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# Sample Solution Paper 5 - Term- 1 Mathematics, Class 6 Class 6 Notes | EduRev

## Class 6 : Sample Solution Paper 5 - Term- 1 Mathematics, Class 6 Class 6 Notes | EduRev

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CBSE VI | Mathematics
Sample Paper 5 – Solution

CBSE Board
Class VI Mathematics
Term I
Sample Paper 5 – Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

1 – 2 + 3 – 4 + 5 = 1 + 3 + 5 – 2 – 4 = 9 – 6 = 3

932, 923, 239, 293, 329, 392.

Out of the given numbers, 139 is a prime number as it has only two factors, namely, 1
and 139.

The HCF of 15 and 27 is 3. Hence,

To add 4 on number line, move 4 steps to the right of 0.

Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest
hundreds.
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its
nearest hundreds as 200.
So, the required estimation of the product is 50 × 200 = 10000.
Page 2

CBSE VI | Mathematics
Sample Paper 5 – Solution

CBSE Board
Class VI Mathematics
Term I
Sample Paper 5 – Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

1 – 2 + 3 – 4 + 5 = 1 + 3 + 5 – 2 – 4 = 9 – 6 = 3

932, 923, 239, 293, 329, 392.

Out of the given numbers, 139 is a prime number as it has only two factors, namely, 1
and 139.

The HCF of 15 and 27 is 3. Hence,

To add 4 on number line, move 4 steps to the right of 0.

Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest
hundreds.
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its
nearest hundreds as 200.
So, the required estimation of the product is 50 × 200 = 10000.

CBSE VI | Mathematics
Sample Paper 5 – Solution

Thus, 36 = 2 × 2 × 3 × 3

13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7

NO and PQ can be extended indefinitely on both sides, so they are lines. On extending,
it can be seen that they would meet at a point. Hence, they are intersecting lines.
One and only one line passes through any two given points.

267 can be estimated as 270.
132 can be estimated as 130.
Thus the required estimated sum = 270 + 130 = 400

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

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. The number of vertices in the given shapes:
i. Sphere : 0
ii. Cylinder : 0
Page 3

CBSE VI | Mathematics
Sample Paper 5 – Solution

CBSE Board
Class VI Mathematics
Term I
Sample Paper 5 – Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

1 – 2 + 3 – 4 + 5 = 1 + 3 + 5 – 2 – 4 = 9 – 6 = 3

932, 923, 239, 293, 329, 392.

Out of the given numbers, 139 is a prime number as it has only two factors, namely, 1
and 139.

The HCF of 15 and 27 is 3. Hence,

To add 4 on number line, move 4 steps to the right of 0.

Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest
hundreds.
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its
nearest hundreds as 200.
So, the required estimation of the product is 50 × 200 = 10000.

CBSE VI | Mathematics
Sample Paper 5 – Solution

Thus, 36 = 2 × 2 × 3 × 3

13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7

NO and PQ can be extended indefinitely on both sides, so they are lines. On extending,
it can be seen that they would meet at a point. Hence, they are intersecting lines.
One and only one line passes through any two given points.

267 can be estimated as 270.
132 can be estimated as 130.
Thus the required estimated sum = 270 + 130 = 400

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

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. The number of vertices in the given shapes:
i. Sphere : 0
ii. Cylinder : 0

CBSE VI | Mathematics
Sample Paper 5 – Solution

15. 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.

16. (–13) + (–19) + (+15) + (–10)
= –13 – 19 + 15 – 10
= –13 – 19 – 10 + 15
= –42 + 15
= –27

17. 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.

18. Two lines in the same plane which never intersect are called parallel lines.
Parallel lines maintain the same distance apart over their entire length.

19. Population of the village = 13295
Increase in population= Average growth – 1 = 399
Population in the successive year = 13295 + 399 = 13694

20. Prime factorisation of 455 =  5 × 7 × 13
Therefore, the dimensions of the cuboid are
5 cm, 7 cm, 13 cm.

Page 4

CBSE VI | Mathematics
Sample Paper 5 – Solution

CBSE Board
Class VI Mathematics
Term I
Sample Paper 5 – Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

1 – 2 + 3 – 4 + 5 = 1 + 3 + 5 – 2 – 4 = 9 – 6 = 3

932, 923, 239, 293, 329, 392.

Out of the given numbers, 139 is a prime number as it has only two factors, namely, 1
and 139.

The HCF of 15 and 27 is 3. Hence,

To add 4 on number line, move 4 steps to the right of 0.

Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest
hundreds.
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its
nearest hundreds as 200.
So, the required estimation of the product is 50 × 200 = 10000.

CBSE VI | Mathematics
Sample Paper 5 – Solution

Thus, 36 = 2 × 2 × 3 × 3

13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7

NO and PQ can be extended indefinitely on both sides, so they are lines. On extending,
it can be seen that they would meet at a point. Hence, they are intersecting lines.
One and only one line passes through any two given points.

267 can be estimated as 270.
132 can be estimated as 130.
Thus the required estimated sum = 270 + 130 = 400

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

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. The number of vertices in the given shapes:
i. Sphere : 0
ii. Cylinder : 0

CBSE VI | Mathematics
Sample Paper 5 – Solution

15. 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.

16. (–13) + (–19) + (+15) + (–10)
= –13 – 19 + 15 – 10
= –13 – 19 – 10 + 15
= –42 + 15
= –27

17. 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.

18. Two lines in the same plane which never intersect are called parallel lines.
Parallel lines maintain the same distance apart over their entire length.

19. Population of the village = 13295
Increase in population= Average growth – 1 = 399
Population in the successive year = 13295 + 399 = 13694

20. Prime factorisation of 455 =  5 × 7 × 13
Therefore, the dimensions of the cuboid are
5 cm, 7 cm, 13 cm.

CBSE VI | Mathematics
Sample Paper 5 – Solution

21. L.C.M. of 24, 45 and 54 = 2 × 2 × 2 × 3 × 3 × 3 × 5 = 1080

22.
21
53
72
?
37 7
72
37 2 7 7
7 2 2 7
74 49
14 14
123
14
11
8
14
??
??
??
??
??
?
?

23. Subtracting a negative number is the same as adding its positive integer.
? 7 – (–3) = 7 + 3 = 10

24. To find the H.C.F of 77, 105 and 231.

77 = 7 × 11
105 = 3 × 5 × 7
231 = 3 × 7 × 11
Hence, H.C.F of 77, 105 and 231 is = 7
Page 5

CBSE VI | Mathematics
Sample Paper 5 – Solution

CBSE Board
Class VI Mathematics
Term I
Sample Paper 5 – Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

1 – 2 + 3 – 4 + 5 = 1 + 3 + 5 – 2 – 4 = 9 – 6 = 3

932, 923, 239, 293, 329, 392.

Out of the given numbers, 139 is a prime number as it has only two factors, namely, 1
and 139.

The HCF of 15 and 27 is 3. Hence,

To add 4 on number line, move 4 steps to the right of 0.

Estimate the product by rounding off 52 to its nearest tens and 188 to its nearest
hundreds.
52 can be rounded off to its nearest tens as 50 and 188 can be rounded off to its
nearest hundreds as 200.
So, the required estimation of the product is 50 × 200 = 10000.

CBSE VI | Mathematics
Sample Paper 5 – Solution

Thus, 36 = 2 × 2 × 3 × 3

13 + (12 – 6 × 3) = 13 + (12 – 18) = 13 – 6 = 7

NO and PQ can be extended indefinitely on both sides, so they are lines. On extending,
it can be seen that they would meet at a point. Hence, they are intersecting lines.
One and only one line passes through any two given points.

267 can be estimated as 270.
132 can be estimated as 130.
Thus the required estimated sum = 270 + 130 = 400

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

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. The number of vertices in the given shapes:
i. Sphere : 0
ii. Cylinder : 0

CBSE VI | Mathematics
Sample Paper 5 – Solution

15. 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.

16. (–13) + (–19) + (+15) + (–10)
= –13 – 19 + 15 – 10
= –13 – 19 – 10 + 15
= –42 + 15
= –27

17. 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.

18. Two lines in the same plane which never intersect are called parallel lines.
Parallel lines maintain the same distance apart over their entire length.

19. Population of the village = 13295
Increase in population= Average growth – 1 = 399
Population in the successive year = 13295 + 399 = 13694

20. Prime factorisation of 455 =  5 × 7 × 13
Therefore, the dimensions of the cuboid are
5 cm, 7 cm, 13 cm.

CBSE VI | Mathematics
Sample Paper 5 – Solution

21. L.C.M. of 24, 45 and 54 = 2 × 2 × 2 × 3 × 3 × 3 × 5 = 1080

22.
21
53
72
?
37 7
72
37 2 7 7
7 2 2 7
74 49
14 14
123
14
11
8
14
??
??
??
??
??
?
?

23. Subtracting a negative number is the same as adding its positive integer.
? 7 – (–3) = 7 + 3 = 10

24. To find the H.C.F of 77, 105 and 231.

77 = 7 × 11
105 = 3 × 5 × 7
231 = 3 × 7 × 11
Hence, H.C.F of 77, 105 and 231 is = 7

CBSE VI | Mathematics
Sample Paper 5 – Solution

Section C

25. To solve using number line start with –8, move 12 steps right and then back 2 steps as
shown below:

So, we reach at 2, therefore (–8 + 12 – 2) = 2

26. AC, AD, BD, BE and CE are diagonals.

27. Temperature in the morning = –14
o
F
Drop in temperature is written as –7
o
F
Temperature at present = –14
o
F + (–7
o
F) = –14
o
F – 7
o
F = –21
o
F

28. Rs. 13550 estimated to nearest thousands = Rs. 14000
Rs. 26788 estimated to nearest thousands = Rs. 27000
Total estimated money (to be received) = Rs. (14000 + 27000) = Rs. 41000
He has to pay Rs. 37000.
And 41000 > 37000
Therefore, he will be able to pay to his supplier with the money received.

29. First we find the LCM of 48, 60, 72.
48 = 2 × 2 × 2 × 2 × 3
60 = 2 × 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720
Hence, they will meet after = 2 rounds.

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## Mathematics (Maths) Class 6

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