Page 1 CBSE IX | Mathematics Sample Paper 3 â€“ Solution CBSE Board Class IX Mathematics Sample Paper 3 â€“ Solution Time: 3 hrs Total Marks: 80 Section A 1. ? ? ? ? ? ? ? ? 2 2 2 2 22 L.H.S 5 6 5 2 5 6 5 a b a 2ab b 5 2 30 6 11 2 30 On compairing with R.H.S, a + b 30 = 11 + 2 30 a 11 and b 2 ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? 2. Since, p(x) is a zero polynomial, ? p(x) = 0 ? cx + d = 0 ? cx = â€“d ? x = d c ? 3. In ?PTQ , base = PQ = 6 cm and height = 6 cm ? Area of ?PTQ = 1 2 ? base ? height = 1 2 × 6 × 6 = 18 cm 2 OR ar ( ? APB) = 1 2 x ar(parallelogram ABCD) (The area of a triangle is half that of a parallelogram on the same base and between the same parallels) ar( ? APB) = 1 2 x 60 cm 2 = 30 cm 2 Page 2 CBSE IX | Mathematics Sample Paper 3 â€“ Solution CBSE Board Class IX Mathematics Sample Paper 3 â€“ Solution Time: 3 hrs Total Marks: 80 Section A 1. ? ? ? ? ? ? ? ? 2 2 2 2 22 L.H.S 5 6 5 2 5 6 5 a b a 2ab b 5 2 30 6 11 2 30 On compairing with R.H.S, a + b 30 = 11 + 2 30 a 11 and b 2 ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? 2. Since, p(x) is a zero polynomial, ? p(x) = 0 ? cx + d = 0 ? cx = â€“d ? x = d c ? 3. In ?PTQ , base = PQ = 6 cm and height = 6 cm ? Area of ?PTQ = 1 2 ? base ? height = 1 2 × 6 × 6 = 18 cm 2 OR ar ( ? APB) = 1 2 x ar(parallelogram ABCD) (The area of a triangle is half that of a parallelogram on the same base and between the same parallels) ar( ? APB) = 1 2 x 60 cm 2 = 30 cm 2 CBSE IX | Mathematics Sample Paper 3 â€“ Solution 4. Substituting x = 1 2 and y = 0 in the equation 2x + y = 1 ? 2 ? 1 2 + 0 = 1 ? 1 = 1 Since, L.H.S = R.H.S The values of x and y are satisfying the given equation. Therefore, ( 1 2 , 0) is the solution of 2x + y = 1. OR (4, 19) is a solution of the equation y = ax + 3 then (4, 19) must satisfy the equation. ? 19 = 4x + 3 ? 4a = 16 ? a = 4 5. Median of a Triangle: A median of a triangle is the line segment that joins any vertex of the triangle with the midpoint of its opposite side. In the given Triangle, the Median from A meets the midpoint of the opposite side, BC at point D. 6. In a parallelogram, the sum of consecutive angles are Supplementary. Here ABCD is a parallelogram, ? ?A + ?B = 180° o 11 A B 90 22 ? ? ? ? ? o o oo o In AOB, AOB OBA OAB 180 BA AOB 180 ....( OA and OB are the angle bisectors of A and B) 22 AOB 90 180 AOB 90 ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Page 3 CBSE IX | Mathematics Sample Paper 3 â€“ Solution CBSE Board Class IX Mathematics Sample Paper 3 â€“ Solution Time: 3 hrs Total Marks: 80 Section A 1. ? ? ? ? ? ? ? ? 2 2 2 2 22 L.H.S 5 6 5 2 5 6 5 a b a 2ab b 5 2 30 6 11 2 30 On compairing with R.H.S, a + b 30 = 11 + 2 30 a 11 and b 2 ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? 2. Since, p(x) is a zero polynomial, ? p(x) = 0 ? cx + d = 0 ? cx = â€“d ? x = d c ? 3. In ?PTQ , base = PQ = 6 cm and height = 6 cm ? Area of ?PTQ = 1 2 ? base ? height = 1 2 × 6 × 6 = 18 cm 2 OR ar ( ? APB) = 1 2 x ar(parallelogram ABCD) (The area of a triangle is half that of a parallelogram on the same base and between the same parallels) ar( ? APB) = 1 2 x 60 cm 2 = 30 cm 2 CBSE IX | Mathematics Sample Paper 3 â€“ Solution 4. Substituting x = 1 2 and y = 0 in the equation 2x + y = 1 ? 2 ? 1 2 + 0 = 1 ? 1 = 1 Since, L.H.S = R.H.S The values of x and y are satisfying the given equation. Therefore, ( 1 2 , 0) is the solution of 2x + y = 1. OR (4, 19) is a solution of the equation y = ax + 3 then (4, 19) must satisfy the equation. ? 19 = 4x + 3 ? 4a = 16 ? a = 4 5. Median of a Triangle: A median of a triangle is the line segment that joins any vertex of the triangle with the midpoint of its opposite side. In the given Triangle, the Median from A meets the midpoint of the opposite side, BC at point D. 6. In a parallelogram, the sum of consecutive angles are Supplementary. Here ABCD is a parallelogram, ? ?A + ?B = 180° o 11 A B 90 22 ? ? ? ? ? o o oo o In AOB, AOB OBA OAB 180 BA AOB 180 ....( OA and OB are the angle bisectors of A and B) 22 AOB 90 180 AOB 90 ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? CBSE IX | Mathematics Sample Paper 3 â€“ Solution Section B 7. Let x = 0.975 0.975975975 ....(1) ? On multiplying both sides of equation (1) by 1000: 1000x = 975.975975 ....(2) On subtracting equation (1) from equation (2), 999x = 975 ? ? ? 975 325 x 999 333 8. 2 2 2 2 2 12 x 2 2x xx 11 x 2 2 x xx 11 x 2 x xx 11 x x 2 xx ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 9. Here ? ADC = y = ? ACD Ext. ? ACD = ? ABC + ? BAC ? 2 ? BAC = ? ACD = y ? ? BAC = y 2 ? y 2 + (180° â€“ 2y) = 180° â€“ 75° ? y 2 + 180° â€“ 2y = 180° â€“ 75 ? y 2 â€“ 2y = â€“ 75° ? â€“ 3y 2 = â€“75° ? y = 50° Page 4 CBSE IX | Mathematics Sample Paper 3 â€“ Solution CBSE Board Class IX Mathematics Sample Paper 3 â€“ Solution Time: 3 hrs Total Marks: 80 Section A 1. ? ? ? ? ? ? ? ? 2 2 2 2 22 L.H.S 5 6 5 2 5 6 5 a b a 2ab b 5 2 30 6 11 2 30 On compairing with R.H.S, a + b 30 = 11 + 2 30 a 11 and b 2 ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? 2. Since, p(x) is a zero polynomial, ? p(x) = 0 ? cx + d = 0 ? cx = â€“d ? x = d c ? 3. In ?PTQ , base = PQ = 6 cm and height = 6 cm ? Area of ?PTQ = 1 2 ? base ? height = 1 2 × 6 × 6 = 18 cm 2 OR ar ( ? APB) = 1 2 x ar(parallelogram ABCD) (The area of a triangle is half that of a parallelogram on the same base and between the same parallels) ar( ? APB) = 1 2 x 60 cm 2 = 30 cm 2 CBSE IX | Mathematics Sample Paper 3 â€“ Solution 4. Substituting x = 1 2 and y = 0 in the equation 2x + y = 1 ? 2 ? 1 2 + 0 = 1 ? 1 = 1 Since, L.H.S = R.H.S The values of x and y are satisfying the given equation. Therefore, ( 1 2 , 0) is the solution of 2x + y = 1. OR (4, 19) is a solution of the equation y = ax + 3 then (4, 19) must satisfy the equation. ? 19 = 4x + 3 ? 4a = 16 ? a = 4 5. Median of a Triangle: A median of a triangle is the line segment that joins any vertex of the triangle with the midpoint of its opposite side. In the given Triangle, the Median from A meets the midpoint of the opposite side, BC at point D. 6. In a parallelogram, the sum of consecutive angles are Supplementary. Here ABCD is a parallelogram, ? ?A + ?B = 180° o 11 A B 90 22 ? ? ? ? ? o o oo o In AOB, AOB OBA OAB 180 BA AOB 180 ....( OA and OB are the angle bisectors of A and B) 22 AOB 90 180 AOB 90 ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? CBSE IX | Mathematics Sample Paper 3 â€“ Solution Section B 7. Let x = 0.975 0.975975975 ....(1) ? On multiplying both sides of equation (1) by 1000: 1000x = 975.975975 ....(2) On subtracting equation (1) from equation (2), 999x = 975 ? ? ? 975 325 x 999 333 8. 2 2 2 2 2 12 x 2 2x xx 11 x 2 2 x xx 11 x 2 x xx 11 x x 2 xx ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 9. Here ? ADC = y = ? ACD Ext. ? ACD = ? ABC + ? BAC ? 2 ? BAC = ? ACD = y ? ? BAC = y 2 ? y 2 + (180° â€“ 2y) = 180° â€“ 75° ? y 2 + 180° â€“ 2y = 180° â€“ 75 ? y 2 â€“ 2y = â€“ 75° ? â€“ 3y 2 = â€“75° ? y = 50° CBSE IX | Mathematics Sample Paper 3 â€“ Solution 10. AD is the bisector of ?A ? ?BAD = ?CAD Exterior ?BDA > ?CAD ? ?BDA > ?BAD ? AB > BD (side opposite the bigger angle is longer) OR In ?PQT, we have PT = PQ â€¦ (1) In ?PQR, PQ + QR > PR PQ + QR > PT + TR PQ + QR > PQ + TR [Using (1)] QR > TR Hence, proved. 11. Let â€˜lâ€™ be the length of the cube. Now, T.S.A. of the cube = 294 cm 2 â€¦(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 OR Given that: Diagonal of a cube = 48 cm i.e., 3 x l = 48 [ Diagonal of cube = 3 x l ] l = 48 3 l = 48 3 = 16 = 4 cm ? Side (l) = 4 cm Page 5 CBSE IX | Mathematics Sample Paper 3 â€“ Solution CBSE Board Class IX Mathematics Sample Paper 3 â€“ Solution Time: 3 hrs Total Marks: 80 Section A 1. ? ? ? ? ? ? ? ? 2 2 2 2 22 L.H.S 5 6 5 2 5 6 5 a b a 2ab b 5 2 30 6 11 2 30 On compairing with R.H.S, a + b 30 = 11 + 2 30 a 11 and b 2 ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? 2. Since, p(x) is a zero polynomial, ? p(x) = 0 ? cx + d = 0 ? cx = â€“d ? x = d c ? 3. In ?PTQ , base = PQ = 6 cm and height = 6 cm ? Area of ?PTQ = 1 2 ? base ? height = 1 2 × 6 × 6 = 18 cm 2 OR ar ( ? APB) = 1 2 x ar(parallelogram ABCD) (The area of a triangle is half that of a parallelogram on the same base and between the same parallels) ar( ? APB) = 1 2 x 60 cm 2 = 30 cm 2 CBSE IX | Mathematics Sample Paper 3 â€“ Solution 4. Substituting x = 1 2 and y = 0 in the equation 2x + y = 1 ? 2 ? 1 2 + 0 = 1 ? 1 = 1 Since, L.H.S = R.H.S The values of x and y are satisfying the given equation. Therefore, ( 1 2 , 0) is the solution of 2x + y = 1. OR (4, 19) is a solution of the equation y = ax + 3 then (4, 19) must satisfy the equation. ? 19 = 4x + 3 ? 4a = 16 ? a = 4 5. Median of a Triangle: A median of a triangle is the line segment that joins any vertex of the triangle with the midpoint of its opposite side. In the given Triangle, the Median from A meets the midpoint of the opposite side, BC at point D. 6. In a parallelogram, the sum of consecutive angles are Supplementary. Here ABCD is a parallelogram, ? ?A + ?B = 180° o 11 A B 90 22 ? ? ? ? ? o o oo o In AOB, AOB OBA OAB 180 BA AOB 180 ....( OA and OB are the angle bisectors of A and B) 22 AOB 90 180 AOB 90 ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? CBSE IX | Mathematics Sample Paper 3 â€“ Solution Section B 7. Let x = 0.975 0.975975975 ....(1) ? On multiplying both sides of equation (1) by 1000: 1000x = 975.975975 ....(2) On subtracting equation (1) from equation (2), 999x = 975 ? ? ? 975 325 x 999 333 8. 2 2 2 2 2 12 x 2 2x xx 11 x 2 2 x xx 11 x 2 x xx 11 x x 2 xx ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 9. Here ? ADC = y = ? ACD Ext. ? ACD = ? ABC + ? BAC ? 2 ? BAC = ? ACD = y ? ? BAC = y 2 ? y 2 + (180° â€“ 2y) = 180° â€“ 75° ? y 2 + 180° â€“ 2y = 180° â€“ 75 ? y 2 â€“ 2y = â€“ 75° ? â€“ 3y 2 = â€“75° ? y = 50° CBSE IX | Mathematics Sample Paper 3 â€“ Solution 10. AD is the bisector of ?A ? ?BAD = ?CAD Exterior ?BDA > ?CAD ? ?BDA > ?BAD ? AB > BD (side opposite the bigger angle is longer) OR In ?PQT, we have PT = PQ â€¦ (1) In ?PQR, PQ + QR > PR PQ + QR > PT + TR PQ + QR > PQ + TR [Using (1)] QR > TR Hence, proved. 11. Let â€˜lâ€™ be the length of the cube. Now, T.S.A. of the cube = 294 cm 2 â€¦(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 OR Given that: Diagonal of a cube = 48 cm i.e., 3 x l = 48 [ Diagonal of cube = 3 x l ] l = 48 3 l = 48 3 = 16 = 4 cm ? Side (l) = 4 cm CBSE IX | Mathematics Sample Paper 3 â€“ Solution 12. 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. Section C 13. ? ? ?2 3 343 = (343) -2/3 = [(7) 3 } -2/3 = = = 1 49 OR 12 23 1 0.01 27 4 = 1 2 1 2 2 1 0.1 3 2 = 1 2 1 0.1 3 2 = 11 9 2 0.1 = 1 10 9 21 = 1 1 2 = 3 2Read More

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