Category: Class 9

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:

	Dew		Frost
(i)	When moisture-laden air near the earth’s surface comes into contact with very cold objects, condensation of vapour takes place into small water droplets called dew.	(i)	If the temperature of the air is less then CPC, the water droplets stuck to the surface of cold objects, freezes forming frost.
(ii)	Water vapour condenses and forms droplets of water.	(ii)	Water droplets stuck to cold surface turns to frozen water droplets.
(iii)	Dew sticks to the cold object but does not freeze.	(iii)	It sticks to the cold object and freeze.

Question 2.
Snow and hail
Answer:

	Snow		Hail
(i)	Precipitation in the form of solid particles of snow is known as snow fall.	(i)	Precipitation in the form of frozen water droplets falling rapidly to the ground is know as hail.
(ii)	The fall of temperature in the atmosphere below the freezing point causes snow fall	(ii)	Extreme heat on the surface of the earth initiates the process of hail formation.
(iii)	Heavy accumulation of snow can collapse the transportation and communication system of the area.	(iii)	It destroys crops and causes loss of life an

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.

Column A	Column B
(1) Snowflakes	(a) upward air flow
(2) Hailstones	(b) sublimation
(3) Dew	(c) microscopic water particles floating in the air
(4) Fog	(d) condensation4 on cold objects

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

Question 2.

Column A	Column B
(1) Orographic rainfall (2) Convectional Rainfall (3) Cyclonic Rainfall	(a) Daily in equatorial areas (b) More in temperate zones (c) Mountain barrier

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 9 English Solutions Kumarbharati Chapter 4.1 Please Listen Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 English Kumarbharati Solutions Chapter 4.1 Please Listen!

English Kumarbharati 9th Solutions Chapter 4.1 Please Listen! Textbook Questions and Answers

Warming Up:

1. Form groups of 6-8. One person (leader) chooses one item – a picture, a paragraph or a lesson from any one of the 9th standard textbooks and writes the reference on a piece of paper and folds it. Others ask him/her questions and try to guess what it is, from his/her answers. Example: (The leader thinks and writes something on a piece of paper. He folds it and keeps it on the table. Others ask questions.)

Question 1.
Is it something from social studies?
Answer:
No.

Question 2.
From Science or Maths?
Answer:
No.

Question 3.
Languages?
Answer:
Yes.

Question 4.
From the English Kumarbharati?
Answer:
Yes.

Question 5.
A prose lesson?
Answer:
No.

Question 6.
A nature poem?
Answer:
No.

Question 7.
Is it in the first unit?
Answer:
No.

Question 8.
Second unit?
Answer:
Yes.

Question 9.
Inspirational poem?
Answer:
Yes.

Invictus!!
Leader shows the folded paper; answer is Invictus!

Agreements and Disagreements.

2. Form pairs. List the things on which you ! have the same opinion and also the ones on which you have different opinions. Prepare a list of ten things in all and see how far you agree or disagree with your friend. Some useful phrases are given alongside.

Question 1.
Form pairs. List the things on which you ! have the same opinion and also the ones on which you have different opinions. Prepare a list of ten things in all and see how far you agree or disagree with your friend. Some useful phrases are given alongside.
You could talk about:

1. Clothes, latest fashion.
2. Performance of sportsmen and women.
3. Trafficll’ransport in your area.
4. Cleanliness and hygiene in your area.
5. Future occupations/Careers.
6. Latest news items.
7. TV programmes
8. Mobile Apps
9. Any subject of your choice.

English Workshop:

1. The poet uses a free, conversational style in his poem. It is also called Colloquial style. Pick out and write down such lines or expressions that support the above statement.

Question 1.
The poet uses a free, conversational style in his poem. It is also called Colloquial style. Pick out and write down such lines or expressions that support the above statement.
(a) ………………………. You have not done what I asked.
(b) ……………………………………………………………………..
(c) ……………………………………………………………………..
(d) ……………………………………………………………………..
(e) ……………………………………………………………………..
(f) ……………………………………………………………………..
Answer:
(a) You have not done what I asked.
(b) You are trampling on my feelings.
(c) All that I ask is that you listen.
(d) When you do something for me that I can and need to do for myself, you contribute to my fear and inadequacy.
(e) But when you accept as a simple fact that I feel what I feel.
(f) then I can stop trying to convince you have not done what I asked.

2. Put the following expressions in a table of Dos and Don’ts as expressed by the poet.

Question 1.
Put the following expressions in a table of Dos and Don’ts as expressed by the poet.
(a) Please listen.
(b) give me advice.
(c) tell me why.
(d) solve my problem
(e) just hear me.
(f) accept as a simple fact
(g) contribute to my fear
(h) wait a minute

Answer:

Dos	Don’ts
1. please listen	1. give me advice
2. just hear me	2. tell me why
3. accept my feelings as a simple fact	3. solve my problem
4. wait a minute	4. contribute to my fear

3. Write in your own words.

Question a.
What does the listener do when the poet asks him to just listen? Give 3 points.
Answer:
When the poet asks the listener to just listen, the listener either:

1. gives advice
2. tells him why he shouldn’t feel that way, or
3. tries to solve the; poet’s problem.

Question b.
Why does the poet remark that advice is cheap?
Answer:
The poet remarks that advice is cheap because everyone is willing to give advice. You only had to write to the agony column of any newspaper with your problems and you could get advice from the columnists. You had to spend just 20 cents to buy the paper.

Question c.
Which two facts show that the poet is confident of overcoming his irrational feelings?
Answer:
The two facts that show that the poet ; is confident of overcoming his irrational feelings 5 are : first, he says he will understand what is behind these feelings. Second, he says that when the answers become obvious, he will not need any advice.

4. According to the poet, how does God help people when they pray to Him for help?

Question 1.
According to the poet, how does God help people when they pray to Him for help?
Answer:
According to the poet, God doesn’t try to give advice or fix things. He is silent and let’s one work things out by oneself. That is how God helps people when they pray to Him for help.

5. Why should one learn to tackle one’s problems by one’s own self?

Question 1.
Why should one learn to tackle one’s problems by one’s own self?
Answer:
When one tackles one’s problems by oneself, one overcomes feelings of fear and inadequacy. Hence one should learn to tackle one’s problems by one’s own self.

6. Why should seniors not over-protect or over-pamper juniors?

Question 1.
Why should seniors not over-protect or over-pamper juniors?
Answer:
If seniors over-protect or over-pamper juniors, the juniors will be unable to stand on their own feet and face the world. They will not be able to overcome their feelings of fear and inadequacy. They will not become confident. Hence, seniors should not over-protect or over-pamper juniors.

7. Find out and write down some proverbs/ axioms/quotations that convey a message similar to ‘Self-help is the best help.’

Question 1.
Find out and write down some proverbs/ axioms/quotations that convey a message similar to ‘Self-help is the best help.’
Answer:
Examples:

1. God helps those who help themselves.
2. Self-help is the root of all success.

8. Maintain a diary for at least a week. Write about your interactions with other people in your surroundings in 3-4 lines. Also write whether you find the interactions happy – unhappy, satisfactory – unsatisfactory, enjoyable – stressful, etc.

Question 1.
Maintain a diary for at least a week. Write about your interactions with other people in your surroundings in 3-4 lines. Also write whether you find the interactions happy – unhappy, satisfactory – unsatisfactory, enjoyable – stressful, etc.
Answer:
Jan. 1: Today is New Year’s day. Holiday for J everyone. Mummy baked a yummy cake. I helped her. Papa and I went for a long walk in the park. Have made heaps of resolutions. Intend to keep every one of them. Am full of determination. Have decided to ‘ ask Seemadidi for help. A wonderful day!

Jan. 2: Not such a good day. Broke my first resolution. Overslept and had to rush to school with my shoes in my hand! Luckily my homework was all done. Kept my resolution of doing my studies before going out to play football. Had an argument with Sahil on the ground. But it ended well.

9. Write an informal letter from a teenager to his/her parent expressing a few thoughts from the poem.

Question 1.
Write an informal letter from a teenager to his/her parent expressing a few thoughts from the poem.
Answer:
Room No. 3
Ladies’ Hostel
Aundh Road
Pune – 411 007.
14th October, 2020

Dearest Mom,

Hi! Sorry, sorry, dear Mom. Are you surprised to see this letter? I wanted to talk to you about this, but then I thought I will be able to express myself better in a letter.

Mom, when I told you about the problems I was having 1 with my roommate here at the hostel, I only wanted j you to listen. I knew I was being irrational at times, and I was trying to understand this feeling. It would have taken me some time, but I would have finally got things sorted out on my own. I’m sorry, but I did not need advice, or any solutions. Now I’ve solved my problem and I am at peace.

I’m sorry, Mom, if I’ve hurt you by snapping at your good intentions. But I’m learning. I still need your support, but I have to make my own way in this world, and this is the first step. Please try to understand this.

Love you always.
Your loving daughter,
Ananya

10. Read aloud a couple of stanzas of the poem j ‘Invictus’ and ‘Please Listen’. In what ways do j they differ? Think and fill up the table with a ‘Yes’ or ‘No’.

Question 1.
Read aloud a couple of stanzas of the poem ‘Invictus’ and ‘Please Listen’. In what ways do j they differ? Think and fill up the table with a ‘Yes’ or ‘No’.

‘Invictus’ is an example of Traditional Poetry. ‘Please Listen’ is an example of Free Verse.
Answer:

	Invictus	Please Listen
1. Rhyming lines	Yes	No
2. Steady rhythm	Yes	No
3. Uniformity in length of lines	Yes	No
4. Uniformity in number of lines in each stanza	Yes	No
5. Figurative language	Yes	No
6. Example of	Traditional poetry	Free verse

English Kumarbharati 9th Digest Chapter 4.1 Please Listen Additional Important Questions and Answers

Simple Factual Activity:

Question 1.
Say whether the following statements are True or False:
Answer:

1. The poet wants advice from the listener – False
2. The poet wants sympathy from the listener – False
3. The poet wants solutions from the listener – False
4. The poet wants the listener to listen to him silently – True

Complex Factual Activity.

Question 1.
How old is the speaker in the poem?
Answer:
The speaker must be a teenager or a young person who is impatient and does not want advice or solutions.

Question 2.
Who is he/she talking to?
Answer:
He /She is talking to someone older a parent or a friend.

Activities based on Poetic Devices.

Question 1.
Pick out the two verbs that are repeated ; throughout the extract.
Answer:
The verbs ‘ask’ and ‘listen’ are repeated throughout the extract.

Question 2.
What style is the poem written in?
Answer:
The poem is written in the conversational or colloquial style.

Simple Factual Activity.

Question 1.
Fill in the blanks with the correct words:
Answer:
1. The poet says that he may be discouraged and faltering, but he is not helpless.
2. The poet accepts that he has irrational feelings.

Complex Factual Activities.

Question 1.
Is the speaker aware of his shortcomings?
Answer:
Yes, the speaker is aware of his/her shortcomings.

Question 2.
What must have happened before the speaker says all this?
Answer:
The speaker must have had some emotional stress and problems, some feelings that he/she could not find reasonable explanations for.

Activities based on Poetic Devices.

Question 1.
Is there any figurative language in the poem?
Answer:
No, there is not much figurative language in the poem.

Simple Factual Activity.

Question 1.
Complete the web:
(The answers are given directly and underlined.)

Think discuss and answer:

Question a.
Does he/she want to improve?
Answer:
Yes, he/she wants to improve.

Question b.
Is he/she confident that he/she can improve?
Answer:
Yes, he/she is confident that he/she can improve.

Question c.
How does he/she want to work it out?
Answer:
He/She wants to work it out on his/her own.

Question d.
Is the poet willing to listen to others?
Answer:
Yes, the poet is willing to listen to others.

Answer the following questions:

Question 1.
Discuss how you can be a good listener.
Answer:
To be a good listener, we must pay attention to what a person is saying without interruption. We must only listen and not try to form replies and counter-arguments in our own mind. We must also nod and show correct expressions to encourage the speaker.

Activities based on Poetic Devices:

Question 1.
The poet uses a free, conversational style in his poem. It is also called Colloquial style. Pick out and write down such lines or expressions that support the above statement.
Answer:
1. So please listen, and just hear me
2. And if you want to talk, wait a minute for your turn – and I will listen to you.

Question 2.
Comment upon the length of lines and the number of lines in each stanza.
Answer:
There is no uniformity in the length of lines or the number of lines in each stanza. It appears as if the poem has first been written in prose and then broken up into separate lines.

Point Format (for understanding).

• Title: Please Listen!
• Poet: Unknown (anonymous)
• Rhyme Scheme: The poem is in free verse. Hence there is no rhyme scheme.
• Figure of Speech: Repetition. ‘When I ask you to listen to me’. This line has been repeated in the poem for emphasis.
• Theme/Central idea: The poet tells us how he just wishes that people would listen to a person’s problems silentiy, instead of giving advice.

The poet is confident that he will find out the reasons behind his irrational feelings and then he will be able to solve his problems by himself. He feels that prayers probably work because God just listens to us silently when we pray and then lets us work things out ourselves. Through the poem, the poet implies that we too should work things out ourselves when we are faced with difficulties or have some irrational feelings.

Paragraph Format.

The poem ‘Please Listen!’ is written by an unknown poet (anonymous).

The poem is in free verse. Hence there is no rhyme scheme. There are not many figures of speech. The important Figure of Speech is Repetition: ‘When I ask you to listen to me’. This line has been repeated in the poem for emphasis.

The poet tells us how he just wishes that people would listen to a person’s problems silently, instead of giving advice. The poet is confident that he will find out the reasons behind his irrational feelings and then he will be able to solve his problems by himself. He feels that prayers probably work because God just listens to us silently when we pray and then lets us work things out ourselves. Through the poem, the poet implies that we too should work things out ourselves when we are faced with difficulties or have some irrational feelings.

Vocabulary Focus.

Question 1.
What is the difference between ‘hear’ and ‘listen’?
Answer:

1. ‘To hear’ means to receive with the ear the sound made by someone or something; to be aware of some sound.
2. ‘To listen’ means to consciously give one’s attention to a sound and try to understand it.

Maharashtra State Board Class 9 English Solutions

శ్రీ ఆంజనేయ సహస్రనామ స్తోత్రమ్

Balbharti Maharashtra State Board Class 9 Sanskrit Solutions Aamod Chapter 13 सरमाया: शीलम् Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Sanskrit Aamod Solutions Chapter 13 सरमाया: शीलम्

Sanskrit Aamod Std 9 Digest Chapter 13 सरमाया: शीलम् Textbook Questions and Answers

भाषाभ्यास:

माध्यमभाषया उत्तरत।

1. सरमायाः कर्तव्यपालने के विघ्नाः अभवन् ?

प्रश्न 1.
सरमायाः कर्तव्यपालने के विघ्नाः अभवन् ?
उत्तरम् :
देवांची कुत्री सरमा तिच्या स्वामिनिष्ठेमुळे वेदवाङ्मयात प्रसिद्ध आहे. पूर्वी पणी नामक असुरांनी देवांच्या गायी पळवल्या, त्या शोधण्यासाठी देवांचा गरुड सुपर्ण पणींच्या देशात गेला. मात्र पणींनी त्याला भेटवस्तू देऊन आपल्याकडे वळवले आणि गोधनाविषयी सुपर्ण याने देवांना काहीच माहिती दिली नाही. गायी शोधण्यासाठी देवांनी त्यांची कुत्री सरमा हिला पाठवले.

पणींच्या देशात जाण्यासाठी सरमेला रसा नदी पार करावी लागणार होती, मात्र ती रसा नदी अतिशय वेगवान होती. ती पार करणे सरमेला शक्य नव्हते; मात्र सोपवलेले काम पूर्ण करायच्या निश्चयाने सरमेने नदीला विनंती केली की तिने तिचा वेग कमी करावा तसेच पाणी उथळ करावे. ज्यामुळे नदी पोहून दुसऱ्या काठावर जाणे शक्य होईल.

मात्र रसा स्वत:ला सर्वश्रेष्ठ समजत असल्याने तिने सरमेची विनंती धुडकावली. रसासारखी वेगवान नदी असूनही सरमेने जराही न डगमगता तिच्यात उडी मारली आणि धैर्याने दुसऱ्या तीरावर पोहोचली. तिथे पोहचल्या नंतर सुद्धा गायी शोधणे सोपे नव्हते. तरीसुद्धा सरमेने चिकाटीने गोधनाचा शोध लावला. इंद्राकडे परतताना पणींनी तिला लालूच दाखवली, तिची निंदा केली. पण सरमेने त्यांना दाद दिली नाही. मार्गात आलेल्या सर्व अडचणींना नेटाने तोंड देऊन सरमेने आपले कर्तव्य बजावले.

In Vedic literature, thestory of सरमा, Indra’s dog is very famous. She is famous for her loyalty towards the Gods. Formerly, the demons named for stole the cattle-wealth of Gods. In order to search the cattle-wealth, the eagle सुपर्ण was sent but the पणिs bribed him. सुपर्ण did not inform Gods about the cattle though he had found it hidden in the cave. Hence, was appointed to search the cattle wealth.

सरमा had to cross the river रसा to reach the country of पणि, रसा was flowing swiftly. It was very difficult for HRAT to cross over; but she did not give up. She requested रसा modestly to slow down her speed so that she would reach the other bank; but the who considered herself superior declined her request.

Yet, without caring for life descended into the current and reached the other bank courageously. She also faced great difficulty in searching the cattle-welath. ufus tried to test her loyalty by trying to bribe her.

Though there were so many difficulties in fulfilling her duty, सरमा overcame those difficulties with determination and accomplished her task.

2. सरमा कर्तव्यपालने विघ्नान् कथं तरति ?

प्रश्न 2.
सरमा कर्तव्यपालने विघ्नान् कथं तरति ?
उत्तरम् :
देवांची कुत्री सरमा तिच्या स्वामिनिष्ठेमुळे वेदवाङ्मयात प्रसिद्ध आहे. पूर्वी पणी नामक असुरांनी देवांच्या गायी पळवल्या. त्या शोधण्यासाठी देवांचा गरुड सुपर्ण पणींच्या देशात गेला. मात्र पणींनी त्याला भेटवस्तू देऊन आपल्याकडे वळवले आणि गोधनाविषयी सुपर्ण याने देवांना काहीच माहिती दिली नाही. गायी शोधण्यासाठी देवांनी त्यांची कुत्री सरमा हिला पाठवले.

पणींच्या देशात जाण्यासाठी सरमेला रसा नदी पार करावी लागणार होती, मात्र ती रसा नदी अतिशय वेगवान होती. ती पार करणे सरमेला शक्य नव्हते; मात्र सोपवलेले काम पूर्ण करायच्या निश्ययाने सरमेने नदीला विनंती केली की तिने तिचा वेग कमी करावा तसेच पाणी उथळ करावे. ज्यामुळे नदी पोहून दुसऱ्या काठावर जाणे शक्य होईल.

मात्र रसा स्वत:ला सर्वश्रेष्ठ समजत असल्याने तिने सरमेची विनंती धुडकावली. रसासारखी वेगवान नदी असूनही सरमेने जराही न डगमगता तिच्यात उडी मारली आणि धैर्याने दुसऱ्या तीरावर पोहोचली. गोधनाच्या शोधासाठी सुद्धा सरमा पणींच्या देशात खूप भटकली. अनेकांना विचारले मात्र कुणीच काही बोलले नाही. शेवटी जंगलातल्या गुहेत शिरुन पणींनी लपवलेल्या गायींचा शोध लावण्यात ती यशस्वी झाली. इंद्राकडे परतताना पणीनी तिला लालूच दाखवली, तिची निंदा केली. पण सरमेने त्यांना दाद दिली नाही.

मार्गात आलेल्या सर्व अडचणींना नेटाने तोंड देऊन सरमेने आपले कर्तव्य बजावले.

In Vedic literature, the story of सरमा, Indra’s dog is very famous. She is famous for her loyalty towards the Gods. Formerly, the demons named for stole the cattle-wealth of Gods. In order to search the cattle-wealth, the eagle सुपर्ण was sent but the पणिs bribed him. सुपर्ण did not inform Gods about the cattle though he had found it hidden in the cave. Hence, सरमा was appointed to search the cattle-wealth.

सरमा had to cross the river रसा to reach the country of पणि. रसा was flowing swiftly. It was very difficult for 1 to cross over; but she did not give up. She requested the modestly to slow down her speed so that she would reach the other bank, but tai who considered herself superior declined her request.

Yet, it without caring for life descended the current and reached the other bank courageously. सरमा wandered here and there to find out the cattlewelath in the country of us. She asked many people there but no one was ready to help her. At the end, she was successful in searching the cattle wealth. wus tried to stop her in every possible manner but सरमा did not step back even a little.

सरमा overcame obstaclescourageously. She was extremly loyal and determined to accomplish her task by defeating all obstacles.

3. पणयः सरमाया: निन्दा कदा अकुर्वन् ?

प्रश्न 3.
पणयः सरमाया: निन्दा कदा अकुर्वन् ?
उत्तरम् :
देवांची कुवी सरमा तिच्या स्वामिनिष्ठेमुळे वेदवाङ्मयात प्रसिद्ध आहे. पूर्वी पणी नामक असुरांनी देवांच्या गायी पळवल्या. त्या शोधण्यासाठी देवांचा गरुड सुपर्ण पणींच्या देशात गेला. मात्र पणींनी त्याला भेटवस्तू देऊन आपल्याकडे वळवले आणि गोधनाविषयी सुपर्ण याने देवांना काहीच माहिती दिली नाही. गायी शोधण्यासाठी देवांनी त्यांची कुत्री सरमा हिला पाठवले.

पणींच्या देशात पोहोचल्यावर सरमैने अतिशय कष्टाने गायींना शोधले. गायींना पाहिल्यावर आनंद झालेली सरमा इंद्राला सांगण्यासाठी निघाली. हे पाहिल्यावर पणींनी तिला अनेक प्रकारे फितवण्याचा, आपल्याकडे वळवण्याचा प्रयत्न केला. तिला दूध, तूप, दही खाण्याचा आग्रह केला. इतकेच नाही तर गोधनातील काही भागही देऊ केला, मात्र सरमा बधली नाही. तेव्हा पणींनी तिची निंदा करण्यास सुरुवात केली. जेणेकरून सरमा इंद्राकडे जाण्यापासून परावृत्त होईल.

In Vedic literature, the story of सरमा, Indra’s dog is very famous. She is famous for her loyalty towards the Gods. Formerly, the demons named for stole the cattle-wealth of Gods. In order to search the cattle-wealth, the eagle you was sent but the पणी bribed him. सुपर्ण did not inform Gods about the cattle though he had found it hidden in the cave. Hence, सरमा was appointed to search the cattle-wealth.

सरमा searched for the cattle-wealth rigorously. There was no help available. She succeeded in finding out the cattle-wealth hidden in one of the caves in the forest. Immediately she started to return to inform the gods. quit tried to bribe her. They offered her milk, butter, ghee. They were ready to share some of the cattle-welath with her; but she refused everything and remained loyal to her duty. Then the Panis started to criticise her to stop from going back to Indra.

4. पणयः सरमां किमर्थं निन्दन्ति ?

प्रश्न 4.
पणयः सरमां किमर्थं निन्दन्ति ?
उत्तरम् :
देवांची कुत्री सरमा तिच्या स्वामिनिष्ठेमुळे वेदवाङ्मयात प्रसिद्ध आहे. पूर्वी पणी नामक असुरांनी देवांच्या गायी पळवल्या. त्या शोधण्यासाठी देवांचा गरुड सुपर्ण पणींच्या देशात गेला. मात्र पणींनी त्याला भेटवस्तू देऊन आपल्याकडे वळवले आणि गोधनाविषयी सुपर्ण याने देवांना काहीच माहिती दिली नाही. गायी शोधण्यासाठी देवांनी त्यांची कुत्री सरमा हिला पाठवले.

देवांच्या गायी पळवून आणल्यानंतर पणींनी त्या गुहेत दडवून ठेवल्या. देवांची कुत्री सरमा हिने अतिशय कष्टाने त्या शोधल्या आणि त्यांची माहिती देण्यासाठी की इंद्राकडे निघाली. सुपर्ण गरुडाप्रमाणे सरमेला आपल्याकडे वळवून घेण्याचा प्रयत्न पणींनी केला मात्र सरमा त्याला बदली नाही. तिने मी लोभाला बळी पडणार नाही. मी इंद्राची स्वामिनिष्ठ सेविका आहे असे ठणकावले.

हे ऐकताच पणी खवळले आणि सरमेचा अपमान केला. सरमा ही मांस खाते असा अपप्रचार केला. यामुळे सरमेचे मनोबल बळेल, असे पणींना वाटले, मात्र अविचल कार्यनिष्ठा असलेली सरमा पणींकडे दुर्लक्ष करून निघाली.

In Vedic literature, the story of सरमा, Indra’s dog is very famous. She is famous for her loyalty towards the Gods. Formerly, the demons named for stole the cattle-wealth of Gods. The was appointed to search the cattle wealth.

सरमा searched for the cattle-wealth rigorously. There was no help available. She succeeded in finding out the cattle-wealth hidden in one of the caves in the forest. Immediately she started to return to inform ths gods. for tried to bribe her.

They offered her milk, butter, ghee. They were ready to share some of the cattle-welath with her; but सरमा with undisturbed mind igonored पणि, After that पणि criticized सरमा that she consumes meat. for thought that he would get discouraged due to insult and would not inform gods.

Sanskrit Aamod Class 9 Textbook Solutions Chapter 13 सरमाया: शीलम् Additional Important Questions and Answers

धातुसाधितविशेषणानि।

धातुसाधित – विशेषणम्	विशेष्यम्
दृष्टः	किङ्करः
नियुक्ता, आज्ञप्ता, प्रस्थिता, प्राप्ता, प्रमुदिता, अवरोधिता, निन्दिता	सरमा
दृष्टा, पराभूता	रसा
रुद्धम्	गोधनसर्वस्वम्
समाप्तम्	अन्वेषणकार्यम्
अन्विष्टाः	धेनवः
अपिहितम्	कर्णद्वयम्

सरमाया: शीलम् Summary in Marathi and English

प्रस्तावना :

प्राण्यांच्या चातुर्यकथा, नीतीकथा जगभरातील साहित्यात खूप प्रसिद्ध आहेत. रामायणातसुद्धा सागरी सेतू बांधण्यासाठी श्रीरामाला मदत करणाऱ्या खारीची गोष्ट येते. सरमा नामक कुत्रीची, तिच्या अविचल स्वामिनिष्ठेची कथा प्रसिद्ध आहे. पूर्वी यणी नामक असुरांनी देवांचे गोधन चोरले. ते शोधण्यासाठी सुपर्ण गरुडास देवांनी पणींच्या देशात पाठवले मात्र पणींनी सुपर्णाला लाच देऊन आपल्याकडे वळवले त्यामुळ देवांना गोधनाविषयी माहिती मिळाली नाही.

देवांनी या कामी सरमा नावाच्या कुत्रीला पाठवले. मार्गात येणाऱ्या वेगवान रसा नदीला ओलांडून, पणींच्या देशात निर्भयतेने शोध घेऊन सरमेने गोधन शोधले. पणींच्या लोभाला बळी न पडता सरमेने इंद्राला सर्व वृत्तांत सांगितला. कामात यशस्वी झालेल्या सरमेला इंद्राने वर दिला तर द्रोही गरुडाला शाप. सरमा ही कुत्री असूनही कार्यनिष्ठा, स्वामिनिष्ठा, धैर्य, स्वाभिमान या गुणांच्या जोरावर आदर्श ठरली तिच्या या कर्तृत्वाचे वर्णन पं. प्रभाकर भातखंडे यांनी संस्कृत पोवाड्यातून केले आहे. पोवाडा हा महाराष्ट्रातील लोककलाप्रकार असून त्यात वीररसाचे महत्त्वाचे स्थान आहे. शूरांच्या गाथा सांगण्यासाठी पोवाड्याची रचना केली जाते.

Stories advising morals through animals is very famous idea in literature all over the world. Infact several values are taught through the medium of animals and birds. This is dates back to the vedic period.

In the Ramayana too a story of squirrel helping the monkeys while building the bridge is known. The story of सरमा a dog from the ऋग्वेद is famous for her firm loyalty. It is said that some demons called ufuis once stole the cattle wealth of gods. Indra sent their eagle yup to look for the cattle. But the ufus bribed it and turned it on their side.

So Indra did not get any information about their cattle. Then Indra sent सरमा for the same task. सरमा accomplished her task fighting against all odds. Indra blessed ERCAT with a boon and cursed the eagle सुपर्ण..

सरमा set an example of courage, loyalty and commitment even if she was a bitch (female dog). This पोवाडा composed by पं. प्रभाकर भातखंडे celebrates T’s courageous behaviour. पोवाडाis a type of Marathi folk songs. The essence of bravery (a) is highlighted in these kind of songs. This present day creation is a beautiful example of Vedic story in a form that can reach the modern generation is very attractive manner.

शाहिरः – सभायां ………………… हे जी जी जी जी जी ।।

शाहिरः – सभायां देवराज इन्द्रः
चिन्तयति कः प्रेषितव्यः।
किकरो नैकोऽपि दृष्टः
यो भवेत् सदा स्वामिनिष्ठः।।

गायकवृन्दः – तदा?
शाहिरः – सेवायां तत्परा, देवगणप्रिया,
नियुक्ता गोधनप्राप्त्यर्थम्।
मघवता आज्ञप्ता सरमा,
प्रस्थिता देवशुनी सरमा। हे जी जी जी जी।।

गायकवृन्दः – प्रस्थिता देवशुनी सरमा। हे जी जी जी जी जी।
शाहिरः – तदा मार्गे दृष्टा, तटरुहतरुत्पाटिनी रसा
जलौघसमृद्ध्या पृथुलरभसा याति सरसा।।

रसां दृष्टा सरमया चिन्तितम् –
“अवश्यं गच्छेयं नदी यदि भवेत् अद्य सुसहा।
अतस्तां याचेऽहं विनयवचसा नम्रशिरसा।।
रसे, कुरु कृपाम्।
जलं तवेदम् । महागभीरम्।
कुरुष्व गाधम् । येन तरेयम्।
परं च तीरम् । अहं व्रजेयम् ।
स्वामिनः शोभनकार्यार्थम्,
आर्यगणमङ्गलकार्यार्थम्।

गायकवृन्दः – आर्यगणमङ्गलकार्यार्थम्। हे जी जी जी जी।।
रसा उवाच – “अहं च श्रेष्ठा, त्वं तु क्षुद्रा
न हि त्वदर्थ, जलप्रवाहम्।
करोमि अल्पम्, अपसर अपसर।।
श्रुत्वा तद् वाक्यं सरमया गर्जितम्।
“पश्य मे सामर्थ्य, कार्यनिष्ठासामर्थ्यम्
इति उक्त्वा सा जले अपतत्

शाहिरः – सा गता परं तीरम्।
रसा सा अधोमुखी भूता।
लज्जया म्लानमुखी जाता।
क्षुद्रया कथं पराभूता।
पश्यति क्षुद्रशुनीं सरमाम्।

गायकवृन्दः – पश्यति क्षुद्रशुनीं सरमाम्।
हे जी जी जी जी जी।।

गद्यम्: – एवं सरमा परतीरं प्राप्ता।
इतः पश्यति, ततः पश्यति
वलं पृच्छति, पणीन् पृच्छति।।
वनं गच्छति, गुहां प्रविशति।
गुहायां पणिभिर्यद्रुद्धं प्रभोः गोधनसर्वस्वम्।
समाप्तमन्वेषणकार्य, समाप्तमन्वेषणकार्यम्।।

गायकवृन्दः – प्रमुदिता इन्द्रशुनी सरमा हे जी जी जी जी जी।।
गद्यम्: – सरमया धेनवः अन्विष्टाः।
एतां वार्ताम् इन्द्रं वक्तुं प्रस्थिता सा।
परं वलैः सा अवरोधिता।

गायकवन्दः – हे सरमे, हे सरमे, दुग्धं पिबतु, दधि खादतु।
घृतं पिबतु, अत्र तिष्ठतु
गोधनस्य भागं तुभ्यं दद्यः।।

सरमा – सेविका नाहं लोभस्य,सेविका नाहं मोहस्य।
गायकवृन्दः – सेविका स्वामिनः इन्द्रस्य।
हे जी जी जी जी जी।।
सेविका स्वामिनः इन्द्रस्य।
हे जी जी जी जी जी।।

गद्यम्: – अन्ते पणिभि: सरमा निन्दिता।
“एषा सरमा जारु खादति।
“एषा सरमा जारु खादति।
सरमया किश्चत् तन्त्र श्रुतम्।
अपिहितं दीर्घ कर्णद्वयम्।
“वृत्तान्तं वक्तुं गमनमारभे।
प्रस्थिता सत्यप्रिया सरमा।

गायकवृन्दः – प्रस्थिता देवशुनी सरमा।
हे जी जी जी जी जी।।

अनुवादः

शाहीर – सभेमध्ये देवराज इंद्र (बसला आहे) कोणाला पाठवावे याचा विचार करत आहे. त्याला एकही सेवक दिसला
नाही – जो सदैव स्वामिनिष्ठ असेल.
गायकगण – तेव्हा मग पुढे?
शाहीर – सेवेमध्ये तत्पर असलेली, देवांची लाडकी, गोधन मिळवण्यासाठी नियुक्त झालेली, इंद्राने आशा दिलेली, सरमा देवांची कुत्री सरमा निघाली.
गायकगण – देवांची कुत्री सरमा निघाली.
शाहीर – तेव्हा तिने वाटेत काठावर वाढलेल्या झाडांना उन्मळून (पाडून) टाकणारी रसा (नदी) पाहिली. पाण्याच्या समृद्ध प्रवाहने (ती) रसा वेगाने वाहत होती. रसेला पाहून सरमेने विचार केला, “आज नदी सुसहा (तरुन जाण्यायोग्य झाली) झाली तर मी अवश्य जाईन. म्हणून मी तिला नम्र बोलण्याने, डोके टेकवून विनंती करते. रसे, कृपा कर, तुझे पाणी खूप खोल आहे. ते उथळ कर. जेणेकरुन मी पोहून जाईन. दुसऱ्या तीरावर मी जाईन. स्वामींच्या (इंद्राच्या) कार्यासाठी आर्यांच्या (देवांच्या) मंगल कामासाठी (मी जात आहे.)
गायकगण – आर्यांच्या (देवांच्या) मंगल कामासाठी (मी जात आहे.)
रसा म्हणाली – मी श्रेष्ठ, तू क्षुद्र जलप्रवाह तुझ्यासाठी नाही करणार उथळ! (कमी) मागे हो, मागे हो. ते वाक्य ऐकून सरमा गर्जली. माझे सामर्थ्य बघ, कार्य निष्ठेची शक्ती बघ असे म्हणून ती पाण्यात उतरली.
शाहीर – ती दुसऱ्या तीरावर गेली. त्या रसेने तोंड झुकवले. लाजेने तिचे तोंड उतरले. मी क्षुद्र कुत्रीकडून कशी पराभूत झाले? (असे म्हणत ती) क्षुद्र सरमेला पाहू लागली.
गायकगण – क्षुद्र सरमेला पाहू लागली.
गद्य – अशा रीतीने सरमा दुसऱ्या तीरावर गेली. इथे पाहिले, तिथे पाहिले. वलांना विचारले, पणींना विचारले, वनात गेली, गुहेत शिरली, गुहेमध्ये देवांचे गोधन पणींनी दडवून ठेवले होते. शोधकार्य संपले, शोधकार्य संपले.
गायकगण – इंद्राची कुत्री सरमा आनंदली.
गद्य – सरमेने गायी शोधल्या ही बातमी इंद्राला सांगण्यासाठी ती निघाली मात्र वलांनी तिला अडवले.
गायकगण – अगं सरमे, दूध पी, दही खा. तूप पी, इथेच थांब गोधनाचा भाग तुलाही देतो आम्ही.
सरमा – मी लोभाची सेविका नाही. मी मोहाची सेविका नाही. (मी लोभामोहाने बधणार नाही.)
गायकगण – इंद्राची सेविका.
इंद्राची सेविका.
गद्य – शेवटी पणींनी सरमेची निंदा केली ही सरमा मांस (क्षुद्रान) खाते. सरमेने ते काहीएक ऐकून घेतले नाही. तिने आपले दोन मोठे कान बंद केले. वृत्तांत सांगण्यासाठी मी जात आहे. सत्यप्रिय सरमा निघाली.
गायकगण – देवांची कुत्री सरमा निघाली.

Shahir : In the assembly, Indra the king of Gods thought who should be sent. He didn’t see any servant who would be ever loyal.
Chorus : Then?
Shahir : Sarama, who was prompt in service, liked by the gods was appointed to bring the cows. Sarama the dog of the Gods who was ordered by Indra set off.
Chorus : The dog of the Gods Sarama set off.
Shahir : Then on the way she saw Rasa (river) who uprooted trees that grow on the banks, flowing with force with a strong current. Seeing Rasa Sarama thought, I shall surely go today if the river becomes tolerable (less forceful). So, I shall ask her modestly with modest words) and head bent in modestly. ‘Oh, Rasa, please do a favour. Your water is very deep. Please make it shallow. By which I shall cross to the other bank. I shall go for the good task of the master, for the auspicious task of the noble ones.
Chorus : For the auspicious task of the noble ones.
Rasa said : “I am superior and you are inferior. I shall not reduce the water current, move away, move away.” Listening to those words, Sarama roared, “See my capacity, the power of my dedication.” saying this she fell (jumped) into the water.
Shahir : She went to the other bank. Rasa was ashamed. She became sad due to shame. How was she defeated by some one inferior. Looks at that lowly dog Sarama.
Chorus : (She) looks at that lowly dog sarama.
Prose : Thus Sarama reached the other bank. She looks here, she looks there. She asks the Vala, she asks the demons. She goes to the forest, she enters the cave. In the cave is all the Lord’s cattle wealth trapped by the demons. The search was over, the search was over.
Chorus: Sarama the God’sdog was delighted.
Prose : Sarama had found the cattle. She set off to inform this news to Indra. But she was stopped by the Valas.
Chorus: O Sarama, O Sarama. Drink milk, eat curds, drink ghee, wait here. We shall give you a part of the cattle wealth.
Sarama : I am not servant of greed. I am not servant of illusion
Chorus : Servant of Lord Indra.
Servant of Lord Indra.
Prose : Finally Sarama was criticised by the demons. “This Sarama eats flesh” But. Sarama paid no heed to all that. She closed the two long ears and started to depart to narrate the account. The truth lover Sarama set off.
Chorus : Sarama the god’s dog set off.

शब्दार्थाः

1. उपहार: – gift – बक्षीस, नजराणा
2. प्रेषितव्यः – should be sent – पाठवण्याजोगा
3. किड्करः – servant – नोकर, सेवक
4. गोधनम् – cattle wealth – गोधन
5. मघवान् – Lord Indra – इंद्रदेव
6. ‘तटरुहतरुः – tree that grows on the banks – झाड
7. जलौघः – water-current – पाण्याचा प्रवाह
8. पृथुलरभसा – with tremendous speed – अत्यंत वेगाने
9. विनयवचसा – with modest speech – विनयपूर्ण वाणीने
10. गभीरम् – deep – खोल
11. म्लानमुखी – sad – खिन्न
12. परतीरम् – to other bank – पलीकडच्या काठावर
13. वलम् – community of demons Pani – पणि नावाचे असूर
14. अन्विष्टाः – found – सापडल्या
15. घृतम् – ghee – तूप
16. जारुः – flesh/bad food – मांस, निकृष्ट खाणे

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Practice Set 4.5 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion

Question 1.
Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
Solution:
Let the number to be subtracted be x.
∴ (12 – x), (16 – x) and (21 – x) are in continued proportion.

∴ 84 – 4x = 80 – 5x
∴ 5x – 4x = 80 – 84
∴ x = -4
∴ -4 should be subtracted from 12,16 and 21 so that the resultant numbers in continued proportion.

Question 2.
If (28 – x) is the mean proportional of (23 – x) and (19 – x), then find the value ofx.
Solution:
(28 – x) is the mean proportional of (23 – x) and (19-x). …[Given]

∴ -5(19 – x) = 9(28 – x)
∴ -95 + 5x = 252 – 9x
∴ 5x + 9x = 252 + 95
∴ 14x = 347
∴ x = \(\frac { 347 }{ 14 }\)

Question 3.
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
Solution:
Let the first number be x.
∴ Third number = 26 – x
12 is the mean proportional of x and (26 – x).
∴ \(\frac { x }{ 12 }\) = \(\frac { 12 }{ 26 – x }\)
∴ x(26 – x) = 12 x 12
∴ 26x – x² = 144
∴ x² – 26x + 144 = 0
∴ x² – 18x – 8x + 144 = 0
∴ x(x – 18) – 8(x – 18) = 0
∴ (x – 18) (x – 8) = 0
∴ x = 18 or x = 8
∴ Third number = 26 – x = 26 – 18 = 8 or 26 – x = 26 – 8 = 18
∴ The numbers are 18, 12, 8 or 8, 12, 18.

Question 4.
If (a + b + c)(a – b + c) = a² + b² + c², show that a, b, c are in continued proportion.
Solution:
(a + b + c)(a – b + c) = a² + b² + c² …[Given]
∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a2 + b2 + c2
∴ a² – ab + ac + ab – b² + be + ac – be + c² = a² + b² + c²
∴ a² + 2ac – b² + c² = a² + b² + c²
∴ 2ac – b² = b²
∴ 2ac = 2b²
∴ ac = b²
∴ b² = ac
∴ a, b, c are in continued proportion.

Question 5.
If \(\frac { a }{ b }\) = \(\frac { b }{ c }\) and a, b, c > 0, then show that,
i. (a + b + c)(b – c) = ab – c²
ii. (a² + b²)(b² + c²) = (ab + be)²
iii. \(\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}\)
Solution:
Let \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = k
∴ b = ck
∴ a = bk =(ck)k
∴ a = ck² …(ii)

i. (a + b + c)(b – c) = ab – c²
L.H.S = (a + b + c) (b – c)
= [ck² + ck + c] [ck – c] … [From (i) and (ii)]
= c(k² + k + 1) c (k – 1)
= c² (k² + k + 1) (k – 1)
R.H.S = ab – c²
= (ck²) (ck) – c² … [From (i) and (ii)]
= c²k³ – c²
= c²(k³ – 1)
= c² (k – 1) (k² + k + 1) … [a³ – b³ = (a – b) (a² + ab + b²]
∴ L.H.S = R.H.S
∴ (a + b + c) (b – c) = ab – c²

ii. (a² + b²)(b² + c²) = (ab + bc)²
b = ck; a = ck²
L.H.S = (a² + b²) (b² + c²)
= [(ck²) + (ck)²] [(ck)² + c²] … [From (i) and (ii)]
= [c²k⁴ + c²k²] [c²k² + c²]
= c²k² (k² + 1) c² (k² + 1)
= c4k² (k² + 1)²
R.H.S = (ab + bc)²
= [(ck²) (ck) + (ck)c]² …[From (i) and (ii)]
= [c²k³ + c²k]²
= [c²k (k² + 1)]2 = c⁴(k² + 1)²
∴ L.H.S = R.H.S
∴ (a² + b²) (b² + c²) = (ab + bc)²

iii. \(\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}\)

9th Standard Algebra Practice Set 4.5 Question 6. Find mean proportional of \(\frac{x+y}{x-y}, \frac{x^{2}-y^{2}}{x^{2} y^{2}}\).
Solution:
Let a be the mean proportional of \(\frac{x+y}{x-y}\) and \(\frac{x^{2}-y^{2}}{x^{2} y^{2}}\)

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Practice Set 4.1 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion

Question 1.
From the following pairs of numbers, find the reduced form of ratio of first number to second number.
i. 72,60
ii. 38,57
iii. 52,78
Solution:
i. 72, 60
\(\text { Ratio }=\frac{72}{60}=\frac{12 \times 6}{12 \times 5}=\frac{6}{5}=6 : 5\)

ii. 38, 57
\(\text { Ratio }=\frac{38}{57}=\frac{19 \times 2}{19 \times 3}=\frac{2}{3}=2 : 3\)

iii. 52, 78
\(\text { Ratio }=\frac{52}{78}=\frac{26 \times 2}{26 \times 3}=\frac{2}{3}=2 : 3\)

Question 2.
Find the reduced form of the ratio of the first quantity to second quantity.
i. ₹ 700, ₹ 308
ii. ₹ 14, ₹ 12 and 40 paise
iii. 5 litres, 2500 ml
iv. 3 years 4 months, 5 years 8 months
v. 3.8 kg, 1900 gm
vi. 7 minutes 20 seconds, 5 minutes 6 seconds
Solution:
i. ₹ 700, ₹ 308
\( \text { Ratio }=\frac{700}{308}=\frac{28 \times 25}{28 \times 11}=\frac{25}{11}=25 : 11\)

ii. ₹ 14, ₹12 and 40 paise
₹ 14 = 14 x 100 paise = 1400 paise
₹ 12 and 40 paise = 12 x 100 paise + 40 paise
= (1200 + 40) paise
= 1240 paise

iii. 5 litres, 2500 ml
5 litres = 5 x 1000 ml = 5000ml

iv. 3 years 4 months, 5 years 8 months
3 years 4 months = 3×12 months + 4 months
= (36 + 4) months
= 40 months
5 years 8 months = 5 x 12 months + 8 months
= (60 + 8) months
= 68 months

v. 3.8 kg, 1900 gm
3.8 kg = 3.8 x 1000 gm = 3800 gm

vi. 7 minutes 20 seconds, 5 minutes 6 seconds
7 minutes 20 seconds = 7 x 60 seconds + 20 seconds
= (420 + 20) seconds
= 440 seconds
5 minutes 6 seconds = 5 x 60 seconds + 6 seconds
= (300 + 6) seconds
= 306 seconds

Question 3.
Express the following percentages as ratios
i. 75 : 100
ii. 44 : 100
iii. 6.25%
iv. 52: 100
v. 0.64%
Solution:

Question 4.
Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?
Solution:
Let the persons required to build a house in 6 days be x.
Days required to build a house and number of persons are in inverse proportion.
∴ 6 × x = 8 × 3
∴ 6 x = 24
∴ x = 4
∴ 4 persons are required to build the house in 6 days.

Question 5.
Convert the following ratios into percentages.
i. 15 : 25
ii. 47 : 50
iii. \(\frac { 7 }{ 10 }\)
iv. \(\frac { 546 }{ 600 }\)
v. \(\frac { 7 }{ 16 }\)
Solution:
Let 15 : 25 = x %

∴ 15 : 25 = 60 %

ii. Let 47 : 50 = x%

∴ 47 : 50 = 94 %

iii. Let \(\frac { 7 }{ 10 }\) = x %

iv. Let \(\frac { 546 }{ 600 }\) = x %

v. Let \(\frac { 7 }{ 16 }\) = x %

Question 6.
The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha’s birth her mothers age was 27 years. Find the present ages of Abha and her mother.
Solution:
The ratio of ages of Abha and her mother is 2 : 5.
Let the common multiple be x.
∴ Present age of Abha = 2x years and
Present age of Abha’s mother = 5x years
According to the given condition, the age of Abha’s mother at the time of Abha’s birth = 27 years
∴ 5x – 2x = 27
∴ 3x = 27
∴ x = 9
∴ Present age of Abha = 2x = 2 x 9 = 18 years
∴ Present age of Abha’s mother = 5x =5 x 9 = 45 years
The present ages of Abha and her mother are 18 years and 45 years respectively.

Question 7.
Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4?
Solution:
Present age of Vatsala = 14 years
Present age of Sara = 10 years
After x years,
Vatsala’s age = (14 + x) years
Sara’s age = (10 + x) years
According to the given condition,
After x years the ratio of their ages will become 5 : 4
∴ \(\frac { 14 + x }{ 10 + x }\) = \(\frac { 5 }{ 4 }\)
∴ 4(14 + x) = 5(10 + x)
∴ 56 + 4x = 50 + 5x
∴ 56 – 50 = 5x – 4x
∴ 6 = x
∴ x = 6
∴ After 6 years, the ratio of their ages will become 5 : 4.

Question 8.
The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana’s present age ?
Solution:
The ratio of present ages of Rehana and her mother is 2 : 7
Let the common multiple be x.
∴ Present age of Rehana = 2x years and Present age of Rehana’s mother = 7x years
After 2 years,
Rehana’s age = (2x + 2) years
Age of Rehana’s mother = (7x + 2) years
According to the given condition,
After 2 years, the ratio of their ages will be 1 : 3
∴ \(\frac { 2x + 2 }{ 7x + 2 }\) = \(\frac { 1 }{ 3 }\)
∴ 3(2x + 2) = 1(7x + 2)
∴ 6x + 6 = 7x + 2
∴ 6 – 2 = 7x – 6x
∴ 4 = x
∴ x = 4
∴ Rehana’s present age = 2x = 2 x 4 = 8 years
∴ Rehana’s present age is 8 years.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 7 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

Problem Set 7 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
Write the correct alternative answer for each of the following questions.

i. Which of the following data is not primary ?
(A) By visiting a certain class, gathering information about attendance of students.
(B) By actual visit to homes, to find number of family members.
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.
(D) Review the cleanliness status of canals by actually visiting them.
Answer:
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

ii. What is the upper class limit for the class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
Answer:
(B) 35

iii. What is the class-mark of class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
Answer:
(D) 30

iv. If the classes are 0 – 10, 10 – 20, 20 – 30, …, then in which class should the observation 10 be included?
(A) 0 – 10
(B) 10 – 20
(C) 0 – 10 and 10-20 in these 2 classes
(D) 20 – 30
Answer:
(B) 10 – 20

v. If \(\overline { x }\) is the mean of x₁, x₂, ……. , x_n and \(\overline { y }\) is the mean of y₁, y₂, ….. y_n and \(\overline { z }\) is the mean of x₁,x₂, …… , x_n , y₁, y₂, …. y_n , then z = ?

Answer:
x₁, x₂, x₃, ……. , x_n
∴ \(\overline{x}=\frac{\sum x}{\mathrm{n}}\)
∴ n\(\overline{x}\) = ∑x
Similarly, n\(\overline{y}\) = ∑y
Now,

\(\text { (A) } \frac{\overline{x}+\overline{y}}{2}\)

vi. The mean of five numbers is 50, out of which mean of 4 numbers is 46, find the 5th number.
(A) 4
(B) 20
(C) 434
(D) 66
Answer:
5^th number = Sum of five numbers – Sum of four numbers
= (5 x 50) – (4 x 46)
= 250 – 184
= 66
(D) 66

vii. Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean.
(A) 40.6
(B) 40.4
(C) 40.3
(D) 40.7
Answer:
New mean = \(\frac { 4000-30+70 }{ 100 }\)
= 40.4
(B) 40.4

viii. What is the mode of 19, 19, 15, 20, 25, 15, 20, 15?
(A) 15
(B) 20
(C) 19
(D) 25
Answer:
(A) 15

ix. What is the median of 7, 10, 7, 5, 9, 10 ?
(A) 7
(B) 9
(C) 8
(D) 10
Answer:
(C) 8

x. From following table, what is the cumulative frequency of less than type for the class 30 – 40?

(A) 13
(B) 15
(C) 35
(D) 22
Answer:
Cumulative frequency of less than type for the class 30 – 40 = 7 + 3 + 12 + 13 = 35
(C) 35

Question 2.
The mean salary of 20 workers is ₹10,250. If the salary of office superintendent is added, the mean will increase by ₹ 750. Find the salary of the office superintendent.
Solution:
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations
The mean salary of 20 workers is ₹ 10,250.
∴ Sum of the salaries of 20 workers
= 20 x 10,250
= ₹ 2,05,000 …(i)
If the superintendent’s salary is added, then mean increases by 750
new mean = 10, 250 + 750 = 11,000
Total number of people after adding superintendent = 20 + 1 = 21
∴ Sum of the salaries including the superintendent’s salary = 21 x 11,000 = ₹ 2,31,000 …(ii)
∴ Superintendent salary = sum of the salaries including superintendent’s salary – sum of salaries of 20 workers
= 2, 31,00 – 2,05,000 …[From (i) and (ii)]
= 26,000
∴ The salary of the office superintendent is ₹ 26,000.

Question 3.
The mean of nine numbers is 77. If one more number is added to it, then the mean increases by 5. Find the number added in the data.
Solution:
∴ \( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations mean of nine numbers is 77
∴ sum of 9 numbers = 11 x 9 = 693 …(i)
If one more number is added, then mean increases by 5
mean of 10 numbers = 77 + 5 = 82
∴ sum of the 10 numbers = 82 x 10 = 820 …(ii)
∴ Number added = sum of the 10 numbers – sum of the 9 numbers = 820 – 693 … [From (i) and (ii)]
= 127
∴ The number added in the data is 127.

Question 4.
The monthly maximum temperature of a city is given in degree Celsius in the following data. By taking suitable classes, prepare the grouped frequency distribution table
29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5.29.0. 29.5.29.9.33.2.30.2
From the table, answer the following questions.
i. For how many days the maximum temperature was less than 34°C?
ii. For how many days the maximum temperature was 34°C or more than 34°C?
Solution:

i. Number of days for which the maximum temperature was less than 34°C
= 8 + 8 + 8 = 24
ii. Number of days for which the maximum temperature was 34°C or more than 34°C
= 5 + 1 = 6

Question 5.
If the mean of the following data is 20.2, then find the value of p.

Solution:

∴ 20.2 (30 + p) = 610 + 20p
∴ 606 + 20.2p = 610 + 20p
∴ 20.2p – 20p = 610 – 606
∴ 0.2p = 4
∴ p = \(\frac { 4 }{ 0.2 }\) = \(\frac { 40 }{ 2 }\) = 20
∴ p = 20

Question 6.
There are 68 students of 9th standard from Model Highschool, Nandpur. They have scored following marks out of 80, in written exam of mathematics.
70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68,
80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43,
36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55,
56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78,
80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37,
45, 42, 70, 37,45, 66, 56, 47
By taking classes 30 – 40, 40 – 50, …. prepare the less than type cumulative frequency table. Using the table, answer the following questions:

i. How many students have scored marks less than 80?
ii. How many students have scored marks less than 40?
iii. How many students have scored marks less than 60?
Solution:
Class

i. 66 students have scored marks less than 80.
ii. 14 students have scored marks less than 40.
iii. 45 students have scored marks less than 60.

Question 7.
By using data in example (6), and taking classes 30 – 40, 40 – 50,… prepare equal to or more than type cumulative frequency table and answer the following questions based on it.
i. How many students have scored marks 70 or more than 70?
ii. How many students have scored marks 30 or more than 30?
Solution:

i. 11 students have scored marks 70 or more than 70.
ii. 68 students have scored marks 30 or more than 30.

Question 8.
There are 10 observations arranged in ascending order as given below.
45, 47, 50, 52, JC, JC + 2, 60, 62, 63, 74. The median of these observations is 53.
Find the value of JC. Also find the mean and the mode of the data.
Solution:
i. Given data in ascending order:
45,47, 50, 52, x, JC+2, 60, 62, 63, 74.
∴ Number of observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
∴
∴ 106 = 2x + 2
∴ 106 – 2 = 2x
∴ 104 = 2x
∴ x = 52
∴ The given data becomes:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.

∴ The mean of the given data is 55.9.

iii. Given data in ascending order:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.
∴ The observation repeated maximum number of times = 52
∴ The mode of the given data is 52.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Problem Set 7 Intext Questions and Activities

Question 1.
To show following information diagrammatically, which type of bar- diagram is suitable?
i. Literacy percentage of four villages.
ii. The expenses of a family on various items.
iii. The numbers of girls and boys in each of five divisions.
iv. The number of people visiting a science exhibition on each of three days.
v. The maximum and minimum temperature of your town during the months from January to June.
vi. While driving a two-wheeler, number of people wearing helmets and not wearing helmet in 100 families.
(Textbook pg. no. 112)
Solution:
i. Percentage bar diagram
ii. Sub-divided bar diagram
iii. Sub-divided bar diagram
iv. Sub-divided bar diagram
v. Sub-divided bar diagram
vi. Sub-divided bar diagram

Question 2.
You gather information for several reasons. Take a few examples and discuss whether the data is primary or secondary.
(Textbook pg. no, 113)
[Students should attempt the above activity on their own.]

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Practice Set 4.2 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion

Question 1.
Using the property \(\frac { a }{ b }\) = \(\frac { ak }{ bk }\), fill in the blanks by substituting proper numbers in the following.

Solution:

Question 2.
Find the following ratios.
i. The ratio of radius to circumference of the circle.
ii. The ratio of circumference of circle with radius r to its area.
iii. The ratio of diagonal of a square to its side, if the length of side is 7 cm.
iv. The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of numbers denoting its perimeter to area.
Solution:
i. Let the radius of circle be r.
then, its circumference = 2πr
Ratio of radius to circumference of the circle

The ratio of radius to circumference of the circle is 1 : 2π.

ii. Let the radius of the circle is r.
∴ circumference = 2πr and area = πr²
Ratio of circumference to the area of circle

∴ The ratio of circumference of circle with radius r to its area is 2 : r.

iii. Length of side of square = 7 cm
∴ Diagonal of square = √2 x side
= √2 x 7
= 7 √2 cm
Ratio of diagonal of a square to its side

∴ The ratio of diagonal of a square to its side is √2 : 1.

iv. Length of rectangle = (l) = 5 cm,
Breadth of rectangle = (b) = 3.5 cm
Perimeter of the rectangle = 2(l + b)
= 2(5 + 3.5)
= 2 x 8.5
= 17 cm
Area of the rectangle = l x b
= 5 x 3.5
= 17.5 cm²
Ratio of numbers denoting perimeter to the area of rectangle

∴ Ratio of numbers denoting perimeter to the area of rectangle is 34 : 35.

Question 3.
Compare the following

Solution:

Question 4.
Solve.
ABCD is a parallelogram. The ratio of ∠A and ∠B of this parallelogram is 5 : 4. FInd the measure of ∠B. [2 Marksl
Solution:
Ratio of ∠A and ∠B for given parallelogram is 5 : 4
Let the common multiple be x.

m∠A = 5x°and m∠B=4x°
Now, m∠A + m∠B = 180° …[Adjacent angles of a parallelogram arc supplementary]
∴ 5x° + 4x°= 180°
∴ 9x° = 180°
∴ x° = 20°
∴ m∠B=4x°= 4 x 20° = 80°
∴ The measure of ∠B is 800.

ii. The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages.
Solution:
The ratio of present ages of Albert and Salim is 5 : 9
Let the common multiple be x.
∴ Present age of Albert = 5x years and
Present age of Salim = 9x years
After 5 years,
Albert’s age = (5x + 5) years and
Salim’s age = (9x + 5) years
According to the given condition,
Five years hence ratio of their ages will be 3 : 5
\(\frac{5 x+5}{9 x+5}=\frac{3}{5}\)
∴ 5(5x + 5) = 3(9x + 5)
∴ 25x + 25 = 27x + 15
∴ 25 – 15 = 27 x – 25 x
∴ 10 = 2x
∴ x = 5
∴ Present age of Albert = 5x = 5 x 5 = 25 years
Present age of Salim = 9x = 9 x 5 = 45 years
∴ The present ages of Albert and Salim are 25 years and 45 years respectively.

iii. The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle.
Solution:
The ratio of length and breadth of a rectangle is 3 : 1
Let the common multiple be x.
Length of the rectangle (l) = 3x cm
and Breadth of the rectangle (b) = x cm
Given, perimeter of the rectangle = 36 cm
Since, Perimeter of the rectangle = 2(l + b)
∴ 36 = 2(3x + x)
∴ 36 = 2(4x)
∴ 36 = 8x
∴ \(x=\frac{36}{8}=\frac{9}{2}=4.5\)
Length of the rectangle = 3x = 3 x 4.5 = 13.5 cm
∴ The length of the rectangle is 13.5 cm and its breadth is 4.5 cm.

iv. The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers.
Solution:
The ratio of two numbers is 31 : 23
Let the common multiple be x.
∴ First number = 31x and
Second number = 23x
According to the given condition,
Sum of the numbers is 216
∴ 31x + 23x = 216
∴ 54x = 216
∴ x = 4
∴ First number = 31x = 31 x 4 = 124
Second number = 23x = 23 x 4 = 92
∴ The two numbers are 124 and 92.

v. If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers.
Solution:
Ratio of two numbers is 10 : 9
Let the common multiple be x.
∴ First number = 10x and
Second number = 9x
According to the given condition,
Product of two numbers is 360
∴ (10x) (9x) = 360
∴ 90x² = 360
∴ x² = 4
∴ x = 2 …. [Taking positive square root on both sides]
∴ First number = 10x = 10x² = 20
Second number = 9x = 9x² = 18
∴ The two numbers are 20 and 18.

Question 5.
If a : b = 3 : 1 and b : c = 5 : 1, then find the value of [3 Marks each]

Solution:
Given, a : b = 3 : 1
∴ \(\frac { a }{ b }\) = \(\frac { 3 }{ 1 }\)
∴ a = 3b ….(i)
and b : c = 5 : 1
∴ \(\frac { b }{ c }\) = \(\frac { 5 }{ 1 }\)
b = 5c …..(ii)
Substituting (ii) in (i),
we get a = 3(5c)
∴ a = 15c …(iii)

Ratio and Proportion 9th Class Practice Set 4.1 Question 6. If \(\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}\) , then find the ratio \(\frac { a }{ b }\).
Solution:
\(\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}\) … [Given]
∴ 0.04 x 0.4 x a = (0.4)² x (0.04)² x b … [Squaring both sides]

9th Algebra Practice Set 4.2 Question 7. (x + 3) : (x + 11) = (x – 2) : (x + 1), then find the value of x.
Solution:
(x + 3) : (x + 11) = (x- 2) : (x+ 1)
\(\quad \frac{x+3}{x+11}=\frac{x-2}{x+1}\)
∴ (x + 3)(x +1) = (x – 2)(x + 11)
∴ x(x +1) + 3(x + 1) = x(x + 11) – 2(x + 11)
∴ x² + x + 3x + 3 = x² + 1 lx – 2x – 22
∴ x² + 4x + 3 = x² + 9x – 22
∴ 4x + 3 = 9x – 22
∴ 3 + 22 = 9x – 4x
∴ 25 = 5x
∴ x = 5

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 2.3 8th Std Maths Answers Solutions Chapter 2 Parallel Lines and Transversals.

Practice Set 2.3 8th Std Maths Answers Chapter 2 Parallel Lines and Transversals

Question 1.
Draw a line l. Take a point A outside the line. Through point A draw a line parallel to line l.
Solution:
Steps of construction:

1. Draw a line l and take any point A outside the line.
2. Place a set-square, such that one arm of the right angle passes through A and the other arm is on line l.
3. Place the second set-square as shown in the figure such that the vertex of the right angle is at point A.
4. Hold the two set-squares in place and draw a line parallel to line l through the edge of the second set-square. Name the line as m.

Line m is the required line parallel to line l and passing through point A.

Question 2.
Draw a line l. Take a point T outside the line. Through point T draw a line parallel to line l.
Solution:
Steps of construction:

1. Draw a line l and take any point T outside the line.
2. Place a set-square, such that one arm of the right angle passes through T and the other arm is on line l.
3. Place the second set-square as shown in the figure such that the vertex of the right angle is at point T.
4. Hold the two set-squares in place and draw a line parallel to line l through the edge of the second set-square. Name the line as m.

Line m is the required line parallel to line l and passing through point T.

Question 3.
Draw a line m. Draw a line n which is parallel to line m at a distance of 4 cm from it.
Solution:
Steps of construction:

1. Draw a line m and take any two points M and N on the line.
2. Draw perpendiculars to line m at points M and N.
3. On the perpendicular lines take points S and T at a distance 4 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 m at a distance of 4 cm from it.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.2 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

Practice Set 1.2 8th Std Maths Answers Chapter 1 Rational and Irrational Numbers

Question 1.
Compare the following numbers.
i. 7, -2
ii. 0, \(\frac { -9 }{ 5 }\)
iii. \(\frac { 8 }{ 7 }\), 0
iv. \(-\frac{5}{4}, \frac{1}{4}\)
v. \(\frac{40}{29}, \frac{141}{29}\)
vi. \(-\frac{17}{20},-\frac{13}{20}\)
vii. \(\frac{15}{12}, \frac{7}{16}\)
viii. \(-\frac{25}{8},-\frac{9}{4}\)
ix. \(\frac{12}{15}, \frac{3}{5}\)
x. \(-\frac{7}{11},-\frac{3}{4}\)
Solution:
i. 7, -2
If a and b are positive numbers such that a < b, then -a > -b.
Since, 2 < 7 ∴ -2 > -7

ii. 0, \(\frac { -9 }{ 5 }\)
On a number line, \(\frac { -9 }{ 5 }\) is to the left of zero.
∴ 0 > \(\frac { -9 }{ 5 }\)

iii. \(\frac { 8 }{ 7 }\), 0
On a number line, zero is to the left of \(\frac { 8 }{ 7 }\) .
∴ \(\frac { 8 }{ 7 }\) > 0

iv. \(-\frac{5}{4}, \frac{1}{4}\)
We know that, a negative number is always less than a positive number.
∴ \(-\frac{5}{4}<\frac{1}{4}\)

v. \(\frac{40}{29}, \frac{141}{29}\)
Here, the denominators of the given numbers are the same.
Since, 40 < 141
∴ \(\frac{40}{29}<\frac{141}{29}\)

vi. \(-\frac{17}{20},-\frac{13}{20}\)
Here, the denominators of the given numbers are the same.
Since, 17 < 13
∴ -17 < -13
∴ \(-\frac{17}{20}<-\frac{13}{20}\)

vii. \(\frac{15}{12}, \frac{7}{16}\)
Here, the denominators of the given numbers are not the same.
LCM of 12 and 16 = 48

Alternate method:
15 × 16 = 240
12 × 7 = 84
Since, 240 > 84
∴ 15 × 16 > 12 × 7

viii. \(-\frac{25}{8},-\frac{9}{4}\)
Here, the denominators of the given numbers are not the same.
LCM of 8 and 4 = 8

ix. \(\frac{12}{15}, \frac{3}{5}\)
Here, the denominators of the given numbers are not the same.
LCM of 15 and 5 = 15

x. \(-\frac{7}{11},-\frac{3}{4}\)
Here, the denominators of the given numbers are not the same.
LCM of 11 and 4 = 44

Maharashtra Board Class 8 Maths Solutions Chapter 1 Rational and Irrational Numbers Practice Set 1.2 Questions and Activities

Question 1.
Verify the following comparisons using a number line. (Textbook pg. no, .3)
i. 2 < 3 but – 2 > – 3
ii. \(\frac{5}{4}<\frac{7}{4}\) but \(\frac{-5}{4}<\frac{-7}{4}\)
Solution:

We know that, on a number line the number to the left is smaller than the other.
∴ 2 < 3 and -3 < -2
i.e. 2 < 3 and -2 > -3