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Class 10 English Chapter 4.4 Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 English Solutions Unit 4.4 The Height of the Ridiculous Notes, Textbook Exercise Important Questions and Answers.

The Height of the Ridiculous Poem 10th Std Question Answer

Maharashtra Board Class 10 English Solutions Unit 4.4 Warming Up Questions and Answers

The Height Of The Ridiculous Appreciation Question 1.
The teacher writes incomplete sentences on the board. He/She asks the students to complete them in their notebooks.
(a) Today, I am happy because ……………………………… .
(b) Today after the class, I wish ……………………………… .
(c) Tomorrow, I feel that ……………………………… .
(d) I want to laugh because ……………………………… .
(e) Today, the class seems to be cheerful about ……………………………… .
Answer:
(a) my grandparents are coming for a holiday.
(b) to eat an ice cream.
(c) I will go for a movie.
(d) I am very happy.
(e) the forthcoming football match.

The Height Of The Ridiculous Question 2.
The teacher writes an incomplete sentence and asks the students to complete it in a funny way.
Answer:
(1) Mother gave me cheese but the cat ate it.
(2) I went to the market and bought an elephant.

Appreciation Of Poem The Height Of Ridiculous Question 3.
Give the words related to:

Syllable
A syllable is a unit of spoken language made up of a single uninterrupted sound formed by a vowel and consonants. For example, single syllable : ant, two syllables – water, three syllables : Inferno.
Answer:

The Height Of The Ridiculous Theme Question 4.
Pick out the word from the given box and write it in the correct columns below.

Here the focus is not on the spellings but the pronunciation of the words.

Words with one syllable	Words with two syllables

Answer:

The Height Of The Ridiculous Notes Question 5.
Count the syllables and circle the appropriate number in the box.

Answer:

The Height Of The Ridiculous Question 6.
Write the names of any five of your friends and mention the number of syllables in each name.

Name	Number of syllables

Answer:

The Height of the Ridiculous Class 10 English Workshop Questions and Answers Maharashtra Board

The Height Of The Ridiculous Question 1.
Find out expressions from the poem that indicate funny moments.
For example, I laughed as I would die.
…………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………
Answer:
(1) was all upon the grin
(2) the grin grew broad
(3) and shot from ear to ear
(4) He read the third; a chuckling noise
(5) The fourth; he broke into a roar
(6) The fifth; his waistband split;
(7) The sixth; he burst five buttons off;
(8) And tumbled in a fit.

Appreciation Of The Poem The Height Of Ridiculous Question 2.
Order of sequence : Arrange the following reactions in their proper order, as per the poem.
(a) His waistband split
(b) The grin grew broad.
(c) Sleepless eye.
(d) Was all upon the grin.
(e) He broke into a roar.
(f) He burst five buttons off.
Answer:
(d) Was all upon the grin
(b) The grin grew bro^d
(e) He broke into a roar
(a) His waistband split
(f) He burst five buttons
(c) Sleepless eye

Height Of Ridiculous Appreciation Question 3.
Form pairs and find out the various rhyming words in the poem and two of your own. Complete the following table.

Answer:

The Height Of Ridiculous Appreciation Question 4.
Match the lines with the Figures of Speech.

Answer:

Appreciation Of The Poem The Height Of The Ridiculous Question 5.
Copy any two stanzas of the poem in the lines below. Using a coloured pen underline the stressed syllables in each line and put a stress-mark ( ) over each.
Answer:
I wrote some lines once on a time
In wondrous merry mood,
And thought, as usual, men would say
They were exceeding good.

The Height Of The Ridiculous Question 6.
Complete the lines of the poem by choosing proper pairs of rhyming words and make it meaningful.
– We returned home late, one ………………………. ,
In the window, there glowed a ………………………. .
Burglars !! was our very first ………………………. ;
For defence, sticks ‘n stones we ………………………. .
”Let’s grab the loot and ………………………. ,”
was uttered soft, by ………………………..
The door we softly ……………………….,
And then we were truly ………………………..
Oops! Before, outside, we’d ……………………….,
The television had been left ………………………..
(run, shocked, gone, night, sought, on, someone, thought, light, unlocked)
Answer:
We returned home late, one night,
In the window there glowed a light.
Burglars! Was our very first thought,
For defence, sticks ‘n stones we sought.
“Let’s grab the loot and run.”
Was uttered soft, by someone.
The door we softly unlocked.
And then we were truly shocked.
Oops! Before outside we’d gone,
The television had been left on!

The Height Of Ridiculous Poem Appreciation Question 7.
Form goups in your class and together compose a short humorous poem. Use jokes, experiences, etc. and convert it to a poetic form. Write and decorate it on chart-paper and put it up in your class, in turns.

Appreciation Of The Height Of The Ridiculous Question 8.
Go through the poem and write an appreciation of the poem in a paragraph format.
Answer:
Point Format
(for understanding)
The title of the poem: ‘The Height of the Ridiculous’
The poet: Oliver Wendell Holmes
Rhyme scheme: abcb.
Figures of speech: Transferred Epithet, Hyperbole, Onomatopoeia, Tautology, Alliteration, etc.
Theme/Central idea: A funny poem to simply entertain the audience; written for Enjoyment.

Paragraph Format
The poem ‘The Height of the Ridiculous’ is written by Oliver Wendell Holmes.

The rhyme scheme of the poem is abcb. There are many figures of speech, like Hyperbole, Tautology, Onomatopoeia, Alliteration, etc. but the one that stands out is Transferred Epithet. In the line ‘Ten days and nights, with sleepless eye’, the adjective ‘sleepless’ should be for the man and not for the eye.

The poem is a humorous one written for enjoyment, with plenty of funny expressions. The main purpose of the poet is to simply entertain the reader.

Appreciation Of Poem The Height Of The Ridiculous Question 9.
Project :
Reading a poem.
Arrange the poetry reading competition. Select the poem of your choice.

• Read the poem silently.
• Repeat the reading of the poem.
• Focus on the pauses, stresses, intonation etc.
• Pay attention to the proper pronunciations.

Poem Appreciation Of The Height Of Ridiculous Question 10.
Choose the correct alternatives: (The answers are given directly and underlined.)
(1) The poet was in a very …………….. mood when he wrote the lines.
(a) tired
(b) happy
(c) bored
(d) wondering
Answer:
(b) happy

(2) The poet was generally a ……………… man.
(a) humorous
(b) wonderful
(c) serious
(d) good
Answer:
(c) serious

Question 11.
Explain:
(a) the contrast between the poet and his servant.
Answer:
The poet was a thin and slender man while his servant was strong and muscular.

(b) the poet’s reaction when he read the lines.
Answer:
The poet laughed heartily when he read the lines. He laughed so hard he thought he would die.

Question 12.
Find out the expression from the extract that indicates funny moments:
Answer:
‘I laughed as I would die’.

Question 13.
Match the lines with the figures of speech:
Lines – Figures of Speech
(a) A sober man am I – (c) Tautology
(b) To mind a slender man like me – (d) Inversion
Answer:
(a) A sober man am I – Inversion
(b) To mind a slender man like me – Alliteration

Question 14.
Complete the following:
(1) There was a grin on the servant’s face when …………………………
(2) The chuckling noise was heard when ……………………..
(3) When he read the fifth line ………………….
(4) The grin grew from ear to ear when the servant ………………….
Answer:
(1) he read the first line.
(2) the servant read the third line.
(3) his waistband split.
(4) read the second line.

Question 15.
Describe the outcome of this experience on the poet.
Answer:
After this experience, the poet has never dared to write any more funny poems.

Question 16.
Which line suggests that the servant was totally out of control?
Answer:
The line ‘And tumbled into a fit’ suggests that the servant was totally out of control.

English Kumarbharati 10th Digest PDF

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Balbharti Maharashtra State Board Class 9 Geography Solutions Chapter 5 Precipitation Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Geography Solutions Chapter 5 Precipitation

Class 9 Geography Chapter 5 Precipitation Textbook Questions and Answers

1. Identify the precipitation type with the help of the description given:

(a) It is the main source of the water that you use. Sometimes it is torrential and sometimes continuous. Most of the agriculture in India is dependent on it.
(b) It seems as if water droplets are floating in the atmosphere. In London, one cannot see the Sun till the afternoon during winters because of this phenomenon.
(c) It never precipitates like this in equatorial areas. Precipitation in the solid form sometimes causes damage to the crops.
(d) A white cotton-like layer spreads on the earth’s surface. Because of this form of precipitation, the State of Jammu and Kashmir has to change its capital in winters. In Maharashtra, it does not precipitate like this.
Answer:
(a) rainfall
(b) fog
(c) hail
(d) snow

2. Look at the following pictures and identify the correct rainfall type.

Answer:
Convectional rainfall

Answer:
Orographic rainfall

Answer:
Cyclonic rainfall

3. Look at the figures above and answer the following questions:

Question 1.
In fig B, on which side of the mountain is it raining more?
Answer:
The windward side is receiving more rainfall.

Question 2.
Shade the rain shadow region in fig B and name it.
Answer:
Students to show the leeward side in the picture.

Question 3.
What is the difference between A and C?
Answer:
In figure 5.4 i.e. convectional rainfall the hot air rises upwards and then the air cools and begins to condense and due to continuous condensation rainfall occurs. Here rainfall is accompanied by lightning and thunder.

In figure 5.6 , i.e. cyclonic rainfall, air from surrounding regions comes towards the centre of the cyclone and starts moving upwards. As it rises, the temperature of the air reduces, condensation occurs and rainfall takes place.

Question 4.
Stormy winds and floods are associated with which rainfall type?
Answer:
Stormy winds and floods are associated with Cyclonic rainfall.

Question 5.
What type of rainfall occurs in Singapore?
Answer:
Cyclonic rainfall occurs in Singapore.

4. Identify the odd man out:

Question 1.
Orographic rainfall, acid rain, cyclonic rainfall, convectional rainfall
Answer:
Acid rain

Question 2.
Snowfall, rainfall, hailstones, dew
Answer:
Dew

Question 3.
Thermometer, rain gauge, anemometer, measuring jar
Answer:
Measuring jar

5. Answer in brief:

Question 1.
In what ways does precipitation occur on the earth?
Answer:
Precipitation means water falls in the solid or liquid state from the clouds to the earth surface. Snow, hailstorms, rainfall are the major forms of precipitation.

(i) Snow:
Answer:

• When the temperature in the atmosphere falls below the freezing point the water vapour directly turns into snowflakes. This is called sublimation.
• Hence the vapour in the form of gas transform into solid snow. Precipitation in the form of solid particles is known as snowfall.
• As snow is in the solid form. It does not run like water and layers of the snow get deposited on the top of the others and when the snow melts the region gets fresh water.

(ii) Hail:

• When there is lot of heat on the earth’s surface, the upward air flow blows at a greater speed. Because of this upward flow, the temperature of air reduces and the condensation of the water vapour takes place, and dark clouds are formed.
• Because of the upward movement of the air, these water droplets go at higher altitude and solidify forming hailstones.
• As the hailstones are heavy, they fall toward the earth’s surface because of gravity. The crops may get destroyed and loss of life and property may occur.
• Hailstones occurs in summer in India, Africa and in some parts of south east Asia.

(iii) Rainfall:

• We get water generally in the form of rainfall. The temperature of the air with water vapour reduces when it goes higher and condensation of the vapour occurs.
• Clouds formed with the condensed water droplets and dust particles accumulate.
• As these water droplets increase in the size, they cannot float in the air anymore because of their weight. They come down as rainfall
• The different types of rainfall are: Convectional rainfall, Orographic rainfall and Cyclonic rainfall.

(iv) Fog, dew and frost:

• When the condensation or solidification of the water vapour in the atmosphere occurs near the earth’s surface, it leads to the formation of fog, dew and frost.

Question 2.
Comment on the rainfall occurring in the rain shadow area.
Answer:

• The winds coming from lakes or seas are moisture-laden and they are obstructed by the high mountain ranges coming in their way
• They start going upwards along the slope of the moutains. The temperature of these winds drop and condensation occurs and rainfall takes place.
• This rainfall takes place because of the obstruction of the mountains which results in the condensation of water vapour.
• The windward side of the mountain gets more rain; the amount of vapour in the air reduces after crossing the mountain and the moisture-holding capacity of the air increases.
• The leeward side of the mountain gets lesser rainfall as compared to windward side.
• Thus, the leeward side area is identified as rain shadow area as it recieves meagre rainfall.

Question 3.
Which type of rainfall occurs in most of the world? Why?
Answer:

• Orographic rainfall occurs in most parts of the world.
• Convectional rainfall is regional in nature.
• There is a certainty in the convectional rainfall occurring in the equatorial areas.
• Comparatively, the orographic and cyclonic rainfall is less certain.
• And therefore, such areas are prone to very heavy rainfall, floods or droughts frequently.

Question 4.
If condensation occurs closer to the earth’s surface, what types of forms become visible?
Answer:
If condensation and solidification of the water vapour in the atmosphere closer to the earth surface are visible, they are in the form of fog, dew or frost.

(i) Fog:

• The temperature of the layers of the air near the surface of the earth reduces. As the temperature reduces, water vapour condenses.
• In this process the water vapour turns into microscopic water particles and float in the air.
• When the density of these droplets in the air increases it leads to the formation of fog

(ii) Dew:

• When moisture-laden air near the earth surface comes in contact with very cold objects condensation of water vapour takes place.
• They turn into very small water droplets and stick to the surface of cold objects, e.g. eg: leaves and this is called dew.

(iii) Frost:

• When the temperature of the air reaches less then 0 degree Celcius the water droplet stuck to the surface of the cold objects and freezes.
• This frozen water droplet is called as frost.

Question 5.
What precautions should be taken while measuring rainfall?
Answer:

• Rainfall is an important source of water on planet earth and rainfall is formed because of changes in the temperature of the air with water vapour.
• The instrument that is used to measure rainfall is called rain gauge.
• The funnel i.e. used for measuring rain has a specific diameter and the rain falling in this funnel is collected in bottle fitted in the gauge.
• The collected water is then measured with the help of measuring jar. In the areas of heavy rainfall, the reading of the rain with rain gauge should be taken every three hours. The measuring jar reads rain in millimetres
• The gauge has to be kept on open ground on 30cm high flat-mount.
• So that the rain water is collected without any obstruction.

6. Distinguish between

Question 1.
Dew and frost
Answer:

Question 2.
Snow and hail
Answer:

Class 9 Geography Chapter 5 Precipitation Intext Questions and Answers

Can you tell?

Question 1.
The blade of grass look like this in winter mornings. From where does the water on the blades of grass come?

Answer:

1. The blade of grass looks like this in winter mornings because of dew. These are small water droplets.
2. The dew is formed in winter because moisture-laden air near the earth surface comes in contact with cold objects due to which condensation of vapour takes place, turning into small water droplets.

Question 2.
Snow is found everywhere in the winters in Kashmir.
Answer:
Snow is found everywhere in winters of Kashmir because Kashmir is located at a higher altitude where the temperature falls below freezing point. Hence water vapour directly turn into snowflakes leading to snowfall.

Question 3.
Why isn’t snow found in our surroundings?
Answer:
Because we have a moderate temperature and we are closer to the sea, snow is not found in our surroundings.

Question 4.
Generally, it rains between June and September in our region.
Answer:
We get rainfall between June and September, in our region when the moisture-laden south-west monsoon winds are obstructed by the Western Ghats leading to orographic rainfall.

Question 5.
How do the rain droplets form?
Answer:
Clouds form when condensed water droplets and dust particles accumulate forming large rain droplets.

Question 6.
In London, there is a fog like this till the afternoon in the winters.
Answer:
In London there is fog till the afternoon in winters because London is far away from equator and it has temperate oceanic climate and they have cool summers.

Question 7.
We do not have fog until afternoons in summers.
Answer:
We do not have fog until afternoons in summer because we are near to equator and we have tropical climate and hot summers.

Question 8.
Sometimes hailstones destroy the standing crops in the field.
Answer:
Hailstones are solid and heavy in nature and they hit the earth due to gravity and this is the reason they destroy the crops in the field.

Question 9.
Why don’t we get hailstones frequently?
Answer:
For the formation of hailstones the following 2 conditions are required:

• Intense heating which results in upwards air flow.
• The decrease in air temperature at higher layers of the atmosphere.
• As India is a tropical country, we do not find cooler air at higher levels because of the intense heating of land.

Think about it.

Question 1.
We use a raincoat or umbrella to protect ourselves from rainfall. What will you use to protect yourself from severe hailstorms?
Answer:
If a person is outside without any coverage, he needs to seek shelter immediately, making sure to protect his head from hailstones.

Question 2.
Because of the conventional processes, convectional rainfall occurs in the afternoon in equational areas. But why doesn’t it rain in afternoons in the oceanic areas of the equatorial belt?
Answer:
One of the necessary conditions of convectional rainfall is intense heating of surface which causes air to expand and rise. Since land heats up faster than water, it rains only on the land in the equatorial regions and not in the oceanic areas.

Question 3.
Why are the areas of high rainfall situated in tropical areas?
Answer:

• Tropical areas receive direct rays of the Sun almost throughout the year. Hence the rate of evaporation1 is high here.
• The tropical region receives convectional rainfall throughout the year and also orographic rainfall is experienced here.
• Thus areas of high rainfall are situated in the tropical area.

Class 9 Geography Chapter 5 Precipitation Additional Important Questions and Answers

Complete the statements choosing the correct option from the bracket:

Question 1.
……………. part of the earth’s surface is full of water.
(a) 30.7%
(b) 4.09%
(c) 60.5%
(d) 70.8%
Answer:
(d) 70.8%

Question 2.
When the temperature in the atmosphere falls below the freezing point and the water vapour directly turns into snowflakes, the process is called as ……………..
(a) sublimation²
(b) frostbite³
(c) carbonation
(d) convection
Answer:
(a) sublimation

Question 3.
In areas located at higher altitudes and high latitudes, where the temperatures are below 0°C get precipitation in the form of ………….
(a) dew
(b) rain
(c) snow
(d) hail
Answer:
(c) snow

Question 4.
Because of ………………. crops may get destroyed and loss of life and property may occur.
(a) dew
(b) rain
(c) snow
(d) hail
Answer:
(d) hail

Question 5.
Hails do not occur in ……………… areas.
(a) temperate
(b) equatorial
(c) landlocked
(d) mountainous
Answer:
(b) equatorial

Question 6.
In equatorial areas, …………. type of rainfall occurs almost daily in the afternoons.
(a) frontal
(b) convectional
(c) cyclonic
(d) orographic
Answer:
(b) convectional

Question 7.
…………… rainfall occurs because of obstruction from high mountain ranges.
(a) Frontal
(b) Convectional
(c) Cyclonic
(d) Orographic
Answer:
(d) orographic

Question 8.
Cyclonic rainfall occurs more in …………… zones.
(a) temperate
(b) equatorial
(c) torrid
(d) polar
Answer:
(a) temperate

Question 9.
…………… rainfall occurs in most of the parts in the world.
(a) Frontal
(b) Convectional
(c) Orographic
(d) Cyclonic
Answer:
(c) orographic

Question 10.
Snowfall can also be measured with the help of ……………
(a) hygrometer
(b) rain gauge
(c) barometer
(d) anemometer
Answer:
(b) rain gauge

Question 11.
A layer of ice is equivalent to 10mm of rainfall.
(a) 10mm
(b) 50mm
(c) 100mm
(c) 120mm
Answer:
(c) 120mm

Question 12.
When moisture-laden air near the earth’s surface comes in contact of very cold objects and form water droplets which stick to the surface of the cold objects is formed.
(a) dew
(b) frost
(c) hail
(d) fog
Answer:
(a) dew

Question 13.
If the temperature of the air reaches less than 0°C, the water droplets stuck to the surfaces of cold objects freeze and form
(a) dew
(b) frost
(c) hail
(d) fog
Answer:
(b) frost

Question 14.
If precipitation does not take place, then conditions of arise.
(a) floods
(b) hail
(c) snowstorm
(d) drought
Answer:
(d) drought

Question 15.
Visibility reduces because of
(a) floods
(b) drought
(c) fog
(d) dew
Answer:
(c) fog

Match the column:

Question 1.

Answer:
(1 – b),
(2 – a),
(3 – d),
(4 – c)

Question 2.

Answer:
(1 – c),
(2 – a),
(3 – b)

Answer in one sentence:

Question 1.
What percentage of the earth’s surface is covered with water?
Answer:
70.8% of the earth’s surface is covered with water.

Question 2.
Why do we see different forms of condensation?
Answer:
Different forms of condensation are seen due to changes in atmospheric conditions.

Question 3.
What is precipitation?
Answer:
When water falls in the solid or liquid state from the clouds to the ground, it is called as precipitation.

Question 4.
Name the major forms of precipitation.
Answer:
Snow, hailstones and rainfall are the major forms of precipitation.

Question 5.
Explain the process of sublimation.
Answer:
When the temperature in the atmosphere falls below the freezing point, water vapour directly turns into snowflakes this process is called sublimation.

Question 6.
In India, hails occur in which season?
Answer:
Hails occur in summer reason in India.

Question 7.
Why don’t hails occur in cold zones?
Answer:
Hails do not occur in cold zones because of lack of upward flow.

Question 8.
Why don’t hails occur in equatorial areas?
Answer:
Hails do not occur in equatorial areas because of the heat in the atmosphere.

Question 9.
Which type of rainfall occurs because of obstruction of mountain?
Answer:
Orographic rainfall occurs because of obstruction of mountains.

Question 10.
Convectional rainfall is mainly experienced in which region?
Answer:
Convectional rainfall is mainly experienced in equatorial region.

Question 11.
What is a Cyclone?
Answer:
Cyclone is a specific air formation when the pressure at an area is less than the surrounding regions.

Question 12.
What is acid rain?
Answer:
Precipitation of water with dissolved acids is called acid rain.

Study the rainfall map of the world given below and answer the following question:

Question 1.
Which region experiences more rainfall?
Answer:
The tropical region experiences more rainfall.

Question 2.
What is the reason for low rainfall in the Central Peninsular India?
Answer:
The Central Peninsular India falls on the leeward side of the Western Ghats and hence a rain shadow region is formed here.

Question 3.
Why does the eastern part of Central African continent gets less rainfall than the western part despite its location close to the equator?
Answer:

• Eastern part of the African Continent is a rain shadow region of westerly monsoon winds whereas the western part lies on the windward side and gets more rain.
• The eastern part of Africa also comes under the influence of the North east trade winds but still receive less rains as they are dry winds originating from the land.

Question 4.
Why does the amount of high rainfall in the western part of the European continent reduce in the eastern part?
Answer:
There are many mountain ranges in the western part of Europe. These obstruct the rain-bearing clouds coming from the west and therefore the amount of rainfall received is high in the west and it reduces towards the east.

Question 5.
Why is rainfall more only in the eastern coast of Australia?
Answer:
The eastern part of Australia is a mountainous region. The winds blowing from the Pacific Ocean are obstructed by these mountains resulting in orographic rainfall towards the east and the formation of a rain shadow zone towards the west.

Observe the horizontal profile of Maharashtra in the following figure and answer the following questions:

Question 1.
What type of rainfall occurs in Maharashtra?
Answer:
Orographic rainfall occurs in Maharashtra.

Question 2.
Where will the rain shadow area lie in Maharashtra?
Answer:
The rain shadow area lies to the Leeward side of Sahyadri hills (Maharashtra plateau).

Question 3.
Think about the figure and estimate the rainfall of your district.
Answer:
The answer may vary.

Give reasons:

Question 1.
Crops may get destroyed due to hailstones.
Answer:

• As hailstones are heavy they fall towards the earth’s surface, but because of the frequent upward flow of air, they are repeatedly taken upwards.
• Here, a new layer of snow encapsulates the hail. This happens quite a few times.
• Hence, concentric layers are formed while the hail grows in size.
• These big heavy hailstones fall rapidly to the ground because of gravity. This type of precipitation is called as hail.
• Hence due to hail, crops may get destroyed.

Question 2.
There is a difference between ice and snow.
Answer:

• In areas located at higher altitudes and high- latitudes, where the temperatures are below 0°C get precipitation in the form of snow.
• Snow is friable and opaque. This snow accumulates in the form of layers on top of each other.
• Because of the pressure from the upper layers, the lower layers of the snow become homogeneous, massive and transparent.
• Massive transparent snow formed in such a way is called ice.

Thus, there is a difference between ice and snow.

Question 3.
In equatorial areas, convectional rainfall occurs almost daily in the afternoons.
Answer:

• In equatorial areas, the surface gets heated because of the sun’s heat and the air near it also gets heated.
• As it gets heated, it spreads and becomes lighter and moves upwards. It cools down when it goes upward. The moisture-holding capacity of cold air is less.
• Consequently, condensation of the water vapour occurs and rainfall occurs in equatorial areas.
• Thus in equatorial areas, convectional rainfall occurs almost daily in the afternoons.

Question 4.
A rain shadow area is formed on the leeward side of the Western Ghats.
Answer:

• Winds coming from Arabian sea are moisture-laden. They are obstructed by the Western Ghats coming in their way.
• According to the slope of the Western Ghats, the moisture-laden winds start going upwards.
• The temperature of these winds drop and condensation occurs and rainfall takes place. Thus, because of the obstruction of the Western Ghats, orographic rainfall occurs.
• The windward side of the mountains gets more rain; amount of vapour in the air reduces after crossing the mountain and the water vapour carrying capacity of the air increases.
• The leeward side of the mountain gets lesser rainfall and hence a rain-shadow area is formed here.

Question 5.
Snowfall is not experienced in Maharashtra.
Answer:

• Solid snow particles are formed in regions where the temperature falls below the freezing point leading to the process of sublimation.
• In the sublimation process, the water vapour directly turns into snowflakes.
• In Maharashtra, during winters the temperature never falls below the freezing point.
• Hence snowflakes are never formed in the atmosphere.
• Thus snowfall is not experienced in Maharashtra.

Question 6.
Hailstones do not occur frequently.
Answer:

• Strong vertical movements of air with very high difference in temperature are an ideal condition for the formation of hailstones.
• Presence of moisture is also necessary in the air.
• Such conditions do not exist frequently.
• Hence hailstones are not experienced frequently.

Question 7.
Dew and frost occur on a large scale in winters.
Answer:

• During winters when moisture-laden air near the earth’s surface comes in contact very cold objects, condensation of the vapour takes place.
• They turn into very small water droplets. These water droplets get stick to the surface of the cold objects. This is called dew.
• If the temperature of the air is less than 0°C, the water droplets stuck to the surfaces of cold objects freeze.
• This frozen water droplet is called frost.
• Thus dew and frost occur on a large scale in winters.

Draw diagram of Rain Gauge:
Answer:

Question 6.
Explain the effects of precipitation.
Answer:

• The main source of potable water available on the earth is precipitation.
• As extreme rainfall is destructive so is the absence of rainfall.
• Floods may occur because of heavy rainfall and causes loss to life and property.
• If precipitation does not take place then conditions of drought arise. It causes a shortage of food and food may have to be imported and farmers’ conditions becomes grave.
• The economy of an agrarian1 country like India is dependent on agriculture. The agriculture in India to a large extent is dependent on monsoons. Hence rainfall in India is important to the whole country.
• A good rainfall at the right time increases crop production while untimely rain can damage the crope.
• Acid rains which is a combination of harmful gases and rainwater is harmful to the living organisms as well as non-living objects.

Explain:

Question 1.
Snowfall
Answer:

• When the temperature in the atmosphere falls below the freezing point, the water vapour directly turns into snowflakes. This is called sublimation.
• Here, the vapour in the form of gas transforms into solid snow.
• Precipitation in the form of solid particles is known as snowfall.
• In high latitudinal and temperate regions, snowfall occurs at the mean sea level while in tropical areas, snowfall occurs at places located higher than the snowline altitude.

Question 2.
Formation of hailstones.
Answer:

• When there is a lot of heat on the earth’s surface, the upward air flow blows at a great speed.
• Because of this upward flow, the temperature of the air reduces and the condensation of the water vapour takes place.
• Dark clouds are formed. Because of the upward movement of air, these water droplets go at a higher altitude.
• Here, solidification of these droplets occur and hailstones are formed.

Question 3.
Cyclonic rainfall
Answer:

• Cyclone is the specific air formation when the pressure at an area is less than the surrounding regions.
• Air from the surrounding region comes toward the center of the cyclone and starts moving upwards.
• As it rises, the temperature of the air reduces, condensation occurs and rainfall takes place.
• It rains in areas over which the cyclone passes. Cyclonic rainfall occurs more in temperate zones and its area is also quite extensive.
• Comparatively, cyclonic rainfall occurring in tropical regions is limited in extent and is stormy in nature.

Question 4.
Rain Gauge.
Answer:

• The instrument that is used to measure rainfall is called rain gauge.
• The funnel that is used for measuring rain has a specific diameter. The rain falling in this funnel is collected in a bottle fitted in the gauge.
• The collected water is then measured with the help of measuring jar. The measuring jar reads in millimetres.
• In areas of heavy rainfall, the reading of the rain is taken every three hours.
• The gauge is kept on open ground on a 30cm flat-mount. Hence, the rainwater is collected without any obstruction.

Question 5
Fog, dew and frost
Answer:
(i) Fog:

• The temperature of the layers of the air near the surface of the earth reduces. As temperature reduces, water vapour condenses.
• In this process, vapour turns into microscopic water particles and float in the air.
• When the density of these droplets in the air increases, fog occurs.

(ii) Dew:

• When moisture-laden air near the earth’s surface comes in contact with very cold objects, condensation of the vapour takes place. They turn into very small water droplets.
• These water droplets get stick to the surface of the cold objects. This is called dew.

(iii) Frost:

• If the temperature of the air reaches less than 0°C, the water droplets stuck to the surfaces of cold objects and freeze.
• This frozen water droplet is called frost.

Question 6.
Acid Rain
Answer:

• Because of air pollution in industrial areas, various gases get mixed in the air.
• Different adds are created when the water vapour in the air reacts chemically with these gases. For example, nitric add, sulphuric add, etc.
• Acids dissolved in rainwater fall with the rain j during precipitation. Such a type of rain which has acids dissolved in it is called acid rain.
• Such type of rainfall is harmful to the living organisms and the non-living objects.

Question 7.
Convectional Rainfall
Answer:

1. In equatorial areas, the surface gets heated because of the sun’s heat and the air near it also gets heated. As it gets heated, it spreads and becomes lighter and moves upwards.
2. It cools down when it goes upward & as the j moisture-holding capacity of cold air is less, condensation and rainfall occurs.
3. This type of rainfall is called as Convectional! Rainfall.
4. In equatorial areas, such a type of rainfall occurs almost daily in the afternoons. Rainfall is accompanied by lightning and thunder.
5. The Congo basin of the Africa and the Amazon basin in the South America experience convectional rainfall.
6. Such a rainfall has a very limited area on the earth.

Question 8.
Orographic rainfall
Answer:

• Winds coming from lakes or seas are moisture-laden. They are obstructed by the high mountain ranges coming in their way.
• They start going upwards along the slope of the mountains.
• The temperature of these winds drop and condensation occurs and rainfall takes place. Thus because of the obstruction of the mountains, this type of rainfall occurs.
• The windward side of the mountains gets; more rain; the amount of vapour in the air reduces after crossing the mountain and the moisture-holding capacity of the air increases.
• The leeward side of the mountain gets lesser rainfall and hence this area is identified as rain- shadow area.

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.

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)

Angles and Pairs of Angles Class 7 Maths Chapter 4 Practice Set 15 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 15 Answers Solutions Chapter 4 Angles and Pairs of Angles.

Std 7 Maths Practice Set 15 Solutions Answers

Question 1.
Observe the figure and complete the table for ∠AWB.

Solution:

Question 2.
Name the pairs of adjacent angles in the figures below.

Solution:
i. ∠ANB and ∠ANC
∠BNA and ∠BNC
∠ANC and ∠BNC

ii. ∠PQR and ∠PQT

Question 3.
Are the following pairs adjacent angles? If not, state the reason.

1. ∠PMQ and ∠RMQ
2. ∠RMQ and ∠SMR
3. ∠RMS and ∠RMT
4. ∠SMT and ∠RMS

Solution:

1. ∠PMQ and ∠RMQ are adjacent angles.
2. ∠RMQ and ∠SMR not adjacent angles since they do not have separate interiors.
3. ∠RMS and ∠RMT not adjacent angles since they do not have separate interiors.
4. ∠SMT and ∠RMS are adjacent angles.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 15 Intext Questions and Activities

Question 1.
Observe the figure alongside and write the answers. (Textbook pg. no. 24)

1. Write the name of the angle shown alongside___.
2. Write the name of its vertex___.
3. Write the names of its arms___.
4. Write the names of the points marked on its arms___.

Solution:

1. ∠ABC
2. Point B
3. Ray BA, ray BC
4. Points A, B, C

Class 7 Maths Solution Maharashtra Board

• Angles and Pairs of Angles Practice Set 15 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 16 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 17 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 18 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 19 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 20 Class 7 Maths Solution
• Angles and Pairs of Angles Practice Set 21 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 22 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 23 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 24 Class 7 Maths Solution
• Operations on Rational Numbers Practice Set 25 Class 7 Maths Solution

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Miscellaneous Exercise 2 8th Std Maths Answers Solutions.

Miscellaneous Exercise 2 8th Std Maths Answers

Question 1.
Questions and their alternative answers are given. Choose the correct alternative answer.
i. Find the circumference of a circle whose area is 1386 cm²? [Chapter 15]
(A) 132 cm²
(B) 132 cm
(C) 42 cm
(D) 21 cm²
Solution:
(B) 132 cm

Hint:
i. Area of the circle = πr²
1386 = \(\frac { 22 }{ 7 }\) x r²
r² = 1386 x \(\frac { 7 }{ 22 }\)
= 63 x 7
= 441
r = √441 … [Taking square root of both sides]
= 21 cm
Circumference of the circle = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 21
= 132 cm

ii. The side of a cube is 4 m. If it is doubled, how many times will be the volume of the new cube, as compared with the original cube? [Chapter 16]
(A) Two times
(B) Three times
(C) Four times
(D) Eight times
Solution:
(D) Eight times

Hint:
ii. Original volume = (4)³ = 64 cu.m
New side = 8 m
∴ New volume = (8)² = 512 cu.m
Now, \(\frac{\text { new volume }}{\text { original volume }}=\frac{512}{64}\) = 8
original volume 64
∴ volume of new cube will increase 8 times as compared to the volume of original cube.

Question 2.
Pranalee was practicing for a 100 m running race. She ran 100 m distance 20 times. The time required, in seconds, for each attempt was as follows. [Chapter 11]
18, 17, 17, 16,15, 16, 15, 14,16, 15, 15, 17, 15, 16,15, 17, 16, 15, 14,15
Find the mean of the time taken for running.
Solution:

∴ The mean of the time taken for running 100 m race is 15.7 seconds.

Question 3.
∆DEF and ∆LMN are congruent in the correspondence EDF ↔ LMN. Write the pairs of congruent sides and congruent angles in the correspondence. [Chapter 13]
Solution:

∆EDF ≅ ∆LMN
∴side ED ≅ side LM
side DF ≅ side MN
side EF ≅ side LN
∠E ≅∠L
∠D ≅∠M
∠F ≅∠N

Question 4.
The cost of a machine is Rs 2,50,000. It depreciates at the rate of 4% per annum. Find the cost of the machine after three years. [Chapter 14]
Solution:
Here, P = Cost of the machine = Rs 2,50,000
R = Rate of depreciation = 4%
N = 3 Years
A = Depreciated price of the machine

∴The cost of the machine after three years will be Rs 2,21,184.

Question 5.
In ☐ABCD, side AB || side DC, seg AE ⊥ seg DC. If l(AB) = 9 cm, l(AE) = 10 cm, A(☐ABCD) = 115 cm² , find l(DC). [Chapter 15]
Solution:
Given, side AB || side DC.
∴ ☐ABCD is a trapezium.

Given, l(AB) = 9 cm, l(AE) = 10 cm,
A(☐ABCD) = 115 cm²
Area of a trapezium
= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
∴ A(☐ABCD) = \(\frac { 1 }{ 2 }\) x [l(AB) + l(DC) x l(AE)]
∴ 115 = \(\frac { 1 }{ 2 }\) x [9 + l(DC)] x 10
∴ \(\frac { 115 \times 2 }{ 10 }\) = 9 + l(DC)
∴ 23 = 9 + l(DC)
∴ l(DC) = 23 – 9
∴ l(DC) = 14cm

Question 6.
The diameter and height of a cylindrical tank is 1.75 m and 3.2 m respectively. How much is the capacity of tank in litre?
[π = \(\frac { 22 }{ 7 }\)] [Chapter 16]
Solution:
Given: For cylindrical tank:
diameter (d) = 1.75 m, height (h) = 3.2 m
To Find: Capacity of tank in litre
diameter (d) = 1.75 m
= 1.75 x 100
….[∵ 1 m = 100cm]
= 175 cm
∴ radius (r) = \(=\frac{\mathrm{d}}{2}=\frac{175}{2}\) cm
h = 3.2 cm
= 3.2 x 100
= 320 cm
Capacity of tank = Volume of the cylindrical tank
= πr²h

∴ The capacity of the tank is 7700 litre.

Question 7.
The length of a chord of a circle is 16.8 cm, radius is 9.1 cm. Find its distance from the centre. [Chapter 17]
Solution:

Let CD be the chord of the Circle with centre O.
Draw seg OP ⊥ chord CD
∴l(PD) = \(\frac { 1 }{ 2 }\) l(CD)
…[Perpendicular drawn from the centre of a circle to its chord bisects the chord]
∴l(PD) = \(\frac { 1 }{ 2 }\) x 16.8 …[l(CD) = 16.8cm]
∴l(PD) = 8.4 cm …(i)
∴In ∆OPD, m∠OPD = 90°
∴[l(OD)]² = [l(OP)]² + [l(PD)]² …..[Pythagoras theorem]
∴(9.1)² = [l(OP)]² + (8.4)² … [From (i) and l(OD) = 9.1 cm]
∴(9.1)² – (8.4)² = [l(OP)]²
∴(9.1 + 8.4) (9.1 – 8.4) = [l(OP)]²
…[∵ a² – b² = (a + b) (a – b)]
∴17.5 x (0.7) = [l(OP)]²
∴12.25 = [l(OP)]²
i.e., [l(OP)]² = 12.25
∴l(OP) = √12.25
…[Taking square root of both sides]
∴l(OP) = 3.5 cm
∴The distance of the chord from the centre is 3.5 cm.

Question 8.
The following tables shows the number of male and female workers, under employment guarantee scheme, in villages A, B, C and D.

i. Show the information by a sub-divided bar-diagram.
ii. Show the information by a percentage bar diagram. [Chapter 11]
Solution:
i.

ii.

Question 9.
Solve the following equations.
i. 17 (x + 4) + 8 (x + 6) = 11 (x + 5) + 15 (x + 3)
ii. \(\frac{3 y}{2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
iii. 5(1 – 2x) = 9(1 -x)
[Chapter 12]
Solution:
i. 17 (x + 4) + 8 (x + 6) = 11 (x + 5) + 15 (x + 3)
∴ 17x + 68 + 8x + 48 = 11x + 55 + 15x + 45
∴ 17x + 8x + 68 + 48 = 11x + 15x + 55 + 45
∴ 25x + 116 = 26x + 100
∴ 25x + 116 – 116 = 26x + 100 – 116
… [Subtracting 116 from both the sides]
∴ 25x = 26x – 16
∴ 25x – 26x = 26x – 16 – 26x
… [Subtracting 26x from both the sides]
∴ -x = -16
∴ \(\frac{-x}{-1}=\frac{-16}{-1}\)
∴ x = 16

ii. \(\frac{3 y}{2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{3 y \times 2}{2 \times 2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{6 y}{4}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{6 y}{4} \times 4+\frac{y+4}{4} \times 4=5 \times 4-\frac{y-2}{4} \times 4\)
……[Multiplying both the sides by 4]
∴ 6y + y + 4 = 20 – (y – 2)
∴ 7y + 4 = 20 – y + 2
∴ 7y + 4 = 22 – y
∴ 7y + 4 – 4 = 22 – y – 4
…..[Subtracting 4 from both the sides]
∴ 7y = 18 – y
∴ 7y + y = 18 – y + y
…[Adding y on both the sides]
∴ 8y = 18
∴ \(\frac{8 y}{8}=\frac{18}{8}\) … [Dividing both the sides by 8]
∴ \(y=\frac { 9 }{ 4 }\)

iii. 5(1 – 2x) = 9(1 – x)
∴ 5 – 10x = 9 – 9x
∴ 5 – 10x – 5 = 9 – 9x – 5
….[Subtracting 5 from both the sides]
∴ -10x = 4 – 9x
∴ -10x + 9x = 4 – 9x + 9x
… [Adding 9x on both the sides]
∴ -x = 4
∴ -x x (- 1) = 4 x (- 1)
… [Multiplying both the sides by – 1]
∴ x = – 4

Question 10.
Complete the activity according to the given steps.
i. Draw rhombus ABCD. Draw diagonal AC.
ii. Show the congruent parts in the figure by identical marks.
iii. State by which, test and in which correspondence ∆ADC and ∆ABC are congruent.
iv. Give reason to show ∠DCA ≅ ∠BCA, and ∠DAC ≅ ∠BAC
v. State which property of a rhombus is revealed from the above steps. [Chapter 13]
Solution:
a.

b. In ∆ADC and ∆ABC,
side AD ≅ side AB …..[Sides of a rhombus]
side DC ≅ side BC …..[Sides of a rhombus]
side AC ≅ side AC … [Common side]
∆ADC ≅ ∆ABC … [By SSS test]
∠DCA ≅ ∠BCA …[Corresponding angles of congruent triangles]
∠DAC ≅ ∠BAC …[Corresponding angles of congruent triangles]
From the above steps, property of rhombus revealed is ‘diagonal of a rhombus bisect the opposite angles’.

Question 11.
The shape of a farm is a quadrilateral. Measurements taken of the farm, by naming its corners as P, Q, R, S in order are as follows. l(PQ) = 170 m,
l(QR) = 250 m, l(RS) = 100 m, l(PS) = 240 m, l(PR) = 260 m.
Find the area of the field in hectare (1 hectare = 10,000 sq.m). [Chapter 15]
Solution:

Area of the field = A(∆PQR) + A(∆PSR)
In ∆PQR, a = 170 m, b = 250 m, c = 260 m
Semiperimeter of ∆PQR = s

Area of the field = A(∆PQR) + A(∆PSR)
= 20400 + 12000
= 32400 sq.m
= \(\frac { 32400 }{ 10000 }\)
…[1 hectare = 10,000 sq.m]
= 3.24 hectares
∴ The area of the field is 3.24 hectares.

Question 12.
In a library, 50% of total number of books is of Marathi. The books of English are \(\frac { 1 }{ 3 }\) of Marathi books. The books on Mathematics are 25% of the English books. The remaining 560 books are of other subjects. What is the total number of books in the library? [Chapter 12]
Solution:
Let the total number of books in the library be x
50% of total number of books is of Marathi.
Number of Marathi books = 50% of x
= \(\frac { 50 }{ 100 }x\)
= \(\frac { x }{ 2 }\)
The books of English are \(\frac { 1 }{ 3 }\) of Marathi books.
Number of books of English = \(\frac{1}{3} \times \frac{x}{2}\)
= \(\frac { x }{ 6 }\)
The books on Mathematics are 25% of the English books.
Number of books of Mathematics
= 25% of \(\frac { x }{ 6 }\)
= \(\frac{25}{100} \times \frac{x}{6}\)
= \(\frac { x }{ 24 }\)
Since, there are 560 books of other subjects, the total number of books in the library are


∴ 24x – 17x = 17x + 13440 – 17x
∴ 7x = 13440
∴ \(\frac{7 x}{7}=\frac{13440}{7}\)
∴ x = 1920
∴ The total number of books in the library are 1920.

Question 13.
Divide the polynomial (6x³ + 11x² – 10x – 7) by the binomial (2x + 1). Write the quotient and the remainder. [Chapter 10]
Solution:

∴ Quotient = 3x² + 4x – 7,
remainder = 0
Explanation:

Maharashtra Board Class 8 Maths Solutions

HCF-LCM Class 6 Maths Chapter 9 Practice Set 23 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 9 HCF-LCM Class 6 Practice Set 23 Answers Solutions.

Std 6 Maths Practice Set 23 Solutions Answers

Question 1.
Write all the factors of the given numbers and list their common factors:
i. 12, 16
ii. 21, 24
iii. 25, 30
iv. 24, 25
v. 56, 72
Solution:
i. Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 16 = 1, 2, 4, 8, 16
∴ Common factors of 12 and 16 = 1, 2, 4

ii. Factors of 21 = 1, 3, 7, 21
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
∴ Common factors of 21 and 24 = 1, 3

iii. Factors of 25 = 1, 5, 25
Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
∴ Common factors of 25 and 30 = 1, 5

iv. Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 25 = 1,5, 25
∴ Common factor of 24 and 25 = 1

v. Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56
Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
∴ Common factors of 56 and 72 = 1, 2, 4, 8

Maharashtra Board Class 6 Maths Chapter 9 HCF-LCM Practice Set 23 Intext Questions and Activities

Question 1.
In the empty boxes, write the proper words: dividend, divisor, quotient, remainder. (Textbook pg. no. 46)

Solution:

When we divide 36 by 4, the remainder is zero. Therefore, 4 is a factor of 36 and 36 is a multiple of 4. But, when we divide 65 by 9, the remainder is not zero. Therefore, 9 is not a factor of 65. Also, 65 is not a multiple of 9.

Question 2.
Write all the factors of the numbers 36 and 48. Also, list their common factors. (Textbook pg. no. 46)
Solution:
36 = 1 × 36
= 2 × 18
= 3 × 12
= 4 × 9
= 6 × 6

48 = 1 × 48
= 2 × 24
= 3 × 16
= 4 × 12
= 6 × 8

∴ Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Common factors of 36 and 48: [1] ,[2], [3], [4], [6], [12]

6th Std Maths Digest Pdf Download

• Bar Graphs Practice Set 18 Class 6 Maths Solution
• Bar Graphs Practice Set 19 Class 6 Maths Solution
• Symmetry Practice Set 20 Class 6 Maths Solution
• Symmetry Practice Set 21 Class 6 Maths Solution
• Divisibility Practice Set 22 Class 6 Maths Solution
• HCF-LCM Practice Set 23 Class 6 Maths Solution
• HCF-LCM Practice Set 24 Class 6 Maths Solution
• HCF-LCM Practice Set 25 Class 6 Maths Solution

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Miscellaneous Exercise 1 8th Std Maths Answers Solutions.

Miscellaneous Exercise 1 8th Std Maths Answers

Question 1.
Choose the correct alternative answer for each of the following questions.
i. In ₹PQRS, m∠P = m∠R = 108°, m∠Q = m∠S = 72°. State which pair of sides of those given below is parallel. [Chapter 8]
(A) side PQ and side QR
(B) side PQ and side SR
(C) side SR and side SP
(D) side PS and side PQ
Solution:
(B) side PQ and side SR

Hint:

In ₹PQRS,
m∠P + m∠S = 108°+ 72
= 180°
Since, interior angles are supplementary.
∴ side PQ || side SR

ii. Read the following statements and choose the correct alternative from those given below them. [Chapter 8]
a. Diagonals of a rectangle are perpendicular bisectors of each other.
b. Diagonals of a rhombus are perpendicular bisectors of each other.
c. Diagonals of a parallelogram are perpendicular bisectors of each other.
d. Diagonals of a kite bisect each other.
(A) Statements (b) and (c) are true
(B) Only statement (b) is true
(C) Statements (b) and (d) are true
(D) Statements (a), (c) and (d) are true.
Solution:
(B) Only statement (b) is true

iii. If 19³ = 6859, find \(\sqrt[3]{0.006859}\). [Chapter 3]
(A) 1.9
(B) 19
(C) 0.019
(D) 0.19
Solution:
(D) 0.19

Hint:
\(\sqrt[3]{0.006859}\)

Question 2.
Find the cube roots of the following numbers. [Chapter 3]
i. 5832
ii. 4096
Solution:
i. 5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
= (2 × 3 × 3) × (2 × 3 × 3) × (2 × 3 × 3)
= (2 × 3 × 3)³
= (18)³
\(\sqrt[3]{5832}=18\)

ii. 4096 = (4 × 4) × (4 × 4) × (4 × 4)
= (4 × 4)
= 16³
\( \sqrt[3]{4096}=16\)

Question 3.
m∝n,n = 15 when m = 25. Hence
i. Find m when n = 87,
ii. Find n when m = 155. [Chapter 7]
Solution:
Given that, m ∝ n
∴ m = kn …(i)
where, k is the constant of variation.
When m = 25, n = 15
∴ Substituting, m = 25 and n = 15 in (i), we get
m = kn
∴ 25 = k × 15
∴ k = \(\frac { 25 }{ 15 }\)
∴ k = \(\frac { 5 }{ 3 }\)
Substituting k = \(\frac { 5 }{ 3 }\) in (i), we get
m = kn
∴ m = \(\frac { 5 }{ 3 }n\) …(ii)

i. When n = 87, m = ?
Substituting n = 87 in (ii), we get
m = \(\frac { 5 }{ 3 }n\)
m = \(\frac { 5 }{ 3 }\) × 87
m = 5 × 29
m = 145

ii. When m = 155, n = ?
∴ Substituting m = 155 in (ii), we get
m = \(\frac { 5 }{ 3 }n\)
∴ 155 = \(\frac { 5 }{ 3 }n\)
∴ \(\frac{155 \times 3}{5}=n\)
∴ n = 31 × 3
∴ n = 93

Question 4.
y varies inversely with x. If y = 30 when x = 12, find [Chapter 7]
i. y when x = 15,
ii. x when y = 18.
Solution:
Given that,
\(y \propto \frac{1}{x}\)
∴ \(y=k \times \frac{1}{x}\)
where, k is the constant of variation.
∴ y × x = k …(i)
When x = 12, y = 30
∴ Substituting, x = 12 and y = 30 in (i), we get
y × x = k
∴ 30 × 12 = k
∴ k = 360
Substituting, k = 360 in (i), we get
y × x = k
∴ y × x = 360 ….(ii)

i. When x = 15,y = ?
∴ Substituting x = 15 in (ii), we get
y × x = 360
∴ y × 15 = 360
∴ y = \(\frac { 360 }{ 15 }\)
∴ y = 24

ii. When y = 18, x = ?
∴ Substituting y = 18 in (ii), we get
y × x = 360
∴18 × x = 360
∴ x = \(\frac { 360 }{ 18 }\)
∴ x = 20

Question 5.
Draw a line l. Draw a line parallel to line l at a distance of 3.5 cm. [Chapter 2]
Solution:
Steps of construction:

1. Draw a line l and take any two points M and N on the line.
2. Draw perpendiculars to line l at points M and N.
3. On the perpendicular lines take points S and T at a distance 3.5 cm from points M and N respectively.
4. Draw a line through points S and T. Name the line as n.

Line n is parallel to line l at a distance of 3.5 cm from it.

Question 6.
Fill in the blanks in the following statement.
The number \((256)^{\frac{5}{7}}\) is __ of __ power of __. [Chapter 3]
Solution:
The number \((256)^{\frac{5}{7}}\) is 7th root of 5th power of 256.

Question 7.
Expand.
i. (5x – 7) (5x – 9)
ii. (2x – 3y)³
iii. \(\left(a+\frac{1}{2}\right)^{3}\) [Chapter 5]
Solution:
i. (5x – 7) (5x – 9)
= (5x)² + (-7 -9) 5x + (-7) × (-9).
…[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= 25x² + (-16) × 5x + 63
= 25x² – 80x + 63

ii. Here, a = 2x and b = 3y
(2x – 3y)³
= (2x)³ – 3 (2x)² (3y) + 3 (2x) (3y)² – (3y)³
…[∵ (a – b)³ = a³ – 3a²b + 3ab² – b³]
= 8x³ – 3 (4x²) (3y) + 3 (2x) (9y²) – 27y³
= 8x³ – 36x²y + 54xy² – 27p³

iii. Here, A= a and B = \(\frac { 1 }{ 2 }\)
\(\left(a+\frac{1}{2}\right)^{3}=(a)^{3}+3(a)^{2}\left(\frac{1}{2}\right)+3(a)\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{3}\)
…[(A + B)³ = A³ + 3A²B + 3AB² + B³]
\(=\mathbf{a}^{3}+\frac{3 \mathbf{a}^{2}}{2}+\frac{3 \mathbf{a}}{4}+\frac{1}{8}\)

Question 8.
Draw an obtuse angled triangle. Draw all of its medians and show their point of concurrence. [Chapter 4]
Solution:

The point of concurrence of the medians PS, RU and QV is G.

Question 9.
Draw ∆ABC such that l(BC) = 5.5 cm, m∠ABC = 90°, l(AB) = 4 cm. Show the orthocentre of the triangle. [Chapter 4]
Solution:

Here, point B is the orthocentre of ∆ABC.

Question 10.
Identify the variation and solve.
It takes 5 hours to travel from one town to the other if speed of the bus is 48 km/hr. If the speed of the bus is reduced by 8 km/hr, how much time will it take for the same travel? [Chapter 7]
Solution:
Let, v represent the speed of the bus and t represent the time required to travel from one town to the other.
The speed of the bus varies inversely with the time required to travel from one town to the other.
∴ \(\mathbf{v} \propto \frac{1}{\mathbf{t}}\)
∴ \(\mathbf{v}=\mathbf{k} \times \frac{1}{\mathbf{t}}\)
where, k is the constant of variation.
∴ v × t = k …(i)
It takes 5 hours to travel from one town to the other if speed of the bus is 48 km/hr.
i.e., when v = 48, t = 5
∴ Substituting v = 48 and t = 5 in (i), we get
v × t = k
∴ 48 × 5 = k
∴ k = 240
Substituting k = 240 in (i), we get
v × t = k
∴ v × t = 240 …(ii)
Since, the speed of the bus is reduced by 8 km/hr,
∴ Speed of the bus in second case (v)
= 48 – 8 = 40 km/hr
∴ When v = 40, t = ?
∴ Substituting v = 40 in (ii), we get
v × t = 240
∴ 40 × t = 240
∴ \(t=\frac { 240 }{ 40 }\)
∴ t = 6
∴ The problem is of inverse variation and the bus would take 6 hours to travel the distance if its speed is reduced by 8 km/hr.

Question 11.
Seg AD and seg BE are medians of ∆ABC and point G is the centroid. If l(AG) = 5 cm, find l(GD). If l(GE) = 2 cm, find l(BE). [Chapter 4]
Solution:
The centroid of a triangle divides each median in the ratio 2:1.
i. Point G is the centroid and seg AD is the median.

ii. Point G is the centroid and seg BE is the median.

∴ l(BG) × 1 = 2 × 2
∴ l(BG) = 4 cm
Now, l(BE) = l(BG) + l(GE)
∴ l(BE) = 4 + 2
∴ l(BE) = 6 cm

Question 12.
Convert the following rational numbers into decimal form. [Chapter 1]
i. \(\frac { 8 }{ 13 }\)
ii. \(\frac { 11 }{ 7 }\)
iii. \(\frac { 5 }{ 16 }\)
iv. \(\frac { 7 }{ 9 }\)
Solution:
i. \(\frac { 8 }{ 13 }\)

ii. \(\frac { 11 }{ 7 }\)

iii. \(\frac { 5 }{ 16 }\)

iv. \(\frac { 7 }{ 9 }\)

Question 13.
Factorize.
i. 2y² – 11y + 5
ii. x² – 2x – 80
iii. 3x² – 4x + 1
Solution:
i. 2y² – 11y + 5
= 2y² – 10y – y + 5
= 2y(y – 5) – 1(y – 5)
= (y – 5)(2y – 1)

ii. x² – 2x – 80
= x² – 10x + 8x – 80
= x (x – 10) + 8 (x – 10)
= (x – 10)(x + 8)

iii. 3x² – 4x + 1
= 3x² – 3x – x + 1
= 3x(x – 1) – 1(x – 1)
= (x – 1) (3x – 1)

Question 14.
The marked price of a T.V. set is Rs 50,000. The shopkeeper sold it at 15% discount. Find the price of it for the customer. [Chapter 9]
Solution:
Here, marked price = Rs 50,000,
discount = 15%
Let the discount percent be x
∴x = 15%
i. Discount

= 500 × 15
= Rs 7,500

ii. Selling price = Marked price – Discount
= 50,000 – 7,500
= Rs 42,500
∴The price of the T.V. set for the customer is Rs 42,500.

Question 15.
Rajabhau sold his flat to Vasantrao for Rs 88,00,000 through an agent. The agent charged 2 % commission for both of them. Find how much commission the agent got. [Chapter 9]
Solution:
Here, selling price of the flat = Rs 88,00,000
Rate of commission = 2%
Commission = 2% of selling price
= \(\frac { 2 }{ 100 }\) × 88,00,000
= 2 × 88,000
= Rs 1,76,000
∴ Total commission = Commission from Rajabhau + Commission from Vasantrao
= Rs 1,76,000 + Rs 1,76,000
= Rs 3,52,000
∴ The agent got a commission of Rs 3,52,000.

Question 16.
Draw a parallelogram ABCD such that l(DC) = 5.5 cm, m∠D = 45°, l(AD) = 4 cm. [Chapter 8]
Solution:
Opposite sides of a parallelogram are congruent.
∴ l(AD) = l(BC) = 4 cm and
l(DC) = l(AB) = 5.5 cm

Question 17.
In the figure, line l || line m and line p || line q. Find the measures of ∠a, ∠b, ∠c and ∠d. [Chapter 2]

Solution:
i. line l|| line m and line p is a transversal.
∴m∠a = 78° …(i) [Corresponding angles]

ii. line p || line q and line m is a transversal.
∴m∠d = m∠a …[Corresponding angles]
∴m∠d = 78° …(ii)[From (i)]

iii. m∠b = m∠d …[Vertically opposite angles]
∴m∠b = 78° …[From (ii)]

iv. line l|| line m and line q is a transversal.
∴m∠c + m∠d = 180° …[Interior angles]
∴m∠c + 78° = 180° … [From (ii)]
∴m∠c =180° – 78°
∴m∠c = 102°
∴m∠a = 78°, m∠b = 78°, m∠c = 102°, m∠d = 78°

Maharashtra Board Class 8 Maths Solutions

Balbharti Maharashtra State Board Class 9 Geography Solutions Chapter 7 International Date Line Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Geography Solutions Chapter 7 International Date Line

Class 9 Geography Chapter 7 International Date Line Textbook Questions and Answers

1. Two boxes in different hemispheres are given in the following diagram. The IDL passes through both the boxes. In one box, the meridian, day and date is given. Find the day and date for the other box.

Answer:

2. Select the correct option :

Class 9 Geography Chapter 7 International Date Line Question 1.
While crossing the IDL, a person will have to add one day when traveling from
(a) East to West
(b) West to East
(c) South to North
(d) North to South
Answer:
(a) East to West

International Date Line Class 9 Question 2.
If it is Wednesday 10 a.m. at 150 E meridian, then what will be the time at IDL?
(a) Wednesday at 6 a.m.
(b) Wednesday 9 p.m.
(c) Thursday 2 p.m.
(d) Thursday at 6 p.m.
Answer:
(b) Wednesday 9 p.m.

International Date Line Questions And Answers Question 3.
According to the international convention, at which meridian does the day and date change occurs?
(a) 0°
(b) 90° E
(c) 90° W
(d) 180°
Answer:
(d) 180°

International Date Line 9th Std Geography Question 4.
At which direction of the IDL does a new day start immediately?
(a) East
(b) West
(c) North
(d) South
Answer:
(b) West

9 Std Geography International Date Line Question 5.
IDL brings coordination in which of the following?
(a) GPS system
(b) Defence departments
(c) Transportation schedules
(d) Determining the hemisphere
Answer:
(c) Transportation schedules

3. Give geographical reasons

Std 9 Geography Chapter 7 International Date Line Question 1.
IDL is proving to be very useful in today’s times
Answer:
IDL is proving to be very useful in today’s times because:

• The IDL brings coordination between international airlines, transportation services, economic and trade activities.
• The IDL has been carved out of the necessity of coordinating time and date.
• It is also important in today’s modern era and rapidly happening global developments.
• We can keep a track of all the calculations of a day and time accurately with the help of IDL in case of global transportation especially, with respect to airways.
• It is only because of the IDL that the schedules of the traffic worldwide are organised properly.

Class 9th Geography Chapter 7 Question Answer Question 2.
The day starts in the Pacific Ocean on the earth
Answer:

• The IDL passes through the Pacific Ocean.
• With reference to the IDL, the day on Earth starts in the West and ends in the East.
• It is one and the same day only at 12 o’clock midnight at the 180° meridian.
• For the countries lying to the east of it i.e. the USA, Chile, etc., it is the end of the day while for the countries lying to the west of it i.e. Japan, Australia, etc. it is the start of a new day.

Thus, a new day starts in the Pacific Ocean on Earth.

4. Write in brief:

International Date Line Std 9 Question 1.
What considerations have been made while deciding the IDL?
Answer:
The following points were taken into consideration while drawing the International Date Line. (IDL):

1. The direction of travel
2. The current day and date

• • Accordingly, while traveling from East to West of IDL, a day is added. For example, while travelling from America to Japan, if it is Thursday 25th December, then it will be Friday 26th December in Japan.
  • While from West to East of IDL, the day remains the same. For example, while travelling from Japan to America, if it is Thursday 25th December, then Thursday 25th December, only should be considered.

Question 2.
While crossing the IDL, what changes will you make?
Answer:

• When you cross the 180° meridian, some precautions need to be taken.
• There is a difference of 12 hours from Prime Meridian, if you go from East or West.
• According to the convention, the start (and end) of a date is considered to be at 180° Meridian.
• Accordingly, an adjustment or change in day and time is made. Thus, while travelling from east to west of IDL , a day is added whereas while travelling from west to east of IDL, the day remains the same.

Question 3.
Why is the IDL not a straight line like the 1800 meridian?
Answer:

• Attempt has been made to make the IDL pass through the Pacific Ocean completely.
• Had it passed through a land or some islands, the people there would have had to follow different dates and timings because dates would have been different on their Eastern and Western parts.
• Also, it would have been difficult to know when one crosses the IDL on land, and when the date changes on the calendar.
• Therefore, the IDL is not a straight line like the 1 180° Meridian. At places, it turns East while at other places, it turns West.

Question 4.
Why doesn’t the IDL pass through land?
Answer:

• If the IDL had passed through a land or some islands, the people there would have had to follow different dates and timings because dates would have been different on their Eastern and Western parts.
• Also, it would have been difficult to know when one crosses the IDL on land and when the date changes on the calendar.
• Hence, the IDL does not pass through land

Question 5.
Why is the IDL considered with respect to the 1800 meridian only?
Answer:

• Our 24-hour day starts at 12 midnight. Because of the earth’s rotation, the midnight occurs at different times, at different places.
• As the Earth is spherical in shape, every place has a place to its East.
• So, it was necessary to determine at what location to the East does the day start on Earth.
• Hence, representatives of many nations got together, under the leadership of an American Professor, Davidson, in the year 1884 and decided the International Date Line.
• The line was drawn opposite to the Greenwich Prime Meridian i.e. with reference to the 180° Meridian.

5. Using an atlas, tell in which of the following routes the IDL will be crossed and show them on the map.

(1) Mumbai- London- New York- Los Angeles- Tokyo
(2) Delhi- Kolkata- Singapore- Melbourne
(3) Kolkata- Hong Kong- Tokyo- San Francisco
(4) Chennai-Singapore- Tokyo- Sydney-Santiago
(5) Delhi-London-New York

Answer:
(1) Mumbai – London – New York – Los Angeles – Tokyo – IDL will be crossed.
(2) Delhi – Kolkata – Singapore – Melbourne – IDL will not be crossed
(3) Kolkata – Hong Kong – Tokyo – San Fransico – IDL will be crossed.
(4) Chennai – Singapore – Tokyo – Sydney – Santiago – IDL will be crossed
(5) Delhi – London – New York – IDL will not be crossed.

Class 9 Geography Chapter 7 International Date Line Intext Questions and Answers

Use your Brain Power:

Question 1.
You are traveling from the Kamchatka Peninsula (in the northern hemisphere) to New Zealand (in the southern hemisphere) along the IDL. It is Monday, 22nd June in the northern hemisphere. What will the day and date in the southern hemisphere?
Answer:
The day will be Monday, 22nd June as we are not crossing the IDL.

Examine a ticket of UA 876 Boeing 787-9 Dreamliner closely and find the answers to the following questions:

Question 1.
From which country will the plane take off and where will it go?
Answer:
The plane will take off from Tokyo (Japan) and will go to San Francisco, California (USA)

Question 2.
What is the duration of the flight?
Answer:
The duration of the flight is 9hrs and 15 min.

Question 3.
What is the day, date and time given at the starting point and destination of the flight?
Answer:
At the starting point of Tokyo (Japan) it is Friday, 1st April 12.30 am and at the destination of San Francisco California (USA) it will arrive on Thursday March 31st at 5.45 pm.

Question 4.
What special note is given on the air ticket?
Answer:
A special note on the ticket states that the flight involves a date change.

Question 5.
What could be the reason behind giving such a note?
Answer:
The reason behind giving such a note is to make us understand that the flight will cross the IDL

Question 6.
During this flight, will the plane cross the IDL? If yes, then from which direction to which direction?
Answer:
During this flight, the plane will cross the IDL from west to east.

Question 7.
What did you understand by reading the ticket?
Answer:
We understand that a day is deducted and the time is fixed backward by 1 day while crossing the IDL from west to east.

Try this:

Question 1.
Complete the following table to understand the time at different meridians: (In this activity we are not taking into consideration the rotation of the earth.)
Answer:

• All places to east of Prime Meridian are ahead of GMT and all places to west of Prime Meridian are behind GMT.
• Earth rotates on its axis and covers 1° longitude in 4 minutes.
• As Sunil is moving 30° east, time will move ahead by 120 min, i.e. 2 hours (30° x 4 min.) and as Minal is moving 30° west, time will move backwards by 120 min. i.e. 2 hours.

Question 1.
What is the day at 00 meridians at Greenwich after completing the table ‘A’?
Answer:
The day at 0° meridian at Greenwich after completing the table ‘A’ is Wednesday.

Question 2.
What is the day at 00 meridians at Greenwich after completing the table ‘B’?
Answer:
The day at 0° meridian at Greenwich after completing the table B is Friday.

Question 3.
Though both were at the same place, why were they experiencing different days? How did this happen?
Answer:
Because Sunil was moving eastward, so time will move ahead whereas Meenal was moving westward, so time is moving backward.

Question 4.
How many days occurred while doing this activity? Name them.
Answer:
While doing this activity, 3 days came in reference as Wednesday, Thurday and Friday.

Question 5.
Which day is correct: Wednesday in Table ‘A’ or Friday in Table ‘B’? Why?
Answer:
None of the given days are correct. The correct day is Thursday after considering IDL.

Explanation:
(i) When Sunil reached 180° meridian i.e. IDL, according to him it is Thursday midnight (24.00). When IDL is crossed from west to east we need to deduct a day (24 hrs.). Thus when 24 hrs are deducted from Thursday 24.00 (midnight) we get Thursday 00.00. So after crossing the IDL the day to be considered is Thursday i.e. the same day.

(ii) When Minal reached 180° meridian i.e. IDL, according to her it is Wednesday midnight (24.00). When IDL is crossed from east to west we need to add a day (24 hrs.). Thus when 24 hrs are added to Wednesday 24.00 (midnight) we get Thursday 24.00 (midnight). So after crossing the IDL the day to be considered is Thursday.

Give it a try:

Your are now aware of the changes required to be made while crossing the ¡DL. Now redo the activity given on Page 59. Tell us the changes that you will have to make while crossing the IDL i.e. 1800 meridian. Your travel will start on Sunday, 21st May 2016 at 10 a.m.

Let’s Recall:

Question 1.
Which meridian is used to determine world standard Time (GMT)?
Answer:
World Standard Time (GMT) is determined using Prime Meridian at 0° longitude.

Question 2.
Which meridian determines Indian Standard Time (1ST)?
Answer:
82°30′ E is the standard meridian of India.

Question 3.
What is the time difference between the GMT and the IST?
Answer:
IST is 5hrs and 30 min ahead of GMT.

Class 9 Geography Chapter 7 International Date Line Additional Important Questions and Answers

Complete the Statements choosing correct option:

Question 1.
Man has studied the rotation speed, direction and shape of the earth and prepared the system.
(a) monometric
(b) chronometric
(c) topographic
(d) GIS
Answer:
(b) Chronometric

Question 2.
Earth rotates from
(a) north to south
(b) south to north
(c) east to west
(d) west to east
Answer:
(d) west to east

Question 3.
In terms of time the part of the earth is ahead of the part.
(a) northern, southern
(b) southern, northern
(c) eastern, western
(d) western, eastern
Answer:
(c) eastern, western

Question 4.
is reached after 12 hours from Prime Meridian
(a) 90° W
(b) 90° E
(c) 180°
(d)120°W
Answer:
(c) 180°

Question 5.
According to IDL, while traveling from east to west
(a) a day is added
(b) a day is deducted
(c) keep the same day
(d) None of these
Answer:
(a) a day is added

Question 6.
Many nations got together under the leadership of in 1884 and decided on International Date Line.
(a) Professor Davidson
(b) Professor Samuelson
(c) Professor Adam Smith
(d) Professor Richardson
Answer:
(a) Professor Davidson

Answer in one sentence:

Question 1.
If it is 12noon at the Prime Meridian then what will be the time at 60°E longitude?
Answer:
If its is 12noon at the Prime Meridian, then it will be 16.00hrs (4pm) at 60°E longitude.

Question 2.
Which part of the earth is ahead in terms of time?
Answer:
In terms of time the eastern part of the earth is ahead of the western part.

Question 3.
When was it decided to draw the International Date line?
Answer:
When representatives of many nations got together under the leadership of American professor Davidson in the year of 1884 they decided on the International Date line.

Question 4.
A new day starts on the earth in which Ocean?
Answer:
A new day starts in the Pacific Ocean on the earth.

The following table shows the days and time to welcome New Year in different countries and cities according to 1ST. Observe the table and answer the questions gives below:

Question 1.
Which location was the first to welcome the New Year in the World? What day was it then?
Answer:
Samoa islands in Apia was the first to welcome the New Year in the world on Saturday.

Question 2.
Which location bid farewell to the year 2016, the last of all?
Answer:
Baker islands in USA was last to bid farewell to the year 2016.

Question 3.
On which day did that location welcome 2017?
Answer:
Baker islands in USA welcomed 2017 on Sunday.

Question 4.
What could be the reason behind the change in the day of Sydney and London?
Answer:
According to 1ST, India welcomed the New Year at 12 midnight on Saturday. Sydney being 5 hours and 30 minutes ahead of IST and has already welcomed the New year before India on Saturday itself. But London is 5 hours and 30 minutes behind IST will welcome the New Year the next day i.e. on Sunday.

Balbharti Maharashtra State Board Class 9 Geography Solutions Chapter 1 Distributional Maps Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Geography Solutions Chapter 1 Distributional Maps

Class 9 Geography Chapter 1 Distributional Maps Textbook Questions and Answers

1. Give reasons why the following sentences are right or wrong:

Geography Class 9 Chapter 1 Distributional Maps Question 1.
The main aim of distributional maps is to show location.
Answer:
Wrong The main aim of distributional maps is to show the distribution of variables like temperature, rainfall, population, etc.

• Distribution maps are type of thematic maps.
• These maps are prepared with special themes like the distribution of various variables in a particular region.
• They make use of different methods to show the distribution of variables like temperature, rainfall, population, etc.

Thus, the main aim of distributional maps is to show the distribution of variables in a particular region.

Distributional Maps Class 9 Question 2.
In choropleth maps, only one value is assigned to the sub-administrative unit.
Answer:
Right

1. In choropleth maps, the data regarding various geographical variables is shown by shading or using tints of various colours.
2. After gathering the data for a selected variable, five to seven classes are made considering the smallest and the largest values.
3. Each class is assigned a tint of the same colour or black-and-white patterns.
4. A particular value shown by a predecided tint of colour can represent only a particular region or group of region.

Thus, in choropleth maps, only one value is assigned to sub administrative unit.

Distributional Maps Questions And Answers Question 3.
In choropleth maps, colours/tints do not change according to the values of the variables.
Answer:
Wrong.
In choropleth maps, colours/tints change according to the values of variables.

• In choropleth maps, the smallest and the largest values are taken into consideration and after that 5-7 classes are made.
• Each class is assigned a unique tint of the same colour or black and white pattern.
• In other words, each class is used to show the distribution of a particular variable in sub- administrative unit.

Distributional Map Question 4.
Choropleth maps are used to show altitudes.
Answer:
Wrong.
Isopleth maps and not the choropleth maps are used to show altitude.

• When the distribution of a variable is continuous like the altitude, the isopleth method map is used.
• Isopleth maps use lines joining the locations having equal values and hence are useful to show the change in altitude in the region.
• When the distribution of a variable is discontinuous, the choropleth method map is used to show its distribution, eg. distribution of population, domestic animals, etc.

Question 5.
Isopleth maps are used to show population distribution.
Answer:
Wrong.
Dot method maps or choropleth method maps are used to show population distribution

• Distribution of a population is discontinuous.
• Isopleth method maps are used to show the distribution of variables which are continuous. For eg. altitude, temperature etc.
• Thus, isopleth method maps are not used to show population distribution.

Question 6.
In dot method, every dot should have an appropriate scale.
Answer:
Right.
The size of the dot is decided as per the value of a variable that is freely scattered across the region.

Question 7.
Isopleth maps are not made using isolines.
Answer:
Wrong.
Isopleth maps are made using isolines.

• Isolines are lines joining places of same values of a given variable in a map.
• Since, isopleth maps are also prepared in the same method, we can say that they are made using isolines.

Question 8.
Distribution of various geographical elements can be shown using dot method.
Answer:
Wrong.
Distribution of various geographical elements cannot be shown using dot method.

• Distribution of geographical elements which are freely scattered across a region can be shown using a dot method.
• Distribution of variables like population, minerals, cattles etc., are freely scattered hence can be shown on a dot method map.
• Whereas the distribution of variables like rainfall, altitude, temperature etc. which are continuous, cannot be shown using a dot method map.

2. Answer in brief.

Question 1.
Explain the use and types of distributional maps.
Answer:
(a) Use of distributional maps: The distributional maps are useful for explaining the patterns of distribution of a particular geographical variable.
(b) The types of distributional maps : The following are the types of distributional maps.

1. Dot Method: The distributional map based on statistical data, made by this method, make use of dots of a pre-decided size to show the distribution of a particular variable, e.g. Dot method map is used to show the population of a region, distribution of cattle etc.
2. Choropleth Method: The distributional map made by this method makes use of shades or tints of various colours, to show the distribution of a particular variable, e.g. Choropleth method map is used to show population density, types of forest cover, uses of land.
3. Isopleth Method: The distributional map made by this method makes use of lines joining locations or places having same or equal values of a particular variable, e.g. Isopleth method map is used to show altitude, temperature, rainfall.

Question 2.
Differentiate between choropleth and isopleths methods.
Answer:

Question 3.
Explain with reasons the method which is best suited to show the distribution of population in a region.
Answer:

• The dot method of making a distributional map is best suited to show the distribution of the population in a region.
• The dots used in this method, are given a particular value which is determined by selecting a proper scale considering the spread of the values of the variables in the given region.
• This method is best suited to show the distribution of the variables which are freely scattered as in the case of population distribution.

3. Which method will you use for the following information?

Question 1.
Talukawise wheat production in the district
Answer:
Dot method

Question 2.
Distribution of the altitude of the land in the district.
Answer:
Isopleth method

Question 3.
Distribution of domestic animals in the State
Answer:
Dot method

Question 4.
The distribution of population density in India
Answer:
Choropleth method

Question 5.
Temperature distribution in Maharashtra State.
Answer:
Isopleth method

4. Study the population distribution map of Kolhapur district and answer the following questions:

Question 1.
Which method has been used to show the distribution of population in the district?
Answer:
The dot method has been used to show the distribution of population in the district.

Question 2.
Explain the direction wise distribution of the population from dense to sparse.
Answer:
The distribution of the population is dense in the Eastern part of the district, whereas the western part of the district has low density of population.

Question 3.
What is the population shown by the largest circle? Which place is that?
Answer:
The population shown by the largest circle is nearly 20 lakhs. That place is Kolhapur city (Karvir Taluka)

Question 4.
Which Taluka has the least population?
Answer:
Gaganbavada Taluka has the least population.

Class 9 Geography Chapter 1 Distributional Maps Intext Questions and Answers

Make friends with maps!
Read the map given below and answer the Questions:

Question 1.
What is the population of the Amravati town?
Answer:
The population of the Amravati town is nearly 5 lakhs.

Question 2.
Name the place having population of 1 lakh on the map.
Answer:
Achalpur has a population of 1 lakh

Question 3.
Which part of the map shows sparse distribution of population?
Answer:
The northwestern part of the map shows sparse distribution of population

Read the map given below and answer the Questions:

Question 1.
Name the talukas having population density between 301 and 400 persons per sq km.
Answer:
Anjangao Surji has population density between 301 and 400 persons per sq km

Question 2.
What is the density of the Amravati taluka?
Answer:
Amravati taluka has the population density of more than 400 persons per sq km

Question 3.
Name the talukas having population density less than 300 persons per sq km.
Answer:
Dharini, Chikaldhara, Daryapur, Chandur Bazar, Morshi, Varud, Tivsa, Dhamangaon Railway, Nandgaon Khandeshwar and Bhatkuli have population density less than 300 persons per sq km.

Question 4.
Which taluka has a population density of less than 100 persons per sq km?
Answer:
Chikaldhara has population density of less than 100 persons per sq km.

Question 5.
Which Talukas have a population density of more than 400 persons per sq km?
Answer:
Amravati and Achalpur have the population density of more than 400 persons per sq km.

Read the map given below and answer the Questions:

Question 1.
In which part of the district is the rainfall more?
Answer:
Rainfall is more in the Northern part of the district.

Question 2.
In which direction is the rainfall decreasing?
Answer:
Rainfall is decreasing towards the west

Question 3.
What is the lowest value of the rainfall in the district?
Answer:
The lowest value of the rainfall in the district is less than 800 mm.

Question 4.
What is the highest value of the rainfall in the district?
Answer:
The highest value of the rainfall in the district is more than 1300 mm.

Question 5.
What is the amount of rainfall in the central part of the district?
Answer:
The amount of rainfall in the central part of the district is 900 mm.

Read the map given below and answer the Questions:

Question 1.
In which direction is the density of population decreasing?
Answer:
The density of population is decreasing towards the North and South direction.

Question 2.
Name the talukas with a population density less than 200.
Answer:
Shahuwadi, Gaganbavada, Ajra and Chandgad have a population density of less than 200.

Question 3.
Name the talukas having population density between 200 and 400.
Answer:
Radhanagri and Gargoti have population density between 200 and 400.

Question 4.
Name the talukas with a population density more than 400.
Answer:
Kolhapur, Panhala, Hatkanangale, Shirol, Gadhinglaj and Kagal have population density more than 400.

Question 5.
In which direction are the talukas having a higher density of population located in the district?
Answer:
The Eastern part of Kolhapur has a higher density of population.

Question 6
Which method has been used in the making of this map?
Answer:
Choropleth method is used in making this map.

Read the map given below and answer the Questions:

Question 1.
In which direction is the rainfall more in the district?
Answer:
The rainfall is more in the Western part of the district.

Question 2.
In which direction is the rainfall decreasing?
Answer:
The rainfall is decreasing towards the East.

Question 3.
Which class shows low rainfall category in the district?
Answer:
O to 600 mm class shows low rainfall category in the district.

Question 4.
Which class shows high rainfall category in the district?
Answer:
More than 2400 (2400 to 3000 mm) class shows high rainfall category in the district.

Question 5.
Which method has been used in making this map?
Answer:
The Isopleth method has been used in making this map

Study the population distribution map of Kolhapur district and answer the following questions:

Question 1.
A field visit is an important study method in Geography.
Answer:
Right.

• Geographical concepts and elements can be directly experienced through field-visits.
• Field-visits are extremely useful for understanding the correlation between human beings and the environment.

Class 9 Geography Chapter 1 Distributional Maps Additional Important Questions and Answers

Complete the statements choosing the correct option:

Question 1.
The main aim of distributional maps is to show the location of the place and of variables.
(a) population
(b) statistics
(c) consumption
(d) distribution
Answer:
(d) distribution

Question 2.
Distributional maps are of types.
(a) two
(b) three
(c) four
(d) five
Answer:
(b) three

Question 3.
To draw distributional maps, we need data.
(a) statistical
(b) population
(c) colourful
(d) distance
Answer:
(a) statistical

Question 4.
The distribution of population is shown by method.
(a) dot
(b) choropleth
(c) isopleth
(d) isolines
Answer:
(a) dot

Question 5.
In maps, the data regarding various geographical variables are shown by shading or tints of various colours.
(a) dot
(b) choropleth,
(c) isopleth
(d) thematic
Answer:
(b) choropleth

Question 6.
When the distribution of a variable is discontinuous the map is used.
(a) isopleth
(b) physical
(c) choropleth
(d) dot
Answer:
(c) choropleth

Question 7.
In method, lines showing equal values are used.
(a) choropleth
(b) isopleth
(c) dot
(d) planar
Answer:
(b) isopleth

Question 8.
When the distribution of a variable is the isopleth method is used.
(a) continuous
(b) discontinuous
(c) sparse
(d) disjoint
Answer:
(a) continuous

Question 9.
While showing population distribution, urban population is shown by
(a) circles
(b) dots
(c) squares
(d) lines
Answer:
(a) circles

Question 10.
is an important study method in Geography.
(a) Maps
(b) Field visit
(c) Picnic
(d) farm visit
Answer:
(b) Field visit

Match the column.

Question 1.

Answer:
(1 – b),
(2 – c),
(3 – a)

Question 2.

Answer:
(1 – a),
(2 – c),
(3 – b)

Answer in one sentence:

Question 1.
What are distributional maps?
Answer:
Maps that show the distribution of various geographical variables are called ‘distributional maps’.

Question 2.
What are thematic maps?
Ans,
Maps that are prepared with special themes are called ‘thematic maps’.

Question 3.
What is a dot-method map?
Answer:
A dot method map is a map type, that shows the distribution of a variable using dot symbols.

Question 4.
What is a choropleth method map?
Answer:
A choropleth method map is a map type, in which the distribution of a variable is shown by shades or tint of various colours.

Question 5.
When do we use the isopleth method for showing the distribution of a variable on a map?
Answer:
We use the isopleth method for showing the distribution of a variable on a map, when the distribution of the variable is continuous.

Question 6.
What is an isopleth method map?
Answer:
An isopleth method map, is a map type, that shows distribution of a variable with the help of lines showing equal values.

Question 7.
What is point related data?
Answer:
The statistical data of particular variables with respect to various places is called ‘point-related data’.

Question 8.
Which elements are generally studied in a geographical field visit?
Answer:
The various elements which can be studied during geographical field visits include physical landforms, river banks, dams, coastal areas, tourism sites, offices or museums related to geographical elements, villages, forest areas etc.

Question 9.
Mention the list of things to be carried while going on a geographical field visit?
Answer:
The various things which one should carry to a geographical field visit include notebook, specimen Questionnaire, pen pencil, scale, tape, compass, bag for collecting samples, maps, camera, etc.

Question 10.
What is field report?
Answer:
The report written on the basis of the information obtained from a field visit is known as field report.

Which method will you use for the following information:

Question 1.
The distribution of rainfall in the district of Nandurbar.
Answer:
Isopleth Method

Question 2.
Cotton crop distribution in Maharashtra.
Answer:
Dot method

Question 3.
Draw a map using the Dot Method:
Answer:

Explain:

Question 1.
Choropleth method
Answer:

1. In the choropleth method various geographical variables are shown by shades or tints of various colours.
2. While making choropleth maps, the data used for different variables is obtained through various processes such as measurement, surveying and so on.
3. In this method, only one value is given to one sub – administrative unit in a region
4. The smallest and the largest values of the given data of the variables are taken into consideration and after that 5-7 classes are made.
5. Each class is assigned a tint of the same colour or black-and-white patterns. The shades or the patterns become darker with the increasing values of the given variable.

Question 2.
Preparations for field visit
Answer:

• Before going for a field-visit, decide the place and the purpose of the visit.
• The elements which will be observed in field visit should be decided.
• For a field visit, you should carry a notebook, specimen Questionnaire
• , pen, pencil, scale, tape, compass, a bag for collecting samples, maps, camera and any other item needed.

Question 3.
Selection of the study area
Answer:

1. Geographical field-visit is organised for study of various elements, for example physical 1 landforms, river banks, dams, coastal areas, tourism sites, offices or museums related to ! geographical elements, villages or forest areas.
2. Elements should be selected after studying : the local conditions and necessary permission letters should be obtained.

Question 4.
Report writing
Answer:

1. A field-report should be written on the basis of information obtained after the visit is complete.
2. Students should add photographs wherever necessary.
3. The following points should be used to make a field report:
   • Introduction
   • Location map and Route Map
   • Physiography¹
   • Climate
   • Population
   • Environmental problems and measures
   • Land utilisation
   • Conclusions