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ICSE Class X Sample / Model Paper 2025 : Mathematics

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ICSE EXAMINATION PAPER - 2024 MATHEMATICS Class-10th (Solved) Maximum Marks: 80 Time allowed: Two and half hours Answer to this Paper must be written on the paper provided separately. You will not be allowed to write during first 15 minutes. This time is to be spent in reading the question paper. The time given at the head of this Paper is the time allowed for writing the answers. Attempt all question from Section A and any four questions from Section B. All working, including rough work, must be clearly shown, and must be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. The intended marks for questions or parts questions are given in brackets [ ] 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) For an Intra-state sale, the CGST paid by a dealer to the Central government is ` 120. If the marked price of the article is ` 2000, the rate of GST is: (a) 6% (b) 10% (c) 12% (d) 16.67% (ii) What must be subtracted from the polynomial x3 + x2 2x +1, so that the result is exactly divisible by (x 3)? (a) 31 (b) 30 (c) 30 (d) 31 (iii) The roots of the quadratic equation px2 qx + r = 0 are real and equal if: (a) p2 = 4qr (b) q2 = 4pr 2 (c) q = 4pr (d) p2 > 4qr 4 x 2 2 and A2 = , (iv) If matrix A = 0 2 0 4 then the value of x is: (a) 2 (b) 4 (c) 8 (d) 10 (v) The median of the following observations arranged in ascending order is 64. Find the value of x: 27, 31, 46, 52, x, x + 4, 71, 79, 85, 90 (a) 60 (b) 61 (c) 62 (d) 66 (vi) Points A (x, y), B (3, 2) and C (4, 5) are collinear. The value of y in terms of x is: (a) 3x 11 (b) 11 3x (c) 3x 7 (d) 7 3x (vii) The given table shows the distance covered and the time taken by a train moving at a uniform speed along a straight track. Distance (in m) 60 90 y Time (in s) 2 x 5 The values of x and y are: (a) x = 4, y = 150 (b) x = 3, y = 100 (c) x = 4, y = 100 (d) x = 3, y = 150 (viii) The 7th term of the given Arithmetic Progression (A.P.): 1 1 1 , 1 , 2 ....is : a a a 1 (a) 6 a 1 (c) 8 a 1 (b) 7 a 1 7 (d) 7 a (ix) The sum invested to purchase 15 shares of a company of nominal value ` 75 available at a discount of 20% is: (a) ` 60 (b) ` 90 (c) ` 1350 (d) ` 900 (x) 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. (xi) Statement 1: sin2 q + cos2 q = 1 Statement 2: cosec2 q + cot2 q = 1 Which of the following is valid? (a) only 1 (b) only 2 (c) both 1 and 2 (d) neither 1 nor 2 Oswaal ICSE, MATHEMATICS, Class-X 2 7 cm 7 cm (xii) In the given diagram, PS and PT are the tangents to the circle. SQ || PT and SPT = 80 . The value of QST is: (a) 140 (b) 90 (c) 80 (d) 50 (xiii) Assertion (A): A die is thrown once and the prob2 ability of getting an even number is . 3 Reason (R): The sample space for even numbers on a die is {2, 4, 6}. (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 and false. (xiv) A rectangular sheet of paper of size 11 cm 7 cm is first rotated about the side 11 cm and then about the side 7 cm to form a cylinder, as shown in the diagram. The ratio of their curved surface areas is: 11 cm (b) 7:11 11p 7 p (c) 11:7 (d) : 7 11 (xv) In the given diagram, DABC ~ DPQR. If AD and PS are bisectors of BAC and QPR respectively then: 11 cm (a) 1:1 A P (iii) 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? [4] Question 3 (i) Factorize: sin3q + cos3q Hence, prove the following identity: sin 3 cos3 sin cos 1 sin cos [4] (ii) In the given diagram, O is the centre of the circle. PR and PT are R two tangents drawn from Q the external N point P and 42 touching the O circle at Q and S respectively. M MN is a 25 P T S diameter of the circle. Given PQM = 42 and PSM = 25 . [4] Find: (a) OQM (b) QNS (c) QOS (d) QMS (iii) Use graph sheet for this question. Take 2 cm = 1 unit along the a [5] (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 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. SECTION-B (40 MARKS) (Attempt any four questions from this Section.) Question 4 B D (a) DABC ~ DPQS (c) DABD~ DPSR Question 2 x 0 4 ,B (i) A 1 1 y C Q S R (b) DABD~ DPQS (d) DABC ~ DPSR 4 0 0 and C [4] x 1 1 Find the values of x and y, if AB = C. (ii) A solid metallic cylinder is cut into two identical halves along its height (as shown in the diagram). The diameter of the cylinder is 7 cm and the height is 10 cm. Find: [4] (a) the total surface area (both the halves). (b) the total cost of painting the two halves at the 22 rate of 30 per cm2 Use p 7 (i) Suresh has a recurring deposit account in a bank. He deposits ` 2000 per month and the bank pays interest at the rate of 8% per annum. If he gets ` 1040 as interest at the time of maturity, find in years total time for which the account was held. [3] (ii) The following table gives the duration of movies in minutes. [3] Duration (in minutes) 100 - 110 110 - 120 120 - 130 130 - 140 140 - 150 150 - 160 No. of movies 5 10 17 8 6 4 Using step - deviation method, find the mean duration of the movies. ( a b )3 64 (iii) If [4] ( a b )3 27 a b (a) Find a b (b) Hence using properties of proportion, find a : b. Solved Paper - 2024 Question 5 (i) The given graph with a histogram represents the number of plants of different heights in a school campus. Study the graph carefully and answer the following questions: [5] 3 (a) Prove that DAPD ~ DBPC (b) Find the length of BD and PB (c) Hence, find the length of PA (d) Find area DAPD: area DBPC Question 7 (i) In the given diagram, an isosceles DABC is inscribed 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 P M 65 65 O B (a) Make a frequency table with respect to the class boundaries and their corresponding frequencies. (b) State the modal class. (c) Identify and note down the mode of the distribution. (d) Find the number of plants whose height range is between 80 cm to 90 cm. (ii) The angle of elevation of the top of a 100 m high tree from two points A and B on the opposite side of the tree are 52 and 45 respectively. Find the distance AB, to the nearest metre. [5] C Q (a) ABC (b) BAC (c) BCQ [3] (ii) Solve the following in equation, write down the solution set and represent it on the real number line. 7x 3 x 2 8 2 x , x I [3] 2 (iii) In the given diagram, ABC is a triangle, where B(4, 4) and C( 4, 2). D is a point on AC. [4] y 7 6 A 5 4 Question 6 (i) Solve the following quadratic equation for x and give your answer correct up to three significant figures: 2x2 10x + 5 = 0 [3] (Use mathematical tables if necessary) (ii) The nth term of an Arithmetic Progression (A.P.) is given by the relation Tn = 6(7 n). [3] Find: (a) its first term and common difference (b) sum of its first 25 terms (iii) In the given diagram DADB and DACB are two right angled triangles with ADB = BCA = 90 . If AB = 10 cm, AD = 6 cm, BC= 2.4 cm and DP = 4.5 cm [4] 3 2 1 D _6 _5 _4 C( _ 4 _ 2) _3 _2 _1 _1 x 1 2 3 4 5 6 _2 _3 _4 B(4, _ 4) _5 _6 (a) Write down the coordinates of A and D. (b) Find the coordinates of the centroid of DABC. (c) If D divides AC in the ratio k : 1, find the value of k. (d) Find the equation of the line BD. 4 Oswaal ICSE, MATHEMATICS, Class-X ANSWERS SECTION A 1. (i) Option (c) is correct. Explanation: Given that, Marked Price, MP = ` 2000 For intra-state sale, the CGST paid is ` 120 Total GST = CGST + SGST As it s an intra-state sale, CGST = SGST So, Total GST = 2 CGST Total GST = 2 120 = ` 240 Marked Price Rate of GST Total GST = 100 240 100 Rate of GST = = 12% 2000 So, the rate of GST for this intra-state sale of goods is 12%. (ii) Option (d) is correct Explanation: Given the polynomial, P(x) = x3 + x2 2x + 1, For the polynomial to be exactly divisible by (x 3), Putting x = 3 in the polynomial, P(3) = (3)3 + (3)2 2(3) + 1 = 27 + 9 6 + 1 = 31 So, 31 must be subtracted from the polynomial. (iii) Option (b) is correct Explanation: Given quadratic equation is px2 qx +r=0 On comparing with ax2 + bx + c = 0, We get, a = p, b = q and c = r For real and equal roots, Discriminant, D = b2 4ac = 0 ( q)2 4(p)(r) = 0 q2 = 4pr (iv) Option (c) is correct Explanation: Given that, 2 2 A = 0 2 We have, 2 2 2 2 A2 = 0 2 0 2 4 + 0 4 + 4 4 8 = = 0 4 0 + 0 0 + 4 On comparing with, 4 8 A2 = 0 4 We get, x = 8 (v) Option (c) is correct Explanation: Given that, Median of observations 27, 31, 46, 52, x, x + 4, 71, 79, 85, 90 is 64 We know that, For even number of observations, N = 10 (here) 5 th term + 6 th term Median = 2 x+x+4 64 = 2 128 = 2x + 4 124 = 2x x = 62 (vi) Option (d) is correct Explanation: Given that, Points A(x, y), B(3, 2) and C(4, 5) are collinear. For collinear points the area of figure formed by these points is zero, So, 1 [x1(y2 y3) + x2(y3 y1) + x3(y1 y2)] = 0 2 On putting values, x[( 2) ( 5)] + 3[( 5) (y)] + 4[(y) ( 2)] = 0 3x 15 3y + 4y + 8 = 0 3x + y = 7 y = 7 3x (vii) Option (d) is correct Explanation: We know that, Distance Time y 60 90 So, = = = k (constant) 2 x 5 60 90 = 2 x x = 3 y 60 Also, = 2 5 y = 150 Hence, the values of x and y are 3 and 150 respectively. (viii) Option (a) is correct 1 1 1 Explanation: Given AP is , + 1 , + 2 ... a a a First term, A = 1 a Common difference, 1 1 d = + 1 = 1 a a So, 7th term (A7) of given AP is: A7 = A + (7 1)d 1 A7 = + 6(1) a 1 A7 = + 6 a (ix) Option (d) is correct Explanation: Given that, Number of shares, n = 15 Nominal value per share = ` 75 Discount = 20% Discounted price per share = Nominal value per share (1 Discount rate) = ` 75 (1 0.20) = ` 75 0.80 = ` 60 So, each share is being sold for ` 60 after the discount. Sum invested = Price per share Number of shares = ` 60 15 = ` 900 Solved Paper - 2024 (x) Option (b) is correct Explanation: We know that, Circumcentre of a triangle is equidistant from all three vertices of the triangle. This means that the distance from the circumcentre to each vertex is the same. (xi) Option (a) is correct Explanation: Statement 1: sin2 q + cos2 q = 1 This is true. Statement 2: cosec2 q + cot2 q = 1 This statement is invalid because cosec2 q cot2 q = 1 Hence, only statement 1 is valid. (xii) Option (d) is correct Explanation: Given that, PS and PT are tangents and SQ || PT SPT = 80 5 B = Q 1 1 Or A = P 2 2 BAD = QPS In DABD and DPQS, B = Q BAD = QPS So, By AA similarity criteria DABD ~ DPQS. 4 0 x 0 2. (i) Given that, A = , B = , and y 1 1 1 4 0 C= x 1 Also, AB = C x 0 4 0 AB = 1 1 y 1 4x + 0 0 + 0 = 4 + y 0 + 1 4x 0 = 4 + y 1 As PS = PT (Tangents from an external point to a circle are equal) Also, PST = PTS (Angles opposite to equal sides) PST + PTS + SPT = 180 (Sum of all angles of triangle) PST + PTS + 80 = 180 So, PST = PTS = 50 PTS = QST = 50 (alternate angles) (xiii) Option (b) is correct Explanation: When a dice is thrown the probability 1 of getting an even number is . 2 Sample space for even numbers on a dice is {2, 4, 6} So, Assertion is false and reason is true. (xiv) Option (a) is correct Explanation: Given that, Size of rectangular sheet = 11 cm 7 cm When it is rotated about side 11 cm then, Height of cylinder formed, h = 7 cm Let the radius of cylinder formed be R, Circumference, 2 R = 11 cm Curved Surface area of cylinder = 2 Rh = 11 7 = 77 cm2 Now, When it is rotated about side 7 cm then, Height of cylinder formed, H = 11 cm Let the radius of cylinder formed be r, Circumference, 2 r = 7 cm Curved Surface area of cylinder = 2 rH = 7 11 = 77 cm2 So, Ratio of their CSA = 77 : 77 = 1 : 1 (xv) Option (b) is correct Explanation: Given that, DABC ~ DPQR AD and PS are bisectors of BAC and QPR, respectively Since ABC PQR we have, A = P On comparing with C, We get, 4x = 4 x = 1 4 + y = x 4 + y = 1 y = 3 So, x = 1 and y = 3 (ii) Given that, Height of cylinder, h = 10 cm Diameter of cylinder, d = 7 cm 7 Radius of cylinder, r = cm 2 (a) Total surface area of both halves = Total surface area of cylindrical part + Area of two rectangular parts = 2 r(h + r) + 2dh 22 7 7 = 2 10 + + 2 7 10 7 2 2 27 = 22 + 140 2 = 297 + 140 = 437 cm2 (b) Rate of painting = ` 30/cm2 Total cost of painting = Total surface area Rate of painting = 437 30 = ` 13,100 (iii) Given that, 15, 30, 60, 120 are in G.P. We have, First term, a = 15 30 Common ratio, r = =2 15 (a) nth term of GP is an an = a(r)n 1 an = 15 (2)n 1 (b) Given that, Sum, S = 945 6 Oswaal ICSE, MATHEMATICS, Class-X Let the number of terms taken be n (d) As QMSN is a cyclic quadrilateral, So, QMS + QNS = 180 a(r n 1) S = (Sum of opposite angles r 1 of cyclic quadrilateral) On substituting the values, QMS + 67 = 180 QMS = 180 67 = 113 15(2 n 1) 945 = 2 1 (iii) (a), (b) 63 = 2n 1 are in graph 2n = 64 n 6 2 = 2 n = 6 So, 6 terms of the GP will give a sum of 945. 3. (i) To factorize: sin3 q + cos3 q We have, sin3 q + cos3 q [Using a3 + b3 = (a + b) (a2 ab + b2)] (sin q + cos q) (sin2 q sin q cos q + cos2 q) (sin q + cos q) (1 sin q cos q) (i) [As sin2 q + cos2 q = 1] 3 3 sin + cos To prove: + sin .cos = 1 sin + cos Taking LHS, sin 3 + cos3 + sin .cos sin + cos Using (i), (sin + cos )(1 sin .cos ) + sin .cos = sin + cos = 1 sin q.cos q + sin q.cos q LHS = RHS Hence proved. (ii) Given that, PR and PT are tangents MN is diameter of circle PQM = 42 PSM = 25 (a) We have, OQP = 90 (Radius is perpendicular to tangent at point of contact) Also, OQM + PQM = OQP = 90 OQM + 42 = 90 OQM = 48 Similarly, OSM = 90 25 = 65 (b) We know that, QNM = 42 and SNM = 25 (Alternate segment theorem) QNS = QNM + SNM = 42 + 25 = 67 (c) QOS = 2 QNS (As angle at the center is twice the angle at circumference) QOS = 2 67 = 134 (c) Coordinates of point B are ( 2, 1) and C are ( 4, 1) (d) Coordinates of point A are (0, 1) (e) The figure ABA B is a square. SECTION B 4. (i) Given that, Suresh deposits amount, P = ` 2000 per month Rate of interest, r = 8% Interest, I = ` 1040 Let the duration of deposit be n months P n(n + 1) r We know that, I = 2 12 100 2000 n(n + 1) 8 1040 = 2 12 100 n(n + 1) = 156 12 13 = 156 So, n = 12 So, total time of deposit is 12 months or 1 year. (ii) We have, xi A h fiui 105 2 10 115 1 10 17 125 = A 0 0 130-140 8 135 1 8 140-150 6 145 2 12 150-160 4 155 3 12 Duration (in mins) No. of movies (f) xi 100-110 5 110-120 10 120-130 ui = For the given data, Let Assumed mean, A = 125 Class interval, h = 10 Using step-deviation method, Mean = A + fi ui h fi Solved Paper - 2024 12 10 50 Mean = 125 + 2.4 Mean = 127.4 ( a + b )3 64 (iii) Given that, = 3 ( a b) 27 (a) Taking cube root both sides, 4 ( a + b) = 3 ( a b) (b) Applying componendo and dividendo, ( a + b) + ( a b) 4+3 = ( a + b) ( a b) 4 3 2a 7 = 2b 1 a 7 = b 1 So, a : b = 7 : 1 5. (i) (a) Mean = 125 + Height (in cms) Number of plants(frequency) 30-40 4 40-50 2 50-60 8 60-70 12 70-80 6 80-90 3 90-100 4 (b) The modal class is 60 70. (c) From the given histogram, Mode = 64 (d) Number of plants whose height range between 80 90 cm is 3. (ii) From the given figure, CD be the tower of height 100 m In DADC, DC tan 52 = AC 100 1.279 = AC 100 AC = 1.279 AC = 78.18 m Now, In DBDC, DC tan 45 = BC 1 = 100 BC BC = 100 m We know that, AB = AC + BC = 78.18 + 100 = 178.18 m or 178 m (approx) 6. (i) Given quadratic equation is 2x2 10x + 5 = 0 On comparing with ax2 + bx + c = 0, We get, a = 2, b = 10 and c = 5 Also, Discriminant, D = b2 4ac = ( 10)2 4(2)(5) = 100 40 = 60 7 Using Quadratic formula, b D x = 2a 10 60 4 10 7.7459 x = 4 x = 4.43,0.563 x = (ii) Given that, Tn = 6(7 n) (a) Putting n = 1, T1 = 6(7 1) = 36 T2 = 6(7 2) = 30 T3 = 6(7 3) = 24 So, First term, a = 36 Common difference, d = T2 T1 = 30 36 = 6 n (b) S25 = [2a + (n 1)d] 2 25 S25 = [2 36 + (25 1) ( 6)] 2 25 S25 = [72 144] 2 25 S25 = ( 72 ) 2 S25 = 900 So, sum of its first 25 terms is 900. (iii) Given that, DABD and DACB are right angled triangles, ADB = BCA = 90 AB = 10 cm, AD = 6 cm, BC = 2.4 cm and DP = 4.5 cm (a) In DAPD and DBPC, We have, ADP = BCP = 90 APD = BPC (Vertically opposite angles) So, By AA similarity DAPD ~ DBPC AD BC Also, = DP CP 6 2.4 = 4 . 5 CP CP = 1.8 cm (b) In right DBPC, BP2 = BC2 + CP2 BP2 = (2.4)2 + (1.8)2 BP = 3 cm And, BD = DP + PB = 4.5 + 3 = 7.5 cm (c) In right DADP, PA2 = AD2 + DP2 PA2 = 62 + (4.5)2 8 Oswaal ICSE, MATHEMATICS, Class-X PA = 7.5 cm 1 4.5 6 25 2 (d) Area DAPD : Area DBPC = = 1 4 2.4 1.8 2 Required ratio is 25 : 4 7. (i) Given that, DABC is isosceles OM is perpendicular to AC COM = 65 We have, AOC = AOM + COM = 65 + 65 = 130 In DMOC, MCO = 180 (90 + 65 ) = 25 (a) We know that, 1 ABC = AOC 2 (Angle at the center is twice the angle at circumference) 1 ABC = 130 2 ABC = 65 (b) As DABC is isosceles, AB = AC and ABC = ACB = 65 (Angles opposite to equal sides) In DABC, ABC + ACB + BAC = 180 65 + 65 + BAC = 180 BAC = 180 130 BAC = 50 (c) Also, As BAC = 50 So, BCQ = 50 (Angle made with chord is equal to angle made with tangent) 7x (ii) Given that, 3 + x + 2 < 8 + 2x, 2 x belongs to Integer Now, 7x 3 + x +2 2 (7 x + 4 ) 3 + x 2 6 + 2x 7x + 4 6 4 7x 2x 10 5x 2 x x 2 7x Also, + 2 < 8 + 2x 2 7x + 4 < 8 + 2x 2 7x + 4 < 16 + 4x 7x 4x < 16 4 3x < 12 x < 4 From (i) and (ii), Required Solution set is { 2, 1, 0, 1, 2, 3} (iii) Given that, B(4, 4) and C( 4, 2) (a) Coordinates of A are (0, 6) Coordinates of D are ( 3, 0) (b) Centroid (x, y) of DABC is 0 + 4 3 (x, y) = , 3 ...(i) ...(ii) 6 4 + 0 3 1 2 (x, y) = , 3 3 (c) D divides AC in Ratio k : 1 Using section formula, 4 k + 0 3 = k +1 3k 3 = 4k 4k 3k = 3 k = 3 (d) Equation of line BD is 4 0 (y 0) = ( x + 3) 4 ( 3) 4 ( x + 3) 7 7y = 4x 12 4x + 7y + 12 = 0 y =

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