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ICSE Class X Sample / Model Paper 2025 : Mathematics (Solved with Model Answers)

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ICSE EXAMINATION PAPER - 2025 MATHEMATICS Class-10th (Solved) Maximum Marks: 80 Time Allotted: Three Hours Instructions to Candidates: 1. Answers to this Paper must be written on the paper provided separately. 2. You will not be allowed to write during first 15 minutes. 3. This time is to be spent in reading the question paper. 4. The time given at the head of this Paper is the time allowed for writing the answers. 5. Attempt all questions from Section A and any four questions from Section B. 6. All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer. 7. Omission of essential working will result in loss of marks. 8. The intended marks for questions or parts of questions are given in brackets [] 9. Mathematical tables and graph papers are to be provided by the school. SECTION-A (40 MARKS) (Attempt all questions from this Section.) Question 1 [15] Choose the correct answers to the questions from the given options. (Do not copy the questions, write the correct answers only.) (i) The given quadratic equation 3x2 + 7x + 2 = 0 has (a) two equal real roots. (b) two distinct real roots. (c) more than two real roots. (d) no real roots. (ii) Mr. Anuj deposits ` 500 per month for 18 months in a recurring deposit account at a certain rate. If he earns ` 570 as interest at the time of maturity, then his matured amount is (a) ` (500 18 + 570) (b) ` (500 19 + 570) (c) ` (500 18 19 + 570) (d) ` (500 9 19 + 570) (iii) Which of the following cannot be the probability of any event? 5 (a) (b) 0.25 4 1 (c) (d) 67% 33 (iv) The equation of the line passing through origin and parallel to the line 3x + 4y + 7 = 0 is (a) 3x + 4y + 5 = 0 (b) 4x - 3y - 5 = 0 (c) 4x - 3y = 0 (d) 3x + 4y = 0 0 1 2 (v) If A = , then A is equal to 1 0 1 1 (a) 0 0 0 0 (b) 1 1 0 1 1 0 (c) (d) 1 0 0 1 (vi) In the given diagram, chords AC and BC are equal. If angle ACD = 120 , then angle AEC is A E 120 60 B C D (a) 30 (b) 60 (c) 90 (d) 120 (vii) The factor common to the two polynomials x2 - 4 and x3 - x2 - 4x + 4 is (a) (x + 1) (b) (x - 1) (c) (x + 2) (d) (x - 2) (viii) A man invested in a company paying 12% dividend on its share. If the percentage return on his investment is 10%, then the shares are (a) at par (b) below par (c) above par (d) cannot be determined (ix) Statement 1: The point which is equidistant from three non-collinear points D, E and F is the circumcentre of the DEF Statement 2: The incentre of a triangle is the point where the bisector of the angles intersects. (a) Both the statements are true. (b) Both the statements are false. (c) Statement 1 is true and Statement 2 is false. (d) Statement 1 is false and Statement 2 is true. (x) Assertion(A): If sin2 A + sin A = 1 then cos4 A + cos2 A = 1 Solved Paper - 2025 Reason(R): 1 - sin2 A = cos2 A (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). (xi) In the given diagram ABC ~ EFG. If ABC= EFG = 60 , then the length of the side FG is F 60 75 cm B E E 25 A 60 3 cm 5 cm cm 25 cm 15 cm D cm 3 H Sum of all the observations Total number of observations (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). Question 2 (i) Solve the following quadratic equation 2x2 - 5x - 4 = 0 Give your answer correct to three significant figures. (Use mathematical tables for this question) [4] (ii) Mrs. Rao deposited ` 50 per month in a recurring deposit account for a period of 3 years. She received ` 10,110 at the time of maturity. Find: [4] (a) the rate of interest, (b) how much more interest Mrs. Rao will receive if she had deposited ` 250 more per month at the same rate of interest and for the same time. (iii) 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. [4] A Reason(R): Mean = 20 cm G 2 C D (a) 15 cm (b) 20 cm (c) 25 cm (d) 30 cm (xii) If the volume of two spheres is in the ratio 27 : 64 then the ratio of their radii is (a) 3 : 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9 (xiii) The marked price of an article is ` 1375. If the CGST is charged at a rate of 4%, then the price of the article including GST is (a) ` 55 (b) ` 110 (c) ` 1430 (d) ` 1485 x (xiv) The solution set for 0 < < 2, x z is 3 (a) {- 5, - 4, - 3, - 2, - 1} (b) {- 6, - 5, - 4, - 3, - 2, - 1} (c) {- 5, - 4, - 3, - 2, - 1, 0} (d) {- 6, - 5, - 4, - 3, - 2, - 1, 0} (xv) Assertion(A): The mean of first 9 natural numbers is 4.5. B C (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. Question 3 (i) Use ruler and compass for the following construction. Construct a ABC where AB = 6 cm, AC = 4.5 cm and BAC = 120 . Construct a circle circumscribing the ABC. Measure and write down the length of the radius of the circle. [4] 5 1 1 2 2 1 (ii) If A = = , B and C = 7 4 3 4 4 2 Find: (a) A + C (b) B(A + C) (c) 5B (d) B(A + C) - 5B [4] 3 Oswaal ICSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-10 (iii) In the given graph ABCD is a parallelogram. Y D [5] Scale: X-axis, 2 cm = 1 unit Y-axis, 2 cm = 1 unit 3 A 2 1 (iii) A solid wooden capsule is shown in Figure 1. The capsule is formed of a cylindrical block and two hemispheres. Find the sum of total surface area of the three parts as shown in Figure 2. Given, the radius of the capsule is 3.5 cm and the length of the cylindrical block is 14 cm. [4] 22 (Use = ) 7 P X 4 3 2 O 1 X 1 2 3 1 C 2 B 3 Y Using the graph, answer the following: (a) write down the coordinates of A, B, C and D. (b) calculate the coordinates of P , the point of intersection of the diagonals AC and BD. (c) find the slope of sides CB and DA and verify that they represent parallel lines. (d) find the equation of the diagonal AC. Figure 1 3.5 cm Question 4 (i) Solve the following inequation, write the solution set and represent it on the real number line. [3] 5 3x 4x + 10 + 11; x R 2x < 5 3 5 (ii) The first term of an Arithmetic Progression (A.P.) is 5, the last term is 50 and their sum is 440. Find: [3] (a) the number of terms (b) common difference (iii) Prove that: [4] ( cot A + tan A 1)( sin A + cos A ) = sec A. cosec A sin 3 A + cos3 A Question 5 (i) Using properties of proportion, find the value of x : [3] 6 x 2 + 3x 5 9 x 2 + 2 x + 5 = ;x 0 3x 5 2x + 5 (ii) It is given that (x - 2) is a factor of polynomial 2x3 - 7x2 + kx - 2. [3] Find: (a) the value of k . (b) hence, factorise the resulting polynomial completely. 14 cm SECTION-B (40 MARKS) (Attempt any four questions from this Section.) 3.5 cm 3.5 cm Figure 2 Question 6 (i) Use a graph paper for this question taking 2 cm = 1 unit along both axes. [5] (a) Plot A(1, 3), B(1, 2) and C(3, 0). (b) Reflect A and B on the x-axis and name their images as E and D respectively. Write down their coordinates. (c) Reflect A and B through the origin and name their images as F and G respectively. (d) Reflect A, B and C on the y-axis and name their images as J, I and H respectively. (e) Join all the points A, B, C, D, E, F, G, H, I and J in order and name the closed figure so formed. (ii) 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. [5] (Use mathematical table for the required values rounded off correct to two places of decimals only) Find the height of the: 4 Solved Paper - 2025 C 35 (iii) Refer to the given bill. [4] A customer paid ` 2000 (rounded off to the nearest ` 10) to clear the bill. Note: 5% discount is applicable on an article if 10 or more such articles are purchased. BILL P D 100 m (a) tower AB (b) building CD Question 7 (i) Use a graph paper for this question. [3] (Take 2 cm = 10 Marks along one axis and 2 cm = 10 students along another axis). Draw a Histogram for the following distribution which gives the marks obtained by 164 students in a particular class and hence find the Mode. 30-40 40-50 Number of Students 10 26 50-60 40 60-70 70-80 54 34 (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 and y-axis respectively. Line RS is perpendicular to PQ and passes through the origin. Find the: [3] (a) coordinates of P and Q (b) equation of line RS Y R 5 units O G.S.T. A 190 06 12% B 50 12 18% P Check whether the total amount paid by the customer is correct or not. Justify your answer with necessary working. Question 8 (i) A man bought ` 200 shares of a company at 25% premium. If he received a return of 5% on his investment. Find the: [3] (a) market value (b) dividend percent declared (c) number of shares purchased, if annual dividend is ` 1000. (ii) For the given frequency distribution, find the: [3] (a) mean, to the nearest whole number (b) median x 10 11 12 13 14 15 16 y 3 2 2 6 3 5 3 (iii) Mr. and Mrs. Das were travelling by car from Delhi to Kasauli for a holiday. Distance between Delhi and Kasauli is approximately 350 km (via NH 152D). Due to heavy rain they had to slow down. The average speed of the car was reduced by 20 km/h and time of the journey increased by 2 hours. Find: [4] (a) the original speed of the car. (b) with the reduced speed, the number of hours they took to reach their destination. Question 9 (i) A hollow sphere of external diameter 10 cm and internal diameter 6 cm is melted and made into a solid right circular cone of height 8 cm. Find the radius of the cone so formed. [3] 22 [Use = ] 7 X 10 cm 3 units X Quantity E B Marks M.P. (`) 14 A Article 6 cm S Q Y 5 Oswaal ICSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-10 (a) the value of n (b) hence find the sum of the n terms of the G.P. (ii) In the given diagram O is the centre of the circle. Chord SR produced meets the tangent XTP at P. [3] 8 cm R S P O (ii) Ms. Sushmita went to a fair and participated in a game. The game consisted of a box having number cards with numbers from 01 to 30. The three prizes were as per the given table: [3] Prize Number on the card drawn at random is a Wall Clock perfect square Water Bottle even number which is also a multiple of 3 X (a) Prove that PTR ~ PST (b) Prove that PT2 = PR PS (c) If PR = 4 cm and PS = 16 cm, find the length of the tangent PT. (iii) The given graph represents the monthly salaries (in ` ) of workers of a factory. [4] Scale: along X-axis, 2 cm = `2000 along Y-axis, 2 cm = 10 workers Y prime number 80 Find the probability of winning a: (a) Wall Clock (b) Water Bottle (c) Purse (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 = 66 . Find: [4] (a) BOX (b) CYX (c) ZYX (d) OXY (12000,75) 70 (10000,70) 60 Number of workers Purse T (8000,55) 50 40 (6000,35) 30 20 (4000,20) Z Y 10 O 0 C 66 A 32 X (2000,10) B Question 10 (i) If 1701 is the nth term of the Geometric Progression (G.P.) 7, 21, 63 ......., find: [3] 2 4 6 8 10 12 Monthly salary (in `) ( 103) X Using graph answer the following: (a) the total number of workers. (b) the median class. (c) the lower-quartile class. (d) number of workers having monthly salary more than or equal to ` 6,000 but less than ` 10,000. Solved Paper - 2025 6 ANSWERS SECTION-A (65 MARKS) (Attempt all questions from this Section.) Answer 1 (i) Option (d) is correct. Explanation: Given: 3x 2 7 x 2 0 On comparing with ax +bx+c=0 we get, a = 3 , b= 7 and c = 2 Using the discriminant formula D = b2 4ac D = 7 2 4 3 2 D = 7 - 24 D = - 17 D < 0 So, the given quadratic equation has no real roots (ii) Option (a) is correct. Explanation: Given P = ` 500 per month n = 18 I = 570 M.V. =? Using the formula M.V. = P n + I = ` (500 18 + 570) (iii) Option(a) is correct. Explanation: Probability of any event is always between 0 to 1 5 As, = 1.25 which is greater than 1. 4 So, option (a) cannot be the probability of any event. (iv) Option(d) is correct Explanation: Given line 3x+4y+7=0 4y = -3x-7 -3 7 x y = 4 4 On comparing with y = mx + c -3 m = 4 For the parallel line slope is same. Equation of line passing through the origin with -3 slope . 4 -3 y-0 = x -0 3x+4y = 0 4 (v) Option (c) is correct. 0 1 Explanation: Since, A = 1 0 0 1 0 1 0 0 1 1 0 1 1 0 A A = 1 0 1 0 1 0 0 1 1 1 0 0 0 1 0 0 = 0 0 1 0 1 0 = 0 1 (vi) Option (d) is correct Explanation: Given AC = BC then CAB = ABC (angles opposite to equal sides are equal) CAB + ABC = ACD (exterior angle property) So, 2 CAB = ACD Thus, 2 CAB = 120 Therefore, CAB = 60 AEC + ABC =180 (As, opposite angles of a cyclic quadrilateral is 180 ) AEC + 60 = 180 AEC = 120 (vii) Option (c & d) is correct. Explanation: P(x) = x2 - 4 Q(x) = x 3 -x2 -4x+4 Factorise P(x) = (x+2)(x-2) Q(x) = x3 -x2 -4x+4 = x2 (x-1)-4(x-1) = (x2 -4) (x-1) = (x+2) (x-1) (x-2) Common Factor = (x+2)(x-2) (viii) Option (c) is correct Explanation: Given dividend = 12% of face value Let Face value be ` 100 then dividend ` 12 Return on investment = 10% Let market price be x Using the yield Formula Dividend Return = 100 M.V 12 10 100 x x = ` 120 As, Market Price > Face Value So, shares are above per. (ix) Option (a) is correct. (x) Option (c) is correct. Explanation: Assertion : sin2 A + sin A = 1 1 cos2 A +sin A = 1 sin A = cos2 A ....(i) Now, sin2A + cos2 A = 1 (cos2 A)2 + cos2 A = (sin A)2 + cos2 A = sin2 A + cos2 A = 1 (using the given reason) Thus, cos4 A + cos2 A = 1 Hence, both Assertion and reason are true and reason is the correct explanation of assertion. (xi) Option (a) is correct. Explanation: Given ABC ~ EFG , EF = 75 cm, AB = 15 cm, BC = 3 cm B = F = 60 Now, using the Basic proportionality theorem. 7 Oswaal ICSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-10 AB BC = EF FG 15 3 = 75 FG 3 75 FG= =15cm 15 (xii) Option (a) is correct. V 27 Explanation: Given 1 = V2 64 4 3 r 3 1 = 27 4 r 3 64 3 2 r1 3 = r2 4 ( xiii) Option (d) is correct. Explanation: Marked Price = ` 1375 CGST = 4% Since GST is divided into CGST and SGST , SGST is also 4%. 8 1375 GST = 100 8 1375 110 = 100 Final Price = ` 1375 + ` 110 = ` 1485 (xiv) Option (a) is correct. x 2 Explanation: Given 0 3 0 < -x < 6 0 > x > -6 -6 < x < 0 The solution set is {-5 , -4, -3, -2, -1} Principle (P) = ` 250 Period n = 3 12 = 36 month M.V. = P n + I Now 10,110 = 250 36 + I I = 10,110 - 9,000 I = 1,110 I = P n (n + 1) R 2 12 100 250 36 37 R I = 2,400 1, 110 2 , 400 250 36 37 R = R = 8% (b) Principle = 300, R = 8%, n = 36 I = P n( n 1) R 12 2 100 I = 300 36 37 8 2400 I = ` 1332 More interest received by Mrs. Rao = 1332 - 1110 = ` 222 (iii) Given: ABC = 90 , DEA = 90 AB = 20 cm , AC = 25 cm, DE = 3 cm (a) In AED and ABC AED = ABC = 90 EAD = CAB = common angle AED ~ ABC (AA Similarity) A (xv) Option (b) is correct. E Explanation: Assertion D cm 3 20 cm cm 25 1+ 2+3+ 4 + 5+6+7 +8+9 Mean = 9 45 =5 = 9 Assertion is false. and Reason is true. Answer 2 (i) Since , 2x2 -5x -4 = 0 On comparing with ax2 + bx+c = 0 , a = 2, b = -5, c = -4 b b 2 4 ac Now, using Formula x 2a 2 5 5 4 2 4 x= 2 2 x = 5 25 32 = 5 57 4 4 5 7.549 x= 4 x = 3.137 , -0.637 (ii) (a) Given Maturity Value (M.V) = ` 10,110 B C (b) Now, In right angled triangle ABC, B = 90 By using pythagoras theorem AC2 = AB2 + BC2 252 = 202 + BC2 625 = 400 + BC2 BC2 = 625 - 400 BC = 225 BC = 15 cm AC2 = 252 - 202 BC2 = 625 - 400 Solved Paper - 2025 BC = 225 = 15 By using BPT theorem AB BC CA = = AE ED DA (using pythagoras theorem) 20 15 AB BC = = AE 3 AE ED AE = 4 cm BC CA 15 25 = = ED DA 3 AD AD = 5 cm Areaof BCED (c) Scale factor of map k2 = Actual area of ground ar ABC AED = 576 m 2 1 1 20 15 3 4 2 2 = 576 m 2 (150 6 ) cm 2 = 576 m 2 = Thus, 144 1 = 576 100 100 40000 1 40000 k = = 1 200 Answer 3 Step to construction: 1. Draw line segment AB = 6 cm. 2. At point A, construct an angle BAX of 120 3. With A as centre and radius 4.5cm draw an arc cutting AX at point C. 4. Join B and C, thus ABC is a required triangle. 5. Draw perpendicular bisectors of AB & AC which intersest at O. 6. With O as centre and OC as radius, draw a circle touching points A,B and C of triangle ABC. 7. Thus, the length of radius is approx 5.2 cm. 1 2 2 1 5 1 (ii) Given: A = 3 4 , B and C 7 4 4 2 1 2 5 1 (a) A + C 3 4 7 4 1 5 1 2 = 3 7 4 4 3 4 = 0 10 2 1 4 3 (b) B(A + C) = 4 2 10 0 2 4 1 10 2 3 1 0 4 4 2 10 4 3 2 0 8 10 6 2 6 16 20 12 4 12 2 1 10 5 (c) 5B 5 4 2 20 10 2 6 10 5 (d) B( A C ) 5 B 4 12 20 10 (i) X C O 4.5 cm A 6 cm B 8 2 10 6 5 4 20 12 10 8 1 16 2 (iii) (a) Coordinate A = (3, 3) B = (0, -2) C = (-4, -2) D = (-1, 3) 9 Oswaal ICSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-10 Y Scale: X-axis, 2 cm = 1 unit Y-axis, 2 cm = 1 unit D ( 1,3) 3 A (3,3) 2 1 P X 4 3 2 O 1 1 2 3 4 X 1 C ( 4, 2) 2 B (0, 2) 3 Y (b) Coordinate of P = Mid point of BD 1 0 3 2 , 2 2 3x 4x 10 11 5 5 3x 50 4 x 55 5 x ....(ii) From (i) and (ii) we get, 1 5 x 8 3 y 2 y1 3 3 0 0 x 2 x1 3 ( 1) 4 Slope of CB = DA then they are parallel line y y1 (d) Eq of AC =y y1 2 x x1 x 2 x1 When x1 = -4 3 ( 2 ) {x ( 4 )} y ( 2 ) 3 ( 4 ) 5 y 2 ( x 4 ) 7 7y + 14 = 5x +20 7y 5x - 6 = 0 5x 7y + 6 = 0 1 Solution set = 5 x < 8 , x R 3 1 1 , 2 2 y y1 2 ( 2 ) 0 = (c) Slope of CB = 2 = =0 x 2 x1 0 ( 4 ) 4 Slope of DA = SECTION-B Answer 4 (i) Since, 2 x 5 3x 10 4 x 11; x R 3 5 5 5 3x 2x 10 3 5 30 x 25 9 x 150 21x 175 175 x< 21 1 x < 8 ....(i) 3 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 (ii) (a) Given a = 5 last term = 50 Sum Sn = 440 n Now Sn = (a+l) 2 n 440 = (5+50) 2 440 2 16 n= 55 (b) l = a + (n - 1)d 50 = 5 + (16-1)d 45 =d 15 d=3 (iii) Prove that (cot A tan A 1)(sin A cos A ) sec A.cosec A sin 3 A cos3 A cos A sin A sin A cos A 1 sin A cos A L.H.S. sin 2 A cos2 A sin A. cos A sin A cos A Solved Paper - 2025 - + 0 -3x2 + 7x - 2 -3x2 + 6x + 0 x - 2 x - 2 - + 0 Alternate Method: Factorisation of the given polynomial is, 2x3 7x2 + 7x 2 = 2x3 4x2 3x2 + 6x + x 2 = 2x2(x 2) 3x(x 2) + 1 (x 2) = (x 2) (2x2 3x + 1) = (x 2) (2x2 2x x + 1) = (x 2) [2x(x 1) 1(x 1)] = (x 2) (x 1) (2x 1) (iii) Given: Height of Cylinder = 14 cm Radius of Cylinder = 3.5 cm Radius of hemisphere = 3.5 cm Total Surface of Area of three Part = total Surface area of Cylinder + 2 total Surface area of hemisphere = 2 r (r+h)+ 2 3 r2 = 2 r(r+h+3r) = 2 r(4r + h) 22 =2 3.5 (4 3.5 + 14) 7 cos2 A sin 2 A sin A.cos A sin A cos A 2 2 sin A cos A sin A.cos A 1 = sin A.cos A = sec A.cosec A = R.H.S Answer 5 2 2 (i) 6 x 3x 5 9 x 2 x 5 ; x 0 3x 5 2x 5 Using componendo and dividendo 6 x 2 6 x 10 9 x 2 4 x 10 6x 2 9x 2 2 2 6 x 6 x 10 9 x 4 x 10 2 3 2 18x +18x 30 =18x2+8x+20 10x = 50 x = 5 (ii) (a) P(x) = 2x3 - 7x2 + kx - 2 Since (x-2) is a factor ; Put x = 2 in the polynomial P(2) = 2(2)3 - 7(2)2 + k(2)-2 0 = 16 28 + 2k - 2 k=7 (b) The Polynomial is 2x3 -7x2 + 7x -2 The factor of polynomial = (x - 2) (2x-1)(x-1) By Long Division method 2x2 - 3x + 1 x-2 2x3 - 7x2 + 7x - 2 2x3 - 4x2 =2 22 3.5 28 7 = 616 cm2 Answer 6 Y (i)(a) J ( 1, 3) ( 1, 2) I 3 A (1, 3) 2 B (1, 2) 1 H( 3, 0) X 4 3 C (3, 0) 2 O 1 1 2 3 4 X 1 2 D (1, 2) 3 F ( 1, 3) E (1, 3) ( 1, 2) G 10 Y (b) Reflection of A and B on the x-axis (c) Reflection of A and B through the origin A(1,3) E(1,-3) A(1,3) F(-1,-3) B(1,2) D(1,-2) B(1,2) G(-1,-2) 11 Oswaal ICSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-10 (d) Reflection of A , B and C on the y-axis = 90 m AB = CD - CP = 90 - 70 = 20 m A(1,3) J(-1,3) B(1,2) I(-1,2) C(3,0) H(-3,0) (e) Joint the Point A B C D E F G H I J A C 35 The figure obtained is Decagon (ii) Given, BD = 100 m = AP As, angle of depression = angle of elevation. Thus, EAP =14 and CAP = 35 In right angle AEP, P = 90 EP tan14 = AP EP 0.25 = 100 25 = EP EP = 25 m A P D B 100 m In APC, P = 90 E 14 CP AP CP 0.70 = AP CP = 70 cm tan 35 = Now, CE = CP - EP = 70 - 25 = 45 m E is the mid-point of CD Therefore, CD = 2CE = 2 45 (given) Answer 7 To find mode. (a) Firstly identify the rectangle with highest frequency (Modal class) which is 60-70. (b) Join the top corners of the modal rectangle with immediate next corners of the adjacent rectangles. (c) Let the point where the joining lines cut each other as P. (d) Draw a perpendicular from A to x -axis. (e) The point A , where the perpendicular meet the X-axis will give the mode. Scale: On Y-axis: 2cm = 10 Marks On X-axis: 2cm = 10 students Y 50 40 30 20 10 X 0 30 40 50 60 70 80 mode = 64.5 cm (approx) (ii)(a) Coordinate of P = (5,0) Coordinate of Q = (0,-3) y y1 0 ( 3) (b) Slope of PQ = 2 x 2 x1 5 0 3 m1 = 5 Now, RS is perpendicular PQ and Slope of RS be m2 m1 m2 = -1 m2 = -5 3 Equation of line RS passing (0,0) -5 y- y1 = (x x1) 3 y-0= -5 (x - 0) 3 12 Solved Paper - 2025 3y = -5x 5x +3y = 0 Y R ) 5,0 X O 3 units X P( 5 units S Q (0, 3) Y (iii) Total cost of Article A = 190 6 = 1140 12 = 136.8 100 Total cost with GST = ` 1276.80 Total Cost of Article B = 50 12 = ` 600 5 5% discount is applicable = 600 = 30 100 Total Cost = 600 - 30 = ` 570 18 GST = 570 = ` 102.6 100 Total Cost with GST = ` 672.60 GST = 1140 (c) Total dividend = Number of Share Dividend per Share 1000 = N 12.5 1000 = 80 Number of Share (N) = 12.5 (ii) Total Bill = ` 1276.80 + 672.60 =` 1949.40 = ` 1950 The customer paid ` 2000 but actual bill ` 1950 So the amount paid by the customer is wrong as total amount is not ` 2000. Answer 8 (i) Given face value of each Share = ` 200 Premium = 25% Return = 5% Annual dividend received =` 1000 (a) Market value = Face value + Premium 25 =200 + 200 100 = 200 + 50 = ` 250 (r% N.V) 100 (b) Return% = M.V 5 = (r 200 100 ) 100 250 100 5 250 100 = r 200 r = 6.25%p.a (a) C.I x f x.f c.f 9.5-10.5 10 3 30 3 10.5-11.5 11 2 22 5 11.5-12.5 12 2 24 7 12.5-13.5 13 6 78 13 13.5-14.5 14 3 42 16 14.5-15.5 15 5 75 21 15.5-16.5 16 3 48 24 Total 24 319 Mean x x. f 319 13.29 13 f 24 24 2 7 1 M = 12.5+ 6 5 =12.5 + 6 = 12.5 + 0.83 =13.33 = 13 (b) Where l = 12.5 n =24 c.f. =7 f =6 h =1 n c. f . 2 h Median = l + f 13 Oswaal ICSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-10 (iii)(a) Let the original speed of car be x km/h The original time taken to travel 350 km d 350 t1 = = s x New time be t2 when speed reduce by 20 km/h = 1 58 2 ( at the center is double at remaining part) Z Average time speed t2 - t1 =2 Now, 350 x 350 x 7000 =2 x ( x 20 ) 350 ( x x + 20 ) = 2 x 2 20 x 3500 = x2 -20x x2 (70 - 50)x 3500 = 0 x2 -70x + 50x -3500 = 0 (x - 70)(x + 50) = 0 x = 70 x = -50 (Not Possible) As, speed cannot be negative. so original speed = 70km/h 350 (b) New time t2 = =7h 70 - 20 Answer 9 (i) Given Internal Radius of sphere = 3 cm External Radius of sphere= 5 cm Height af cone = 8 cm Now, Volume af Hollow Sphere = Volume of Cone 4 3 4 3 1 r1 r2 = r 2 h 3 3 3 4 1 r 53 33 = R 2 h 3 3 ( Y 350 t2 = x 2 350 350 = 2 x 2 x ) 4(125 - 27) = R2 8 98 = R2 2 R = 7 cm ( ii)(a) Perfect square between 1 to 30 P(E) = 1, 4 , 9 ,16 25 P(E) = 5 P(E) 5 1 P(Wall Clock) = = = P(S) 30 6 (b) Multiple of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 Even among them = 6, 12, 18, 24, 30 5 1 = P(Water Bottle) = 30 6 (c) Prime Number = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 10 1 = P(Purse) = 30 3 (iii) (a) In BOX OXB = 90 (point of contact) OXB + OBX + BOX = 180 90 +32 + BOX = 180 BOX = 58 1 (b) CYX = COX = 29 2 O C 66 32 X B (c) Since ZY = XY , then YXZ = XZY AXZ = ZYX (alternate segment theorem) ZYX = 66 In XYZ XYZ + YZX + ZXY = 180 66 + x + x = 180 2x = 114 x = 57 So YZX = ZXY = 57 (d) AXZ + ZXY + YXB = 180 (straight line property) 66 + 57 + YXB = 180 YXB = 180 - 123 YXB = 57 Now, OXY = OXB - YXB = 90 - 57 Thus, OXY = 33 A Answer 10 (i) Given Tn = 1701 a=7 T2 21 r= = = 3 T1 7 (a) Using the formula Tn = arn-1 1701 = 7(3)n-1 3n-1 = 243 3n-1 = 35 n = 6 (b) Using the formula Sn = S6 = S6 = ( a rn 1 ) r 1 ( 7 36 1 ) 3 1 7 ( 729 1) 2 7 728 S6 = 2 S6 = 2548 (ii) (a) In PTR and PST TPR = TPS (Common) Solved Paper - 2025 (b) Using basic proportionality theorem PT PR = PS PT PTR = PST ( Alternate Segment Theorem) PTR ~ PST (AA Similarity) R S 14 PT2 = PR PS Hence Proved. (c) Put PR = 4 and PS = 16 in PT2 = PR PS PT2 = 4 16 PT2 = 64 PT = 8 cm P O T X Y (iii) Scale: along X-axis, 2 cm = `2000 along Y-axis, 2 cm = 10 workers 80 70 Number of workers 60 50 40 30 20 1800 0 2 (a) Total Number of Workers = 75 n + 1 75 + 1 (b) Median = = = 38 workers 2 2 Median = 6400 Median Class = 6000 8000 (c) Lower Quartile Class n + 1 75 + 1 76 = = = 19 workers = 4 4 4 6400 10 4 6 8 10 12 Monthly salary (in `) ( 103) X Lower Quartile Class = 2000 4000 (d) Number of workers having monthly salary more than or equal 6000 = 35 workers Number of workers having salary less than 10000 = 70 So, Number of workers = 70 35 more than equal 6000 and less than 10000 = 35

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Additional Info : ICSE Class X Prelims 2024 :Public School (PPS), Hadapsar, Pune)
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