Class 7

Balbharti Maharashtra State Board Class 7 Geography Solutions Chapter 2 The Sun, the Moon and the Earth Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 7 Geography Solutions Chapter 2 The Sun, the Moon and the Earth

Class 7 Geography Chapter 2 The Sun, the Moon and the Earth Textbook Questions and Answers

1. Correct the wrong statements. Write down the corrected ones.

Question 1.
The moon revolves around the sun.
Answer:
Wrong – The moon revolves around the earth.

Question 2.
On a full moon day, the moon, the sun and the earth are positioned in this sequence.
Answer:
Wrong – On a full moon day, the sun, the earth and the moon are positioned in this sequence.

Question 3.
The revolutionary orbits of the earth and the moon are in the same plane.
Answer:
Wrong – The revolutionary orbits of the earth and the moon are not in the same plane. The moon’s revolutionary orbit makes an angle of about 5° with that the earth.

Question 4.
In one revolution of the moon, its orbit intersects the earth’s orbit only once.
Answer:
Wrong – In one revolution of the moon, its orbit intersects the earth’s orbit twice.

Question 5.
It is alright to observe a solar eclipse without protecting the eyes.
Answer:
Wrong – It is necessary to view the sun disc through dark glasses or through special goggles made for that purpose, otherwise the intense light of the sun can be harmful to the naked eye.

Question 6.
An annular solar eclipse occurs when the moon is in the perigee position.
Answer:
Wrong – An annular solar eclipse occurs when the moon is in the apogee position.

2. Select the correct option.

Question 1.
Solar eclipse

Answer:
(b)

Question 2.
The shape of sun disc at the time of annular solar eclipse.

Answer:
(a)

Question 3.
Apogee position of the moon.

Answer:
(c)

3. Complete the following table.

Details	Lunar Eclipse	Solar Eclipse
Phase of the moon	…………	…………….
Sequence	Moon-Earth- Sun	………….
Type of Eclipse	………….	…………..
Maximum duration of total eclipse	107 minutes	…………….

Answer:

Details	Lunar Eclipse	Solar Eclipse
Phase of the moon	Full Moon Day	New Moon Day
Sequence	Moon-Earth-Sun	Moon-Earth-Sun
Type of Eclipse	Total and Partial	Total, Partial, Annular
Maximum duration of total eclipse	107 minutes	440 Seconds

4. Draw and label the diagrams.

Question 1.
Total and partial solar eclipse:
Answer:

Question 2.
Total and partial lunar eclipse:
Answer:

5. Answer the following:

Question 1.
Why do the sun, the moon and the earth not lie in one and the same line on every full moon and new moon day?
Answer:
(i) The orbital path of the earth and that of moon are not in the same plane.
(i) The moon’s revolutionary orbit makes an angle of about 5° with that of the earth.
(iii) On each new moon day, the lines joining the earth and the sun and the moon make an angle of 0° whereas on each full moon day, this angle is 180°.
(iv) So, the sun, the earth and the moon may not be in one straight line in the same plane on every new moon or full moon day.

Question 2.
When a total solar eclipse occurs why is the partial eclipse also seen from the earth?
Answer:
(i) On a new moon day if the sun, the moon & the earth fall in one line & are in the same plane, the shadow of the moon falls on the earth
(ii) This shadow is of two types – the central portion of the shadow is darker & the periphery is light.
(iii) In the area where the dark shadow falls, the sun becomes completely invisible. Such an area experiences a total solar eclipse.
(iv) However during the same period, at the places where the shadow is lighter, the sun disc appears partially covered. Such an area experience partial solar eclipse. Thus when a total solar eclipse occurs a partial eclipse is also seen from the earth.

Question 3.
Suggest measures that can be taken to eradicate the superstitions related to the eclipses.
Answer:
The following measures can be taken to eradicate superstition related to eclipses:

• Use of media to create awareness.
• Parental guidance to help think logically.
• Teachers guidance to help students develop a scientific outlook.
• Campaigns, public meetings and lectures especially in rural areas to eradicate superstitions.

Question 4.
What precautions should we take while observing a solar eclipse?
Answer:
The precautions to be taken while observing a solar eclipse are:

• We should not observe a solar eclipse with naked eyes as the intense light of the sun can harm them.
• We must use dark glasses or goggles that are specially designed for viewing the solar eclipse.

Question 5.
What types of Solar eclipses will occur in perigee conditions?
Answer:
Total and Partial solar eclipse will occur in perigee condition.
(1) Total solar eclipse :

• On a new moon day, the sun, the moon the earth are in a straight line & the shadow of the moon falls on the earth
• The area of dark shadow falls on the earth, the sun becomes completely invisible. This condition is known as total solar eclipse.

(2) Partial solar eclipse:

• However during the same period at places where the shadow is lighter, the sun disc appears partially covered,
• This condition is described as partial solar eclipse.

Activities:

1. Collect paper cuttings about eclipses and paste them in a notebook.
2. Write a note on an eclipse that you have seen.
3. Using the internet, ‘Panchanga’ and calendar collect information about the eclipses that are likely to occur this year.

Class 7 Geography Chapter 2 The Sun, the Moon and the Earth InText Questions and Answers

Use your brain power:

Question 1.
On the day of solar eclipse, in which part of the earth will it not be seen?
Answer:
Solar eclipse will not be seen where there is night.

Question 2.
Can we see total and annular solar eclipses on the same occasion?
Answer:
No, total and annular solar eclipses cannot be seen on the same occasion.

Question 3.
Why is an annular lunar eclipse not seen?
Answer:
An annular lunar eclipse cannot be seen because:

• The size of the earth is bigger than that of the moon.
• As compared to the sun, the moon is close to the earth.
• Therefore, it is not possible that a dark shadow of the earth is cast in space & does not reach the moon.

Question 4.
Which eclipses will you see from the moon?
Answer:
Solar eclipses can be seen from the moon.

Question 5.
Why are solar eclipses caused by the other planets not seen from the earth?
Answer:
(i) A Solar eclipse occurs when the moon comes in between the sun & the earth & all of them are in the same plane & fall in one line on a new moon day.
(ii) Similarly, when Venus & Mercury come in between the earth & the sun, they make a small speck against the surface of the sun as they are too far away.
(iii) The other planets have orbits outside the earth’s & cannot fall in line between the earth & the sun.
(iv) Hence, the solar eclipse caused by other planets are not seen from the earth.

Think about it:

Question 1.
Figure shows positions of the moon as seen from the space and as seen from the earth. How will you identify which are which?

Answer:
Half of the moon’s portion is illuminated by the sun, and the other half remains dark. However, only some part of this illuminated portion of the moon can be seen from the earth. The position of the moon from the Earth is as follows:

• New moon
• Waxing crescent
• 1st Quarter
• Waxing Gibbous
• Full moon
• Waning Gibbous
• Last quarter
• Waning crescent

Question 2.
Like sunlight and moonlight, is there anything called the earth light? If yes, where do you think it is found?
Answer:
Yes, Earthlight is the partial illumination of the dark portion of the moon’s surface by light reflected from the earth’s airglow.

Question 3.
When solar eclipses do not occur on a new moon day, does it mean that the moon does not have any shadow at all?
Answer:
The moon casts a shadow on every new moon day, but solar eclipse occurs only when the moon’s shadow falls on the earth.

Class 7 Geography Chapter 2 The Sun, the Moon and the Earth Additional Important Questions and Answers

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

Question 1.
The shape of the moon’s orbit is _________. (round, elliptical, square)
Answer:
elliptical

Question 2.
The moon has _____ and ______ motions. (axial, orbital, elliptical)
Answer:
axial, orbital

Question 3.
When the moon is closest to the earth it is said to be in ________. (apogee, perigee, orbit)
Answer:
perigee

Question 4.
When the moon is farthest to the earth it is said to be in ______. (apogee, perigee, arial)
(round, elliptical, square)
Answer:
apogee

Question 5.
Solar Eclipse occurs on the _______ day. (full moon, new moon, quarter moon)
Answer:
new moon

Question 6.
Lunar Eclipse occurs on the ______ day. (full moon, new moon, quarter)
Answer:
full moon

Question 7.
_______ solar eclipse is a rare phenomenon. (Total, Partial, Annular)
Answer:
Annular

Question 8.
When the shadow of the moon on the central portion of the earth is darker while the periphery is lighter it is known as a _______ solar eclipse.
(partial, total, lunar)
Answer:
total

Question 9.
When the shadow of the moon on the central portion of the earth is lighter and the sun disc appears to be partially covered, it is known _________ solar eclipse.
(total, partial, solar)
Answer:
partial

Question 10.
The moon is ________ when the earth, the moon and the sun make an angle of 90°. (full, semicircular, crescent)
Answer:
semicircular

Question 11.
Occultation occurs with reference to the _____. (sun, moon, earth)
Answer:
moon

Question 12.
Transit is associated with the _______ .(sun, moon, earth)
Answer:
sun

Question 13.
During the period of _______ eclipse, a large number of birds and animals get confused. (solar, lunar, annular)
Answer:
solar

Correct the wrong statements. Write down the correct ones:

Question 1.
When the moon is closest to the earth it is said to be in apogee and when it is the farthest, the position is called perigee.
Answer:
Wrong – When the moon is closest to the earth it is said to be in perigee and when it is the farthest, the position is called apogee.

Question 2.
The moon wanes from the new moon day to the full moon day and waxes from the full moon to the new moon day.
Answer:
Wrong – The moon waxes from the new moon day to the full moon day and wanes from the full moon to the new moon day.

Select the correct options:

Question 1.

Answer:
(c)

Question 2.

Answer:
(c)

Complete the following table:

Question 1.

Details	Numerical facts
1. Perigee distance	……….
2. Apogee distance	……….
3. Angular distance between the moon’s and the earth’s orbit	………..
4. Maximum duration of a total solar eclipse	……….
5. Maximum duration of a total lunar eclipse	…………
6. Number of solar eclipse (types)	………….
7. Number of lunar eclipse (types)	……………

Answer:

Details	Numerical facts
1. Perigee distance	3,56,000 km
2. Apogee distance	4,07,000 km
3. Angular distance between the moon’s and the earth’s orbit	5°
4. Maximum duration of a total solar eclipse	7 minutes 20 seconds
5. Maximum duration of a total lunar eclipse	107 minutes
6. Number of solar eclipse (types)	3
7. Number of lunar eclipse (types)	2

Name the following:
Question 1.
Motions of the moon.
Answer:
Axial and orbital motion.

Question 2.
Types of solar eclipse.
Answer:
Total solar eclipse, Partial solar eclipse, Annular solar eclipse.

Question 3.
Types of lunar eclipse.
Answer:
Total lunar eclipse and Partial lunar eclipse.

Question 4.
Eclipse that confuses animals and birds.
Answer:
Solar eclipse.

Question 5.
An example of occultation.
Answer:
Total solar eclipse.

Question 6.
The event of the occurrence of solar eclipse or lunar eclipse.
Answer:
Astronomical event.

Define the following:

Question 1.
Perigee:
Answer:
When the moon is the closest to the earth in its orbit, it is perigee.

Question 2.
Apogee:
Answer:
When the moon is at the farthest from the earth in its orbit, it is apogee.

Question 3.
Occultation:
Answer:
The moon revolves around the earth. While doing so, it obscures1 a star or a planet and that celestial2 body appears to hide behind the moon. This is called occultation.

Question 4.
Transit:
Answer:
If an inner planet like Mercury or Venus comes in between the line of the earth and the sun, a transit occurs. At that time a small dot appears to move across the sun’s disc.

Question 5.
Phases of the moon:
Answer:
The illuminated portion of the moon disc observed from the earth that keeps on changing every day within a lunar month are called the phases of the moon.

Question 6.
Solar Eclipse:
Answer:
If the moon comes between the earth and the sun in a straight line on new moon day, the shadow of the moon falls on the earth and the sun becomes totally or partially invisible in the shadow zone. This phenomenon is called the solar eclipse.

Question 7.
Lunar Eclipse:
Answer:
On a full moon day, the . moon’s path of revolution passes through the thick shadow of the earth and the moon becomes totally or partially invisible. This phenomenon is called lunar eclipse.

Question 8.
Annular Solar eclipse
Answer:
Sometimes when the moon is in apogee position, a deep shadow of the moon is cast in space & does not reach the earth. From a very small region of the earth, only an illuminated edge of the sun disc is seen in the form of a ring. This is called annular solar eclipse.

Answer in one sentence:

Question 1.
What is the period of waxing moon?
Answer:
The period of waxing moon is the fortnight from new moon to full moon in which the illuminated part of the moon from the earth keeps on increasing (waxing)

Question 2.
What is the period of waning moon?
Answer:
The period of waning moon is the fortnight from full moon to new moon in which the illuminated part of the moon seen from the earth keeps on decreasing (waning)

Question 3.
Why can we see the phases of the moon from the earth?
Answer:
We can see the phases of the moon from the earth due to the sunlight reflected from the moon.

Question 4.
When does the moon appear semi-circular in shape in the sky?
Answer:
On the first and the third quarter days, the moon, the earth & the sun make an angle of 90° due to which we see half the portion of the illuminated moon in a semi-circular shape in the sky.

Question 5.
What types of solar eclipses occur in perigee position?
Answer:
Total & Partial Solar eclipses occur in perigee position.

Question 6.
When does a lunar eclipse occur?
Answer:
A lunar eclipse occurs when the moon enters the shadow of the earth while revolving around it.

Give geographical reasons for the following statements:

Question 1.
We constantly see only one and the same side of the moon.
Answer:

• The time, the moon takes to make one revolution around the earth and one rotation around itself is the same.
• As the axial and the orbital motions of the moon are almost of the same duration, only one, the same side of the moon faces the earth.
• Therefore, we constantly see only one and the same side of the moon.

Question 2.
The distance between the moon and the earth keeps on changing.
Answer:

• The moon’s orbit of the revolution is elliptical to that of the earth.
• Hence the distance of the moon from the earth is not the same everywhere along its orbit while revolving.
• When it is the closest to the earth it is said to be in perigee and when it is at the farthest, the position is called apogee.

Question 3.
Eclipses do not occur on each new moon or full moon day.
Answer:

• The orbital path of the earth and that of the moon are not in the same plane.
• The moon’s revolutionary orbit makes an angle of about 5° with that of the earth.
• So, the sun, the earth and the moon may not be in one straight line in the same plane on every new moon or full moon day.
• Hence, eclipses do not occur on each new moon or full moon day.

Question 4.
The Annular solar eclipse is rarely seen.
Answer:
The Annular solar eclipse is rarely seen because

• The moon is in apogee position.
• The deep shadow of the moon is cast in space and does not reach the earth.
• From a very small region of the earth, only an illuminated edge of the sun disc is seen in the form of a ring.

Question 5.
Birds and animals are confused during the solar eclipse.
Answer:
Birds and animals are confused during the solar eclipse because –

• Untimely darkness sets in.
• The event does not suit their biological clock.
• Their response to the event is also unusual.

Question 6.
On the first and third quarter days, the moon appears semicircular in shape.
Answer:

• On the first and third quarter days, the moon, the earth and the sun make an angle of 90°
• At these positions, we see half the portion of an illuminated moon.
• Hence, in the sky, the moon appears semi-circular in shape.

Answer the following:

Question 1.
What precautions should we take while observing a solar eclipse?
Answer:
The precautions to be taken while observing a solar eclipse are:

• We should not observe a solar eclipse with naked eyes as the intense light of the sun can harm them.
• We must use dark glasses or goggles that are specially designed for viewing the solar eclipse.

Question 2.
What types of Solar eclipses will occur in perigee condition?
Answer:
Total and Partial solar eclipse will occur in perigee condition.
(i) Total solar eclipse :

• On a new moon day, the sun, the moon the earth are in a straight line & the shadow of the moon falls on the earth
• The area of dark shadow falls on the earth, the sun becomes completely invisible. This condition is known as total solar eclipse.

(ii) Partial solar eclipse:

• However during the same period at places where the shadow is lighter, the sun disc appears partially covered,
• This condition is described as partial solar eclipse.

Question 3.
How are lunar eclipses formed?
Answer:
(i) A lunar eclipse occurs when the moon enters the shadow of the earth while revolving around it.
(ii) On a full moon day if the sun, the earth & the moon, are in a straight line, the orbital path of the moon passes through the dark shadow of the earth.
(iii) If the moon is totally hidden within the shadow, a total lunar eclipse is seen & if only a part of the moon is in the shadow, a partial lunar eclipse is seen.

Write short notes on:

Question 1.
Characteristics of solar eclipse.
Answer:

• A solar eclipse occurs on a new moon day, but not on every new moon day.
• If and only if, the sun, the moon and the earth are in the same plane and fall in one line, the solar eclipses occur.
• The maximum duration of a total solar eclipse is 7 minutes and 20 seconds Question 40 seconds).

Question 2.
Characteristics of lunar eclipse.
Answer:

• A lunar eclipse occurs on a full moon day, but not on every full moon day.
• A lunar eclipse occurs if and only if the sun, the moon and the earth are in the same plane and fall in one line.
• The maximum duration of a total lunar eclipse is 107 minutes.

Question 3.
Occupation.
Answer:

• This is a typical event occurring in space.
• The moon revolves around the earth, while doing so, it obscures a star or a planet and that celestial body appears to hide behind the moon. This is called occultation.

Question 4.
What is a Transit?
Answer:

• If an inner planet like a Mercury or Venus comes in between the line of the earth and the sun, a transit occurs.
• At that time, a small dot appears to move across the sun’s disc.
• Transit is a type of solar eclipse.

Distinguish between the following:

Question 1.
Solar eclipse and Lunar eclipse :
Answer:

Solar eclipse	Lunar eclipse
(i) A solar eclipse occurs on a new moon day. (ii) If and only if the sun, the moon and the earth are in the same plane and fall in one line, the solar eclipse occurs. (iii) The maximum duration of total solar eclipse is 7 minutes and 20 seconds (440 seconds).	(i) A lunar eclipse occurs on a full moon day. (ii) A lunar eclipse occurs if and only if the sun, the earth and the moon are in the same plane and fall in one line. (iii) The maximum duration of a total lunar eclipse is 107 minutes.

Question 2.
Total Solar eclipse and Annular Solar eclipse.
Answer:

Total Solar eclipse	Annular Solar eclipse
(i) When the moon is between the sun and the earth all the three celestial objects are on the same plane and fall in one line.	(i) moon is in apogee position. This means it is farthest from the earth.
(ii) The central portion of the shadow of the moon is darker and the periphery is light.	(ii) A deep shadow of the moon is cast in space and does not reach the earth.
(iii) In the area of dark shadow on the earth, the sun becomes completely invisible. This condition is known as total solar eclipse	(iii) From a very small region of the earth, only an illuminated edge of the sun as a disc is seen in the form of a ring. This is called an annular eclipse.

Draw and label the diagrams:

Question 1.
Moon’s Positions:
Answer:

Question 2.
Angle between the planes of orbit.
Answer:

Question 3.
Consider the relative positions of the sun, the moon, and the earth on the new moon day and both the quarters. What will be the angle between the lines joining the earth and the moon as well as the earth and the sun? How many times will this angle be formed in a month?
Answer:
The angle between the lines joining the earth and the moon as well as the earth and the sun is 90°. This angle will be formed twice a month.

Balbharti Maharashtra State Board Class 7 Geography Solutions Part 1 Chapter 1 How Seasons Occur Part 1 Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 7 Geography Solutions Chapter 1 How Seasons Occur Part 1

Class 7 Geography Chapter 1 How Seasons Occur Part 1 InText Questions and Answers

1. Answer the following questions in one sentence:

Question 1.
How do day and nigbt occur on the earth?
Answer:
Day and night occur on the earth because of the earths rotation.

Question 2.
What term is used to describe the earth’s revolution around the sun?
Answer:
The term orbital motion is used to describe the earth’s revolution around the sun.

Question 3.
How long does the earth take to do so?
Answer:
The earth takes 365 days and 6 hours to complete one revolution.

Question 4.
In which hemispheres is our country located?
Answer:
Our country lies in the northeastern hemisphere.

Question 5.
Why don’t the sun’s rays fall perpendicularly at all places on the earth?
Answer:
The sun’s rays do not fall perpendicularly at all the places on the earth because the earth is not flat but spherical in shape.

2. Can you tell?

Records of the entries of sunrise and sunset for the month of June are given below. Answer the questions, that follow.

Date	Sunrise	Sunset	Day	Night	Source of Information
19th June	06.01	19:18	13 Hrs. 16 Min.	10 Hrs. 44 Min.	Time and date
20th June	06.02	19:18	13 Hrs. 16 Min.	10 Hrs. 44 Min.	Time and date
21st June	06.02	19:18	13 Hrs. 16 Min.	10 Hrs. 44 Min.	Time and date
22nd June	06.02	19:18	13 Hrs. 16 Min.	10 Hrs. 44 Min.	Time and date
23rd June	06.02	19:19	13 Hrs. 16 Min.	10 Hrs. 44 Min.	Time and date
24th June	06.02	19:19	13 Hrs. 16 Min.	10 Hrs. 44 Min.	Time and date
25th June	06.03	19:19	13 Hrs. 16 Min.	10 Hrs. 44 Min.	Time and date
26th June	06.03	19:19	13 Hrs. 16 Min.	10 Hrs. 44 Min.	Time and date
27th June	06.03	19:19	13 Hrs. 15 Min.	10 Hrs. 45 Min.	Time and date
28th June	06.04	19:19	13 Hrs. 15 Min.	10 Hrs. 45 Min.	Time and date

Question 1.
Refer to the table above and state which are the longest days?
Answer:
The longest days are from 19th June to 26th June.

Question 2.
What difference do you notice in the duration of the nights from 19th June to 28th June?
Answer:
The duration of the nights keeps on increasing day by day.

Question 3.
What is the reason behind it?
Answer:
The sun is now moving southwards.

Question 4.
How did you find out the duration of the night?
Answer:
The duration of the day is subtracted from 24 hours to get the duration of the night.

Question 5.
Which two dates have days and nights of the same duration?
Answer:
March 22 and September 23 have the same duration of days and nights.

Question 6.
With the help of the table, you saw how the duration of the day and the night changes. Do you think such a change occurs everywhere on the earth?
Answer:
Yes, duration of day and night changes everywhere on the earth.

Question 7.
Use the above format to record the duration of daytime from 19th to 28th of every month from September to December.
Answer:
Students should expected to attempt the activity & answer on their own.

3. Think about it:

Question 1.
If the position of the shadow on the wall moves towards the north, in which direction does the location of sunrise or sunset appear to shift?
Answer:
If the position of the shadow on the wall moves towards the north, the location of sunrise or sunset will shift southwards.

Class 7 Geography Chapter 1 How Seasons Occur Part 1 Additional Important Questions and Answers

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

Question 1.
The earth’s _______ has enabled us to measure time in terms of days. (revolution, rotation, orbital motion)
Answer:
rotation

Question 2.
The earth rotates from _________ (north to south, east to west, west to east)
Answer:
west to east

Question 3.
The Earth takes _______ hours to rotate around itself. (365,144, 24)
Answer:
24

Question 4.
India lies in the ____ hemisphere. (northeastern, northwestern, southeastern)
Answer:
northeastern

Question 5.
The ________of sunrise and the sunset on the horizon³ for the whole year keep on changing. (position, shadow, rotation)
Answer:
position

Balbharti Maharashtra State Board Class 7 Geography Solutions Chapter 4 Air Pressure Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 7 Geography Solutions Chapter 4 Air Pressure

Class 7 Geography Chapter 4 Air Pressure Textbook Questions and Answers

1. Give reasons:

Question 1.
Air pressure decreases with increasing altitude.
Answer:

• The proportion of dust in the air, water vapour, heavy gases, etc. is higher in the air and closer to the surface of the earth.
• This proportion decreases with increasing altitude.
• As one moves higher and higher from the surface of the earth, the air becomes thinner and thinner.
• As a result, the air pressure decreases with increasing altitude.

Question 2.
Pressure belts oscillate.
Answer:

• The duration and the intensity of sunrays varies during particular periods of the year in both the hemispheres.
• So, the locations of the temperature zones and the pressure belts dependent on the sun’s heat also vary.
• This change is of the order of 5° to 7° towards the north in Uttarayan1 and 5° to 7° south in Dakshinayan2.
• In this way pressure belts oscillate.

2. Give short answers to the following questions.

Question 1.
What effect does temperature have on air pressure?
Answer:

• Temperature and air pressure are closely related. Wherever the temperature is high, the air pressure is low.
• As the temperature rises, the air gets heated, expands, and becomes lighter.
• This lighter air in the vicinity of the earth’s surface starts moving up towards the sky
• As a result, the air pressure in such areas decreases.

Question 2.
Why is the subpolar low pressure belt formed?
Answer:

• Due to earth’s curvature, the area between two parallels gets reduced as we move towards the poles.
• This results in lesser friction1 of the air with the earth’s surface.
• Air in this region is thrown out because of this reduced friction and also because of the earth’s rotational motion.
• This leads to the development of a low pressure belt in the sub polar region i.e. in area between 55° & 65° parallels in both the hemispheres.

3. Write notes on:

Question 1.
Mid-latitudinal high pressure belts.
Answer:

• The heated air from the equatorial region becomes lighter, starts ascending and after reaching higher altitudes, moves towards the polar region, i.e., towards the North and the South Pole.
• Due to low temperatures at the higher altitudes, the air cools down and becomes heavier.
• This heavier air descends down in both the hemispheres in the region between 25° to 35° parallels.
• This leads to the formation of high pressures, belts in these parallels of latitudes in both the hemispheres.
• This air is dry, hence the region does not get rainfall.
• Consequently, most of the hot deserts on the earth are found in these regions.

Question 2.
Horizontal distribution of air pressure.
Answer:

• The heat received from the sun is uneven in different regions.
• The distribution of temperature is uneven from the equator to the poles.
• This difference can lead to difference in air pressure.
• There are four air pressure belts formed on the earth surface.
• Equatorial low-pressure belt between 5°N and 5°S parallels as the temperature is high here.
• Mid latitudinal high-pressure belt between 25° and 35° parallels due to descending heavier air.
• Subpolar low-pressure belt formed between 55° and 65° parallels due to friction and rotation.
• Polar high-pressure belt formed between 80° and 90° parallels due to low temperatures.

4. Fill in the gaps with the appropriate option.

Question 1.
At higher altitudes air becomes ________. (thicker, thinner, hotter, more humid)
Answer:
thinner

Question 2.
Air pressure is expressed in _______ .(millibars, millimeters, milliliters, milligrams)
Answer:
millibars

Question 3.
On the earth, air pressure is _______.(uniform, uneven, high, low)
Answer:
uneven

Question 4.
The ______ pressure belt spreads between 5° North and 5° South parallel. (equatorial low, polar high, subpolar low, mid-latitudinal high)
Answer:
equatorial low

5. How does a high-pressure belt get formed near 30 ° parallel? Why does this region have hot deserts?
Answer:
(i) The heated air from the equatorial region becomes lighter, starts ascending and after reaching higher altitudes, moves towards the polar region, i.e., towards the North and the South Pole.

(ii) Due to low temperatures at the higher altitudes, the air cools down and becomes heavier. This heavier air descends down in both the hemispheres in the region between 25° to 35° parallels.

(iii) This leads to the formation of high pressures belts in these parallels of latitudes in both the hemispheres.

(iv) This air is dry, hence the region does not get rainfall. Consequently, most of the hot deserts on the earth are found in these regions.

6. Draw a neat diagram showing pressure belts. Label the diagram.
Answer:

Class 7 Geography Chapter 4 Air Pressure InText Questions and Answers

Formative Assessment
Can you tell?

Observe the diagram Fig. (a) and (b) carefully and answer the following questions:

Question 1.
Which pressure belt is mainly found in the Tropics?
Answer:
Equatorial low pressure belt is mainly found in the tropics.

Question 2.
With which pressure belt are the polar winds associated? In which temperature zone are they observed?
Answer:
The polar winds are associated with polar high pressure belt and sub polar low pressure belt. It is observed in the frigid zone.

Question 3.
What could be the reason behind a low pressure belt in the Tropics?
Answer:
Low pressure belt is formed in the tropics because the temperature is high.

Question 4.
With which pressure belts are the winds in the Temperate zone associated?
Answer:
The winds in temperate zone are associated with mid latitudinal high pressure belt.

Question 5.
Write the latitudinal extent of the low pressure belts.
Answer:
The latitudinal extent of Equatorial low pressure belt is between 50°N & 50°S parallel & the latitudinal extent of the sub polar low pressure belt is between 55° & 65° parallel in both the hemispheres.

Observe the map given above and study the distribution of air pressure and answer the following

Question 1.
The nature of the isobars.
Answer:
The isobars are joining places of equal air pressure on the map.

Question 2.
High and low pressure belts and their latitudinal position.
Answer:
(i) The latitudinal position of high pressure belt is between 25° & 35° parallels and between 80° & 90° parallels in both the hemisphere
(ii) The latitudinal position of low pressure belt is between 0° & 5° parallels and between 55° & 65° parallels in both the hemispheres.

Question 3.
The direction of the isobars and the distance between successive isobars over the oceans and continents.
Answer:
(i) In the northern hemisphere most of the isobars are in southwest to north east direction over the continents. Also the distance between the isobars varies.

(ii) Closely spaced isobars indicate large pressure changes over a small area. Widely spaced isobars indicate gentle or gradual pressure change.

(iii) In the southern hemisphere, the isobars extend  in east-west direction. The distance between the isobars is fairly constant over the oceans & so the isobars are fairly parallel to each other.

Question 4.
Comparison of the isobars in the northern and the southern hemispheres.
Answer:
In the northern hemisphere the isobars are far spaced and uneven, whereas in the southern hemispheres it is closely spaced and parallel.

Use your brainpower !

Question 1.
If there is low pressure at the equator, what will be the condition of air pressure in the Arctic Zone?
Answer:
The Arctic zone will experience high pressure as the temperature is lower than 0°C.

Try this:

Question 1.

• Take a flying lantern.
• Tie an approximately 5m long thread to the flying lantern so that you can bring the lantern down whenever required.
• After carefully reading the instructions given on the package of the lantern open it and light the candle placed in it.
• After some time, bring the lantern down with the help of the thread and put off the candle.

Question 2.
Did the flying lantern start ascending immediately after the candle was lit?
Answer:
Yes

Question 2.
What would have happened to the flying lantern had the candle got extinguished after the lantern had gone up in the air?
Answer:
The lantern would have fallen back on the earth.

Give it a try: 

Question 1.
Study the temperative distribution map given in your std VI textbook and the pressure distribution map in this lesson to find the correlation between air temperature and air pressure.
Answer:

• The temperature deceases continuously from the equator to the poles but the air pressure varies alternately.
• In the equatorial region the average temperature is high. Hence, the air pressure is low.
• In the polar regions, the temperature is low & hence the air pressure is comparative high.

Class 7 Geography Chapter 4 Air Pressure Additional Important Questions and Answers

Fill in the blanks choosing the correct option from the bracket:

Question 1.
Pressure belts oscillates between ______ parallels. (5° to 7°, 10° to 20°, 80° and 90°, 25° to 30°)
Answer:
5° to 7°

Question 2.
The instrument used to measure air pressure is ______ .(barometer, thermometer, hygrometer, seismometer)
Answer:
barometer

Question 3.
In both the polar regions, the temperature is below ________ throughout the year. (5°C, 0°C, 7°C, 6°C)
Answer:
0°C

Question 4.
The line that joins the places of equal pressure on the map is called an _______. (isotherm, isohytes, millibars, isobar)
Answer:
isobar

Match the following:

Question 1.

A (Pressure Belt)	B (Parallels)
(1) Sub Polar low pressure (2) Mid latitudinal high pressure (3) Polar high pressure (4) Equatorial low pressure	(a) 25° to 35° (b) 5°N and 5°S (c) 55° to 65° (d) 80° to 90°

Answer:
1 – c
2 – a
3 – d
4 – b

Fill in the blanks:

Question 1.
The latitudinal extent of the temperate zones is much _____ while belts are narrower.
Answer:
Larger

Question 2.
The extent of air pressure belt is upto ______ parallel.
Answer:
10°

Question 3.
Pressure belts are formed between the _______ and the pole.
Answer:
Equator

Question 4.
The sun rays fall perpendicular between the ______and ________.
Answer:
Tropic of Cancer, Tropic of Capricorn

Question 5.
Due to the earths curvature, the area betweentwo parallels gets ______ as we move towards the poles.
Answer:
reduced

Question 6.
The air pressure at sea level is ______ millibars.
Answer:
1013.2

State whether the following statements are true or false:

Question 1.
Air pressure is uniform on all places on the earth’s surface.
Answer:
false

Question 2.
Whenever the temperature is high, the air pressure is also high.
Answer:
false

Question 3.
The heat received from the sun, is uneven in different regions.
Answer:
true

Question 4.
In both the polar regions, the temperature is low throughout the year.
Answer:
true

Question 5.
Air pressure is measured in units of millimetres.
Answer:
false

Answer the following questions in one to two sentence:

Question 1.
Which factors influence air pressure?
Answer:
The altitude of a region, temperature of the air and the amount of water vapour in the air are some factors influencing air pressure.

Question 2.
What is the extent of air pressure belt?
Answer:
The extent of air pressure belt is generally upto 10° parallel.

Question 3.
What is the latitudinal extent of temperate zone?
Answer:
The latitudinal extent of temperate zone is from 23°30’N to 66°30’N and 23° 30’S to 66° 30’S.

Question 4.
What is the temperature in the polar region?
Answer:
In both the polar regions, the temperature is below 0°C throughout the year.

Question 5.
Why do all things in and on the earth stay earth bound?
Answer:
All things in and on the earth stay bound due to the earth’s gravity.

Question 6.
Why are temperature zones created on the surface of the earth?
Answer:
The heat received from the sun is uneven in different regions. Hence the distribution of the temperature is uneven from the equator to the poles. As a result, temperature zones are created.

Question 7.
Most of the hot deserts on the earth are found in which region?
Answer:
Most of the hot deserts on the earth are found in the mid latitudinal high pressure belt ie; between 25° to 30° parallels in both hemispheres.

Give reasons:

Question 1.
Air pressure is maximum at sea level.
Answer:

• All things in and on the earth stay earthbound because of the earth’s gravity. This includes air which is in the gaseous form.
• Due to the earth’s gravity, air is pulled to the earth’s surface.
• Also as one moves higher & higher from the earth’s surface the air becomes thinner & thinner.
• Therefore, the air pressure is maximum at sea- level.

Question 2.
A low pressure belt is formed near the equator.
Answer:

• The sunrays can be perpendicular between the Tropic of Cancer and the Tropic of Capricorn.
• The temperature is higher in this region.
• Hence air in this region gets heated, expands and becomes lighter and moves towards the sky.
• As this process operates continuously, a low pressure belt gets formed in the central part of this region between the parallels 5° N and 5° S, near the equator.

Question 3.
High pressure belt is formed near the polar region.
Answer:

• In both the polar regions, the temperature is below 0° throughout the year.
• The air is cold.
• Hence, high pressure belt is formed in the polar region.

Question 4.
Low pressure belt is observed between 55° and 65° parallels in both the hemispheres.
Answer:

• Due to the earth’s curvature, the area between two parallels gets reduced as we move towards the poles. This results in lesser friction of the air with the earth’s surface.
• The air between 55° and 65° parallels is thrown out because of the reduced friction and also due to the earth’s rotation.
• Therefore, a low pressure belt is observed between 55° and 65° parallels in both the hemispheres.

Question 5.
Temperature and air pressure are closely related.
Answer:

• Wherever the temperature is high, the air pressure is low. As the temperature rises the air gets heated, expands, and become lighter.
• Thin, lighter air in the vicinity3 of the earth’s surface starts moving up towards the sky. As a result the air pressure in such area decreases.
• Hence, temperature and air pressure are closely related.

Question 6.
Most of the hot deserts on the earth are found in mid-latitudinal high pressure belts.
Answer:

• The air in mid-latitudinal high pressure belt (between 25° to 35° parallels in both hemisphere) is found to be dry.
• The amount of water vapour is very low & hence this region gets extremely scarce or no rainfall.
• Consequently, most of the hot deserts on the earth are found in mid latitudinal high pressure belts.

Give short answers to the following questions:

Question 1.
What are the effects of air pressure?
Answer:
Air pressure has the following effects.

• Origin of winds.
• Generation of storms
• Convectional type of rain.

Question 2.
What is the difference between the temperature zones and pressure belts?
Answer:
(i) The difference between the temperature zones & pressure belts is that the latitudinal extent of temperature zones is much larger while pressure belts are narrower.

(ii) For example, the Temperate zone extends from 23°30′ to 66°30′ in both hemisphere. Compared to this the pressure belt has limited extent which is generally upto 10° parallel.

(iii) Also the temperature zones are continuous & spread from the equator to the poles from Torrid to Frigid.

(iv) Pressure belts are not continuous & areas of high & low pressure are found in different regions from the equator to the poles.

Question 3.
How does a high pressure belt get formed near 30° parallel? Why does this region have hot deserts?
Answer:
(i) The heated air from the equatorial region becomes lighter, starts ascending and after reaching higher altitudes, moves towards the polar region, i.e., towards the North and the South Pole.

(ii) Due to low temperatures at the higher altitudes, the air cools down and becomes heavier. This heavier air descends down in both the hemispheres in the region between 25° to 35° parallels.

(iii) This leads to the formation of high pressures belts in these parallels of latitudes in both the hemispheres.

(iv) This air is dry, hence the region does not get rainfall. Consequently, most of the hot deserts on the earth are found in these regions.

Think about it:

Question 1.
What will be the effect on air pressure if the temperature drops? Why?
Answer:
If the temperature drops, the air pressure will increase as the air becomes heavy.

Maharashtra State Board Class 7 Science Solutions

• Chapter 1 The Living World : Adaptations and Classification
• Chapter 2 Plants : Structure and Function
• Chapter 3 Properties of Natural Resources
• Chapter 4 Nutrition in Living Organisms
• Chapter 5 Food Safety
• Chapter 6 Measurement of Physical Quantities
• Chapter 7 Motion, Force and Work
• Chapter 8 Static Electricity
• Chapter 9 Heat
• Chapter 10 Disaster Management
• Chapter 11 Cell Structure and Micro-organisms
• Chapter 12 The Muscular System and Digestive System in Human Beings
• Chapter 13 Changes – Physical and Chemical
• Chapter 14 Elements, Compounds and Mixtures
• Chapter 15 Materials we Use
• Chapter 16 Natural Resources
• Chapter 17 Effects of Light
• Chapter 18 Sound : Production of Sound
• Chapter 19 Properties of a Magnetic Field
• Chapter 20 In the World of Stars

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 50 Answers Solutions Chapter 14 Algebraic Formulae – Expansion of Squares.

Algebraic Formulae – Expansion of Squares Class 7 Practice Set 50 Answers Solutions Chapter 14

Question 1.
Expand:
i. (5a + 6b)²
ii. \(\left(\frac{\mathrm{a}}{2}+\frac{\mathrm{b}}{3}\right)^{2}\)
iii. (2p – 3q)²
iv. \(\left(x-\frac{2}{x}\right)^{2}\)
v. (ax + by)²
vi. (7m – 4)²
vii. \(\left(x+\frac{1}{2}\right)^{2}\)
viii. \(\left(a-\frac{1}{a}\right)^{2}\)
Solution:
i. (5a + 6b)²
Here, A = 5a and B = 6b
(5a + 6b)² = (5a)² + 2 × 5a × 6b + (6b)²
…. [(A + B)² = A² + 2AB + B²]
∴ (5a + 6b)² = 25a² + 60ab + 36b²

ii. \(\left(\frac{\mathrm{a}}{2}+\frac{\mathrm{b}}{3}\right)^{2}\)
Here A = \(\frac { a }{ 2 }\) and B = \(\frac { b }{ 3 }\)

iii. (2p – 3q)²
Here, a = 2p and b = 3q
(2p – 3q)² = (2p)² – 2 × (2p) × (3q) + (3q)²
…. [(a – b)² = a² – 2ab + b²]
∴ (2p – 3q)² = 4p² – 12pq + 9q²

iv. \(\left(x-\frac{2}{x}\right)^{2}\)
Here a = x and b = \(\frac { 2 }{ x }\)

v. (ax + by)²
Here, A = ax and B = by
(ax + by)² = (ax)² + 2 × ax × by + (by)²
…. [(A + B)² = A² + 2AB + B²]
∴ (ax + by)² = a²x² + 2abxy + b²y²

vi. (7m – 4)²
Here, a = 7m and b = 4
(7m – 4)² = (7m)² – 2 × 7m × 4 + 4²
…. [(a – b)² = a² – 2ab + b²]
∴ (7m – 4)² = 49m² – 56m + 16

vii. \(\left(x+\frac{1}{2}\right)^{2}\)
Here a = x and b = \(\frac { 1 }{ 2 }\)

viii. \(\left(a-\frac{1}{a}\right)^{2}\)
Here A = a and B = \(\frac { 1 }{ a }\)

Question 2.
Which of the options given below is the square of the binomial
(A) \(64-\frac{1}{x^{2}}\)
(B) \(64+\frac{1}{x^{2}}\)
(C) \(64-\frac{16}{x}+\frac{1}{x^{2}}\)
(D) \(64+\frac{16}{x}+\frac{1}{x^{2}}\)
Solution:
(C) \(64-\frac{16}{x}+\frac{1}{x^{2}}\)

Hint:
= \(\left(8-\frac{1}{x}\right)^{2}=8^{2}-2 \times 8 \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}\) …[(a – b)² = a² – 2ab + b²]
= \(64-\frac{16}{x}+\frac{1}{x^{2}}\)

Question 3.
Of which of the binomials given below is the m²n² + 14mnpq + 49p²q² the expansion?
(A) (m + n) (p + q)
(B) (mn – pq)
(C) (7mn + pq)
(D) (mn + 7pq)
Solution:
(D) (mn + 7pq)

Hint:
Here, square root of the first term = mn
Square root of the last term = 7pq
∴ Required binomial = (mn + 7pq)²

Question 4.
Use an expansion formula to find the values of:
i. (997)²
ii. (102)²
iii. (97)²
iv. (1005)²
Solution:
i. (997)² = (1000 – 3)²
Here, a = 1000 and b = 3
(1000 – 3)² = (1000)² – 2 x 1000 x 3 + 3²
…. [(a – b)² = a² – 2ab + b²]
= 1000000 – 6000 + 9
= 994009
∴ (997)² = 994009

ii. (102)² = (100 + 2)²
Here, a = 100 and b = 2
(100 + 2)² = (100)² + 2 x 100 x 2 + 2²
…. [(a + b)² = a² + 2ab + b²]
= 10000 + 400 + 4
= 10404
∴ (102)² = 10404

iii. (97)² = (100 – 3)²
Here, a = 100 and b = 3
(100 – 3)² = (100)² – 2 x 100 x 3 + 3²
…. [(a – b)² = a² – 2ab + b²]
= 10000 – 600 + 9
= 9409
∴ (97)² = 9409

iv. (1005)² = (1000 + 5)²
Here, a = 1000 and b = 5 (1000 + 5)²
= (1000)² + 2 x 1000 x 5 + 5²
…. [(a + b)² = a² + 2ab + b²]
= 1000000+ 10000 + 25
= 1010025
∴ (1005)² = 1010025

Maharashtra Board Class 7 Maths Chapter 14 Algebraic Formulae – Expansion of Squares Practice Set 50 Intext Questions and Activities

Question 1.
Use the given values to verify the formulae for squares of binomials. (Textbook pg. no. 92)
i. a = -7, b = 8
ii. a = 11,b = 3
iii. a = 2.5,b = 1.2
Solution:
i. (a + b)² = (-7 + 8)²
= 1²
= 1
a² + 2ab + b² = (-7)² + 2 x (-7) x 8 + 8²
= 49 – 112 + 64
= 1
∴(a + b)² = a² + 2ab + b²
(a – b)² = (-7 – 8)²
= (-15)²
= 225
a² – 2ab + b² = (-7)² – 2 x (-7) x 8 + (8)²
= 49 + 112 + 64
= 225
∴(a – b)² = a² – 2ab + b²

ii. (a + b)² = (11 + 3)²
= 14²
= 196
a² + 2ab + b² = 11² + 2 x 11 x 3 + 3²
= 121 + 66 + 9
= 196
∴(a + b)² = a² + 2ab + b²
(a – b)² = (11 – 3)² = 8²
= 64
a² – 2ab + b² = 11² – 2 x 11 x 3 + 3²
= 121 – 66 + 9
= 64
∴(a – b)² = a² – 2ab + b²

iii. (a + b)² = (2.5 + 1.2)²
= 3.7²
= 13.69
a² + 2ab + b² = (2.5)² + 2 x 2.5 x 1.2 + (1.2)²
= 6.25 + 6 + 1.44
= 13.69
∴(a + b)² = a² + 2ab + b²
(a – b)² = (2.5 – 1.2)²
= 1.32
= 1.69
a² – 2ab + b² = (2.5)² – 2 x 2.5 x 1.2 + (1.2)²
= 6.25 – 6 + 1.44
= 1.69
∴(a – b)² = a² – 2ab + b²-

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 40 Answers Solutions Chapter 10 Bank and Simple Interest.

Bank and Simple Interest Class 7 Practice Set 40 Answers Solutions Chapter 10

Question 1.
If Rihanna deposits Rs 1500 in the school fund at 9 p.c.p.a for 2 years, what is the total amount she will get?
Solution:
Here, P = Rs 1500, R = 9 p.c.p.a , T = 2 years
∴ Total interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{1500 \times 9 \times 2}{100}\)
= 1500 x 9 x 2
= Rs 270
∴ Total amount = Principal + Interest
= 1500 + 270
= Rs 1770
∴ Rihanna will get a total amount of Rs 1770.

Question 2.
Jethalal took a housing loan of Rs 2,50,000 from a bank at 10 p.c.p.a. for 5 years. What is the yearly interest he must pay and the total amount he returns to the bank?
Solution:
Here, P = Rs 250000, R = 10 p.c.p.a., T = 5 years
∴ Total interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{250000 \times 10 \times 5}{100}\)
= 2500 x 10 x 5
= Rs 1,25,000
∴ Yearly interest = Total interest ÷ Time = 1,25,000 ÷ 5 = Rs 25000
Total amount to be returned = Principal + Total interest
= 250000 + 125000 = Rs 375000
∴ The yearly interest is Rs 25,000 and Jethalal will have to return Rs 3,75,000 to the bank.

Question 3.
Shrikant deposited Rs 85,000 for \(2\frac { 1 }{ 2 }\) years at 7 p.c.p.a. in a savings bank account. What is the total
interest he received at the end of the period?
Solution:
Here, P = Rs 85000, R = 7 p.c.p.a., T = \(2\frac { 1 }{ 2 }\) years = 2.5 years
∴ Total interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{85000 \times 7 \times 2.5}{100}\)
= \(\frac{85000 \times 7 \times 25}{100 \times 10}\)
= 85 x 7 x 25
= Rs 14875
∴ The total interest received by Shrikant at the end of the period is Rs 14875.

Question 4.
At a certain rate of interest, the interest after 4 years on Rs 5000 principal is Rs 1200. What would be the interest on Rs 15000 at the same rate of interest for the same period?
Solution:
The interest on Rs 5000 after 4 years is Rs 1200.
Let us suppose the interest on Rs 15000 at the same rate after 4 years is Rs x.
Taking the ratio of interest and principal, we get
∴ \(\frac{x}{15000}=\frac{1200}{5000}\)
∴ \(x=\frac{1200 \times 15000}{5000}\)
= Rs 3600
∴ The interest received on Rs 15000 is Rs 3600.

Question 5.
If Pankaj deposits Rs 1,50,000 in a bank at 10 p.c.p.a. for two years, what is the total amount he will get from the bank?
Solution:
Here, P = 150000, R = 10 p.c.p.a., T = 2 years
∴ Total interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{150000 \times 10 \times 2}{100}\)
= Rs 30000
∴ Total amount = Principal + Total Interest
= 150000 + 30000
= Rs 180000
∴ Pankaj will receive Rs 180000 from the bank.

Maharashtra Board Class 7 Maths Chapter 10 Banks and Simple Interest Practice Set 40 Intext Questions and Activities

Question 1.
Observe the entries made in the page of a passbook shown below and answer the following questions. (Textbook pg. no. 70)

1. On 2.2.16 the amount deposited was Rs__and the balance Rs__.
2. On 12.2.16, Rs__were withdrawn by cheque no. 243965. The balance was Rs__
3. On 26.2.2016 the bank paid an interest of Rs__

Solution:

1. 1500, 7000
2. 3000, 9000
3. 135

Practice Set 40 Class 7 Question 2.
Suvidya borrowed a sum of Rs 30000 at 8 p.c.p.a. interest for a year from her bank to buy a computer. At the end of the period, she had to pay back an amount of Rs 2400 over and above what she had borrowed.
Based on this information fill in the blanks below. (Textbook pg. no. 70)

1. Principal = Rs__
2. Rate of interest =__%
3. Interest = Rs__
4. Time =__year.
5. The total amount returned to the bank = 30,000 + 2,400 = Rs__

Solution:

1. 30000
2. 8
3. 2400
4. 1
5. Rs 32400

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 1 Answers Solutions.

Maharashtra Board Miscellaneous Problems Set 1 Class 7 Maths Solutions

Question 1.
Solve the following:
i. (-16) × (-5)
ii. (72) ÷ (-12)
iii. (-24) × (2)
iv. 125 ÷ 5
v. (-104) ÷ (-13)
vi. 25 × (-4)
Solution:
i. (-16) × (-5) = 80

ii. 72 ÷ (-12) = \(\frac { 72 }{ -12 }\)
= \(\frac{1}{(-1)} \times \frac{72}{12}\)
(-1) × 12
= -6

iii. (-24) × 2 = -48

iv. 125 ÷ 5 = \(\frac { 125 }{ 5 }\)
= 25

v. (-104) ÷ (-13) = \(\frac { -104 }{ -13 }\)
= \(\frac { 104 }{ 13 }\)
= 8

vi. 25 × (-4) = -100

Question 2.
Find the prime factors of the following numbers and find their LCM and HCF:
i. 75,135
ii. 114,76
iii. 153,187
iv. 32,24,48
Solution:
i. 75 = 3 × 25
= 3 × 5 × 5
135 = 3 × 45
= 3 × 3 × 15
= 3 × 3 × 3 × 5
∴ HCF of 75 and 135 = 3 × 5
= 15
LCM of 75 and 135 = 3 × 5 × 5 × 3 × 3
= 675

ii. 114 = 2 × 57
= 2 × 3 × 19
76 = 2 × 38
= 2 × 2 × 19
∴ HCF of 114 and 76 = 2 × 19
= 38
LCM of 114 and 76 = 2 × 19 × 3 × 2
= 228

iii. 153 = 3 × 51
= 3 × 3 × 17
187 = 11 × 17
∴ HCF of 153 and 187 = 17
LCM of 153 and 187 = 17 × 3 × 3 × 11
= 1683

iv. 32 = 2 × 16
= 2 × 2 × 8
= 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2
24 = 2 × 12
= 2 × 2 × 6
= 2 × 2 × 2 × 3
48 = 2 × 24
= 2 × 2 × 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
∴ HCF of 32, 24 and 48 = 2 × 2 × 2
= 8
LCM of 32,24 and 48 = 2 × 2 × 2 × 2 × 2 × 3
= 96

Question 3.
Simplify:
i. \(\frac { 322 }{ 391 }\)
ii. \(\frac { 247 }{ 209 }\)
iii. \(\frac { 117 }{ 156 }\)
Solution:
i. \(\frac { 322 }{ 391 }\)

ii. \(\frac { 247 }{ 209 }\)

iii. \(\frac { 117 }{ 156 }\)

Question 4.
i. 784
ii. 225
iii. 1296
iv. 2025
v. 256
Solution:
i. 784

∴ 784 = 2 × 2 × 2 × 2 × 7 × 7
∴ √784 = 2 × 2 × 7
= 28

ii. 225

∴ 225 = 3 × 3 × 5 × 5
∴ √225 = 3 × 5
= 15

iii. 1296

∴ 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
∴ √1296 = 2 × 2 × 3 × 3
= 36

iv. 2025

∴ 2025 = 3 × 3 × 3 × 3 × 5 × 5
∴ √2025 = 3 × 3 × 5
= 45

v. 256

∴ 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ √256 = 2 × 2 × 2 × 2
= 16

Question 5.
There are four polling booths for a certain election. The numbers of men and women who cast their vote at each booth is given in the table below. Draw a joint bar graph for this data.

Polling Booths	Navodaya Vidyalaya	Vidyaniketan School	City High School	Eklavya School
Women	500	520	680	800
Men	440	640	760	600

Solution:

Question 6.
Simplify the expressions:
i. 45 ÷ 5 + 120 × 4 – 12
ii. (38 – 8) × 2 ÷ 5 + 13
iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)
iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}
Solution:
i. 45 ÷ 5 + 120 × 4 – 12
= 9 + 80 – 12
= 89 – 12
= 77

ii. (38 – 8) × 2 ÷ 5 + 13
= 30 × 2 ÷ 5 + 13
= 60 ÷ 5 + 13
= 12 + 13
= 25

iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)
\(\frac{5}{3}+\frac{4}{7} \times \frac{21}{32}\)
\(\frac{5}{3}+\frac{3}{8}=\frac{40}{24}+\frac{9}{24}\)
\(\frac{49}{24}\)

iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}
= 3 × {4[90 – 5] + 2}
= 3 × {4 × 85 + 2}
= 3 × (340 + 2)
= 3 × 342
= 1026

Question 7.
Solve:
i. \(\frac{5}{12}+\frac{7}{16}\)
ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)
iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)
iv. \(4 \frac{3}{8} \div \frac{25}{18}\)
Solution:
i. \(\frac{5}{12}+\frac{7}{16}\)

ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)

iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)
= 4 × (-2)
= -8

iv. \(4 \frac{3}{8} \div \frac{25}{18}\)
= \(\frac{7}{4} \times \frac{9}{5}\)
= \(\frac { 63 }{ 20 }\)

Question 8.
Construct ∆ABC such that m∠A = 55°, m∠B = and l(AB) = 5.9 cm.
Solution:

Question 9.
Construct ∆XYZ such that, l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.
Solution:

Question 10.
Construct ∆PQR such that, m∠P = 80°, m∠Q = 70°, l(QR) = 5.7 cm.
Ans:
In ∆PQR,
m∠P + m∠Q + m∠R = 180° …. (Sum of the measures of the angles of a triangle is 180°)
∴ 80 + 70 + m∠R = 180
∴ 150 + m∠R = 180
∴ m∠R = 180 – 150
∴ m∠R = 30°

Question 11.
Construct ∆EFG from the given measures. l(FG) = 5 cm, m∠EFG = 90°, l(EG) = 7 cm.
Solution:

Question 12.
In ∆LMN, l(LM) = 6.2 cm, m∠LMN = 60°, l(MN) 4 cm. Construct ∆LMN.
Solution:

Question 13.
Find the measures of the complementary angles of the following angles:
i. 35°
ii. a°
iii. 22°
iv. (40 – x)°
Solution:
i. Let the measure of the complementary
angle be x°.
35 + x = 90
∴35 + x-35 = 90 – 35
….(Subtracting 35 from both sides)
∴x = 55
∴The complementary angle of 35° is 55°.

ii. Let the measure of the complementary angle be x°.
a + x = 90
∴a + x – a = 90 – a
….(Subtracting a from both sides)
∴x = (90 – a)
∴The complementary angle of a° is (90 – a)°.

iii. Let the measure of the complementary angle be x°.
22 + x = 90
∴22 + x – 22 = 90 – 22
….(Subtracting 22 from both sides)
∴x = 68
∴The complementary angle of 22° is 68°.

iv. Let the measure of the complementary angle be a°.
40 – x + a = 90
∴40 – x + a – 40 + x = 90 – 40 + x
….(Subtracting 40 and adding x on both sides)
∴a = (50 + x)
∴The complementary angle of (40 – x)° is (50 + x)°.

Question 14.
Find the measures of the supplements of the following angles:
i. 111°
ii. 47°
iii. 180°
iv. (90 – x)°
Solution:
i. Let the measure of the supplementary
angle be x°.
111 + x = 180
∴ 111 + x – 111 = 180 – 111
…..(Subtracting 111 from both sides)
∴ x = 69
∴ The supplementary angle of 111° is 69°.

ii. Let the measure of the supplementary angle be x°.
47 + x = 180
∴47 + x – 47 = 180 – 47
….(Subtracting 47 from both sides)
∴x = 133
∴The supplementary angle of 47° is 133°.

iii. Let the measure of the supplementary angle be x°.
180 + x = 180
∴180 + x – 180 = 180 – 180
….(Subtracting 180 from both sides)
∴x = 0
∴The supplementary angle of 180° is 0°.

iv. Let the measure of the supplementary angle be a°.
90 – x + a = 180
∴90 – x + a – 90 + x = 180 – 90+ x
….(Subtracting 90 and adding x on both sides)
∴a = 180 – 90 + x
∴a = (90 + x)
∴The supplementary angle of (90 – x)° is (90 + x)°.

Question 15.
Construct the following figures:
i. A pair of adjacent angles
ii. Two supplementary angles which are not adjacent angles.
iii. A pair of adjacent complementary angles.
Solution:
i.

ii.

iii.

Question 16.
In ∆PQR the measures of ∠P and ∠Q are equal and m∠PRQ = 70°, Find the measures of the following angles.

1. m∠PRT
2. m∠P
3. m∠Q

Solution:
Here, ∠PRQ and ∠PRT are angles in a linear pair.
m∠PRQ + m∠PRT = 180°
∴70 + m∠PRT = 180
∴m∠PRT = 180 – 70
∴m∠PRT = 110°
Now, ∠PRT is the exterior angle of ∆PQR.
∴m∠P + m∠Q = m∠PRT
∴m∠P + m∠P = m∠PRT ….(The measures of ∠P and ∠Q is same)
∴2m∠P = 110
∴m∠P = \(\frac { 110 }{ 2 }\)
∴m∠P = 55°
∴m∠Q =

Question 17.
Simplify
i. 5⁴ × 5³
ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)
iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)
iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)
Solution:
Simplify
i. 5⁴ × 5³
= 5⁴⁺³
= 5⁷

ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)

iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)

iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)

Question 18.
Find the value:
i. 17¹⁶  ÷ 17¹⁶
ii. 10^-3
iii. (2³)²
iv. 4⁶ × 4^-4
Solution:
i. 17¹⁶  ÷ 17¹⁶
= 17⁰
= 1

ii. 10^-3
= \(\frac{1}{10^{3}}\)
= \(\frac{1}{1000}\)

iii. (2³)²
= 2^3×2
= 2⁶
= 2 × 2 × 2 × 2 × 2 × 2
= 64

iv. 4⁶ × 4^-4
= 4^6+(-4)
= 4²
= 4 × 4
= 16

Question 19.
Solve:
i. (6a – 5b – 8c) + (15b + 2a – 5c)
ii. (3x + 2y) (7x – 8y)
iii. (7m – 5n) – (-4n – 11m)
iv. (11m – 12n + 3p) – (9m + 7n – 8p)
Solution:
i. (6a – 5b – 8c) + (15b + 2a – 5c)
= (6a + 2a) + (-5b + 15b) + (-8c – 5c)
= 8a + 10b – 13c

ii. (3x + 2y) (7x – 8y)
= 3x × (7x – 8y) + 2yx (7x – 8y)
= 21x² – 24xy + 14xy – 16y²
= 21x² – 10xy – 16y²

iii. (7m – 5n) – (-4n – 11m)
= 7m – 5n + 4n + 11m
= (7m + 11m) + (-5n + 4n)
= 18m – n

iv. (11m – 12n + 3p) – (9m + 7n – 8p)
= 11m – 12n + 3p – 9m – 7n + 8p
= (11m – 9m) + (-12n – 7n) + (3p + 8p)
= 2m – 19n + 11p

Question 20.
Solve the following equations:
i 4(x + 12) = 8
ii. 3y + 4 = 5y – 6
Solution:
i. 4(x + 12) = 8
∴4x + 48 = 8
∴4x + 48 – 48 = 8 – 48
….(Subtracting 48 from both sides)
∴ 4x = -40
∴ x = \(\frac { -40 }{ 4 }\)
∴ x = -10

ii. 3y + 4 = 5y – 6
∴ 3y + 4 + 6 = 5y – 6 + 6
….(Adding 6 on both sides)
∴ 3y + 10 = 5y
∴ 3y + 10 – 3y = 5y – 3y
….(Subtracting 3y from both sides)
∴ 10 = 2y
∴ 2y = 10
∴ y = \(\frac { 10 }{ 2 }\)
∴ y = 5

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 2 Answers Solutions.

Maharashtra Board Miscellaneous Problems Set 2 Class 7 Maths Solutions

Question 1.
Angela deposited Rs 15000 in a bank at a rate of 9 p.c.p.a. She got simple interest amounting to Rs 5400. For how many years had she deposited the amount?
Solution:
Here, P = Rs 15000, R = 9 p.c.p.a., I = Rs 5400

∴ T = 4
∴ Angela had deposited the amount for 4 years.

Question 2.
Ten men take 4 days to complete the task of tarring a road. How many days would 8 men take?
Solution:
Let us suppose that 8 men require x days to tar the road.
Number of days required by 10 men to tar the road = 4
The number of men and the number of days required to tar the road are in inverse proportion.
∴ 8 × x = 10 x 4
∴ \(x=\frac{10 \times 4}{8}\)
∴ x = 5
∴ 8 men will require 5 days to tar the road.

Question 3.
Nasruddin and Mahesh invested Rs 40,000 and Rs 60,000 respectively to start a business. They made a profit of 30%. How much profit did each of them make?
Solution:
Total amount invested = Rs 40,000 + Rs 60,000
= Rs 1,00,000
Profit earned = 30%
∴ Total profit = 30% of 1,00,000
= \(\frac { 30 }{ 100 }\) × 100000
= Rs 30000
Proportion of investment = 40000 : 60000
= 2:3 …. (Dividing by 20000)
Let Nasruddin’s profit be Rs 2x and Mahesh’s profit be Rs 3x.
∴ 2x + 3x = 30000
∴ 5x = 30000
∴ x = \(\frac { 30000 }{ 5 }\).
∴ x = 6000
∴ Nasruddin’s profit = 2x = 2 × 6000 = Rs 12000
Mahesh’s profit = 3x = 3 × 6000 = Rs 18000
∴ The profits of Nasruddin and Mahesh are Rs 12000 and Rs 18000 respectively.

Question 4.
The diameter of a circle is 5.6 cm. Find its circumference.
Solution:
Diameter of the circle (d) = 5.6 cm
Circumference = πd
= \(\frac{22}{7} \times 5.6\)
= \(\frac{22}{7} \times \frac{56}{10}\)
= 17.6 cm
∴ The circumference of the circle is 17.6 cm.

Question 5.
Expand:
i. (2a – 3b)²
ii. (10 + y)²
iii. \(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}\)
iv. \(\left(y-\frac{3}{y}\right)^{2}\)
Solution:
i. Here, A = 2a and B = 3b
∴ (2a – 3b)² = (2a)² – 2 × 2a × 3b + (3b)²
…. [(A – B)² = A² – 2AB + B²]
= 4a² – 12ab + 9b²

ii. Here, a = 10 and b = y
(10 + y)² = 102 + 2 × 10xy + y²
…. [(a + b)² = a² + 2ab + b²]
= 100 + 20y + y²

iii. Here, a = \(\frac { p }{ 3 }\) and b = \(\frac { q }{ 4 }\)
\(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}=\left(\frac{p}{3}\right)^{2}+2 \times \frac{p}{3} \times \frac{q}{4}+\left(\frac{q}{4}\right)^{2}\)
…. [(a + b)² = a² + 2ab + b²]
\(\frac{p^{2}}{9}+\frac{p q}{6}+\frac{q^{2}}{16}\)

iv. Here, a = y and b = \(\frac { 3 }{ y }\)
\(\left(y-\frac{3}{y}\right)^{2}=y^{2}-2 \times y \times \frac{3}{y}+\left(\frac{3}{y}\right)^{2}\)
…. [(a – b)² = a² – 2ab + b²
= \(y^{2}-6+\frac{9}{y^{2}}\)

Question 6.
Use a formula to multiply:
i. (x – 5)(x + 5)
ii. (2a – 13)(2a + 13)
iii. (4z – 5y)(4z + 5y)
iv. (2t – 5)(2t + 5)
Solution:
i. Here, a = x and b = 5
(x – 5)(x + 5) = (x)² – (5)²
…. [(a + b)(a – b) = a² – b²]
= x² – 25

ii. Here, A = 2a and B = 13
(2a – 13)(2a + 13) = (2a)² – (13)²
…. [(A + B)(A – B) = A² – B²]
= 4a² – 169

iii. Here, a = 4z and b = 5y
(4z – 5y)(4z + 5y) = (4z)² – (5y)²
…. [(a + b)(a – b) = a² – b²]
= 16z² – 25y²

iv. Here, a = 2t and b = 5
(2t – 5)(2t + 5) = (2t)² – (5)²
…. [(a + b)(a – b) = a² – b²]
= 4t² – 25

Question 7.
The diameter of the wheel of a cart is 1.05 m. How much distance will the cart cover in 1000 rotations of the wheel?
Solution:
Diameter of the wheel (d) = 1.05 m
∴ Distance covered in 1 rotation of wheel = Circumference of the wheel
= πd
= \(\frac{22}{7} \times 1.05\)
= 3.3 m
∴ Distance covered in 1000 rotations = 1000 x 3.3 m
= 3300 m
= \(\frac { 3300 }{ 1000 }\) km …[1m = \(\frac { 1 }{ 1000 }\)km]
= 3.3 km
∴ The distance covered by the cart in 1000 rotations of the wheel is 3.3 km.

Question 8.
The area of a rectangular garden of length 40 m, is 1000 sq m. Find the breadth of the garden and its perimeter. The garden is to be enclosed by 3 rounds of fencing, leaving an entrance of 4 m. Find the cost of fencing the garden at a rate of Rs 250 per metre.
Solution:
Length of the rectangular garden = 40 m
Area of the rectangular garden = 1000 sq. m.
∴ length × breadth = 1000
∴ 40 × breadth = 1000
∴ breadth = \(\frac { 1000 }{ 40 }\)
= 25 m
Now, perimeter of the rectangular garden = 2 × (length + breadth)
= 2 (40 + 25)
= 2 × 65
= 130 m
Length of one round of fence = circumference of garden – width of the entrance
= 130 – 4
= 126 m
∴ Total length of fencing = length of one round of wire × number of rounds = 126 × 3
= 378 m
∴ Total cost of fencing = Total length of fencing × cost per metre of fencing
= 378 × 250
= 94500
∴ The cost of fencing the garden is Rs 94500.

Question 9.
From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.
Solution:

In ∆ABC, ∠B = 90°, and l(BC) = 21, and l(AB) = 20
∴ According to Pythagoras’ theorem,
∴ l(AC)² = l(BC)² + l(AB)²
∴ l(AC)² = 21² + 20²
∴ l(AC)² = 441 + 400
∴ l(AC)² = 841
∴ l(AC)² = 29²
∴ l(AC) = 29
Perimeter of ∆ABC = l(AB) + l(BC) + l(AC)
= 20 + 21 + 29
= 70
∴ The length of hypotenuse AC is 29 units, and the perimeter of ∆ABC is 70 units.

Question 10.
If the edge of a cube is 8 cm long, find its total surface area.
Solution: ,
Total surface area of the cube = 6 × (side)²
= 6 × (8)²
= 6 × 64
= 384 sq. cm
The total surface area of the cube is 384 sq.cm.

Question 11.
Factorize: 365y⁴z³ – 146y²z⁴
Solution:
= 365y⁴z³ – 146y²z⁴
= 73 (5y⁴z³ – 2y²z⁴)
= 73y² (5y²z³ – 2z⁴)
= 73y²z³(5y² – 2z)

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 12 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Practice Set 12 Answers Solutions Chapter 3

Question 1.
i. 25, 40
ii. 56, 32
iii. 40, 60, 75
iv. 16, 27
v. 18, 32,48
vi. 105, 154
vii. 42, 45, 48
viii. 57, 75, 102
ix. 56, 57
x. 777, 315, 588
Solution:
i. 25, 40

∴ 25 = 5 × 5
40 = 2 × 2 × 2 × 5
∴ HCF of 25 and 40 = 5

ii. 56, 32

∴ 56 = 2 × 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
∴ HCF of 56 and 32 = 2 × 2 × 2
∴ HCF of 56 and 32 = 8

iii. 40, 60, 75

∴ 40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
∴ HCF of 40, 60 and 75 = 5

iv. 16, 27

∴ 16 = 2 × 2 × 2 × 2 × 1
27 = 3 × 3 × 3 × 1
∴ HCF of 16 and 27 = 1

v. 18, 32,48

∴ 18 = 2 × 3 × 3
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 18, 32 and 48 = 2

vi. 105, 154

∴ 105 = 3 × 5 × 7
154 = 2 × 2 × 11
∴ HCF of 105 and 154 = 7

vii. 42, 45, 48

∴ 42 = 2 × 3 × 7
45 = 3 × 3 × 5
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 42,45 and 48 =3

viii. 57, 75, 102

∴ 57 = 3 × 19
75 = 3 × 5 × 5
102 = 2 × 3 × 17
∴ HCF of 57, 75 and 102 = 3

ix. 56, 57

∴ 56 = 2 × 2 × 2 × 7 × 1
57 = 3 × 19 × 1
∴ HCF of 56 and 57 = 1

x. 777, 315, 588

∴ 777 = 3 × 7 × 37
315 = 3 × 3 × 5 × 7
588 = 2 × 2 × 3 × 7 × 7
∴ HCF of 777, 315 and 588 = 3 × 7
HCF of 777, 315 and 588 = 21

Question 2.
Find the HCF by the division method and reduce to the simplest form:
i. \(\frac { 275 }{ 525 }\)
ii. \(\frac { 76 }{ 133 }\)
iii. \(\frac { 161 }{ 69 }\)
Solution:
i. \(\frac { 275 }{ 525 }\)

ii. \(\frac { 76 }{ 133 }\)

iii. \(\frac { 161 }{ 69 }\)

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 12 Intext Questions and Activities

Question 1.
In each of the following examples, write all the factors of the numbers and find the greatest common divisor. (Textbook pg. no. 17)
i. 28, 42
ii. 51, 27
iii. 25, 15, 35
Solution:
i. Factors of 28 = 1,2,4, 7, 14, 28
Factors of 42 = 1,2, 3, 6, 7, 14, 21, 42
∴ HCF of 28 and 42 = 14

ii. Factors of 51 = 1, 3, 17, 51
Factors of 27 = 1, 3, 9, 27
∴ HCF of 51 and 27 = 3

iii. Factors of 25 = 1, 5, 25
Factors of 15 = 1, 3, 5, 15
Factors of 35 = 1, 5, 7, 35
∴ HCF of 25, 15 and 35 = 5

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 1 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Practice Set 1 Answers Solutions Chapter 1

Question 1.
Draw line segments of the lengths given below and draw their perpendicular bisectors:
i. 5.3 cm
ii. 6.7 cm
iii. 3.8 cm
Solution:
i.

Line AB is the perpendicular bisector of seg PQ.

ii.

Line UV is the perpendicular bisector of seg ST.

iii.

Line ST is the perpendicular bisector of seg LM.

Question 2.
Draw angles of the measures given below and draw their bisectors:
i. 105°
ii. 55°
iii. 90°
Solution:
i. 105°

ii. 55°

iii. 90°

Question 3.
Draw, an obtuse-angled triangle and a right-angled triangle. Find the points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie?
Solution:

The points of concurrence of the angle bisectors of both the triangles lie in the interior of the triangles.

Question 4.
Draw a right-angled triangle. Draw the perpendicular bisectors of its sides. Where does the point of concurrence lie?
Solution:

The point of concurrence of the perpendicular bisectors of the sides of the right angled triangle lies on the hypotenuse.

Question 5.
Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.
Solution:
Since, Maithili, Shaila and Ajay live in three different places, lines joining their houses will form a triangle.
The position of the toy shop which is equidistant from three houses can be found out by drawing the perpendicular bisector of the sides of the triangle joining the three houses.
The shop will be at the point of concurrence of the perpendicular bisectors.

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 1 Intext Questions and Activities

Question 1.
Draw a line segment PS of length 4cm and draw its perpendicular bisector. (Textbook pg. no. 1)

1. How will your verify that CD is the perpendicular bisector? m∠CMS = __°
2. Is l(PM) = l(SM)?

Solution:

1. Here, m∠CMS = 90°
2. Also, l(PM) = l(SM) = 2cm
   ∴ line CD is the perpendicular bisector of seg PS.