Page 1 CBSE IX | Mathematics Sample Paper 2 – Solution CBSE Board Class IX Mathematics Sample Paper 2 – Solution Time: 3 hrs Total Marks: 80 Section A 1. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 27 3 3 1 2 3 6 3 3 3 3 1 2 3 4 2. Given polynomial is x 2 – x – 1 Substituting x = –1 in x 2 – x – 1, we have (–1) 2 – (–1) – 1 = 1 + 1 – 1 = 1 3. Two lines l and m in a plane are said to be parallel lines if they do not have common point, i.e. they do not intersect. 4. Substituting x = 2 and y = 0 in x – 2y = 4, we get L.H.S = 2 – 2 × 0 = 2 ? 4 i.e. L.H.S. ? R.H.S Therefore, (2, 0) is not a solution of x – 2y = 4. OR Put y = 0 to find the coordinate of x-axis. ? 2x – y = 4 ? 2x = 4 ? x = 2 Hence, the graph of the linear equation 2x – y = 4 cut x-axis at (2, 0). 5. ?PSR = ?RQP = 125° (opposite angles will be equal since PQRS is a parallelogram) ?PQT = 180° (PQT is a straight line) ? ?PQR + ?RQT = 180° ? 125° + ?RQT = 180° ? ?RQT = 55° 6. Class size is the difference between two successive class marks. ? Class size = 10 – 6 = 4 OR According to the given distribution, Class size = 52 – 47 = 5. Page 2 CBSE IX | Mathematics Sample Paper 2 – Solution CBSE Board Class IX Mathematics Sample Paper 2 – Solution Time: 3 hrs Total Marks: 80 Section A 1. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 27 3 3 1 2 3 6 3 3 3 3 1 2 3 4 2. Given polynomial is x 2 – x – 1 Substituting x = –1 in x 2 – x – 1, we have (–1) 2 – (–1) – 1 = 1 + 1 – 1 = 1 3. Two lines l and m in a plane are said to be parallel lines if they do not have common point, i.e. they do not intersect. 4. Substituting x = 2 and y = 0 in x – 2y = 4, we get L.H.S = 2 – 2 × 0 = 2 ? 4 i.e. L.H.S. ? R.H.S Therefore, (2, 0) is not a solution of x – 2y = 4. OR Put y = 0 to find the coordinate of x-axis. ? 2x – y = 4 ? 2x = 4 ? x = 2 Hence, the graph of the linear equation 2x – y = 4 cut x-axis at (2, 0). 5. ?PSR = ?RQP = 125° (opposite angles will be equal since PQRS is a parallelogram) ?PQT = 180° (PQT is a straight line) ? ?PQR + ?RQT = 180° ? 125° + ?RQT = 180° ? ?RQT = 55° 6. Class size is the difference between two successive class marks. ? Class size = 10 – 6 = 4 OR According to the given distribution, Class size = 52 – 47 = 5. CBSE IX | Mathematics Sample Paper 2 – Solution Section B 7. ky 2 – 6ky + 8k = k(y 2 – 6y + 8) = k(y 2 – 4y – 2y + 8) = k(y – 4)(y – 2) Thus, the dimensions of the cuboid are given by the expressions k, (y – 4) and (y – 2). 8. 9. Here ? ADC = y = ? ACD Ext. ? ACD = ? ABC + ? BAC ? 2 ? BAC = ? ACD = y ? ? BAC = y 2 ? y 2 + (180° – 2y) = 180° – 75° ? y 2 + 180° – 2y = 180° – 75 ? y 2 – 2y = – 75° ? – 3y 2 = –75° ? y = 50° 10. Let ‘l’ be the length of the cube. Now, T.S.A. of the cube = 294 cm 2 ….(given) ? 6l 2 = 294 2 294 l 49 6 ? ? ? ? Side (l) = 7 cm. Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm 3 Page 3 CBSE IX | Mathematics Sample Paper 2 – Solution CBSE Board Class IX Mathematics Sample Paper 2 – Solution Time: 3 hrs Total Marks: 80 Section A 1. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 27 3 3 1 2 3 6 3 3 3 3 1 2 3 4 2. Given polynomial is x 2 – x – 1 Substituting x = –1 in x 2 – x – 1, we have (–1) 2 – (–1) – 1 = 1 + 1 – 1 = 1 3. Two lines l and m in a plane are said to be parallel lines if they do not have common point, i.e. they do not intersect. 4. Substituting x = 2 and y = 0 in x – 2y = 4, we get L.H.S = 2 – 2 × 0 = 2 ? 4 i.e. L.H.S. ? R.H.S Therefore, (2, 0) is not a solution of x – 2y = 4. OR Put y = 0 to find the coordinate of x-axis. ? 2x – y = 4 ? 2x = 4 ? x = 2 Hence, the graph of the linear equation 2x – y = 4 cut x-axis at (2, 0). 5. ?PSR = ?RQP = 125° (opposite angles will be equal since PQRS is a parallelogram) ?PQT = 180° (PQT is a straight line) ? ?PQR + ?RQT = 180° ? 125° + ?RQT = 180° ? ?RQT = 55° 6. Class size is the difference between two successive class marks. ? Class size = 10 – 6 = 4 OR According to the given distribution, Class size = 52 – 47 = 5. CBSE IX | Mathematics Sample Paper 2 – Solution Section B 7. ky 2 – 6ky + 8k = k(y 2 – 6y + 8) = k(y 2 – 4y – 2y + 8) = k(y – 4)(y – 2) Thus, the dimensions of the cuboid are given by the expressions k, (y – 4) and (y – 2). 8. 9. Here ? ADC = y = ? ACD Ext. ? ACD = ? ABC + ? BAC ? 2 ? BAC = ? ACD = y ? ? BAC = y 2 ? y 2 + (180° – 2y) = 180° – 75° ? y 2 + 180° – 2y = 180° – 75 ? y 2 – 2y = – 75° ? – 3y 2 = –75° ? y = 50° 10. Let ‘l’ be the length of the cube. Now, T.S.A. of the cube = 294 cm 2 ….(given) ? 6l 2 = 294 2 294 l 49 6 ? ? ? ? Side (l) = 7 cm. Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm 3 CBSE IX | Mathematics Sample Paper 2 – Solution OR Given that: Side (S) of metal cube = 4 cm Length (l) of the tank = 8 cm Breadth (b) of the tank = 4 cm Since the volume of cube immersed = volume of the cuboidal tank We get, S x S x S = l x b x h 4 3 = 8 x 4 x h h = 64 32 ? h = 2 cm Thus, rise in water level = 2 cm 11. Number of students born in August = 6 Total number of students = 40 Number of students born in August 6 3 Required probability= Total number of students 40 20 ?? 12. Let the angles of a quadrilateral be 2x, 5x, 8x and 9x respectively. By the angle sum property of a quadrilateral, we have 2x + 5x + 8x + 9x = 360° ? 24x = 360° ? x = 15° Now, First angle = 2x = 2 × 15 = 30°, Second angle = 5x = 5 × 15 = 75°, Third angle = 8x = 8 × 15 = 120° and Fourth angle = 9x = 9 × 15 = 135°. Thus, the angles of a quadrilateral are 30°, 75°, 120° and 135°. OR Let x be the fourth angle of a quadrilateral. According to the question, The sum of the angles of a quadrilateral is 360°. ? 50° + 110° + 40° + x = 360° ?200° + x = 360° ? x = 160° Hence, the fourth angle of a quadrilateral is 160°. Page 4 CBSE IX | Mathematics Sample Paper 2 – Solution CBSE Board Class IX Mathematics Sample Paper 2 – Solution Time: 3 hrs Total Marks: 80 Section A 1. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 27 3 3 1 2 3 6 3 3 3 3 1 2 3 4 2. Given polynomial is x 2 – x – 1 Substituting x = –1 in x 2 – x – 1, we have (–1) 2 – (–1) – 1 = 1 + 1 – 1 = 1 3. Two lines l and m in a plane are said to be parallel lines if they do not have common point, i.e. they do not intersect. 4. Substituting x = 2 and y = 0 in x – 2y = 4, we get L.H.S = 2 – 2 × 0 = 2 ? 4 i.e. L.H.S. ? R.H.S Therefore, (2, 0) is not a solution of x – 2y = 4. OR Put y = 0 to find the coordinate of x-axis. ? 2x – y = 4 ? 2x = 4 ? x = 2 Hence, the graph of the linear equation 2x – y = 4 cut x-axis at (2, 0). 5. ?PSR = ?RQP = 125° (opposite angles will be equal since PQRS is a parallelogram) ?PQT = 180° (PQT is a straight line) ? ?PQR + ?RQT = 180° ? 125° + ?RQT = 180° ? ?RQT = 55° 6. Class size is the difference between two successive class marks. ? Class size = 10 – 6 = 4 OR According to the given distribution, Class size = 52 – 47 = 5. CBSE IX | Mathematics Sample Paper 2 – Solution Section B 7. ky 2 – 6ky + 8k = k(y 2 – 6y + 8) = k(y 2 – 4y – 2y + 8) = k(y – 4)(y – 2) Thus, the dimensions of the cuboid are given by the expressions k, (y – 4) and (y – 2). 8. 9. Here ? ADC = y = ? ACD Ext. ? ACD = ? ABC + ? BAC ? 2 ? BAC = ? ACD = y ? ? BAC = y 2 ? y 2 + (180° – 2y) = 180° – 75° ? y 2 + 180° – 2y = 180° – 75 ? y 2 – 2y = – 75° ? – 3y 2 = –75° ? y = 50° 10. Let ‘l’ be the length of the cube. Now, T.S.A. of the cube = 294 cm 2 ….(given) ? 6l 2 = 294 2 294 l 49 6 ? ? ? ? Side (l) = 7 cm. Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm 3 CBSE IX | Mathematics Sample Paper 2 – Solution OR Given that: Side (S) of metal cube = 4 cm Length (l) of the tank = 8 cm Breadth (b) of the tank = 4 cm Since the volume of cube immersed = volume of the cuboidal tank We get, S x S x S = l x b x h 4 3 = 8 x 4 x h h = 64 32 ? h = 2 cm Thus, rise in water level = 2 cm 11. Number of students born in August = 6 Total number of students = 40 Number of students born in August 6 3 Required probability= Total number of students 40 20 ?? 12. Let the angles of a quadrilateral be 2x, 5x, 8x and 9x respectively. By the angle sum property of a quadrilateral, we have 2x + 5x + 8x + 9x = 360° ? 24x = 360° ? x = 15° Now, First angle = 2x = 2 × 15 = 30°, Second angle = 5x = 5 × 15 = 75°, Third angle = 8x = 8 × 15 = 120° and Fourth angle = 9x = 9 × 15 = 135°. Thus, the angles of a quadrilateral are 30°, 75°, 120° and 135°. OR Let x be the fourth angle of a quadrilateral. According to the question, The sum of the angles of a quadrilateral is 360°. ? 50° + 110° + 40° + x = 360° ?200° + x = 360° ? x = 160° Hence, the fourth angle of a quadrilateral is 160°. CBSE IX | Mathematics Sample Paper 2 – Solution Section C 13. Given: a = 3 + b ? a – b = 3 Applying the cubic identity on both the sides, we have (a – b) 3 = 3 3 ? a 3 – b 3 – 3(a)(b)(a – b) = 27 ? a 3 – b 3 – 3ab(3) = 27 (? a – b = 3) ? a 3 – b 3 – 9ab = 27 OR (3x + 2y) 2 = 12 2 9x 2 + 12xy + 4y 2 = 144 9x 2 + 4y 2 + 12xy = 144 9x 2 + 4y 2 + 12 × 6 = 144 ? xy = 6 9x 2 + 4y 2 = 144 – 72 9x 2 + 4y 2 = 72 14. Since AB ? DC, ?x = 30° [Alternate angles] In ?ABD, 80° + 30° + ?y = 180° ? ?y = 180° – 110° = 70° In ?BDC, 30° + (70° – 30°) + ?z = 180° ? ?z = 110° OR Since, PQRS is a square. PS = SR and ? PSR = 90°. In ?PSR, PS = SR ? SRP = ?QRP angles opposite to equal sides ? SRP + ?QRP + ? PSR = 180° 2?SRP + 90° = 180° ?SRP = 45° Page 5 CBSE IX | Mathematics Sample Paper 2 – Solution CBSE Board Class IX Mathematics Sample Paper 2 – Solution Time: 3 hrs Total Marks: 80 Section A 1. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 27 3 3 1 2 3 6 3 3 3 3 1 2 3 4 2. Given polynomial is x 2 – x – 1 Substituting x = –1 in x 2 – x – 1, we have (–1) 2 – (–1) – 1 = 1 + 1 – 1 = 1 3. Two lines l and m in a plane are said to be parallel lines if they do not have common point, i.e. they do not intersect. 4. Substituting x = 2 and y = 0 in x – 2y = 4, we get L.H.S = 2 – 2 × 0 = 2 ? 4 i.e. L.H.S. ? R.H.S Therefore, (2, 0) is not a solution of x – 2y = 4. OR Put y = 0 to find the coordinate of x-axis. ? 2x – y = 4 ? 2x = 4 ? x = 2 Hence, the graph of the linear equation 2x – y = 4 cut x-axis at (2, 0). 5. ?PSR = ?RQP = 125° (opposite angles will be equal since PQRS is a parallelogram) ?PQT = 180° (PQT is a straight line) ? ?PQR + ?RQT = 180° ? 125° + ?RQT = 180° ? ?RQT = 55° 6. Class size is the difference between two successive class marks. ? Class size = 10 – 6 = 4 OR According to the given distribution, Class size = 52 – 47 = 5. CBSE IX | Mathematics Sample Paper 2 – Solution Section B 7. ky 2 – 6ky + 8k = k(y 2 – 6y + 8) = k(y 2 – 4y – 2y + 8) = k(y – 4)(y – 2) Thus, the dimensions of the cuboid are given by the expressions k, (y – 4) and (y – 2). 8. 9. Here ? ADC = y = ? ACD Ext. ? ACD = ? ABC + ? BAC ? 2 ? BAC = ? ACD = y ? ? BAC = y 2 ? y 2 + (180° – 2y) = 180° – 75° ? y 2 + 180° – 2y = 180° – 75 ? y 2 – 2y = – 75° ? – 3y 2 = –75° ? y = 50° 10. Let ‘l’ be the length of the cube. Now, T.S.A. of the cube = 294 cm 2 ….(given) ? 6l 2 = 294 2 294 l 49 6 ? ? ? ? Side (l) = 7 cm. Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm 3 CBSE IX | Mathematics Sample Paper 2 – Solution OR Given that: Side (S) of metal cube = 4 cm Length (l) of the tank = 8 cm Breadth (b) of the tank = 4 cm Since the volume of cube immersed = volume of the cuboidal tank We get, S x S x S = l x b x h 4 3 = 8 x 4 x h h = 64 32 ? h = 2 cm Thus, rise in water level = 2 cm 11. Number of students born in August = 6 Total number of students = 40 Number of students born in August 6 3 Required probability= Total number of students 40 20 ?? 12. Let the angles of a quadrilateral be 2x, 5x, 8x and 9x respectively. By the angle sum property of a quadrilateral, we have 2x + 5x + 8x + 9x = 360° ? 24x = 360° ? x = 15° Now, First angle = 2x = 2 × 15 = 30°, Second angle = 5x = 5 × 15 = 75°, Third angle = 8x = 8 × 15 = 120° and Fourth angle = 9x = 9 × 15 = 135°. Thus, the angles of a quadrilateral are 30°, 75°, 120° and 135°. OR Let x be the fourth angle of a quadrilateral. According to the question, The sum of the angles of a quadrilateral is 360°. ? 50° + 110° + 40° + x = 360° ?200° + x = 360° ? x = 160° Hence, the fourth angle of a quadrilateral is 160°. CBSE IX | Mathematics Sample Paper 2 – Solution Section C 13. Given: a = 3 + b ? a – b = 3 Applying the cubic identity on both the sides, we have (a – b) 3 = 3 3 ? a 3 – b 3 – 3(a)(b)(a – b) = 27 ? a 3 – b 3 – 3ab(3) = 27 (? a – b = 3) ? a 3 – b 3 – 9ab = 27 OR (3x + 2y) 2 = 12 2 9x 2 + 12xy + 4y 2 = 144 9x 2 + 4y 2 + 12xy = 144 9x 2 + 4y 2 + 12 × 6 = 144 ? xy = 6 9x 2 + 4y 2 = 144 – 72 9x 2 + 4y 2 = 72 14. Since AB ? DC, ?x = 30° [Alternate angles] In ?ABD, 80° + 30° + ?y = 180° ? ?y = 180° – 110° = 70° In ?BDC, 30° + (70° – 30°) + ?z = 180° ? ?z = 110° OR Since, PQRS is a square. PS = SR and ? PSR = 90°. In ?PSR, PS = SR ? SRP = ?QRP angles opposite to equal sides ? SRP + ?QRP + ? PSR = 180° 2?SRP + 90° = 180° ?SRP = 45° CBSE IX | Mathematics Sample Paper 2 – Solution 15. 27p 3 + 8q 3 + 54p 2 q + 36pq 2 = (3p) 3 + (2q) 3 + 18pq(3p+2q) = (3p) 3 + (2q) 3 + 3 × 3p × 2q (3p + 2q) = (3p + 2q) 3 [(a + b) 3 = a 3 + b 3 + 3ab (a + b) [where a = 3p and b = 2q ] = (3p + 2q) (3p + 2q) (3p + 2q) 16. b 2 + c 2 + 2(ab + bc + ca) = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca – a 2 [Adding and subtracting a 2 ] = [a 2 + b 2 + c 2 + 2ab + 2bc + 2ca] – a 2 = (a + b + c) 2 – (a) 2 [Using x 2 + y 2 + 2xy + 2yz + 2zx = (x + y + z) 2 ] = (a + b + c + a)(a + b + c – a) [Because a 2 – b 2 = (a+ b) (a – b)] = (2a + b +c )(b + c) 17. 2x = z (Alternate angles, as l1 || l2) y = z (Alternate angles, as a1 || a2) So, 2x = y Now, y + 4x – 15 = 180° (linear pair) 2x + 4x – 15 = 180° ? 6x = 195° ? x = 32.5 18. Number of white balls = x Total number of balls = 12 If 6 white balls are added, we have Total number of balls = 18 Number of white balls = x + 6 x6 Now, P(getting a white ball) 18 ? ? According to the given information, ? ?? ? ?? ?? ? ?? ? ? ? ?? ?? x 6 x 2 18 12 x 6 x 18 6 6x 36 18x 12x 36 x3 x P(white ball) 12 ??Read More

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