Class 6

Balbharti Maharashtra State Board Class 6 Geography Solutions Chapter 1 The Earth and the Graticule Notes, Textbook Exercise Important Questions and Answers.

Maharashtra Board Class 6 Geography Solutions Chapter 1 The Earth and the Graticule

Class 6 Geography Chapter 1 The Earth and the Graticule Textbook Questions and Answers

A. Place a tick mark (✓) against the correct option:

Question 1.
What term is used for the imaginary east-west horizontal lines on the earth?
(i) Meridians
(ii) International Date Line
(iii) Parallels
Answer:
Parallels

Question 2.
What is the shape of the meridians?
(i) Circular
(ii) Semicircular
(iii) Points
Answer:
Semicircular

Question 3.
What do the parallels of latitude and meridians of longitude together form on the globe?
(i) Angular distance
(ii) Hemisphere
(iii) Graticule
Answer:
Graticule

Question 4.
How many parallels are there in the northern hemisphere?
(i) 90
(ii) 89
(iii) 91
Answer:
90

Question 5.
Which circles form the eastern and western hemispheres?
(i) 0° parallel and 180° meridian
(ii) 0° Prime Meridian and 180° meridian
(iii) North and South Polar circles
Answer:
0° Prime Meridian and 180° meridian

Question 6.
Which circle appears as a point on the globe?
(i) Equator
(ii) North/South Pole
(iii) Prime Meridian
Answer:
North/South Pole

Question 7.
How many places on the earth may be located on 45° N parallel?
(i) One
(ii) Many
(iii) Two
Answer:
Many

B. Observe a globe and examine the following statements. Correct the wrong ones:

Question 1.
Parallels of latitude lie parallel to the Prime Meridian,
Answer:
Wrong: Parallels of latitude lie parallel to the equator.

Question 2.
All parallels of latitude converge at the equator.
Answer:
Wrong: All meridians of longitude converge at the poles.

Question 3.
Parallels and meridians are imaginary lines.
Answer:
Right.

Question 4.
8° 4′ 65″ N is a north meridian.
Answer:
Wrong: 8° 4′ 65″ N is a north parallel.

Question 5.
Meridians are parallel to each other.
Answer:
Wrong: Latitudes are parallel to each other.

C. Find the correct graticule out of the following and put a tick mark against it.

Answer:
Figure (a) is correct.

D. Answer the following:

Question 1.
How will you express the latitude and longitude of the North Pole?
Answer:
The latitude of the North Pole would be 90°N. All the meridians of longitudes pass through the North Pole. So the North pole would be 0° longitude.

Question 2.
How much is the angular distance between the Tropic of Cancer and Tropic of Capricorn?
Answer:
The angular distance between the tropic of Cancer and tropic of Capricorn is 23°30′ + 23°30′ = 47°

Question 3.
Using a globe, write down the names of the countries through which the equator passes.
Answer:
The countries through which the equator passes are Ecuador, Colombia, Brazil, Sao tome and Principe, Gabon, Republic of the Congo, Democratic Republic of the Congo, Uganda, Kenya, Somalia, Maldives, Indonesia and Kiribati.

Question 4.
Write down the main uses of the graticule.
Answer:
Graticules help us to determine the locations on the earth. In the modern age, Geographical Information Systems, Global Positioning System, Google Maps, Wikimapia, Bhuvan of ISRO also make use of graticules.

E. Complete the following table:

Characteristics	Parallels of latitude	Meridians of longitude
Shape
	Size of each parallel is different
Distance		Distance between two meridians is larger on the equator and the same decreases towards the Poles.

Answer:

Characteristics	Parallels of latitude	Meridians of longitude
Shape	Circular	Semicircular
Size	Size of each parallel is different	Size of each meridian is same
Distance	Distance between two parallels is the same everywhere	Distance between two meridians is larger on the equator and the same decreases towards the Poles.

Class 6 Geography Chapter 1 The Earth and the Graticule Textbook Questions and Answers

Think a little!

Question 1.
A game of reading the meridians on the world map is going on. Shaheen and Sanket are asking each other to locate places on specific meridians and are making notes of the same. Shaheen asks Sanket to locate Wrangel Island on 180° meridians. Sanket could locate the island in the map but both are confused while making a note of it. They are puzzled whether to write 180° E or 180° W? What would be the precise answer? Please help them. Can we use a similar logic with reference to 0° meridian as well?
Answer:
The 0° and the 180° meridians lie opposite to each other and form a circle around the earth. This circle divides the earth in the eastern and western hemisphere. Shaheen and Sanket can write the Wrangled island to be loacted on 180° meridian.

Do it yourself!

Use figure 1.4 of the geography textbook:

In the upper portion of the circle, at the centre X, draw angles of 30°, V₁ X K₁ and V₂ X K₂; K₁ and K₂ being the points on the circle. Draw an ellipse joining K₁ and K₂.
In the lower half of the circle, mark angles of 60° and name the points on the circle as P₁ and P₂.

Question 1.
Draw an ellipse joining P₁ and P₂.
Answer:

Can you tell?

Question 1.
Is the distance between K₁ K₂ and P₁ P₂ the same?
Answer:
No, the distance between K₁ K₂ and P₁ P₂ are not the same.

Question 2.
Compare the distances XK₁ and XP₂. Are these distances the same or are they different?
Answer:
Yes, the distances are the same.

Question 3.
Now compare the ellipses you have drawn. Which is the larger ellipse? Why?
Answer:
The ellipse through K₁ and K₂ is larger than the ellipse through P₁ and P₂ . This is because the distance between K₁ K₂ is greater than the distance between P₁ P₂.

Observe the picture (fig) on page 5 of the textbook and answer the following questions:

Let the line AM be 0°.
Draw the line MB. Measure the angle it makes with the line AM and write it near B. Note the semicircle that passes through B and joins the North and South Poles. Trace it.
Now join MC. Measure ∠AMC and write it next to C. Draw a semicircle that passes through ‘C’ and joins the North and South Poles.
Draw a line that passes through point A at 0°, and joins the North and South Poles.

Answer:
∠AMB = 70°
∠AMC = 85°

Use your brain power!

Question 1.
How many parallels and meridians can be drawn on a globe at an interval of 10°?
Answer:
19 parallels and 36 meridians can be drawn on a globe at an interval of 10°.

Class 6 Geography Chapter 1 The Earth and the Graticule Textbook Questions and Answers

Fill in the blanks choosing the correct options from the brackets:

Question 1.
A miniature model of the earth is called a _____.(ball, globe, sphere)
Answer:
globe

Question 2.
The location of any place on the earth is determined with reference to the ______ of the earth. (poles, circle, centre)
Answer:
centre

Question 3.
The _______ is considered as 0° parallel.(equator, poles, circles)
Answer:
equator

Question 4.
The ______ bisects the earth into north and south parts. (poles, equator, circles)
Answer:
equator

Question 5.
One can draw _____ parallels on the earth at the interval of 1°. (90,181,360)
Answer:

Question 6.
The 0° meridian is known as the _____. (Central meridian, Equator, Prime Meridian)
Answer:
Prime Meridian

Question 7.
Each degree is divided into 60 ______.(minutes, hours, seconds)
Answer:
minutes

Question 8.
One can draw _____ meridians each at a distance of 1°. (90,181, 360)
Answer:
360

Question 9.
All meridians are _______ in size.(unequal, equal, uneven)
Answer:
equal

Question 10.
Exact location of a place on the earth can be located using ______.(equator, latitude and longitude, Prime Meridian)
Answer:
latitude and longitude

Question 11.
The distance between any two adjacent parallels is _______ on the surface of the earth. (111 km, 102 km, 44 km)
Answer:
111 km

Question 12.
The parallels and meridians on the globe form a net that is called a ______.(latitude,graticule, longitude)
Answer:
graticule

Match the pairs correctly:

Question 1.

Different parallels	Distance between meridians
(1) Poles	(a) 111 km
(2) Tropic of Cancer	(b) 102 km
(3) Polar Circles	(c) 0 km
(4) Equator	(d) 44 km
(5) Meridians	(e) 360
	(f) 1°

Answer:
1 – c
2 – b
3 – d
4 – a
5 – e

Place a tick mark (✓) against the correct option:

Question 1.
Which meridian is considered as the Prime Meridian?
(i) 0°
(ii) 80°
(iii) 90°
Answer:
0°

Answer the following questions in one sentence:

Question 1.
What are parallels of latitudes?
Answer:
Ellipses that are created at some angular distance from the centre of the earth and are parallel to one another are called parallels of latitudes.

Question 2.
How many parallels are there in the northern hemisphere?
Answer:
There are 90 parallels in the northern hemisphere.

Question 3.
Which circle divides the earth in the eastern and western hemisphere?
Answer:
The Prime Meridian divides the earth in the eastern and western hemisphere.

Question 4.
What is the distance between any two adjacent parallels on the surface of the earth?
Answer:
The distance between any two adjacent parallels on the surface of the earth is 111 km.

Question 5.
What is a graticule?
Answer:
The parallels and meridians on the globe form a net that is called a graticule.

Question 6.
What is used to determine the location on the earth?
Answer:
Latitude and longitude is used to determine the location on the earth.

Write the full forms of:

Question 1.
GIS
Answer:
Geographical Information System

Question 2.
GPS
Answer:
Global Positioning System

Question 3.
IRNSS
Answer:
Indian Regional Navigation Satellite System

Question 4.
ISRO
Answer:
Indian Space Research Organisation

Give geographical reasons for the following statements:

Question 1.
Parallels and meridians are imaginary lines on the earth.
Answer:
Parallels and meridians can be drawn on a globe though not on the earth. That is why parallels and meridians are imaginary lines on the earth.

Question 2.
Geographers developed a miniature model of the earth in the form of a globe.
Answer:
Oceanic waters, uneven nature of the land, forest, innumerable islands of different sizes and buildings make it impossible to draw lines on the earth. In order to overcome this difficulty, geographers developed a miniature model of the earth in the form of a globe.

Question 3.
Latitude and longitudes are expressed into degree, minutes and seconds.
Answer:
To locate the places within the distance of 111 km exactly, the unit degree is divided into smaller units. Degrees are divided into minutes and seconds.

Answer the following questions in short:

Question 1.
Explain the meaning of the term ‘equator’.
Answer:

• The equator is considered as 0° parallel.
• It is the largest parallel and great circle.
• It bisects the earth into two equal hemispheres viz, the northern and southern hemisphere.

Question 2.
What are Poles of the earth?
Answer:
On the globe and also on the earth, at the north and south ends of the earth’s axis, Poles appear as points. These are called the North Pole and the South Pole respectively.

Observe a globe and examine the following statements. Correct the wrong ones:

Question 1.
Parallels of latitude lie parallel to the Prime Meridian,
Answer:
Wrong: Parallels of latitude lie parallel to the equator.

Question 2.
All parallels of latitude converge at the equator.
Answer:
Wrong: All meridians of longitude converge at the poles.

Question 3.
Parallels and meridians are imaginary lines.
Answer:
Right.

Question 4.
80° 4′ 65″ N is a north meridian.
Answer:
Wrong: 80° 4′ 65″ N is a north parallel.

Question 5.
Meridians are parallel to each other.
Answer:
Wrong: Latitudes are parallel to each other.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 16 Quadrilaterals Class 6 Practice Set 37 Answers Solutions.

6th Standard Maths Practice Set 37 Answers Chapter 16 Quadrilaterals

Question 1.
Observe the figures below and find out their names:

Solution:
i. Pentagon (5 sides)
ii. Hexagon (6 sides)
iii. Heptagon (7 sides)
iv. Octagon (8 sides)

Maharashtra Board Class 6 Maths Chapter 16 Quadrilaterals Practice Set 37 Intext Questions and Activities

Question 1.
Observe the figures given below and say which of them are quadrilaterals. (Textbook pg. no. 81)

Solution:
Is a quadrilateral: (i)

Question 2.
Draw a quadrilateral. Draw one diagonal of this quadrilateral and divided it into two triangles. Measures all the angles in the figure. Is the sum of the measures of the four angles of the quadrilateral equal to the sum of the measures of the six angles of the two triangles? Verity that this is so with other quadrilaterals. (Textbook pg. no. 84)
Solution:

m∠PQR = 104°
m∠QRP = 26°
m∠RPQ = 50°
m∠PRS = 34°
m∠RSP = 106°
m∠SPR = 40°
∴ Sum of the measures of the angles of quadrilateral = m∠PQR + m∠QRP + m∠RPQ + m∠PRS + m∠RSP + m∠SPR
= 104° + 26° + 50° + 34° + 106° + 40°
= 360°
Also, we observe that
Sum of the measures of the angles of quadrilateral = Sum of the measures of angles of the two triangles (PQR and PRS)
= (104°+ 26°+ 50°)+ (34° + 106° + 40°)
= 180° + 180°
= 360°
[Note: Students should drew different quadrilaterals and verify the property.]

Question 3.
For the pentagon shown in the figure below, answer the following: (Textbook pg. no. 84)

1. Write the names of the five vertices of the pentagon.
2. Name the sides of the pentagon.
3. Name the angles of the pentagon.
4. See if you can sometimes find players on a field forming a pentagon.

Solution:

1. The vertices of the pentagon are points A, B, C, D and E.
2. The sides of the pentagon are segments AB, BC, CD, DE and EA.
3. The angles of the pentagon are ∠ABC, ∠BCD, ∠CDE, ∠DEA and ∠EAB.
4. The players shown in the above figure form a pentagon. The players are standing on the vertices of

Question 4.
Cut out a paper in the shape of a quadrilateral. Make folds in it that join the vertices of opposite angles. What can these folds be called? (Textbook pg. no. 83)

Solution:
The folds are called diagonals of the quadrilateral.

Question 5.
Take two triangular pieces of paper such that . one side of one triangle is equal to one side of the other. Let us suppose that in ∆ABC and ∆PQR, sides AC and PQ are the equal sides. Join the triangles so that their equal sides lie B side by side. What figure do we get? (Textbook pg. no. 83)

Solution:
If we place the triangles together such that the equal sides overlap, the two triangles form a quadrilateral.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 2 Angles Class 6 Practice Set 3 Answers Solutions.

6th Standard Maths Practice Set 3 Answers Chapter 2 Angles

Question 1.
Use the proper geometrical instruments to construct the following angles. Use the compass and the ruler to bisect them:

1. 50°
2. 115°
3. 80°
4. 90°

Solution:

Maharashtra Board Class 6 Maths Chapter 2 Angles Practice Set 3 Intext Questions and Activities

Question 1.
Construct an angle bisector to obtain an angle of 30°. (Textbook pg. no. 11)
Solution: .
In order to get a bisected angle of a given measure, the student has to draw the angle having twice the measurement of required bisected angle.

For getting measurement of 30° (for the bisected angle), one has to make an angle of 60° (i.e. 30° × 2).

Step 1:
Draw ∠ABC of 60°.

Step 2:
Cut arcs on the rays BA and BC. Name these points as D and E respectively.

Step 3:
Place the compass point on point D and draw an arc inside the angle.
Without changing the distance of the compass, place the compass point on point E and cut the previous arc. Name the point of intersection as O

Step 4:
Draw ray BO.
Ray BO is the angle bisector of ∠ABC.
i.e. m∠ABO = m∠CBO = 30°

Question 2.
Construct an angle bisector to draw an angle of 45°. (Textbook pg. no. 11)
Solution:
For getting measurement of 45° (for the bisected angle), one has to make an angle of 90° (i.e. 45° × 2).
Step 1:
Draw ∠PQR of 90°.

Step 2:
Cut arcs on the rays QP and QR.
Name these points as M and N respectively.

Step 3:
Place the compass point on point M and draw an arc inside the angle.
Without changing the distance of the compass, place the compass point on point N and cut the

Step 4:
Draw ray QO.
Ray QO is the angle bisector of ∠PQR.
i.e. m∠PQO = m∠RQO = 45°

Question 3.
Ask three or more children to stand in a straight line. Take two long ropes. Let the child in the middle hold one end of each rope. With the help of the ropes, make the children on either side stand along a straight line. Tell them to move so as to form an acute angle, a right angle, an obtuse angle, a straight angle, a reflex angle and a full or complete angle in turn. Keeping the rope stretched will help to ensure that the children form straight lines. (Textbook pg. no. 6)
Solution:

Question 4.
Look at the pictures below and identify the different types of angles. (Textbook pg. no. 8)

Solution:
i. Complete angle
ii. Reflex and Acute angle
iii. Acute and Obtuse angle

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 15 Answers Solutions.

6th Standard Maths Practice Set 15 Answers Chapter 5 Decimal Fractions

Question 1.
Write the proper number in the empty boxes.

Solution:

Question 2.
Convert the common fractions into decimal fractions:
i. \(\frac { 3 }{ 4 }\)
ii. \(\frac { 4 }{ 5 }\)
iii. \(\frac { 9 }{ 8 }\)
iv. \(\frac { 17 }{ 20 }\)
v. \(\frac { 36 }{ 40 }\)
vi. \(\frac { 7 }{ 25 }\)
vii. \(\frac { 19 }{ 200 }\)
Solution:
i. \(\frac { 3 }{ 4 }\)

ii. \(\frac { 4 }{ 5 }\)

iii. \(\frac { 9 }{ 8 }\)

iv. \(\frac { 17 }{ 20 }\)

v. \(\frac { 36 }{ 40 }\)

vi. \(\frac { 7 }{ 25 }\)

vii. \(\frac { 19 }{ 200 }\)

Question 3.
Convert the decimal fractions into common fractions:
i. 27.5
ii. 0.007
iii. 90.8
iv. 39.15
v. 3.12
vi. 70.400
Solution:
i. 27.5
= \(\frac { 275 }{ 10 }\)

ii. 0.007
= \(\frac { 7 }{ 1000 }\)

iii. 90.8
= \(\frac { 908 }{ 10 }\)

iv. 39.15
= \(\frac { 3915 }{ 100 }\)

v. 3.12
= \(\frac { 312 }{ 100 }\)

vi. 70.400
= 70.4
= \(\frac { 704 }{ 10 }\)

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 10 Equations Class 6 Practice Set 27 Answers Solutions.

6th Standard Maths Practice Set 27 Answers Chapter 10 Equations

Question 1.
Rewrite the following using a letter:
i. The sum of a certain number and 3.
ii. The difference is obtained by subtracting 11 from another number.
iii. The product of 15 and another number.
iv. Four times a number is 24.
Solution:
i. Let the number be x.
∴ x + 3 represents the sum of a certain number x and 3.

ii. Let the number be x.
∴ x – 11 represents the number obtained by subtracting 11 from another number x.

iii. Let the number be x.
∴ 15x represents the product of 15 and another number x.

iv. Let the number be x.
∴ 4x = 24 represents four the product of a number x four times.

Question 2.
Find out which operation must be done on both sides of these equations in order to solve them:

1. x + 9 = 11
2. x – 4 = 9
3. 8x = 24
4. \(\frac { x }{ 6 }\) = 3

Solution:

1. Subtract 9 from both sides.
2. Add 4 to both sides.
3. Divide both sides by 8.
4. Multiply both sides by 6.

Question 3.
Given below are some equations and the values of the variables. Are these values the solutions to those equations?

No.	Equation	Value of the Variable	Solution (Yes/No)
i.	y – 3 = 11	y = 3	No
ii.	17 = n + 7	n = 10
iii.	30 = 5x	x = 6
iv.	\(\frac { m }{ 2 }\) = 14	m = 7

Solution:

No.	Equation	Value of the Variable	Solution (Yes/No)
i.	y – 3 = 11	y = 3	No
ii.	17 = n + 7	n = 10	Yes
iii.	30 = 5x	x = 6	Yes
iv.	\(\frac { m }{ 2 }\) = 14	m = 7	No

i. y – 3 = 11
∴ y – 3 + 3 = 11 + 3
…. (Adding 3 to both sides)
∴ y + 0 = 14
∴ y = 14

ii. 17 = n + 7
∴ 17 – 7 = n + 7 – 7
…. (Subtracting 7 from both sides)
∴ 17 + (-7) = n + 7 – 7
∴ 10 = n
∴  n = 10

iii. 30 = 5x
∴ \(\frac{30}{5}=\frac{5x}{5}\)
…. (Dividing both sides by 5)
∴  6 = 1x
∴ 6 = x
∴  x = 6

iv. \(\frac { m }{ 2 }\) = 14
∴ \(\frac { m }{ 2 }\) × 2 = 14 × 2
…. (Multiplying both sides by 2)
\(\frac { m\times2 }{ 2\times1 }\) = 28
∴ m = 28

Question 4.
Solve the following equations:
i. y – 5 = 1
ii. 8 = t + 5
iii. 4x = 52
iv. 19 = m – 4
v. \(\frac { p }{ 4 }=9\)
vi. x + 10 = 5
vi. m – 5 = -12
vii. p + 4 = -1
Solution:
i. y – 5 = 1
∴y – 5 + 5 = 1 + 5
…. (Adding 5 to both sides)
∴y + 0 = 6
∴y = 6

ii. 8 = t + 5
∴8 – 5 = t + 5 – 5
……(Subtracting 5 from both sides)
∴8 + (-5) = t + 0
∴ 3 = t
∴t = 3

iii. 4x = 52
∴\(\frac{4x}{4}=\frac{52}{4}\)
…. (Dividing both sides by 4)
∴ 1x = 13
∴ x = 13

iv. 19 = m -4
∴ 19 + 4 = m – 4 + 4
…. (Adding 4 to both sides)
∴ 23 = m + 0
∴ m = 23

v. \(\frac { p }{ 4 }\) = 9
∴ \(\frac { p }{ 4 }\) × 4 = 9 × 4 …. (Multiplying both sides by 4)
∴ \(\frac { p\times4 }{ 4\times1 }=36\)
∴ 1p = 36
∴ p = 36

vi. x + 10 = 5
∴ x + 10 – 10 = 5 – 10
…. (Subtracting 10 from both sides)
∴ x + 0 = 5 + (-10)
∴ x = -5

vii. m – 5 = -12
∴m – 5 + 5 = – 12 + 5
…. (Adding 5 to both sides)
∴m + 0 = -7
∴m = -7

viii. p + 4 = – 1
∴p + 4 – 4 = -1 – 4
…. (Subtracting 4 from both sides)
∴p + 0 = (-1) + (-4)
∴P = -5

Question 5.
Write the given information as an equation and find its solution:
i. Haraba owns some sheep. After selling 34 of them in the market, he still has 176 sheep. How many sheep did Haraba have at first?

ii. Sakshi prepared some jam at home and filled it in bottles. After giving away 7 of the bottles to her friends she still has 12 for herself. How many bottles had she made in all? If she filled 250g of jam in each bottle, what was the total weight of the jam she made?

iii. Archana bought some kilograms of wheat. She requires 12 kg per month and she got enough wheat milled for 3 months. After that, she had 14 kg left. How much wheat had Archana bought altogether?
Solution:
i. Let the number of sheep before selling be x.
∴ x – 34 = 176
∴ x – 34 + 34 = 176 + 34 ….(Adding 34 to both sides)
∴ x + 0 = 210
∴ x = 210
The number of sheep with Haraba before selling is 210.

ii. Let the total number of bottles be x.
∴ x – 7 = 12
∴ x – 7 + 7 = 12 + 7 ….(Adding 7 to both sides)
∴ x + 0 = 19
∴ x = 19
Weight of jam in each bottle = 250g
∴ Total weight of jam = 19 × 250g = 4750 g = \(\frac { 4750 }{ 1000 }\)kg = 4.75 kg
∴ The total number of bottles of jam made by Sakshi is 19, and the total weight of jam made is 4.75 kg.

iii. Let the total wheat bought by Archana be x kg.
Wheat used in 1 month = 12 kg
∴ Wheat used in 3 months = 3 × 12 = 36 kg
∴ x – 36 = 14
∴ x – 36 + 36 = 14 + 36 ….(Adding 36 to both sides)
∴ x + 0 = 50
∴ x = 50
∴ The total amount of wheat bought by Archana was 50 kg.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 15 Triangles and their Properties Class 6 Practice Set 36 Answers Solutions.

6th Standard Maths Practice Set 36 Answers Chapter 15 Triangles and their Properties

Question 1.
Observe the figures below and write the type of the triangle based on its angles:

Solution:
i. right angled
ii. Obtuse angled
iii. acute angled

Question 2.
Observe the figures below and write the type of the triangle based on its sides:

Solution:
i. equilateral
ii. scalene
iii. isosceles

Question 3.
As shown in the figure, Avinash is standing near his house. He can choose from two roads to go to school. Which way is shorter? Explain why.

Solution:
The two roads which Avinash can choose to go to school are

1. Road AB + Road BC
2. Road AC

The three roads together form ∆ABC.
Road AC is shorter because the sum of the lengths of any two sides (side AB + side BC) of a triangle is always greater than the third side (side AC).

Question 4.
The lengths of the sides of some triangles are given. Say what types of triangles they are.

1. 3 cm, 4 cm, 5 cm
2. 3.4 cm, 3.4 cm, 5 cm
3. 4.3 cm, 4.3 cm, 4.3 cm
4. 3.7 cm, 3.4 cm, 4 cm

Solution:

1. Since, no two sides have equal lengths, the given triangle is a scalene triangle.
2. Since, two sides have equal length, the given triangle is an isosceles triangle.
3. Since, all the three sides have equal lengths, the given triangle is an equilateral triangle.
4. Since, no two sides have equal lengths, the given triangle is a scalene triangle.

Question 5.
The lengths of the three segments are given for constructing a triangle. Say whether a triangle with these sides can be drawn. Give the reason for your answer.
i. 17 cm, 7 cm, 8 cm
ii. 7 cm, 24 cm, 25 cm
iii. 9 cm, 6 cm, 16 cm
iv. 8.4 cm, 16.4 cm, 4.9 cm
v. 15 cm, 20 cm, 25 cm
vi. 12 cm, 12 cm, 16 cm
Solution:
i. The lengths of the three sides are 17 cm, 7 cm, 8 cm.
a. 7 cm + 17 cm = 24 cm, greater than 8 cm
b. 8 cm +17 cm = 25 cm, greater than 7 cm
c. 7 cm + 8 cm =15 cm, not greater than 17 cm
The sum of lengths of two sides in (c) is not greater than the length of the third side.
∴ Triangle cannot be drawn with sides 17 cm, 7 cm, 8 cm.

ii. The lengths of the three sides are 7 cm, 24 cm, 25 cm.
a. 7 cm + 24 cm = 31 cm, greater than 25 cm
b. 25 cm + 7 cm = 32 cm, greater than 24 cm
c. 24 cm + 25 cm = 49 cm, greater than 7 cm
The sum of lengths of two sides is greater than the length of the third side.
∴ Triangle can be drawn with sides 7 cm, 24 cm, 25 cm.

iii. The lengths of the three sides are 9 cm, 6 cm, 16 cm.
a. 9 cm + 16 cm = 25 cm, greater than 6 cm
b. 6 cm + 16 cm = 22 cm, greater than 9 cm
c. 9 cm+ 6 cm =15 cm, not greater than 16 cm
The sum of lengths of two sides in (c) is not greater than the length of the third side.
∴ Triangle cannot be drawn with sides 9 cm, 6 cm, 16 cm.

iv. The lengths of the three sides are 8.4 cm, 16.4 cm, 4.9 cm.
a. 8.4 cm + 16.4 cm = 24.8 cm, greater than 4.9 cm
b. 4.9 cm + 16.4 cm = 21.3 cm, greater than 8.4 cm
c. 8.4 cm + 4.9 cm = 13.3 cm, not greater than 16.4 cm
The sum of lengths of two sides in (c) is not greater than the length of the third side.
∴ Triangle cannot be drawn with sides 8.4 cm, 16.4 cm, 4.9 cm.

v. The lengths of the three sides are 15 cm, 20 cm, 25 cm.
a. 15 cm + 20 cm = 35 cm, greater than 25 cm
b. 25 cm + 20 cm = 45 cm, greater than 15 cm
c. 15 cm + 25 cm = 40 cm, greater than 20 cm
The sum of lengths of two sides is greater than the length of the third side.
∴ Triangle can be drawn with sides 15 cm, 20 cm, 25 cm.

vi. The lengths of the three sides are 12 cm, 12 cm, 16 cm.
a. 12 cm + 12 cm = 24 cm, greater than 16 cm
b. 12 cm + 16 cm = 28 cm, greater than 12 cm
c. 12 cm + 16 cm = 28 cm, greater than 12 cm
The sum of lengths of two sides is greater than the length of the third side.
∴ Triangle can be drawn with sides 12 cm, 12 cm, 16 cm.

Maharashtra Board Class 6 Maths Chapter 15 Triangles and their Properties Practice Set 36 Intext Questions and Activities

Question 1.
In the given figure, some points and some line segments joining them have been drawn. Which of these figures is a triangle? Which figure is not a triangle? Why not? (Textbook pg. no. 77)

Solution:
ABC it is a closed figure with three sides. Hence, ABC is a triangle.
PQRS has three sides but it is not a closed figure. Hence, PQRS is not a triangle.

Question 2.
As seen above, ∆ABC has three sides. Line segment AB is one side. Write the names of the other two sides. ∆ABC has three angles. ∠ABC is one among them. Write the names of the other angles. (Textbook pg. no. 77)
Solution:
The names of other two sides are: seg BC and seg AC
The names of other angles are: ∠BCA and ∠CAB

Question 3.
Measure the sides of the following triangles in centimeters, using a divider and ruler. Enter the lengths in the table below. What do you observe? (Textbook pg. no. 77)

In ∆ABC	In ∆PQR	In ∆XYZ
l (AB) =       cm	l (QR) =       cm	l (XY) =       cm
l (BC) =       cm	l (PQ) =       cm	l (YZ) =       cm
l (AC) =       cm	l (PR) =        cm	l (XZ) =       cm

Solution:

In ∆ABC	In ∆PQR	In ∆XYZ
l (AB) = 2.6 cm	l (QR) = 2.8 cm	l (XY) = 2.8 cm
l (BC) = 2.6 cm	l (PQ) = 3.8 cm	l (YZ) = 2.6 cm
l (AC) = 2.6 cm	l (PR) = 3.8 cm	l (XZ) = 4.3 cm

We observe that,

1. ∆ABC is an equilateral triangle,
2. ∆PQR is an isosceles triangle, and
3. ∆XYZ is a scalene triangle.

Question 4.
Measure all the angles of the triangles given below. Enter them in the following table. (Textbook pg. no. 78)

In ∆DEF	In ∆PQR	In ∆LMN
Measure of ∠D = m ∠D =___	Measure of ∠P = m ∠P =___	Measure of ∠L =__
Measure of ∠E = m ∠E =___	Measure of ∠Q =___=___	Measure of ∠M =___
Measure of ∠F = ___=___	Measure of ∠R =___=___	Measure of ∠N =___
Observation: All three angles are acute angles.	Observation: One angle is right angle and two are acute angles.	Observation: One angle is an obtuse angle and two are acute.

Solution:

In ∆DEF	In ∆PQR	In ∆LMN
Measure of ∠D = m ∠D = 60º	Measure of ∠P = m ∠P = 45º	Measure of ∠L = 30º
Measure of ∠E = m ∠E = 68º	Measure of ∠Q = m = 90º	Measure of ∠M = 116º
Measure of ∠F = m = 52º	Measure of ∠R = m ∠R = 45º	Measure of ∠N = 34º

1. ADEF is an acute angled triangle,
2. APQR is a right angled triangle,
3. ALMN is an obtuse angled triangle.

Question 5.
Observe the set squares in your compass box. What kind of triangles are they? (Textbook pg. no. 78)

Solution:
The first set square is a scalene triangle and also a right angled triangle.
The second set square is an isosceles triangle and also a right angled triangle.

Question 6.
Properties of a triangle. (Textbook pg. no. 79)
Take a triangular piece of paper. Choose three different colors or signs to mark the three comers of the triangle on both sides of the paper. Fold the paper at the midpoints of two sides as observe?

Solution:
The three angles of the triangle form a straight angle.
∴ m∠A + m∠B + m∠C = 180°
Hence, the sum of the measures of the angles of a triangle is 180°.

Question 7.
Properties of a triangle (Textbook pg. no. 79)
Take a triangular piece of paper and make three different types of marks near the three angles. Take a point approximately at the center of the triangle. From this point, draw three lines that meet the three sides. Cut the paper along those lines. Place the three angles side by side as shown. See how the three angles of a triangle together form a straight angle, or, an angle that measures 180°.

Solution:
The three angles of the triangle form a straight angle.
Hence, the sum of the measures of the angles of a triangle is 180°.

Question 8.
Draw any triangle on a paper. Name its vertices A, B, C. Measure the lengths of its three sides using a divider and scale and enter them in the table. (Textbook pg. no. 79)

Length of side	Sum of the lengths of two sides	Length of the third side
l (AB) =         cm	l (AB) + l (BC) =         cm	l (AC) =         cm
l (BC) =         cm	l (BC) + l (AC) =         cm	l (AB) =         cm
l (AC) =         cm	l (AC) + l (AB) =        cm	l (BC) =         cm

Solution:

Length of side	Sum of the lengths of two sides	Length of the third side
l (AB) = 2.7 cm	l (AB) + l (BC) = 6.6 cm	l (AC) = 5.6 cm
l (BC) = 2.9 cm	l (BC) + l (AC) = 9.5 cm	l (AB) = 2.7 cm
l (AC) = 5.6 cm	l (AC) + l (AB) = 8.3 cm	l (BC) = 3.9 cm

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 14 Answers Solutions.

6th Standard Maths Practice Set 14 Answers Chapter 5 Decimal Fractions

Question 1.
In the table below, write the place value of each of the digits in the number 378.025.

Place	Hundreds	Tens	Units	Tenths	Hundredths	Thousandths
	100	10	1	\(\frac { 1 }{ 10 }\)	\(\frac { 1 }{ 100 }\)	\(\frac { 1 }{ 1000 }\)
Digit	3	7	8	0	2	5
Place value	300			\(\frac { 0 }{ 10 }=0\)		\(\frac { 5 }{ 1000 }\) = 0.005

Solution:

Place	Hundreds	Tens	Units	Tenths	Hundredths	Thousandths
	100	10	1	\(\frac { 1 }{ 10 }\)	\(\frac { 1 }{ 100 }\)	\(\frac { 1 }{ 1000 }\)
Digit	3	7	8	0	2	5
Place value	300	7 × 10 = 70	8 × 1 = 8	\(\frac { 0 }{ 10 }=0\)	\(\frac { 2 }{ 100 }\) = 0.02	\(\frac { 5 }{ 1000 }\) = 0.005

Question 2.
Solve :
i. 905.5 + 27.197
ii. 39 + 700.65
iii. 40 + 27.7 + 2.451
Solution:
i. 905.5 + 27.197

ii. 39 + 700.65

iii. 40 + 27.7 + 2.451

Question 3.
Subtract:
i. 85.96 – 2.345
ii. 632.24 – 97.45
iii. 200.005 – 17.186
Solution:
i. 85.96 – 2.345

ii. 632.24 – 97.45

iii. 200.005 – 17.186

Question 4.
Avinash traveled 42 km 365 m by bus, 12 km 460 in by car and walked 640 m. How many kilometers did he travel altogether? (Write your answer in decimal fractions)
Solution:
Distance traveled in bus = 42 km 365 m
= 42 km + \(\frac { 365 }{ 1000 }\) km
= 42 km + 0.365 km

= 42.365 km
Distance travelled in car = 12 km 460 m
= 12 km + \(\frac { 460 }{ 1000 }\) km
= 12 km + 0.460 km

= 12.460 km
Distance walked = 640 m
= \(\frac { 640 }{ 1000 }\) = 0.640 km
∴ Total distance travelled = Distance travelled in bus + Distance travelled in car + Distance walked
= 42.365 + 12.460 + 0.640

= 55.465 km
∴ Distance travelled altogether by Avinash is 55.465 km.

Question 5.
Ayesha bought 1.80 m of cloth for her salwaar and 2.25 for her kurta. If the cloth costs Rs 120 per metre, how much must she pay the shopkeeper?
Solution:
Total length of cloth bought = 1.80 m + 2.25 m
= 4.05 m

Cost of 1 m of cloth = Rs 120
∴ Cost of 4.05 m of cloth = 4.05 x 120

∴ Amount to be paid to the shopkeeper is Rs 486.

Question 6.
Sujata bought a watermelon weighing 4.25 kg and gave 1 kg 750 g to the children in her neighbourhood. How much of it does she have left?
Solution:
Total weight of watermelon = 4.25 kg
Weight of watermelon given to children = 1 kg 750 g
= 1 kg + \(\frac { 750 }{ 1000 }\) kg
= 1 kg + 0.75 kg

= 1.75 kg
∴ Weight of watermelon left = Total weight of watermelon – Weight of watermelon given to children
= 4.25 kg – 1.75 kg

= 2.5 kg
∴ Weight of watermelon left with Sujata is 2.5 kg.

Question 7.
Anita was driving at a speed of 85.6 km per hour. The road had a speed limit of 55 km per hour. By how much should she reduce her speed to be within the speed limit?
Solution:
Speed at which Anita is driving = 85.6 km per hr.
Speed limit = 55 km per hr.
∴ Anita should reduce her speed by 85.6 km per hr – 55 km per hr.

= 30.6 km per hr.
∴ Anita should reduce her speed by 30.6 km per hour to be within the speed limit.

Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Practice Set 14 Intext Questions and Activities

Question 1.
Nandu went to a shop to buy a pen, notebook, eraser and a paint box. The shopkeeper told him the prices. A pen costs four and a half rupees, an eraser one and a half, a notebook six and a half and a paintbox twenty-five rupees and fifty paise. Nandu bought one of each article. Prepare his bill.
If Nandu gave a 100 rupee note, how much money does he get back? (Textbook pg. no. 29)

Nandu will get __ rupees back.
Solution:
100 – 38 = 62.00
Nandu will get Rs 62 rupees back.

Question 2.
Take a pen and notebook with you when you go to the market with your parent. Note the weight of every vegetable your mother buys. Find out the total weight of those vegetables. (Textbook pg. no. 30)
Solution:
(Students should attempt this activity on their own.)

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 2 Angles Class 6 Practice Set 2 Answers Solutions.

6th Standard Maths Practice Set 2 Answers Chapter 2 Angles

Question 1.
Match the following:

Measure of the angle		Type of the angle
i.	180°	a.	Zero angle
ii.	240°	b.	Straight angle
iii.	360°	c.	Reflex angle
iv.	0°	d.	Complete angle

Solution:
(i – Straight Angle),
(ii – Reflex Angle),
(iii – Complete Angle),
(iv – Zero Angle).

Question 2.
The measures of some angles are given below. Write the type of each angle:

1. 75°
2. 0°
3. 215°
4. 360°
5. 180°
6. 120°
7. 148°
8. 90°

Solution:

1. Acute angle
2. Zero angle
3. Reflex angle
4. Complete angle
5. Straight angle
6. Obtuse angle
7. Obtuse angle
8. Right angle

Question 3.
Look at the figures below and write the type of each of the angles:

Solution:
a. Acute angle
b. Right angle
c. Reflex angle
d. Straight angle
e. Zero angle
f. Complete angle

Question 4.
Use a protractor to draw an acute angle, a right angle and an obtuse angle:
Solution:

[Note: Students may draw acute and obtuse angles of measure other than the ones given.]

Maharashtra Board Class 6 Maths Chapter 2 Angles Practice Set 2 Intext Questions and Activities

Question 1.
Look at the angles shown in the pictures below. Identify the type of angle and write its name below the picture: (Textbook pg. no. 6)

Solution:

Question 2.
Complete the following table: (Textbook pg. no. 6)

Solution:

Sr. No.	i.	ii.	iii.
Name of the angle	∠PYR or ∠RYP	∠LMN or ∠NML	∠BOS or ∠SOB
Vertex of the angle	Y	M	O
Arms of the angle	YP and YR	ML and MN	OB and OS

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 11 Ratio-Proportion Class 6 Practice Set 29 Answers Solutions.

6th Standard Maths Practice Set 29 Answers Chapter 11 Ratio-Proportion

Question 1.
If 20 metres of cloth costs Rs 3600, find the cost of 16 m of cloth.
Solution:
Cost of 20 metres of cloth = Rs 3600
∴ Cost of 1 metre of cloth = \(\frac{\text { cost of } 20 \text { metres of cloth }}{20}=\frac{3600}{20}\)
= Rs 180
∴ Cost of 16 metres of cloth = Cost of 1 metre of a cloth × 16
= 180 x 16 = Rs 2880
∴ The cost of 16 metres of cloth is Rs 2880.

Question 2.
Find the cost of 8 kg of rice, if the cost of 10 kg is Rs 325.
Solution:
Cost of 10 kg rice = Rs 325
∴ Cost of 10 kg rice

Cost of 8 kg rice = Cost of 1 kg rice x 8

∴ The cost of 8 kg rice is Rs 260.

Question 3.
If 14 chairs cost Rs 5992, how much will have to be paid for 12 chairs?
Solution:
Cost of 14 chairs = Rs 5992
∴ Cost of 1 chairs

= Rs 428
∴ Cost of 12 chairs = Cost of 1 chair x 12
= 428 x 12 = Rs 5136
∴ The amount to be paid for 12 chairs is Rs 5136.

Question 4.
The weight of 30 boxes is 6 kg. What is the weight of 1080 such boxes?
Solution:
Weight of 30 boxes = 6 kg
∴ Weight of 1 box

∴ Weight of 1080 boxes = Weight of 1 box x 1080

∴ The weight of 1080 boxes is 216 kg.

Question 5.
A car travelling at a uniform speed covers a distance of 165 km in 3 hours. At that same speed,
a. How long will it take to cover a distance of 330 km?
b. How far will it travel in 8 hours?
Solution:
Distance covered in 3 hours = 165 km
Distance covered in 1 hour

= 55 km
a. Time required to covered a distance of 330 km

= 6 hours
∴ The time required to cover a distance of 330 km is 6 hours.

b. Distance traveled in 8 hours = Distance covered in 1 hour x 8
= 55 x 8 = 440 km
∴ The distance traveled in 8 hours is 440 km.

Question 6.
A tractor uses up 12 litres of diesel while ploughing 3 acres of land. How much diesel will be needed to plough 19 acres of land?
Solution:
Diesel required to plough 3 acres of land =12 litres
∴ Diesel required to plough 1 acre of land

= 4 liters
∴ Diesel required to plough 19 acres of land = Diesel required to plough 1 acre of land x 19
= 4 x 19 = 76 litres
∴ Diesel needed to plough 19 acres of land is 76 litres.

Question 7.
At a sugar factory, 5376 kg of sugar can be obtained from 48 tonnes of sugarcane. If Savitatai has grown 50 tonnes of sugarcanes, how much sugar will it yield?
Solution:
Sugar obtained from 48 tonnes of sugarcane = 5376 kg
∴ Sugar obtained from 48 tonnes of sugarcane

∴ Sugar obtained from 50 tonnes of sugarcane = Sugar obtained from 1 tonne of sugarcane x 50
= 112 x 50 = 5600 kg
∴ 50 tonnes of sugarcane will yield 5600 kg of sugar.

Question 8.
In an orchard, there are 128 mango trees in 8 rows. If all the rows have an equal number of trees, how many trees would there be in 13 rows?
Solution:
Number of mango trees in 8 rows =128
Number of mango trees in 1 row

∴ Number of mango trees in 13 rows = Number of mango trees in 1 row x 13
= 16 x 13 = 208
∴ The number of mango trees in 13 rows are 208.

Question 9.
A pond in a field holds 120000 litres of water. It costs Rs 18000 to make such a pond. How many ponds will be required to store 480000 litres of water, and what would be the expense?
Solution:
Capacity of 1 pond = 1,20,000 litres
Total quantity of water = 4,80,000 litres
∴ Number of ponds required

Amount required to make 1 pond = Rs 18,000
∴ Amount required to make 4 ponds = Amount required to make 1 pond x 4
= 18,000 x 4 = Rs 72,000
∴ The number of ponds required to store 4,80,000 litres of water is 4, and the expense incurred in making the ponds is Rs 72,000.

Maharashtra Board Class 6 Maths Chapter 11 Ratio-Proportion Practice Set 29 Intext Questions and Activities

Question 1.
Vijaya wanted to gift pens to seven of her friends on her birthday. When she went to a shop to buy them, the shopkeeper told her the rate for a dozen pens.
i. Can you help Vijaya to find the cost of 7 pens?
ii. If you find the cost of one pen, you can also find the cost of 7, right? (Textbook pg. no. 59)

Solution:
Cost of 12 pens = Rs 84.
∴ Cost of 12 pens

∴ Cost of 7 pens = Cost of one pen x Number of pens = 7 × 7
∴ Cost of 7 pens = Rs 49
∴ The cost of 7 pens (Rs 49) can be found by unitary method.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 10 Equations Class 6 Practice Set 26 Answers Solutions.

6th Standard Maths Practice Set 26 Answers Chapter 10 Equations

Question 1.
Different mathematical operations are given in the two rows below. Find out the number you get in each operation and make equations.

1. 1. 16 ÷ 2,
   2. 5 × 2,
   3. 9 + 4,
   4. 72 ÷ 3,
   5. 4 + 5,

1. 8 × 3,
2. 19 – 10,
3. 10 – 2,
4. 37 – 27,
5. 6 + 7

Solution:

1. 16 ÷ 2 = 8
2. 5 × 2 = 10
3. 9 + 4 = 13
4. 72 ÷ 3 = 24
5. 4 + 5 = 9
6. 8 × 3 = 24
7. 19 – 10 = 9
8. 10 – 2 = 8
9. 37 – 27 = 10
10. 6 + 7 = 13

∴ The equations are

1. 16 ÷ 2 = 10 – 2
2. 5 × 2 = 37 – 27
3. 9 + 4 = 6 + 7
4. 72 ÷ 3 = 8 x 3
5. 4 + 5 = 19 – 10