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

## Sample Papers For Class 7

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

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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

-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

Since 11 has highest frequency, the mode is 11.

The given equation can be written in the form of statement as "One third of p  is q".

There are 3 acute angles, AOB, BOC and AOC.

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

-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

Since 11 has highest frequency, the mode is 11.

The given equation can be written in the form of statement as "One third of p  is q".

There are 3 acute angles, AOB, BOC and AOC.

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

Thus, each part equals 0.12543.

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.

The two triangles can be proved to be congruent by using SAS congruency criterion.
The corresponding equal parts in triangles ABC and ADE are

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

-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

Since 11 has highest frequency, the mode is 11.

The given equation can be written in the form of statement as "One third of p  is q".

There are 3 acute angles, AOB, BOC and AOC.

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

Thus, each part equals 0.12543.

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.

The two triangles can be proved to be congruent by using SAS congruency criterion.
The corresponding equal parts in triangles ABC and ADE are

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

-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

Since 11 has highest frequency, the mode is 11.

The given equation can be written in the form of statement as "One third of p  is q".

There are 3 acute angles, AOB, BOC and AOC.

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

Thus, each part equals 0.12543.

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.

The two triangles can be proved to be congruent by using SAS congruency criterion.
The corresponding equal parts in triangles ABC and ADE are

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
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

-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

Since 11 has highest frequency, the mode is 11.

The given equation can be written in the form of statement as "One third of p  is q".

There are 3 acute angles, AOB, BOC and AOC.

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

Thus, each part equals 0.12543.

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.

The two triangles can be proved to be congruent by using SAS congruency criterion.
The corresponding equal parts in triangles ABC and ADE are

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
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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