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

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

``` Page 1

CBSE VII | Mathematics
Sample Paper 1 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 1 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

According to the distributive property of integers, we have
a ? (b + c) = a ? b + a ? c

A number is chosen from numbers 1 to 5.
Odd numbers are 1, 3, 5.
Required probability
= number of ways to choose an odd number /total number of numbers
= 3/5

3x + 4 = 25
Transposing 4 to R.H.S, we get
3x = 25 - 4
3x = 21
Dividing both sides by 3, we get
x = 7

Since, the angle measuring 150
o
and y are corresponding angles. Therefore, y = 150
o
.
(As the lines are parallel, corresponding angles are equal)

We know that the measure of an exterior angle of a triangle is equal to the sum of its
two opposite interior angles.
So, x + 90
o
= 125
o

Therefore, x = 35
o

Negative of a negative integer is a positive integer.
So, - (-5) = +5
So, -2 - (-5) = -2 + 5

Page 2

CBSE VII | Mathematics
Sample Paper 1 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 1 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

According to the distributive property of integers, we have
a ? (b + c) = a ? b + a ? c

A number is chosen from numbers 1 to 5.
Odd numbers are 1, 3, 5.
Required probability
= number of ways to choose an odd number /total number of numbers
= 3/5

3x + 4 = 25
Transposing 4 to R.H.S, we get
3x = 25 - 4
3x = 21
Dividing both sides by 3, we get
x = 7

Since, the angle measuring 150
o
and y are corresponding angles. Therefore, y = 150
o
.
(As the lines are parallel, corresponding angles are equal)

We know that the measure of an exterior angle of a triangle is equal to the sum of its
two opposite interior angles.
So, x + 90
o
= 125
o

Therefore, x = 35
o

Negative of a negative integer is a positive integer.
So, - (-5) = +5
So, -2 - (-5) = -2 + 5

CBSE VII | Mathematics
Sample Paper 1 - Solution

Since, the measure of an exterior angle of a triangle is equal to the sum of
measures of its two opposite interior angles.

The given observations are 11, 10, 12, 12, 9, 10, 14, 12, 9.

Angle x is alternate angle to angle 30
o
.
And, since alternate interior angles are equal, x = 30
o
.

Since, given triangle is a right angled triangle, we have
x
2
= 3
2
+ 4
2
= 9 + 16 = 25 = 5
2

Therefore, x = 5 cm

Given, AB = RQ, AC = RP, ?BAC = ?QRP

Hence, ?ABC ? ?RQP
Page 3

CBSE VII | Mathematics
Sample Paper 1 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 1 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

According to the distributive property of integers, we have
a ? (b + c) = a ? b + a ? c

A number is chosen from numbers 1 to 5.
Odd numbers are 1, 3, 5.
Required probability
= number of ways to choose an odd number /total number of numbers
= 3/5

3x + 4 = 25
Transposing 4 to R.H.S, we get
3x = 25 - 4
3x = 21
Dividing both sides by 3, we get
x = 7

Since, the angle measuring 150
o
and y are corresponding angles. Therefore, y = 150
o
.
(As the lines are parallel, corresponding angles are equal)

We know that the measure of an exterior angle of a triangle is equal to the sum of its
two opposite interior angles.
So, x + 90
o
= 125
o

Therefore, x = 35
o

Negative of a negative integer is a positive integer.
So, - (-5) = +5
So, -2 - (-5) = -2 + 5

CBSE VII | Mathematics
Sample Paper 1 - Solution

Since, the measure of an exterior angle of a triangle is equal to the sum of
measures of its two opposite interior angles.

The given observations are 11, 10, 12, 12, 9, 10, 14, 12, 9.

Angle x is alternate angle to angle 30
o
.
And, since alternate interior angles are equal, x = 30
o
.

Since, given triangle is a right angled triangle, we have
x
2
= 3
2
+ 4
2
= 9 + 16 = 25 = 5
2

Therefore, x = 5 cm

Given, AB = RQ, AC = RP, ?BAC = ?QRP

Hence, ?ABC ? ?RQP

CBSE VII | Mathematics
Sample Paper 1 - Solution

Section B

13. To find the complement of each of the given angle, we have to subtract them from
90
o
, since the sum of two complementary angles is 90
o
.
(a) Complementary angle of 45
o
= 90
o
- 45
o
= 45
o

(b) Complementary angle of 54
o
= 90
o
- 54
o
= 36
o

(c) Complementary angle of 65
o
= 90
o
- 65
o
= 25
o

14.
(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".

(b)
y
7
- 3 = 9
Statement:
For
y
7
, one-seventh of a number y
For
y
7
- 3, 3 subtracted from one-seventh of a number y
Thus, for
y
7
- 3 = 9, the final statement is
"3 subtracted from one-seventh of a number y gives 9".

15.  part of the exercise is solved by Raju.
When  is converted into lowest form, we get

which is same as part of exercise solved by Sameer.
Thus, both have solved same part of the exercise.

Page 4

CBSE VII | Mathematics
Sample Paper 1 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 1 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

According to the distributive property of integers, we have
a ? (b + c) = a ? b + a ? c

A number is chosen from numbers 1 to 5.
Odd numbers are 1, 3, 5.
Required probability
= number of ways to choose an odd number /total number of numbers
= 3/5

3x + 4 = 25
Transposing 4 to R.H.S, we get
3x = 25 - 4
3x = 21
Dividing both sides by 3, we get
x = 7

Since, the angle measuring 150
o
and y are corresponding angles. Therefore, y = 150
o
.
(As the lines are parallel, corresponding angles are equal)

We know that the measure of an exterior angle of a triangle is equal to the sum of its
two opposite interior angles.
So, x + 90
o
= 125
o

Therefore, x = 35
o

Negative of a negative integer is a positive integer.
So, - (-5) = +5
So, -2 - (-5) = -2 + 5

CBSE VII | Mathematics
Sample Paper 1 - Solution

Since, the measure of an exterior angle of a triangle is equal to the sum of
measures of its two opposite interior angles.

The given observations are 11, 10, 12, 12, 9, 10, 14, 12, 9.

Angle x is alternate angle to angle 30
o
.
And, since alternate interior angles are equal, x = 30
o
.

Since, given triangle is a right angled triangle, we have
x
2
= 3
2
+ 4
2
= 9 + 16 = 25 = 5
2

Therefore, x = 5 cm

Given, AB = RQ, AC = RP, ?BAC = ?QRP

Hence, ?ABC ? ?RQP

CBSE VII | Mathematics
Sample Paper 1 - Solution

Section B

13. To find the complement of each of the given angle, we have to subtract them from
90
o
, since the sum of two complementary angles is 90
o
.
(a) Complementary angle of 45
o
= 90
o
- 45
o
= 45
o

(b) Complementary angle of 54
o
= 90
o
- 54
o
= 36
o

(c) Complementary angle of 65
o
= 90
o
- 65
o
= 25
o

14.
(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".

(b)
y
7
- 3 = 9
Statement:
For
y
7
, one-seventh of a number y
For
y
7
- 3, 3 subtracted from one-seventh of a number y
Thus, for
y
7
- 3 = 9, the final statement is
"3 subtracted from one-seventh of a number y gives 9".

15.  part of the exercise is solved by Raju.
When  is converted into lowest form, we get

which is same as part of exercise solved by Sameer.
Thus, both have solved same part of the exercise.

CBSE VII | Mathematics
Sample Paper 1 - Solution

16. When two line intersect the following figure is formed. This shows that 4 angles are
formed.

17. Divide the total length of ribbon by the length of each strip of ribbon that is cut from
it to get the total number of ribbon strips.
So, divide by
=
=
=
= = 6
Thus, 6 strips can be cut from the ribbon.

18. In the figure, we can see that POR and ROT are right angles.

Thus,  ?ROS + ?2 = 90
o

?ROS = 67
o

So, ?2 = 90
o
- 67
o
= 23
o

Also, ?POQ + ?1 = 90
o

?POQ = 52
o

So, ?1 = 90
o
- 52
o
= 38
o

19.
Multiply both sides by 6 (L.C.M of 3 and 2)
2 (2x -1) = 3 (x + 2)
4x - 2 = 3x + 6
Transpose 3x to L.H.S and -2 to R.H.S
4x -3x = 6 + 2
x = 8

Page 5

CBSE VII | Mathematics
Sample Paper 1 - Solution

CBSE Board
Class VII Mathematics
Term I
Sample Paper 1 - Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

According to the distributive property of integers, we have
a ? (b + c) = a ? b + a ? c

A number is chosen from numbers 1 to 5.
Odd numbers are 1, 3, 5.
Required probability
= number of ways to choose an odd number /total number of numbers
= 3/5

3x + 4 = 25
Transposing 4 to R.H.S, we get
3x = 25 - 4
3x = 21
Dividing both sides by 3, we get
x = 7

Since, the angle measuring 150
o
and y are corresponding angles. Therefore, y = 150
o
.
(As the lines are parallel, corresponding angles are equal)

We know that the measure of an exterior angle of a triangle is equal to the sum of its
two opposite interior angles.
So, x + 90
o
= 125
o

Therefore, x = 35
o

Negative of a negative integer is a positive integer.
So, - (-5) = +5
So, -2 - (-5) = -2 + 5

CBSE VII | Mathematics
Sample Paper 1 - Solution

Since, the measure of an exterior angle of a triangle is equal to the sum of
measures of its two opposite interior angles.

The given observations are 11, 10, 12, 12, 9, 10, 14, 12, 9.

Angle x is alternate angle to angle 30
o
.
And, since alternate interior angles are equal, x = 30
o
.

Since, given triangle is a right angled triangle, we have
x
2
= 3
2
+ 4
2
= 9 + 16 = 25 = 5
2

Therefore, x = 5 cm

Given, AB = RQ, AC = RP, ?BAC = ?QRP

Hence, ?ABC ? ?RQP

CBSE VII | Mathematics
Sample Paper 1 - Solution

Section B

13. To find the complement of each of the given angle, we have to subtract them from
90
o
, since the sum of two complementary angles is 90
o
.
(a) Complementary angle of 45
o
= 90
o
- 45
o
= 45
o

(b) Complementary angle of 54
o
= 90
o
- 54
o
= 36
o

(c) Complementary angle of 65
o
= 90
o
- 65
o
= 25
o

14.
(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".

(b)
y
7
- 3 = 9
Statement:
For
y
7
, one-seventh of a number y
For
y
7
- 3, 3 subtracted from one-seventh of a number y
Thus, for
y
7
- 3 = 9, the final statement is
"3 subtracted from one-seventh of a number y gives 9".

15.  part of the exercise is solved by Raju.
When  is converted into lowest form, we get

which is same as part of exercise solved by Sameer.
Thus, both have solved same part of the exercise.

CBSE VII | Mathematics
Sample Paper 1 - Solution

16. When two line intersect the following figure is formed. This shows that 4 angles are
formed.

17. Divide the total length of ribbon by the length of each strip of ribbon that is cut from
it to get the total number of ribbon strips.
So, divide by
=
=
=
= = 6
Thus, 6 strips can be cut from the ribbon.

18. In the figure, we can see that POR and ROT are right angles.

Thus,  ?ROS + ?2 = 90
o

?ROS = 67
o

So, ?2 = 90
o
- 67
o
= 23
o

Also, ?POQ + ?1 = 90
o

?POQ = 52
o

So, ?1 = 90
o
- 52
o
= 38
o

19.
Multiply both sides by 6 (L.C.M of 3 and 2)
2 (2x -1) = 3 (x + 2)
4x - 2 = 3x + 6
Transpose 3x to L.H.S and -2 to R.H.S
4x -3x = 6 + 2
x = 8

CBSE VII | Mathematics
Sample Paper 1 - Solution

20. Eggs produced by the poultry farm = 600
Eggs delivered to each shop = 600 ? 10 = 60
Money earned by a particular shopkeeper = Rs. 276
Money earned if all eggs were good = 60 × 5 = Rs. 300
Money lost due to rotten eggs = 300 - 276 = Rs. 24
Cost that shopkeeper will give for one rotten egg = Rs. 2
Number of rotten eggs = 24 ? 2 = 12
Hence, 12 eggs were rotten.

21. Material required to make 1 shirt = yards
Material required to make 6 shirts =
Thus, to make 6 shirts, yards of material will be required.

22. Given, CDE QPR
We have to find the angle in ?PQR that corresponds to ?D.
Now, since corresponding parts of congruent triangles are congruent, ?D  ?P.
Hence, ?D = 60
o
.

23. Average score = mean score
Mean

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