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Page 1
CBSE VIII | Mathematics
Sample Paper 1 – Solution
CBSE Board
Class VIII Mathematics
Term II
Sample Paper 1 – Solution
Time: 2 ½ hours Total Marks: 80
Section A
1.
Correct option: B
? ? ? ? ? ? ? ? ? ? ?
333 33
1000 5 2 5 2 5 2 5 2 5 2 10
2.
Correct option: B
For a hexagonal prism, V = 12, E = 18
Applying Euler's formula: F + V - E = 2
F = 2 + E – V = 2 + 18 – 12 = 8
3. Correct option: C
? ? ? ? ? ? ? ? ?
22
484 2 2 11 11 2 11 2 11 22
4. Correct option: D
Area of square = 49 cm
2
= 7
2
cm
2
Therefore, side of square is = 7 cm
So the perimeter of square = 7 ? 4 = 28 cm
5. Correct answer: C
6. Correct option: C
480 15
x5
480 5
x 160
15
?
?
? ? ?
7. Correct option: A
6k 17 29
k2
12k 34 29k
17k 34 k 2
?
?
? ? ?
? ? ? ?
Page 2
CBSE VIII | Mathematics
Sample Paper 1 – Solution
CBSE Board
Class VIII Mathematics
Term II
Sample Paper 1 – Solution
Time: 2 ½ hours Total Marks: 80
Section A
1.
Correct option: B
? ? ? ? ? ? ? ? ? ? ?
333 33
1000 5 2 5 2 5 2 5 2 5 2 10
2.
Correct option: B
For a hexagonal prism, V = 12, E = 18
Applying Euler's formula: F + V - E = 2
F = 2 + E – V = 2 + 18 – 12 = 8
3. Correct option: C
? ? ? ? ? ? ? ? ?
22
484 2 2 11 11 2 11 2 11 22
4. Correct option: D
Area of square = 49 cm
2
= 7
2
cm
2
Therefore, side of square is = 7 cm
So the perimeter of square = 7 ? 4 = 28 cm
5. Correct answer: C
6. Correct option: C
480 15
x5
480 5
x 160
15
?
?
? ? ?
7. Correct option: A
6k 17 29
k2
12k 34 29k
17k 34 k 2
?
?
? ? ?
? ? ? ?
CBSE VIII | Mathematics
Sample Paper 1 – Solution
8. Correct option: D
Each of the three expressions are with minimum power of x is 2; power of y is 2 and
no power of z, so the greatest common factor of the three expressions is x
2
y
2
.
9. Correct answer: A
Central angle =
o o o
Value of component 1
360 360 90
Total value 4
? ? ? ?
10. Correct answer: C
a
2
- 2ab + b
2
= a
2
- ab - ab + b
2
= a(a - b) - b(a - b) = (a - b)
2
11. Correct answer: C
(2, 2) lies on the graph of y = x as x = 2 and y = 2 i.e., x = y
12. Correct answer: C
If the unit place of a number is zero then it is divisible by both 5 and 10.
Section B
13.
(i)
(ii)
Page 3
CBSE VIII | Mathematics
Sample Paper 1 – Solution
CBSE Board
Class VIII Mathematics
Term II
Sample Paper 1 – Solution
Time: 2 ½ hours Total Marks: 80
Section A
1.
Correct option: B
? ? ? ? ? ? ? ? ? ? ?
333 33
1000 5 2 5 2 5 2 5 2 5 2 10
2.
Correct option: B
For a hexagonal prism, V = 12, E = 18
Applying Euler's formula: F + V - E = 2
F = 2 + E – V = 2 + 18 – 12 = 8
3. Correct option: C
? ? ? ? ? ? ? ? ?
22
484 2 2 11 11 2 11 2 11 22
4. Correct option: D
Area of square = 49 cm
2
= 7
2
cm
2
Therefore, side of square is = 7 cm
So the perimeter of square = 7 ? 4 = 28 cm
5. Correct answer: C
6. Correct option: C
480 15
x5
480 5
x 160
15
?
?
? ? ?
7. Correct option: A
6k 17 29
k2
12k 34 29k
17k 34 k 2
?
?
? ? ?
? ? ? ?
CBSE VIII | Mathematics
Sample Paper 1 – Solution
8. Correct option: D
Each of the three expressions are with minimum power of x is 2; power of y is 2 and
no power of z, so the greatest common factor of the three expressions is x
2
y
2
.
9. Correct answer: A
Central angle =
o o o
Value of component 1
360 360 90
Total value 4
? ? ? ?
10. Correct answer: C
a
2
- 2ab + b
2
= a
2
- ab - ab + b
2
= a(a - b) - b(a - b) = (a - b)
2
11. Correct answer: C
(2, 2) lies on the graph of y = x as x = 2 and y = 2 i.e., x = y
12. Correct answer: C
If the unit place of a number is zero then it is divisible by both 5 and 10.
Section B
13.
(i)
(ii)
CBSE VIII | Mathematics
Sample Paper 1 – Solution
14. A number is divisible by 4 if the number formed by its last two digits is divisible by
4.
(i) Consider 45748
Number formed by last two digits is 48.
48 is divisible by 4.
Hence, 45748 is also divisible by 4.
(ii) Consider 21404
Number formed by last two digits is 04.
Now, 04 is divisible by 4.
Hence, 21404 is also divisible by 4.
15. 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
Hence, the perimeter of the given trapezium is 28 cm.
16. 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.
17. Let x be the greater part.
Then, 64 x ? is the smaller part.
Now, according to condition given,
??
??
3 5(64 )
3 320 5
xx
xx
3x + 5x = 320
8x = 320
x = 40
Hence, the two parts are 40 and 24.
18. Here,
Total number of outcomes = 10 + 25 = 35
Let E be the event of getting a prize.
Number of outcomes favourable to event E = 10
P (E) =
Number of favourable outcomes 10 2
Total number of outcomes 35 7
??
Probability of getting a prize =
Page 4
CBSE VIII | Mathematics
Sample Paper 1 – Solution
CBSE Board
Class VIII Mathematics
Term II
Sample Paper 1 – Solution
Time: 2 ½ hours Total Marks: 80
Section A
1.
Correct option: B
? ? ? ? ? ? ? ? ? ? ?
333 33
1000 5 2 5 2 5 2 5 2 5 2 10
2.
Correct option: B
For a hexagonal prism, V = 12, E = 18
Applying Euler's formula: F + V - E = 2
F = 2 + E – V = 2 + 18 – 12 = 8
3. Correct option: C
? ? ? ? ? ? ? ? ?
22
484 2 2 11 11 2 11 2 11 22
4. Correct option: D
Area of square = 49 cm
2
= 7
2
cm
2
Therefore, side of square is = 7 cm
So the perimeter of square = 7 ? 4 = 28 cm
5. Correct answer: C
6. Correct option: C
480 15
x5
480 5
x 160
15
?
?
? ? ?
7. Correct option: A
6k 17 29
k2
12k 34 29k
17k 34 k 2
?
?
? ? ?
? ? ? ?
CBSE VIII | Mathematics
Sample Paper 1 – Solution
8. Correct option: D
Each of the three expressions are with minimum power of x is 2; power of y is 2 and
no power of z, so the greatest common factor of the three expressions is x
2
y
2
.
9. Correct answer: A
Central angle =
o o o
Value of component 1
360 360 90
Total value 4
? ? ? ?
10. Correct answer: C
a
2
- 2ab + b
2
= a
2
- ab - ab + b
2
= a(a - b) - b(a - b) = (a - b)
2
11. Correct answer: C
(2, 2) lies on the graph of y = x as x = 2 and y = 2 i.e., x = y
12. Correct answer: C
If the unit place of a number is zero then it is divisible by both 5 and 10.
Section B
13.
(i)
(ii)
CBSE VIII | Mathematics
Sample Paper 1 – Solution
14. A number is divisible by 4 if the number formed by its last two digits is divisible by
4.
(i) Consider 45748
Number formed by last two digits is 48.
48 is divisible by 4.
Hence, 45748 is also divisible by 4.
(ii) Consider 21404
Number formed by last two digits is 04.
Now, 04 is divisible by 4.
Hence, 21404 is also divisible by 4.
15. 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
Hence, the perimeter of the given trapezium is 28 cm.
16. 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.
17. Let x be the greater part.
Then, 64 x ? is the smaller part.
Now, according to condition given,
??
??
3 5(64 )
3 320 5
xx
xx
3x + 5x = 320
8x = 320
x = 40
Hence, the two parts are 40 and 24.
18. Here,
Total number of outcomes = 10 + 25 = 35
Let E be the event of getting a prize.
Number of outcomes favourable to event E = 10
P (E) =
Number of favourable outcomes 10 2
Total number of outcomes 35 7
??
Probability of getting a prize =
CBSE VIII | Mathematics
Sample Paper 1 – Solution
19. (a + b)
2
= a
2
+ 2ab + b
2
Here, a = 3 and b = -4
20. The generalized form is as below:
(i) 45 = 40 + 5
= 4 × 10 + 5 × 1
(ii) 123 = 100 + 20 + 3
= 1 × 100 + 2 × 10 + 3 × 1
21. Since, volume of cuboid = l ? b ? h
Here, V = 800 cm
3
, l = 20 cm and b = 10 cm
22. We know that two quantities x and y vary directly if the ratio
x
y
remains constant, so
we find the ratio in each case.
x 5 7 9 11
y 15 21 36 33
Ratio
x
y
Clearly, the ratio
x
y
is not same for all the cases.
Thus, x and y are not varying directly.
23.
1.
Hence, the cost of one pen is Rs. (2x - 6).
Page 5
CBSE VIII | Mathematics
Sample Paper 1 – Solution
CBSE Board
Class VIII Mathematics
Term II
Sample Paper 1 – Solution
Time: 2 ½ hours Total Marks: 80
Section A
1.
Correct option: B
? ? ? ? ? ? ? ? ? ? ?
333 33
1000 5 2 5 2 5 2 5 2 5 2 10
2.
Correct option: B
For a hexagonal prism, V = 12, E = 18
Applying Euler's formula: F + V - E = 2
F = 2 + E – V = 2 + 18 – 12 = 8
3. Correct option: C
? ? ? ? ? ? ? ? ?
22
484 2 2 11 11 2 11 2 11 22
4. Correct option: D
Area of square = 49 cm
2
= 7
2
cm
2
Therefore, side of square is = 7 cm
So the perimeter of square = 7 ? 4 = 28 cm
5. Correct answer: C
6. Correct option: C
480 15
x5
480 5
x 160
15
?
?
? ? ?
7. Correct option: A
6k 17 29
k2
12k 34 29k
17k 34 k 2
?
?
? ? ?
? ? ? ?
CBSE VIII | Mathematics
Sample Paper 1 – Solution
8. Correct option: D
Each of the three expressions are with minimum power of x is 2; power of y is 2 and
no power of z, so the greatest common factor of the three expressions is x
2
y
2
.
9. Correct answer: A
Central angle =
o o o
Value of component 1
360 360 90
Total value 4
? ? ? ?
10. Correct answer: C
a
2
- 2ab + b
2
= a
2
- ab - ab + b
2
= a(a - b) - b(a - b) = (a - b)
2
11. Correct answer: C
(2, 2) lies on the graph of y = x as x = 2 and y = 2 i.e., x = y
12. Correct answer: C
If the unit place of a number is zero then it is divisible by both 5 and 10.
Section B
13.
(i)
(ii)
CBSE VIII | Mathematics
Sample Paper 1 – Solution
14. A number is divisible by 4 if the number formed by its last two digits is divisible by
4.
(i) Consider 45748
Number formed by last two digits is 48.
48 is divisible by 4.
Hence, 45748 is also divisible by 4.
(ii) Consider 21404
Number formed by last two digits is 04.
Now, 04 is divisible by 4.
Hence, 21404 is also divisible by 4.
15. 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
Hence, the perimeter of the given trapezium is 28 cm.
16. 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.
17. Let x be the greater part.
Then, 64 x ? is the smaller part.
Now, according to condition given,
??
??
3 5(64 )
3 320 5
xx
xx
3x + 5x = 320
8x = 320
x = 40
Hence, the two parts are 40 and 24.
18. Here,
Total number of outcomes = 10 + 25 = 35
Let E be the event of getting a prize.
Number of outcomes favourable to event E = 10
P (E) =
Number of favourable outcomes 10 2
Total number of outcomes 35 7
??
Probability of getting a prize =
CBSE VIII | Mathematics
Sample Paper 1 – Solution
19. (a + b)
2
= a
2
+ 2ab + b
2
Here, a = 3 and b = -4
20. The generalized form is as below:
(i) 45 = 40 + 5
= 4 × 10 + 5 × 1
(ii) 123 = 100 + 20 + 3
= 1 × 100 + 2 × 10 + 3 × 1
21. Since, volume of cuboid = l ? b ? h
Here, V = 800 cm
3
, l = 20 cm and b = 10 cm
22. We know that two quantities x and y vary directly if the ratio
x
y
remains constant, so
we find the ratio in each case.
x 5 7 9 11
y 15 21 36 33
Ratio
x
y
Clearly, the ratio
x
y
is not same for all the cases.
Thus, x and y are not varying directly.
23.
1.
Hence, the cost of one pen is Rs. (2x - 6).
CBSE VIII | Mathematics
Sample Paper 1 – Solution
24. 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 is 6.
Section C
25. Let B ’s income be Rs. 100.
Then, A ’s income = Rs. 160
If A ’s income is Rs. 160, then B ’s income = Rs. 100.
If A ’s income is Rs 100, then B ’s income = Rs.
100
100 Rs. 62.50
160
??
??
??
??
Therefore, B ’s income is less than A ’s income by (100 – 62.50)%, i.e. by 37.5%.
26. Converting 5.832 into fraction, we get
5832
1000
Now,
3
3
3
3
3
5832
1000
5832
1000
2 2 2 3 3 3 3 3
2 2 2 5 5 5
233
25
18
10
1.8
?
? ? ? ? ? ? ?
?
? ? ? ? ?
??
?
?
?
?
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FAQs on Class 8 Math: CBSE Sample Question Paper Solutions Term II – 1 - Mathematics (Maths) Class 8
| 1. What are some important topics covered in the CBSE Class 8 Math Sample Question Paper? |  |
Ans. The CBSE Class 8 Math Sample Question Paper covers topics such as algebraic expressions, linear equations, mensuration, geometry, statistics, and probability.
| 2. How can I prepare for the CBSE Class 8 Math Term II exam? | |
Ans. To prepare for the CBSE Class 8 Math Term II exam, you can start by thoroughly studying the textbook and understanding the concepts. Practice solving sample question papers, revise important formulas and theorems, and seek help from your teacher or classmates if needed.
| 3. Are there any tips for solving algebraic expressions in the Class 8 Math exam? | |
Ans. Yes, here are some tips for solving algebraic expressions in the Class 8 Math exam:
- Simplify the expressions by combining like terms.
- Follow the order of operations (PEMDAS) while evaluating expressions.
- Substitute the given values to find the value of the expression if possible.
- Pay attention to negative signs and use brackets where necessary.
| 4. How can I improve my understanding of geometry for the Class 8 Math exam? | |
Ans. To improve your understanding of geometry for the Class 8 Math exam, you can:
- Practice drawing and labeling geometric figures accurately.
- Learn and apply the properties of different geometric shapes.
- Solve a variety of problems involving angles, triangles, quadrilaterals, and circles.
- Use visual aids, such as diagrams and models, to understand concepts better.
| 5. Is it important to solve sample question papers for the Class 8 Math exam? | |
Ans. Yes, solving sample question papers is important for the Class 8 Math exam as it helps you familiarize yourself with the exam pattern, types of questions, and time management. It also allows you to identify your weak areas and work on improving them before the actual exam.
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