Class 6

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 6 Answers Solutions.

Integers Class 6 Maths Chapter 3 Practice Set 6 Solutions Maharashtra Board

Std 6 Maths Practice Set 6 Solutions Answers

Question 1.
Write the opposite number of each of the numbers given below.

Number	47	+52	-33	-84	-21	+16	-26	80
Opposite number

Solution:

Number	47	+52	-33	-84	-21	+16	-26	80
Opposite number	-47	-52	+33	+84	+21	-16	+26	-80

Std 6 Maths Digest

• Practice Set 1 Class 6 Answers
• Practice Set 2 Class 6 Answers
• Practice Set 3 Class 6 Answers
• Practice Set 4 Class 6 Answers
• Practice Set 5 Class 6 Answers
• Practice Set 6 Class 6 Answers
• Practice Set 7 Class 6 Answers
• Practice Set 8 Class 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 11 Ratio-Proportion Class 6 Practice Set 28 Answers Solutions.

Ratio-Proportion Class 6 Maths Chapter 11 Practice Set 28 Solutions Maharashtra Board

Std 6 Maths Practice Set 28 Solutions Answers

6th Standard Maths Practice Set 28 Question 1.
In each example below, find the ratio of the first number to the second:
i. 24, 56
ii. 63,49
iii. 52, 65
iv. 84, 60
v. 35, 65
vi. 121, 99
Solution:
i. 24, 56
\(\frac{24}{56}=\frac{24 \div 8}{56 \div 8}=\frac{3}{7}\)
= 3:7

ii. 63,49
\(\frac{63}{49}=\frac{63 \div 7}{49 \div 7}=\frac{9}{7}\)
= 9:7

iii. 52, 65
\(\frac{52}{65}=\frac{52 \div 13}{65 \div 13}=\frac{4}{5}\)
= 4:5

iv. 84, 60
\(\frac{84}{60}=\frac{84 \div 12}{60 \div 12}=\frac{7}{5}\)
= 7:5

v. 35, 65
\(\frac{35}{65}=\frac{35 \div 5}{65 \div 5}=\frac{7}{13}\)
= 7:13

vi. 121, 99
\(\frac{121}{99}=\frac{121 \div 11}{99 \div 11}=\frac{11}{9}\)
= 11:9

6th Maths Practice Set 28 Question 2.
Find the ratio of the first quantity to the second.
i. 25 beads, 40 beads
ii. Rs 40, Rs 120
iii. 15 minutes, 1 hour
iv. 30 litres, 24 litres
v. 99 kg, 44000 grams
vi. 1 litre, 250 ml
vii. 60 paise, 1 rupee
viii. 750 grams, \(\frac { 1 }{ 2 }\) kg
ix. 125 cm, 1 metre
Solution:
i. Required Ratio = \(\frac{25}{40}=\frac{25 \div 5}{40 \div 5}=\frac{5}{8}\)

ii. Required Ratio = \(\frac{40}{120}=\frac{40 \div 40}{120 \div 40}=\frac{1}{3}\)

iii. 1 hour = 60 minutes
Required Ratio = \(\frac{15}{60}=\frac{15 \div 15}{60 \div 15}=\frac{1}{4}\)

iv. Required Ratio = \(\frac{30}{24}=\frac{30 \div 6}{24 \div 6}=\frac{5}{4}\)

v. 99 kg = 99 x 1000 grams = 99000 grams
Required Ratio = \(\frac{99000}{44000}=\frac{99000 \div 1000}{44000 \div 1000}=\frac{99}{44}\)
= \(\frac{99}{44}=\frac{99 \div 11}{44 \div 11}=\frac{9}{4}\)

vi. 1 litre, 250 ml
1 litre = 1000 ml
Required Ratio = \(\frac{1000}{250}=\frac{1000 \div 10}{250 \div 10}=\frac{100}{25}\)
= \(\frac{100}{25}=\frac{100 \div 25}{25 \div 25}=\frac{4}{1}\)

viii. 750 grams, \(\frac { 1 }{ 2 }\) kg
\(\frac { 1 }{ 2 }\) kg = \(\frac { 1000 }{ 2 }\) grams = 500 grams
Required Ratio = \(\frac{750}{500}=\frac{750 \div 10}{500 \div 10}=\frac{75}{50}\)
= \(\frac{75}{50}=\frac{75 \div 25}{50 \div 25}=\frac{3}{2}\)

ix. 125 cm, 1 metre
1 metre = 100 cm
Required Ratio = \(\frac{125}{100}=\frac{125 \div 25}{100 \div 25}=\frac{5}{4}\)

6th Std Maths Practice Set 28 Question 3.
Reema has 24 notebooks and 18 books. Find the ratio of notebooks to books.
Solution:
Ratio of notebooks to books

∴ The ratio of notebooks to books with Reema is \(\frac { 4 }{ 3 }\)

Practice Set 28 Question 4.
30 cricket players and 20 kho-kho players are training on a field. What is the ratio of cricket players to the total number of players?
Solution:
Total number of players = Cricket players + Kho-kho players
= 30 + 20 = 50
Ratio of cricket players to the total number of players

∴ The ratio of cricket players to the total number of players is \(\frac { 3 }{ 5 }\).

Question 5.
Snehal has a red ribbon that is 80 cm long and a blue ribbon 2.20 m long. What is the ratio of the length of the red ribbon to that of the blue ribbon?
Solution:
1 metre =100 cm
Length of the red ribbon = 80 cm
Length of the blue ribbon = 2.20 m = 2.20 x 100 cm
\(=\frac{220}{100} \times \frac{100}{1}=\frac{220 \times 100}{100 \times 1}\)
= 220 cm
∴ Ratio of length of the red ribbon to that of the blue ribbon

∴ The ratio of the length of the red ribbon to that of the blue ribbon is \(\frac { 4 }{ 11 }\).

11 Ratio Question 6.
Shubham’s age today is 12 years and his father’s is 42 years. Shubham’s mother is younger than his father by 6 years. Find the following ratios.
i. Ratio of Shubham’s age today to his mother’s age today.
ii. Ratio of Shubham’s mother’s age today to his father’s age today.
iii. The ratio of Shubham’s age to his mother’s age when Shubham was 10 years old.
Solution:
Shubham’s age today = 12 years
Shubham’s father’s age = 42 years
Shubham’s mother age = Shubham’s father’s age – 6 years
= 42 years – 6 years = 36 years

i. Ratio of Shubham’s age today to his mother’s age today

∴ The ratio of Shubham’s age today to his mother’s age today is \(\frac { 1 }{ 3 }\).

ii. Ratio of Shubham’s mother age today to his father’s age today

∴ The ratio of Shubham’s mother’s age today to his father’s age today is \(\frac { 6 }{ 7 }\).

iii. Shubham’s age today is 12 years and his mothers age is 36 years.
Hence when Shubham’s age was 10 years, his mother’s age was 34 years (i.e. 36 – 2 years).
Ratio of Shubham’s age to his mother’s age when Shubham was 10 years old

∴ The ratio of Shubham’s age to his mother’s age when Shubham was 10 years old is \(\frac { 5 }{ 17 }\)

Maharashtra Board Class 6 Maths Chapter 11 Ratio-Proportion Practice Set 28 Intext Questions and Activities

Question 1.
In the figure, colour some squares with any colour you like and leave some blank. (Textbook pg. no. 57)

i. Count all the boxes and write the number.
ii. Count the colored ones and write the number.
iii. Count the blank ones and write the number.
iv. Find the ratio of the colored boxes to the blank ones.
v. Find the ratio of the colored boxes to the total boxes.
vi. Find the ratio of the blank boxes to the total boxes.
Solution:
i. The number of all boxes is 16.
ii. The number of colored boxes is 7.
iii. The number of blank boxes is 9.
iv. Ratio of the colored boxes to the blank ones

v. Ratio of the colored boxes to the total boxes

vi. Ratio of the blank boxes to the total boxes

Std 6 Maths Digest

• Practice Set 26 Class 6 Answers
• Practice Set 27 Class 6 Answers
• Practice Set 28 Class 6 Answers
• Practice Set 29 Class 6 Answers
• Practice Set 30 Class 6 Answers
• Practice Set 31 Class 6 Answers
• Practice Set 32 Class 6 Answers
• Practice Set 33 Class 6 Answers
• Practice Set 34 Class 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 5 Answers Solutions.

Integers Class 6 Maths Chapter 3 Practice Set 5 Solutions Maharashtra Board

Std 6 Maths Practice Set 5 Solutions Answers

Question 1.
Add:

1. 8 + 6
2. 9 + (-3)
3. 5 + (-6)
4. – 7 + 2
5. – 8 + 0
6. – 5 + (-2)

Solution:

1. 8 + 6 = (+8) + (+6) = +14	2. 9 + (-3) = (+9) + (- 3) = +6	3. 5 + (-6) = (+5) + (-6) = -1
4. -7 + 2 = (-7) + (+2) = -5	5. -8 + 0 = (-8) + 0 = -8	6. -5 + (-2) = (-5) + (-2) = -7

Question 2.
Complete the table given below:

+	8	4	-3	-5
-2	-2 + 8 = +6
6
0
-4

Solution:

+	8	4	-3	-5
-2	(-2) + (+8) = +6	(-2) +(+4) = 2	(-2) +(-3) =-5	(-2) +(-5) =-7
6	(+6) + (+8) = 14	(+6) + (+4) = 10	(+6) + (-3) = 3	(+6) + (-5) = 1
0	0 + (+8) = 8	0 + (+4) = 4	0 + (-3) = -3	0 + (-5) = -5
-4	(-4) + (+8) = 4	(-4) +(+4) = 0	(-4) + (-3) = -7	(-4) + (-5) = -9

Maharashtra Board Class 6 Maths Chapter 3 Integers Practice Set 5 Intext Questions and Activities

Question 1.
On the playground, mark a timeline showing the years from 2000 to 2024. With one child standing at the position of the 2017, ask the following questions: (Textbook pg. no. 15)

1. While playing this game, what is his/her age?
2. Five years ago, which year was it? And what was his / her age then?
3. In which year will he / she go to Std X? How old will he / she be then?

The child should find answers to such questions by walking the right number of units and in the right direction on the timeline.
[Assume child born year is 2009]
Solutions:

1. Age of child is 8 years.
2. Five years ago, year was 2012. His/her age is 3 years.
3. In 2024, he/she will go the Std X. His/her age is 15 years.

Question 2.
On a playground mark a timeline of 100 years. This will make it possible to count the years from 0 to 2100 on it. Important historical events can then be shown in proper centuries. (Textbook pg. no. 16)
Solution:
(Students should attempt this activity on their own)

Question 3.
Observe the figures and write appropriate number in the boxes given below. (Textbook pg. no. 16 and 17)
i.

a. At first the rabbit was at the number ____
b. It hopped ___ units to the right.
c. It is now at the number ___
Solution:
i.
a. +1
b. 5
c. +6

ii.

a. At first the rabbit was at the number ___
b. It hopped ____ units to the right.
c. It is now at the number ____
Solution:
ii.
a. -2
b. 5
c. +3

iii.

a. At first the rabbit was at the number ___
b. It hopped ____ units to the left.
c. It is now at the number ___
Solution:
iii.
a. -3
b. 4
c. -7

iv.

a. At first the rabbit was at the number ___
b. It hopped___units to the left.
c. It is now at the number ____
Solution:
iv.
a. +3
b. 4
c. -1

Std 6 Maths Digest

• Practice Set 1 Class 6 Answers
• Practice Set 2 Class 6 Answers
• Practice Set 3 Class 6 Answers
• Practice Set 4 Class 6 Answers
• Practice Set 5 Class 6 Answers
• Practice Set 6 Class 6 Answers
• Practice Set 7 Class 6 Answers
• Practice Set 8 Class 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 17 Geometrical Constructions Class 6 Practice Set 39 Answers Solutions.

Geometrical Constructions Class 6 Maths Chapter 17 Practice Set 39 Solutions Maharashtra Board

Std 6 Maths Practice Set 39 Solutions Answers

Question 1.
Draw line l. Take any point P on the line. Using a set square, draw a line perpendicular to line l at the point P.
Solution:
Step 1:

Step 2:

line PQ ⊥ line l

Question 2.
Draw a line AB. Using a compass, draw a line perpendicular to AB at point B.
Solution:
Step 1:

Step 2:

Step 3:

line BC ⊥ line AB.

Question 3.
Draw line CD. Take any point M on the line. Using a protractor, draw a line perpendicular to line CD at the point M.
Solution:
Step 1:

Step 2:

line MN ⊥ line CD

Maharashtra Board Class 6 Maths Chapter 17 Geometrical Constructions Practice Set 39 Questions and Activities

Question 1.
When constructing a building, what is the method used to make sure that a wall is exactly upright? What does the mason in the picture have in his hand? What do you think is his purpose for using it? (Textbook pg. no. 87)

Solution:
When constructing a building, a weight (usually with a pointed tip at the bottom) suspended from a string called as plummet or plump bob is aligned from the top of the wall to make sure that the wall is built exactly upright.
The mason in the picture is holding a plumb bob.
The string of the plumb bob is suspended from the top of the wall, such that plumb bob hangs freely. By observing whether the vertical wall is parallel to the string we can check if the constructed wall is vertical.

Question 2.
Have you looked at lamp posts on the roadside? How do they stand? (Textbook pg. no. 87)
Solution:
The lamp posts on the road side are standing erect or vertical.

Question 3.
For the above explained construction, why must we take a distance greater than half of the length of AB? What will happen if we take a smaller distance? (Textbook pg. no. 88)
Solution:
For the above construction, in step-3 we take distance greater than half of the length of AB, so that the arcs drawn by keeping the compass on points A and B intersect each other at point Q.
If the distance in compass is less than half of the length of AB, then the arcs drawn by keeping the compass at A and B will not intersect each other.

Std 6 Maths Digest

• Practice Set 35 Class 6 Answers
• Practice Set 36 Class 6 Answers
• Practice Set 37 Class 6 Answers
• Practice Set 38 Class 6 Answers
• Practice Set 39 Class 6 Answers
• Practice Set 40 Class 6 Answers
• Practice Set 41 lass 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 6 Bar Graphs Class 6 Practice Set 18 Answers Solutions.

Bar Graphs Class 6 Maths Chapter 6 Practice Set 18 Solutions Maharashtra Board

Std 6 Maths Practice Set 18 Solutions Answers

Question 1.
This bar graph shows the maximum temperatures in degrees Celsius in different cities on a certain day in February. Observe the graph and answer the questions:

1. What data is shown on the vertical and the horizontal lines?
2. Which city had the highest temperature?
3. Which cities had equal maximum temperatures?
4. Which cities had a maximum temperature of 30 °C?
5. What is the difference between the maximum temperatures of Panchgani and Chandrapur?

Solution:

1. Temperature is shown on the vertical line and cities are shown on the horizontal line.
2. The city Chandrapur had the highest temperature.
3. Pune and Nashik had the equal maximum temperature of 30°C and Panchgani and Matheran had the equal maximum temperature of 25°C.
4. Pune and Nashik had a maximum temperature of 30 °C.
5. The difference between the maximum temperatures of Panchgani and Chandrapur can be calculated as Difference in temperature = Temperature of Chandrapur – Temperature of Panchgani
   = 35°C – 25°C
   = 10°C

Maharashtra Board Class 6 Maths Chapter 6 Bar Graphs Practice Set 18 Intext Questions and Activities

Question 1.
Observe the picture alongside: (Textbook pg. no. 35)

1. To which sport is this data related?
2. How many things does the picture tell us about?
3. What shape has been used in the picture to represent runs?

Ans:

1. The given data is related to cricket.
2. The picture tells about runs scored in different overs by India and Srilanka. The represents the wickets fallen in that over.
3. To represent runs, rectangular or bar shape is used.

Question 2.
A pictogram of the types and numbers of vehicles in a city is given below.
Taking 1 picture = 5 vehicles, write the numbers in the pictogram. (Textbook pg. no.35)

Solution:

Drawing pictograms is time consuming.
Sometimes, it is practically not possible to draw pictures for the given values (for example population of villages etc). In such cases, representing the data by making use of graphs can serve the purpose. Such data can be represented by using graphs.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 18 Three Dimensional Shapes Class 6 Practice Set 41 Answers Solutions.

Three Dimensional Shapes Class 6 Maths Chapter 18 Practice Set 41 Solutions Maharashtra Board

Std 6 Maths Practice Set 41 Solutions Answers

Question 1.
Write the number of faces, edges and vertices of each shape in the table.

Solution:

Name	Cylinder	Cone	Pentagonal pyramid	Hexagonal pyramid	Hexagonal prism	Pentagonal prism
Faces	3 (2 flat + 1 curved)	2 (1 flat + 1 curved)	6 (5 triangles + 1 pentagon)	7 (6 triangles + 1 hexagon)	8 (6 rectangles + 2 hexagons)	7 (5 rectangles + 2 pentagons)
Vertices	0	1	6	7	12	10
Edges	2 (circular)	1 (circular)	10	12	18	15

Maharashtra Board Class 6 Maths Chapter 18 Three Dimensional Shapes Practice Set 41 Questions and Activities

Question 1.

1. Take a rectangular sheet.
2. Bring together its opposite sides. What shape does it form? (Textbook pg. no. 94)

Solution:
It forms a hollow cylinder.

Question 2.

1. Take a cylindrical tin.
2. Take a rectangular sheet with one side equal to the height of the tin.
3. Wrap it around the tin to cover it completely and cut away the extra paper.
4. Then unfold it and spread it out on a table.
5. Take another sheet. Place the tin on it and draw its circular outline.
6. Cut away the paper around it. Cut out another circle like this one.
7. Place these discs next to the rectangular paper as shown in the given figure. Which figure is obtained? (Textbook pg. no. 94)

Solution:
The figure obtained is the net of the closed cylinder.

Question 3.
Can you tell? (Textbook pg. no. 95)
When playing carom, you make a pile of the pieces as shown in the picture. What is the shape of this pile?
If you place a number of CD’s or round biscuits one on top of the other, what shape do you get?

Solution:
In all the cases, it will form a cylindrical shape (2 circular faces and 1 curved surface).

Question 4.

1. Draw a net as shown in figure (a) on a card sheet and cut it out.
2. Fold along the dotted lines of the square and bring the sides together so that the vertices A, B, C and D meet at a point.

What shape does it form? (Textbook pg. no. 95)

Solution:
The given net forms a quadrangular pyramid.

Question 5.

1. Draw a net as shown in figure (a) on a card sheet and cut it out.
2. Fold along the dotted lines of the triangle and bring the sides together so that the vertices A, B and C meet at a point.

What shape does it form? (Textbook pg. no. 95)

Solution:
The given net forms a triangular pyramid.

Question 6.

1. Using a compass draw a circle with centre C on a paper.
2. Draw two radii CR and CS.
3. Cut out the circle.
4. Cut along the radii and obtain two pieces of the circle.
5. Bring together the sides CR and CS of each piece.

On completing the activity, what shapes did you get? (Textbook pg. no. 95)

Solution:
On completing the activity, we get an open cone.

Read Also:

• Shapes Worksheets

Std 6 Maths Digest

• Practice Set 35 Class 6 Answers
• Practice Set 36 Class 6 Answers
• Practice Set 37 Class 6 Answers
• Practice Set 38 Class 6 Answers
• Practice Set 39 Class 6 Answers
• Practice Set 40 Class 6 Answers
• Practice Set 41 lass 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 17 Answers Solutions.

Decimal Fractions Class 6 Maths Chapter 5 Practice Set 17 Solutions Maharashtra Board

Std 6 Maths Practice Set 17 Solutions Answers

Question 1.
Carry out the following divisions.
i. 4.8÷2
ii. 17.5÷5
iii. 20.6÷2
iv. 32.5÷25
Solution:
i. 4.8÷2

ii. 17.5÷5

iii. 20.6÷2

iv. 32.5÷25

Question 2.
A road is 4 km 800 m long. If trees are planted on both its sides at intervals of 9.6 m, how many trees were planted?
Solution:
Length of road = 4 km 800 m
= 4 × 1000 m + 800 m
= 4000 m + 800 m
= 4800 m
Number of trees on one side = 4800 ÷ 9.6

= 500
∴ Number of trees on both sides = 2 x number of trees on one side
= 2 x 500 = 1000
If the trees are planted at the beginning of the road, then
Total number of trees = 1000 + 2 = 1002
∴ Total number of trees planted is 1000 or 1002.

Question 3.
Pradnya exercises regularly by walking along a circular path on a field. If she walks a distance of 3.825 km in 9 rounds of the path, how much does she walk in one round?
Solution:
Total distance walked in 9 rounds = 3.825 km
∴Distance walked in 1 round = 3.825 4 ÷ 9

= 0.425 km
∴ Total distance walked in 1 round is 0.425 km.

Question 4.
A pharmaceutical manufacturer bought 0.25 quintal of hirada, a medicinal plant, for Rs 9500. What is the cost per quintal of hirada? (1 quintal = 100 kg)
Solution:
Cost of 0.25 quintal of hirada = Rs 9500
∴ Cost of 1 quintal of hirada = 9500 ÷ 0.25

= Rs 38,000
∴ Cost per quintal of hirada is Rs 38,000.

Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Practice Set 17 Intext Questions and Activities

Question 1.
Maths is fun! (Textbook pg. no. 34)

1. Consider any three-digit number (say 527).
2. Multiply the number by 7. Then multiply the product obtained by 13, and this product by 11.
3. The found product is 5,27,527.

Take two or three other numbers. Do the same multiplication and find out how it is done.
Solution:
7 × 13 × 11 = 1001
∴ 527 × 1001 = 527 × (1000+ 1)
= (527 × 1000) + (527 × 1)
= 527000 + 527 = 527527
Thus, when any three-digit number is multiplied with 1001, the product obtained is a six-digit number in which the original three-digit number is written back to back twice.
(Students may consider any other three-digit numbers and verify the property.)

Std 6 Maths Digest

• Practice Set 9 Class 6 Answers
• Practice Set 10 Class 6 Answers
• Practice Set 11 Class 6 Answers
• Practice Set 12 Class 6 Answers
• Practice Set 13 Class 6 Answers
• Practice Set 14 Class 6 Answers
• Practice Set 15 Class 6 Answers
• Practice Set 16 Class 6 Answers
• Practice Set 17 Class 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 16 Quadrilaterals Class 6 Practice Set 38 Answers Solutions.

Quadrilaterals Class 6 Maths Chapter 16 Practice Set 38 Solutions Maharashtra Board

Std 6 Maths Practice Set 38 Solutions Answers

Question 1.
Draw ₹XYZW and answer the following:
i. The pairs of opposite angles.
ii. The pairs of opposite sides.
iii. The pairs of adjacent sides.
iv. The pairs of adjacent angles.
v. The diagonals of the quadrilateral.
vi. The name of the quadrilateral in different ways.
Solution:

i. a. ∠XYZ and ∠XWZ
b. ∠YXW and ∠YZW

ii. a. side XY and side WZ
b. side XW and side YZ

iii. a. side XY and side XW
b. side WX and side WZ
c. side ZW and side ZY
d. side YZ and side YX

iv. a. ∠XYZ and ∠YZW
b. ∠YZW and ∠ZWX
c. ∠ZWX and ∠WXY
d. ∠WXY and ∠XYZ

v. Seg XZ and seg YW

vi. ₹XYZW
₹YZWX
₹ZWXY
₹WXYZ
₹XWZY
₹WZYX
₹ZYXW
₹YXWZ

Question 2.
In the table below, write the number of sides the polygon has.

Names	Quadrilateral	Octagon	Pentagon	Heptagon	Hexagon
Number of sides

Solution:

Names	Quadrilateral	Octagon	Pentagon	Heptagon	Hexagon
Number of sides	4	8	5	7	6

Question 3.
Look for examples of polygons in your surroundings. Draw them.
Solution:

Question 4.
We see polygons when we join the tips of the petals of various flowers. Draw these polygons and write down the number of sides of each polygon.
Solution:

Question 5.
Draw any polygon and divide it into triangular parts as shown here. Thus work out the sum of the measures of the angles of the polygon.

Solution:

Hexagon ABCDEF can be divided in 4 triangles namely ∆BAF, ∆BFE, ∆BED and ∆BCD
Sum of the measures of the angles of a triangle = 180°
∴ Sum of measures of the angles of the polygon ABCDEF = Sum of the measures of all the four triangles
= 180° + 180° + 180°+ 180°
= 720°
∴ The sum of the measures of the angles of the given polygon (hexagon) is 720°.

Maharashtra Board Class 6 Maths Chapter 16 Quadrilaterals Practice Set 38 Intext Questions and Activities

Question 1.
From your compass boxes, collect set squares of the same shapes and place them side by side in all possible different ways. What figures do you get? Write their names. (Textbook pg. no. 85)
a. Two set squares
b. Three set squares
c. four set squares
Solution:
a. Two set squares

b. Three set squares

c. four set squares

Question 2.
Kaprekar Number. (Textbook pg. no. 86)
i. Take any 4-digit number in which all the digits are not the same.
ii. Obtain a new 4-digit number by arranging the digits in descending order.
iii. Obtain another 4-digit number by arranging the digits of the new number in ascending order.
iv. Subtract the smaller of these two new numbers from the bigger number. The difference obtained will be a 4-digit number. If it is a 3-digit number, put a 0 in the thousands place. Repeat the above steps with the difference obtained as a result of the subtraction.
v. After some repetitions, you will get the number 6174. If you continue to repeat the same steps you will get the number 6174 every time. Let us begin with the number 8531.
8531 → 7173 → 6354 → 3087 → 8352 → 6174 → 6174
This discovery was made by the mathematician, Dattatreya Ramchandra Kaprekar. That is why the number 6174 was named the Kaprekar number.
Solution:

Std 6 Maths Digest

• Practice Set 35 Class 6 Answers
• Practice Set 36 Class 6 Answers
• Practice Set 37 Class 6 Answers
• Practice Set 38 Class 6 Answers
• Practice Set 39 Class 6 Answers
• Practice Set 40 Class 6 Answers
• Practice Set 41 lass 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 16 Answers Solutions.

Decimal Fractions Class 6 Maths Chapter 5 Practice Set 16 Solutions Maharashtra Board

Std 6 Maths Practice Set 16 Solutions Answers

Question 1.
If, 317 × 45 = 14265, then 3.17 × 4.5 = ?
Solution:
3.17 × 4.5
= 14.265

Question 2.
If, 503 × 217 = 109151, then 5.03 × 2.17 = ?
Solution:
5.03 x 2.17
= 10.9151

Question 3.
i. 2.7 × 1.4
ii. 6.17 × 3.9
iii. 0.57 × 2
iv. 5.04 × 0.7
Solution:
i. 2.7 × 1.4

= 3.75

ii. 6.17 × 3.9

= 24.063

iii. 0.57 × 2

= 1.14

iv. 5.04 × 0.7

= 3.528

Question 4.
Virendra bought 18 bags of rice, each bag weighing 5.250 kg. How much rice did he buy altogether? If the rice costs Rs 42 per kg, how much did he pay for it?
Solution:
Weight of one bag of rice = 5.250 kg
Number of bags of rice = 18
∴ Total Weight = 18 × 5.250

Cost of 1 kg of rice = Rs 42
∴ Cost of 94.5 kg of rice = 42 × 94.5

∴ Total rice bought by Virendra is 94.5 kg, and the amount paid for it is Rs 3969.

Question 5.
Vedika has 23.5 metres of cloth. She used it to make 5 curtains of equal size. If each curtain required 4 metres 25 cm to make, how much cloth is left over?
Solution:
We know, that 1 m = 100 cm
Cloth required to make 1 curtain = 4 m 25 cm
= 4 m + \(\frac { 25 }{ 100 }\) m
= 4 m + 0.25 m

= 4.25 m
∴ Cloth required to make 5 curtains = 5 × 4.25

= 21.25 m
Cloth remaining with Vedika = Total cloth with Vedika – Cloth used
= 23.5 m – 21.25 m

= 2.25 m
∴ The length of cloth remaining with Vedika is 2.25 m.

Std 6 Maths Digest

• Practice Set 9 Class 6 Answers
• Practice Set 10 Class 6 Answers
• Practice Set 11 Class 6 Answers
• Practice Set 12 Class 6 Answers
• Practice Set 13 Class 6 Answers
• Practice Set 14 Class 6 Answers
• Practice Set 15 Class 6 Answers
• Practice Set 16 Class 6 Answers
• Practice Set 17 Class 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 7 Answers Solutions.

Integers Class 6 Maths Chapter 3 Practice Set 7 Solutions Maharashtra Board

Std 6 Maths Practice Set 7 Solutions Answers

Question 1.
Write the proper signs >, < or = in the boxes below:

1. -4 __ 5
2. 8 __ -10
3. +9 __ +9
4. -6 __ 0
5. 7 __ 4
6. 3 __ 0
7. -7 __ 7
8. -12 __ 5
9. -2 __ -8
10. -1 __ -2
11. 6 __ -3
12. -14 __ -14

Solution:

1. -4 < 5
2. 8 > -10
3. +9 = +9
4. -6 < 0
5. 7 > 4
6. 3 > 0
7. -7 < 7
8. -12 < 5
9. -2 > -8
10. -1 > -2
11. 6 > -3
12. -14 = -14

Std 6 Maths Digest

• Practice Set 1 Class 6 Answers
• Practice Set 2 Class 6 Answers
• Practice Set 3 Class 6 Answers
• Practice Set 4 Class 6 Answers
• Practice Set 5 Class 6 Answers
• Practice Set 6 Class 6 Answers
• Practice Set 7 Class 6 Answers
• Practice Set 8 Class 6 Answers