Prasanna

MP Board Class 7th Science Solutions Chapter 18 Wastewater Story

Activities

Activity – 1
We have given one example of the use of clean water. You can add many more?

Answer:

Activity – 2
Locates an open drain near your home, school or on the roadside and inspect water flowing through it? Record color, odour and any other observation. Discuss with your friends and your teacher and fill up the following Table?
Table: Contaminant survey

Wastewater Story Text Book Exercises

Question 1.
Fill in the blanks:

1. Cleaning of water is a process of removing ……………………….
2. Waste – water released by houses is called ………………………
3. Dried ……………………… is used as manure.
4. Drains get blocked by …………………….. and ………………………..

Answer:

1. Contaminants
2. Sewage
3. Dung
4. Plastic, sludge.

Question 2.
What is sewage? Explain why it is harmful to discharge untreated sewage into rivers or seas?
Answer:
Sewage is waste – water released by homes, hospitals, offices, industries and other users. It also includes rainwater that has run down the street during a heavy rain or storm. The water that washes off roads and roof tops carries harmful substances with it. Basically sewage is a liquid waste. Most of it is water, which has dissolved and suspended impurities which are called contaminants. That is why it is harmful to discharge untreated sewage in to rivers or seas.

Question 3.
Why should oils and fats be not released in the drain? Explain.
Answer:
Oils and fats should not be released in the drains because they harden the soil in the pipes and block them. Fats get clogged in the holes of the soil in the drain and block it. It does not allow the waste – water to flow and thus the whole sewer system is blocked.

Question 4.
Describe the steps involved in getting clarified water from waste – water?
Answer:
Treatment of waste – water involves physical, chemical, and biological processes, which remove physical, chemical and biological matter that contaminates the waste – water.

1. Waste – water is passed through bar screens. Large objects like rags, sticks, cans, plastic packets, napkins are removed.

2. Water goes to a grit and sand removal tank. The speed of the incoming waste-water is decreased to allow sand, grit and pebbles to settle down.

3. The water is then allowed to settle in a large tank which is sloped towards the middle. Solids like faces settle at the bottom and are removed with a scraper. This is the sludge. A skimmer removes the float able solids like oil and grease. Water so cleared is called clarified water.

4. The sludge is transferred to a separate tank where it is decomposed by the anaerobic bacteria. The bio gas produced in the process can be used as fuel or can be used to produce electricity.

5. Air is pumped into the clarified water to help aerobic bacteria to grow. Bacteria consume human waste, food waste, soaps and other unwanted matter still remaining in clarified water. After several hours, the suspended microbes settle at the bottom of the tank as activated sludge. The water is then removed from the top.

Question 5.
What is sludge? Explain how it is treated?
Answer:
Sludge is the collected solid waste from the waste – water during the treatment in water treatment plant. Sludge is decomposed in a separate tank by the anaerobic bacteria. The activated sludge is about 97% water. The water is removed by sand drying beds or machines. Dried sludge is used as manure, returning organic matter and nutrients to the soil.

The treated water has a very low level of organic material and suspended matter. It is discharged into a sea, a river or into the ground. Nature cleans it up further. Sometimes it may be necessary to disinfect water with chemicals like chlorine and ozone before releasing it into the distribution system.

Question 6.
Untreated human excreta is a health hazard? Explain?
Answer:
Untreated human excreta is a health hazard. It may cause water pollution and soil pollution. Both the surface water and groundwater get polluted. Groundwater is a source of water for wells, tubewells, springs and many rivers. Thus, it becomes the most common route for water borne diseases. They include chlorea, typhoid, polio, meaningities, hepatitis and dysentery.

Question 7.
Name two chemicals used to disinfect water?
Answer:
Ozone, chlorine.

Question 8.
Explain the function of bar screens in a waste – water treatment plant?
Answer:
Bar screens clear the waste – water of all the physical impurities. Large waste objects like napkins, plastics, can sticks, rags etc. are removed from the waste – water through the bar screens.

Question 9.
Explain the relationship between sanitation and disease?
Answer:
Untreated human excreta is a health hazard. It may cause water pollution and soil pollution. Both the surface water and groundwater get polluted. Groundwater is a source of water for walls, tubewells, springs and many rivers. Thus, it becomes the most common route for water borne diseases. They include cholera, typhoid, polio, meningitis, hepatitis and dysentery. That is proper sanitation is must to avoid some of the deadliest diseases.

Question 10.
Outline your role as an active citizen in relation to sanitation?
Answer:
As active citizen we should take care of our personal and environmental sanitation. We should make people around us, aware of the benefits of sanitation we should help the municipal corporations and gram panchayats to cover all the open drains and remove the unhygienic and disease causing substances thrown in open.

Question 11.
Here is a crossword puzzle: Good luck!

Across

1. Liquid waste products
2. Solid waste extracted in sewage treatment
3. A word related to hygiene
4. Waste matter discharged from human body.

Down

1. Used water
2. A pipe carrying sewage.
3. Micro – organisms which causes cholera.
4. A chemical to disinfect water.

Answer:

Question 12.
Study the following statements about ozone:

1. It is essential for breathing of living organisms.
2. It is used to disinfect water.
3. It absorbs ultraviolet rays.
4. Its proportion in air is about 3%.

Which of these statements are correct?

1. (a), (b) and (c)
2. (b) and (c)
3. (a) and (d)
4. All four.

Answer:

2. (b) and (c).

Extended Learning – Activities and Projects

Question 1.
Construct a crossword puzzle of your using the keywords?
Answer:

Across:

1. Mixing with air.
2. Decomposed product of leaves.
3. Does not read oxygen
4. Needs oxygen.

Down:

1. Necessary for hygiene
2. Solid waste
3. Pipes to carry sewage

Question 2.
Then and now: Talk to your grand parents and other elderly people in the neighbourhood? Find out the sewage disposal systems available to them. You can also write letters to people living in far off places to get more information. Prepare a brief report on the information you collected?
Answer:
Do yourself.

Question 3.
Visit a sewage treatment plant?
It could be as exciting and enriching as a visit to a zoo, a museum, or a park? To guide your observation here are a few suggestions?

Record in your notepad:
Place ……………….. Date …………………… Time
Name of the official at the plant ………………………. Guide/Teacher …………………………

1. The location of the sewage plant.
2. Treatment capacity.
3. The purpose of screening as the initial process?
4. How is air bubbled through the aeration tank?
5. How safe is the water at the end of the treatment? How is it tested?
6. Where is the water discharged after treatment?
7. What happens to the plant during heavy rains?
8. Is bio gas consumed within the plant or sold to other consumers?
9. What happens to the treated sludge?
10. Is there any special effort to protect nearby houses from the plant?
11. Other observations.

Answer:
Do with the help of your subject teacher.

Wastewater Story Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative :

Question (a)
This is not the cause of water pollution?
(a) Floods
(b) Rains
(c) Chemicals
(d) Open defecation.
Answer:
(b) Rains

Question (b)
To improve sanitation following new technique is being used?
(a) Vermi – processing toilets
(b) Sewer system
(c) Onsite sewage disposal
(d) Both (a) and (b).
Answer:
(d) Both (a) and (b).

Question (c)
Following should not be disposed off in the drains?
(a) Tissue Papers
(b) Excreta
(c) Oils and Fats
(d) Waste water.
Answer:
(c) Oils and Fats

Question (d)
Following process is not a part of waste – water treatment?
(a) Decomposition
(b) Grit and Sand Removal
(c) Evaporation
(d) Chlorination.
Answer:
(c) Evaporation

Question (e)
Polluted water causes disease likes –
(a) Hepatitis
(b) Typhopid
(c) Cholera
(d) Diarrhoea.
Answer:
(c) Cholera

Question (f)
In a public sewerage system, the largest sewers are –
(a) Filters
(b) Interceptors
(c) Waterways
(d) None of these.
Answer:
(b) Interceptors

Question 2.
Fill in the blanks:

1. …………………… bacteria is used to treat sludge.
2. Sewage is a liquid waste which causes water and soil
3. Waste – water is treated in a sewage treatment ………………………
4. …………………… and …………………….. are the products of water clarification.
5. By – products of waste – water treatment are sludge and ………………………
6. Addition of disease causing organisms in water is called water …………………….
7. …………………………… in open cause health hazards.
8. ……………………….. is used as manure.
9. To improve sanitation, low cost ……………………….. sewage disposal systems are being encouraged.
10. Adopting good sanitation practices should be our way of ……………………………….

Answer:

1. Anaerobic
2. Pollution
3. Plant
4. Sludge and bio gas
5. Bio – gas
6. Contamination
7. Defecation
8. Activated sludge
9. Onsite
10. Life.

Question 3.
Which of the following statements are true (T) or false(F):

1. Sewage contains pure water for drinking.
2. Used water is waste – water.
3. Waste – water could not be reused.
4. Open drain system is a breeding place for flies.
5. Manholes are located at every 50 m to 60 m in the sewerage.
6. Waste – water is passed through for screens.
7. Eucalyptus trees absorb all surplus waste – water rapidly and release pure water vapour into the atmosphere.

Answer:

1. False
2. True
3. False
4. True
5. True
6. True
7. True.

Wastewater Story Very Short Answer Type Questions

Question 1.
Define sewage?
Answer:
Sewage is water that contains waste products produced by human beings. It is also called waste water.

Question 2.
Which is the world water day?
Answer:
22nd March.

Question 3.
Which is proclaimed as the International Decade for action on water for life?
Answer:
United Nations proclaimed the period 2005 – 2015 as the international Decade for action on “Water for life”.

Question 4.
What do you mean by cleaning of water?
Answer:
Cleaning of water is a process of removing pollutants before it enters a water body or is reused.

Question 5.
What is sewage treatment?
Answer:
This process of wastewater treatment is commonly known as “Sewage Treatment”.

Question 6.
In how many steps sewage treatment divided?
Answer:
The sewage treatment in most cities involve two main steps primary and secondary treatment. Some cities also require an additional step called tertiary treatment.

Question 7.
What do you mean by primary treatment?
Answer:
The primary treatment removes the heaviest solid material from sewage. This process removes about half the suspended solids and bacteria in sewage. Sometimes chlorine gas is added to kill most of the remaining bacteria.

Question 8.
What do you mean by secondary treatment?
Answer:
The secondary treatment removes from 85% to 90% of the solids and oxygen consuming wastes remaining in sewage after it has undergone primary treatment. The most common methods of secondary treatment are the activated sludge process and the trickling filtration process.

Question 9.
Why should we plant eucalyptus along sewage ponds?
Answer:
These trees absorb all surplus waste – water rapidly and release pure water vapour into the atmosphere.

Question 10.
Write certain inorganic impurities in the waste – water?
Answer:
Metals, phosphates and nitrates.

Question 11.
Name certain disease causing micro – organism?
Answer:
Bacterias, Viruses etc.

Question 12.
Which process removes the solids like faces and other substances from the waste – water?
Answer:
Grit and sand removal tank.

Question 13.
How is sand, grit on pebbles settled down?
Answer:
Water goes to a grit and sand removal tank. The speed of the incoming waste-water is decreased to allow sand, grit and pebbles to settle down.

Question 14.
How is dry sludge used?
Answer:
Dried sludge is used as manure, returning organic matter and nutrients to the soil.

Question 15.
Who decomposes the sludge?
Answer:
Anaerobic bacteria decompose the sludge.

Question 16.
Which instrument is used to remove floatable solids from the waste – water?
Answer:
A skimmer is used to remove floatable impurities.

Question 17.
What helps to clean the clarified water?
Answer:
Aerobic bacteria helps to clean the clarified water.

Question 18.
Why is ozone and chlorine used?
Answer:
Ozone and chlorine is used to kill the bacteria etc. present in the clarified water.

Question 19.
Why is air pumped to clarified water?
Answer:
Air is pumped into the clarified water to help aerobic bacteria to grow. Bacteria consume human waste, food waste, soaps and other unwanted matter still remaining in clarified water.

Wastewater Story Short Answer Type Questions

Question 1.
Explain sewage?
Answer:
Waste water including human excreta which flows from our homes into the drains is called domestic sewage. This contains microbes which cause water – borne diseases. This waste water is often dumped into water bodies.

Question 2.
How is water polluted?
Answer:
Water is used various purposes in homes, industries and agriculture. When water is used for cleaning, bathing, washing, dying etc. it pollutes the water. Unwanted waste materials and chemicals etc get added in the water and this wastes the water.

Question 3.
What is done to improve sanitation?
Answer:
To improve sanitation, low cost onsite sewage disposal stem are being encouraged. Examples are septic tanks, chemical toilets, composting pits. Septic tanks are suitable for places where there is no sewerage system, for hospitals isolated buildings or a Luster of 4 to 5 houses.

Question 4.
What is vermi process toilet?
Answer:
A design of a toilet in which humans excreta is treated by earthworms has been tested in India. It has been found to be a novel, j low water – use toilet for safe processing of human waste. The operation of the toilet is very simple and hygienic. The human excreta is completely converted to vermi cakes resource much needed for soil.

Wastewater Story Long Answer Type Questions

Question 1.
How defection in open cause health hazards?
Answer:
Due to lack of proper sewage disposal system a large amount of people in India defecates in open. They use riverbeds, railway lines, fields and drains for this purpose. These excreta dries down and percolate in soil with rain water. It pollutes the ground water.

Excreate along river bed pollutes the river water. In this way water on the ground and under the ground get polluted. This polluted water contains the micro – organisms of various communicable diseases like cholera, typhoid, hepatitis and meaning it is dysentery etc.

Question 2.
Suggest some better house keeping practices?
Answer:

1. Cooking oil and fats should not be thrown down the drain. They can harden and block the pipes. In an open drain the fats dog the soil pores redunt. Its effectiveness in filtering water. Oil and fats should he thrown in the dustbin.

2. Chemicals like paints, vents, insecticides, motor oil, medicines may kill microbes it help purity water. So they should not be thrown in the drain.

3. Used ten – leaves, solid food remains, soft toys, cotton, sanitary towels, etc. should also be thrown in the dustbin. These waste choke the drains. They do not allow free flow of oxygen. This hampers the degradation process.

Question 3.
What is the composition of sewage?
Answer:
Sewage is a complex mixture containing suspended solids, organic and inorganic impurities. nutrients, saprotrophic and disease causing bacteria and other microbes.

1. Organic impurities:
Human faces, animal waste, oil, urea (urine), pesticides, herbicides, fruit and vegetable waste, etc.

2. Inorganic impurities:
Nitrates, phosphates, metals. Nutrients Phosphorus and nitrogen.

3. Bacteria:
Such as which cause cholera and typhoid.

4. Other microbes:
Such as which cause dysentery.

Question 4.
How is sewage treated?
Answer:
Domestic sewage should be treated before being discharged into the river. Sewage is treated by first separating the solid material by sedimentation and filtration. Compressed air is then passed through the liquid which is then chlorinated to kill micro – organisms. The solid matter separated from sewage can be used to generated bio gas which can be used as fuel. The sludge that is left can be used as manure in the fields to grow organic foods.

Question 5.
What are the different ways in which solid waste can be disposed off?
Answer:
The different ways are as:

1. Domestic wastes like fruit and vegetable waste, leftover food, leaves of potted plants can be converted into compost and used as manure.
2. Most of the solid waste is buried in low lying areas to level uneven land. This is called landfill.
3. Wastes coming from industries such as metals can be recycled and used again.
4. Broken plastic articles like plastic bags, buckets, bowls, cups, plates, etc. can be melted and remolded to make new articles.
5. The waste disposal on a large scale is done by the municipality of a city using incinerators. The solid waste is burnt at high temperature. Ash is removed from time to time.

Question 6.
Explain sewage treatment plant.
Answer:
Sewage is water that contains waste products by human beings. It is also known as waste water. In fact sewage comes from the sinks and toilets of homes, restaurants, factories and office buildings. The sewage mainly consists of dissolved material that cannot be seen and bits of such solid matter as human waste and ground up rubbish.

It also contains harmful chemical and disease producting bacteria. Most sewage ultimately goes into lakes, rivers and oceans. However, in many western nations, the sewage is treated in some way before it goes into the waterways as a semi – clear liquid called effluent. Most methods used to treat sewage convert organic sewage into inorganic compounds viz. nitratres, sulphates and phosphates. Some of these compounds serve as food for algae. As the algae decay using excess of oxygen from water, the fish and plants in water will ultimately die.

Question 7.
Explain rural sewerage system?
Answer:
Many rural areas not served by public sewers. In such areas, most home owners use septic tanks to treat their sewage. These tanks are concrete or steel containers buried underground at home and buildings. Sewage flows into a septic tank through a pipe connecting the tank with a building.

Solids in the sewage sink to the bottom of the tank as sludge or float to the surface as scum. Effluent then flows from the tank into a system of pipes with open joints that allow sewage effluent to be gradually distributed into the soil. The soil bacteria then destroy the remaining organic material in the influent.

In a septic tank, bacteria in the sewage attack and digest the sludge and scum. The digestion process changes most to the wastes into gas and a harmless substance called humus. The gas then escapes into the air. The humus in the tank should be pumped our periodically and taken to a sewage treatment plant.

Question 8.
Explaining urban sewerage system?
Answer:
In a public sewerage system, the largest sewers, known as interceptors, carry the sewage to a wastewater treatment plant. The sewage treatment in most cities involves two main steps, primary treatment and secondary treatment. Some cities also require an additional step called tertiary treatment. At a treatment plant, sewage first passes through a screen that traps the largest pieces of matter. It then flows through a grit chamber, where heavy inorganic matter, such as sand, settles down.

The liquid next flows into a large primary sedimentations tank. Many suspended solids sink to the bottom of this tank and form a muddy material called sludge. Grease floats to the surface, where it is removed by a process called skimming. The effluent is then released into waterways.

Primary treatment removes bout half of the suspended solids and bacteria in sewage. Sometimes chlorine gas is added after primary or secondary treatment to kill most of the remaining bacteria. The secondary treatment removes about 85 to 90 percent of the solids and oxygen consuming wastes remaining in sewage after it has undergone primary treatment.

Question 9.
Explain the most common methods of secondary treatment in urban sewerage system?
Answer:
The most common methods of secondary treatment are:

1. The activated sludge process.
2. The trickling filtration process.

In case of activated sludge process, the influent from the primary sedimentation tank flows into a second tank called an aeration tank. The useful bacteria move through the liquid and change the organic matter into less harmful substances. The liquid then flows into a final sedimentation tank, where the sludge settles down to the bottom.

The influent is then discharged into waterways. In case of trickling filtration process, the filters are filled with crushed rocks. As sewage is distributed over the rocks, it reacts with slime that develops on the rocks. The slime contains useful bacteria that change organic material in the sewage into less harmful substances. These substances are removed in a final sedimentation tank, where they fall to the bottom as sludge.

Sometimes tertiary treatment is also used after primary and secondary treatment to produce purer effluent. The tertiary treatment methods include chemical treatment, microscopic screening, radiation treatment, etc. Tertiary treatment makes effluent safer to discharge into waterways.

MP Board Class 7th Science Solutions

MP Board Class 6th General English Essay Writing

The Cow (द काउ)
गाय

1. The cow is a pet animal.
2. She is four footed animal.
3. She has two eyes and two horns.
4. She has a long tail and two big ears.
5. She eats grass, leaves, corn and hay.
6. She gives us milk.
7. She gives us calves.
8. Her calves are used for agriculture.
9. Curd, butter and ghee is made from her milk.
10. Hindus worship the cow.
11. We call her ‘Gau Mata’.

Meanings of Difficult Words (कठिन शब्दों के अर्थ) :
Pet (पैट) = पालतू, animal (एनिमल) = जानवर, four footed (फोर फुटेड) = चौपाया, eyes (आइज़) = आँखें, horns (हॉर्स) = सींग, long (लौंग) = लम्बा, tail (टेल) = पूँछ, big (बिग) = बड़ा, ears (ईयर्स) = कान, eat (ईट) = खाना, grass (ग्रास) = घास, leaves (लीव्ज़) = पत्तियाँ, corn (कॉर्न) = अनाज, hay (हे) = भूसा, give (गिव) = देना, milk (मिल्क) = दूध, calves (काव्ज़) = बछड़े, used (यूज्ड) = प्रयोग किया जाता है, agriculture (एग्रीकल्चर) = कृषि, curd (कर्ड) = दही, butter (बटर) = मक्खन, ghee (घी) = घी, Hindu (हिन्दू) = हिन्दू, worship (वर्शिप) = पूजा करते हैं, Gau Mata (गऊ माता) = गऊ माता।

The Dog (द डॉग)
कुत्ता

1. The dog is a pet animal.
2. It is found all over the world.
3. It is found in many colors such as black, white, brown and black and white.
4. It has four legs, a tail, two eyes and two ears.
5. Its has sharp teeth also.
6. It eats bread, milk and meat.
7. It watches its master’s house.
8. It barks when its sees a stranger.
9. The dog is a good friend of man.
10. It is very useful and faithful animal.

Meanings of Difficult Words (कठिन शब्दों के अर्थ) :
Found (फउण्ड) = पाया जाता है, world (वर्ल्ड) = विश्व, संसार, many (मैनी) = बहुत, colors (कलर्स) = रंगों, black (ब्लैक) = काला, white (व्हाइट) = सफेद, brown (ब्राउन) = कत्थई, black and white (ब्लैक एण्ड ह्वाइट) = चितकबरा, sharp (शार्प) = तेज, bread (ब्रेड) = रोटी, meat (मीट) = माँस, watch (वॉच) = रखवाली करना, master (मास्टर) = मालिक, house (हाउस) = घर, bark (बार्क) = भौंकना, when (व्हेन) = जब, see (सी) = देखना, stranger (स्ट्रेन्जर) = अजनबी, friend (फ्रेण्ड) = मित्र, useful (यूज़फुल) = उपयोगी, faithful (फेथफुल) = स्वामी भक्त।

My School (माई स्कूल)
मेरा स्कूल

1. I read in Sarafa School, Lashkar.
2. Mr. P.C. Dubey is my Head Master.
3. There are five hundred students in my school.
4. There are nine teachers in my school.
5. All teachers are very hard working.
6. My school has a beautiful building.
7. There are nine rooms and an office.
8. There is a big playground also.
9. There is a big library in my school.
10. The result of my school is always very good.
11. like my school very much.

Meanings of Difficult Words (कठिन शब्दों के अर्थ) :
Head Master (हैड मास्टर) = प्रधान अध्यापक, five hundred (फाइव हण्ड्रेड) = पाँच सौ, hard working (हार्ड वर्किंग) = कठिन परिश्रमी, beautiful (ब्युटीफुल) = सुन्दर, building (बिल्डिंग) = भवन, rooms (रूम्स) = कमरे, office (ऑफिस) = कार्यालय, playground (प्ले-ग्राउण्ड) = खेल का मैदान, big (बिग) = बड़ी, library (लायब्रेरी) = पुस्तकालय, result (रिज़ल्ट) = परिणाम, always (ऑलवेज़) = हमेशा, like (लाइक) = पसन्द करना।

The Postman (द पोस्टमैन)
डाकिया

1. The Postman is a government servant.
2. He wears khaki uniform.
3. He goes door to door.
4. He has a leather bag on his shoulder.
5. He delivers letters and money-orders.
6. We all wait for him every day.
7. His duty is very hard.
8. But his pay is very low.
9. We should be kind to him.
10. He must be paid well.

Meanings of Difficult Words (कठिन शब्दों के अर्थ) :
Government (गवर्नमेण्ट) = सरकारी, servant (सर्वेण्ट) = नौकर, wears (वेअर्स) = पहनता है, khaki (खाकी) = खाकी, uniform (यूनीफॉर्म) = वर्दी, door (डोर) = द्वार, leather (लैदर) = चमड़ा, bag बैग = थैला, shoulder (शोल्डर) = कन्धा, delivers (डेलीवर्स) = बाँटता है, letter (लैटर) = पत्र, moneyorder (मनी-ऑडर्र) = धनादेश, wait (वेट) = इन्तजार, every day (एवरी डे) = रोजाना, duty (ड्यूटी) = कर्त्तव्य, hard (हार्ड) = कठिन, but (बट) = किन्तु, pay (पे) = वेतन, low (लो) = कम, नीचा, kind (काइण्ड) = दयालु, well (वैल) = अच्छा

My Daily Life (माई डेली लाइफ)
मेरी दिनचर्या

1. I get up at 5 O’clock in the morning.
2. After attending to nature call I go for a walk.
3. I return and take breakfast.
4. I study from 7 a.m. to 9 a.m.
5. Then I take bath and lunch and go to school.
6. I study at school from 10.30 a.m. to 4:30 p.m.
7. I return home and have tea at 5 p.m.
8. I play games for an hour, then I study for three hours and take dinner at 9:00 p.m.
9. I watch television for an hour.
10. I go to bed at 10 p.m.
11. On Sundays I watch films and serials on T.V. and help my parents in domestic work.

Meanings of Difficult Words (कठिन शब्दों के अर्थ) :
Nature call (नेचर कॉल) = नित्य कर्म, breakfast (ब्रेकफॉस्ट) = सुबह का नाश्ता, lunch (लन्च) = दोपहर का नाश्ता, dinner (डिनर) = रात का भोजन, study (स्टडी) = अध्ययन करना, domestic (डोमेस्टिक) = घरेलू, serials (सीरियल्स) = धारावाहिक।

My Best Friend (माई बेस्ट फ्रेण्ड)
मेरा प्रिय मित्र

1. I have several friends but Ram is my best friend.
2. He is my neighbour and also my classmate.
3. He always stands first in the class.
4. But he is not proud.
5. He always helps his classmates.
6. His father is a businessman and his mother is a teacher.
7. We go to school together and study at home together.
8. He is good at sports and games too.
9. He faces problems boldly.
10. I like him for his qualities and good nature.

Meanings of Difficult Words (कठिन शब्दों के अर्थ) :
Several (सेवरल) = अनेक, neighbour (नेबर) = पड़ोसी, classmate (क्लासमेट), = सहपाठी, proud (प्राउड) = घमण्ड, businessman (बिजेनेसमैन) = व्यापारी, face (फेस) = सामना करना, boldly (बोल्डली) = हिम्मत से, qualities (क्वालिटीज़) = अच्छाइयाँ, nature (नेचर) = व्यवहार।

MP Board Class 6 English Solutions

MP Board Class 9th Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.3

Question 1.
In Fig. given below E is any point on median AD of a ∆ABC. Show that ar (ABE) = ar (ACE).

Solution:
Given.
E is any point on median AD of ∆ABC.
To prove
ar (ABE) = ar (ACE)

Proof:
In ∆ABC, AD is the median
ar (ABD) = ar(ACD) ….(1)
In ∆EBC, ED is the median
∴ ar (BDE) = ar (CDE) ….(2)
Subtracting (2) from (1), we get
ar (ABD) – ar (BDE) = ar (ACD) – ar (CDE)
ar (ABE) = ar (ACE)

Question 2.
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = \(\frac{1}{4}\) ar (ABC).

Solution:
Given
E is the mid point of – median AD of ∆ABC.
To prove:
ar (BED) = \(\frac{1}{4}\) ar (ABC)
Proof:
In ∆ABC, AD is the media
ar (ABD) = \(\frac{1}{2}\) ar (ABC) …..(1)
In ∆ABD, BE is the median

ar (BED) = \(\frac{1}{2}\) ar (ABD)
= \(\frac{1}{2}\) [\(\frac{1}{2}\) ar (ABC)]
= \(\frac{1}{4}\) ar (ABC)

Question 3.
Show that the diagonals of parallelogram divide it into four triangles of equal area.
Solution:
Given.
ABCD is a parallelogram.
To prove:
ar (AOB) = ar (BOC) = ar (COD) = ar (AOD)

Proof:
In ∆ABC, OA is the median
∴ ar (AOB) = ar (AOD) …..(1)
In AABC, BO is the median
∴ ar (AOB) = ar (BOC) …..(2)
In ABCD, CO is the median
ar (BOC) = ar (COD) …(3)
From (1), (2) and (3), we get
ar (AOB) ar (BOC) = ar (COD) = ar (AOD)

Question 4.
In Fig. given below, ABC and ABC are two triangles on the same base AB. If line – segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

Solution:
Given.
∆ABC and ∆ABD have a common base AB.
OC = OD
To prove
ar (ABC) = ar (ABD)

Proof:
As OC = OD,
O is the mid-point of CD
In ∆ACD, O is the median
ar (AOC) = ar(AOD)
In ABCD, BO is the median
ar (BOC) = ar (BOD)
Adding (1) and (2), we get
ar (AOC) + (BOC) = ar (AOD) + ar (BOD)
ar (ABC) = ar (ABC)

Question 5.
D, E and F are respectively the mid-points of the sides BC, CA and AB of a AABC. Show that

1. BDEF is a parallelogram.
2. ar (DEF) = \(\frac{1}{2}\) ar (ABC)
3. ar (BDEF) = \(\frac{1}{4}\) ar (ABC).

Solution:
D, E, F are the mid points of sides BC, CA and AB of ∆ABC.
To prove:

1. BDEF is a parallelogram.
2. ar (DEF) = \(\frac{1}{4}\) ar (ABC)
3. ar (BDEF) = \(\frac{1}{2}\) ar (ABC)

Proof:
In ∆ABC, F is the mid-point of AB, E is the mid point of AC.
∴ FE ∥ BC and FE = – BC (By MPT)
⇒ FE ∥ BD and FE =BD

1. BDEF is a ∥^gm
Similarly CDFE and AEDF are ∥^gm
In parallelogram BDEF, DF is the diagonal
∴ ax (BDF) = ar (DEF)
[In a ∥^gm diagonal divides it into 2∆s of equal areas]
Similarly, In ∥^gm CDFE, DE is a diagoilal …(1)
ar (CDE) = (DEF) …(2)
In ∥^gm AEDF, FE is a diagonal
ar (AEF) = ar (DEF) …(3)
From (1), (2) and (3), we get
ar (BDF) = ar (CDF) = ar (AEF) = (DEF) …(4)

2. ar (ABC) = ar (AEF) + ar (ADF) + ar (CDE) + ar (DEF)
ar (ABC) = 4 ar (DEF) [Using (4)]
ar (DEF) = \(\frac{1}{4}\) ar (ABC)

3. ar (BDEF) = ar (BDF) + ar (DEF)
ar (BDEF) = 2ar (DEF) [∴ ar (BDF) = ar (DEF)]
= 2 x \(\frac{1}{4}\) ar (ABC) = \(\frac{1}{2}\) ar (ABC)

Question 6.
In the Fig. diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB – CD, then show that:

1. ar (DOC) = ar (AOB)
2. ar DCB = ar (ACB)
3. DA ∥ CB or ABCD is a parallelogram.

[Hint: From D and B, draw perpendiculars to AC.]

Solution:
Given
ABCD is a quadrilateral in which OB = OD and AB = CD.
To prove:

1. ar (DOC) = ar (AOB)
2. ar (DCB) = ar (ACB)
3. DA ∥ CB or ABCD is a ∥^gm

Construction:
Draw DE and BF perpendicular from point D and B on AC.
Proof:
1. In ∆OED and ∆OFB
∠1 = ∠2 (V.O.A.’s)
OD = OB (given)
∠E = ∠F (each 90°)
∆ OED = ∆ OFB (by AAS)
and so DE = BF (by CPCT)
In ∆DEC and ∆BFA, DE = BF (proved)
∠E = ∠F (each 90°)
DC = BA (given)
∆DEC = ∆BFA (by RHS)
and so ∠3 = ∠4 (by CPCT)
ar (OED) = ar (OFB) [∴ AOED = AOFB] …(1)
ar (DEC) = ar (BFB) [∴ ADEC = ABEA] …..(2)
Adding (1) and (2), we get
ar (OED) + ar (DEC) = ar (OFB) = ar (BFA)
ar (OCD) = ar (OAB) …..(3)

2. Adding ar (OBC) on both sides of equation (3)
ar (OCD) + (OBC) = ar (OAB) + ar (OBC)
ar (DCB) = ar (ACB)

3. ∆DCB and ∆ACB have the same base BC and have equal area
∴ they will lie between the same parallels BC and AD
and so BC ∥ AD
∠3 and ∠4 are A.I.A’s and are equal
∴ AB ∥ DC
In quadrilateral ABCD, AB ∥ DC and AB = DC (given)
ABCD is a parallelogram.

Question 7.
D and E are points on sides AB and AC respectively of ∆ABC such that ar (DBC) = ar (EBQ. Prove that DE ∥ BC.
Solution:
Given
ar (DBC) = ar (EBC)
To prove:
DE ∥ BC

Proof:
∆ DBC and ∆EBC have the same base BC and
ar (DBC) = ar (EBC)
∴ They will lie between the same parallel lines DE and BC.
and so DE ∥ BC

Question 8.
XY is a line parallel to side BC of a triangle ABC. If BE ∥ AC and CF ∥ AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).
Solution:
Given
XY ∥ BC, BE ∥ AC and CF ∥ AB.
To prove
ar (ABE) = ar (ACF)
Proof:
In quadrilateral ACBE, AE ∥ CB (∴ BC ∥ XY)
and AC ∥ EB
ACBE is a ∥^gm

Similarly ABCF is a parallelogram.
∥^gm ACBE and ABCF have the same base BC and are between the same parallels BC and AY.
∴ ar (ACBE) = ar (ABCF)
ar (ABE) + ar (ABC) = ar (ABC) + ar (ACF)
∴ ar (ABE) = ar (ACF)

Question 9.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. below). Show that ar (ABCD) = ar (PBQR).
[Hint: Joint AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Solution:
Given
ABCD and BPRQ are parallelograms. CP ∥ AQ
To prove:
ar (ABCD) = ar (PBQR)
Construcion:
Join AC and PQ
Proof:
∆ACQ and ∆APQ lie on the same base AQ and are between the same parallels AQ and CP.
∴ ar (ACQ) = ar (APQ)
⇒ ar (ABQ) + ar (ABC) = ar (ABQ) + ar (BQP)
∴ ar (ABC) = ar (BQP) …..(i)

Multiplying (1) by 2 on both sides
2 ar (ABC) = 2 ar (BQP)
∴ ar (ABCD) = ar (PBQR)
[∴ ABCD and PBQR are parallelogram]

Question 10.
Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect each other at C. Prove that ar (AOD) = ar (BOC).
Solution:
Given
ABCD is a trapezium in which AB ∥ DC.
To prove:
ar (AOD) = ar (BOC).

Proof:
∆ADC and ∆BCD lie on the same base DC and between the same parallels AB and CD.
∴ ar (ADC) = ar (BCD)
Subtracting ar (DOC) from both sides
ar (ADC) – ar (DOC) = ar (BCD) – ar (DOC)
ar (AOD) = ar (BOC)

Question 11.
In Fig. ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that:

1. ar (ACB) = ar (ACF)
2. ar (AEDF) = ar (ABCDE).

Solution:
Given
ABCDE is a pentagon.
To prove:
1. ar (ACB) = ar (ACF)

2. ar (AEDF) = ar (ABCDE)
Proof:
∆ACB and ∆ACF lie on the same base AC and between the same parallels AC and BE.
ar (ACB) = ar (ACF)

3. Adding ar (AEDC) on both sides
ar (ACB) + ar (AEDC) = ar (ACF) + ar (AEDC)
ar (ABCDE) = ar (AEDF)

Question 12.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Solution:
Given.
ABCD is a quadrilateral.
Construction:
Join AC. Draw DE ∥ CA which intersect BA produced at E.

Proof:
∆ADC and ∆ACE both lie on the same base AC and between the same parallels AC and DE.
ar (ADC) = ar (ACE)
Adding ar (ABC) on both sides
ar (ADC) + ar (ABC) = ar (ACE) + ar (ABC)
ar (ABCD) = ar (EBC)

Question 13.
ABCD is a trapezium with AB ∥ DC. Aline parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
Solution:
Given
ABCD is trapezium with AB ∥ DC and AC ∥ XY.
To prove:
ar (ADX) = ar (ACY).
Construction:
Join CX.
Proof:
∆ADX and ∆ACX both lie on the same base AX and between the same parallel AN and DC.

∴ ar (ADX) = ar (ACX) …(1)
∆ACX and ∆ACT both lie on the same base AC and between the same parallels AC and AT.
ar (ADY) = ar (ACX) …(2)
From (1) and (2), we get
ar (ADX) = ar(ACT)

Question 14.
In Fig. below AP ∥ BQ ∥ CR. Prove that ar (AQC) = ar (PBR).

Solution:
Given:
AP∥BQ∥CR.
To prove:
ar (AQC) = ar (PBR)
Proof:
∆ABQ and ∆PBQ both lie on the same base BQ and between the same parallels AP and BQ.
∴ ar (ABQ) = ar (PBQ) …(1)
∆BCQ and ∆BRQ both lie on the same base BQ and between the same parallels BQ and CR.

∴ ar (BCQ) = ar (BRQ) …(2)
Adding (1) and (2), we get
ar (ABQ) tar (BCQ) = ar (PBQ) + ar (BRQ)
∴ ar (AQC) = ar (PBR)

Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at 0 in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
Solution:
Given:
ar (AOD) = ar (BOC)
To prove:
ABCD is a trapezium.

Proof:
ar ∠AOD = ar (BOC) (given)
Adding ar (AOB) on both sides
ar (AOD) + ar (AOB) = ar (BOC) + ar (AOB)
ar (ABD) = ar (ABC)
∆ABD and ∆ABC both lie on same base AB and have equal area.
∴ they will lie between the same parallels.
⇒ AB ∥ DC
∴ ABCD is a trapezium.

Question 16.
In Fig. below, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:
Given
ar (DRC) = ar (DPC)
ar (BDP) = ar (ARC)
To prove:
ABCD and DCPR are trapezium.

Proof:
∆DRC and ∆DPC both lie on the same base DC and have equal area .•. They will lie between the same parallels.
⇒ DC ∥ BP
and so DCPR is a trapezium.
ar (BDP) = ar (ARC) [given] …(1)
ar (DPC) = ar (DRC) [given] …(2)
Subtracting (1) and (2), we get
ar (BDP) – ar (DPC) = ar (ARC) – (DRC)
ar (BDC) = ar (ADC)
∆ADC and ∆BDC both lie on same base DC and have equal area.
∴ they will lie between the same parallels
⇒ AB ∥ DC
and so ABCD is a trapezium.

MP Board Class 9th Maths Solutions

MP Board Class 6th General English Letters and Applications

Applications
(प्रार्थना-पत्र)

Question 1.
Write an application to your head master for four days sick leave.
(अपने प्रधानाध्यापक को चार दिन की बीमारी की छुट्टी के लिए प्रार्थना-पत्र लिखिए।)
Answer:
To,
The Head Master
Janta Junior School,
Gwalior

Sir,
I beg to say that I am suffering from fever since last night. So I cannot come to school. Kindly grant me leave for four days only. Thanks.

Yours obedient pupil
Mohan Rawat
Class VI-A
November 5,20……

Meanings of the Difficult Words :
sir (सर) = महोदय, beg (बेग) = प्रार्थना, suffering (सफरिंग) = पीड़ित, fever (फीवर) = बुखार, since (सिन्स) = से, night (नाइट) = रात्रि, cannot (कैननॉट) = नहीं सकता, kindly (काइण्डली) = कृपया, grant (ग्राण्ट) = स्वीकृत करना, leave (लीव) = अवकाश, thank (र्थंक) = धन्यवाद, obedient (ओबेडिएण्ट) = आज्ञाकारी, pupil (प्यूपिल) = शिष्य।

Question 2.
Write an application to your head-master for seven days leave for the marriage of your brother.
(अपने भाई के विवाह में सम्मिलित होने के लिए 7 दिन के अवकाश हेतु प्रार्थना-पत्र लिखिए।)
Answer:
To.
The Headmaster
Govt. Middle School,
Raipur (M.P.)

Sir,
I beg to say that I have to attend the marriage of my brother. So I cannot come to school.
Please grant me leave for seven days from today.
Thanks.

Yours obedient
John Paul
Class VI-A
Date 8.12.20………

Meaning of Difficult Words :
attend (अटैण्ड) = शामिल होना, marriage (मैरिज) = विवाह, brother (ब्रदर) = भाई, grant (ग्राण्ट) = स्वीकृत, please (प्लीज़) = कृपया, seven (सेवन) = सात, days (डेज़) = दिन, today (टुडे) = आज, from (फ्रॉम) = से।

Question 3.
Write an application for fee concession.
(शुल्क मुक्ति के लिए एक प्रार्थना-पत्र लिखिए।)
Answer:
To,
The Headmaster
V.J. Middle School,
Guna (M.P.)

Sir,
I beg to say that my father is very poor and unable to pay the school fee. Kindly grant me freeship. I shall be highly obliged.
Thanks.

Your’s obediently
Ranjan Dubey
Class VI-B
Date 15.7.20……

Meaning of Difficult words :
father (फादर) = पिता, very (वैरी) = बहुत, poor (पूअर) = गरीब, unable (अनेबल) = असमर्थ, to pay (टू पे) = धन जमा करना, fee (फी) = शुल्क, freeship (फ्रीशिप) = शुल्क मुक्ति, shall (शैल) = गा, गी, गे, highly (हाइली) = बहुत ज्यादा, obliged (ओबलाइज्ड) = कृतज्ञ।

Letters
(पत्र)

Question 1.
Write a letter to your father to send you Rs 100/- by money-order.
(अपने पिता को ₹ 100 का धनादेश भेजने के लिए एक पत्र लिखिए।)
Answer:

1/10. P.T. Road,
Gwalior
6th October, 20 …..

Dear Father,
I am quite well here. I hope you will be quite well there. I have to buy some books so I need one hundred rupees. Kindly send it by money-order early.
With best regards to mother and love to brother and sister.

Yours loving son
Manish Sharma

Meaning of Difficult Words :
dear (डियर) = प्रिय, father (फादर) = पिता, quite (क्वाइट) = बिल्कुल, पूर्णतः, well (वैल) = अच्छा, ठीक, here (हिअर) = यहाँ, buy (बाय) = खरीदना, some (सम) = कुछ, books (बुक्स) = किताबें, need (नीड) = आवश्यकता, kindly (काइण्डली) = कृपया, दयापूर्वक, send (सेण्ड) = भेजना, money-order (मनी-ऑडर्र) = धनादेश, early (अर्ली) = शीघ्र, with (विद) = साथ, regard (रिगार्ड) = सम्मान आदर, love (लव) = प्यार।

Question 2.
Write a letter to your friend inviting him on your birthday.
(अपने जन्मदिन पर अपने मित्र को आमन्त्रित करने के लिए पत्र लिखो।)
Answer:

206, Sahil Colony.
Gwalior,
7th May, 20 …….

Dear Richa,
I am happy to inform that my birthday falls on 15th May. I invite you to attend my birthday party at 8 p.m. at my residence. Please try your best to come. Please convey my best regards to dear aunt and uncle. Waiting eagerly for you.

Yours affectionately
Divya.

Meaning of Difficult Words :
invite (इनवाइट) = न्यौता, inform (इन्फॉर्म) = सूचित करना, attend (अटैण्ड) = शामिल होना, convey (कन्वे) = देना, eagerly (ईगर्ली) = बेसब्री से, affectionately (अफेक्शनेट्ली) = स्नेही।

Question 3.
Write a letter to your friend inviting him to spend the vacations with you in your village.
(अपने मित्र को, अपने गाँव में छुट्टियाँ बिताने के लिए पत्र लिखो।)
Answer:

Vishakha Enclave
Sagar
16th Oct., 20 …….

Dear Sohan,
Our school has been closed for Diwali holidays. I want to spend holidays with you in my village about 30 kilometers from here. I invite you to spend some days with me in my village. You will feel very good here. I hope you will come here soon. Waiting for you.

Yours affectionately
Rohan

Meanings of Difficult Words :
spend (pas) = बिताना, invite (इन्वाइट) = न्यौता, soon (सून) = जल्दी, waiting (वेटिंग) = इन्तजार में, village (विलेज) = गाँव, affectionately (अफेक्शनेट्ली) = स्नेही।

MP Board Class 6 English Solutions

MP Board Class 6th General English Paragraph Writing

एक ही विषय पर या उससे सम्बन्धित पर कहे गये वाक्यों के समूह को गद्यांश कहते हैं।

गद्यांश लेखन के नियम :

1. प्रत्येक गद्यांश एक ही विषय से सम्बन्धित होना चाहिए एवं उसका प्रत्येक वाक्य मुख्य विषय से सम्बन्धित होना चाहिए।
2. गद्यांश का प्रथम वाक्य पाठक की उत्सुकता जगाने वाला व अन्तिम वाक्य उसकी जिज्ञासा शान्त करने वाला होना चाहिए।
3. गद्यांश में विचारों या घटनाओं का क्रम उनके घटित होने के क्रम में व मुख्य विषय से सम्बन्धित होना चाहिए।

Question 1.
Write a short paragraph on the importance of trees.
Answer:
The trees are very important. They supply us with food and shelter. They give us oxygen. We also get fuel from them. They maintain the balance of carbon dioxide and oxygen in the air. Every part of the trees is very useful for us. Several trees have oil in their seeds and they supply medicines also. We get wood from trees for making furniture, houses and paper.

कठिन शब्द :
important = महत्वपूर्ण, supply = उपलब्ध करना, shelter = आश्रय, oxygen = ऑक्सीजन, fuel = ईंधन, maintain = बनाये रखते हैं, balance = सन्तुलन, carbon dioxide = कार्बन डाइ-ऑक्साइड, useful! = लाभदायक, several = अनेक, oil = तेल, seeds = बीज, medicines = औषधियाँ, wood = लकड़ी।

Question 2.
Write a paragraph on Mahatma Gandhi
Answer:
Mahatma Gandhi was a saint. He wore only Khadi. He generally walked on foot. He fought for the independence of India and freed it on 15th of August, 1947. He was against social evils like untouchability, caste system, illiteracy and poverty. Many people did not like his social reforms and opposed him. He was shot dead by Nathuram Godse when he was going in a prayer meeting. He will always be remembered as Father of the Nation.

कठिन शब्द :
Saint = साधू, untouchability = छुआछूत, illiteracy = अशिक्षा, poverty = गरीबी, reforms = सुधार, opposed = विरोध किया।

Question 3.
Write a paragraph on the Himalayas.
Answer:
The Himalayas are beautiful mountains in the north of India. Some of the world’s highest peaks are in Himalayas. The highest peak is the Mount Everest. The tops of the mountains are covered with snow throughout the year. Many rivers flow from these mountains. The climate and the scenery are most charming. Many visitors go to the hill stations for pleasure.

कठिन शब्द :
peak = चोटी, climate = जलवायु, charming = आकर्षक, visitors = दर्शकगण, pleasure! = 3717561

Question 4.
My Friend
उपर्युक्त शीर्षक के आधार पर निम्न लेख को दी गई Aids की सहायता से पूर्ण कीजिए-
Answer:
His Name and Age :
Ajay is my ……… He is about 11 years ……… He lives in ……… He is a …….. boy. He reads in my ……… He is the son of a ……..

His Health and Habits :
Ajay is a ……. boy. He is very …….. He gets up early in the ……… He goes for a ……… daily. He works very ……… He is good at …….. He is a …….. boy. He reads in my ……….

His Dress :
He wears …….. dress. He wears a shirt and a ……….. His ………… are always laced.

His Behaviour :
Ajay always speaks …….. His ……… is good to all. He is loving to all.

कठिन शब्द :
Old = आयु का , simple = साधारण , class = कक्षा, healthy = तन्दुरुस्त, active = चुस्त, behaviour = व्यवहार, shoes = जूते, shorts = नेकर, morning = प्रातः, walk = घूमना, studies = अध्ययन, truth = सच्चाई, obey = आज्ञा पालन करना, clean = साफ करना, friend = मित्र, hard = कठिन।

Question 5.
My Classroom

My class-room :
(मेरी कक्षा) पर गद्यांश निम्न शब्दों की सहायता से पूर्ण कीजिए।
Aids :
Maps, picture, study, room, walls, blackboard, bags, ceiling, tables, clean, windows, chairs, desks, stones.
Answer:
I ……. in room number four. It is my class ……. Its ……. are white. There are two doors and two ……. in it. There is a …..on the front wall.

There are forty ……. and forty …….. in it. There is also a …….. and ……… for the teacher. My classroom is neat and ………

Its floor is made of …….. It is swept daily. There are two ……… fans in it. There are ……. big almirahs in it. We put our …… in them. There is a ……..of Mahatma Gandhi in it. There are also some ……. hanging on the walls.

We like our class-room very much.

MP Board Class 6 English Solutions

MP Board Class 7th Science Solutions Chapter 17 Forests: Our Lifeline

Forests: Our Lifeline Intex Questions

Question 1.
What would happen if forests disappear?
Answer:

1. If forests disappear, the amount of carbon dioxide in air will increase, resulting in the increase of earth’s temperature.
2. In the absence of trees and plants, the animals will not get food and shelter.
3. In the absence of trees, the soil will not hold water which will cause floods.
4. Deforestation will ending our life and environment. Think, what, we can do to preserve our forests.

Forests: Our Lifeline Text Book Exercises

Question 1.
Explain how animals dwelling in the forest help it grow and regenerate?
Answer:
Animals help in growing and regenerating forests in many ways. Animals help in dispersing, seeds of certain plants. The decaying animal dung also provides nutrients to the seedling to plant. By this number of plants are growing which serve as a food for a number of herbivorous animals.

Herbivorous helps the carnivores to grow as they serve as food for them. So, the flora and fauna grows in the forests. Also, the animals work as the cleaning agents in the forest micro – organisms work on the death bodies of plants and animals and regenerate them.

Question 2.
Explain how forest prevent floods?
Answer:
If trees or widely we can say that forests are not present, rain will directly hit the ground and may cause the flood the area around it. Also, the heavy rain may damage the soil. Roots of trees and grasses basically bind the soil together, but in the absence of soil, they will be washed away or eroded.

The washed soil will get deposited in river and thus reduce the water carrying capacity of rivers. Which is the major causes of flood. Also, forests can absorb a lot of water. This also helps in preventing floods.

Question 3.
What are decomposers? Name any two of them. What do they do in the forest?
Answer:
The micro – organisms which convert the dead plants and animals to humus are known as decomposers. These micro – organisms play an important role in the forest. The any two decomposers are grubs and beetles. They clean the forests of the decaying dead bodies and replenishes the nutrient back to the forest soil.

Question 4.
Explain the role of forest in maintaining the balance between oxygen and carbon – dioxide in the atmosphere?
Answer:
Plants release oxygen through the C0₂ photosynthesis. The Plants help to provide oxygen for animal respiration. They also maintain the balance of oxygen and carbon dioxide in the atmosphere. That is why forests are called lungs.

Question 5.
Explain why there is no waste in a forest?
Answer:
The decomposers decompose the dead organisms. The decomposed matter is absorbed by plants as nutrients. Thus, there is no waste in a forest.

Question 6.
List five products we get from forests?
Answer:
We get following products from forests:

1. Medicine
2. Spice
3. Wood
4. Honey.
5. Gum

Question 7.
Fill in the blanks:

1. The insects, butterflies, honeybees and birds help flowering plants in ………………………..
2. A forest is a purifier of ……………………………. and …………………………
3. Herbs form the ……………………………… layer in the forest.
4. The decaying leaves and animal droppings in a forest enrich the ………………………….

Answer:

1. Pollination
2. Water, air
3. Lowest
4. Humus.

Question 8.
Why should we worry about the conditions and issues related to forests far from us?
Answer:
We should be worried about deforestation, as it would lead to floods, increase in earth’s temperature, depriving animals of their habitats and soil erosion.

Question 9.
Explain why there is a need of variety of animals and plants in a forest?
Answer:
Variety of animals are necessary for their survival and maintenance of food chain. For example, grass is eaten by insects, which in turn, are eaten by the frog. The frog is consumed by snakes and snakes is eaten by eagles.

Thus, it forms a food chain. There are so many food chains in the forest. All of these are linked. If any one food chain is disturbed, it affects other food chains. Also, animals convert the death plants and animals into humus and increase the fertility of soil.

Question 10.
In figure, the artist has forgotten to put the labels and directions on the arrows. Mark the directions on the arrows and label the diagram using the following labels: clouds, rain, atmosphere, carbon dioxide, oxygen, plants, animals, soil roots, water table?
Answer:

Question 11.
Which of the following is not a forest product?

1. Gum
2. Plywood
3. Sealing wax
4. Kerosene

Answer:
4. Kerosene.

Question 12.
Which of the following statements is not correct?

1. Forests protect the soil from erosion.
2. Plants and animals in a forest are not dependent on one another.
3. Forests influence the climate and water cycle.
4. Soil helps forests to grow and regenerate.

Answer:
2. Plants and animals in a forest are not dependent on one another.

Question 13.
Micro – organisms act upon the dead plants to produce?

1. Sand
2. Mushrooms
3. Humus
4. wood.

Answer:
3. humus.

Extended Learning – Activities and Projects

Question 1.
The Department of Environment is to decide whether some portion of a forest in your area could be cleared for a housing complex. Write a letter to the department explaining your point of view as a concerned citizen?
Answer:
Do yourself.

Question 2.
Visit a forest? Here is a list of points that would make your visit more fruitful?

1. Make sure that you have permission to go into the forest.
2. Make sure that you can find your way around. Get a map and go along with some one who is familiar with the area.
3. Keep a record of the things you see and do. Observations make the visit interesting. Sketches and photographs are useful.
4. You may record bird calls.
5. Collect different kinds of seeds or hard fruits like nuts.
6. Try to recognise various types of trees, shrubs, herbs, etc. Make lists of plants from different places in the forest and of different layers. You may not be able to name all the plants, but it is worth recording and seeing where they grow. Make a record of approximate heights of plants, crown shape, bark texture, leaf size, and flower colour.
7. Learn to recognise the animal’s droppings.
8. Interview the forest officials and the people of surrounding villages and other visitors.
9. You must never collect birds eggs, and their nests should never be disturbed.

Answer:
Do yourself.

Forests: Our Lifeline Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (a)
Plants give out during photosynthesis?
(a) carbon dioxide
(b) oxygen
(c) humus
(d) none of these.
Answer:
(b) oxygen

Question (b)
Interdependence of animals for food is called –
(a) food – order
(b) food – cycle
(c) food – chain
(d) none of these.
Answer:
(c) food – chain

Question (c)
Small trees and grass and shrubs from the –
Answer:
(a) Canopy
(b) Under stories
(c) Lowest canopy
(d) All of these
Answer:
(b) Under stories

Question (d)
Deforestation leads to
(a) Soil erosion
(b) Floods
(c) Both of these
(d) None of these
Answer:
(c) Both of these

Question 2.
Fill in the blanks:

1. The branch part of a tree above the stem is known as the ……………………… of the tree.
2. The micro – organisms which convert the dead plants and animals to humus are known as …………………………
3. The plants help to provide ………………………….. for animal respiration.
4. The plants maintain the balance of oxygen and carbon dioxide in the …………………………
5. The covered ground with decaying material acts like a ……………………………
6. The root system helps water to seep down in the ……………………………
7. Forests protect the soil from ………………………..
8. Forests are the lifeline for the forest dwelling …………………………….

Answer:

1. Crown
2. Decomposers
3. Oxygen
4. Atmosphere
5. Sponge
6. Ground
7. Erosion
8. Communities.

Question 3.
Which of the following statements are true (T) or false (F):

1. Forests cause rain on the earth.
2. In a forests, the herbs form the lowest layer of the vegetation.
3. The various components of the forest are not interdependent on one another.
4. The forests keeps on growing and changing, and can regenerate.
5. Soils helps forests to grow and regenerate.
6. Forests influence climate, water – cycle and air quality.
7. Forests is a renewable natural resources.

Answer:

1. True
2. Ture
3. False
4. True
5. True
6. True
7. True.

Forests: Our Lifeline Very Short Answer Type Questions

Question 1.
What do you mean by forest?
Answer:
Forest is a place hosting a number of animals and plants.

Question 2.
Name any two trees in forest?
Answer:
Sheesham, Neem.

Question 3.
Write any three things we get from forests?
Answer:
Gum, medicine and wood.

Question 4.
Which plants constitute canopy in forests?
Answer:
Tall and giant trees.

Question 5.
What is the basic unit of any food chain?
Answer:
Plants.

Question 6.
Write one food chain in forest?
Answer:
Grass → Insects → Frog → Snake → Eagle

Question 7.
How is forest environment?
Answer:
Forest environment is peaceful and a cool breeze is blowing.

Question 8.
Name some animals that live deep inside forest?
Answer:
Bison, Jackals, Boar, Elephants.

Question 9.
What is humus?
Answer:
Humus is a dark coloured substance which is formed from the dead bodies of plants and animals.

Question 10.
What happens if an animal dies in the forest?
Answer:
The dead animals become food for vultures, jackals, insects and crows.

Question 11.
How does forest help in preventing floods?
Answer:
Forests absorb the rain .water in the soil and prevent the floods.

Question 12.
What is seed dispersal?
Answer:
The method by which plants scatter their seeds to reduce competition between off springs.

Question 13.
Define soil erosion?
Answer:
The weaving away of the soil by wind or water.

Question 14.
What do you mean by canopy?
Answer:
The branches of the tall trees look like a roof over the other plants in the forest is called canopy.

Question 15.
Define crown?
Answer:
Branch part of a tree above the stem is known as the crown.

Question 16.
What is deforestation?
Answer:
Cutting and destroying the forest is called deforestation.

Question 17.
What are the effects of deforestation?
Answer:
Floods, rising temperature, scarcity of flood and wood and disturbed flood chains are the results of deforestation.

Question 18.
What do you mean by afforestation?
Answer:
Planting more trees is called afforestation.

Question 19.
What would happen if it rains heavily in your town?
Answer:
There will be water logging or even flood. Everything will be in a mesh. No food, no water. No place, no stay.

Forests: Our Lifeline Short Answer Type Questions

Question 1.
Define soil erosion?
Answer:
Removal of fertile top layers of earth (soil) is called soil erosion. The soil erosion can be brought about by various agents like – water, wind, glaciers and ocean waves. Erosion occurs in both wet and arid regions. The various human activities like falling of trees, over – cropping and improper tilling accelerate soil erosion. The soil erosion can lead to floods.

Question 2.
What are the major factors that lead to soil erosion? Name them?
Answer:
The major factors that lead to soil erosion are the following:

1. Overgrazing by livestock
2. Deforestation
3. Over cultivation
4. Poor irrigation facilities
5. Water logging
6. Shifting cultivation.

Question 3.
Write about the dependence of animals on plants?
Answer:
Dependence of Animals on Plants:

1. Plants produce food by the process of photosynthesis and animals depend directly or indirectly on them for food.
2. Plants produce oxygen during photosynthesis and animals use this oxygen for respiration.
3. Plants act as habitat for the animals dwelling on trees.

Question 4.
Write about the dependence of plants on animals?
Answer:
Dependence of Plants on Animals:

1. Animals produce carbon dioxide during respiration which is released in the atmosphere. Plants use this oxygen for preparing food.
2. A number of insects, birds and bats help in pollination.
3. Some animals help in dispersal of fruits and seeds.

Question 5.
How forests help in bringing rain?
Answer:
Trees absorbs water from the soil through their roots. They then release the excess of water through the process of transpiration in the form of water vapours. These water vapours form clouds and clouds bring rain.

Question 6.
How are forests useful to us?
Answer:
The forests are very useful to us for the following reasons:

1. They conserve soil and water.
2. They are helpful in causing rains.
3. They give us food and fruit.
4. They are the natural habitats of animals.
5. They help in the maintenance of ecological balance.

Question 7.
Write a short note on deforestation?
Answer:
Deforestation is another important cause which brings about soil erosion. Deforestation leads to depletion of water vapour in the atmosphere so which in turn adversely affects the formation of rain bearing clouds and consequently the rainfall. In the absence of adequate rainfall, the soil dries up and its nutrient value is affected.

The decrease in rainfall affects the growth of plants and trees, thus leads to the formation of deserts. When there is no plant cover over the earth, it becomes naked. When rain falls over this naked earth, the water does not percolate down the earth and washes away. Similarly, strong winds also take away the top soil, thus causing soil erosion.

Question 8.
In which layer of the soil would you find humus? What is its importance to the soil?
Answer:
Humus is found in the top of layer of soil. The presence of humus ensures that the nutrients of the dead plants and animals are released into the soil. From there, these nutrients are again absorbed by the roots of the living plants.

Forests: Our Lifeline Long Answer Type Questions

Question 1.
How we can conserved the forests?
Answer:

1. Unplanned and indiscriminate deforestation must be stopped.
2. Forest should not be over – exploited.
3. Use of wood as fuel should be discouraged.
4. Plant new fast growing trees.
5. Protect forest from fires, insects and diseases.
6. Protect national parks, wildlife sanctuaries or biosphere reserves.
7. Follow the guidelines of international organizations like W.W.F. and UNESCO for forest conservation.

Question 2.
Write the uses of forests?
Answer:
Forests are useful to us for the following reasons:

1. Forests provide timber for building and furniture.
2. Forests keep the environment cool and increase the chances of rainfall.
3. The roots of the trees hold the soil firmly thereby checking soil erosion.
4. Forests provide habitat to a variety of wild plants and animals.
5. The plants of forests use CO₂ and release oxygen in photosynthesis, thereby maintaining balance of gases in atmosphere.
6. Forests provide animal products such as honey, lac, wax, etc.
7. Forests provide plant products such as fruits, nuts, gums, resins, fibres, medicines, essential oil (like sandal wood), etc.
8. They are involved in the absorption of water which percolates in the soil, thereby becoming a part of the groundwater.

Question 3.
What will happen if forests disappear?
Answer:

1. Soil erosion will increase which will affect soil fertility and productivity.
2. Air pollution will increase which will ultimately threaten thirty existence of life on the Earth.
3. Unpredictable changes will take place in weather and climate.
4. Reduction in rainfall.
5. Reduction in recharging of existing water resources
6. Wildlife may also disappear along with forest.
7. Shortage of forest products.
8. Disturbance in ecological balance.
9. Increase in temperature
10. Global warming may reach in flooding of low lying area.

Question 4.
Draw a figure to show the difference between canopy and under stories?
Answer:

Question 5.
Draw a figure to show in the dependence of plant, soil and decomposers in a forest?
Answer:

MP Board Class 7th Science Solutions

MP Board Class 6th Maths Solutions Chapter 10 क्षेत्रमिति Intext Questions

पाठ्य-पुस्तक पृष्ठ संख्या # 222-223

प्रयास कीजिए

प्रश्न 1.
अपनी अध्ययन टेबल के ऊपरी चारों सिरों की लम्बाइयों को मापिए तथा उन्हें लिखिए।

AB = _ सेमी, BC = _ सेमी, CD = _ सेमी, DA = _ सेमी
अब चारों भुजाओं की लम्बाइयों का योगफल
= AB + BC + CD + DA
= _ सेमी + _ सेमी + _ सेमी + _ सेमी
= _ सेमी
क्या आप बता सकते हैं कि परिमाप कितना है?
हल :
AB = 130 सेमी, BC = 80 सेमी, CD = 130 सेमी, DA = 80 सेमी
∴ AB + BC + CD + DA
= 130 सेमी + 80 सेमी + 130 सेमी + 80 सेमी
= 420 सेमी
हाँ, इसका परिमाप = 420 सेमी

प्रश्न 2.
अपनी नोटबुक के एक पृष्ठ की चारों भुजाओं की लम्बाइयों को मापिए और उन्हें लिखिए। चारों भुजाओं की लम्बाइयों का योगफल ।
= AB + BC + CD + DA
= _ सेमी + _ सेमी + _ सेमी + _ सेमी
= _ सेमी
पृष्ठ का परिमाप कितना है?
हल : AB + BC + CD + DA = 20 सेमी + 16 सेमी + 20 सेमी + 16 सेमी = 72 सेमी
अतः पृष्ठ का परिमाप = 72 सेमी

प्रश्न 3.
मीरा 150 मी लम्बाई तथा 80 मी चौड़ाई वाले एक पार्क में जाती है। वह इस पार्क का एक पूरा चक्कर लगाती है। उसके द्वारा तय की गई दूरी ज्ञात कीजिए।
हल :
पार्क की लम्बाई = 150 मी,
पार्क की चौड़ाई = 80 मी
पार्क का परिमाप = 2 x (लम्बाई + चौड़ाई)
= 2 x (150 मी + 80 मी)
= 2 x (230 मी)
= 460 मी
∴ मीरा द्वारा तय की गई दूरी = 460 मी

प्रश्न 4.
निम्न आकृतियों का परिमाप ज्ञात कीजिए :


हल :
(a) ∵ AB = 40 सेमी, BC = 10 सेमी, CD = 40 सेमी, DA = 10 सेमी
∴ परिमाप = AB + BC + CD + DA
= 40 सेमी + 10 सेमी + 40 सेमी + 10 सेमी
= 100 सेमी

(b) ∵ AB = BC = CD = DA = 5 सेमी
∴ परिमाप = AB + BC + CD + DA
= 5 सेमी + 5 सेमी + 5 सेमी + 5 सेमी
= 20 सेमी

(c) परिमाप = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= 1 सेमी + 3 सेमी + 3 सेमी + 1 सेमी + 3 सेमी + 3 सेमी + 1 सेमी + 3 सेमी + 3 सेमी + 1 सेमी + 3 सेमी + 3 सेमी
= 28 सेमी

(d) परिमाप = AB + BC + CD + DE + EF + FA
= 100 मी + 120 मी + 90 मी + 45 मी + 60 मी + 80 मी
= 495 मी

पाठ्य-पुस्तक पृष्ठ संख्या # 224

प्रश्न 1. निम्नलिखित आयतों के परिमाप ज्ञात कीजिए :
हल :

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 9 आँकड़ों का प्रबंधन Ex 9.4

पाठ्य-पुस्तक पृष्ठ संख्या # 219-220

प्रश्न 1.
एक स्कूल के 120 विद्यार्थियों का इस आशय से सर्वेक्षण किया गया है कि वे अपने खाली समय में किस क्रियाकलाप को पसन्द करते हैं। निम्न आँकड़े प्राप्त हुए

1 इकाई लम्बाई = 5 विद्यार्थी का पैमाना लेकर एक दण्ड आलेख बनाइए। खेलने के अतिरिक्त कौन-सा क्रियाकलाप अधिकांश विद्यार्थियों द्वारा पसन्द किया जाता हैं?
हल :
दण्ड आलेख

खेलने के अतिरिक्त अधिकांश विद्यार्थियों द्वारा कहानी की पुस्तक पढ़ना पसन्द किया जाता है।

प्रश्न 2.
छः क्रमागत दिनों में किसी दुकानदार द्वारा बेची गई गणित की पुस्तकों की संख्या आगे दी गई है

अपनी पसन्द का पैमाना चुनते हुए, उपर्युक्त सूचना के लिए एक दण्ड आरेख खींचिए।
हल :
दण्ड आलेख

प्रश्न 3.
वर्ष 1998 से 2002 के बीच एक फैक्टरी द्वारा निर्मित साइकिलों की संख्या निम्नलिखित सारणी द्वारा दर्शाई गई है

इन आँकड़ों को एक दण्ड आलेख द्वारा प्रदर्शित कीजिए। अपनी पसन्द का पैमाना चुनिए।
(a) किस वर्ष में अधिकतम संख्या में साइकिलें निर्मित की गईं ?
(b) किस वर्ष में न्यूनतम संख्या में साइकिलें निर्मित की गईं ?
हल :

(a) वर्ष 2002 में अधिकतम संख्या में साइकिलें निर्मित की गईं।
(b) वर्ष 1999 में न्यूनतम संख्या में साइकिलें निर्मित की गईं।

प्रश्न 4.
किसी शहर के व्यक्तियों की संख्या विभिन्न आयु समूहों के अनुसार नीचे सारणी में दी हुई हैं।

इन आँकड़ों को एक दण्ड आलेख द्वारा निरूपित कीजिए। (1 इकाई लम्बाई = 1 हजार लीजिए।)
निम्नलिखित प्रश्नों के उत्तर दीजिए :
(a) किन दो आयु समूहों में जनसंख्या बराबर है?
(b) 60 वर्ष और उससे अधिक आयु के सभी व्यक्ति वरिष्ठ नागरिक कहलाते हैं इस शहर में कितने वरिष्ठ नागरिक हैं ?
हल :
दण्ड आलेख

(a) आयु समूह 30-44 और 45-59 की जनसंख्या बराबर है।
(b) इस शहर में वरिष्ठ नागरिकों की संख्या 80,000 + 40,000 = 1,20,000 है।

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 9 आँकड़ों का प्रबंधन Ex 9.3

पाठ्य-पुस्तक पृष्ठ संख्या # 215-216

प्रश्न 1.
आगे दिया हुआ दण्ड आलेख वर्ष 1998-2002 में सरकार द्वारा खरीदे गए गेहूँ की मात्रा दर्शाता है :

इस दण्ड आलेख को पढ़िए और अपने प्रेक्षणों को लिखिए :
(a) किस वर्ष में गेहूँ का अधिकतम उत्पादन हुआ ?
(b) किस वर्ष में गेहूँ का न्यूनतम उत्पादन हुआ ?
उत्तर-
(a) वर्ष 2002 में गेहूँ का अधिकतम उत्पादन हुआ।
(b) वर्ष 1998 में गेहूँ का न्यूनतम उत्पादन हुआ।

प्रश्न 2.
इस दण्ड आलेख को देखिए जो एक रेडीमेड कपड़ों की दुकान में सोमवार से शनिवार तक हुई कमीजों की बिक्री को दर्शाता है :

अब निम्नलिखित प्रश्नों के उत्तर दीजिए :
(a) उपर्युक्त दण्ड आलेख में क्या सूचना दर्शाई गई है ?
(b) कमीजों की संख्या को निरूपित करने के लिए क्षैतिज रेखा पर क्या पैमाना लिया गया है ?
(c) किस दिन अधिकतम कमीजें बेची गईं और कितनी संख्या में बेची गईं, लिखें?
(d) किस दिन न्यूनतम संख्या में कमीजें बेची गईं ?
(e) बृहस्पतिवार को कितनी कमीजें बेची गईं ?
उत्तर-
(a) यह दण्ड आलेख सोमवार से शनिवार तक बेची गई कमीजों की संख्या दर्शाता है।
(b) कमीजों की संख्या को निरूपित करने के लिए क्षैतिज रेखा पर पैमाना 1 सेमी = 5 कमीजें ली गई हैं।
(c) शनिवार को अधिकतम कमीजें बेची गईं। इस दिन 60 कमीजें बेची गईं।
(d) मंगलवार को न्यूनतम कमीजें बेची गईं।
(e) बृहस्पतिवार को 35 कमीजें बेची गईं।

प्रश्न 3.
इस दण्ड आलेख को देखिए जो अजीज द्वारा अर्धवार्षिक परीक्षा में विभिन्न विषयों में प्राप्त किए गए अंकों को प्रदर्शित करता है:

निम्नलिखित प्रश्नों के उत्तर दीजिए :
(a) यह दण्ड आलेख क्या सूचना प्रदर्शित करता है ?
(b) किस विषय में अजीज ने अधिकतम अंक प्राप्त किए ?
(c) किस विषय में उसने न्यूनतम अंक प्राप्त किए ?
(d) विषयों के नाम लिखिए और उनमें से प्रत्येक में प्राप्त किए गए अंक भी लिखिए।
उत्तर-
(a) यह दण्ड आलेख अजीज द्वारा विभिन्न विषयों में प्राप्त अंकों को प्रदर्शित करता है।
(b) अजीज ने हिन्दी में अधिकतम अंक प्राप्त किए।
(c) उसने सामाजिक विषय में न्यूनतम अंक प्राप्त किए।
(d) हिन्दी – 80, अंग्रेजी – 60, गणित – 70, विज्ञान – 50 और सामाजिक विषय – 40.

पाठ्य-पुस्तक पृष्ठ संख्या # 218

इन्हें कीजिए

अपने मित्रों के साथ पाँच और ऐसी स्थितियों के बारे में सोचिए, जहाँ हम आँकड़े प्राप्त कर सकते हैं। संख्याओं का प्रयोग करके सारणियाँ बनाइए और उन्हें दण्ड आलेखों द्वारा प्रदर्शित कीजिए।

प्रश्न 1.
निम्नांकित दी गई सारणी में छात्रवृत्तियों का विभिन्न वर्षों में वितरण निम्नानुसार है :

दण्ड आलेख द्वारा निरूपित कीजिए।
हल:

प्रश्न 2.
नीचे दिए गए दण्ड आलेख को पढ़कर प्रश्नों के उत्तर दीजिए :

(i) इस दण्ड आलेख से आपको कौन-सी जानकारी प्राप्त होती है ?
(ii) अलग-अलग शिक्षण सत्र में छात्राओं की संख्या के परिवर्तन का क्रम क्या है?
(iii) 2011-12 में छात्राओं की संख्या क्या थी?
(iv) किस सत्र में छात्राओं की संख्या सबसे अधिक है?
(v) 2012-13 की छात्राओं की संख्या कौन-से सत्र की तुलना में दो गुनी है ?
उत्तर-
(i) इस दण्ड आलेख से हमें विभिन्न शिक्षण सत्रों में कक्षा VI की छात्राओं की संख्या की जानकारी प्राप्त होती है।
(ii) अलग-अलग शिक्षण सत्र के लिए दण्डों की ऊँचाई निरन्तर बढ़ रही है, जो दर्शाती है कि छात्राओं की संख्या में लगातार वृद्धि हुई है।
(iii) 2011-12 में छात्राओं की संख्या 200 थी।
(iv) सत्र 2014-15 का दण्ड सर्वाधिक ऊँचाई (400 छात्राएँ) दर्शा रहा है। अतः सत्र 2014-15 में छात्राओं की संख्या सबसे अधिक है।
(v) 2012-13 की छात्राओं की संख्या (300), सत्र 201011 की तुलना में दोगुनी है।

प्रश्न 3.
नीचे दिए गए दण्ड आलेख को पढ़कर प्रश्नों के उत्तर दीजिए :
विभिन्न साधनों से स्कूल आने वाले विद्यार्थियों की संख्या

(i) दण्डालेख से हमें कौन-सी जानकारी मिलती है ?
(ii) स्कूल आने के साधन कौन-से अक्ष पर दर्शाए गए हैं ?
(iii) Y-अक्ष पर क्या दर्शाया गया है ?
(iv) पैदल स्कूल आने वाले विद्यार्थियों की संख्या कितनी है ?
(v) सरकारी बस से कितने विद्यार्थी स्कूल आते हैं ?
(vi) स्कूल में कुल कितने विद्यार्थी हैं ?
(vii) स्कूल बस कितने विद्यार्थियों को स्कूल लाती है ?
(viii) स्कूटर से आने वाले विद्यार्थियों की संख्या कितनी है ?
उत्तर-
(i) इस दण्डालेख से विभिन्न साधनों से स्कूल आने वाले विद्यार्थियों की संख्या से सम्बन्धित जानकारी मिलती है।
(ii) स्कूल आने के साधन X-अक्ष पर दर्शाए गए हैं।
(iii) Y-अक्ष पर विद्यार्थियों की संख्या दर्शाई गयी है।
(iv) पैदल स्कूल आने वाले विद्यार्थियों की संख्या 50 है।
(v) सरकारी बस से 40 विद्यार्थी स्कूल आते हैं।
(vi) स्कूल में कुल 300 विद्यार्थी हैं।
(vii) स्कूल बस 120 विद्यार्थियों को स्कूल लाती है।
(viii) स्कूटर से स्कूल आने वाले विद्यार्थियों की संख्या 20 है।

प्रश्न 4.
एक दुकानदार द्वारा छः क्रमागत दिनों में बेचे गए बल्बों की संख्या इस प्रकार है

इस सूचना को दण्ड आलेख द्वारा निरूपित कीजिए।
हल:

प्रश्न 5.
एक प्राथमिक स्कूल में रखी अलमारी में विभिन्न प्रकार की पुस्तकों की संख्या निम्न प्रकार है:

इन आँकड़ों को दण्ड आलेख द्वारा दर्शाइए और निम्न प्रश्नों के उत्तर दीजिए
(i) अलमारी में सर्वाधिक पुस्तकें किस विषय की हैं ?
(ii) अलमारी में किस विषय की पुस्तकें सबसे कम हैं ?
(ii) अलमारी में बराबर पुस्तकें कौन-सी हैं ?
(iv) अलमारी में विज्ञान और गणित की कुल कितनी पुस्तकें हैं ?
हल :
दण्ड आलेख

(i) अलमारी में सर्वाधिक पुस्तकें हिन्दी की हैं।
(ii) अलमारी में सबसे कम पुस्तकें अंग्रेजी की हैं।
(iii) अलमारी में विज्ञान और लघु कथाओं की पुस्तकें बराबर हैं।
(iv) अलमारी में विज्ञान और गणित की कुल पुस्तकें 75 हैं।

MP Board Class 6th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 13 सीधा और प्रतिलोम समानुपात Ex 13.1

प्रश्न 1.
एक रेलवे स्टेशन के निकट कार पार्किंग शुल्क इस प्रकार है –

• 4 घण्टों तक – ₹ 60
• 8 घण्टों तक – ₹ 100
• 12 घण्टों तक – ₹ 140
• 24 घण्टों तक – ₹180

जाँच कीजिए कि क्या कार पार्किंग शुल्क पार्किंग समय के प्रत्यक्ष अनुपात में है?
हल:
∴ \(\frac{4}{60}\) ≠ \(\frac{8}{100}\) ≠ \(\frac{12}{140}\) ≠ \(\frac{24}{180}\)
∴ कार पार्किंग शुल्क पार्किंग समय के प्रत्यक्ष अनुपात में नहीं है।

प्रश्न 2.
एक पेंट के मूल मिश्रण (base) के 8 भागों में लाल रंग के पदार्थ का 1 भाग मिलाकर मिश्रण तैयार किया जाता है। निम्नलिखित सारणी में, मूल मिश्रण के वे भाग ज्ञात कीजिए जिन्हें मिलाए जाने की आवश्यकता है –

हल:
यहाँ, माना कि लाल रंग के पदार्थ का भाग x, मूल मिश्रण के ” भाग में मिलाया जाता है, तथा x और y प्रत्यक्ष अनुपात में हैं।
अतः x और y का अनुपात अचर रहेगा।
तब \(\frac{x}{y}\) = \(\frac{1}{8}\)
अतः अभीष्ट अनुपात = \(\frac{4}{32}\), \(\frac{7}{56}\), \(\frac{12}{96}\), \(\frac{20}{160}\)
सारणी में रखने पर,

प्रश्न 3.
प्रश्न 2. में यदि लाल रंग पदार्थ के 1 भाग के लिए 75 mL मूल मिश्रण की आवश्यकता है, तो मूल मिश्रण के 1800 mL में हमें कितना लाल रंग का पदार्थ मिलाना चाहिए?
हल:
माना कि 1800 mL में x भाग लाल रंग का पदार्थ मिलाना चाहिए।
सारणी में रखने पर,

अतः मूल मिश्रण में हमें 24 भाग लाल रंग का पदार्थ मिलाना चाहिए।

प्रश्न 4.
किसी सॉफ्ट ड्रिंक फैक्ट्री में एक मशीन 840 बोतलें 6 घण्टे में भरती है। वह मशीन पाँच घण्टे में कितनी बोतलें भरेगी?
हल:
माना कि 5 घण्टे में x बोतलें भरी जा सकती हैं। तब सारणी के रूप में रखने पर,

अतः मशीन 5 घण्टे में 700 बोतलें भरेगी।

प्रश्न 5.
एक बैक्टीरिया या जीवाणु के फोटोग्राफ (चित्र) को 50,000 गुना आवर्धित करने पर उसकी लम्बाई 5 cm हो जाती है, जैसा कि संलग्न चित्र में दिखाया गया है। इस बैक्टीरिया की वास्तविक लम्बाई क्या है ? यदि फोटोग्राफ को केवल 20,000 गुना आवर्धित किया जाए, तो उसकी आवर्धित लम्बाई क्या होगी?

हल:
बैक्टीरिया की आवर्धित लम्बाई = 50,000 गुना
बैक्टीरिया की लम्बाई = 5 cm

अतः बैक्टीरिया की वास्तविक लम्बाई = 10_-4 cm
यहाँ, फोटोग्राफ की लम्बाई तथा आवर्धित लम्बाई अनुक्रमानुपाती हैं।
माना कि आवर्धित लम्बाई x cm है
सारणी के रूप में लिखने पर,

यहाँ आवर्धित लम्बाई तथा आवर्धित फोटोग्राफ परस्पर अनुक्रमानुपाती हैं।

अतः आवर्धित लम्बाई = 2 cm

प्रश्न 6.
एक जहाज के मॉडल में उसका मस्तूल (mast) 9 cm ऊँचा है, जबकि वास्तविक जहाज का मस्तूल 12 m ऊँचा है। यदि जहाज की लम्बाई 28 m है तो उसके मॉडल की लम्बाई कितनी है?
हल:

स्पष्ट है, मॉडल की लम्बाई और वास्तविक लम्बाई परस्पर – अनुक्रमानुपाती हैं।
\(\frac{9}{12}\) = \(\frac{x}{28}\)
x = \(\frac{9×28}{12}\) = 21
अतः जहाज के मॉडल की लम्बाई = 21 cm

प्रश्न 7.
मान लीजिए 2 kg चीनी में 9x 10 क्रिस्टल हैं। निम्नलिखित चीनी में कितने क्रिस्टल होंगे?

1. 5 kg
2. 1.2 kg.

हल:
माना कि 5 kg चीनी और 1.2 kg चीनी में क्रमशः x और y क्रिस्टल हैं। इन्हें सारणी रूप में लिखने पर,

स्पष्ट है कि यहाँ क्रिस्टल की संख्या और चीनी की मात्रा परस्पर अनुक्रमानुपाती हैं।
1.

अतः 5 kg चीनी में 2.25 x 10₇ क्रिस्टल होंगे।
2.

अतः 1.2 kg चीनी में 5.4 x 10₆ क्रिस्टल होंगे।

प्रश्न 8.
रश्मि के पास एक सड़क का मानचित्र है, जिसके पैमाने में 1 cm की दूरी 18 km निरूपित करती है। वह उस सड़क पर अपनी गाड़ी से 72 km की दूरी तय करती है। उसके द्वारा तय की गई दूरी मानचित्र में क्या होगी?
हल:
यहाँ, 1 cm की दूरी = 18 km.
माना कि रश्मि के द्वारा तय की गई दूरी मानचित्र में x cm है, तब दी हुई सूचना को सारणी के रूप में निरूपित करने पर,

यहाँ, मानचित्र में दूरी तथा वास्तविक दूरी परस्पर अनुक्रमानुपाती हैं।
\(\frac{1}{18}\) = \(\frac{x}{72}\)
या x = \(\frac{1×72}{18}\) cm = 4 cm
अतः मानचित्र में दूरी = 4 cm

प्रश्न 9.
एक 5 m 60 cm ऊँचे ऊर्ध्वाधर खम्भे की छाया की लम्बाई 3 m 20 cm है। उसी समय पर ज्ञात कीजिए –

1. 10 m 50 cm ऊँचे एक अन्य खम्भे की छाया की लम्बाई।
2. उस खम्भे की ऊँचाई जिसकी छाया की लम्बाई 5 m है।

हल:
1. माना कि x m उस खम्भे की लम्बाई है जिसकी छाया की लम्बाई 10 m 50 cm है। y m उस खम्भे की ऊँचाई है जिसकी छाया 5 m है।
इन्हें सारणी के रूप निरूपित करने पर,

यहाँ, यह स्पष्ट है कि खम्भे की ऊँचाई और छाया परस्पर अनुक्रमानुपाती हैं।

अतः छाया की लम्बाई = 6 m
2.

अतः खम्भे की ऊँचाई = 8 m 75 cm

प्रश्न 10.
माल से लदा हुआ एक ट्रक 25 मिनट में 14 km चलता है। यदि चाल वही रहे, तो वह 5 घण्टे में कितनी दूरी तय कर पाएगा?
हल:
माना कि ट्रक 5 घण्टे में x km दूरी तय करता है। तब इन्हें सारणी के रूप में निरूपित करने पर,

यहाँ, दूरी तथा समय परस्पर अनुक्रमानुपाती हैं।
\(\frac{14}{25/60}\) = \(\frac{x}{5}\)
या x = \(\frac{5x14x60}{25}\) = 168 km
अतः ट्रक द्वारा 5 घण्टे में तय की गई दूरी = 168 km

पाठ्य-पुस्तक पृष्ठ संख्या # 216-217

इन्हें कीजिए (क्रमांक 13.3)

प्रश्न 1.
एक वर्गांकित कागज पर भिन्न-भिन्न भुजाओं के पाँच वर्ग खींचिए। निम्नलिखित सूचना को एक सारणी के रूप में लिखिए –

ज्ञात कीजिए कि क्या भुजा की लम्बाई –

1. वर्ग के परिपाम के अनुक्रमानुपाती है।
2. वर्ग के क्षेत्रफल के अनुक्रमानुपाती है।

हल:

सूचना को सारणी के रूप में निरूपित करने पर,

यहाँ स्पष्ट है कि,

1. वर्ग की भुजा की लम्बाई वर्ग के परिमाप के अनुक्रमानुपाती है।
2. वर्ग की भुजा की लम्बाई वर्ग के क्षेत्रफल के अनुक्रमानुपाती नहीं है।

प्रश्न 2.
पाँच व्यक्तियों के लिए हलवा बनाने के लिए, निम्नलिखित सामग्री की आवश्यकता होती है: सूजी/रवा = 250 g, चीनी = 300 g, घी = 200 g, पानी = 200 g. समानुपात की अवधारणा का प्रयोग करते हुए, अपनी कक्षा के लिए हलवा बनाने के लिए इन सामग्रियों की मात्राओं में होने वाले परिवर्तनों का आकलन (estimate) कीजिए।
हल:
माना कि कक्षा में विद्यार्थियों की संख्या = 20 है।
स्पष्ट है कि यहाँ अनुक्रमानुपाती की स्थिति है।
अत: वांछित सामग्री \(\frac{20}{5}\) = 4 गुना होगी –
अर्थात् सूजी/रवा = 250 x 4g = 1000 g = 1 kg
चीनी = 300 x 4g = 1200 g = 1.200kg
घी = 200 x 4g = 800 g
पानी = 200 x 4g = 800 g.

प्रश्न 3.
एक पैमाने का चुनाव करते हुए, अपनी कक्षा के कमरे का मानचित्र खींचिए, जिसमें खिड़कियाँ, दरवाजे, ब्लैकबोर्ड इत्यादि दर्शाए गए हों (एक उदाहरण यहाँ दिया है।

हल: माना की पैमाना 1 : 40 है, तब
मानचित्र इस प्रकार है –

सोचिए, चर्चा कीजिए और लिखिए (क्रमांक 13.2)

प्रश्न 1.
सीधा समानुपात (विचरण) की अब तक हल की गई समस्याओं में से कुछ को लीजिए। क्या आप सोचते हैं कि इन समस्याओं को इकाई की विधि या ऐकिक विधि (unitary method) से हल किया जा सकता है?
हल:
हाँ, इन समस्याओं को ऐकिक विधि से हल किया जा सकता है।
हम यहाँ प्रश्नावली 13.1 से प्रश्न 4 एवं प्रश्न 10 को हल करते हैं –
प्रश्न 4 का हल:
∴ 6 घण्टे में भरी जाने वाली बोतलें = 840
∴ 1 घण्टे में भरी जाने वाली बोतलें = \(\frac{840}{6}\)
∴ 5 घण्टे में भरी जाने वाली बोतलें = \(\frac{840×5}{6}\)
= 140 x 5 = 700 बोतलें।
प्रश्न 10 का हल:
25 मिनट = \(\frac{25}{60}\) घण्टे
∴ \(\frac{25}{60}\) घण्टे में तय की गई दूरी = 14 km
∴ 1 घण्टे में तय की गई दूरी = \(\frac{14}{25/60}\) km
∴ 5 घण्टे में तय की गई दूरी = \(\frac{14x60x5}{25}\) km
= 168 km

पाठ्य-पुस्तक पृष्ठ संख्या # 219

इन्हें कीजिए (क्रमांक 13.4)

प्रश्न 1.
एक वर्गांकित कागज लीजिए और उस पर 48 काउंटरों (counters) को पंक्तियों की विभिन्न संख्याओं में नीचे दर्शाए अनुसार व्यवस्थित कीजिए –


आप क्या देखते हैं? जब R में वृद्धि होती है, तो C में कमी होती है।

1. क्या R₁ : R₂ = C₂ : C₁ है?
2. क्या R₃ : R₄ = C₄ : C₃
3. क्या R और C परस्पर व्युत्क्रमानुपाती है?

इस क्रियाकलाप को 36 काउंटरों के साथ प्रयास कीजिए।
हल:
C₁ → 24
C₂ → 16
C₅ → 6
हम यहाँ देखते हैं कि R में वृद्धि होती है तो C में कमी होती

1. R₁ : R₂ = 2 : 3 और
C₂ : C₁ = 16 : 24 = 2 : 3
R₁ : R₂ = C₂ : C₁

2. R₃ : R₄ = 4 : 6 = 2 : 3 और
C₄ : C₃ = 8 : 12 = 2 : 3
R₃ : R₄ = C₄ : C₃

3. हाँ, R और C परस्पर व्युत्क्रमानुपाती हैं। उत्तर अब, 36 काउण्टरों के साथ क्रियाकलाप।

सूचना को सारणी के रूप में लिखने पर, हम प्राप्त करते हैं –

हम देखते हैं कि R में वृद्धि होती है, तो C में कमी होती हैं।

1. R₁ : R₂ = 2 : 3 और C₂ : C₁ = 12 : 18 = 2 : 3
R₁ : R₂ = C₂ : C₁

2. R₃ : R₄ = 4 : 6 = 2 : 3 और C₄ : C₃ = 6 : 9 = 2 : 3
R₃ : R₄ = C₄ : C₃

3. हाँ, R और C परस्पर व्युत्क्रमानुपाती हैं।

प्रयास कीजिए (क्रमांक 14.2)

प्रश्न 1.
निम्नलिखित सारणियों को देखिए तथा ज्ञात कीजिए कि कौन-कौन से चरों (यहाँ x और y) के युग्म परस्पर प्रतिलोम समानुपात में हैं –

हल:
(i) हम देखते हैं कि
x × y = 50 x 5 ≠ 40 x 6 ≠ 30 x 7 ≠ 20 x 8
अतः x और y परस्पर प्रतिलोम समानुपात में नहीं हैं।

(ii) हम देखते हैं कि
x × y = 100 x 6 = 200 x 30 = 300 x 20
= 400 x 15 = अचर
अतः x और y परस्पर प्रतिलोम समानुपात में हैं।

(iii) हम देखते हैं कि
x × y = 90 x 10 = 60 x 15 = 45 x 20 ≠ 30 x 25 ≠ 20 x 30 ≠ 5 x 35
अतःx और , परस्पर प्रतिलोम समानुपात में नहीं हैं।

MP Board Class 8th Maths Solutions