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# Sample Solution Paper 1 - Term- 2 Mathematics, Class 6 Class 6 Notes | EduRev

## Mathematics (Maths) Class 6

Created by: Praveen Kumar

## Class 6 : Sample Solution Paper 1 - Term- 2 Mathematics, Class 6 Class 6 Notes | EduRev

``` Page 1

CBSE VI | Mathematics
Sample Paper 1 – Solution

CBSE Board
Class VI Mathematics
Term II
Sample Paper 1 – Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

1.35 =

Data collected from a group of 40 students is an example of primary data.

Maximum marks = 47, Minimum marks = 21
Difference = Maximum – Minimum = 47 – 21 = 26

Perimeter of a square = 4 × length of a side

Area of a square = side × side = p × p = p
2

15 : 19 is not equivalent to 50 : 90.

An isosceles triangle has exactly one line of symmetry.

Protractor is used to draw and measure angles.

Out of the four the one-tenth part of 0.7 is the greatest. Hence, 0.7 has the highest
value.

n
8
5
?

is the correct expression.

Letter H has a horizontal line of symmetry.
Page 2

CBSE VI | Mathematics
Sample Paper 1 – Solution

CBSE Board
Class VI Mathematics
Term II
Sample Paper 1 – Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

1.35 =

Data collected from a group of 40 students is an example of primary data.

Maximum marks = 47, Minimum marks = 21
Difference = Maximum – Minimum = 47 – 21 = 26

Perimeter of a square = 4 × length of a side

Area of a square = side × side = p × p = p
2

15 : 19 is not equivalent to 50 : 90.

An isosceles triangle has exactly one line of symmetry.

Protractor is used to draw and measure angles.

Out of the four the one-tenth part of 0.7 is the greatest. Hence, 0.7 has the highest
value.

n
8
5
?

is the correct expression.

Letter H has a horizontal line of symmetry.

CBSE VI | Mathematics
Sample Paper 1 – Solution

= 5 + 5 + 5 + 5 + 5 + 3 = 28

Section B

13. Cost of a book = Rs. 165.35
Cost of a pen = Rs. 72.00
Cost of a notebook = Rs. 14.85
Total Cost is given by,

Total money to be paid by Preeta = Rs. 252.20

14. (a) Secondary
(b) Primary

15. We know that a regular pentagon has 5 sides, so we can divide the perimeter by 5 to
get the measure of one side.
One side of pentagon = 25 cm ÷ 5 cm = 5 cm

16. Let the number of rows be 'n'.
Since there are 11 students in a row and number of rows are n, the Rule is given as,
Number of students in the parade = 11n.

17. Three symmetrical objects are
(i) An electric tube-light
(ii) A water glass
(iii) A fan

18. Steps of construction:
(1) Draw a line l. Mark a point A on a line l.
(2) Place the compass ’ pointer on the 0 mark of the ruler. Open it to place the pencil
point up to the 4.5 cm mark.
(3) Taking caution that the opening of the compass has not changed, place the
pointer on A and swing an arc to cut l at B.
(4) AB is a line segment of required length.

Page 3

CBSE VI | Mathematics
Sample Paper 1 – Solution

CBSE Board
Class VI Mathematics
Term II
Sample Paper 1 – Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

1.35 =

Data collected from a group of 40 students is an example of primary data.

Maximum marks = 47, Minimum marks = 21
Difference = Maximum – Minimum = 47 – 21 = 26

Perimeter of a square = 4 × length of a side

Area of a square = side × side = p × p = p
2

15 : 19 is not equivalent to 50 : 90.

An isosceles triangle has exactly one line of symmetry.

Protractor is used to draw and measure angles.

Out of the four the one-tenth part of 0.7 is the greatest. Hence, 0.7 has the highest
value.

n
8
5
?

is the correct expression.

Letter H has a horizontal line of symmetry.

CBSE VI | Mathematics
Sample Paper 1 – Solution

= 5 + 5 + 5 + 5 + 5 + 3 = 28

Section B

13. Cost of a book = Rs. 165.35
Cost of a pen = Rs. 72.00
Cost of a notebook = Rs. 14.85
Total Cost is given by,

Total money to be paid by Preeta = Rs. 252.20

14. (a) Secondary
(b) Primary

15. We know that a regular pentagon has 5 sides, so we can divide the perimeter by 5 to
get the measure of one side.
One side of pentagon = 25 cm ÷ 5 cm = 5 cm

16. Let the number of rows be 'n'.
Since there are 11 students in a row and number of rows are n, the Rule is given as,
Number of students in the parade = 11n.

17. Three symmetrical objects are
(i) An electric tube-light
(ii) A water glass
(iii) A fan

18. Steps of construction:
(1) Draw a line l. Mark a point A on a line l.
(2) Place the compass ’ pointer on the 0 mark of the ruler. Open it to place the pencil
point up to the 4.5 cm mark.
(3) Taking caution that the opening of the compass has not changed, place the
pointer on A and swing an arc to cut l at B.
(4) AB is a line segment of required length.

CBSE VI | Mathematics
Sample Paper 1 – Solution

19. Ratio of 30 cm to 4 m

Ratio of 20 sec to 6 minutes

Since,
,
Therefore the given ratio do not form a proportion.

20. To get this answer subtract 27.84 from 84.5

Hence, 56.66 must be subtracted from 84.5 to get 27.84

21. The shaded portion is made up of line segments. It is covered by full and half
squares. We have to calculate the number of fully filled and half filled squares.
? Fully filled squares = 6
? Half-filled squares = 2
Area covered by fully filled square = 6 × 1 = 6 sq. units.
Area covered by half-filled square = 2 ×
1
2
= 1
Therefore, total area = 6 + 1 = 7 sq. units

22. Drawing the relational part the images become as follows:

23. (a) Butterscotch is liked by 5 + 5 + 2 = 12 students.
(b) Chocolate flavor is less favourite.

Page 4

CBSE VI | Mathematics
Sample Paper 1 – Solution

CBSE Board
Class VI Mathematics
Term II
Sample Paper 1 – Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

1.35 =

Data collected from a group of 40 students is an example of primary data.

Maximum marks = 47, Minimum marks = 21
Difference = Maximum – Minimum = 47 – 21 = 26

Perimeter of a square = 4 × length of a side

Area of a square = side × side = p × p = p
2

15 : 19 is not equivalent to 50 : 90.

An isosceles triangle has exactly one line of symmetry.

Protractor is used to draw and measure angles.

Out of the four the one-tenth part of 0.7 is the greatest. Hence, 0.7 has the highest
value.

n
8
5
?

is the correct expression.

Letter H has a horizontal line of symmetry.

CBSE VI | Mathematics
Sample Paper 1 – Solution

= 5 + 5 + 5 + 5 + 5 + 3 = 28

Section B

13. Cost of a book = Rs. 165.35
Cost of a pen = Rs. 72.00
Cost of a notebook = Rs. 14.85
Total Cost is given by,

Total money to be paid by Preeta = Rs. 252.20

14. (a) Secondary
(b) Primary

15. We know that a regular pentagon has 5 sides, so we can divide the perimeter by 5 to
get the measure of one side.
One side of pentagon = 25 cm ÷ 5 cm = 5 cm

16. Let the number of rows be 'n'.
Since there are 11 students in a row and number of rows are n, the Rule is given as,
Number of students in the parade = 11n.

17. Three symmetrical objects are
(i) An electric tube-light
(ii) A water glass
(iii) A fan

18. Steps of construction:
(1) Draw a line l. Mark a point A on a line l.
(2) Place the compass ’ pointer on the 0 mark of the ruler. Open it to place the pencil
point up to the 4.5 cm mark.
(3) Taking caution that the opening of the compass has not changed, place the
pointer on A and swing an arc to cut l at B.
(4) AB is a line segment of required length.

CBSE VI | Mathematics
Sample Paper 1 – Solution

19. Ratio of 30 cm to 4 m

Ratio of 20 sec to 6 minutes

Since,
,
Therefore the given ratio do not form a proportion.

20. To get this answer subtract 27.84 from 84.5

Hence, 56.66 must be subtracted from 84.5 to get 27.84

21. The shaded portion is made up of line segments. It is covered by full and half
squares. We have to calculate the number of fully filled and half filled squares.
? Fully filled squares = 6
? Half-filled squares = 2
Area covered by fully filled square = 6 × 1 = 6 sq. units.
Area covered by half-filled square = 2 ×
1
2
= 1
Therefore, total area = 6 + 1 = 7 sq. units

22. Drawing the relational part the images become as follows:

23. (a) Butterscotch is liked by 5 + 5 + 2 = 12 students.
(b) Chocolate flavor is less favourite.

CBSE VI | Mathematics
Sample Paper 1 – Solution

24. Let the isosceles triangle be ABC, in which AB = AC = 18 cm.
Also perimeter is given 50 cm, we need to find BC.
Perimeter of triangle = 50 cm
AB + AC + BC = 50 cm
? 18 + 18 + x = 50 cm
? 36 + x = 50 cm
? x = 50 – 36 = 14 cm
Therefore, length of third side is 14 cm.

Section C

25. Let the marks of Rohit, Ajay and Vipul be 4x, 5x and 6x respectively.
Given that Ajay's marks = 75
? 5x = 75

75
x 15
5
? ? ?
Hence, marks of Rohit = 4x = 4 × 15 = 60 marks
Marks of Vipul = 6x = 6 × 15 = 90 marks

26.
(a) Economics has the most students enrolled.
(b) From lowest to highest: Physics, Chemistry, Psychology, Political Science,
Economics.
(c) From ratio of the number of students enrolled in Economics to the number
enrolled in Chemistry we can state that enrollment in Economics is 2 times
larger than in Chemistry.

27. Steps of construction:
(1) Given AB whose length is not known.
(2) Fix the compasses pointer on A and the pencil end on B. The opening of the
instrument now gives the length of AB .
(3) Draw any line l. Choose a point C on l. Without changing the compasses setting,
place the pointer on C.
(4) Swing an arc that cuts l at a point, say D. Now CD is a copy of AB .

Page 5

CBSE VI | Mathematics
Sample Paper 1 – Solution

CBSE Board
Class VI Mathematics
Term II
Sample Paper 1 – Solution
Time: 2 ½ hours                          Total Marks: 80

Section A

1.35 =

Data collected from a group of 40 students is an example of primary data.

Maximum marks = 47, Minimum marks = 21
Difference = Maximum – Minimum = 47 – 21 = 26

Perimeter of a square = 4 × length of a side

Area of a square = side × side = p × p = p
2

15 : 19 is not equivalent to 50 : 90.

An isosceles triangle has exactly one line of symmetry.

Protractor is used to draw and measure angles.

Out of the four the one-tenth part of 0.7 is the greatest. Hence, 0.7 has the highest
value.

n
8
5
?

is the correct expression.

Letter H has a horizontal line of symmetry.

CBSE VI | Mathematics
Sample Paper 1 – Solution

= 5 + 5 + 5 + 5 + 5 + 3 = 28

Section B

13. Cost of a book = Rs. 165.35
Cost of a pen = Rs. 72.00
Cost of a notebook = Rs. 14.85
Total Cost is given by,

Total money to be paid by Preeta = Rs. 252.20

14. (a) Secondary
(b) Primary

15. We know that a regular pentagon has 5 sides, so we can divide the perimeter by 5 to
get the measure of one side.
One side of pentagon = 25 cm ÷ 5 cm = 5 cm

16. Let the number of rows be 'n'.
Since there are 11 students in a row and number of rows are n, the Rule is given as,
Number of students in the parade = 11n.

17. Three symmetrical objects are
(i) An electric tube-light
(ii) A water glass
(iii) A fan

18. Steps of construction:
(1) Draw a line l. Mark a point A on a line l.
(2) Place the compass ’ pointer on the 0 mark of the ruler. Open it to place the pencil
point up to the 4.5 cm mark.
(3) Taking caution that the opening of the compass has not changed, place the
pointer on A and swing an arc to cut l at B.
(4) AB is a line segment of required length.

CBSE VI | Mathematics
Sample Paper 1 – Solution

19. Ratio of 30 cm to 4 m

Ratio of 20 sec to 6 minutes

Since,
,
Therefore the given ratio do not form a proportion.

20. To get this answer subtract 27.84 from 84.5

Hence, 56.66 must be subtracted from 84.5 to get 27.84

21. The shaded portion is made up of line segments. It is covered by full and half
squares. We have to calculate the number of fully filled and half filled squares.
? Fully filled squares = 6
? Half-filled squares = 2
Area covered by fully filled square = 6 × 1 = 6 sq. units.
Area covered by half-filled square = 2 ×
1
2
= 1
Therefore, total area = 6 + 1 = 7 sq. units

22. Drawing the relational part the images become as follows:

23. (a) Butterscotch is liked by 5 + 5 + 2 = 12 students.
(b) Chocolate flavor is less favourite.

CBSE VI | Mathematics
Sample Paper 1 – Solution

24. Let the isosceles triangle be ABC, in which AB = AC = 18 cm.
Also perimeter is given 50 cm, we need to find BC.
Perimeter of triangle = 50 cm
AB + AC + BC = 50 cm
? 18 + 18 + x = 50 cm
? 36 + x = 50 cm
? x = 50 – 36 = 14 cm
Therefore, length of third side is 14 cm.

Section C

25. Let the marks of Rohit, Ajay and Vipul be 4x, 5x and 6x respectively.
Given that Ajay's marks = 75
? 5x = 75

75
x 15
5
? ? ?
Hence, marks of Rohit = 4x = 4 × 15 = 60 marks
Marks of Vipul = 6x = 6 × 15 = 90 marks

26.
(a) Economics has the most students enrolled.
(b) From lowest to highest: Physics, Chemistry, Psychology, Political Science,
Economics.
(c) From ratio of the number of students enrolled in Economics to the number
enrolled in Chemistry we can state that enrollment in Economics is 2 times
larger than in Chemistry.

27. Steps of construction:
(1) Given AB whose length is not known.
(2) Fix the compasses pointer on A and the pencil end on B. The opening of the
instrument now gives the length of AB .
(3) Draw any line l. Choose a point C on l. Without changing the compasses setting,
place the pointer on C.
(4) Swing an arc that cuts l at a point, say D. Now CD is a copy of AB .

CBSE VI | Mathematics
Sample Paper 1 – Solution

28. From the figure the smaller rectangles are equal.
Therefore,
Area of one smaller rectangle = length × breadth = 3 × 4 = 12 sq. m
Area of 3 smaller rectangles = 3 × 12 = 36 sq. m
Area of bigger rectangle = length × breadth = 10 × 8 = 80 sq. m
Therefore, area of remaining part = 80 – 36 = 44 sq. m

29. Rate percent per annum is interest given on Rs. 100 for a year.
Let the interest for Rs. 100 per annum be Rs. x.
Principal : Principal :: Interest : Interest
5250 : 100 :: 420 : x
Product of extreme terms = 5250x
Product of the middle terms = 100 × 420 = 42000
5250x 100 420
42000
x8
5250
??
? ? ?

30.
(a) In this case time is unknown and distance is known. Therefore, we proceed as
follows:
6 hours = 6 × 60 minutes = 360 minutes
300 km is covered in 360 minutes
Time required to cover 1 km distance =
360 6
min
300 5
?

Therefore, 280 km can be covered in
6
280 336 minutes 5 hours 36 minutes
5
? ? ?

(b) In this case distance is unknown and time is known. Therefore, we  proceed as
follows:
Distance covered in 6 hours = 300 km
Distance covered in 1 hour =
300
50 km
6
?
Therefore, distance covered in 10 hours = 50 × 10 = 500 km.

31. Area of a square wall = side × side = 10 × 10 = 100 sq. m
Area of four square walls = 4 × 100 = 400 sq. m
Cost of painting 1 sq. m of wall = Rs. 20
Therefore, total cost of painting = Rs. 20 × 400 = Rs. 8000

32.    (a) In 2009-April, May and in 2010-May, June.
(b) In 2009-May and in 2010-May.
(c) In 2009-June and in 2010-March and May

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## Mathematics (Maths) Class 6

221 videos|105 docs|43 tests

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