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CBSE Sample Paper Solutions 1 Term 1 - Class 6 Maths
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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).
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FAQs on CBSE Sample Paper Solutions 1 Term 1 - Class 6 Maths
| 1. What is CBSE? |  |
Ans. CBSE stands for Central Board of Secondary Education. It is a national level board of education in India that conducts examinations for Class 10 and Class 12 students in affiliated schools across the country.
| 2. What is the syllabus for Class 6 Math under CBSE? | |
Ans. The syllabus for Class 6 Math under CBSE includes topics such as knowing numbers, whole numbers, playing with numbers, basic geometrical ideas, understanding elementary shapes, integers, fractions, decimals, data handling, mensuration, algebra, ratio and proportion, symmetry, practical geometry, and more.
| 3. How can I prepare for my Class 6 Math exam? | |
Ans. To prepare for your Class 6 Math exam, you can follow these tips:
- Understand the concepts thoroughly by referring to your textbook and class notes.
- Practice solving different types of problems to improve your problem-solving skills.
- Make a study schedule and allocate specific time for Math practice.
- Take help from your teacher or classmates if you have any doubts or queries.
- Solve previous years' question papers to get familiar with the exam pattern.
- Revise regularly and solve sample papers to assess your preparation level.
| 4. What is the marking scheme for the Class 6 Math exam? | |
Ans. The marking scheme for the Class 6 Math exam may vary from school to school. Generally, it includes a combination of theory questions and practical problems. The marks distribution for different sections may also vary. It is best to refer to your school's guidelines or consult your teacher for the specific marking scheme.
| 5. Are there any recommended reference books for Class 6 Math CBSE exam preparation? | |
Ans. While the NCERT textbook is the primary resource for Class 6 Math CBSE exam preparation, there are several additional reference books available in the market. Some popular ones include "Mathematics for Class 6" by R.S. Aggarwal, "Mathematics for Class 6" by R.D. Sharma, and "Mathematics Olympiad Class 6: Practice Book" by Arihant Experts. It is advisable to consult your teacher or school for any recommended reference books.
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