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# Sample Solution Paper 3 - Term- 1 Math, Class 8 Class 8 Notes | EduRev

## Class 8 : Sample Solution Paper 3 - Term- 1 Math, Class 8 Class 8 Notes | EduRev

``` Page 1

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

CBSE
Class VIII Mathematics
Sample Paper – 3 Solution
Time: 3 hours                     Total Marks: 80

Section A
A rational number is a number of the form
p
q
where q? 0. So, in case of reciprocal of 0,
denominator will be 0.

20
20% of x=
100 5
x
x ?

Five vertices.

4. Correct option : A
?
196 2
484
?
1
7
2
?
?
1
7
11
11

0 and 1

1
3
= 1
Thus, the natural number 1 is equal to its cube.

A rational number is a number of the form
p
q
where q? 0. So, in case of reciprocal of 0,
denominator will be 0.

8. Correct option: A
6 17 29
2
12 34 29
17 34
2
k
k
kk
k
k
?
?
? ? ?
??
??

Page 2

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

CBSE
Class VIII Mathematics
Sample Paper – 3 Solution
Time: 3 hours                     Total Marks: 80

Section A
A rational number is a number of the form
p
q
where q? 0. So, in case of reciprocal of 0,
denominator will be 0.

20
20% of x=
100 5
x
x ?

Five vertices.

4. Correct option : A
?
196 2
484
?
1
7
2
?
?
1
7
11
11

0 and 1

1
3
= 1
Thus, the natural number 1 is equal to its cube.

A rational number is a number of the form
p
q
where q? 0. So, in case of reciprocal of 0,
denominator will be 0.

8. Correct option: A
6 17 29
2
12 34 29
17 34
2
k
k
kk
k
k
?
?
? ? ?
??
??

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

The given shape has 12 edges.

Sales tax =
4
380 15.20
100
??

11. Correct option: C
? ? ? ? ? ? ? ? ?
22
484 2 2 11 11 2 11 2 11 22

12. Correct option: B
?
?
? ? ?
14 441
22
22 441
.693
14
x
x Rs

Section B
(Questions 8 to 11 carry 2 marks each)

13. The number
2
5
lies between 0 and 1.
Draw a number line. Mark points O and A to represent 0 and 1, respectively.
Divide OA into 5 equal parts (equal to the denominator of
2
5
).The second point, Q,
represents the rational number
2
5
.

14. Steps of construction:
a) Draw AC = 8 cm.
b) Draw perpendicular bisector XY of AC meeting AC at O.
c) From O cut off OD =
1
2
× 6 cm = 3 cm along OX and OB =
1
2
× 6 cm =3 cm along OY.
d) Join AB, BC, CD and DA.
Page 3

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

CBSE
Class VIII Mathematics
Sample Paper – 3 Solution
Time: 3 hours                     Total Marks: 80

Section A
A rational number is a number of the form
p
q
where q? 0. So, in case of reciprocal of 0,
denominator will be 0.

20
20% of x=
100 5
x
x ?

Five vertices.

4. Correct option : A
?
196 2
484
?
1
7
2
?
?
1
7
11
11

0 and 1

1
3
= 1
Thus, the natural number 1 is equal to its cube.

A rational number is a number of the form
p
q
where q? 0. So, in case of reciprocal of 0,
denominator will be 0.

8. Correct option: A
6 17 29
2
12 34 29
17 34
2
k
k
kk
k
k
?
?
? ? ?
??
??

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

The given shape has 12 edges.

Sales tax =
4
380 15.20
100
??

11. Correct option: C
? ? ? ? ? ? ? ? ?
22
484 2 2 11 11 2 11 2 11 22

12. Correct option: B
?
?
? ? ?
14 441
22
22 441
.693
14
x
x Rs

Section B
(Questions 8 to 11 carry 2 marks each)

13. The number
2
5
lies between 0 and 1.
Draw a number line. Mark points O and A to represent 0 and 1, respectively.
Divide OA into 5 equal parts (equal to the denominator of
2
5
).The second point, Q,
represents the rational number
2
5
.

14. Steps of construction:
a) Draw AC = 8 cm.
b) Draw perpendicular bisector XY of AC meeting AC at O.
c) From O cut off OD =
1
2
× 6 cm = 3 cm along OX and OB =
1
2
× 6 cm =3 cm along OY.
d) Join AB, BC, CD and DA.

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

ABCD is the required rhombus.

15. We have,
Number of vertices (V) = 8
Number of edges (E) = 12
Let, number of faces = F
Since every polyhedron satisfy Euler's formula, therefore
F + V = E + 2
Or, F + 8 = 12 + 2
Or, F = 14 - 8 = 6
Hence, the number of faces are 6.
16. We have:
(a + b)
2
= a
2
+ b
2
+ 2abTaking, a = -20 and b = -5, we get
(-25)
2
= {-20 + (-5)}
2

= (-20)
2
+ (-5)
2
+ 2(-20)(-5)
= 400 + 25 + 200 = 625
17. From the graph, it is clear that 5 students favoured orange and 1 student favoured
green.
Now, 5 – 1 = 4
Therefore, 4 more students favoured orange colour than green.

18. Since ABCD is an isosceles trapezoidal, we have AD = BC
Therefore, AD = BC = 4 cm.
Now the perimeter of given trapezium
= AB + BC + CD + DA
= 12 + 4 + 8 + 4
= 28
Page 4

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

CBSE
Class VIII Mathematics
Sample Paper – 3 Solution
Time: 3 hours                     Total Marks: 80

Section A
A rational number is a number of the form
p
q
where q? 0. So, in case of reciprocal of 0,
denominator will be 0.

20
20% of x=
100 5
x
x ?

Five vertices.

4. Correct option : A
?
196 2
484
?
1
7
2
?
?
1
7
11
11

0 and 1

1
3
= 1
Thus, the natural number 1 is equal to its cube.

A rational number is a number of the form
p
q
where q? 0. So, in case of reciprocal of 0,
denominator will be 0.

8. Correct option: A
6 17 29
2
12 34 29
17 34
2
k
k
kk
k
k
?
?
? ? ?
??
??

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

The given shape has 12 edges.

Sales tax =
4
380 15.20
100
??

11. Correct option: C
? ? ? ? ? ? ? ? ?
22
484 2 2 11 11 2 11 2 11 22

12. Correct option: B
?
?
? ? ?
14 441
22
22 441
.693
14
x
x Rs

Section B
(Questions 8 to 11 carry 2 marks each)

13. The number
2
5
lies between 0 and 1.
Draw a number line. Mark points O and A to represent 0 and 1, respectively.
Divide OA into 5 equal parts (equal to the denominator of
2
5
).The second point, Q,
represents the rational number
2
5
.

14. Steps of construction:
a) Draw AC = 8 cm.
b) Draw perpendicular bisector XY of AC meeting AC at O.
c) From O cut off OD =
1
2
× 6 cm = 3 cm along OX and OB =
1
2
× 6 cm =3 cm along OY.
d) Join AB, BC, CD and DA.

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

ABCD is the required rhombus.

15. We have,
Number of vertices (V) = 8
Number of edges (E) = 12
Let, number of faces = F
Since every polyhedron satisfy Euler's formula, therefore
F + V = E + 2
Or, F + 8 = 12 + 2
Or, F = 14 - 8 = 6
Hence, the number of faces are 6.
16. We have:
(a + b)
2
= a
2
+ b
2
+ 2abTaking, a = -20 and b = -5, we get
(-25)
2
= {-20 + (-5)}
2

= (-20)
2
+ (-5)
2
+ 2(-20)(-5)
= 400 + 25 + 200 = 625
17. From the graph, it is clear that 5 students favoured orange and 1 student favoured
green.
Now, 5 – 1 = 4
Therefore, 4 more students favoured orange colour than green.

18. Since ABCD is an isosceles trapezoidal, we have AD = BC
Therefore, AD = BC = 4 cm.
Now the perimeter of given trapezium
= AB + BC + CD + DA
= 12 + 4 + 8 + 4
= 28

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

Hence, the perimeter of the given trapezium is 28 cm.
19. Original price of the book = Rs. 50
Increased % = 6%
Thus, New Price = Original Price + 6% of Original price
New price = Rs.
6
50 50 .(50 3) .53
100
Rs Rs ? ? ? ? ?
20. Area of square = (5a – 2b)
2

We use the identity (x – y )
2
= x
2
– 2xy + y
2
Taking, x = 5a and y = 2b
We get, (5a – 2b)
2
= (5a)
2
– 2(5a)(2b ) + (2b)=25a
2
– 20ab + 4b
2

21. 82
2
= (80 + 2)
2

In the property (a + b)
2
= a
2
+ b
2
+ 2ab, putting a = 80 and b = 2, we get:
(80 + 2)
2
= 80
2
+ 2
2
+ 2 × 80 × 2
= 6400 + 4 + 320
= 6724
22. Let x be the ten’s digit of the two-digit number.
Then its unit's digit = x + 7.
Number = unit’s digit + 10(ten’s digit) = (x + 7) + 10(x) = 11x + 7.
Sum of the digits = (x + 7) + (x) = 2x + 7.
Given, sum of the digits is half of the whole number.
Therefore,
1
(2 7) (11 7)
2
xx ? ? ?
4x + 14 = 11x + 7
? ? ?
? ? ?
4 11 7 14
77
xx
x

i.e. x = 1
Ten’s digit of the two-digit number is x = 1.
Unit’s digit of the two-digit number = x + 7 = 1 + 7 = 8
Thus, the required number is 18.
23.  Let the required number be x. Then,
26% of x = 65
Page 5

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

CBSE
Class VIII Mathematics
Sample Paper – 3 Solution
Time: 3 hours                     Total Marks: 80

Section A
A rational number is a number of the form
p
q
where q? 0. So, in case of reciprocal of 0,
denominator will be 0.

20
20% of x=
100 5
x
x ?

Five vertices.

4. Correct option : A
?
196 2
484
?
1
7
2
?
?
1
7
11
11

0 and 1

1
3
= 1
Thus, the natural number 1 is equal to its cube.

A rational number is a number of the form
p
q
where q? 0. So, in case of reciprocal of 0,
denominator will be 0.

8. Correct option: A
6 17 29
2
12 34 29
17 34
2
k
k
kk
k
k
?
?
? ? ?
??
??

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

The given shape has 12 edges.

Sales tax =
4
380 15.20
100
??

11. Correct option: C
? ? ? ? ? ? ? ? ?
22
484 2 2 11 11 2 11 2 11 22

12. Correct option: B
?
?
? ? ?
14 441
22
22 441
.693
14
x
x Rs

Section B
(Questions 8 to 11 carry 2 marks each)

13. The number
2
5
lies between 0 and 1.
Draw a number line. Mark points O and A to represent 0 and 1, respectively.
Divide OA into 5 equal parts (equal to the denominator of
2
5
).The second point, Q,
represents the rational number
2
5
.

14. Steps of construction:
a) Draw AC = 8 cm.
b) Draw perpendicular bisector XY of AC meeting AC at O.
c) From O cut off OD =
1
2
× 6 cm = 3 cm along OX and OB =
1
2
× 6 cm =3 cm along OY.
d) Join AB, BC, CD and DA.

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

ABCD is the required rhombus.

15. We have,
Number of vertices (V) = 8
Number of edges (E) = 12
Let, number of faces = F
Since every polyhedron satisfy Euler's formula, therefore
F + V = E + 2
Or, F + 8 = 12 + 2
Or, F = 14 - 8 = 6
Hence, the number of faces are 6.
16. We have:
(a + b)
2
= a
2
+ b
2
+ 2abTaking, a = -20 and b = -5, we get
(-25)
2
= {-20 + (-5)}
2

= (-20)
2
+ (-5)
2
+ 2(-20)(-5)
= 400 + 25 + 200 = 625
17. From the graph, it is clear that 5 students favoured orange and 1 student favoured
green.
Now, 5 – 1 = 4
Therefore, 4 more students favoured orange colour than green.

18. Since ABCD is an isosceles trapezoidal, we have AD = BC
Therefore, AD = BC = 4 cm.
Now the perimeter of given trapezium
= AB + BC + CD + DA
= 12 + 4 + 8 + 4
= 28

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

Hence, the perimeter of the given trapezium is 28 cm.
19. Original price of the book = Rs. 50
Increased % = 6%
Thus, New Price = Original Price + 6% of Original price
New price = Rs.
6
50 50 .(50 3) .53
100
Rs Rs ? ? ? ? ?
20. Area of square = (5a – 2b)
2

We use the identity (x – y )
2
= x
2
– 2xy + y
2
Taking, x = 5a and y = 2b
We get, (5a – 2b)
2
= (5a)
2
– 2(5a)(2b ) + (2b)=25a
2
– 20ab + 4b
2

21. 82
2
= (80 + 2)
2

In the property (a + b)
2
= a
2
+ b
2
+ 2ab, putting a = 80 and b = 2, we get:
(80 + 2)
2
= 80
2
+ 2
2
+ 2 × 80 × 2
= 6400 + 4 + 320
= 6724
22. Let x be the ten’s digit of the two-digit number.
Then its unit's digit = x + 7.
Number = unit’s digit + 10(ten’s digit) = (x + 7) + 10(x) = 11x + 7.
Sum of the digits = (x + 7) + (x) = 2x + 7.
Given, sum of the digits is half of the whole number.
Therefore,
1
(2 7) (11 7)
2
xx ? ? ?
4x + 14 = 11x + 7
? ? ?
? ? ?
4 11 7 14
77
xx
x

i.e. x = 1
Ten’s digit of the two-digit number is x = 1.
Unit’s digit of the two-digit number = x + 7 = 1 + 7 = 8
Thus, the required number is 18.
23.  Let the required number be x. Then,
26% of x = 65

CBSE VIII | Mathematics
Sample Paper – 3 Solutions

26
65
100
100
65
26
250
x
x
x
??
??
? ? ?
??
??
??

Hence, the required number is 250.

24.
(by commutativity)

(by distributivity)

Section C
25. Here, at unit's place A × 5 = A
That is we need to search for a number, which when multiplied to 5 gives the same unit
place.
Such a number is 5 or 0, as 5 × 5 = 25 or 0 × 5 = 0
Case (1) Taking 5 in place of A, we get
2
B   A
5
C B 5
?

Here, B can take the value 2, which satisfy the condition, taking, B = 2, we get
2
2   5
5
1 2 5
?

Thus, A = 5, B = 2 and C = 1
```
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## Mathematics (Maths) Class 8

135 videos|321 docs|48 tests

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