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

Mathematics (Maths) Class 7

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

 Page 1


  
 
CBSE VII | Mathematics 
Sample Paper 2 – Solution 
 
     
CBSE Board 
Class VII Mathematics 
Term II 
Sample Paper 2 – Solution 
Time: 2 ½ hours                          Total Marks: 80 
 
Section A 
 
1. Correct answer: B 
 
 
2. Correct answer: B 
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.  
 
3. Correct answer: D 
6 components - three angles and three sides 
 
4. Correct answer: A 
Cylinder 
 
5. Correct answer: B 
-x + 1 is an example of binomial as it contains two terms. 
 
6. Correct answer: C 
 
 
7. Correct answer: A 
 
Hence, order of rotational symmetry is 2.  
  
 
Page 2


  
 
CBSE VII | Mathematics 
Sample Paper 2 – Solution 
 
     
CBSE Board 
Class VII Mathematics 
Term II 
Sample Paper 2 – Solution 
Time: 2 ½ hours                          Total Marks: 80 
 
Section A 
 
1. Correct answer: B 
 
 
2. Correct answer: B 
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.  
 
3. Correct answer: D 
6 components - three angles and three sides 
 
4. Correct answer: A 
Cylinder 
 
5. Correct answer: B 
-x + 1 is an example of binomial as it contains two terms. 
 
6. Correct answer: C 
 
 
7. Correct answer: A 
 
Hence, order of rotational symmetry is 2.  
  
 
  
 
CBSE VII | Mathematics 
Sample Paper 2 – Solution 
 
     
8. Correct answer: C 
The length of given cuboid is 6 units. 
 
9. Correct answer: D 
 
 
10.  Correct answer: A 
                Perimeter of rhombus = 28 cm 
 Thus, 4 x (side) = 28 cm 
 Side = 
28
4
 = 7 cm 
 
11. Correct answer: A 
(128 ÷32) ÷ (-4) = 4 ÷ (-4)  = -1 
 
12. Correct answer: A 
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 – 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 – Solution 
 
     
CBSE Board 
Class VII Mathematics 
Term II 
Sample Paper 2 – Solution 
Time: 2 ½ hours                          Total Marks: 80 
 
Section A 
 
1. Correct answer: B 
 
 
2. Correct answer: B 
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.  
 
3. Correct answer: D 
6 components - three angles and three sides 
 
4. Correct answer: A 
Cylinder 
 
5. Correct answer: B 
-x + 1 is an example of binomial as it contains two terms. 
 
6. Correct answer: C 
 
 
7. Correct answer: A 
 
Hence, order of rotational symmetry is 2.  
  
 
  
 
CBSE VII | Mathematics 
Sample Paper 2 – Solution 
 
     
8. Correct answer: C 
The length of given cuboid is 6 units. 
 
9. Correct answer: D 
 
 
10.  Correct answer: A 
                Perimeter of rhombus = 28 cm 
 Thus, 4 x (side) = 28 cm 
 Side = 
28
4
 = 7 cm 
 
11. Correct answer: A 
(128 ÷32) ÷ (-4) = 4 ÷ (-4)  = -1 
 
12. Correct answer: A 
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 – 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 – 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 – Solution 
 
     
CBSE Board 
Class VII Mathematics 
Term II 
Sample Paper 2 – Solution 
Time: 2 ½ hours                          Total Marks: 80 
 
Section A 
 
1. Correct answer: B 
 
 
2. Correct answer: B 
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.  
 
3. Correct answer: D 
6 components - three angles and three sides 
 
4. Correct answer: A 
Cylinder 
 
5. Correct answer: B 
-x + 1 is an example of binomial as it contains two terms. 
 
6. Correct answer: C 
 
 
7. Correct answer: A 
 
Hence, order of rotational symmetry is 2.  
  
 
  
 
CBSE VII | Mathematics 
Sample Paper 2 – Solution 
 
     
8. Correct answer: C 
The length of given cuboid is 6 units. 
 
9. Correct answer: D 
 
 
10.  Correct answer: A 
                Perimeter of rhombus = 28 cm 
 Thus, 4 x (side) = 28 cm 
 Side = 
28
4
 = 7 cm 
 
11. Correct answer: A 
(128 ÷32) ÷ (-4) = 4 ÷ (-4)  = -1 
 
12. Correct answer: A 
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 – 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 – 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 – 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 – Solution 
 
     
CBSE Board 
Class VII Mathematics 
Term II 
Sample Paper 2 – Solution 
Time: 2 ½ hours                          Total Marks: 80 
 
Section A 
 
1. Correct answer: B 
 
 
2. Correct answer: B 
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.  
 
3. Correct answer: D 
6 components - three angles and three sides 
 
4. Correct answer: A 
Cylinder 
 
5. Correct answer: B 
-x + 1 is an example of binomial as it contains two terms. 
 
6. Correct answer: C 
 
 
7. Correct answer: A 
 
Hence, order of rotational symmetry is 2.  
  
 
  
 
CBSE VII | Mathematics 
Sample Paper 2 – Solution 
 
     
8. Correct answer: C 
The length of given cuboid is 6 units. 
 
9. Correct answer: D 
 
 
10.  Correct answer: A 
                Perimeter of rhombus = 28 cm 
 Thus, 4 x (side) = 28 cm 
 Side = 
28
4
 = 7 cm 
 
11. Correct answer: A 
(128 ÷32) ÷ (-4) = 4 ÷ (-4)  = -1 
 
12. Correct answer: A 
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 – 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 – 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 – 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 – 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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