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# Sample Solution Paper 1 - Term- 1 Mathematics, Class 6 Class 6 Notes | EduRev

## Class 6 : Sample Solution Paper 1 - Term- 1 Mathematics, Class 6 Class 6 Notes | EduRev

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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

On the given number line, from 8, five steps are moved towards the left.
Thus, the number line represents 8 - 5 = 3.

According to distributive law of multiplication over addition, we have:
12 × (45 + 30) = (12 × 45) + (12 × 30)
267 can be estimated as 270.
132 can be estimated as 130.
Thus the required estimated sum = 270 + 130 = 400

We have
10 = 2 × 5
18 = 2 × 3 × 3
HCF of 10 and 18 is 2.
Thus, 2 is the required number.

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

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

On the given number line, from 8, five steps are moved towards the left.
Thus, the number line represents 8 - 5 = 3.

According to distributive law of multiplication over addition, we have:
12 × (45 + 30) = (12 × 45) + (12 × 30)
267 can be estimated as 270.
132 can be estimated as 130.
Thus the required estimated sum = 270 + 130 = 400

We have
10 = 2 × 5
18 = 2 × 3 × 3
HCF of 10 and 18 is 2.
Thus, 2 is the required number.

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

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

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.

There are 1000 + 1 = 1001 whole numbers upto 1000.
i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ........., 1000
( –42) + ( –35) = –42 – 35 = –77

Fifth multiple of 18 = 18 × 5 = 90

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

On the given number line, from 8, five steps are moved towards the left.
Thus, the number line represents 8 - 5 = 3.

According to distributive law of multiplication over addition, we have:
12 × (45 + 30) = (12 × 45) + (12 × 30)
267 can be estimated as 270.
132 can be estimated as 130.
Thus the required estimated sum = 270 + 130 = 400

We have
10 = 2 × 5
18 = 2 × 3 × 3
HCF of 10 and 18 is 2.
Thus, 2 is the required number.

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

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

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.

There are 1000 + 1 = 1001 whole numbers upto 1000.
i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ........., 1000
( –42) + ( –35) = –42 – 35 = –77

Fifth multiple of 18 = 18 × 5 = 90

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

On the given number line, from 8, five steps are moved towards the left.
Thus, the number line represents 8 - 5 = 3.

According to distributive law of multiplication over addition, we have:
12 × (45 + 30) = (12 × 45) + (12 × 30)
267 can be estimated as 270.
132 can be estimated as 130.
Thus the required estimated sum = 270 + 130 = 400

We have
10 = 2 × 5
18 = 2 × 3 × 3
HCF of 10 and 18 is 2.
Thus, 2 is the required number.

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

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

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.

There are 1000 + 1 = 1001 whole numbers upto 1000.
i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ........., 1000
( –42) + ( –35) = –42 – 35 = –77

Fifth multiple of 18 = 18 × 5 = 90

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

On the given number line, from 8, five steps are moved towards the left.
Thus, the number line represents 8 - 5 = 3.

According to distributive law of multiplication over addition, we have:
12 × (45 + 30) = (12 × 45) + (12 × 30)
267 can be estimated as 270.
132 can be estimated as 130.
Thus the required estimated sum = 270 + 130 = 400

We have
10 = 2 × 5
18 = 2 × 3 × 3
HCF of 10 and 18 is 2.
Thus, 2 is the required number.

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

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

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.

There are 1000 + 1 = 1001 whole numbers upto 1000.
i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ........., 1000
( –42) + ( –35) = –42 – 35 = –77

Fifth multiple of 18 = 18 × 5 = 90

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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