Page 1 CBSE VI | Mathematics Sample Paper 1 – Solution CBSE Board Class VI Mathematics Term I Sample Paper 1 – Solution Time: 2 ½ hours Total Marks: 80 Section A 1. Correct answer: A On the given number line, from 8, five steps are moved towards the left. Thus, the number line represents 8 - 5 = 3. 2. Correct answer: A According to distributive law of multiplication over addition, we have: 12 × (45 + 30) = (12 × 45) + (12 × 30) 3. Correct answer: B 267 can be estimated as 270. 132 can be estimated as 130. Thus the required estimated sum = 270 + 130 = 400 4. Correct answer: B We have 10 = 2 × 5 18 = 2 × 3 × 3 HCF of 10 and 18 is 2. Thus, 2 is the required number. 5. Correct answer: A To convert into mixed fraction first divide numerator by denominator. The quotient is taken as the whole number part of mixed fraction. Remainder obtained is taken as the numerator and divisor as the denominator of the fractional part of the mixed fraction. Therefore, ? 52 1 33 6. Correct answer: D A region in the interior of a circle enclosed by an arc on one side and a pair of radii on the other two sides is called a sector of the circle. Page 2 CBSE VI | Mathematics Sample Paper 1 – Solution CBSE Board Class VI Mathematics Term I Sample Paper 1 – Solution Time: 2 ½ hours Total Marks: 80 Section A 1. Correct answer: A On the given number line, from 8, five steps are moved towards the left. Thus, the number line represents 8 - 5 = 3. 2. Correct answer: A According to distributive law of multiplication over addition, we have: 12 × (45 + 30) = (12 × 45) + (12 × 30) 3. Correct answer: B 267 can be estimated as 270. 132 can be estimated as 130. Thus the required estimated sum = 270 + 130 = 400 4. Correct answer: B We have 10 = 2 × 5 18 = 2 × 3 × 3 HCF of 10 and 18 is 2. Thus, 2 is the required number. 5. Correct answer: A To convert into mixed fraction first divide numerator by denominator. The quotient is taken as the whole number part of mixed fraction. Remainder obtained is taken as the numerator and divisor as the denominator of the fractional part of the mixed fraction. Therefore, ? 52 1 33 6. Correct answer: D A region in the interior of a circle enclosed by an arc on one side and a pair of radii on the other two sides is called a sector of the circle. CBSE VI | Mathematics Sample Paper 1 – Solution 7. Correct answer: D One crore can be written as 1,00,00,000. One thousand can be written as 1000. So, 10000 times one thousand would make one crore. 8. Correct answer: A There are 1000 + 1 = 1001 whole numbers upto 1000. i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ........., 1000 9. Correct answer: C ( –42) + ( –35) = –42 – 35 = –77 10. Correct answer: B Fifth multiple of 18 = 18 × 5 = 90 11. Correct answer: A 12. Correct answer: B The English alphabet Z represents an open curve. Section B 13. Place value of 9 at the Ten Lakhs place = 9000000 Place value of 9 at the hundreds place = 900 Difference = 9000000 – 900 = 8999100 14. Radius of a circle is a line joining the center of circle to any point on the circle. So, the radii drawn in the given figure are OP, OQ and OR. 15. The number of vertices in the given shapes: (i) Sphere: 0 (ii) Cylinder: 0 (iii) Cone: 1 (iv) Pyramid: 5 Page 3 CBSE VI | Mathematics Sample Paper 1 – Solution CBSE Board Class VI Mathematics Term I Sample Paper 1 – Solution Time: 2 ½ hours Total Marks: 80 Section A 1. Correct answer: A On the given number line, from 8, five steps are moved towards the left. Thus, the number line represents 8 - 5 = 3. 2. Correct answer: A According to distributive law of multiplication over addition, we have: 12 × (45 + 30) = (12 × 45) + (12 × 30) 3. Correct answer: B 267 can be estimated as 270. 132 can be estimated as 130. Thus the required estimated sum = 270 + 130 = 400 4. Correct answer: B We have 10 = 2 × 5 18 = 2 × 3 × 3 HCF of 10 and 18 is 2. Thus, 2 is the required number. 5. Correct answer: A To convert into mixed fraction first divide numerator by denominator. The quotient is taken as the whole number part of mixed fraction. Remainder obtained is taken as the numerator and divisor as the denominator of the fractional part of the mixed fraction. Therefore, ? 52 1 33 6. Correct answer: D A region in the interior of a circle enclosed by an arc on one side and a pair of radii on the other two sides is called a sector of the circle. CBSE VI | Mathematics Sample Paper 1 – Solution 7. Correct answer: D One crore can be written as 1,00,00,000. One thousand can be written as 1000. So, 10000 times one thousand would make one crore. 8. Correct answer: A There are 1000 + 1 = 1001 whole numbers upto 1000. i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ........., 1000 9. Correct answer: C ( –42) + ( –35) = –42 – 35 = –77 10. Correct answer: B Fifth multiple of 18 = 18 × 5 = 90 11. Correct answer: A 12. Correct answer: B The English alphabet Z represents an open curve. Section B 13. Place value of 9 at the Ten Lakhs place = 9000000 Place value of 9 at the hundreds place = 900 Difference = 9000000 – 900 = 8999100 14. Radius of a circle is a line joining the center of circle to any point on the circle. So, the radii drawn in the given figure are OP, OQ and OR. 15. The number of vertices in the given shapes: (i) Sphere: 0 (ii) Cylinder: 0 (iii) Cone: 1 (iv) Pyramid: 5 CBSE VI | Mathematics Sample Paper 1 – Solution 16. Anna is 7 feet above sea level. She jumps 3 feet down and walks another 2 feet down. Total distance travelled downwards = 3 + 2 = 5 feet. 17. ( –13) + ( –19) + (+15) + ( –10) = –13 – 19 + 15 – 10 = –13 – 19 – 10 + 15 = –42 + 15 = –27 18. A 9-digit numeral in Indian system = 94,50,27,983 In International system: 945,027,983 - Nine hundred forty five million twenty seven thousand nine hundred eighty three. 19. (i) If =31 and =11 then, = + = 31+11 = 42 (ii) If =45 and =61 then, = - = 61-45 = 16 20. Given number is 1258. Its unit digit is 8, which is divisible by 2. So, 1258 is divisible by 2. Sum of its digits = 1 + 2 + 5 + 8 = 16, which is not divisible by 3. So, 1258 is not divisible by 3. Since, 1258 is divisible by 2 but not by 3, it is not divisible by 6. 21. Starting from zero, a jump of 8 units is made to the right to reach 8. Then, 3 jumps (each of 1 unit i.e. from 8 to 7, 7 to 6, 6 to 5) are taken to the left to reach 5. So, we conclude that 8 – 3 = 5 Page 4 CBSE VI | Mathematics Sample Paper 1 – Solution CBSE Board Class VI Mathematics Term I Sample Paper 1 – Solution Time: 2 ½ hours Total Marks: 80 Section A 1. Correct answer: A On the given number line, from 8, five steps are moved towards the left. Thus, the number line represents 8 - 5 = 3. 2. Correct answer: A According to distributive law of multiplication over addition, we have: 12 × (45 + 30) = (12 × 45) + (12 × 30) 3. Correct answer: B 267 can be estimated as 270. 132 can be estimated as 130. Thus the required estimated sum = 270 + 130 = 400 4. Correct answer: B We have 10 = 2 × 5 18 = 2 × 3 × 3 HCF of 10 and 18 is 2. Thus, 2 is the required number. 5. Correct answer: A To convert into mixed fraction first divide numerator by denominator. The quotient is taken as the whole number part of mixed fraction. Remainder obtained is taken as the numerator and divisor as the denominator of the fractional part of the mixed fraction. Therefore, ? 52 1 33 6. Correct answer: D A region in the interior of a circle enclosed by an arc on one side and a pair of radii on the other two sides is called a sector of the circle. CBSE VI | Mathematics Sample Paper 1 – Solution 7. Correct answer: D One crore can be written as 1,00,00,000. One thousand can be written as 1000. So, 10000 times one thousand would make one crore. 8. Correct answer: A There are 1000 + 1 = 1001 whole numbers upto 1000. i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ........., 1000 9. Correct answer: C ( –42) + ( –35) = –42 – 35 = –77 10. Correct answer: B Fifth multiple of 18 = 18 × 5 = 90 11. Correct answer: A 12. Correct answer: B The English alphabet Z represents an open curve. Section B 13. Place value of 9 at the Ten Lakhs place = 9000000 Place value of 9 at the hundreds place = 900 Difference = 9000000 – 900 = 8999100 14. Radius of a circle is a line joining the center of circle to any point on the circle. So, the radii drawn in the given figure are OP, OQ and OR. 15. The number of vertices in the given shapes: (i) Sphere: 0 (ii) Cylinder: 0 (iii) Cone: 1 (iv) Pyramid: 5 CBSE VI | Mathematics Sample Paper 1 – Solution 16. Anna is 7 feet above sea level. She jumps 3 feet down and walks another 2 feet down. Total distance travelled downwards = 3 + 2 = 5 feet. 17. ( –13) + ( –19) + (+15) + ( –10) = –13 – 19 + 15 – 10 = –13 – 19 – 10 + 15 = –42 + 15 = –27 18. A 9-digit numeral in Indian system = 94,50,27,983 In International system: 945,027,983 - Nine hundred forty five million twenty seven thousand nine hundred eighty three. 19. (i) If =31 and =11 then, = + = 31+11 = 42 (ii) If =45 and =61 then, = - = 61-45 = 16 20. Given number is 1258. Its unit digit is 8, which is divisible by 2. So, 1258 is divisible by 2. Sum of its digits = 1 + 2 + 5 + 8 = 16, which is not divisible by 3. So, 1258 is not divisible by 3. Since, 1258 is divisible by 2 but not by 3, it is not divisible by 6. 21. Starting from zero, a jump of 8 units is made to the right to reach 8. Then, 3 jumps (each of 1 unit i.e. from 8 to 7, 7 to 6, 6 to 5) are taken to the left to reach 5. So, we conclude that 8 – 3 = 5 CBSE VI | Mathematics Sample Paper 1 – Solution 22. (i) –9 > –15 (ii) –10 < 10 (iii) 0 < 3 (iv) –28 < 17 23. Since the sum of all three angles of a triangle is 180 o . We have, ?X + ?Y + ?Z = 180 o Or, ?X + 60 o + 50 o = 180 o Or, ?X + 110 o = 180 o Or, ?X = 180 o – 110 o Hence, ?X = 70 o 24. Using distributive property of multiplication over addition, we have: 101 × 33 = (100 + 1) × 33 = 3300 + 33 = 3333 101 × 333 = (100 + 1) × 333 = 33300 + 333 = 33633 101 × 3333 = (100 + 1) × 3333 = 333300 + 3333 = 336633 101 × 33333 = (100 + 1) × 33333 = 3333300 + 33333 = 3366633 Section C 25. 3 35 Cost of notebook Rs. 8 Rs. 44 ?? 2 52 Cost of pen Rs. 10 Rs. 55 ?? LCM of 4 and 5 = (2 × 2 × 5) = 20 Total cost of both the items 35 52 Rs. 45 35 5 52 4 Rs. 20 20 175 208 Rs. 20 20 383 Rs. 20 3 Rs. 19 20 ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ? ? Page 5 CBSE VI | Mathematics Sample Paper 1 – Solution CBSE Board Class VI Mathematics Term I Sample Paper 1 – Solution Time: 2 ½ hours Total Marks: 80 Section A 1. Correct answer: A On the given number line, from 8, five steps are moved towards the left. Thus, the number line represents 8 - 5 = 3. 2. Correct answer: A According to distributive law of multiplication over addition, we have: 12 × (45 + 30) = (12 × 45) + (12 × 30) 3. Correct answer: B 267 can be estimated as 270. 132 can be estimated as 130. Thus the required estimated sum = 270 + 130 = 400 4. Correct answer: B We have 10 = 2 × 5 18 = 2 × 3 × 3 HCF of 10 and 18 is 2. Thus, 2 is the required number. 5. Correct answer: A To convert into mixed fraction first divide numerator by denominator. The quotient is taken as the whole number part of mixed fraction. Remainder obtained is taken as the numerator and divisor as the denominator of the fractional part of the mixed fraction. Therefore, ? 52 1 33 6. Correct answer: D A region in the interior of a circle enclosed by an arc on one side and a pair of radii on the other two sides is called a sector of the circle. CBSE VI | Mathematics Sample Paper 1 – Solution 7. Correct answer: D One crore can be written as 1,00,00,000. One thousand can be written as 1000. So, 10000 times one thousand would make one crore. 8. Correct answer: A There are 1000 + 1 = 1001 whole numbers upto 1000. i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ........., 1000 9. Correct answer: C ( –42) + ( –35) = –42 – 35 = –77 10. Correct answer: B Fifth multiple of 18 = 18 × 5 = 90 11. Correct answer: A 12. Correct answer: B The English alphabet Z represents an open curve. Section B 13. Place value of 9 at the Ten Lakhs place = 9000000 Place value of 9 at the hundreds place = 900 Difference = 9000000 – 900 = 8999100 14. Radius of a circle is a line joining the center of circle to any point on the circle. So, the radii drawn in the given figure are OP, OQ and OR. 15. The number of vertices in the given shapes: (i) Sphere: 0 (ii) Cylinder: 0 (iii) Cone: 1 (iv) Pyramid: 5 CBSE VI | Mathematics Sample Paper 1 – Solution 16. Anna is 7 feet above sea level. She jumps 3 feet down and walks another 2 feet down. Total distance travelled downwards = 3 + 2 = 5 feet. 17. ( –13) + ( –19) + (+15) + ( –10) = –13 – 19 + 15 – 10 = –13 – 19 – 10 + 15 = –42 + 15 = –27 18. A 9-digit numeral in Indian system = 94,50,27,983 In International system: 945,027,983 - Nine hundred forty five million twenty seven thousand nine hundred eighty three. 19. (i) If =31 and =11 then, = + = 31+11 = 42 (ii) If =45 and =61 then, = - = 61-45 = 16 20. Given number is 1258. Its unit digit is 8, which is divisible by 2. So, 1258 is divisible by 2. Sum of its digits = 1 + 2 + 5 + 8 = 16, which is not divisible by 3. So, 1258 is not divisible by 3. Since, 1258 is divisible by 2 but not by 3, it is not divisible by 6. 21. Starting from zero, a jump of 8 units is made to the right to reach 8. Then, 3 jumps (each of 1 unit i.e. from 8 to 7, 7 to 6, 6 to 5) are taken to the left to reach 5. So, we conclude that 8 – 3 = 5 CBSE VI | Mathematics Sample Paper 1 – Solution 22. (i) –9 > –15 (ii) –10 < 10 (iii) 0 < 3 (iv) –28 < 17 23. Since the sum of all three angles of a triangle is 180 o . We have, ?X + ?Y + ?Z = 180 o Or, ?X + 60 o + 50 o = 180 o Or, ?X + 110 o = 180 o Or, ?X = 180 o – 110 o Hence, ?X = 70 o 24. Using distributive property of multiplication over addition, we have: 101 × 33 = (100 + 1) × 33 = 3300 + 33 = 3333 101 × 333 = (100 + 1) × 333 = 33300 + 333 = 33633 101 × 3333 = (100 + 1) × 3333 = 333300 + 3333 = 336633 101 × 33333 = (100 + 1) × 33333 = 3333300 + 33333 = 3366633 Section C 25. 3 35 Cost of notebook Rs. 8 Rs. 44 ?? 2 52 Cost of pen Rs. 10 Rs. 55 ?? LCM of 4 and 5 = (2 × 2 × 5) = 20 Total cost of both the items 35 52 Rs. 45 35 5 52 4 Rs. 20 20 175 208 Rs. 20 20 383 Rs. 20 3 Rs. 19 20 ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ? ? CBSE VI | Mathematics Sample Paper 1 – Solution 26. The given fractions are . LCM of 3, 6, 9, 12 = (3 x 2 x 3 x 2) = 36 So, we convert each one of given fractions into an equivalent fraction having 36 as denominator. Now, Clearly, The given fractions in ascending order are . 27. Let the numbers be a and b. Then, a + b = 55 and ab = 5 × 120 = 600 Therefore, the required sum = 28. a) Lines p, q and r are intersecting lines. b) Point at which the lines meet is called the point of intersection. The point O represents the point of intersection. c) Infinite number of lines can pass through the point O (point of intersection).Read More

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