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ICSE Class X Board Exam 2026 : English Paper 1 (English Language)

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Diana
Smt. Sulochanadevi Singhania School, Thane
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Nayak's Tutorials Date: 28/12/2025 Std: X ICSE Preliminary Examination - 2 Mathematics Marks: 80 Duration: 3 Hrs. MODAL ANSWER PAPER SECTION A (Attempt all questions from this Section.) Question 1 [15] Choose the correct answers to the questions from the given options. (Do not copy the question, write the correct answer only.) (i) If 2 + + 6 = ( 2)( 3) for all values of x, then the value of k is: (a) -5 (b) -3 (c) -2 (d) 5 Ans. (a) -5 (ii) 57, 54, 51, 48, are in Arithmetic Progression. The value of the 8th term is: (a) 36 (b) 78 (c) -36 (d) -78 Ans. (a) 36 (iii) In the adjoining figure, AC is a diameter of the circle. AP = 3 cm 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 (b) 49 cm2 (c) 80 cm2 (d) 98 cm2 Ans. (c) 80 cm2 (iv) Given that the sum of the squares of the first seven natural numbers is 140, then their mean is: (a) 20 (b) 70 (c) 280 (d) 980 Ans. (a) 20 (v) The table shows the values of x and y, where x is proportional to y. 16 12 N M 18 6 What are the values of M and N ? (a) M = 4, N = 9 (b) M = 9, N = 3 (c) M = 9, N = 4 (d) M = 12, N = 0 Ans. (c) M = 9, N = 4 (vi) If matrix A = [-1 2] and matrix B[ (a) [-3] (b) [8] 3 ] , then matrix AB is equal to: 4 (c) [5] Ans. (c) [5] 1 2 (d) [ ] 3 4 (vii) Assertion (A): If + = = then ab = 1 Reason (R): 2 2 = 1 (a) (A) is true and (R) is false. (b) (A) is false and (R) is true. (c) Both (A) and (R) are true and (R) is the correct explanation of (A). (d) Both (A) and (R) are true, but (R) is not the correct explanation of (A). Ans. (c) Both (A) and (R) are true and (R) is the correct explanation of (A). (viii) For the given 25 variables: x1, x2, x3 x25 Assertion (A): To find median of the given data, the variate needs to be arranged in ascending or descending order. Reason (R): The median is the central most term of the arranged data. (a) A is true, R is false (b) A is false, R is true (c) both A and R are true (d) both A and R are false Ans. (c) both A and R are true (ix) 1. Shares of company A, paying 12%, `100 shares are at `80. 2. Shares of company B, paying 12%, `100 shares at `100. 3. Shares of company C, paying 12 %, `100 shares are at `120. Shares of which company are at premium? (a) Company A (b) Company B (c) Company C (d) Company A and C Ans. (c) Company C (x) The median of a grouped frequency distribution is found graphically by drawing: (a) a linear graph (b) a histogram (c) a frequency polygon (d) a cumulative frequency curve Ans. (d) a cumulative frequency curve (xi) In the given diagram, the radius of the circle with centre O is 3 cm. PA and PB are the tangents to the circle which are at right angle to each other. The length of OP is: (a) 3 2 cm Ans. (c) cm (b) 3 cm (c) 3 2 cm (d) 6 2 cm (xii) The coordinates of the vertices of ABC are respectively. (-4, -2), (6, 2) and (4, 6). The centroid G of ABC is: (a) (2, 2) (b) (2, 3) (c) (3, 3) (d) (0, -1) Ans. (a) (2, 2) (xiii) Volume of a cylinder of height 3 cm is 48 . Radius of the cylinder is: (a) 48 cm (b) 16 cm (c) 4 cm (d) 24 cm Ans. (c) 4 cm (xiv) The circumcentre of a triangle is the point which is: (a) at equal distance from the three sides of the triangle. (b) at equal distance from the three vertices of the triangle. (c) the point of intersection of the three medians. (d) the point of intersection of the three altitudes of the triangle. Ans. (b) at equal distance from the three vertices of the triangle. (xv) The solution set for < (a) {-5, -4, -3, -2, -1} < 2, is: (b) {-6, -5, -4, -3, -2, -1} (c) {-5, -4, -3, -2, -1, 0} (d) {-6, -5, -4, -3, -2, -1, 0} Ans. (a) {-5, -4, -3, -2, -1} Question 2 (i) 15, 30, 60, 120 are in G.P. (Geometric Progression). (a) Find the nth term of this G.P. in terms of n. (b) How many terms of the above G.P. will give the sum 945 ? Ans. Given that, 15, 30, 60, 120 are in G.P. We have, First term, a = 15 Common ratio, r= (a) nth term of GP is an =2 an = a(r) an = 15 (2)n-1 n-1 (b) Given that, Sum, S = 945 Let the number of terms taken be n S= ( ) On substituting the values, = ( ) 63 = 2n -1 2n = 64 2n = 2 6 n=6 So, 6 terms of the GP will give a sum of 945. [4] (ii) In ABC, ABC = 90 , AB = 20 cm, AC = 25 cm, DE is perpendicular to AC such that DEA = 90 and DE = 3 cm as shown in the given figure. (a) Prove that ABC AED (b) Find the lengths of BC, AD and AE. (c) If BCED represents a plot of land on a map whose actual area on ground is 576 m2, then find the scale factor of the map. Ans. (a) ABC and AED, ABC = AED [Both = 90 ] BAC = DAE [Common angles] ABC AED (By AA similarity postulate) Hence, proved that ABC AED. (b) Given, AB = 20 cm, AC = 25 cm, DE = 3 cm In ABC, By Pythagoras theorem, AB2 + BC2 = AC2 (20)2 + BC2 = (25)2 400 + BC2 = 625 BC2 = 625 - 400 BC2 = 225 BC = BC = 15 cm We know that, Since, corresponding sides of similar triangles are proportional we have: = = Solving, = = = = AE = 4 cm. Substituting values in = = = = we get: [4] = AD = 5 cm. Hence, BC = 15 cm, AE = 4 cm, AD = 6 cm. (c) Given, Area on ground = 576 m2 By formula, Area of triangle = Area of ABC = = base height AB BC 20 15 = 150 cm2. Area of AED = = = AE DE 4 3 12 = 6 cm2. From figure, Area of Quadrilateral (BCED) = Area of ABC Area of AED = 150 6 = 144 cm2. = Actual ground area = 576 m2. = 576 10000 cm2 = 5760000 cm2 Let scale factor be k. By formula, = Substituting values we get: = = = = Hence, scale factor equals 1 : 200. (iii) Find the value of a if x a is a factor of the polynomial 3x3 + x2 ax 81. Ans. p(x) = 3x3 + x2 ax 81 (x-a) is a factor of p(x) p(a) = 0 3(a)3 + (a)2 a(a) 81 = 0 3a3 + a2 a2 81 = 0 a=3 [4] Question 3 (i) In the given diagram, O is the centre of circle circumscribing the ABC. CD is perpendicular to chord AB. OAC = 32 . Find each of the unknown angles x, y, and z. Ans. Step 1: Find the value of x In OAC, OA = OC (radii of the same circle) Therefore, OAC is an isosceles triangle, and the angles opposite the equal sides are equal. OCA = OAC = 32 The sum of angles in a triangle is 180 . In OAC: AOC + OAC + OCA = 180 x + 32 + 32 = 180 x + 64 = 180 x = 180 64 x = 116 Step 2: Find the value of y The angle subtended by an arc at the center is twice the angle subtended by it at the remaining part of the circumference (at point B) AOC = 2 ABC x = 2y 116 = 2y y= y = 58 Step 3: Find the value of z CD is perpendicular to chord AB, so in the right-angles BDC, BDC = 90 The sum of angles in BDC is 180 : BDC + BCD + CBD = 180 90 + z + y = 180 Substitute the value of y from Step 2: 90 + z + 58 = 180 148 + z = 180 z = 180 148 z = 32 The unknown angles are x = 116 , y = 58 , and z = 32 . [4] (ii) From the given figure: [4] (a) Write down the coordinates of A and B. (b) If P divides AB in the ratio 2:3, find the coordinates of point P (c) Find the equation of a line parallel to line AB and passing through origin. Ans. (a) The coordinates of A are (5, 0) and the coordinates of B are (0, 3). (b) The coordinates of point P are (3, 6/5). (c) The equation of the line parallel to line AB and passing through the origin is 3x + 5y = 0 (or y = ) (iii) Use a ruler and compass to answer this questions. (a) Construct a circle of radius 4.5 cm and draw a chord AB of length 6.5 cm. (b) At A, construct CAB = 75 , where C lies on the circumference of the circle. (c) Construct the locus of all points equidistant from A and B. (d) Construct the locus of all points equidistant from CA and BA. (e) Mark the point of intersection of the two loci as P. Measure and write down the length of CP. Ans. Steps of construction: 1. With O as center draw a circle of radius 4.5 cm. 2. Take a point A on the circumference with A as center cut an arc of radius 6.5 cm, intersecting circumference at point B. 3. Construct CAB = 75 , where C lies on the circumference of the circle. 4. Draw XY, the perpendicular bisector of AB. 5. Draw AZ, the angular bisector of angle A. 6. Mark point P as the intersection of AZ and XY. 7. Measure CP. Hence, the length of CP = 5.2 cm. [5] SECTION B Question 4 (i) A man busy 250, ten-rupee shares each at `12.50. If the rate of dividend is 7%, [3] find the: (a) dividend her receives annually. (b) percentage return on his investment. Ans. (a) Nominal Value of 1 share = `10 Market Value of 1 share = `12.50 Number of shares purchased = 250 Nominal Value of 250 shares = 250 10 = `2500 Rate of dividend = 7% Dividend received = 7% of 2500 = = `175 Hence, annual dividend = `175. (b) Amount Invested = No. of shares Market Value = 250 12.50 = `3125 Return percentage = = = Dividend received Investment 100 = 5.6 Hence, return percentage = 5.6% (ii) Solve the following, write the solution set and represent it on the real number line. 5 21 < 5 7 3 6 3 7 + , Ans. Given, inequation: < Solving L.H.S. of the inequation: < < < < 15 < 15 < (1) Solving R.H.S of the inequation: + + + [3] + + (2) From equation (1) and (2) Solution set = { < , } Hence, solution set = { < , } (iii) A manufacturing company prepares spherical ball bearings, each of radius 7 mm and mass 4 gm. These ball bearings are packed into boxes. Each box can have a maximum of 2156 cm3 of ball bearings. Find the: (a) maximum number of ball bearings that each box can have. (b) mass of each box of ball bearings in kg. (Use = Ans. (a) Given, 22 7 ) Radius of ball bearings = 7mm Volume of box = 2156 cm3 = 2156 103 mm3 Number of ball bearings that each box can have (N) = Volume of box Volume of each ball Substituting values we get: = = = = = = 1500 Hence, maximum no. of ball bearings in a box = 1500. (b) Mass of each box = No. of balls Mass of each ball = 1500 4gm = 6000 gm [4] = = 6 kg. Hence, mass of each box = 6 kg. Question 5 (i) Shown below is a table illustrating the monthly income distribution of a company [3] with 100 employees. Monthly Income 0-4 4-8 8-12 12-16 16-20 20-24 55 15 06 08 12 4 (in `10,000) Number of Employees Using step-deviation method, find the mean monthly income of an employee. Ans. In the given table, Class size (i) = 4. Monthly Income No. of employees (f) Class mark d =x A u = d/i fu 0-4 55 2 -4 -1 -55 4-8 15 A=6 0 0 0 8-12 06 10 4 1 06 12-16 08 14 8 2 16 16-20 12 18 12 3 36 20-24 4 22 16 4 16 Total f = 100 fu = 19 By formula, Mean = A + =6+ =6+ = 6 + 0.76 = 6.76 Hence, mean = 6.76 (ii) The following bill shows the GST rates and the marked price of articles: BILL: COMPUTERS Articles Marked price Rate of GST Graphic Card Rs. 15500.00 18% Laptop adapter Rs. 1900.00 28% Find the total amount to be paid for the above bill. [3] Ans. Billed amount of Graphic Card = + = Billed amount of Laptop adapter = + = Total Bill = 18290 + 2432 = 20722 (iii) In the given diagram, ABCD is a cyclic quadrilateral and PQ is a tangent [3] to the smaller circle at E. Given AEP = 70 , BOC = 110 . Find: (a) ECB (b) BEC (c) BFC (d) DAB Ans. (a) ECB = BEP = 70 (Angle in alternate segment) (b) BEC = BOC = = ( at the circumference is half at centre) (c) BFC, BFC + BEC = 180 (Opp. of cyclic quadrilateral) BFC = 180 55 = 125 (d) DAB = ECB = 70 (Ext. of a cyclic quadrilateral) Question 6 (i) In the given figure (drawn no to scale) chords AD and BC intersect at P, where AB = 9 cm, PB 3 cm and PD = 2 cm. (a) Prove that APB CPD (b) Find the length of CD (c) Find area APB : area CPD Ans. (a) In APB and CPD, APB = CPD (Vertically opposite angles are equal) BAP = DCP (Angles in same segment are equal) APB CPD (By A.A axiom) Hence, proved that APB CPD. [3] (b) We know that, Corresponding sides of similar triangles are proportional. = = CD = CD = 6 cm Hence, CD = 6 cm. (c) We know that, Ratio of area of similar triangles is equal to the ratio of square of the corresponding sides. = = = =9:4 Hence, area of APB : area CPD = 9 : 4. (ii) A box contains some green, yellow and white tennis balls. The probability if selecting a green ball is balls, then find: 1 4 1 and yellow ball is . If the box contains 10 white 3 (a) total number of balls in the box. (b) probability of selecting a white ball. Ans. (a) Let total no of balls = . . P (W) = P (W) = = (i) P (G) + P (Y) + P (W) = P (E) = 1 + = + = = Total no. of Balls = 24 (b) P (W) = P (W) = [3] (iii) A (a, b), B (-4, 3) and C (8, -6) are the vertices of a ABC. Point D is on the BC such that BD : DC is 2 : 1 and M (6, 0) is midpoint of AD. Find: (a) coordinates of point D (b) coordinates of point A (c) equation of a line passing through M and parallel to line BC. Ans. (a) Given BD : DC = 2 : 1 Let coordinates of D be (x, y) By section formula, + + , + ) + (x, y) = ( Substituting values, we get: ( , ) = ( + ( ) + ( , ) = ( + ( , ) = ( , + , , ) + ) ) ( , ) = ( , ) Hence, coordinates of D = (4, -3) (b) By mid-point formula, Mid-point = ( + + , Given, ) M (6, 0) is the mid-point of AD (6, 0) = ( + +( ) , ) + (6, 0) = ( + = 6 and , ) =0 a + 4 = 12 and b 3 = 0 a = 12 4 = 8 and b = 3. A = (a, b) = (8, 3) Hence, coordinates of A = (8, 3) (c) By formula, Slope of line = Slope of line BC = ( ) = = [4] We know that, Slope of parallel lines are equal Slope of line parallel to BC = By point-slope form, Equation of line: y y1 = m(x x1) Equation of line parallel to BC and passing through M is: y 0= y= (x 6) (x 6) 4y = -3 (x 6) 4y = -3 + 18 3x + 4y = 18 Hence, equation of the required line is 3x + 4y = 18 Question 7 (i) The table given below shows the runs scored by a cricket team during the overs of a match. Overs Runs Scored 20-30 37 30-40 45 40-50 40 50-60 60 60-70 51 70-80 35 (a) Draw a histogram representing the above distribution. (b) Estimate the modal runs scored. Ans. Steps : 1. Take 2 cm along x-axis = 10 overs and 1 cm along y-axis = 10 runs. 2. Since, the scale on x-axis starts at 20, a break (zig-zag curve) is shown near the origin along x-axis to indicate that the graph is drawn to scale beginning at 20 and not at origin itself. 3. Construct rectangles corresponding to the given data. 4. In highest rectangle, draw two st. lines KN and LI from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let Z be the point of intersection of KN and LI. 5. Through Z, draw a vertical line to meet the x-axis at A. The abscissa of the point A represents 57. Hence, mode = 57. [3] (ii) Suresh has a recurring deposit account in a bank. He deposits `2000 per month [3] and the bank pays interest at the rate of 8% per annum. If he gets `1040 as interest at the time of maturity, find the in years total time for which the account was held. Ans. Given that, Suresh deposits amount, = `2000 per month Rate of interest, r = 8% Interest, = `1040 Let the duration of deposit be n months We know that, = 1040 = ( + ) ( + ) n (n + 1) = 156 12 13 = 156 So, n = 12 So, total time of deposit is 12 months or 1year. (iii) Factorize: sin3 + cos3 [4] Hence, prove the following identity: sin3 + cos3 + + = 1 Ans. To factorize: sin3 + cos3 We have, sin3 + cos3 [Using a3 + b3 = (a+b) (a2 ab + b2)] (sin + cos ) (sin2 - sin cos + cos2 ) (sin + cos ) (1 sin cos ) [As To prove: + + Taking LHS, Using (i), = ... (i) sin2 + cos2 = 1] + . = + + + . ( + )( . ) + + . = 1 sin .cos + sin .cos LHS = RHS Hence proved. Question 8 (i) What number must be added to each of the numbers 4, 6, 8, 11 in order to get the four numbers in proportion ? Ans. Let x be added to get the four numbers in proportion (4 + x) : (6 + x) : : (8 + x) : (11 + x) (4 + x)(11 + x) = (6 + x) (8 + x) x2 + 15x + 44 = x2 + 14x + 48 15x 14x = 48 44 x=4 4 should be added to get the four numbers in proportion. [3] (ii) A man bought `200 shares of a company at 25% premium. If he received a return [3] of 5% on his investment. Find the: (a) market value (b) dividend percent declared (c) number of shares purchased, if annual dividend is `1000. Ans. For one share: Face value = `200 Premium = 25% of Face value = = `50 (a) By formula, M.V. = Face value + Premium = `200 + `50 = `250. Hence, market value = `250. (b) Given, Return = 5% Return on 1 share = = `12.50 By formula, Dividend earned = No. of shares rate of dividend F.V. of 1 share Let rate of dividend be r% Substituting values we get: 12.50 = = . r = 6.25% Hence, dividend percent = 6.25%. (c) By formula, Annual dividend = Number of shares Dividend% Face value of 1 share 1000 = Number of shares . 1000 = Number of shares 12.5 Number of shares = . = 80. Hence, number of shares purchased = 80. (iii) The roots of the equation (q-r)x2 + (r-p)x + (p q) 0 are equal. Prove that: 2q = p + r, that is, p, q and r in A.P. Ans. Given, The roots of the equation (q-r)x2 + (r-p)x + (p q) 0 are equal Discriminant (D) = 0 b2 4ac = 0 (r - p)2 4 (q r) (p q) = 0 r2 + p2 2pr 4 (qp q2 rp + qr) = 0 [4] r2 + p2 2pr 4qp 4q2 4rp + 4qr = 0 r2 + p2 + 2pr 4qp 4qr + 4q2 = 0 (p + r)2 4q (p + r) + 4q2 = 0 Let p + r = y y2 4qy + 4q2 = 0 (y = 2q)2 = 0 y 2q = 0 y = 2q p + r = 2q. Hence, proved that p + r = 2q. Question 9 (i) In the given diagram, AB is a vertical tower 100 m away from the foot of a 30 storied building CD. The angles of depression from the point C and E, (E being the mid-point of CD), are 35 and 14 respectively. (Use mathematical table for the required values rounded off correct to two places of decimals only) Find the height of the: (a) tower AB (b) building CD Ans. From figure, Let height of the building CD be H meters and height of tower AB be h meters. Given, BD = 100 m E is the midpoint of CD CE = = From figure, CAP = XCA = 35 [Alternate angles are equal] EAP = YEA = 14 [Alternate angles are equal] [5] From figure, PD = AB = h AP = BD = 100 m In ACP, tan 35 = Perpendicular tan 35 = Base tan 35 = 0.70 = 0.70 100 = H h H h = 70 (1) In AEP, tan 14 = Perpendicular tan 14 = tan 14 = 0.25 = Base = 0.25 100 = 25 (2) Substituting equation (2) from (1), we get: ( ) = + = = = H = 45 2 = 90 m. H h = 70 90 h = 70 h = 90 - 70 h = 20 m. Hence, height of the tower AB = 20 m and height of the building CD = 90 m. (ii) The line segment joining A (2, -3) and B (-3, 2) is intercepted by the x-axis at the point M and the y axis at the point N. PQ is perpendicular to AB produced at R and meets the y-axis at a distance of 6 units from the origin O as shown in the diagram at S. Find the: [5] (a) coordinates of M and N (b) coordinates of S (c) slope of AB (d) equation of line PQ Ans. (a) From figure, Coordinates of M = (-1, 0) and coordinates of N = (0, -1) (b) From figure, Coordinates of S (0, 6) (c) By formula, Slope = Slope of AB = ( ) = + = = Hence, slope of AB = -1 (d) We know that, Product of slope of perpendicular lines = -1 Slope of AB Slope of PQ = -1 -1 Slope of PQ = -1 Slope of PQ = =1 Since, PQ passes through point S. By point-slope form, Equation of line: y y1 = m(x x1) y 6 = 1 (x 0) y 6=x y=x+6 Hence, equation of line PQ is y = x + 6. Question 10 (i) In the given diagram, an isosceles ABC is inscribed in a circle in a circle with centre O. PQ is a tangent to the circle at C. OM is perpendicular to chord AC and COM = 65 . Find: (a) ABC (b) BAC (c) BCQ Ans. (a) From figure, AOC = AOM + COM = 65 + 65 130 . [3] We know that, Angle at the center is twice the angle formed by the same arc at any other point of the circle. AOC = 2 ABC = = = Hence, ABC = 65 (b) In ABC, AB = AC (Given) ACB = ABC = 65 (Opposite angles of equal sides are equal) By angle sum property of triangle, ACB + ABC + BAC = 180 65 + 65 + BAC = 180 BAC = 180 65 65 = 50 . Hence, BAC = 50 . (c) We know that, The angle formed between the tangent and the chord through the point of contact of the tangent is equal to the angle formed by the chord in the alternate segment. BCQ = BAC = 50 . Hence BCQ = 50 . 1 2 (ii) If = [ 4 1 1 0 3 1 2 ], = [ ], = [ ] = [ ] 1 5 0 1 4 2 4 [3] Find A (B + C) 14 Ans. = [ ], = [ ], = [ ] A(B + C) 14 = [ ] ]+ [ ]} [ =[ ] {[ =[ ][ ] [ ] =[ + + + ] [ + =[ ] [ =[ ] ] ] ] (iii) Use a ruler and compass to answer this question. Construct ABC = 90 , where AB = 6 cm, BC = 8 cm. (a) Construct the locus of points equidistant from B and C. (b) Construct the locus of points equidistant from A and B. [4] (c) Mark the point which satisfies both the conditions (a) and (b) as O. Construct the locus of points keeping a fixed distance OA from the fixed point O. (d) Construct the locus of points which are equidistant from BA and BC. Ans. Steps of construction: 1. Draw a line segment BC = 8 cm 2. Construct ABC = 90 , such that AB = 6 cm. 3. Draw XY, the perpendicular bisector of BC. 4. Draw PQ, the perpendicular bisector of AB. 5. Mark point O, the intersection of segment XY and PQ. 6. Draw BZ, the angle bisector of AB and BC. We know that, Locus of points equidistant from two points is the perpendicular bisector of the line joining the two points segment. (a) Locus of points equidistant from B and C is XY. (b) Locus of points equidistant from A and B is PQ. We know that, Locus of points equidistant from two sides is the angular bisector of angle between them. (d) Locus of points which are equidistant from BA and BC is BZ.

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