Sample Solution Paper 2 - Term- 1 Mathematics, Class 7 Class 7 Notes | EduRev

Sample Papers For Class 7

Created by: Praveen Kumar

Class 7 : Sample Solution Paper 2 - Term- 1 Mathematics, Class 7 Class 7 Notes | EduRev

 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 More

Complete Syllabus of Class 7

Dynamic Test

Content Category

Related Searches

Class 7 Class 7 Notes | EduRev

,

Previous Year Questions with Solutions

,

Sample Paper

,

Free

,

Viva Questions

,

shortcuts and tricks

,

Objective type Questions

,

MCQs

,

mock tests for examination

,

Semester Notes

,

Exam

,

Class 7 Class 7 Notes | EduRev

,

practice quizzes

,

Summary

,

video lectures

,

pdf

,

ppt

,

Sample Solution Paper 2 - Term- 1 Mathematics

,

Important questions

,

Sample Solution Paper 2 - Term- 1 Mathematics

,

past year papers

,

Extra Questions

,

Class 7 Class 7 Notes | EduRev

,

Sample Solution Paper 2 - Term- 1 Mathematics

,

study material

;