Page 1 CBSE IX | Mathematics Sample Paper 5 â€“ Solution CBSE Board Class IX Mathematics Sample Paper 5 â€“ Solution Time: 3 hrs Total Marks: 80 Section A 1. 11 0.011 1000 ? OR Yes, because 0 can be written as 0 1 which is of the form p q , where p and q are integers and q ? 0. 2. x + y = 2 â€¦â€¦(i) and x â€“ y = 4 â€¦â€¦(ii) Adding (i) and (ii), we get x = 3 Putting x = 3 in the equation (i), we get y = â€“1 ? x = 3 and y = â€“1 3. Since AOB is a Straight Line, ? ?AOB = 180° ? ?AOD + ?COD + ?COB = 180° ? x + 10° + x + x + 20° = 180° ? 3x = 150° ? x = 50° 4. If for one of the solutions of the equation ax + by + c = 0, x is negative and y is positive, then a portion of the above line definitely lies in the II Quadrant. As in the II Quadrant the x-axis contains only negative numbers and y-axis contains only positive numbers. OR Let the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y. Hence, the linear equation is 5x = 2y 5x â€“ 2y = 0. Page 2 CBSE IX | Mathematics Sample Paper 5 â€“ Solution CBSE Board Class IX Mathematics Sample Paper 5 â€“ Solution Time: 3 hrs Total Marks: 80 Section A 1. 11 0.011 1000 ? OR Yes, because 0 can be written as 0 1 which is of the form p q , where p and q are integers and q ? 0. 2. x + y = 2 â€¦â€¦(i) and x â€“ y = 4 â€¦â€¦(ii) Adding (i) and (ii), we get x = 3 Putting x = 3 in the equation (i), we get y = â€“1 ? x = 3 and y = â€“1 3. Since AOB is a Straight Line, ? ?AOB = 180° ? ?AOD + ?COD + ?COB = 180° ? x + 10° + x + x + 20° = 180° ? 3x = 150° ? x = 50° 4. If for one of the solutions of the equation ax + by + c = 0, x is negative and y is positive, then a portion of the above line definitely lies in the II Quadrant. As in the II Quadrant the x-axis contains only negative numbers and y-axis contains only positive numbers. OR Let the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y. Hence, the linear equation is 5x = 2y 5x â€“ 2y = 0. CBSE IX | Mathematics Sample Paper 5 â€“ Solution 5. Here the class marks are uniformly spaced, Therefore, Class size is the difference between any two consecutive class marks. ? Class Size = 52 â€“ 47 = 5 6. ?PSR = ?RQP = 125° (since PQRS is a parallelogram, then the opposite angles will be equal) ? ?PQT = 180°â€¦â€¦â€¦â€¦â€¦.. (PQT is a straight line) ? ?PQR + ?RQT = 180° ? 125° + ?RQT = 180° ? ?RQT = 55° Section B 7. a = 2 + 3 ? 11 a 23 ? ? = 1 2 3 2 3 2 3 ? ? ?? = ? ? 2 2 23 23 ? ? = 2 â€“ 3 So, a + 1 a = (2 + 3 ) + (2 â€“ 3 ) = 4 8. (i) I quadrant: (5, 2) (ii) II quadrant: (â€“5, 2) (iii) III quadrant: (â€“5, â€“2) (iv) IV quadrant: (5, â€“2) 9. Let â€˜lâ€™ be the length of the cube. Now, T.S.A. of the cube = 294 cm 2 â€¦(given) ? 6l 2 = 294 ? Side, l = 7 cm Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm 3 OR According to the question, Volume of one matchbox = 4 cm × 2.5 cm × 1.5 cm = 15 cm 3 Volume of 12 such matchboxes = 15 × 12 = 180 cm 3 Page 3 CBSE IX | Mathematics Sample Paper 5 â€“ Solution CBSE Board Class IX Mathematics Sample Paper 5 â€“ Solution Time: 3 hrs Total Marks: 80 Section A 1. 11 0.011 1000 ? OR Yes, because 0 can be written as 0 1 which is of the form p q , where p and q are integers and q ? 0. 2. x + y = 2 â€¦â€¦(i) and x â€“ y = 4 â€¦â€¦(ii) Adding (i) and (ii), we get x = 3 Putting x = 3 in the equation (i), we get y = â€“1 ? x = 3 and y = â€“1 3. Since AOB is a Straight Line, ? ?AOB = 180° ? ?AOD + ?COD + ?COB = 180° ? x + 10° + x + x + 20° = 180° ? 3x = 150° ? x = 50° 4. If for one of the solutions of the equation ax + by + c = 0, x is negative and y is positive, then a portion of the above line definitely lies in the II Quadrant. As in the II Quadrant the x-axis contains only negative numbers and y-axis contains only positive numbers. OR Let the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y. Hence, the linear equation is 5x = 2y 5x â€“ 2y = 0. CBSE IX | Mathematics Sample Paper 5 â€“ Solution 5. Here the class marks are uniformly spaced, Therefore, Class size is the difference between any two consecutive class marks. ? Class Size = 52 â€“ 47 = 5 6. ?PSR = ?RQP = 125° (since PQRS is a parallelogram, then the opposite angles will be equal) ? ?PQT = 180°â€¦â€¦â€¦â€¦â€¦.. (PQT is a straight line) ? ?PQR + ?RQT = 180° ? 125° + ?RQT = 180° ? ?RQT = 55° Section B 7. a = 2 + 3 ? 11 a 23 ? ? = 1 2 3 2 3 2 3 ? ? ?? = ? ? 2 2 23 23 ? ? = 2 â€“ 3 So, a + 1 a = (2 + 3 ) + (2 â€“ 3 ) = 4 8. (i) I quadrant: (5, 2) (ii) II quadrant: (â€“5, 2) (iii) III quadrant: (â€“5, â€“2) (iv) IV quadrant: (5, â€“2) 9. Let â€˜lâ€™ be the length of the cube. Now, T.S.A. of the cube = 294 cm 2 â€¦(given) ? 6l 2 = 294 ? Side, l = 7 cm Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm 3 OR According to the question, Volume of one matchbox = 4 cm × 2.5 cm × 1.5 cm = 15 cm 3 Volume of 12 such matchboxes = 15 × 12 = 180 cm 3 CBSE IX | Mathematics Sample Paper 5 â€“ Solution 10. Given equation is 7x â€“ 5y = â€“3 i. (â€“1, â€“2) Putting x = â€“1 and y = â€“2 in the L.H.S. of the given equation, we get 7x â€“ 5y = 7(â€“1) â€“ 5(â€“2) = â€“7 + 10 = 3 ? R.H.S. ? L.H.S. ? R.H.S. Hence, (â€“1, â€“2) is not a solution of this equation. ii. (â€“4, â€“5) Putting x = â€“4 and y = â€“5 in the L.H.S. of the given equation, we get 7x â€“ 5y = 7(â€“4) â€“ 5(â€“5) = â€“28 + 25 = â€“3 = R.H.S. ? L.H.S. = R.H.S. Hence, (â€“4, â€“5) is a solution of this equation. OR An equation of the form ax + by + c = 0, where a, b, c are real numbers such that a ? 0 and b ? 0, is called a linear equation in two variables x and y. 11. Perimeter = 36 cm = 10 cm + 2(Length of an equal side) Length of an equal side = 13 cm Here, 36 s 18 2 ?? , and the sides are 10, 13 and 13. By Heron's formula, Area = 18 8 5 5 60 ? ? ? ? sq. cm 12. (â€“2x + 5y â€“ 3z) 2 = (â€“2x) 2 + (5y) 2 + (â€“3z) 2 + 2(â€“2x)(5y) + 2(5y)( â€“3z) + 2(â€“2x)( â€“3z) = 4x 2 + 25y 2 + 9z 2 â€“ 20xy â€“ 30yz + 12zx Section C 13. Let x= 0.001 Then, x= 0.001001001â€¦â€¦â€¦.. (i) Therefore, 1000x = 1.001001001â€¦â€¦â€¦â€¦â€¦ (ii) Subtracting (i) from (ii), we get, 999x = 1 ? x = 1 999 Hence, 0.001 = 1 999 Page 4 CBSE IX | Mathematics Sample Paper 5 â€“ Solution CBSE Board Class IX Mathematics Sample Paper 5 â€“ Solution Time: 3 hrs Total Marks: 80 Section A 1. 11 0.011 1000 ? OR Yes, because 0 can be written as 0 1 which is of the form p q , where p and q are integers and q ? 0. 2. x + y = 2 â€¦â€¦(i) and x â€“ y = 4 â€¦â€¦(ii) Adding (i) and (ii), we get x = 3 Putting x = 3 in the equation (i), we get y = â€“1 ? x = 3 and y = â€“1 3. Since AOB is a Straight Line, ? ?AOB = 180° ? ?AOD + ?COD + ?COB = 180° ? x + 10° + x + x + 20° = 180° ? 3x = 150° ? x = 50° 4. If for one of the solutions of the equation ax + by + c = 0, x is negative and y is positive, then a portion of the above line definitely lies in the II Quadrant. As in the II Quadrant the x-axis contains only negative numbers and y-axis contains only positive numbers. OR Let the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y. Hence, the linear equation is 5x = 2y 5x â€“ 2y = 0. CBSE IX | Mathematics Sample Paper 5 â€“ Solution 5. Here the class marks are uniformly spaced, Therefore, Class size is the difference between any two consecutive class marks. ? Class Size = 52 â€“ 47 = 5 6. ?PSR = ?RQP = 125° (since PQRS is a parallelogram, then the opposite angles will be equal) ? ?PQT = 180°â€¦â€¦â€¦â€¦â€¦.. (PQT is a straight line) ? ?PQR + ?RQT = 180° ? 125° + ?RQT = 180° ? ?RQT = 55° Section B 7. a = 2 + 3 ? 11 a 23 ? ? = 1 2 3 2 3 2 3 ? ? ?? = ? ? 2 2 23 23 ? ? = 2 â€“ 3 So, a + 1 a = (2 + 3 ) + (2 â€“ 3 ) = 4 8. (i) I quadrant: (5, 2) (ii) II quadrant: (â€“5, 2) (iii) III quadrant: (â€“5, â€“2) (iv) IV quadrant: (5, â€“2) 9. Let â€˜lâ€™ be the length of the cube. Now, T.S.A. of the cube = 294 cm 2 â€¦(given) ? 6l 2 = 294 ? Side, l = 7 cm Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm 3 OR According to the question, Volume of one matchbox = 4 cm × 2.5 cm × 1.5 cm = 15 cm 3 Volume of 12 such matchboxes = 15 × 12 = 180 cm 3 CBSE IX | Mathematics Sample Paper 5 â€“ Solution 10. Given equation is 7x â€“ 5y = â€“3 i. (â€“1, â€“2) Putting x = â€“1 and y = â€“2 in the L.H.S. of the given equation, we get 7x â€“ 5y = 7(â€“1) â€“ 5(â€“2) = â€“7 + 10 = 3 ? R.H.S. ? L.H.S. ? R.H.S. Hence, (â€“1, â€“2) is not a solution of this equation. ii. (â€“4, â€“5) Putting x = â€“4 and y = â€“5 in the L.H.S. of the given equation, we get 7x â€“ 5y = 7(â€“4) â€“ 5(â€“5) = â€“28 + 25 = â€“3 = R.H.S. ? L.H.S. = R.H.S. Hence, (â€“4, â€“5) is a solution of this equation. OR An equation of the form ax + by + c = 0, where a, b, c are real numbers such that a ? 0 and b ? 0, is called a linear equation in two variables x and y. 11. Perimeter = 36 cm = 10 cm + 2(Length of an equal side) Length of an equal side = 13 cm Here, 36 s 18 2 ?? , and the sides are 10, 13 and 13. By Heron's formula, Area = 18 8 5 5 60 ? ? ? ? sq. cm 12. (â€“2x + 5y â€“ 3z) 2 = (â€“2x) 2 + (5y) 2 + (â€“3z) 2 + 2(â€“2x)(5y) + 2(5y)( â€“3z) + 2(â€“2x)( â€“3z) = 4x 2 + 25y 2 + 9z 2 â€“ 20xy â€“ 30yz + 12zx Section C 13. Let x= 0.001 Then, x= 0.001001001â€¦â€¦â€¦.. (i) Therefore, 1000x = 1.001001001â€¦â€¦â€¦â€¦â€¦ (ii) Subtracting (i) from (ii), we get, 999x = 1 ? x = 1 999 Hence, 0.001 = 1 999 CBSE IX | Mathematics Sample Paper 5 â€“ Solution OR 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a a b b a b a a a b a b b a a b a b b a b b ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 2 2 2 2 2 2 2 a a b a b b ?? ? ?? ? ? ? ? 2 2 2 2 2 2 a a b a b b ?? ? ?? 2 2 2 2 2 2 a a b a b b ?? ? ?? 2 2 b a ? 14. Given: x = 2y + 6 or x - 2y - 6 = 0 We know that if a + b + c = 0, then a 3 + b 3 + c 3 = 3xyz Therefore, we have: (x) 3 + (â€“2y) 3 + (â€“6) 3 = 3x(â€“2y)( â€“6) Or, x 3 â€“ 8y 3 â€“ 36xy â€“ 216 = 0 OR Let p(x) = x 3 + 3x 2 + 3x + 1 and g(x) = x + p Now, g(x) = 0 x + p = 0 x = - p By the remainder theorem, when p(x) is divided by x + p then the remainder is p(-p). p(-p) = (-p) 3 + 3(-p) 2 + 3(-p) + 1 = â€“ p 3 + 3p 2 â€“ 3p + 1 Hence, remainder is â€“ p 3 + 3p 2 â€“ 3p + 1. Page 5 CBSE IX | Mathematics Sample Paper 5 â€“ Solution CBSE Board Class IX Mathematics Sample Paper 5 â€“ Solution Time: 3 hrs Total Marks: 80 Section A 1. 11 0.011 1000 ? OR Yes, because 0 can be written as 0 1 which is of the form p q , where p and q are integers and q ? 0. 2. x + y = 2 â€¦â€¦(i) and x â€“ y = 4 â€¦â€¦(ii) Adding (i) and (ii), we get x = 3 Putting x = 3 in the equation (i), we get y = â€“1 ? x = 3 and y = â€“1 3. Since AOB is a Straight Line, ? ?AOB = 180° ? ?AOD + ?COD + ?COB = 180° ? x + 10° + x + x + 20° = 180° ? 3x = 150° ? x = 50° 4. If for one of the solutions of the equation ax + by + c = 0, x is negative and y is positive, then a portion of the above line definitely lies in the II Quadrant. As in the II Quadrant the x-axis contains only negative numbers and y-axis contains only positive numbers. OR Let the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y. Hence, the linear equation is 5x = 2y 5x â€“ 2y = 0. CBSE IX | Mathematics Sample Paper 5 â€“ Solution 5. Here the class marks are uniformly spaced, Therefore, Class size is the difference between any two consecutive class marks. ? Class Size = 52 â€“ 47 = 5 6. ?PSR = ?RQP = 125° (since PQRS is a parallelogram, then the opposite angles will be equal) ? ?PQT = 180°â€¦â€¦â€¦â€¦â€¦.. (PQT is a straight line) ? ?PQR + ?RQT = 180° ? 125° + ?RQT = 180° ? ?RQT = 55° Section B 7. a = 2 + 3 ? 11 a 23 ? ? = 1 2 3 2 3 2 3 ? ? ?? = ? ? 2 2 23 23 ? ? = 2 â€“ 3 So, a + 1 a = (2 + 3 ) + (2 â€“ 3 ) = 4 8. (i) I quadrant: (5, 2) (ii) II quadrant: (â€“5, 2) (iii) III quadrant: (â€“5, â€“2) (iv) IV quadrant: (5, â€“2) 9. Let â€˜lâ€™ be the length of the cube. Now, T.S.A. of the cube = 294 cm 2 â€¦(given) ? 6l 2 = 294 ? Side, l = 7 cm Volume of cube = l × l × l = 7 × 7 × 7 = 343 cm 3 OR According to the question, Volume of one matchbox = 4 cm × 2.5 cm × 1.5 cm = 15 cm 3 Volume of 12 such matchboxes = 15 × 12 = 180 cm 3 CBSE IX | Mathematics Sample Paper 5 â€“ Solution 10. Given equation is 7x â€“ 5y = â€“3 i. (â€“1, â€“2) Putting x = â€“1 and y = â€“2 in the L.H.S. of the given equation, we get 7x â€“ 5y = 7(â€“1) â€“ 5(â€“2) = â€“7 + 10 = 3 ? R.H.S. ? L.H.S. ? R.H.S. Hence, (â€“1, â€“2) is not a solution of this equation. ii. (â€“4, â€“5) Putting x = â€“4 and y = â€“5 in the L.H.S. of the given equation, we get 7x â€“ 5y = 7(â€“4) â€“ 5(â€“5) = â€“28 + 25 = â€“3 = R.H.S. ? L.H.S. = R.H.S. Hence, (â€“4, â€“5) is a solution of this equation. OR An equation of the form ax + by + c = 0, where a, b, c are real numbers such that a ? 0 and b ? 0, is called a linear equation in two variables x and y. 11. Perimeter = 36 cm = 10 cm + 2(Length of an equal side) Length of an equal side = 13 cm Here, 36 s 18 2 ?? , and the sides are 10, 13 and 13. By Heron's formula, Area = 18 8 5 5 60 ? ? ? ? sq. cm 12. (â€“2x + 5y â€“ 3z) 2 = (â€“2x) 2 + (5y) 2 + (â€“3z) 2 + 2(â€“2x)(5y) + 2(5y)( â€“3z) + 2(â€“2x)( â€“3z) = 4x 2 + 25y 2 + 9z 2 â€“ 20xy â€“ 30yz + 12zx Section C 13. Let x= 0.001 Then, x= 0.001001001â€¦â€¦â€¦.. (i) Therefore, 1000x = 1.001001001â€¦â€¦â€¦â€¦â€¦ (ii) Subtracting (i) from (ii), we get, 999x = 1 ? x = 1 999 Hence, 0.001 = 1 999 CBSE IX | Mathematics Sample Paper 5 â€“ Solution OR 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a a b b a b a a a b a b b a a b a b b a b b ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 2 2 2 2 2 2 2 a a b a b b ?? ? ?? ? ? ? ? 2 2 2 2 2 2 a a b a b b ?? ? ?? 2 2 2 2 2 2 a a b a b b ?? ? ?? 2 2 b a ? 14. Given: x = 2y + 6 or x - 2y - 6 = 0 We know that if a + b + c = 0, then a 3 + b 3 + c 3 = 3xyz Therefore, we have: (x) 3 + (â€“2y) 3 + (â€“6) 3 = 3x(â€“2y)( â€“6) Or, x 3 â€“ 8y 3 â€“ 36xy â€“ 216 = 0 OR Let p(x) = x 3 + 3x 2 + 3x + 1 and g(x) = x + p Now, g(x) = 0 x + p = 0 x = - p By the remainder theorem, when p(x) is divided by x + p then the remainder is p(-p). p(-p) = (-p) 3 + 3(-p) 2 + 3(-p) + 1 = â€“ p 3 + 3p 2 â€“ 3p + 1 Hence, remainder is â€“ p 3 + 3p 2 â€“ 3p + 1. CBSE IX | Mathematics Sample Paper 5 â€“ Solution 15. In ?POR and ?QOS ?QPR = ?PQS (given) OP = OQ (O is the mid-point of PQ) ?POS = ?QOR (given) ?POS + x o = ?QOR + x o ?POR = ?QOS By ASA congruence rule, ?PQR ? ?QOS ? PR = QS (By CPCT) 16. Let p(z) = az 3 + 4z 2 + 3z â€“ 4 and q(z) = z 3 â€“ 4z + a When p(z) is divided by z â€“ 3, the remainder is given by p(3) = a × 3 3 + 4 × 3 2 + 3 × 3 â€“ 4 = 27a + 36 + 9 â€“ 4 = 27a + 41 â€¦.(i) When q(z) is divided by z â€“ 3, the remainder is given by q(3) = 3 3 â€“ 4 × 3 + a = 27 â€“ 12 + a = 15 + a â€¦.(ii) Given that p(3) = q(3). So, from (i) and (ii), we have 27a + 41 = 15 + a 27a â€“ a = â€“41 + 15 26a = â€“26 a = â€“1Read More

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