1 x3 + 3 = 64 12 x 1 x3 + 3 = 52 x 2 1 2x3 + 3 = 2 x3 + 3 = 2 52 = 104 x x Q. 11. (a) 4a3b 44a2b + 112b 4ab a2 11a 28 4ab a2 7a 4a 28 4ab a(a 7) 4(a 7) 4ab a 7 a 4 (b) Construction: Draw TM QS 1 1 Area of RQS QS RN 35 20 350 cm2 2 2 Now, QS QM MS 35 25 MS MS 10 cm In STM, MS2 TM2 ST2 TM2 ST2 MS2 (26)2 (10)2 676 100 576 TM 24 cm PQ 1 1 Area of trapezium PQST (PT QS) PQ (25 35) 24 720 cm2 2 2 Thus, area of given figure Area of RQS Area of trapezium PQST 350 cm2 720 cm2 1070 cm2 (c) Given: In ABC, AB = AC = x, BC = 20 cm, Area of ABC = 250 cm2 To find: x Construction: Draw AD BC Since ABC is an isosceles triangle. AD bisects BC. BD = DC = 20/2 = 10cm 1 Area of ABC = BC AD 250cm2[Given] 2 1 20 AD 250 AD 25cm 2 In rt. ADC, AD2 DC2 AC2 252 102 x2 x2 625 100 725 x 5 29 cm [Pythagoras Theorem]

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