Class 6

Light and the Formation of Shadows Class 6 Science Chapter 14 Questions And Answers Maharashtra Board

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Class 6 Science Chapter 14

Balbharti Maharashtra State Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows Notes, Textbook Exercise Important Questions and Answers.

Std 6 Science Chapter 14 Light and the Formation of Shadows Question Answer Maharashtra Board

Class 6 Science Chapter 14 Light and the Formation of Shadows Question Answer Maharashtra Board

1. Choose an appropriate word and fill in the blanks.

Question a.
A ……………. is a natural source of light.
Answer:
star

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question b.
A ………….. is an artificial source of light.
Answer:
candle

Question c.
When light passes through a prism, it gets separated into ………….. colours.
Answer:
seven

Question d.
The image obtained in the pinhole camera is …………. .
Answer:
inverted

Question e.
A shadow is formed when an …………… object comes in the way of light.
Answer:
opaque

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question f.
When a ………….. object comes in the way of light, light passes …………… it. options : seven, star, through, transparent, opaque, colors, shape, erect, inverted, luminous, candle.
Answer:
transparent, through

2. Write whether the following objects are luminous or non-luminous.

Question a.
Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows 1
Answer:

Object Luminous /Non-luminous
A book Non-luminous
A burning candle Luminous
A wax cloth Non-luminous
A pencil Non-luminous
A pen Non-luminous
A light bulb Luminous
A tyre Non-luminous
A torch Luminous
Stars Luminous
The planets Non-luminous
A satellite Non-luminous
The Moon Non-luminous

3. Match the following.

Question a.

Natural sources of light Man-made sources of light
The Sun Tubelight
Stars in the night sky Light bulb
Fireflies Torch
Anglerfish Burning candle
Honey mushroom Oil lamps

Answer:

Transparent Opaque Translucent
Piece of glass Water White plastic
Tea kettle
Note book
Cloth
Wooden
cupboard		 Tinted glass
Oil paper
Sheet of notebook
Wax paper

4. Write the answers to the following.

Question a.
What things are necessary for the formation of a shadow?
Answer:
Things necessary for the formation of a shadow are:

  1. A source of light
  2. An object
  3. A surface or screen on which the shadow is formed

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question b.
When can an object be seen?
Answer:
We can see an object when reflected rays reach our eyes.

Question c.
What is a shadow?
Answer:

  1. If an opaque object comes in the way of a light source, light does not pass through it.
  2. As a result the light does not reach a wall or any other surface on the other side of the object.
  3. That part remains dark. This dark part is called the shadow of the object.

Project:

Question 1.
Prepare a Newton’s disc.

Question 2.
Find out how to save electricity with the help of the sunlight we receive during the day.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 3.
Read a biography of Sir C. V. Raman and find out about the discoveries he made.

Class 6 Science Chapter 14 Light and the Formation of Shadows Important Questions and Answers

Fill in the blanks.

Question 1.
The ……………. is the main natural source of light.
Answer:
Sun

Question 2.
The light emitted by an electric torch is more …………….. than that obtained from a candle.
Answer:
intense

Question 3.
The left and right sides of the original object appear to be …………….. in the mirror.
Answer:
exchanged

Question 4.
The image is as far behind the mirror as the object is in ……………… of it.
Answer:
front

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 5.
The …………….. of the image is the same as that of the object.
Answer:
height

Question 6.
The materials through which light passes is said to be …………… .
Answer:
transparent

Question 7.
The materials through which light does not pass is said to be ………….. .
Answer:
opaque

Question 8.
The materials through which light passes partially is said to be ………….. .
Answer:
translucent

Question 9.
If an ………….. object comes in the way of a light source, light does not pass it.
Answer:
opaque, through

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 10.
The shadow of an object formed due to sunlight is ………….. in the mornings and evenings and ……………. in the afternoon.
Answer:
long, short

Question 11.
The shadow of an object is formed only when ………….. does not pass through the object.
Answer:
light

Question 12.
Stars are ………….. .
Answer:
luminous

Question 13.
Planets, satellites are ………….. .
Answer:
non-luminous

Question 14.
The largest sundial is at ………….., New Delhi.
Answer:
Jantar Mantar

Question 15.
………….. is celebrated as National Science Day.
Answer:
28th February

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 16.
Light travels in a straight line. This is called ………….. .
Answer:
linear propagation of light

Question 17.
The image formed on the diaphragm of the pinhole camera is ………….. .
Answer:
inverted

Question 18.
The kind of shadow an object forms depends upon the ………….. between the ………….., the object and the ………….. or the ………….. on which the shadow is formed.
Answer:
relative distance, source of light, surface, screen

State whether following statements are True or False.

Question 1.
Light travels in a straight line.
Answer:
True

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 2.
Stars are luminous.
Answer:
True

Question 3.
Image in a pinhole camera is inverted.
Answer:
True

Question 4.
In the afternoon, shadows are long.
Answer:
False

Question 5.
Fireflies are a natural source of light.
Answer:
True

Question 6.
We see the candle clearly when we bend the tube.
Answer:
False

Question 7.
We can see our image clearly in running water.
Answer:
False

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 8.
Tracing paper is transparent.
Answer:
False

Question 9.
The light obtained from an electric torch is more intense than that obtained from a candle.
Answer:
True

Question 10.
28tn February is celebrated as “National Science day” since 1987 in India.
Answer:
True

Question 11.
Classify the following into natural and man-made/artificial sources of light. (tubelight, light bulb, torch, burning candle, the sun, fireflies, anglerfish, honey mushroom, stars in the night sky, oil lamps, lanterns)
Answer:

Natural sources of light Man-made sources of light
The Sun
Stars in the night sky
Fireflies
Anglerfish
Honey mushroom		 Tubelight
Light bulb
Torch
Burning candle
Oil lamps
Lanterns

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 12.
Identify the transparent, opaque and translucent objects from among the following. (piece of glass, wax paper, tinted glass, oil paper, white plastic, a tea kettle, a notebook, cloth, water, a wooden cupboard, sheet of notebook.)
Answer:

Transparent Opaque Translucent
Piece of glass
Water		 White plastic
Tea kettle
Notebook
Cloth
Wooden
cupboard		 Tinted glass
Oil-paper
Sheet of notebook
Wax paper

Question 13.
Classify the following into the type of images they form: Clear image, faint image, no image. (still clear water, cemented wall, wooden surface, new steel dish, flower, glossy granite cladding of a wall, mirror, butter paper).
Answer:

Clear Image Faint Image No Image
Still clear water,
New steel dish,
Glossy granite
cladding of a wall,
Mirror		 Butter paper Wooden surface
Flower
Cemented wall

Question 14.
Relate images formed with the surfaces.
Answer:

  1. The clear images are formed on plane surfaces.
  2. Faint or no images are formed on rough surfaces.

Answer in one sentence.

Question 1.
What are luminous objects?
Answer:
The objects which emit light i.e. which themselves are a source of light, are called luminous objects.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 2.
What determines the intensity of light?
Answer:
The intensity of light is determined by the extent to which the objects emit light.

Question 3.
What are non-luminous objects?
Answer:
The objects that are not sources of light themselves are called as non-luminous objects.

Question 4.
What are artificial sources of light?
Answer:
Man-made objects which emit light are artificial sources of light.

Question 5.
What are natural sources of light?
Answer:
Natural substances, materials which emit light are called natural sources of light.

Question 6.
What is linear propagation of light?
Answer:
Property of light travelling in a straight line is linear propagation of light.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 7.
What is reflection of light?
Answer:
The rays of light falling on an object from a source of light are thrown back from the substance of that object. This is reflection of light.

Question 8.
How do we see objects around us?
Answer:
The rays of light falling on an object from a source of light are thrown back from the surface of that object. This is called reflection of light. We see the object when the reflected rays reach our eyes.

Question 9.
What is moonlight?
Answer:
Sunlight reflected from the surface of the moon reaching us, in which we see the moon is called the moonlight.

Question 10.
What type of image is formed in the mirror?
Answer:
The image formed in the mirror is ‘laterally inverted’ i.e. right side appears as left side and left side appears as right side.

Question 11.
What change do you see in the image if you decrease or increase your distance from the mirror?
Answer:
When the distance between object and mirror is increased the size of image decreases where as, when the distance is decreased the image size increases.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 12.
What difference do you find in the height of the image in the mirror and yourself?
Answer:
The size of the image in the mirror is the same as that of the object.

Question 13.
What is the image on the diaphragm of the pinhole camera?
Answer:
An inverted or an upside down image of the candle is seen on the diaphragm of the pinhole camera.

Question 14.
What do you mean by transparent object?
Answer:
The objects / materials through which light passes are said to be transparent.

Question 15.
What do you mean by opaque materials?
Answer:
The materials through which light does not pass are said to be opaque.

Question 16.
What do you mean by translucent materials?
Answer:
The materials through which light passes partially are said to be translucent.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 17.
How is the shadow in the morning, afternoon and evening?
Answer:
The shadows are long in the mornings and evenings and short in the afternoon.

Question 18.
What is shade of a tree?
Answer:
The shade of a tree is its shadow.

Question 19.
How many colours is sunlight made up of?
Answer:
Sunlight is made up of seven colours.

Give reasons for the following.

Question 1.
When we see in the mirror, we see our image in the mirror.
Answer:
When we see our face in the mirror, the light reflected from our face falls on the mirror and gets reflected back again. Hence, we see our image in the mirror.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 2.
Opaque materials cast shadow.
Answer:
An opaque material does not allow light to pass. Hence, it casts a shadow.

Question 3.
Transparent and translucent object do not cast a shadow.
Answer:
Translucent objects cast a faint shadow whereas transparent objects do not cast a shadow at all because they allow light to pass through them.

Use your brain power!

Question 1.
Why is the image on the diaphragm of the pinhole camera inverted?
Answer:
1. The pinhole camera works on the principle of light travelling in a straight line.
2. The rays of light from the candle flame go in all directions.
3. We consider only two rays that pass through the hole and fall on the screen.
Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows 2
4. The rays intersect at the pinhole.
5. Since the rays cross over at that point, the top of the object appears at the bottom of the image and the bottom of the image appears at the top. Thus, we see an inverted image of the candle.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 2.
How will you light up a dark room using reflected light?
Answer:
Focusing on the wall with torch light. Mirrors or reflectors can be used to get light from outside.

Question 3.
Try to start the TV by operating the remote control from behind it.
Answer:
T.V will not start.

Question 4.
In which step is the flame of the candle seenclearly? Why?
Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadow 3
Answer:
In step 1 the flame of the candle is seen clearly because light travels in straight line.

Can you tell?

Question 1.
Can we see anything in total darkness?
Answer:
No, we cannot see anything in total darkness.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 2.
What helps us to see the objects around us?
Answer:
Reflected light helps us to see objects around us.

Question 3.
What does the light in each one of the pictures originate form?
Answer:
Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadow 4
1. Bulb
2. Firefly
3. Candle
4. Sun

Question 4.
Name the natural sources of light.
Answer:
Sun, Fireflies

Question 5.
In which objects do we see our reflection?
Answer:
All objects reflect light rays, but the best reflectors of light are mirrors, still water in a lake, new steel dish i.e. smooth shiny surfaces.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 6.
What difference do you notice on looking through the windows in the picture? What causes the difference? The picture shows transperant, opaque, translucent window panes. Spot them.
Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows 5
Answer:

  1. Through the first window we can see a clear picture of things outside.
  2. Second window gives a faint image.
  3. Through the third window, we can’t see anything
  4. The difference in the image is due to the material of the window panes.
  5. The first window pane is transparent.
  6. The second window pane is translucent.
  7. The third window pane is opaque.

Answer the following questions in brief.

Question 1.
List factors on which shadow depends.
Answer:
Shadow depends on relative distance between the source of light, the object and the surface on which the shadow is formed.

Question 2.
How we can see that light travels in straight line.
Answer:

  1. In the morning or in the afternoon, rays of light enter a slit in a door, window or a small hole in the roof.
  2. As these rays of light from the slit or the hole move towards the floor, the dust particles in their way are clearly seen.
  3. Due to these particles, the path of light becomes visible to us.
  4. Thus we can see that their path is along straight lines.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 3.
What is the difference between an object and its reflection? What causes the difference?
Answer:

  1. Object and its reflection result in formation of images.
  2. Reflections taking place from highly polished metals, mirrors, still water etc, form clear images.
  3. Reflections taking place from wooden surface, flower, book form dull, blurred images.
  4. The difference in reflections is caused by the surface of the object.
  5. Regular reflections have smooth, polished surfaces, hence, image is clear.
  6. Diffused reflections have hard, rough surfaces, hence, image is dull.

Question 4.
List characteristics of images in a plane mirror.
Answer:

  1. The left and right sides of the original object appear to be exchanged in the mirror image.
  2. The image is as far behind the mirror as the object is in front of it.
  3. The size of the image is the same as that of the object.

Question 8.
State the characteristics of image formed by a pinhole camera.
Answer:
Characteristics of an image formed by a pinhole camera are as follows:

  1. It is inverted/upside down.
  2. It can be obtained on a screen – real image.

Try this.

Question 1.
Make your friend stand in between the torch and the wall. What happens?
Answer:
Friend’s Shadow forms on the wall.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 2.
Place a glass filled with water on a sheet of paper in the window so it receives direct sunlight. What is seen on the paper?
Answer:
We see rainbow colours on the paper.

Question 3.
Can we do the same in a dark room with the help of a prism and a torch? What do we learn from this?
Answer:
yes, we can. Light gets seperated into seven colour. From this we learn that white light contains seven colours.

Question 4.
If you dip the wire loop in the soap water and then blow it, soap bubbles are formed. Are the beautiful colours of the rainbow seen in these bubbles?
Answer:
Yes, splitting of white light into different colours takes place.

Question 5.
What do you see on holding a CD in the sun?
Answer:
CD reflects rainbow colours, and interesting : patterns.

Maharashtra Board Class 6 Science Solutions Chapter 14 Light and the Formation of Shadows

Question 6.
Raise your right hand. In mirror which hand of the mirror image is raised?
Answer:
Left hand of the mirror image is raised.

Question 7.
Is there any difference between your height and height of the mirror image?
Answer:
The height remains the same.

6th Std Science Questions And Answers:

Practice Set 37 Class 6 Answers Maths Chapter 16 Quadrilaterals Maharashtra Board

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

Quadrilaterals Class 6 Maths Chapter 16 Practice Set 37 Solutions Maharashtra Board

Std 6 Maths Practice Set 37 Solutions Answers

Question 1.
Observe the figures below and find out their names:
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 37 1
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)
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 37 2
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:
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 37 3
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.

Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 37 4

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)
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 37 5
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)
Maharashtra Board Class 6 Maths Solutions Chapter 16 Quadrilaterals Practice Set 37 6
Solution:
If we place the triangles together such that the equal sides overlap, the two triangles form a quadrilateral.

Std 6 Maths Digest

Practice Set 3 Class 6 Answers Maths Chapter 2 Angles Maharashtra Board

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

Angles Class 6 Maths Chapter 2 Practice Set 3 Solutions Maharashtra Board

Std 6 Maths Practice Set 3 Solutions Answers

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 Solutions Chapter 2 Angles Practice Set 3 1

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°.
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 3 2

Step 2:
Cut arcs on the rays BA and BC. Name these points as D and E respectively.
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 3 3

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
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 3 4

Step 4:
Draw ray BO.
Ray BO is the angle bisector of ∠ABC.
i.e. m∠ABO = m∠CBO = 30°
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 3 5

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°.
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 3 6

Step 2:
Cut arcs on the rays QP and QR.
Name these points as M and N respectively.
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 3 7

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
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 3 8

Step 4:
Draw ray QO.
Ray QO is the angle bisector of ∠PQR.
i.e. m∠PQO = m∠RQO = 45°
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 3 9

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:
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 3 10

Question 4.
Look at the pictures below and identify the different types of angles. (Textbook pg. no. 8)
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 3 11
Solution:
i. Complete angle
ii. Reflex and Acute angle
iii. Acute and Obtuse angle

Std 6 Maths Digest

Practice Set 15 Class 6 Answers Maths Chapter 5 Decimal Fractions Maharashtra Board

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

Decimal Fractions Class 6 Maths Chapter 5 Practice Set 15 Solutions Maharashtra Board

Std 6 Maths Practice Set 15 Solutions Answers

Question 1.
Write the proper number in the empty boxes.
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 15 1
Solution:
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 15 2

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. \(\frac { 3 }{ 4 }\)
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 15 3

ii. \(\frac { 4 }{ 5 }\)
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 15 4

iii. \(\frac { 9 }{ 8 }\)
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 15 5

iv. \(\frac { 17 }{ 20 }\)
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 15 6

v. \(\frac { 36 }{ 40 }\)
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 15 7

vi. \(\frac { 7 }{ 25 }\)
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 15 8

vii. \(\frac { 19 }{ 200 }\)
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 15 9

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
= \(\frac { 704 }{ 10 }\)

Std 6 Maths Digest

Practice Set 27 Class 6 Answers Maths Chapter 10 Equations Maharashtra Board

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

Equations Class 6 Maths Chapter 10 Practice Set 27 Solutions Maharashtra Board

Std 6 Maths Practice Set 27 Solutions Answers

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..

Std 6 Maths Digest

Practice Set 36 Class 6 Answers Maths Chapter 15 Triangles and their Properties Maharashtra Board

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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.

Triangles and their Properties Class 6 Maths Chapter 15 Practice Set 36 Solutions Maharashtra Board

Std 6 Maths Practice Set 36 Solutions Answers

Question 1.
Observe the figures below and write the type of the triangle based on its angles:
Maharashtra Board Class 6 Maths Solutions Chapter 15 Triangles and their Properties Practice Set 36 1
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:
Maharashtra Board Class 6 Maths Solutions Chapter 15 Triangles and their Properties Practice Set 36 2
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.
Maharashtra Board Class 6 Maths Solutions Chapter 15 Triangles and their Properties Practice Set 36 3
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)
Maharashtra Board Class 6 Maths Solutions Chapter 15 Triangles and their Properties Practice Set 36 4
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)
Maharashtra Board Class 6 Maths Solutions Chapter 15 Triangles and their Properties Practice Set 36 5

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)
Maharashtra Board Class 6 Maths Solutions Chapter 15 Triangles and their Properties Practice Set 36 6

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)
Maharashtra Board Class 6 Maths Solutions Chapter 15 Triangles and their Properties Practice Set 36 7
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?
Maharashtra Board Class 6 Maths Solutions Chapter 15 Triangles and their Properties Practice Set 36 8
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°.
Maharashtra Board Class 6 Maths Solutions Chapter 15 Triangles and their Properties Practice Set 36 10
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:
Maharashtra Board Class 6 Maths Solutions Chapter 15 Triangles and their Properties Practice Set 36 9

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

Std 6 Maths Digest

Practice Set 14 Class 6 Answers Maths Chapter 5 Decimal Fractions Maharashtra Board

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

Decimal Fractions Class 6 Maths Chapter 5 Practice Set 14 Solutions Maharashtra Board

Std 6 Maths Practice Set 14 Solutions Answers

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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 1

ii. 39 + 700.65
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 2

iii. 40 + 27.7 + 2.451
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 3

Question 3.
Subtract:
i. 85.96 – 2.345
ii. 632.24 – 97.45
iii. 200.005 – 17.186
Solution:
i. 85.96 – 2.345
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 4

ii. 632.24 – 97.45
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 5

iii. 200.005 – 17.186
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 6

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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 7
= 42.365 km
Distance travelled in car = 12 km 460 m
= 12 km + \(\frac { 460 }{ 1000 }\) km
= 12 km + 0.460 km
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 8
= 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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 9
= 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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 10
Cost of 1 m of cloth = Rs 120
∴ Cost of 4.05 m of cloth = 4.05 x 120
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 11
∴ 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
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 12
= 1.75 kg
∴ Weight of watermelon left = Total weight of watermelon – Weight of watermelon given to children
= 4.25 kg – 1.75 kg
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 13
= 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.
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 14
= 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)
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 15
Nandu will get __ rupees back.
Solution:
100 – 38 = 62.00
Nandu will get Rs 62 rupees back.
Maharashtra Board Class 6 Maths Solutions Chapter 5 Decimal Fractions Practice Set 14 16

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.)

Std 6 Maths Digest

Practice Set 2 Class 6 Answers Maths Chapter 2 Angles Maharashtra Board

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

Angles Class 6 Maths Chapter 2 Practice Set 2 Solutions Maharashtra Board

Std 6 Maths Practice Set 2 Solutions Answers

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:
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 2 1
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:
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 2 2
[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)
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 2 3
Solution:
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 2 4

Question 2.
Complete the following table: (Textbook pg. no. 6)
Maharashtra Board Class 6 Maths Solutions Chapter 2 Angles Practice Set 2 5
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

Std 6 Maths Digest

Practice Set 29 Class 6 Answers Maths Chapter 11 Ratio-Proportion Maharashtra Board

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

Ratio-Proportion Class 6 Maths Chapter 11 Practice Set 29 Solutions Maharashtra Board

Std 6 Maths Practice Set 29 Solutions Answers

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
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 1
Cost of 8 kg rice = Cost of 1 kg rice x 8
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 2
∴ 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
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 3
= 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
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 4
∴ Weight of 1080 boxes = Weight of 1 box x 1080
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 5
∴ 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
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 6
= 55 km
a. Time required to covered a distance of 330 km
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 7
= 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
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 8
= 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
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 9
∴ 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
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 10
∴ 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
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 11
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)
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 12
Solution:
Cost of 12 pens = Rs 84.
∴ Cost of 12 pens
Maharashtra Board Class 6 Maths Solutions Chapter 11 Ratio-Proportion Practice Set 29 13
∴ 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.

Std 6 Maths Digest

Practice Set 26 Class 6 Answers Maths Chapter 10 Equations Maharashtra Board

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

Equations Class 6 Maths Chapter 10 Practice Set 26 Solutions Maharashtra Board

Std 6 Maths Practice Set 26 Solutions Answers

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. 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

Std 6 Maths Digest