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Sample Solution Paper 2 - Term- 2 Mathematics, Class 7 Class 7 Notes | EduRev

Mathematics (Maths) Class 7

Created by: Praveen Kumar

Class 7 : Sample Solution Paper 2 - Term- 2 Mathematics, Class 7 Class 7 Notes | EduRev

``` Page 1

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

CBSE Board
Class VII Mathematics
Term II
Sample Paper 2 â€“ Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

For the option (B), 1 is the only common factor between numerator and
denominator and the denominator is a positive integer, hence it is in the standard
form.

6 components - three angles and three sides

Cylinder

-x + 1 is an example of binomial as it contains two terms.

Hence, order of rotational symmetry is 2.

Page 2

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

CBSE Board
Class VII Mathematics
Term II
Sample Paper 2 â€“ Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

For the option (B), 1 is the only common factor between numerator and
denominator and the denominator is a positive integer, hence it is in the standard
form.

6 components - three angles and three sides

Cylinder

-x + 1 is an example of binomial as it contains two terms.

Hence, order of rotational symmetry is 2.

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

The length of given cuboid is 6 units.

Perimeter of rhombus = 28 cm
Thus, 4 x (side) = 28 cm
Side =
28
4
= 7 cm

(128 ÷32) ÷ (-4) = 4 ÷ (-4)  = -1

2x + 3 = 7
If we transpose 3 to RHS, then the term with variable will remain on one side and
the constants will be on the other side.
So, the first step is to transpose 3 to RHS.
i.e. 2x = 7 â€“ 3

Section B

13. Percentage of marks scored by Rahul =
40
100 80%
50
??
Percentage of marks scored by Rohan =
75
100 75%
100
??
Hence, Rahul scored more marks than Rohan.

Page 3

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

CBSE Board
Class VII Mathematics
Term II
Sample Paper 2 â€“ Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

For the option (B), 1 is the only common factor between numerator and
denominator and the denominator is a positive integer, hence it is in the standard
form.

6 components - three angles and three sides

Cylinder

-x + 1 is an example of binomial as it contains two terms.

Hence, order of rotational symmetry is 2.

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

The length of given cuboid is 6 units.

Perimeter of rhombus = 28 cm
Thus, 4 x (side) = 28 cm
Side =
28
4
= 7 cm

(128 ÷32) ÷ (-4) = 4 ÷ (-4)  = -1

2x + 3 = 7
If we transpose 3 to RHS, then the term with variable will remain on one side and
the constants will be on the other side.
So, the first step is to transpose 3 to RHS.
i.e. 2x = 7 â€“ 3

Section B

13. Percentage of marks scored by Rahul =
40
100 80%
50
??
Percentage of marks scored by Rohan =
75
100 75%
100
??
Hence, Rahul scored more marks than Rohan.

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

14. In the given number line, the rational number between 0 and -1 would be
1
2
?
.
Rational number between 1 and 2 is
3
2
as 1 can be written as
2
2
.
So, the next point would be
3
2
and 2 can be written as
4
2
, which is the same.
Thus, the number line representing the missing values is as follows:

15.
(a) ?A = 120
o
, ?B = 90
o
and AB = 8 cm
Since, ?A + ?B = 120
o
+ 90
o
= 210
o

That is the sum of two angles is more than 180
o
.
Hence, the triangle is not possible.

(b) ?P = 90
o
, ?Q = 90
o
and PQ = 9 cm
Since, ?P + ?Q = 90
o
+ 90
o
= 180
o

That is the sum of two angles is equal to 180
o
.
And the measure of third angle is zero.
Hence, the triangle is not possible.

16.

Page 4

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

CBSE Board
Class VII Mathematics
Term II
Sample Paper 2 â€“ Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

For the option (B), 1 is the only common factor between numerator and
denominator and the denominator is a positive integer, hence it is in the standard
form.

6 components - three angles and three sides

Cylinder

-x + 1 is an example of binomial as it contains two terms.

Hence, order of rotational symmetry is 2.

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

The length of given cuboid is 6 units.

Perimeter of rhombus = 28 cm
Thus, 4 x (side) = 28 cm
Side =
28
4
= 7 cm

(128 ÷32) ÷ (-4) = 4 ÷ (-4)  = -1

2x + 3 = 7
If we transpose 3 to RHS, then the term with variable will remain on one side and
the constants will be on the other side.
So, the first step is to transpose 3 to RHS.
i.e. 2x = 7 â€“ 3

Section B

13. Percentage of marks scored by Rahul =
40
100 80%
50
??
Percentage of marks scored by Rohan =
75
100 75%
100
??
Hence, Rahul scored more marks than Rohan.

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

14. In the given number line, the rational number between 0 and -1 would be
1
2
?
.
Rational number between 1 and 2 is
3
2
as 1 can be written as
2
2
.
So, the next point would be
3
2
and 2 can be written as
4
2
, which is the same.
Thus, the number line representing the missing values is as follows:

15.
(a) ?A = 120
o
, ?B = 90
o
and AB = 8 cm
Since, ?A + ?B = 120
o
+ 90
o
= 210
o

That is the sum of two angles is more than 180
o
.
Hence, the triangle is not possible.

(b) ?P = 90
o
, ?Q = 90
o
and PQ = 9 cm
Since, ?P + ?Q = 90
o
+ 90
o
= 180
o

That is the sum of two angles is equal to 180
o
.
And the measure of third angle is zero.
Hence, the triangle is not possible.

16.

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

17. Required sum:
(6m - 7n - 5p) + (-4m - 9n + 6p) + (-4m - 9n + 6p)
= 6m - 7n - 5p - 4m - 9n + 6p - 4m - 9n + 6p
= (6m - 4m - 4m) + (-7n - 9n - 9n) + (-5p + 6p + 6p)
= (6 - 4 - 4)m + (-7 - 9 - 9)n + (-5 + 6 + 6)p
= -2m - 25n + 7p

18.

19. The other holes are as below:
i.

ii.

Page 5

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

CBSE Board
Class VII Mathematics
Term II
Sample Paper 2 â€“ Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

For the option (B), 1 is the only common factor between numerator and
denominator and the denominator is a positive integer, hence it is in the standard
form.

6 components - three angles and three sides

Cylinder

-x + 1 is an example of binomial as it contains two terms.

Hence, order of rotational symmetry is 2.

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

The length of given cuboid is 6 units.

Perimeter of rhombus = 28 cm
Thus, 4 x (side) = 28 cm
Side =
28
4
= 7 cm

(128 ÷32) ÷ (-4) = 4 ÷ (-4)  = -1

2x + 3 = 7
If we transpose 3 to RHS, then the term with variable will remain on one side and
the constants will be on the other side.
So, the first step is to transpose 3 to RHS.
i.e. 2x = 7 â€“ 3

Section B

13. Percentage of marks scored by Rahul =
40
100 80%
50
??
Percentage of marks scored by Rohan =
75
100 75%
100
??
Hence, Rahul scored more marks than Rohan.

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

14. In the given number line, the rational number between 0 and -1 would be
1
2
?
.
Rational number between 1 and 2 is
3
2
as 1 can be written as
2
2
.
So, the next point would be
3
2
and 2 can be written as
4
2
, which is the same.
Thus, the number line representing the missing values is as follows:

15.
(a) ?A = 120
o
, ?B = 90
o
and AB = 8 cm
Since, ?A + ?B = 120
o
+ 90
o
= 210
o

That is the sum of two angles is more than 180
o
.
Hence, the triangle is not possible.

(b) ?P = 90
o
, ?Q = 90
o
and PQ = 9 cm
Since, ?P + ?Q = 90
o
+ 90
o
= 180
o

That is the sum of two angles is equal to 180
o
.
And the measure of third angle is zero.
Hence, the triangle is not possible.

16.

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

17. Required sum:
(6m - 7n - 5p) + (-4m - 9n + 6p) + (-4m - 9n + 6p)
= 6m - 7n - 5p - 4m - 9n + 6p - 4m - 9n + 6p
= (6m - 4m - 4m) + (-7n - 9n - 9n) + (-5p + 6p + 6p)
= (6 - 4 - 4)m + (-7 - 9 - 9)n + (-5 + 6 + 6)p
= -2m - 25n + 7p

18.

19. The other holes are as below:
i.

ii.

CBSE VII | Mathematics
Sample Paper 2 â€“ Solution

20.
1. 1. First take an isometric dot sheet.
2. Draw the line segment AB and AD of length 5 units and 3 units respectively.
3. For the height draw the line segment AG, BC and DE of 6 units each.
4. Join EG and GC.
5. Again draw EF and CF of 5 units and 3 units respectively.

21. Mass of earth = 5,970,000,000,000,000,000,000,000 kg
= 597 × 10000000000000000000000 kg
= 597 × 10
22
kg
= 5.97 × 10
24
kg

22. Number of cubes in first layer = 7
Number of cubes in second layer = 2
Hence, total number of cubes = 7 + 2 = 9

23. Average score = mean score
? ? ? ? ? ? ? ? ? ?
?
?
?
Sum of all observations
Mean=
Total number of observations
12 23 10 77 15 78 90 54 23 10 1
11
393
11
35.7

24. 725 × (-35) + (-725) × 65
= 725 × (-35) - 725 × 65
= 725 x (-35 - 65)         [Using distributive property]
= 725 × (-100)
= -72500

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Mathematics (Maths) Class 7

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