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# Sample Solution Paper 4 - Term- 1 Mathematics, Class 7 Class 7 Notes | EduRev

## Mathematics (Maths) Class 7

Created by: Praveen Kumar

## Class 7 : Sample Solution Paper 4 - Term- 1 Mathematics, Class 7 Class 7 Notes | EduRev

``` Page 1

CBSE VII | Mathematics
Sample Paper 4 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 4 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

For example:
(a) 56 + 73 = 129
(b) 113 + 82 = 195

A proper fraction is a fraction that represents a part of a whole.

On a number line when we add a positive integer we move to the right
Towards or away from origin depends on the number to which it is added.

An improper fraction is a combination of whole and a proper fraction.

Average of both
=
60 20
40
2
?
?

A variable takes on different numerical values; its value is not fixed. Variables are
denoted usually by letters of the alphabets, such as x, y, z, l, m, n, p, etc.

Total runs = 36 + 35 + 50 + 46 + 60 + 55 = 282.
To find the mean, we find the sum of all the observations and divide it by the
number of observations.
Therefore, in this case, mean =
282
47
6
?

Page 2

CBSE VII | Mathematics
Sample Paper 4 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 4 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

For example:
(a) 56 + 73 = 129
(b) 113 + 82 = 195

A proper fraction is a fraction that represents a part of a whole.

On a number line when we add a positive integer we move to the right
Towards or away from origin depends on the number to which it is added.

An improper fraction is a combination of whole and a proper fraction.

Average of both
=
60 20
40
2
?
?

A variable takes on different numerical values; its value is not fixed. Variables are
denoted usually by letters of the alphabets, such as x, y, z, l, m, n, p, etc.

Total runs = 36 + 35 + 50 + 46 + 60 + 55 = 282.
To find the mean, we find the sum of all the observations and divide it by the
number of observations.
Therefore, in this case, mean =
282
47
6
?

CBSE VII | Mathematics
Sample Paper 4 - Solution

an equation is a condition on a variable. The condition is that two expressions
should have equal value. Note that at least one of the two expressions must contain
the variable.

When the sum of the measures of two angles is 90°, the angles are called
complementary angles.

A median connects a vertex of a triangle to the mid-point of the opposite side.

The tests of congruency are
SSS
SAS
AAS,SAA,ASA (all same)

First convert both the distances to the same unit.
So, 3 km = 3 × 1000 m = 3000 m.
Thus, the required ratio, 3 km : 300 m is 3000 : 300 = 10 : 1.

Section B

13.
(a)

? ? ? ? ? ? ? ? 8 4 8 4
8 4 8 4
12 4
? ? ? ? ? ?
? ? ? ? ?
? ? ? ?

(b)

? ? ? ? ? ? 3 7 19 15 8 9
3 7 19 15 8 9
15 2
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?

Page 3

CBSE VII | Mathematics
Sample Paper 4 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 4 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

For example:
(a) 56 + 73 = 129
(b) 113 + 82 = 195

A proper fraction is a fraction that represents a part of a whole.

On a number line when we add a positive integer we move to the right
Towards or away from origin depends on the number to which it is added.

An improper fraction is a combination of whole and a proper fraction.

Average of both
=
60 20
40
2
?
?

A variable takes on different numerical values; its value is not fixed. Variables are
denoted usually by letters of the alphabets, such as x, y, z, l, m, n, p, etc.

Total runs = 36 + 35 + 50 + 46 + 60 + 55 = 282.
To find the mean, we find the sum of all the observations and divide it by the
number of observations.
Therefore, in this case, mean =
282
47
6
?

CBSE VII | Mathematics
Sample Paper 4 - Solution

an equation is a condition on a variable. The condition is that two expressions
should have equal value. Note that at least one of the two expressions must contain
the variable.

When the sum of the measures of two angles is 90°, the angles are called
complementary angles.

A median connects a vertex of a triangle to the mid-point of the opposite side.

The tests of congruency are
SSS
SAS
AAS,SAA,ASA (all same)

First convert both the distances to the same unit.
So, 3 km = 3 × 1000 m = 3000 m.
Thus, the required ratio, 3 km : 300 m is 3000 : 300 = 10 : 1.

Section B

13.
(a)

? ? ? ? ? ? ? ? 8 4 8 4
8 4 8 4
12 4
? ? ? ? ? ?
? ? ? ? ?
? ? ? ?

(b)

? ? ? ? ? ? 3 7 19 15 8 9
3 7 19 15 8 9
15 2
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?

CBSE VII | Mathematics
Sample Paper 4 - Solution

(c)

23 41 11 23 41 11
7 29
? ? ? ?
? ? ? ?

(d)

? ? ? ? ? ? ? ? 39 24 15 36 52 36
39 24 15 36 52 36
0 20
? ? ? ? ? ? ?
? ? ? ? ?
??

(e)

231 79 51 399 159 81
101 159
? ? ? ? ? ?
? ? ? ?

14. The numbers in ascending order are:
Team A scored: – 40, 10, 0
Total score = – 40 + 10 + 0 = – 30
Team B scored: 10, 0, – 40
Total score = 10 + 0 + (– 40)= – 30
? The scores of both teams are equal.
Yes, we can add integers in any order. As we had observed that the scores obtained
by both teams in successive rounds were numerically equal but different in order,
yet total score of both teams were equal.

15. (i)
3 2 5 3 10 3 7
2
5 5 5 5 5
??
? ? ? ? ?

(ii)
? ? 4 8 7
7 4 8 7 39 7
44
8 8 8 8 8 8
??
?
? ? ? ? ? ?

Page 4

CBSE VII | Mathematics
Sample Paper 4 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 4 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

For example:
(a) 56 + 73 = 129
(b) 113 + 82 = 195

A proper fraction is a fraction that represents a part of a whole.

On a number line when we add a positive integer we move to the right
Towards or away from origin depends on the number to which it is added.

An improper fraction is a combination of whole and a proper fraction.

Average of both
=
60 20
40
2
?
?

A variable takes on different numerical values; its value is not fixed. Variables are
denoted usually by letters of the alphabets, such as x, y, z, l, m, n, p, etc.

Total runs = 36 + 35 + 50 + 46 + 60 + 55 = 282.
To find the mean, we find the sum of all the observations and divide it by the
number of observations.
Therefore, in this case, mean =
282
47
6
?

CBSE VII | Mathematics
Sample Paper 4 - Solution

an equation is a condition on a variable. The condition is that two expressions
should have equal value. Note that at least one of the two expressions must contain
the variable.

When the sum of the measures of two angles is 90°, the angles are called
complementary angles.

A median connects a vertex of a triangle to the mid-point of the opposite side.

The tests of congruency are
SSS
SAS
AAS,SAA,ASA (all same)

First convert both the distances to the same unit.
So, 3 km = 3 × 1000 m = 3000 m.
Thus, the required ratio, 3 km : 300 m is 3000 : 300 = 10 : 1.

Section B

13.
(a)

? ? ? ? ? ? ? ? 8 4 8 4
8 4 8 4
12 4
? ? ? ? ? ?
? ? ? ? ?
? ? ? ?

(b)

? ? ? ? ? ? 3 7 19 15 8 9
3 7 19 15 8 9
15 2
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?

CBSE VII | Mathematics
Sample Paper 4 - Solution

(c)

23 41 11 23 41 11
7 29
? ? ? ?
? ? ? ?

(d)

? ? ? ? ? ? ? ? 39 24 15 36 52 36
39 24 15 36 52 36
0 20
? ? ? ? ? ? ?
? ? ? ? ?
??

(e)

231 79 51 399 159 81
101 159
? ? ? ? ? ?
? ? ? ?

14. The numbers in ascending order are:
Team A scored: – 40, 10, 0
Total score = – 40 + 10 + 0 = – 30
Team B scored: 10, 0, – 40
Total score = 10 + 0 + (– 40)= – 30
? The scores of both teams are equal.
Yes, we can add integers in any order. As we had observed that the scores obtained
by both teams in successive rounds were numerically equal but different in order,
yet total score of both teams were equal.

15. (i)
3 2 5 3 10 3 7
2
5 5 5 5 5
??
? ? ? ? ?

(ii)
? ? 4 8 7
7 4 8 7 39 7
44
8 8 8 8 8 8
??
?
? ? ? ? ? ?

CBSE VII | Mathematics
Sample Paper 4 - Solution

16.
(i)
3 21 1
74
5 5 5
? ? ?
(ii)
1 4 1
41
3 3 3
? ? ?

17. We may arrange the marks obtained by group of student in a science test in an
ascending order as following –39, 48, 56, 75, 76, 81, 85, 85, 90, 95
(i) Highest marks = 95 Lowest marks = 39
(ii) Range = 95 – 39  = 56
(iii) Mean marks =
? ? 85 76 90 85 39 48 56 95 81 75
10
? ? ? ? ? ? ? ? ?

730
73
10
??

18. Scores of 15 students in mathematics test are –
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
By arranging these scores in an ascending order
5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Mode of a given data is that value of observation which occurs for the most number
of times and the median of the given data is the middle observation while the data is
arranged in an ascending or descending order.
As here are 15 terms in the given data so the median of this data will be 8th
observation.
Hence, median = 20
Also we may find that 20 occurs 4 times (i.e. maximum number of times).
So, mode of this data = 20.
Yes, both are same.

Page 5

CBSE VII | Mathematics
Sample Paper 4 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 4 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

For example:
(a) 56 + 73 = 129
(b) 113 + 82 = 195

A proper fraction is a fraction that represents a part of a whole.

On a number line when we add a positive integer we move to the right
Towards or away from origin depends on the number to which it is added.

An improper fraction is a combination of whole and a proper fraction.

Average of both
=
60 20
40
2
?
?

A variable takes on different numerical values; its value is not fixed. Variables are
denoted usually by letters of the alphabets, such as x, y, z, l, m, n, p, etc.

Total runs = 36 + 35 + 50 + 46 + 60 + 55 = 282.
To find the mean, we find the sum of all the observations and divide it by the
number of observations.
Therefore, in this case, mean =
282
47
6
?

CBSE VII | Mathematics
Sample Paper 4 - Solution

an equation is a condition on a variable. The condition is that two expressions
should have equal value. Note that at least one of the two expressions must contain
the variable.

When the sum of the measures of two angles is 90°, the angles are called
complementary angles.

A median connects a vertex of a triangle to the mid-point of the opposite side.

The tests of congruency are
SSS
SAS
AAS,SAA,ASA (all same)

First convert both the distances to the same unit.
So, 3 km = 3 × 1000 m = 3000 m.
Thus, the required ratio, 3 km : 300 m is 3000 : 300 = 10 : 1.

Section B

13.
(a)

? ? ? ? ? ? ? ? 8 4 8 4
8 4 8 4
12 4
? ? ? ? ? ?
? ? ? ? ?
? ? ? ?

(b)

? ? ? ? ? ? 3 7 19 15 8 9
3 7 19 15 8 9
15 2
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?

CBSE VII | Mathematics
Sample Paper 4 - Solution

(c)

23 41 11 23 41 11
7 29
? ? ? ?
? ? ? ?

(d)

? ? ? ? ? ? ? ? 39 24 15 36 52 36
39 24 15 36 52 36
0 20
? ? ? ? ? ? ?
? ? ? ? ?
??

(e)

231 79 51 399 159 81
101 159
? ? ? ? ? ?
? ? ? ?

14. The numbers in ascending order are:
Team A scored: – 40, 10, 0
Total score = – 40 + 10 + 0 = – 30
Team B scored: 10, 0, – 40
Total score = 10 + 0 + (– 40)= – 30
? The scores of both teams are equal.
Yes, we can add integers in any order. As we had observed that the scores obtained
by both teams in successive rounds were numerically equal but different in order,
yet total score of both teams were equal.

15. (i)
3 2 5 3 10 3 7
2
5 5 5 5 5
??
? ? ? ? ?

(ii)
? ? 4 8 7
7 4 8 7 39 7
44
8 8 8 8 8 8
??
?
? ? ? ? ? ?

CBSE VII | Mathematics
Sample Paper 4 - Solution

16.
(i)
3 21 1
74
5 5 5
? ? ?
(ii)
1 4 1
41
3 3 3
? ? ?

17. We may arrange the marks obtained by group of student in a science test in an
ascending order as following –39, 48, 56, 75, 76, 81, 85, 85, 90, 95
(i) Highest marks = 95 Lowest marks = 39
(ii) Range = 95 – 39  = 56
(iii) Mean marks =
? ? 85 76 90 85 39 48 56 95 81 75
10
? ? ? ? ? ? ? ? ?

730
73
10
??

18. Scores of 15 students in mathematics test are –
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
By arranging these scores in an ascending order
5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Mode of a given data is that value of observation which occurs for the most number
of times and the median of the given data is the middle observation while the data is
arranged in an ascending or descending order.
As here are 15 terms in the given data so the median of this data will be 8th
observation.
Hence, median = 20
Also we may find that 20 occurs 4 times (i.e. maximum number of times).
So, mode of this data = 20.
Yes, both are same.

CBSE VII | Mathematics
Sample Paper 4 - Solution

19.
(i) 5p + 2 = 17
Put p = 1 in L.H.S
(5 × 1) + 2 = 7 ? R.H.S

Put p = 2 in L.H.S
(5 × 2) + 2 = 10 + 2 = 12 ? R.H.S

Put p = 3 in L.H.S
(5 × 3) + 2 = 17 = R.H.S
Hence p = 3 is a solution of the given equation.

(ii) 3m – 14 = 4
Put m = 4,
(3 × 4) – 14 = – 2 ? R.H.S

Put m = 5
(3 × 5)  – 14 = 1 ? R.H.S

Put m = 6
(3 × 6) – 14 = 18 – 14 = 4 = R.H.S
Hence, m = 6 is a solution of the given equation.

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## Mathematics (Maths) Class 7

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