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ICSE Class X Sample / Model Paper 2026 : Mathematics Prelim 3

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Nayak Nayak's Tutorials Date: 25/01/2026 Std: X ICSE Preliminary Examination - 3 Mathematics Marks: 80 Duration: 3 Hrs. MODAL ANSWER PAPER SECTION A (Attempt all questions from this Section.) Question 1 Choose the correct answers to the questions from the given options. (Do not copy the question, write the correct answer only.) (i) In the given figure BAP BAP = DCP = 70 , PC 6 cm and CA = 4 cm, then PD : DB is: (a) 5 : 3 (c) 3 : 2 (b) 3 : 5 (d) 2 : 3 Ans. (c) 3 : 2 (ii) In the given diagram RT is a tangent touching the circle at S. If PST = 30 and SPQ = 60 then PSQ is equal to : (a) 40 (c) 60 (b) 30 (d) 90 Ans. (d) 90 (iii) The nth term of an Arithmetic Progression (A.P) is 2n + 5. The 10th term is: (a) 7 (b) 15 Ans. (c) 25 (iv) If A = [ 1 (c) 25 (d) 45 1 2 2] and B = 0 3 Which of the following operation is possible? (a) A - B Ans. (c) AB (b) A + B (c) AB (d) BA (v) An observer at point E, which is at a certain distance from the lamp post AB, finds the angle of elevation of top of lamp post from positions C, D and E as , and . It is given that B, C, D aSS SSnd E are along a straight line. Which of the following condition is satisfied ? (a) tan > tan (c) tan > tan Ans. (a) tan > tan (b) tan < tan (d) tan < tan [15] (vi) Event A: the sun will rise from east tomorrow. Event B: It will rain on Monday Event C: February month has 29 days in a leap year. Which of the above event(s) has probability equal to 1 ? (a) all events A, B and C (b) both events A and B (c) both events B and C (d) both event A and C Ans. (d) both event A and C (vii) A solid sphere is cut into two identical hemispheres. Statement 1: The total volume of two hemispheres is equal to the volume of the original sphere. Statement 2: The total surface area of two hem hemispheres ispheres together is equal to the surface area of the original sphere. Which of the following is valid ? (a) Both the statements are true. (b) Both the statements are false. (c) Statement 1 is true, and S Statement 2 is false. (d) Statement 1 is false, and Statement 2 is true. Ans. (c) Statement 1 is true, and Statement 2 is false. (viii) Given, x + 2 (a) + 3 and x is a prime number. The solution set for x is: (b) {0} Ans. (a) (c) {1} (d) {0, 1} (ix) A mixture of paint is prepared by mixing 2 parts of red pigments with 5 parts of the base. Using the given information in the following table, find the values of a, b & c to get the required mixture of paint. Parts of red pigment 2 4 b 6 Parts of base 5 a 12.5 c (a) a = 10, b = 10, c = 10 (c) a = 10, b = 5, c = 5 Ans. (d) a = 10, b = 5, c = 15 (b) a = 5, b = 2, c = 5 (d) a = 10, b = 5, c = 15 (x) Assertion (A): If sin2A + sinA = 1 then cos4 A + cos2A = 1 Reason (R): 1 sin2A = cos2A (a) (A) is true, (R) is false. (b) (A) is false, (R) is true. (c) Both (A) and (R) are true, and (R) is the correct reason for (A). (d) Both (A) and (R) are true, and (R) is the incorrect reason for (A). Ans. (c) Both (A) and (R) are true, and (R) is the correct reason for (A). (xi) In the given diagram ABC ABC EFG. If ABC = EFG = 60 , then the length of the side FG is: (a) 15 cm (b) 20 cm (c) 25 cm (d) 30 cm Ans. (a) 15 cm (xii) In the figure given below, AC is a diameter of the circle. AP = 3cm and PB = 4 cm and QP AB. If the area of APQ is 18 cm2, then the area of shaded portion QPBC is: (a) 32 cm2 Ans. (c) 80 cm2 (b) 49 cm2 (c) 80 cm2 (d) 98 cm2 (xiii) When the roots of a quadratic equation are real and equal then the discriminant of the quadratic equation is: (a) Infinite Ans. (c) Zero (b) Positive (c) Zero (d) Negative (xiv) In the given diagram AC is a diameter of the circle and ADB = 35 . The degree measure of x is: (a) 55 Ans. (a) 55 (xv) If 2 0 (b) 35 2 1 8 + 3 = 1 4 0 12 The value of x is: (a) 2 (b) 3 Ans. (d) 5 (c) 45 (d) 70 (c) 4 (d) 5 8 1 Question 2 (i) In a recurring deposit account for 2 years, the total amount deposited by a person is 9600. If the interest earned by him is one one-twelfth twelfth of his total deposit, then t find (a) the interest he earns (b) his monthly deposit Ans. (a) Given, (c) the rate of interest Interest earned by man = One One-twelfth of the total deposit = 9600 = 800 Hence, the interest earned = 800. (b) Given, In a recurring deposit account for 2 years (or 24 months), the total amount deposited by a person is 9600. Money deposited per month = = 400. [4] Hence, monthly deposit = 400. (c) By formula, 800 = 400 ( ) 800 = 400 800 = 400 25 800 = 10000 800 =100 R R= = 8% (ii) In the given diagram, O is the centre of the circle. PR and PT are two tangents drawn from the external point P and touching the circle at Q and S respectively. MN is a diameter of the circle, Given PQM = 42 and PSM = 25 . Find: (a) OQM (b) QNS (c) QOS (d) QMS Ans. (a) From figure, OQP = 90 (Tangent is perpendicular to radius at the point of contact) OQM = OQP - PQM OQM = 90 42 = 48 Hence, OQM = 48 (b) From figure, QNM = PQM = 42 (By alternate segment theorem) SNM = PSM = 25 (By alternate segment theorem) QNS = QNM + SNM QNS = 42 + 25 = 67 Hence, QNS = 67 . (c) We know that, Angle subtended by an arc at the center is twice the angle subtended by the arc at any other point of the circle. QOS = 2 QNS QOS = 2 67 = 134 Hence, QOS = 134 (d) From figure, QMSN is a cyclic quadrilateral. We know that, Sum of opposite angles of a cyclic quadrilateral is 180 . [4] QMS + QNS = 180 QMS + 67 = 180 QMS = 180 67 = 113 . Hence, QMS = 113 . (iii) Prove that: Ans. L.H.S. = + [4] ( = = 2 secA + ) ( ) = ( = ( = ) ) ( ) ( ) = 2 sec A L.H.S. = R.H.S. Question 3 (i) While factorizing a given polynomial, using remainder & factor, theorem, a student finds that (2x + 1) is a factor of 2x3 + 7x2 + 2x 3. (a) Is the student s solution correct stating that (2x + 1) is a factor of the given polynomial ? (b) Give a valid reason for your answer. Also, factorize the given polynomial completely. Ans. 2x + 1 = 0 2x = -1 x=- Substituting x = in 2x3 + 7x2 + 2x 3, we get: + + + ( ) + + Since, remainder is not equal to zero. [4] Hence, (2x + 1) is not a factor of the given polynomial. Substituting = + in 2x3 + 7x2 + 2x 3, we get: + + + + 2-2 0 Since, remainder is equal to zero. is factor of polynomial, = = 2x = 1 2x 1 is factor of polynomial Dividing 2x3 + 7x2 + 2x 3 by 2x 1, we get: 2x3 + 7x2 + 2x 3 by 2x 1 = (2x -1) (x2 + 4x +3) = (2x -1) [x2 + 3x + x + 3] = (2x -1) 1) [x (x+3) + 1 (x + 3)] = (2x -1) 1) (x + 1) (x +3) Hence, 2x3 + 7x2 + 2x -3 = (2x -1) (x +1) (x + 3) (ii) If = Find: 1 3 2 2 1 , = and = 5 1 7 4 4 4 2 (a) A + C (b) B(A+C) Ans. (a) Given, A+C = + = + ( ) + + + ( ) = Hence, A + C = (c) 5B [4] (d) B(A+C) -5B 5B (b) From part (a), we get: B(A + C) = A+C= . = ( + ) ( + ) ( + ) ( + ) = + + = + + Hence, B(A+C) = (c) Solving, 5B = 5. = Hence, 5B = (d) From part b, B(A+C) = From part c, 5B = Solving, B(A+C) -5B = = Hence, B(A+C) 5B = (iii) Use graph sheet for this question. Take 2 cm = 1 unit along the axes. (a) Plot A (0,3), B (2, 1) and C (4, -1). (b) Reflect point B and C in y-axis and name their images as B and C respectively. Plot and write coordinates of the points B and C . (c) Reflect point A and in the line BB and name its images as A . (d) Plot and write coordinates of point A . (e) Join the points ABA B and give the geometrical name of the closed figure so formed. Ans. From figure, [5] Coordinates of point A = (0, 1), B = (-2, 1) and C (-4, -1) The closed figure ABA B B is a square. Question 4 SECTION B (i) The polynomial 3x3 + 8x2 15x + k has (x 1) as a factor. Find the value of k. Hence factorize the resulting polynomial completely. Ans. x - 1 = 0 x = 1. Given, (x 1) is a factor of 3x 3 + 8x2 15x + k. Thus, on substituting x = 1 in 3x 3 + 8x 2 15x + k, the remainder will be zero. 3 (1)3 + 8(1)2 15 (1) + k = 0 3.1 + 8.1 15 + k = 0 3 + 8 15 + k = 0 11 15 + k = 0 k 4=0 k = 4. Polynomial = 3x3 + + 8x2 15x + 4 On dividing (3x3 +8x2 15 + 4) by (x-1), we get: 3x3 + 8x2 15x + 4 = (x -1) (3x2 + 11x 4) = (x 1) [3x2 + 12x x - 4] = (x 1) [3x(x + 4) 1 (x + 4)] = (x 1) (3x 1) (x + 4) Hence, 3x3 + 8x2 15x + 4 = (x -1) (3x -1) (x + 4) [3] (ii) Using ruler and compass construct a triangle ABC in which AB = 6cm, BAC = 120 and AC = 5 cm. Construct a circle passing through A, B and C. Measure and write [3] down the radius of the circle. Ans. Steps of construction: 1. Draw AB = 6 cm. 2. Draw AP so that angle BAP = 120 . 3. From AP cut AC = 5 cm BAC = 120 . 4. Join B and C to get triangle ABC. 5. Draw perpendicular bisectors of AB and AC which cut each other at point O. 6. With O as center and OA (or OB or OC) as r adius draw circle which passes through points A, B and C. Hence, radius of circle obtained = 5cm. (iii) Using step-deviation deviation method, find mean for the following frequency distribution. Class 0-15 15-30 30-45 45-60 60 60-75 75-90 Frequency 3 4 7 6 8 2 Ans. In the given table i is the class interval which is equal to 15. Class Class mark (x) d = (x-A) u = d/i Frequency (f) fu 0-15 7.5 -45 -3 3 -9 15-30 22.5 -30 -2 4 -8 30-45 37.5 -15 -1 7 -7 45-60 A = 52.5 0 0 6 0 60-75 67.5 15 1 8 8 75-90 82.5 30 2 2 4 =30 u = -12 Total Mean = A + = 52.5 = . + = 52.5 6 = 46.50 Hence, mean = 46.50 [4] Question 5 (i) ABCD is a square where B (1,3), D (3,2) are the end points of the diagonal BD. Find: [3] (a) the coordinates of point of intersection of the diagonals AC and BD (b) the equation of the diagonal AC Ans. Diagonal of square bisects each other at 90 O be the midpoint of both of BD and AC, O (a, b) = O (a, b) = , , mBD.mAC = -1 (BD AC) mAC = -1 =2 mAC = 2 Equation of AC is given by (y y1) = mAC (x x1) y- = 2 (x 2) 4x 2y 3 = 0 (ii) Using Componendo and Dividendo solve for x: Ans. Given : [3] =3 = Applying componendo and dividendo, we get: = = = + = Squaring both sides we get: 2x + 2 = 4(2x 1) 2x + 2 = 8x - 4 6x = 6 x= =1 Hence, x = 1. (iii) Oil is stored in a spherical vessel occupying 3/4 of its full capacity. Radius of this Spherical vessel is 28 cm cm.. The oil is then poured into a cylindrical vessel with a radius of 21 cm.. Find the height of the oil in the cylindrical vessel (correct to the nearest cm). Take = [4] Ans. Given, Radius of spherical vessel (r) = 28 cm Volume of spherical vessel (v) = Volume of oil in vessel = Substituting values we get: = = = Radius of cylindrical vessel (R) = 21 cm Let height of oil in cylindrical vessel be h cm. Volume of oil = Volume of cylinder upto which oil is filled ( R2h) = = = = = . . Hence, height of the oil in the cylindrical vessel = 50 cm. Question 6 (i) A bag contain 25 cards, numbered through 1 to 25. A card is drawn at random. What is the probability that the number on the card drawn is: (a) multiple of 5 (b) a perfect square (c) a prime number ? Ans. Total number of outcomes = 25 (a) Number of multiples of 5 between 1 to 25 = 5 (5, 10, 15, 20, 25) P (that number on card is a multiple of 5) = No. of multiples of 5 Total number of outcomes = = Hence, the required probability is (b) Number of perfect squares between 1 to 25 = 5 (1, 4, 9, 16, 25) P (that number on card is a perfect square) [3] = No. of perfect squares Total number of outcomes = = Hence, nce, the required probability is (c) Number of prime numbers between 1 to 25 = 9 (2, 3, 5, 7, 11, 13, 17, 19, 23) P (that number is a prime number) = No. of prime numbers Total number of outcomes = Hence, the require probability is (ii) In the given graph, P and Q are points such that PQ cuts off intercepts of 5 units and 3 units along the x-axis axis and y y-axis axis respectively. Line RS is perpendicular to PQ and passes through the origin. Find the: (a) coordinates of P and Q (b) equation of line RS Ans. (a) Given, PQ cuts off intercepts of 5 units and 3 u units the x-axis and y-axis axis respectively. Thus, Coordinates of P = (5, 0) and coordinates of Q = (0, -3) Hence, coordinates of P = (5, 0) and coordinates of Q = (0, -3). (b) By formula, Slope (m) = Slope of PQ = mPQ = = We know that, = Product of slopes of perpendicular lines = -1 Slope of RS Slope of PQ = -1 Slope of RS = -1 1 Slope of RS = By point-slope slope formula, Equation of line : y y1 = m(x x1) [3] Since, slope of RS = and it passes through the origin. Equation of RS: y 0= y= (x 0) 3y = -5x 5x + 3y = 0 Hence, the equation of line RS is 5x + 3y = 0. (iii) The first, the last term and the common difference of an Arithmetic Progression [4] are 98, 1001 and 7 respectively. Find the following for the given Arithmetic Progression: (a) number of terms n Ans. Find the number of terms n (b) Sum of the n terms The formula for the nth term of an arithmetic progression is an a1 + (n-1)d. We are given a1 = 98, an = 1001, and d = 7. We can solve for n: 1001 = 98 + (n-1)7 Subtract 98 from both sides 903 = (n-1)7 Divide by 7: =n 1 129 = n 1 Add 1 to both sides: n = 130 Find the sum of the n terms The formula for the sum of the first n terms is Sn = (a1 + an). Using the calculated value n = 130 and the given values a1 = 98, an = 1001: S130 = (98 + 1001) S130 = 65 (1099) S130 = 71435 (a) The number of terms n is 130. (b) The sum of the n terms is 71435. Question 7 (i) Solve the following quadratic equation for x and give your answer correct to three significant figures: 2x2 10x + 5 = 0 (Use mathematical tables if necessary) Ans. Comparing equation 2x2 10x + 5 = 0 with ax2 + bx + c = 0, we get: a = 2, b = -10 and c = 5 By formula, x= Substituting values we get: ( ) ( ) = = [3] = = = ( ) = = . = . = . . . , , = 4.44, 0.565 Hence, x = 4.44, 0.565 0.565. (ii) Mr. Gupta invested 33000 in buying 100 shares of a company at 10% premium. [3] The dividend declared by the company is 12%. Find: (a) the number of shares purchased by him. (b) his annual dividend. Ans. Money invested = 33000 N.V. of share = 100 M.V. = N.V + Premium = 100 + = 100 + 10 = 110. (a) Number of shares = Money invested M.V. = = 300. Hence, no. of shares purchased = 300 (b) By formula, Annual dividend = Number of shares Rate of dividend N.V. = 300 100 = 3600 Hence, annual dividend = 3600. (iii) In the given figure, AC//DE//BF. If AC = 24 cm, EG = 8 cm, GB = 16 cm, BF = 30 cm. [4] (a) Prove GED GBF (b) Find DE Ans. (a) Given, (c) DB : AB DE // BF DEG = GBF GBF (Alternate angles area equal) EDG = GFB GFB (Alternate angles are equal) GED GBF GBF (By A.A. axiom) Hence, proved that GED GED GBF. (b) We know that, Corresponding sides of similar triangle are proportional. Since, GED GBF = = DE = 30 DE = 15 cm. Hence, DE = 15 cm. (c) In BAC and BDE, BAC = BDE BDE (Corresponding angles are equal) ABC = DBE DBE (Common angle) BAC BDE (By A.A. axiom) We know that, Corresponding sides of similar triangle are proportional. Since, BAC BDE = = = = BD : AB = 5 : 8 Hence, BD : AB = 5 : 8. Question 8 (i) A vertical tower standing on a horizontal plane is surmounted by a vertical flagstaff. At a point 100m away from the foot of the tower, the angle of elevation of the top and bottom of the flagstaff are 54 54 and 42 respectively. Find the height of the flagstaff. Give your answer correct to nearest metre. Ans. In APB, [5] tan 42 = 0.9004 = AB = 0.9004 100 = 90.04 m In APF, tan 54 = 1.3764 = AF = 1.3764 100 = 137.64 m From figure, BF = AF AB = 137.64 90.04 = 47.60 48 m. Hence, height of flagstaff = 48 m. (ii) The marks of 200 students in a test were recorded as follows: Marks% No. of students [5] 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 5 7 11 20 40 52 36 15 Using a graph sheet draw ogive for the given data and use it to find the: (a) median (b) number of students who obtained more than 65% marks. (c) number of students who did not pass, if the pass percentage was 35. Ans. Marks (%) 0-10 5 5 10-20 7 12 20-30 11 23 30-40 20 43 40-50 40 83 50-60 52 135 60-70 36 171 70-80 15 186 80-90 09 195 90-100 05 200 (a) Median = 53 1 (b) More than 65% = 46 2 (c) Didn t pass = 31 2 9 5 Question 9 (i) If A = 4 4 , find A2. If A2 = p A, then find the value of p. 4 4 [3] Ans. Given, A2 = pA + ( ) ( ) ( ) + ( ) = + ( ) ( ) ( ) + + 32 = = + = = 32 = 4p p= =8 Hence, p = 8 (ii) Solve the following inequation. Write down the solution set and represent it on the real number line. -5 (x 9) 17 9x > x + 2, x R Ans. Given, equation: -5 5 (x 9) 17 9x > x + 2 Solving L.H.S. of the given equation: -5(x 9) 17 9x -5x + 45 17 9x -5x + 9x 17 45 4x -28 [3] x - x - 7 (1) Solving L.H.S. of the given equation: 17 9x > x + 2 x + 9x < 17 - 2 10x < 15 x< x< x < 1.5 (2) From equation (1) and (2), we get: Solution set: {x : -7 x < 1.5} Solution set on the number line: (iii) X, Y, Z and C are the points on the circumference of a circle with centre O . AB is a tangent to the circle at X and ZY = XY. Given OBX = 32 and AXZ AXZ = 66 . 66 Find: (a) BOX (b) CYX (c) ZYX (d) OXY Ans. (a) Given, In BOX, OX BX (radius tangent at its point of contact) OXB = 90 . By angle-sum sum property of triangle, BOX + OBX + OXB OXB = 180 180 BOX + 32 + 90 = 180 180 BOX + 122 = 180 BOX = 180 122 BOX = 58 Hence, BOX = 58 (b) From figure, COX = BOX = 58 We know that, The angle which, an arc of a circle subtends at the centre is double that which it subtends at any point on the remaining part of the circumference. circumference CYX = CYX = COX [4] CYX = 29 . Hence, CYX = 29 . (c) We know that, The angle between a tangent and a chord through the point of contact is equal to an angle in alternate segment. ZYX = AXZ = 66 Hence, ZYX = 66 . (d) From figure, In isosceles ZXY, ZY = XY We know that, The angles opposite to equal of a triangle are equal. ZXY = XZY By angle sum property in XYZ, XZY + ZXY + ZYX = 180 ZYX + 2 XZY = 180 2 XZY = 180 - ZYX XZY = = = = 57 . We know that, The angle between a tangent and a chord through the point of contact is equal to an angle in alternate segment. YXB = XZY = 57 . Also, ZXY = XZY = 57 . From figure, OXY = OXB - YXB OXY = 90 57 = 33 Hence, OXY = 33 . Question 10 (i) The following bill shows the GST rate and marked price of articles: [3] Vidhyut Electronics S. No. Item Marked Price Quantity Rate of GST (a) LED TV set 12000 01 28% (b) MP4 player 5000 01 18% Find the total amount to be paid (including GST) for the above bill. Ans. For LED TV set, M.P. = 12,000 G.S.T. = 28% G.S.T. amount = 28% of 12,000 = , = 28 120 = 3360 Total amount = 12,000 + 3360 = 15,360. For MP4 player, M.P. = 5,000 G.S.T. = 18% G.S.T. amount = 18% of 5,000 = , = 18 50 = 900 Total amount = 5,000 + 900 = 5900 Total bill = 15,360 + 5,900 = 21,260. Hence, total bill = 21,260. (ii) A man covers a distance of 100 km, travelling with a uniform speed of x km/hr. Had the speed been 5 km/hr more it would have taken 1 hour less. Find x the original speed. Ans. In first case: Distance = 100 km Speed = x km/hr Distance Time = Speed = In second case: Distance = 100 km Speed = (x + 5) km/hr Distance Time = Speed = According to question, The difference between the time taken between first and second case is 1 hour. - =1 ( ) ( =1 ) =1 =1 500 = + + = + = ( + ) ( + ) = ( + )( + ) = = + = = Since, speed cannot be negative. [3] x = 20 km/hr. Hence, original speed = 20 km/hr. (iii) A and B are two points on the x x-axis and y-axis respectively. (a) Write down the coordinates of A and B. (b) P is a point on AB such that AP : PB = 3 : 1. Using section formula find the coordinates of point P. (c) Find the equation of a line passing through P and perpendicular to AB. Ans. (a) From figure, A = (4, 0) and B = (0, 4) (b) Let coordinates of P be (x, y) By section formula, (x, y) = , , Substituting values we get: (x, y) = = = , , , = (1, 3) Hence, coordinates of P = (1, 3) (c) By formula, Slope = Substituting values we get: Slope of AB = = = -1 We know that, Product of slope of perpendicular lines = -1 Slope of AB Slope of line perpendicular to AB = -1 - 1 Slope of line perpendicular to AB = -1 Slope of line perpendicular to AB = =1 [4] Line passing through P and perpendicular to AB: y y1 = m (x x1) y 3 = 1 (x 1) y 3=x 1 y=x 1+3 y=x+2 Hence, required equation is y = x + 2.

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