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Class 7 Math: CBSE Sample Question Paper Solutions Term I – 2
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Page 1
CBSE VII | Mathematics
Sample Paper 2 - Solution
CBSE Board
Class VII Mathematics
Term I
Sample Paper 2 - Solution
Time: 2 ½ hours Total Marks: 80
Section A
1. Correct answer: C
-5 + 9 + (-5) + (-10) + (1)
-5 + 9 = 4
4 + (-5) = -1
-1 + (-10) = -1 - 10 = -11
-11 + 1 = -10
So, -5 + 9 + (-5) + (-10) + (1) = -10
2. Correct answer: B
3. Correct answer: B
Since 11 has highest frequency, the mode is 11.
4. Correct answer: B
The given equation can be written in the form of statement as "One third of p is q".
5. Correct answer: D
There are 3 acute angles, AOB, BOC and AOC.
6. Correct answer: A
In triangles ABC and PQC, we have:
AB = PQ
BC = CQ
B = Q
Thus, triangles ABC and PQC are congruent.
Therefore, BAC = CPQ
Now, applying angle sum property in triangle ABC, we get,
BAC = 180
o
- 70
o
- 30
o
= 80
o
Therefore, CPQ = 80
o
Page 2
CBSE VII | Mathematics
Sample Paper 2 - Solution
CBSE Board
Class VII Mathematics
Term I
Sample Paper 2 - Solution
Time: 2 ½ hours Total Marks: 80
Section A
1. Correct answer: C
-5 + 9 + (-5) + (-10) + (1)
-5 + 9 = 4
4 + (-5) = -1
-1 + (-10) = -1 - 10 = -11
-11 + 1 = -10
So, -5 + 9 + (-5) + (-10) + (1) = -10
2. Correct answer: B
3. Correct answer: B
Since 11 has highest frequency, the mode is 11.
4. Correct answer: B
The given equation can be written in the form of statement as "One third of p is q".
5. Correct answer: D
There are 3 acute angles, AOB, BOC and AOC.
6. Correct answer: A
In triangles ABC and PQC, we have:
AB = PQ
BC = CQ
B = Q
Thus, triangles ABC and PQC are congruent.
Therefore, BAC = CPQ
Now, applying angle sum property in triangle ABC, we get,
BAC = 180
o
- 70
o
- 30
o
= 80
o
Therefore, CPQ = 80
o
CBSE VII | Mathematics
Sample Paper 2 - Solution
7. Correct answer: A
8. Correct answer: D
Thus, each part equals 0.12543.
9. Correct answer: C
Let the whole number be x.
Twice of the whole number = 2x
9 added to twice of the whole number = 9 + 2x
From the given information, we have:
9 + 2x = 31
2x = 31 - 9
2x = 22
x = 11
Thus, the required whole number is 11.
10. Correct answer: D
11. Correct answer: A
The two triangles can be proved to be congruent by using SAS congruency criterion.
The corresponding equal parts in triangles ABC and ADE are
12. Correct answer: A
2x + 3 = 7
If we will transpose 3 to RHS, the term with variable will remain on one side and
the constants will be on other side.
So, first step is to Transpose 3 to RHS.
i.e., 2x = 7 – 3
Page 3
CBSE VII | Mathematics
Sample Paper 2 - Solution
CBSE Board
Class VII Mathematics
Term I
Sample Paper 2 - Solution
Time: 2 ½ hours Total Marks: 80
Section A
1. Correct answer: C
-5 + 9 + (-5) + (-10) + (1)
-5 + 9 = 4
4 + (-5) = -1
-1 + (-10) = -1 - 10 = -11
-11 + 1 = -10
So, -5 + 9 + (-5) + (-10) + (1) = -10
2. Correct answer: B
3. Correct answer: B
Since 11 has highest frequency, the mode is 11.
4. Correct answer: B
The given equation can be written in the form of statement as "One third of p is q".
5. Correct answer: D
There are 3 acute angles, AOB, BOC and AOC.
6. Correct answer: A
In triangles ABC and PQC, we have:
AB = PQ
BC = CQ
B = Q
Thus, triangles ABC and PQC are congruent.
Therefore, BAC = CPQ
Now, applying angle sum property in triangle ABC, we get,
BAC = 180
o
- 70
o
- 30
o
= 80
o
Therefore, CPQ = 80
o
CBSE VII | Mathematics
Sample Paper 2 - Solution
7. Correct answer: A
8. Correct answer: D
Thus, each part equals 0.12543.
9. Correct answer: C
Let the whole number be x.
Twice of the whole number = 2x
9 added to twice of the whole number = 9 + 2x
From the given information, we have:
9 + 2x = 31
2x = 31 - 9
2x = 22
x = 11
Thus, the required whole number is 11.
10. Correct answer: D
11. Correct answer: A
The two triangles can be proved to be congruent by using SAS congruency criterion.
The corresponding equal parts in triangles ABC and ADE are
12. Correct answer: A
2x + 3 = 7
If we will transpose 3 to RHS, the term with variable will remain on one side and
the constants will be on other side.
So, first step is to Transpose 3 to RHS.
i.e., 2x = 7 – 3
CBSE VII | Mathematics
Sample Paper 2 - Solution
Section B
13. Total number of balls = 12
It is also given that the bag contains equal number of balls of each of the four colours
(yellow, blue, green and red).
Therefore,
Number of yellow balls = Number of blue balls = Number of green balls =
Number of red balls = 3
P(yellow) =
P(blue) =
P(green) =
P(red) =
14. Let the number be x
Then, Three fifth of a number =
3
5
x
5 added to three-fifth of a number = 5 +
3
5
x
Thus, the linear equation will be
Solving the linear equation to find x.
Transposing 5 to R.H.S., we get
Thus, the required number is
5
9
?
.
Page 4
CBSE VII | Mathematics
Sample Paper 2 - Solution
CBSE Board
Class VII Mathematics
Term I
Sample Paper 2 - Solution
Time: 2 ½ hours Total Marks: 80
Section A
1. Correct answer: C
-5 + 9 + (-5) + (-10) + (1)
-5 + 9 = 4
4 + (-5) = -1
-1 + (-10) = -1 - 10 = -11
-11 + 1 = -10
So, -5 + 9 + (-5) + (-10) + (1) = -10
2. Correct answer: B
3. Correct answer: B
Since 11 has highest frequency, the mode is 11.
4. Correct answer: B
The given equation can be written in the form of statement as "One third of p is q".
5. Correct answer: D
There are 3 acute angles, AOB, BOC and AOC.
6. Correct answer: A
In triangles ABC and PQC, we have:
AB = PQ
BC = CQ
B = Q
Thus, triangles ABC and PQC are congruent.
Therefore, BAC = CPQ
Now, applying angle sum property in triangle ABC, we get,
BAC = 180
o
- 70
o
- 30
o
= 80
o
Therefore, CPQ = 80
o
CBSE VII | Mathematics
Sample Paper 2 - Solution
7. Correct answer: A
8. Correct answer: D
Thus, each part equals 0.12543.
9. Correct answer: C
Let the whole number be x.
Twice of the whole number = 2x
9 added to twice of the whole number = 9 + 2x
From the given information, we have:
9 + 2x = 31
2x = 31 - 9
2x = 22
x = 11
Thus, the required whole number is 11.
10. Correct answer: D
11. Correct answer: A
The two triangles can be proved to be congruent by using SAS congruency criterion.
The corresponding equal parts in triangles ABC and ADE are
12. Correct answer: A
2x + 3 = 7
If we will transpose 3 to RHS, the term with variable will remain on one side and
the constants will be on other side.
So, first step is to Transpose 3 to RHS.
i.e., 2x = 7 – 3
CBSE VII | Mathematics
Sample Paper 2 - Solution
Section B
13. Total number of balls = 12
It is also given that the bag contains equal number of balls of each of the four colours
(yellow, blue, green and red).
Therefore,
Number of yellow balls = Number of blue balls = Number of green balls =
Number of red balls = 3
P(yellow) =
P(blue) =
P(green) =
P(red) =
14. Let the number be x
Then, Three fifth of a number =
3
5
x
5 added to three-fifth of a number = 5 +
3
5
x
Thus, the linear equation will be
Solving the linear equation to find x.
Transposing 5 to R.H.S., we get
Thus, the required number is
5
9
?
.
CBSE VII | Mathematics
Sample Paper 2 - Solution
15. Given: a = -8, b = -7, c = 6
(a+ b) + c = [(-8) + (-7)] + 6 = -15 + 6 = -9
a + (b + c) = (-8) + [(-7) + 6] = -8 - 1 = -9
Hence, (a + b) + c = a + (b + c).
16. First, we need to line up the decimals as follows:
3.25 = 3.250
0.075 = 0.075
5 = 5.000
Now, adding them gives
3.250
+ 0.075
+ 5.000
8.325
17. We know that the sum of two sides of a triangle is always greater than the third.
The given lengths of the sides are 5 cm, 3 cm, 4 cm.
Let us check whether the above stated property holds true. We have:
5 + 3 = 8, which is greater than 4
5 + 4 = 9, which is greater than 3
3 + 4 = 7, which is greater than 5
Thus, it is possible to draw a triangle with given side lengths.
18. Given that, m||p and t is the transversal
We know that, if two parallel lines are cut by a transversal, each pair of alternate
interior angles are equal.
So, ? a = ? z (pair of alternate interior angles)
Thus, ? z = 57
o
19. The numbers in ascending order are:
11, 12, 12, 12, 19, 23, 33, 34, 34, 45, 46, 49, 50, 55, 56, 65, 67, 78, 81, 87, 98
As the number of observations (21) are odd,
Median = middle observation = 11
th
observation = 46
Mode is the observation that appears most often.
Here, 12 appears maximum number of times (thrice). So, 12 is the mode.
20. 725 × (-35) + (-725) × 65
= 725 × (-35) - 725 × 65
= 725 x (-35 - 65) [Using distributive property]
= 725 × (-100)
= -72500
Page 5
CBSE VII | Mathematics
Sample Paper 2 - Solution
CBSE Board
Class VII Mathematics
Term I
Sample Paper 2 - Solution
Time: 2 ½ hours Total Marks: 80
Section A
1. Correct answer: C
-5 + 9 + (-5) + (-10) + (1)
-5 + 9 = 4
4 + (-5) = -1
-1 + (-10) = -1 - 10 = -11
-11 + 1 = -10
So, -5 + 9 + (-5) + (-10) + (1) = -10
2. Correct answer: B
3. Correct answer: B
Since 11 has highest frequency, the mode is 11.
4. Correct answer: B
The given equation can be written in the form of statement as "One third of p is q".
5. Correct answer: D
There are 3 acute angles, AOB, BOC and AOC.
6. Correct answer: A
In triangles ABC and PQC, we have:
AB = PQ
BC = CQ
B = Q
Thus, triangles ABC and PQC are congruent.
Therefore, BAC = CPQ
Now, applying angle sum property in triangle ABC, we get,
BAC = 180
o
- 70
o
- 30
o
= 80
o
Therefore, CPQ = 80
o
CBSE VII | Mathematics
Sample Paper 2 - Solution
7. Correct answer: A
8. Correct answer: D
Thus, each part equals 0.12543.
9. Correct answer: C
Let the whole number be x.
Twice of the whole number = 2x
9 added to twice of the whole number = 9 + 2x
From the given information, we have:
9 + 2x = 31
2x = 31 - 9
2x = 22
x = 11
Thus, the required whole number is 11.
10. Correct answer: D
11. Correct answer: A
The two triangles can be proved to be congruent by using SAS congruency criterion.
The corresponding equal parts in triangles ABC and ADE are
12. Correct answer: A
2x + 3 = 7
If we will transpose 3 to RHS, the term with variable will remain on one side and
the constants will be on other side.
So, first step is to Transpose 3 to RHS.
i.e., 2x = 7 – 3
CBSE VII | Mathematics
Sample Paper 2 - Solution
Section B
13. Total number of balls = 12
It is also given that the bag contains equal number of balls of each of the four colours
(yellow, blue, green and red).
Therefore,
Number of yellow balls = Number of blue balls = Number of green balls =
Number of red balls = 3
P(yellow) =
P(blue) =
P(green) =
P(red) =
14. Let the number be x
Then, Three fifth of a number =
3
5
x
5 added to three-fifth of a number = 5 +
3
5
x
Thus, the linear equation will be
Solving the linear equation to find x.
Transposing 5 to R.H.S., we get
Thus, the required number is
5
9
?
.
CBSE VII | Mathematics
Sample Paper 2 - Solution
15. Given: a = -8, b = -7, c = 6
(a+ b) + c = [(-8) + (-7)] + 6 = -15 + 6 = -9
a + (b + c) = (-8) + [(-7) + 6] = -8 - 1 = -9
Hence, (a + b) + c = a + (b + c).
16. First, we need to line up the decimals as follows:
3.25 = 3.250
0.075 = 0.075
5 = 5.000
Now, adding them gives
3.250
+ 0.075
+ 5.000
8.325
17. We know that the sum of two sides of a triangle is always greater than the third.
The given lengths of the sides are 5 cm, 3 cm, 4 cm.
Let us check whether the above stated property holds true. We have:
5 + 3 = 8, which is greater than 4
5 + 4 = 9, which is greater than 3
3 + 4 = 7, which is greater than 5
Thus, it is possible to draw a triangle with given side lengths.
18. Given that, m||p and t is the transversal
We know that, if two parallel lines are cut by a transversal, each pair of alternate
interior angles are equal.
So, ? a = ? z (pair of alternate interior angles)
Thus, ? z = 57
o
19. The numbers in ascending order are:
11, 12, 12, 12, 19, 23, 33, 34, 34, 45, 46, 49, 50, 55, 56, 65, 67, 78, 81, 87, 98
As the number of observations (21) are odd,
Median = middle observation = 11
th
observation = 46
Mode is the observation that appears most often.
Here, 12 appears maximum number of times (thrice). So, 12 is the mode.
20. 725 × (-35) + (-725) × 65
= 725 × (-35) - 725 × 65
= 725 x (-35 - 65) [Using distributive property]
= 725 × (-100)
= -72500
CBSE VII | Mathematics
Sample Paper 2 - Solution
21. Time taken by Mala to drink a glass of milk =
7
8
mins
Time taken by Varun to drink a glass of milk =
9
16
mins
To compare both the fractions, we have to change them into like fractions.
;
Since, 14 > 9,
7
8
>
9
16
Thus, Mala took longer time to finish the glass of milk.
Now, we have to subtract the time durations of Mala and Varun to calculate how
slow was Mala than Varun.
Thus, Mala took
5
16
mins more than Varun to finish a glass of milk.
22. Sum of 38 and -87 = 38 + (-87) = 38 - 87 = -49
Subtracting (-134) from -49, we get
-49 - (-134) = -49 + 134 = 85
23. Pie filling made in 1 minute = 9.2 kg
Pie filling made in 6 minutes = 6 x 9.2 kg = 55.2 kg
24.
(a) 6n + 4 = 10
Statement:
For 6n, six times of a number n.
For 6n + 4, six times of a number n added to 4.
Thus, for 6n + 4 = 10, the final statement is
‘Six times of a number n added to 4 gives 10’.
Read MoreFAQs on Class 7 Math: CBSE Sample Question Paper Solutions Term I – 2
| 1. How do I solve percentage problems in CBSE Class 7 maths sample papers? |  |
Ans. Percentage problems involve finding what part of a whole equals a specific fraction out of 100. The basic formula is (Part/Whole) × 100 = Percentage. In CBSE sample question papers, these appear as word problems requiring conversions between percentages, decimals, and fractions. Practice calculating discounts, profit-loss, and interest using this ratio method to master this topic efficiently.
| 2. What are the common mistakes students make with integers and rational numbers in term 1 sample papers? | |
Ans. Students often confuse the rules for adding and subtracting negative integers, especially when two negatives appear consecutively. Another frequent error involves incorrect ordering of rational numbers on a number line. Many also struggle with simplifying fractions before operations. Reviewing the sign rules for operations and practising ordering rational numbers systematically helps avoid these mistakes during CBSE sample question paper attempts.
| 3. How should I approach geometry questions involving angles and lines in Class 7 term 1 exams? | |
Ans. Geometry questions test understanding of angle relationships, parallel lines, and transversal properties. Begin by identifying given information and labelling angles clearly. Use properties like vertically opposite angles, corresponding angles, and alternate angles strategically. Drawing accurate diagrams is essential-many students lose marks through careless sketches. Refer to mind maps and flashcards on EduRev to visualise angle theorems clearly before attempting sample papers.
| 4. Why do algebraic expressions seem confusing when solving equations in CBSE sample papers? | |
Ans. Algebraic confusion typically stems from misunderstanding variable representation and coefficient operations. Students often forget that letters represent unknown numbers and apply arithmetic rules incorrectly. The transition from concrete arithmetic to abstract algebra challenges many learners. Practice simple linear equations first, then progress to multi-step problems. Focus on isolating variables systematically rather than rushing through calculations during term 1 sample question solving.
| 5. What's the fastest way to check my answers when practising CBSE Class 7 term 1 sample papers? | |
Ans. Verification involves substituting your solution back into the original equation or problem statement. For percentage and ratio questions, estimate answers mentally first, then check if actual results align with estimates. Sketch geometry diagrams twice independently to confirm angles. Organise working clearly to spot calculation errors quickly. Completing timed practice sessions with sample papers develops speed and accuracy simultaneously for effective exam preparation.
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