Page 1 CBSE VII | Mathematics Sample Paper 2 – Solution CBSE Board Class VII Mathematics Term II Sample Paper 2 – Solution Time: 2 ½ hours Total Marks: 80 Section A 1. Correct answer: B 2. Correct answer: B For the option (B), 1 is the only common factor between numerator and denominator and the denominator is a positive integer, hence it is in the standard form. 3. Correct answer: D 6 components - three angles and three sides 4. Correct answer: A Cylinder 5. Correct answer: B -x + 1 is an example of binomial as it contains two terms. 6. Correct answer: C 7. Correct answer: A Hence, order of rotational symmetry is 2. Page 2 CBSE VII | Mathematics Sample Paper 2 – Solution CBSE Board Class VII Mathematics Term II Sample Paper 2 – Solution Time: 2 ½ hours Total Marks: 80 Section A 1. Correct answer: B 2. Correct answer: B For the option (B), 1 is the only common factor between numerator and denominator and the denominator is a positive integer, hence it is in the standard form. 3. Correct answer: D 6 components - three angles and three sides 4. Correct answer: A Cylinder 5. Correct answer: B -x + 1 is an example of binomial as it contains two terms. 6. Correct answer: C 7. Correct answer: A Hence, order of rotational symmetry is 2. CBSE VII | Mathematics Sample Paper 2 – Solution 8. Correct answer: C The length of given cuboid is 6 units. 9. Correct answer: D 10. Correct answer: A Perimeter of rhombus = 28 cm Thus, 4 x (side) = 28 cm Side = 28 4 = 7 cm 11. Correct answer: A (128 ÷32) ÷ (-4) = 4 ÷ (-4) = -1 12. Correct answer: A 2x + 3 = 7 If we transpose 3 to RHS, then the term with variable will remain on one side and the constants will be on the other side. So, the first step is to transpose 3 to RHS. i.e. 2x = 7 – 3 Section B 13. Percentage of marks scored by Rahul = 40 100 80% 50 ?? Percentage of marks scored by Rohan = 75 100 75% 100 ?? Hence, Rahul scored more marks than Rohan. Page 3 CBSE VII | Mathematics Sample Paper 2 – Solution CBSE Board Class VII Mathematics Term II Sample Paper 2 – Solution Time: 2 ½ hours Total Marks: 80 Section A 1. Correct answer: B 2. Correct answer: B For the option (B), 1 is the only common factor between numerator and denominator and the denominator is a positive integer, hence it is in the standard form. 3. Correct answer: D 6 components - three angles and three sides 4. Correct answer: A Cylinder 5. Correct answer: B -x + 1 is an example of binomial as it contains two terms. 6. Correct answer: C 7. Correct answer: A Hence, order of rotational symmetry is 2. CBSE VII | Mathematics Sample Paper 2 – Solution 8. Correct answer: C The length of given cuboid is 6 units. 9. Correct answer: D 10. Correct answer: A Perimeter of rhombus = 28 cm Thus, 4 x (side) = 28 cm Side = 28 4 = 7 cm 11. Correct answer: A (128 ÷32) ÷ (-4) = 4 ÷ (-4) = -1 12. Correct answer: A 2x + 3 = 7 If we transpose 3 to RHS, then the term with variable will remain on one side and the constants will be on the other side. So, the first step is to transpose 3 to RHS. i.e. 2x = 7 – 3 Section B 13. Percentage of marks scored by Rahul = 40 100 80% 50 ?? Percentage of marks scored by Rohan = 75 100 75% 100 ?? Hence, Rahul scored more marks than Rohan. CBSE VII | Mathematics Sample Paper 2 – Solution 14. In the given number line, the rational number between 0 and -1 would be 1 2 ? . Rational number between 1 and 2 is 3 2 as 1 can be written as 2 2 . So, the next point would be 3 2 and 2 can be written as 4 2 , which is the same. Thus, the number line representing the missing values is as follows: 15. (a) ?A = 120 o , ?B = 90 o and AB = 8 cm Since, ?A + ?B = 120 o + 90 o = 210 o That is the sum of two angles is more than 180 o . Hence, the triangle is not possible. (b) ?P = 90 o , ?Q = 90 o and PQ = 9 cm Since, ?P + ?Q = 90 o + 90 o = 180 o That is the sum of two angles is equal to 180 o . And the measure of third angle is zero. Hence, the triangle is not possible. 16. Page 4 CBSE VII | Mathematics Sample Paper 2 – Solution CBSE Board Class VII Mathematics Term II Sample Paper 2 – Solution Time: 2 ½ hours Total Marks: 80 Section A 1. Correct answer: B 2. Correct answer: B For the option (B), 1 is the only common factor between numerator and denominator and the denominator is a positive integer, hence it is in the standard form. 3. Correct answer: D 6 components - three angles and three sides 4. Correct answer: A Cylinder 5. Correct answer: B -x + 1 is an example of binomial as it contains two terms. 6. Correct answer: C 7. Correct answer: A Hence, order of rotational symmetry is 2. CBSE VII | Mathematics Sample Paper 2 – Solution 8. Correct answer: C The length of given cuboid is 6 units. 9. Correct answer: D 10. Correct answer: A Perimeter of rhombus = 28 cm Thus, 4 x (side) = 28 cm Side = 28 4 = 7 cm 11. Correct answer: A (128 ÷32) ÷ (-4) = 4 ÷ (-4) = -1 12. Correct answer: A 2x + 3 = 7 If we transpose 3 to RHS, then the term with variable will remain on one side and the constants will be on the other side. So, the first step is to transpose 3 to RHS. i.e. 2x = 7 – 3 Section B 13. Percentage of marks scored by Rahul = 40 100 80% 50 ?? Percentage of marks scored by Rohan = 75 100 75% 100 ?? Hence, Rahul scored more marks than Rohan. CBSE VII | Mathematics Sample Paper 2 – Solution 14. In the given number line, the rational number between 0 and -1 would be 1 2 ? . Rational number between 1 and 2 is 3 2 as 1 can be written as 2 2 . So, the next point would be 3 2 and 2 can be written as 4 2 , which is the same. Thus, the number line representing the missing values is as follows: 15. (a) ?A = 120 o , ?B = 90 o and AB = 8 cm Since, ?A + ?B = 120 o + 90 o = 210 o That is the sum of two angles is more than 180 o . Hence, the triangle is not possible. (b) ?P = 90 o , ?Q = 90 o and PQ = 9 cm Since, ?P + ?Q = 90 o + 90 o = 180 o That is the sum of two angles is equal to 180 o . And the measure of third angle is zero. Hence, the triangle is not possible. 16. CBSE VII | Mathematics Sample Paper 2 – Solution 17. Required sum: (6m - 7n - 5p) + (-4m - 9n + 6p) + (-4m - 9n + 6p) = 6m - 7n - 5p - 4m - 9n + 6p - 4m - 9n + 6p = (6m - 4m - 4m) + (-7n - 9n - 9n) + (-5p + 6p + 6p) = (6 - 4 - 4)m + (-7 - 9 - 9)n + (-5 + 6 + 6)p = -2m - 25n + 7p 18. 19. The other holes are as below: i. ii. Page 5 CBSE VII | Mathematics Sample Paper 2 – Solution CBSE Board Class VII Mathematics Term II Sample Paper 2 – Solution Time: 2 ½ hours Total Marks: 80 Section A 1. Correct answer: B 2. Correct answer: B For the option (B), 1 is the only common factor between numerator and denominator and the denominator is a positive integer, hence it is in the standard form. 3. Correct answer: D 6 components - three angles and three sides 4. Correct answer: A Cylinder 5. Correct answer: B -x + 1 is an example of binomial as it contains two terms. 6. Correct answer: C 7. Correct answer: A Hence, order of rotational symmetry is 2. CBSE VII | Mathematics Sample Paper 2 – Solution 8. Correct answer: C The length of given cuboid is 6 units. 9. Correct answer: D 10. Correct answer: A Perimeter of rhombus = 28 cm Thus, 4 x (side) = 28 cm Side = 28 4 = 7 cm 11. Correct answer: A (128 ÷32) ÷ (-4) = 4 ÷ (-4) = -1 12. Correct answer: A 2x + 3 = 7 If we transpose 3 to RHS, then the term with variable will remain on one side and the constants will be on the other side. So, the first step is to transpose 3 to RHS. i.e. 2x = 7 – 3 Section B 13. Percentage of marks scored by Rahul = 40 100 80% 50 ?? Percentage of marks scored by Rohan = 75 100 75% 100 ?? Hence, Rahul scored more marks than Rohan. CBSE VII | Mathematics Sample Paper 2 – Solution 14. In the given number line, the rational number between 0 and -1 would be 1 2 ? . Rational number between 1 and 2 is 3 2 as 1 can be written as 2 2 . So, the next point would be 3 2 and 2 can be written as 4 2 , which is the same. Thus, the number line representing the missing values is as follows: 15. (a) ?A = 120 o , ?B = 90 o and AB = 8 cm Since, ?A + ?B = 120 o + 90 o = 210 o That is the sum of two angles is more than 180 o . Hence, the triangle is not possible. (b) ?P = 90 o , ?Q = 90 o and PQ = 9 cm Since, ?P + ?Q = 90 o + 90 o = 180 o That is the sum of two angles is equal to 180 o . And the measure of third angle is zero. Hence, the triangle is not possible. 16. CBSE VII | Mathematics Sample Paper 2 – Solution 17. Required sum: (6m - 7n - 5p) + (-4m - 9n + 6p) + (-4m - 9n + 6p) = 6m - 7n - 5p - 4m - 9n + 6p - 4m - 9n + 6p = (6m - 4m - 4m) + (-7n - 9n - 9n) + (-5p + 6p + 6p) = (6 - 4 - 4)m + (-7 - 9 - 9)n + (-5 + 6 + 6)p = -2m - 25n + 7p 18. 19. The other holes are as below: i. ii. CBSE VII | Mathematics Sample Paper 2 – Solution 20. 1. 1. First take an isometric dot sheet. 2. Draw the line segment AB and AD of length 5 units and 3 units respectively. 3. For the height draw the line segment AG, BC and DE of 6 units each. 4. Join EG and GC. 5. Again draw EF and CF of 5 units and 3 units respectively. 21. Mass of earth = 5,970,000,000,000,000,000,000,000 kg = 597 × 10000000000000000000000 kg = 597 × 10 22 kg = 5.97 × 10 24 kg 22. Number of cubes in first layer = 7 Number of cubes in second layer = 2 Hence, total number of cubes = 7 + 2 = 9 23. Average score = mean score ? ? ? ? ? ? ? ? ? ? ? ? ? Sum of all observations Mean= Total number of observations 12 23 10 77 15 78 90 54 23 10 1 11 393 11 35.7 24. 725 × (-35) + (-725) × 65 = 725 × (-35) - 725 × 65 = 725 x (-35 - 65) [Using distributive property] = 725 × (-100) = -72500Read More

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