Courses

# Sample Solution Paper 2 - Term- 1 Mathematics, Class 6 Class 6 Notes | EduRev

## Mathematics (Maths) Class 6

Created by: Praveen Kumar

## Class 6 : Sample Solution Paper 2 - Term- 1 Mathematics, Class 6 Class 6 Notes | EduRev

``` Page 1

CBSE VI | Mathematics
Sample Paper 2 – Solution

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

Section A

Arrange the numbers in place-value chart:
Cr T L L T Th Th H T O
4 7 8 9 6 3 0 4
4 7 8 9 6 3 4 0

Clearly both numbers have 8 digits.
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively.

But at tens place, first number has 0 and second number has 4.
Clearly, 0 < 4
Hence
47896304 < 47896340

To add 0 and 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

Since,

Therefore, 36 = 2 × 2 × 3 × 3
Page 2

CBSE VI | Mathematics
Sample Paper 2 – Solution

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

Section A

Arrange the numbers in place-value chart:
Cr T L L T Th Th H T O
4 7 8 9 6 3 0 4
4 7 8 9 6 3 4 0

Clearly both numbers have 8 digits.
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively.

But at tens place, first number has 0 and second number has 4.
Clearly, 0 < 4
Hence
47896304 < 47896340

To add 0 and 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

Since,

Therefore, 36 = 2 × 2 × 3 × 3

CBSE VI | Mathematics
Sample Paper 2 – Solution

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.

The number just before 1000000 is one less than 1000000.
The required number = 1000000 – 1 = 999999.

On a number line, –110 lies next to –111 on the right.
Therefore, the successor of –111 is –110.

The given fraction is
Dividing the numerator and denominator by 3, we get

Thus,
is equivalent to .

The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6.
The number 653 is neither divisible by 3 nor by 2 and hence not by 6.
Now, consider the number 531.
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3.
But it is not an even number, so it is not divisible by 2.
Thus, 531 is divisible by 3 but not by 6.

(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A.

Page 3

CBSE VI | Mathematics
Sample Paper 2 – Solution

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

Section A

Arrange the numbers in place-value chart:
Cr T L L T Th Th H T O
4 7 8 9 6 3 0 4
4 7 8 9 6 3 4 0

Clearly both numbers have 8 digits.
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively.

But at tens place, first number has 0 and second number has 4.
Clearly, 0 < 4
Hence
47896304 < 47896340

To add 0 and 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

Since,

Therefore, 36 = 2 × 2 × 3 × 3

CBSE VI | Mathematics
Sample Paper 2 – Solution

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.

The number just before 1000000 is one less than 1000000.
The required number = 1000000 – 1 = 999999.

On a number line, –110 lies next to –111 on the right.
Therefore, the successor of –111 is –110.

The given fraction is
Dividing the numerator and denominator by 3, we get

Thus,
is equivalent to .

The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6.
The number 653 is neither divisible by 3 nor by 2 and hence not by 6.
Now, consider the number 531.
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3.
But it is not an even number, so it is not divisible by 2.
Thus, 531 is divisible by 3 but not by 6.

(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A.

CBSE VI | Mathematics
Sample Paper 2 – Solution

Section B

13.

T Cr Cr T L L T Th Th H T O
(i)

7 0 7 0 7 5
(ii)

5 3 6 1 8 4 9 3

i. Seven lakh seven thousand seventy five.
ii. Five crore thirty-six lakh eighteen thousand four hundred ninety three.

14.

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

15.
i. Going 6 m to the West
ii. A withdrawal of Rs 100
iii. 10 km below sea level
iv. Spending Rs 500

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

17.

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

Page 4

CBSE VI | Mathematics
Sample Paper 2 – Solution

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

Section A

Arrange the numbers in place-value chart:
Cr T L L T Th Th H T O
4 7 8 9 6 3 0 4
4 7 8 9 6 3 4 0

Clearly both numbers have 8 digits.
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively.

But at tens place, first number has 0 and second number has 4.
Clearly, 0 < 4
Hence
47896304 < 47896340

To add 0 and 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

Since,

Therefore, 36 = 2 × 2 × 3 × 3

CBSE VI | Mathematics
Sample Paper 2 – Solution

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.

The number just before 1000000 is one less than 1000000.
The required number = 1000000 – 1 = 999999.

On a number line, –110 lies next to –111 on the right.
Therefore, the successor of –111 is –110.

The given fraction is
Dividing the numerator and denominator by 3, we get

Thus,
is equivalent to .

The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6.
The number 653 is neither divisible by 3 nor by 2 and hence not by 6.
Now, consider the number 531.
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3.
But it is not an even number, so it is not divisible by 2.
Thus, 531 is divisible by 3 but not by 6.

(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A.

CBSE VI | Mathematics
Sample Paper 2 – Solution

Section B

13.

T Cr Cr T L L T Th Th H T O
(i)

7 0 7 0 7 5
(ii)

5 3 6 1 8 4 9 3

i. Seven lakh seven thousand seventy five.
ii. Five crore thirty-six lakh eighteen thousand four hundred ninety three.

14.

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

15.
i. Going 6 m to the West
ii. A withdrawal of Rs 100
iii. 10 km below sea level
iv. Spending Rs 500

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

17.

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

CBSE VI | Mathematics
Sample Paper 2 – Solution

18. The opposite sides of a parallelogram are parallel and equal.
Therefore, LM = NO

19. 90, 91, 92, 93, 94, 95, 96 are the required numbers.

20. Number of circles in step 1 = 3 = 1 × 2 + 1
Number of circles in step 2 = 5 = 2 × 2 + 1
Thus, we can observe that the number of circles is obtained by multiplying the step
number by 2 and then adding 1.
Therefore, number of circles in the 100
th
step = (100 × 2) + 1 = 201

21.

20570 = 2 × 5 × 11 × 11 × 17

22. 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.

23. (–13) + (–19) + (+15) + (–10)
= –13 – 19 + 15 – 10
= –13 – 19 – 10 + 15
= –42 + 15
= –27

24. 21397 can be estimated as 21000
27807 can be estimated as 28000
42305 can be estimated as 42000
On adding, we get 21000 + 28000 + 42000 = 91000
Page 5

CBSE VI | Mathematics
Sample Paper 2 – Solution

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

Section A

Arrange the numbers in place-value chart:
Cr T L L T Th Th H T O
4 7 8 9 6 3 0 4
4 7 8 9 6 3 4 0

Clearly both numbers have 8 digits.
At crores, ten lakhs, lakhs, ten thousands, thousands and hundeds place both have
the same digits i.e. 4, 7, 8, 9, 6, 3 respectively.

But at tens place, first number has 0 and second number has 4.
Clearly, 0 < 4
Hence
47896304 < 47896340

To add 0 and 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

Since,

Therefore, 36 = 2 × 2 × 3 × 3

CBSE VI | Mathematics
Sample Paper 2 – Solution

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.

The number just before 1000000 is one less than 1000000.
The required number = 1000000 – 1 = 999999.

On a number line, –110 lies next to –111 on the right.
Therefore, the successor of –111 is –110.

The given fraction is
Dividing the numerator and denominator by 3, we get

Thus,
is equivalent to .

The numbers 138 and 432 are divisible by both 2 and 3 and hence by 6.
The number 653 is neither divisible by 3 nor by 2 and hence not by 6.
Now, consider the number 531.
Since, the sum of the digits of the number 531 is divisible by 3, so 531 is divisible by 3.
But it is not an even number, so it is not divisible by 2.
Thus, 531 is divisible by 3 but not by 6.

(i) ? A, ? B (ii) ? B, ? C (iii) ? C, ? D (iv) ? D, ? A.

CBSE VI | Mathematics
Sample Paper 2 – Solution

Section B

13.

T Cr Cr T L L T Th Th H T O
(i)

7 0 7 0 7 5
(ii)

5 3 6 1 8 4 9 3

i. Seven lakh seven thousand seventy five.
ii. Five crore thirty-six lakh eighteen thousand four hundred ninety three.

14.

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

15.
i. Going 6 m to the West
ii. A withdrawal of Rs 100
iii. 10 km below sea level
iv. Spending Rs 500

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

17.

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

CBSE VI | Mathematics
Sample Paper 2 – Solution

18. The opposite sides of a parallelogram are parallel and equal.
Therefore, LM = NO

19. 90, 91, 92, 93, 94, 95, 96 are the required numbers.

20. Number of circles in step 1 = 3 = 1 × 2 + 1
Number of circles in step 2 = 5 = 2 × 2 + 1
Thus, we can observe that the number of circles is obtained by multiplying the step
number by 2 and then adding 1.
Therefore, number of circles in the 100
th
step = (100 × 2) + 1 = 201

21.

20570 = 2 × 5 × 11 × 11 × 17

22. 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.

23. (–13) + (–19) + (+15) + (–10)
= –13 – 19 + 15 – 10
= –13 – 19 – 10 + 15
= –42 + 15
= –27

24. 21397 can be estimated as 21000
27807 can be estimated as 28000
42305 can be estimated as 42000
On adding, we get 21000 + 28000 + 42000 = 91000

CBSE VI | Mathematics
Sample Paper 2 – Solution

Section C
25. C stands for 100
D stands for 500
V stands for 5
I stands for 1
X stands for 10
M stands for 1000
In ascending order, the numbers can be arranged as
1 < 5 < 10 < 100 < 500 < 1000
Thus, the given roman numerals can be arranged in ascending order as
I, V, X, C, D, M

26. 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.

27. The given fractions are .

LCM of 2, 3, 6, 9 = (2 x 3 x 3) = 18
So, we convert each of the given fractions into an equivalent fraction with 18 as the
denominator.
Thus, we have:

Hence, the like fractions are

```

## Mathematics (Maths) Class 6

221 videos|105 docs|43 tests

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;