Category: 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.

6th Standard Maths Practice Set 6 Answers Chapter 3 Integers

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

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

6th Standard Maths Practice Set 28 Answers Chapter 11 Ratio-Proportion

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

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

6th Standard Maths Practice Set 38 Answers Chapter 16 Quadrilaterals

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:

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

6th Standard Maths Practice Set 39 Answers Chapter 17 Geometrical Constructions

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.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 4 Answers Solutions.

6th Standard Maths Practice Set 4 Answers Chapter 3 Integers

Question 1.
Classify the following numbers as positive numbers and negative numbers.
-5, +4, -2, 7, +26, -49, -37, 19, -25, +8, 5, -4, -12, 27
Solution:

Positive Numbers	+4, 7, +26, 19, +8, 5, 27
Negative Numbers	-5, -2, -49, -37, -25, -4, -12

Question 2.
Given below are the temperatures in some cities. Write them using the proper signs.

Place	Shimla	Leh	Delhi	Nagpur
Temperature	7 °C below 0°	12 °C below 0°	22 °C above 0°	31 °C above 0°

Solution:

Place	Shimla	Leh	Delhi	Nagpur
Temperature with proper sign	-7 °C	-12 °C	+22 °C	+31 °C

Question 3.
Write the numbers in the following examples using the proper signs.

1. A submarine is at a depth of 512 meters below sea level.
2. The height of Mt Everest, the highest peak in the Himalayas, is 8848 meters.
3. A kite is flying at a distance of 120 meters from the ground.
4. The tunnel is at a depth of 2 meters under the ground.

Solution:

1. A submarine is at a depth of -512 meters from sea level.
2. The height of Mt Everest, the highest peak in the Himalayas is +8848 meters.
3. A kite is flying at a distance of +120 meters from the ground.
4. The tunnel is at a depth of -2 meters from the ground.

Maharashtra Board Class 6 Maths Chapter 3 Integers Practice Set 4 Intext Questions and Activities

Question 1.
Take warm water in one beaker, some crushed ice in another and a mixture of salt and crushed ice in a third beaker. Ask your teacher for help in measuring the temperature of the substance in each of the beakers using a thermometer. Note the temperatures. (Textbook pg. no. 13)

Solution:
( Students should attempt this activity on their own)

Question 2.
Look at the picture of the kulfi man. Why do you think he keeps the kulfi moulds in a mixture of salt and ice? (Textbook pg. no. 14)

Solution:
Kulfi man keeps the kulfi moulds in a mixture of salt and ice because such a mixture helps in keeping the kulfi cool for a longer period of time. The kulfi kept in the said mixture relatively takes more time to melt. This mixture is Considered ideal as it has the temperature of -4°C as against the temperature of ice i.e. 0°C.

Question 3.
My class, i.e. Std. VI, is a part of my school. My school is in my town. My town is a part of a taluka. In the same way, the taluka is a part of a district, and the district is a part of Maharashtra State. In the same way, what can you say about these groups of numbers? Textbook pg. no. 15)

Solution:
By observing the above given groups of numbers, we can infer that natural numbers are a part of whole numbers. In turn, whole numbers are a part of integers.

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

6th Standard Maths Practice Set 17 Answers Chapter 5 Decimal Fractions

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.)

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

6th Standard Maths Practice Set 16 Answers Chapter 5 Decimal Fractions

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.

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

6th Standard Maths Practice Set 18 Answers Chapter 6 Bar Graphs

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.

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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.

6th Standard Maths Practice Set 41 Answers Chapter 18 Three Dimensional Shapes

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.

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Balbharti Maharashtra State Board Class 6 Marathi Solutions Sulabhbharati Chapter 3 पाऊस आला! पाऊस आला! Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 6 Marathi Solutions Chapter 3 पाऊस आला! पाऊस आला! (कविता)

Marathi Sulabhbharti Class 6 Solutions Chapter 3 पाऊस आला! पाऊस आला! Textbook Questions and Answers

1. खालील प्रश्नांची एक दोन वाक्यांत उत्तरे लिहा.

प्रश्न अ.
विजा केव्हा चमकल्या?
उत्तर:
विजा ऐन दुपारी चमकल्या.

प्रश्न आ.
सुटलेला वारा कसा होता?
उत्तर:
सुटलेला वारा भणाणवारा होता.

प्रश्न इ.
पाऊस आल्यामुळे आजोबांनी काय केले?
उत्तर:
पाऊस आल्यामुळे आजोबांनी छत्री शिवली.

प्रश्न ई.
आलेल्या पावसामुळे बाबांनी चडफड का केली?
उत्तर:
बाबांना आधीच उशीर झाला होता व त्यातच पाऊस पडला म्हणून बाबांनी चडफड केली.

प्रश्न उ.
पावसामुळे आईचे कोणते नुकसान झाले?
उत्तर:
पावसामुळे आईचे पापड भिजले.

2.

प्रश्न अ.
शब्दातील शेवटचे अक्षर सारखे येणारे शब्द कवितेतून शोधा व लिहा. उदा., वारा-गारा
उत्तर:
(अ) कुत्री – छत्री
(आ) गिल्ला – किल्ला
(इ) पापड – चडफड
(ई) पळा – घोटाळा
(उ) कुट्टी – सुट्टी

प्रश्न आ.
कडकड, चडफड, तडफड यांसारखे आणखी शब्द तयार करा.
उत्तर:

1. धडधड
2. गडगड
3. बडबड
4. खडखड
5. रडरड
6. गडबड
7. पडझड

3. खालील शब्दांचे वचन बदला.

प्रश्न 1.
खालील शब्दांचे वचन बदला.
उत्तर:
(अ) माणूस – माणसे
(आ) गाय – गाई
(इ) दप्तर – दप्तरे
(ई) पाणी – पाणी
(उ) वह्या – वही
(ऊ) पत्र – पत्रे

4. अचानक आलेल्या पावसामुळे तुमची कधी फजिती झाली आहे का? तो प्रसंग वर्गात सांगा. 

प्रश्न अ.
अचानक आलेल्या पावसामुळे तुमची कधी फजिती झाली आहे का? तो प्रसंग वर्गात सांगा.
उत्तर:
एकदा ताई आणि मी बाजारात गेलो होतो. अचानक पाऊस आला. छत्र्या नव्हत्या. आम्ही झाडाखाली उभे राहीलो पण पूर्ण भिजलो होतो. रस्त्यात पाणी तुंबले होते. रिक्षा बंद झाल्या होत्या. दोघे कसेबसे घरी पोहोचलो. आईने पापड, मिरच्या वाळवण ठेवले होते. तेही भिजले. बाबांना घरी यायला खूप उशीर झाला. बाहेर वाळत घातलेले कपडेही भिजले. पावसामुळे खूप धावपळ झाली.

5. अचानक आलेल्या पावसामुळे दिवाळीत तुम्ही केलेल्या तयारीचे कोणकोणते नुकसान होते ते चार-पाच वाक्यात लिहा.

प्रश्न अ.
अचानक आलेल्या पावसामुळे दिवाळीत तुम्ही केलेल्या तयारीचे कोणकोणते नुकसान होते ते चार-पाच वाक्यात लिहा.
उत्तर:
दिवाळीत आम्ही सर्व मुलांनी मिळून किल्ला बांधला होता. त्यावर शिपाई, रखवालदार ठेवले होते. मातीच्याच प्रतिमा असल्याने त्या तुटून गेल्या. किल्ला ढासळला. ताईने दारात काढलेली रांगोळी पुसली गेली. आमचे फटाकेही भिजले.

6. सुरुवातीला एक वाक्य दिले आहे. तुमच्या मनाने पुढील वाक्ये लिहा. 

प्रश्न अ.
रिमझिम पाऊस पडत होता.
उत्तर:

1. पाणी झुळझुळ वाहत होते.
2. खारूताई झाडावर सरसर चढली.
3. गांधीजी झरझर चालत जात.
4. आजीचे हात थरथर कापत होते.

प्रश्न आ.
रिक्षा खडखड करत थांबली.
उत्तर:

1. मुले भरभर चालत होती.
2. पंख्याची घरघर सुरू होती.
3. हृदय धडधड करीत होते.
4. पंखांची फडफड थांबली.

7. ‘वारा’ या शब्दाशी संबंधित आलेले शब्द वाचा. त्यांचा वाक्यात उपयोग करा. उदा. भणाणणारा वारा सुटला होता.

प्रश्न अ.
‘वारा’ या शब्दाशी संबंधित आलेले शब्द वाचा. त्यांचा वाक्यात उपयोग करा. उदा. भणाणणारा वारा सुटला होता.
उत्तरः

1. मंदमंद – पहाटे मंदमंद वारा सुटला होता.
2. गिरक्या – मैदानात वाऱ्याच्या गिरक्या येत होत्या.
3. जोरदार – जोरदार मोसमी वारे सुटले होते.
4. थंडगार – हिमालयाचा वारा थंडगार होता.
5. झोंबणारा – हिवाळ्यात नदीकाठचा वारा झोंबणारा होता.

8. पापड कशाकशापासून बनवले जातात याची माहिती आईला विचारून लिहा. 

प्रश्न अ.
पापड कशाकशापासून बनवले जातात याची माहिती आईला विचारून लिहा.
उत्तर:
पापड विविध प्रकारचे असतात. उडदाची डाळ, मिरे यांपासून उडदाचे पापड बनतात. तांदळापासून तांदळाचे पापड बनतात. काही पापड पोयांपासून तर काही नाचणीपासून बनतात. साबूदाण्यापासून व बटाट्यापासूनही उपवासाचे पापड बनतात.

9. उन्हाळ्यांमध्ये वाळवून साठवण्याचे कोणकोणते पदार्थ आई करते ते लिहा. 

प्रश्न अ.
उन्हाळ्यांमध्ये वाळवून साठवण्याचे कोणकोणते पदार्थ आई करते ते लिहा.
उत्तर:
उन्हाळ्यात वाळवून साठवण्याचे पदार्थ – पापड, कुर्डया, सांडगे, शेवया, मिरच्या, आंबापोळी, फणसपोळी, उपवासाच्या चकल्या, उपवासाचे पापड, आमचूर, बोरकूट इ.

10. पावसाळा सुरु होताच तुम्ही पावसातून बचाव करण्यासाठी कोणकोणती पूर्वतयारी करता?
उदा. छत्री खरेदी करणे.

प्रश्न अ.
पावसाळा सुरु होताच तुम्ही पावसातून बचाव करण्यासाठी कोणकोणती पूर्वतयारी करता?
उदा. छत्री खरेदी करणे.
उत्तर:
1. रेनकोट, चपला खरेदी करणे.

2. घराची डागडुजी करणे, दुरुस्ती करणे.
3. वाहने गंजू नये त्यासाठी उपाय करणे.

Marathi Sulabhbharti Class 6 Solutions Chapter 3 पाऊस आला! पाऊस आला! Important Additional Questions and Answers

रिकाम्या जागी कवितेतील योग्य शब्द लिहा.
प्रश्न 1.

1. ……………….. भणाण वारा.
2. दिवाळीतला ………………..
3. आजोबांनी ………………..
4. बाबा गेले ………………..
5. ……………….. पापड
6. हसत म्हणाल्या ………………..
7. ‘………………..’, शाळेला सुट्टी!

उत्तर:

1. कडाड कडकड
2. खचला किल्ला
3. शिवली छत्री
4. करीत चडफड
5. आईचेही भिजले
6. मॅडम कुट्टी
7. चला पळा.

प्रश्न 2.
एका शब्दांत उत्तरे लिहा.

1. ऐन दुपारी चमकल्या
2. जिकडे तिकडे
3. भुंकत सुटली सगळी
4. मुलांनी केला एकच

उत्तर:

1. विजा
2. गारा
3. कुत्री
4. गिल्ला

खालील प्रश्नांची एक दोन वाक्यांत उत्तरे लिहा.

प्रश्न 1.
मॅडम हसत काय म्हणाल्या?
उत्तर:
मॅडम म्हणाल्या, ‘चला पळा, शाळेला सुट्टी!’

खालील प्रश्नांची दोन-तीन वाक्यांत उत्तरे लिहा.

प्रश्न 1.
ऐन दुपारी काय झाले?
उत्तर:
ऐन दुपारी विजा कडाड कडकड चमकल्या. भणाण वारा सुटला. जिकडे तिकडे गारा पडू लागल्या व पाऊस
पडू लागला.

प्रश्न 2.
पावसामुळे काय काय झाले?
उत्तर:
ऐन दुपारी पाऊस पडला. दिवाळीतला किल्ला खचला. सगळी कुत्री भुंकत सुटली. आजोबांनी छत्री शिवली. बाबांना आधीच उशीर झाला होता, त्यातच पावसाने घोटाळा केला. ते चडफड करीतच गेले. आईचे पापड भिजले.

प्रश्न 3.
शाळेतील मुलांनी गिल्ला का केला?
उत्तर:
अचानक दुपारी पाऊस पडू लागला. सर्व मुलांना मजा वाटली. त्यातच मॅडमनी शाळेला सुट्टी दिली. त्या आनंदात मुलांनी गिल्ला केला.

व्याकरण व भाषाभ्यास:

प्रश्न 1.
खालील शब्दांचे वचन बदला.

1. छत्री
2. कुत्री
3. वीज
4. किल्ला

उत्तर:

1. छत्र्या
2. कुत्रा
3. विजा
4. किल्ले

जोड्या जुळवा.

प्रश्न 1.

‘अ’ गट	‘ब’ गट
1. आजोबा	(अ) आई
2. मॅडम	(ब) कुत्रा
3. बाबा	(क) सर
4. कुत्री	(ड) आजी

उत्तर:

‘अ’ गट	‘ब’ गट
1. आजोबा	(ड) आजी
2. मॅडम	(क) सर
3. बाबा	(अ) आई
4. कुत्री	(ब) कुत्रा

शब्दातील शेवटचे अक्षर सारखे येणारे शब्द कवितेतून शोधा व लिहा.
उदा. वारा – गारा

प्रश्न 1.
1. जिकडे
2. आला
उत्तर:
1. तिकडे
2. केला

प्रश्न 2.
‘पाऊस’ या शब्दाशी संबंधित आलेले शब्द वाचा. त्यांचा वाक्यात उपयोग करा.

उत्तर:

1. मुसळधार – मुसळधार पावसाने पूर आला.
2. रिपरिप – दिवसभर रिपरिप पाऊस होता.
3. झराळ – झराळ पावसात आम्ही ओलेचिंब झालो.
4. धोंधों – धोंधों पावसाने गावाचे नुकसान झाले.
5. रिमझिम – मुलांना रिमझिम पावसात भिजायला आवडते.

प्रश्न 3.
खालील शब्दांचे समानार्थी शब्द लिहा.

1. खचणे
2. चडफड
3. गिल्ला
4. ऐन दुपारी
5. भिजणे

उत्तर:

1. ढासळणे
2. राग
3. गोंगाट, गोंधळ
4. भर दुपारी
5. ओले होणे

प्रश्न 4.
खालील शब्दांचे विरूद्धार्थी अर्थाचे शब्द लिहा.

1. भिजणे
2. आला
3. शिवणे
4. खचणे
5. उशीर

उत्तर:

1. वाळणे
2. गेला
3. उसवणे
4. उभारणे
5. लवकर

पाऊस आला! पाऊस आला! Summary in Marathi

काव्य परिचय:

प्रस्तुत कवितेत पडणाऱ्या गारा, पावसामुळे आई बाबांची उडालेली तारांबळ, शाळेला मिळालेल्या सुट्टीचा आनंद वर्णन केला आहे.

शब्दर्थ:

1. ऐन दुपारी – भर दुपारी (in the afternoon)
2. खचला – ढासळला (collapse)
3. किल्ला – दुर्ग (fort)
4. जिकडेतिकडे – सर्वत्र (everywhere)
5. गारा – पावसाच्या पाण्याचे बर्फासारखे खडे (hailstones)
6. भुंकणे – कुत्र्याचा आवाज (to bark)
7. चडफड – आतल्या आत राग करणे (to get angry, restlessness)
8. गिल्ला – आवाज, गोंगाट (noise, shouting)
9. कुत्रा – श्वान (dog)
10. शिवणे – (to stitch)
11. घोटाळा – गडबड, गोंधळ (disorder, chaws)