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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’.
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FAQs on Class 7 Math: CBSE Sample Question Paper Solutions Term I – 2 - Mathematics (Maths) Class 7
| 1. What is the format of the Class 7 Math CBSE Sample Question Paper for Term I? |  |
Ans. The format of the Class 7 Math CBSE Sample Question Paper for Term I usually follows the pattern set by the CBSE board. It consists of multiple-choice questions (MCQs), short answer questions, and long answer questions. The paper is divided into sections or parts, each focusing on different topics or concepts.
| 2. How can I prepare for the Class 7 Math CBSE Term I exam effectively? | |
Ans. To prepare effectively for the Class 7 Math CBSE Term I exam, you can follow these steps:
1. Understand the syllabus and exam pattern.
2. Make a study schedule and allocate time for each topic.
3. Review the concepts thoroughly and practice solving different types of problems.
4. Use sample question papers and previous year papers for practice.
5. Seek help from your teachers or classmates if you have any doubts or difficulties.
6. Revise regularly and solve mock tests to assess your preparation level.
| 3. Are there any specific topics that I should focus on for the Class 7 Math CBSE Term I exam? | |
Ans. Yes, there are certain topics that you should focus on for the Class 7 Math CBSE Term I exam. Some important topics include:
1. Integers
2. Fractions and Decimals
3. Simple Equations
4. Lines and Angles
5. Triangles
6. Perimeter and Area
7. Data Handling
8. Rational Numbers
| 4. How can I manage my time during the Class 7 Math CBSE Term I exam? | |
Ans. Time management is crucial during the Class 7 Math CBSE Term I exam. Here are some tips to help you manage your time effectively:
1. Read the question paper carefully and understand the instructions before starting.
2. Allocate time for each section or part based on the number of questions and their difficulty level.
3. Start with the questions you are most confident about to gain momentum.
4. If you get stuck on a difficult question, move on and come back to it later.
5. Keep an eye on the time and make sure to complete the paper within the given duration.
6. Leave some time at the end for revising your answers and checking for any mistakes.
| 5. Can solving CBSE sample question papers help in scoring better in the Class 7 Math CBSE Term I exam? | |
Ans. Yes, solving CBSE sample question papers can definitely help in scoring better in the Class 7 Math CBSE Term I exam. These sample papers provide you with a clear idea of the exam pattern, marking scheme, and the types of questions asked. By practicing these papers, you can improve your speed and accuracy in solving different types of problems. Additionally, it helps you identify your weak areas and work on them before the actual exam.
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