CBSE Class 7 > Class 7 Notes > Mathematics (Maths) (Old NCERT) > Class 7 Math: CBSE Sample Question Paper Solutions Term II – 2
Class 7 Math: CBSE Sample Question Paper Solutions Term II – 2
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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
Read MoreFAQs on Class 7 Math: CBSE Sample Question Paper Solutions Term II – 2
| 1. What topics are covered in the CBSE Class 7 Maths Term II sample question paper? |  |
Ans. CBSE Class 7 Maths Term II sample papers typically cover algebra, geometry, statistics, and data handling. Key areas include linear equations, properties of triangles, probability concepts, and pictorial representations of information. The question paper tests both computational skills and conceptual understanding across these domains. Students should review practice problems in each section to identify their strengths and weaknesses before the actual examination.
| 2. How do I solve mensuration problems in Class 7 Maths sample papers effectively? | |
Ans. Mensuration problems in Class 7 require understanding formulas for area, perimeter, and volume of 2D and 3D shapes. Begin by identifying the shape, noting given dimensions, then apply the appropriate formula systematically. Common mistakes include forgetting square units for area or mixing up radius with diameter. Students can refer to flashcards and mind maps on EduRev to memorise formulas and visualise geometric concepts clearly before attempting sample question solutions.
| 3. What's the difference between rational and irrational numbers in CBSE Class 7? | |
Ans. Rational numbers can be expressed as fractions (p/q where q ≠ 0), including integers and decimals, while irrational numbers cannot be written as simple fractions-like √2 or π. In Class 7, students work primarily with rational numbers, understanding their properties and operations. Recognising this distinction helps solve algebraic equations and number-based problems in sample papers accurately and builds foundation for advanced mathematics.
| 4. Why am I getting wrong answers in Class 7 geometry problems from sample papers? | |
Ans. Common errors include misidentifying angle relationships, forgetting angle sum properties, or miscalculating using incorrect formulas. Always verify whether angles are complementary (90°), supplementary (180°), or vertically opposite before solving. Draw diagrams clearly and label all given information. Review worked examples in sample question solutions carefully, noting how experts approach each problem type, to avoid repeating calculation mistakes in actual exams.
| 5. How should I prepare using CBSE Term II sample papers to score better in Class 7 Maths? | |
Ans. Attempt complete sample papers under timed conditions to build exam confidence and identify weak areas. After solving, compare your answers with detailed solutions, noting where conceptual gaps exist. Focus revision on topics where you lost marks frequently. Use PPTs and MCQ tests available on EduRev alongside sample papers to reinforce understanding systematically, then re-attempt similar problems to measure improvement before final examinations.
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