# Sample Solution Paper 3 - Term- 1 Mathematics, Class 7 Class 7 Notes | EduRev

## Class 7 : Sample Solution Paper 3 - Term- 1 Mathematics, Class 7 Class 7 Notes | EduRev

``` Page 1

CBSE VII | Mathematics
Sample Paper 3 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 3 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

(128 ÷32)÷ (-4)
= 4 ÷ (-4)
= -1

Total cost = 2.40 × 10
= Rs. 24

The given observations can be arranged in ascending order as
4, 6, 9, 10, 11, 12 and 18
Here, number of observations = 7 (odd)
Median = Middle observation = 10

2x + 3 = 7
If we will transpose 3 to RHS, then the term with variable will remain on one side
and the constants will be on other side.
So, the first step is to transpose 3 to RHS.
i.e. 2x = 7 – 3

?BCA = 180° - 150° = 30°                      (linear pair angles)
Also, ?B = ?BCA = 30°                           (Angles opp. to equal sides are equal)
? ?A = 180° - 30° - 30° = 120°            (Using angle sum property of triangle)

Increased amount =
12
Rs. ×54=Rs. 6.48
100

Page 2

CBSE VII | Mathematics
Sample Paper 3 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 3 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

(128 ÷32)÷ (-4)
= 4 ÷ (-4)
= -1

Total cost = 2.40 × 10
= Rs. 24

The given observations can be arranged in ascending order as
4, 6, 9, 10, 11, 12 and 18
Here, number of observations = 7 (odd)
Median = Middle observation = 10

2x + 3 = 7
If we will transpose 3 to RHS, then the term with variable will remain on one side
and the constants will be on other side.
So, the first step is to transpose 3 to RHS.
i.e. 2x = 7 – 3

?BCA = 180° - 150° = 30°                      (linear pair angles)
Also, ?B = ?BCA = 30°                           (Angles opp. to equal sides are equal)
? ?A = 180° - 30° - 30° = 120°            (Using angle sum property of triangle)

Increased amount =
12
Rs. ×54=Rs. 6.48
100

CBSE VII | Mathematics
Sample Paper 3 - Solution

?
28
2
33

So the multiplicative inverse is
3
8
.

The triangle ABC is a right angled triangle,
By Pythagoras theorem, we have: c
2
= a
2
+ b
2

21b - 32 + 7b - 20b
= 21b + 7b - 20b – 32
= 8b – 32

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

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.

Page 3

CBSE VII | Mathematics
Sample Paper 3 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 3 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

(128 ÷32)÷ (-4)
= 4 ÷ (-4)
= -1

Total cost = 2.40 × 10
= Rs. 24

The given observations can be arranged in ascending order as
4, 6, 9, 10, 11, 12 and 18
Here, number of observations = 7 (odd)
Median = Middle observation = 10

2x + 3 = 7
If we will transpose 3 to RHS, then the term with variable will remain on one side
and the constants will be on other side.
So, the first step is to transpose 3 to RHS.
i.e. 2x = 7 – 3

?BCA = 180° - 150° = 30°                      (linear pair angles)
Also, ?B = ?BCA = 30°                           (Angles opp. to equal sides are equal)
? ?A = 180° - 30° - 30° = 120°            (Using angle sum property of triangle)

Increased amount =
12
Rs. ×54=Rs. 6.48
100

CBSE VII | Mathematics
Sample Paper 3 - Solution

?
28
2
33

So the multiplicative inverse is
3
8
.

The triangle ABC is a right angled triangle,
By Pythagoras theorem, we have: c
2
= a
2
+ b
2

21b - 32 + 7b - 20b
= 21b + 7b - 20b – 32
= 8b – 32

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

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.

CBSE VII | Mathematics
Sample Paper 3 - Solution

Section B

13. 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
.

14. 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.

15. 725 × (-35) + (-725) × 65
= 725 × (-35) - 725 × 65
= 725 x (-35 - 65)         [Using distributive property]
= 725 × (-100)
= -72500

16. Sum of 38 and -87 = 38 + (-87) = 38 - 87 = -49

Subtracting (-134) from -49, we get
-49 - (-134) = -49 + 134 = 85

17. 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

18. Pie filling made in 1 minute = 9.2 kg
Pie filling made in 6 minutes = 6 × 9.2 kg = 55.2 kg

Page 4

CBSE VII | Mathematics
Sample Paper 3 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 3 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

(128 ÷32)÷ (-4)
= 4 ÷ (-4)
= -1

Total cost = 2.40 × 10
= Rs. 24

The given observations can be arranged in ascending order as
4, 6, 9, 10, 11, 12 and 18
Here, number of observations = 7 (odd)
Median = Middle observation = 10

2x + 3 = 7
If we will transpose 3 to RHS, then the term with variable will remain on one side
and the constants will be on other side.
So, the first step is to transpose 3 to RHS.
i.e. 2x = 7 – 3

?BCA = 180° - 150° = 30°                      (linear pair angles)
Also, ?B = ?BCA = 30°                           (Angles opp. to equal sides are equal)
? ?A = 180° - 30° - 30° = 120°            (Using angle sum property of triangle)

Increased amount =
12
Rs. ×54=Rs. 6.48
100

CBSE VII | Mathematics
Sample Paper 3 - Solution

?
28
2
33

So the multiplicative inverse is
3
8
.

The triangle ABC is a right angled triangle,
By Pythagoras theorem, we have: c
2
= a
2
+ b
2

21b - 32 + 7b - 20b
= 21b + 7b - 20b – 32
= 8b – 32

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

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.

CBSE VII | Mathematics
Sample Paper 3 - Solution

Section B

13. 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
.

14. 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.

15. 725 × (-35) + (-725) × 65
= 725 × (-35) - 725 × 65
= 725 x (-35 - 65)         [Using distributive property]
= 725 × (-100)
= -72500

16. Sum of 38 and -87 = 38 + (-87) = 38 - 87 = -49

Subtracting (-134) from -49, we get
-49 - (-134) = -49 + 134 = 85

17. 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

18. Pie filling made in 1 minute = 9.2 kg
Pie filling made in 6 minutes = 6 × 9.2 kg = 55.2 kg

CBSE VII | Mathematics
Sample Paper 3 - Solution

19. Distance travelled with 1 gallon =
2 32
10 =
33
miles
Distance travelled with
1 11
5=
22
gallons.
11 32
= × miles
23
16
=11× miles
3
176
= miles
3
mile
Thus, Sam can go
176
3
miles with
11
2
gallons.

20.  ASA congruence criterion:
The Angle Side Angle (ASA) postulate states that if under correspondence, two
angles and the included side of a triangle is equal to two corresponding angles and
included side of another triangle, then the two triangles are congruent.
Consider the triangles ABC and XYZ as shown below.

Two angles and the included side are congruent.
? ABC = ? XYZ (equal angle)
BC = YZ (equal side)
? ACB = ? XZY (equal angle)
So, ABC XYZ
Therefore, by the ASA congruence criterion, the triangles are congruent.

Page 5

CBSE VII | Mathematics
Sample Paper 3 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 3 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

(128 ÷32)÷ (-4)
= 4 ÷ (-4)
= -1

Total cost = 2.40 × 10
= Rs. 24

The given observations can be arranged in ascending order as
4, 6, 9, 10, 11, 12 and 18
Here, number of observations = 7 (odd)
Median = Middle observation = 10

2x + 3 = 7
If we will transpose 3 to RHS, then the term with variable will remain on one side
and the constants will be on other side.
So, the first step is to transpose 3 to RHS.
i.e. 2x = 7 – 3

?BCA = 180° - 150° = 30°                      (linear pair angles)
Also, ?B = ?BCA = 30°                           (Angles opp. to equal sides are equal)
? ?A = 180° - 30° - 30° = 120°            (Using angle sum property of triangle)

Increased amount =
12
Rs. ×54=Rs. 6.48
100

CBSE VII | Mathematics
Sample Paper 3 - Solution

?
28
2
33

So the multiplicative inverse is
3
8
.

The triangle ABC is a right angled triangle,
By Pythagoras theorem, we have: c
2
= a
2
+ b
2

21b - 32 + 7b - 20b
= 21b + 7b - 20b – 32
= 8b – 32

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

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.

CBSE VII | Mathematics
Sample Paper 3 - Solution

Section B

13. 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
.

14. 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.

15. 725 × (-35) + (-725) × 65
= 725 × (-35) - 725 × 65
= 725 x (-35 - 65)         [Using distributive property]
= 725 × (-100)
= -72500

16. Sum of 38 and -87 = 38 + (-87) = 38 - 87 = -49

Subtracting (-134) from -49, we get
-49 - (-134) = -49 + 134 = 85

17. 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

18. Pie filling made in 1 minute = 9.2 kg
Pie filling made in 6 minutes = 6 × 9.2 kg = 55.2 kg

CBSE VII | Mathematics
Sample Paper 3 - Solution

19. Distance travelled with 1 gallon =
2 32
10 =
33
miles
Distance travelled with
1 11
5=
22
gallons.
11 32
= × miles
23
16
=11× miles
3
176
= miles
3
mile
Thus, Sam can go
176
3
miles with
11
2
gallons.

20.  ASA congruence criterion:
The Angle Side Angle (ASA) postulate states that if under correspondence, two
angles and the included side of a triangle is equal to two corresponding angles and
included side of another triangle, then the two triangles are congruent.
Consider the triangles ABC and XYZ as shown below.

Two angles and the included side are congruent.
? ABC = ? XYZ (equal angle)
BC = YZ (equal side)
? ACB = ? XZY (equal angle)
So, ABC XYZ
Therefore, by the ASA congruence criterion, the triangles are congruent.

CBSE VII | Mathematics
Sample Paper 3 - Solution

21.  Let A and B be the two numbers such that,
40% of A =
2
B
3

Then,
?
??
??
? ? ? ?
??
??
??
40 2
100 3
22
53
2 5 10
3 2 3
: 5:3
AB
AB
A
B
AB

22. Here, AB = PR (= 3.5 cm),
BC = PQ (= 7.1 cm)
And AC = QR (= 5 cm)

This shows that the three sides of one triangle are equal to the three sides
Of the other triangle. So, by SSS congruence rule, the two triangles are
congruent. From the above three equality relations, it can be easily seen
that A ? R, B ? P and C ? Q.
So, we have ?ABC ? ?RPQ

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## Mathematics (Maths) Class 7

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