Class 7

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.4

प्रश्न 1.
ज्ञात कीजिए :
(i) 12 ÷ \(\frac {3}{4}\)
(ii) 14 ÷ \(\frac {5}{6}\)
(iii) 8 ÷ \(\frac {7}{3}\)
(iv) 4 ÷ \(\frac {8}{3}\)
(v) 3 ÷ 2\(\frac {7}{3}\)
(vi) 5 ÷ 3\(\frac {4}{7}\)
हल :

प्रश्न 2.
निम्नलिखित भिन्नों में से प्रत्येक का व्युत्क्रम ज्ञात कीजिए। व्युत्क्रमों को उचित भिन्न, विषम भिन्न एवं पूर्ण संख्या के रूप में वर्गीकृत कीजिए:
(i) \(\frac {3}{7}\)
(ii) \(\frac {5}{8}\)
(iii) \(\frac {9}{7}\)
(iv) \(\frac {6}{5}\)
(v) \(\frac {12}{7}\)
(vi) \(\frac {1}{8}\)
(vii) \(\frac {1}{11}\)
हल :
(i) \(\frac {3}{7}\) का व्युत्क्रम = \(\frac {7}{3}\) है, यह उचित भिन्न नहीं है।
(ii) \(\frac {5}{8}\) का व्युत्क्रम = \(\frac {8}{5}\), यह उचित भिन्न नहीं है।
(iii) \(\frac {9}{7}\) का व्युत्क्रम = \(\frac {7}{9}\), यह उचित भिन्न है।
(iv) \(\frac {6}{5}\) का व्युत्क्रम = \(\frac {5}{6}\), यह उचित भिन्न है।
(v) \(\frac {12}{7}\) का व्युत्क्रम = \(\frac {7}{12}\), यह उचित भिन्न है।
(vi) \(\frac {1}{8}\) का व्युत्क्रम = \(\frac {8}{1}\) = 8, यह उचित भिन्न है।
(vii) \(\frac {1}{11}\) का व्युत्क्रम = \(\frac {11}{1}\) = 11, यह एक पूर्ण संख्या है।

प्रश्न 3.
ज्ञात कीजिए:
(i) \(\frac {7}{3}\) ÷ 2
(ii) \(\frac {4}{9}\) ÷ 5
(iii) \(\frac {6}{13}\) ÷ 7
(iv) 4\(\frac {1}{3}\) ÷ 3
(v) 3\(\frac {1}{2}\) ÷ 4
(vi) 4\(\frac {3}{7}\) ÷ 7
हल :

प्रश्न 4.
ज्ञात कीजिए :

हल :

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 ठोस आकारों का चित्रण Ex 15.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 ठोस आकारों का चित्रण Ex 15.4

प्रश्न 1.
निम्नलिखित ठोसों के ठीक ऊपर एक जलता हुआ बल्ब रखा गया है। प्रत्येक स्थिति में प्राप्त छाया के आकार का नाम बताइए। इस छाया का एक रफ चित्र बनाने का प्रयास कीजिए। (पहले आप प्रयोग करने का प्रयास करें और फिर उत्तर दें।)

हल :
जब बल्ब ठोसों के ऊपर जलता हुआ रखा है :
गेंद : इसकी छाया वृत्त की तरह प्रतीत होगी।
बेलनाकार पाइप : इसकी छाया आयत की तरह प्रतीत होगी।
पुस्तक : इसकी छाया वर्ग की तरह प्रतीत होगी।

प्रश्न 2.
यहाँ कुछ 3-D वस्तुओं की छायाएँ दी गई हैं जो उन्हें एक ओवरहैड प्रोजेक्टर के लैम्प (बल्ब) के अन्तर्गत या नीचे रख कर प्राप्त की गई हैं। प्रत्येक छाया से मिलान वाले ठोस की पहचान कीजिए। (इनमें एक से अधिक उत्तर हो सकते हैं।)

हल :
प्रत्येक छाया से मिलान वाले ठोस निम्न हैं-
(i) एक गेंद, एक थाली आदि।
(ii) एक घन, पुस्तक आदि।
(iii) एक शंकु, आइसक्रीम कोन आदि।
(iv) एक बेलन, एक घनाभ आदि।

प्रश्न 3.
जाँच कीजिए कि क्या ये कथन सत्य हैं-
(i) एकघन एक आयत के आकार की छाया दे सकता है।
(ii) एक घन एक षड्भुज के आकार की छाया दे सकता है।
हल :
(i) सत्य
(ii) असत्य।

Haryana Board HBSE 7th Class Sanskrit Solutions रुचिरा भाग 2

• Chapter 1 सुभाषितानि
• Chapter 2 दुर्बुद्धिः विनश्यति
• Chapter 3 स्वावलम्बनम्
• Chapter 4 हास्यबालकविसम्मेलनम्
• Chapter 5 पण्डिता रमाबाई
• Chapter 6 सदाचारः
• Chapter 7 सड.कल्पः सिद्धिदायकः
• Chapter 8 त्रिवर्णः ध्वजः
• Chapter 9 अहमपि विद्यालयं गमिष्यामि
• Chapter 10 विश्वबंधुत्वम्
• Chapter 11 समवायो हि दुर्जयः
• Chapter 12 विद्याधनम्
• Chapter 13 अमृतं संस्कृतम्
• Chapter 14 अनारिकायाः जिज्ञासा
• Chapter 15 लालनगीतम्

Haryana State Board HBSE 7th Class Science Solutions Chapter 18 Wastewater Story Textbook Exercise Questions and Answers.

Haryana Board 7th Class Science Solutions Chapter 18 Wastewater Story

HBSE 7th Class Science Wastewater Story Textbook Questions and Answers

Question 1.
Fill in the blanks :
(a) Cleaning of water is a process of removing _________.
(b) Wastewater released by houses is called _________.
(c) Dried _________ is used as manure.
(d) Drains get blocked by _________ and _________.
Answer:
(a) pollutants
(b) sewage
(c) sludge
(d) chemicals, kitchenwaste.

Question 2.
What is sewage? Explain why it is harmful to discharge untreated sewage into rivers or seas.
Answer:
Sewage is a liquid containing wastes disposed off by household, industrial and agricultural, activities in water. It is dangerous to release untreated sewage in water because it can pollute the whole source of water. Sewage contain harmful substances and disease causing organisms. It is therefore, dangerous and unsafe to release untreated sewage in water.

Question 3.
Why should oils and fats be not realeased 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 wastewater to flow and thus the whole sewer system is blocked.

Question 4.
Describe the steps involved in getting clarified water from wastewater.
Answer:
Water is treated physically, chemically and biologically in wastewater treatment plant.
Following Steps are involved in the purification of water:
(i) At first stage all the physical impurities like stones, rags, napkins, plastic bags, cans, packets etc. are removed. It is done by passing the water through bar screens.

(ii) Then water is taken to grit and sand removal tank where impurities are removed by sedimentation.

(iii) Solids impurities and faeces etc. are collected from the bottom of the water. These solid impurities collected are called sludge. Water is cleared of floatable solids like oil and grease.

(iv) Clarified water is cleared of other impurities by aerator. All disease causing bacteria are removed by chlorination and water is released in various water bodies.

Question 5.
What is sludge? Explain how it is treated.
Answer:
Sludge is the collected solid waste from the wastewater during the treatment in water treatment plant. Sludge is decomposed in a separate tank by the anaerobic bacteria. Activated sludge is used as manure.

Question 6.
Untreated human excreta is a health hazard. Explain.
Answer:
Untreated excreta can cause a lot of health related problems. It pollutes soil, water and air. The polluted water contain disease causing bacteria, which can spread epidemics like cholera, meningities etc.

Question 7.
Name two chemicals used to disinfect water.
Answer:
Chlorine and ozone are the chemical used to clean the wastewater.

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

Question 9.
Explain the relationship between sanitation and disease.
Answer:
Sanitation and disease are related each other. It sanitation is there no disease will occur, but if the sanitation is not there various types of diseases will occur and spread. So sanitation should be kept to avoid 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 unhygenic and disease causing substances thrown in’open.

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

Across:
3. Liquid waste products
4. Solid waste extracted in sewage treatment
6. A word related to hygiene
8. Waste matter discharged from human body

Down:
1. Used water
2. A pipe carrying sewage
5. Micro-organisms which causes cholera
7. A chemical to disinfect water
Answer:

Question 12.
Study the following statements about ozone:
(a) It is essential for breathing of living organisms.
(b) It is used to disinfect water.
(c) It absorbs ultraviolet rays.
(d) Its- proportion in air is about 3%.
Which of these statements are correct?
(i) (a), (b) and (c)
(ii) (b) and (c)
(iii) (a) and (d)
(iv) All four
Answer:
(b) and (c)

Extended Learning – Activities and projects

Question 1.
Construct a crossword puzzle of your own using the keywords.
Answer:
Do it yourself.

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 it 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 _________.
(a) The location of the sewage plant.
(b) Treatment capacity.
(c) The purpose of screening as the initial process.
(d) How is air bubbled through the aeration tank?
(e) How safe is the water at the end of the treatment? How is it tested?
(f) Where is the water discharged after treatment?
(g) What happens to the plant during heavy rains?
(h) Is biogas consumed within the plant or sold to other consumers?
(i) What happens to the treatment sludge?
(j) Is there any special effort to protect nearby houses from the plant?
(k) Other observations.
Answer:
Do it yourself.

HBSE 7th Class Science Wastewater Story Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Write different sources of wastewater?
Answer:
Household activities, industrial activities and agricultural activities.

Question 2.
Name certain organic impurities in the wastewater.
Answer:
Animal waste, Human faeces, oil and urine, fruits and vegetables.

Question 3.
Write certain Inorganic impurities in, the wastewater.
Answer:
Metals, phosphates and nitrates.

Question 4.
Name certain disease causing micro-organism.
Answer:
Bacterias, Viruses etc.

Question 5.
Which process removes the solids like faeces and other substances from the wastewater?
Answer:
Grit and sand removal tank.

Question 6.
Which instrument is used to remove floatable solids from the wastewater?
Answer:
A skimmer is used to remove floatable impurities.

Question 7.
Who decomposes the sludge?
Answer:
Anaerobic bacteria decompose the sludge.

Question 8.
What helps to clean the clarified water?
Answer:
Aerobic bacteria helps to clean the clarified water.

Question 9.
Why is ozone and chlorine used?
Answer:
Ozone and Chlorine is used to kill the bacteria etc. present in the clarified water.

Short Answer Type Questions

Question 1.
How is water’polluted?
Answer:
Water is used for 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 2.
How “bar screen” and ‘grit and sand removal tank’ help in clarification of water?
Answer:
When wastewater is passed through bar screens it separates big and large objects like plastics, bags, sticks, can, napkins etc. In grit and sand removal tank other solid impurities like pebbless and etc. are removed.

Question 3.
How is sludge treated?
Answer:
Sludge is the solid impurities separated from the sewage. It is removed and treated in a separate tank with anaerobic bacteria. During this process biogas is produced which is used to produce electricity. Dried sludge is used as manure.

Question 4.
What are the problem arising due to open drains and other unsanitary conditions?
Answer:
Open drains and unsanitary conditions produce bad smell. It becomes an idle place of breeding for mosquitoes, files and other harmful insects. These insects spread many harmful diseases and other health hazards.

Question 5.
How the kitchen waste blocks the drains?
Answer:
Kitchen waste like oils and fats clogs the pores in soil and reduce the Alteration process. It also blocks the pipes by hardening the pipes. Wastes like tealeaves solid food remains, cotton etc. also choke the pipes and slows down flow of oxygen. This slows down the decomposition process by the aerobic bacteria.

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, hepatilis and meningiti it is dysentry etc.

Wastewater Story Class 7 HBSE Notes

• Water is a precious natural resource.
• We cannot imagine our lives without water.
• We waste a lot of water daily in various household and industrial activities Such water is called wastewater.
• The wastewater produced during household acitivities, industrial activities and various agricultural processes is also called sewage.
• Sewage is the liquid waste which can cause various’diseases and environmental hazards if not managed.
• Sewage is collected from its sources and treated to destroy its harmful constituent to clean it.
• It is made usable in treatment plants and disposed off in various sources of water.
• Drainage system should be covered to avoid communicable diseases.
• We should not throw waste in open and should not defecate in open.
• Low cost disposal methods can be adopted in the areas where proper sewer system is not available.

Haryana State Board HBSE 7th Class Science Solutions Chapter 4 Heat Textbook Exercise Questions and Answers.

Haryana Board 7th Class Science Solutions Chapter 4 Heat

HBSE 7th Class Science Heat Textbook Questions and Answers

Question 1.
State similarities and differences between the laboratory thermometer and the clinical thermometer.
Answer:
Laboratory Thermometer: While performing experiments involving measurement of temperature in the laboratory, a mercury thermometer is used. It is a thermometer having graduations marked on Celsius scale from 0°C to 100°C.

The difference between the lower and upper fixed points is called range of the thermometer. The range of a mercury thermometer is 100°C. A mercury thermometer having graduations from 0°C to 100°C is shown in figure 4.8.

Clinical Thermometer:
To measure the temperature of a person running fever, doctor uses a thermometer known as Clinicdl Thermometer. It is a specially designed mercury thermometer and is used in the clinics and hospitals by doctors to measure human body temperature.

In construction, it differs from laboratory thermometers, in the sense that a kink or constriction is provided in the stem just above the bulb. Because of this kink, mercury doesn’t fall back into the bulb when clinical thermometer is taken out of the mouth of a patient to know the temperature. A jerk is given to the thermometer so that it is set again for measuring human body temperature. The temperature interval marked on the clinical thermometers ranges from 35°C to 43°C.

Question 2.
Give two examples each of conductors and insulators of heat.
Answer:
Conductors: aluminium, iron
Insulators: plastic, wood

Question 3.
Fill in the blanks:
(a) The hotness of an object is, is determined by its …………… .
(b) Temperature of boiling water cannot be measured by a …………… thermometer.
(c) Temperature is measured in degree ……………. .
(d) No medium is required for transfer of heat by the process of ………….. .
(e) A cold steel spoon is dipped in a cup of hot milk. It transfers heat to its other end by the process of …………. .
(f) Clothes of …………… colours absorb heat better than clothes of light colours.
Answer:
(a) Touching
(b) Clinical
(c) Celsius
(d) Radiation
(e) Conduction
(f) Dark

Question 4.
Match the following:

Answer:

Question 5.
Discuss why wearing more layers of clothing of during winter keeps us warmer tljian wearing just one thick piece of clothing?
Answer:
More layers of clothing keep us warm in winters as they have a lot of space between them. This space gets filled up with air. As air is a bad conductor, it does not allow the body heat to escape out.

Question 6.
Look at Fig. 4.10. Mark where the heat is being transferred by conduction, by convection and by radiation.
Answer:

Question 7.
In places of hot climate it is advised that the outer walls of houses be painted white. Explain.
Answer:
In places of hot climate it is advised that the outer wail of houses be painted white because white colour do not radiate heat easily.

Question 8.
One litre of water at 30°C is mixed with one litre of water at 50°C. The temperature of the mixture will he:
(a) 80°C
(b) More than 50°C but less than 80°C
(d) 20°C
(d) Between 30°C and 50°C
Answer:
(b) More than 50°C but less than 80°C

Question 9.
An iron ball at 40°C is dropped in a mug containing water at 40°C. The heat will:
(a) Flows from iron ball to water.
(b) Not flow from iron ball to water or from water to iron ball.
(c) Flows from water to iron ball.
(d) Increase the temperature of both.
Answer:
(a) flows from iron ball to water.

Question 10.
A wooden spoon is dipped in a cup of ice-cream. Its other end:
(a) Becomes cold by the process of conduction.
(b) Becomes cold by the process of convection.
(c) Becomes cold by the process of radiation.
(d) Does not become cold.
Answer:
(d) does not become cold.

Question 11.
Stainless steel pans are usually provided with copper bottoms. The reason for this could be that:
(a) Copper bottom makes the pan more durable.
(b) Such pans appear colourful.
(c) Copper is a better conductor of heat than the stainless steel.
(d) Copper is easier to clean than the stainless steel.
Answer:
(c) Copper is a better conductor of heat than the stainless steel.

Extended Learning – Activities And Projects

Question 1.
Go to a doctor or your nearest health centre. Observe the doctor taking temperature of patients. Enquire:
(a) Why he/she dips the thermometer in a liquid before use.
(b) Why the thermometer is kept under the tongue.
(c) Whether the body temperature can be measured by keeping the thermometer at some place other than mouth.
(d) Whether the temperature of different parts of the body is the same or different. You can add m.ore questions which come to your mind.
Answer:
Do yourself. You can take help of your family doctor.

Question 2.
Go to a veterinary doctor (a doctor who treats animals). Discuss and find out the normal temperature of domestic animals and birds.
Answer:
Do yourself.

Question 3.
Wrap a thin paper strip tightly around an iron rod. Try to burn the paper with candle while rotating the iron rod continuously. Does it burn? Explain your observation.
Answer:
Do yourself.

Question 4.
Take two similar transparent glass bottles having wide mouths. Put a few crystals of potassium permaganate or pour a few drops of ink in one bottle. Fill this bottle with hot water. Fill the other bottle with cold water. Cover the cold water bottle with a thick piece of paper such as a postcard. Press the postard firmly with one hand and hold the bottle with the other hand. Invert the bottle and place it on top of the hot water bottle. Hold both the bottles firmly. Ask some other person to pull the postcard. Observe what happens. Explain.
Answer:
Do yourself.

HBSE 7th Class Science Heat Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is heat?
Answer:
The energy transferred from one body to another body due to a temperature difference between them is called heat.

Question 2.
Define temperature.
Answer:
Temperature is the degree of hotness or coldness of a body.

Question 3.
Define thermometer.
Answer:
A device used for measuring the temperature of different objects is called a thermometer.

Question 4.
Hot water bottles are used for fermentations. Explain why?
Answer:
Hot water bottles are used for fermentation as the water does not cool quickly due to its large specific heat capacity.

Question 5.
When equal amount of air, iron and oil are heated from 15°C to 25°C. List them in the order of increasing expansion.
Answer:
Iron, oil, air.

Question 6.
Name any four substances which expand on heating.
Answer:
Aluminium, steel, iron and copper.

Question 7.
Name any two substances which have negligible changes in length on heating.
Answer:
Pyrex glass, Invar.

Question 8.
Name two substances which contract on heating?
Answer:
Ice, Bismuth.

Question 9.
Do all liquids expand when heated or is there any exception?
Answer:
Yes, below 4°C when heated, water contract in stead of expanding.

Question 10.
Which substances has the highest heat capacity?
Answer:
Water.

Question 11.
What is the temperature of a normal human being on the Celsius scale?
Answer:
The temperature of a normal human being on the Celsius scale is 37°C.

Question 12.
Name the thermometer used for measuring the temperature of the human body.
Answer:
The thermometer used to measure the temperature of a human body is called clinical thermometer.

Question 13.
Define convection.
Answer:
The phenomenon due to which particles of a medium actually move to the source of heat energy and then move away from it after absorbing heat energy is called convection.

Question 14.
Why is convection not possible in solids?
Answer:
The molecules of a solid are held strongly due to intermolecular forces. As these molecules cannot travel to the source of heat energy, convection is not possible in case of solids.

Question 15.
What is ventilation?
Answer:
The process by which impure and warm air inside a room is continuously replaced by fresh air from outside is called ventilation.

Question 16.
What do you understand from the term lower standard point?
Answer:
The 0°C correspondence to the temperature of pure melting ice and is called the lower standard point.

Question 17.
What do you understand by upper standard point?
Answer:
100°C corresponds to the temperature of pure boiling water and is called the upper standard point.

Question 18.
Define conduction.
Answer:
The process of transmission of heat energy in solids without the actual movement of particles from their position is called conduction.

Question 19.
Give two examples to show that heat caused biological changes.
Answer:
(i) Conversion of ‘gur’ into alcohol.
(ii) Conversion of milk into curd.

Question 20.
Why is ice wrapped in gunny bags?
Answer:
Gunny bags have a number of fine pores, filled with air. Air being a bad conductor of heat does not allow the external heat to go in and melts the ice.

Question 21.
Why is tea generally served in China clay cups and plates?
Answer:
Because the China clay is bad conductor of heat and checks the heat conduction from the tea to the surroundings and thus keeps the tea hot.

Question 22.
By which of the processes (conduction, convection, radiation), it is not possible to transfer heat?
(i) In a vacuum
(ii) Through the solid material.
Answer:
(i) Conduction and convection
(ii) Convection and radiations.

Question 23.
Why do we use cooking utensils made of metals and alloys?
Answer:
Metals and alloys are good conductors of heat so that the heat from the flame is conducted to the food inside quickly and efficiently.

Question 24.
Is medium necessary for the transference of heat by radiations?
Answer:
No, medium is not necessary for the transference of heat by radiation.

Question 25.
Why do we feel warm while standing beside a burning furnace?
Answer:
When we stand beside a burning furnace, we feel warms, because of the heat radiation coming from the furnace.

Question 26.
Firemen wear helmets made from brass polished from outside. Why?
Answer:
Brass polished helmets reflects most of the heat and absorbs only a very little part of it when they fight against a fire.

Question 27.
When does the conduction of heat stop in two bodies in contact?
Answer:
The conduction of heat stops in two bodies in contact when both attain the same (equal) temperature. In this position they have no temperature difference.

Question 28.
Why is the handle of a press made of ebonite or wood?
Answer:
The handle of press is made of ebonite or wood, because ebonite and wood are bad conductors of heat. So the heat does not reach in our body.

Question 29.
Define radiation.
Answer:
The transfer of heat energy from a hot body to a cold body directly, without heating the space in between the two bodies is called radiation.

Question 30.
Define insulators.
Answer:
The materials which do not allow heat to pass through them easily are poor conductors of heat such as plastic and wood. Poor conductors are known as insulators.

Question 31.
What is thermos flask?
Answer:
Thermos flask is a device in which heat losses due to conduction, convention and radiation are minimised.

Short Answer Type Questions

Question 1.
The cooking utensils are made of metals like aluminium or copper while their handles are made of wood. Why?
Answer:
The cooking utensils are made of metals like aluminium or copper because these metals are good conductors of heat so they help in the transfer of heat.

The handles of cooking utensils are made up of bad conductors such as wood, plastic etc. As the handles made of bad conductors do not get heated up while cooking, we can hold them easily.

Question 2.
Kites and eagles fly without flapping their wings. Why?
Answer:
During the day, hot air surrounding the land gradually rises up and a convection air current develops. Kites and eagles move along this rising current without flapping the wings to fly high up in the sky.

Question 3.
Why do the birds puff up their feathers in winter?
Answer:
The birds puff up their feathers in winter, because in doing so they trap large amount of air, which in turn acts as an insulator and does not allow their body heat to flow out.

Question 4.
Explain briefly the formation of air currents.
Answer:
In coastal areas, during the day, the cool air blows from sea towmrds the land called sea breeze. During the night, the cool air blows from land to sea called the land breeze. This happens because of convection air currents. At night, land cools down much faster than sea. So, sea is warmer than land and the air current blows from land to sea. During the day, land is hotter than sea. As a result, the air rises up and cold air from sea blows towards the land to take its place.

Question 5.
Cement or concrete floors are made in pieces with metals or glass strips in between. Give reasons.
Answer:
It is done to allow for the expansion in summer and contraction in winter. If cement or concrete floors are made out in one continuous pieces. It would develop crackes due to contraction and expansion.

Question 6.
Room heaters have shiny reflectors. Why?
Answer:
Room heaters have shiny reflectors as the shiny surface absorbs very little heat. They reflect all the heat which makes the room heaters more effective.

Question 7.
Why is the handle of a pressure cooker covered with thick plastic?
Answer:
The handle of a pressure cooker covered with thick plastic, because plastic is a bad conductor of heat. Hence, the heat from the hot cooker does not flow to the handle, with the result that we can hold easily.

Question 8.
The desert sand is very hot in the day and very cool at night. Why?
Answer:
The desert sand has low specific heat capacity. So sand gets very hot in day time. During the night sand cools more quickly due to cold air which moves towards the sea.

Question 9.
Why is the handle of a metallic kettle covered with strips of cane?
Answer:
Cane strips is bad conductor of heat. Therefore, it cannot get heated when water is boiled in kettle. Therefore, the handle of a metallic kettle is covered with cane strips.

Question 10.
Why does the handle of an iron made of ebonite or wood?
Answer:
Ebonite or wood is a bad conductor of heat. Therefore, the heat from the iron is not conducted to the hand.

Question 11.
How does a blanket keep you warm in a cold night, even though it is not a source of heat?
Answer:
The blanket is made of wool, which is a bad conductor of heat. It has fine pores in which air is trapped which is also a bad conductor of heat. They do not allow heat to escape from our body to the surroundings. Therefore, the blanket keeps our body warm in winter.

Question 12.
Why are cloudy nights warmer than the clear nights?
Answer:
Cloudy nights are warmer than clear nights due to the fact that the heat which is radiated by the earth is again reflected back by the clouds. While on the other hand this phenomenon is not possible when the night is clear.

Question 13.
How are had conductors of heat useful in our daily life?
Answer:
Bad conductors are used for holding hot utensils in the kitchen. Bricks and mud which are bad conductors are used in building houses, especially in the ruler. regions. They keep houses warm in winter and cool in summer. The feathres of birds, hair and fur of animals keep them warm in winter, as they are bad conductor of heat.

Question 14.
Why do we wear woolen clothes in winter?
Answer:
Woolen clothes have fine pores filled with air. Wool and air are bad conductors of heat They do not allow heat of our bodies to escape out. Therefore woolen cloth keep the body warm in winter. And for the same reason we wear woolen cloth in winter.

Long Answer Type Questions

Question 1.
Describe briefly the process of conduction.
Answer:
Heat travels through solids by conduction only. The process of transmission of heat energy in solids without the actual movement of particles from their position is called conduction. Conduction takes place under the following conditions:
(i) A material medium that is compact.
(ii) No movement of any particle in the medium.
Thus, solids transmit heat though conduction. When a solid substance is heated, the particles that become hot start vibrating and pass on the excess energy to the neighbouring particles. This process continues till the whole body is heated. However during such an energy transfer, a particle does not change its position. Hence, heat is carried without the bodily movement of the particles.

There are some solids which allow the heat to pass through them. Such substances are called the conductors. Almost all metals such as iron, copper, silver, aluminium etc. are good conductors. The substances which do not allow the heat to pass through them are called the insulators. Wood, plastic, wool, clay etc. are bad conductors.

Question 2.
Write an experiment to prove that heat radiations can travel through vacuum.
Answer:
Take a flat bottomed flask. In the mouth of the flask fix a rubber stopper having two holes. Pass a thermometer through one hole and a glass tube with stopper through the other hole.

Remove air from the flask, by connecting to a vacuum pump. Record the temperature from the thermometer. Place the flask in bright sunlight or in front of an electric heater. It is observed that the thermometer records a rise in temperature. Thus, the experiment proves that transference of heat by radiaton requires no medium.

Question 3.
Explain the working of a thermos flask.
Answer:
Thermos flask is a special kind of flask for keeping liquids. It prevents loss or gain of heat by any of the three methods. It keeps liquid warm or cold for a long time. It was invented by Sir James Dewan in 1893.

It consists of a double-walled glass-cylinder. A vacuum is created in the space between the double walls. The two inner surfaces of glass walls are silvered. The flask is placed on a spring within a non-conduction cylindrical case to protect it from jerks and breakages. The gaps between flask and outer casing is packed with felt pads. The mouth of the flask is finally closed with a cork.

Since the container is made of glass which is a bad conductor of heat and therefore, the loss or gain of heat due to conduction is prevented. As there is vacuum between the double walls, it prevents transmission of heat by convection. Due to silvering of the inner surfaces, the heat loss by radiation is prevented. The cork and the felt pads also help to preserve the heat inside the flask. So, the hot liquids like tea, coffee, etc. remain hot and cold liquids like water, juice, etc. remain cold for a long time.

Question 4.
Write five applications of radiation.
Answer:
Applications of radiation:
(i) The base of cooking utensils is painted black so that it can absorb more heat and hence cooking is done fast.
(ii) Dark clothes are worn in winters as they absorb more heat from the sunlight and keep our body warm.
(iii) White clothes are worn in summer as they absorb very little heat and keep us cool.
(iv) Room-heaters have shiny reflectors as the shiny surface absorbs very little heat. They reflect all the heat which makes the room- heaters more effective.
(v) Crockeries are polished and painted white from outside so that they do not radiate heat easily.

Question 5.
Describe the process of radiation. How is it different from conduction and convection?
Answer:
Radiation is a process of transmission of heat in which heat energy travels in straight lines from hotter to cooler regions without the help of any intervening medium. It means that this process is different from the other two as it does not need any medium like conduction or convection. Radiation can take place in a vaccum. For example, heat from the sun reaches us through the outer space (vacuum) by means of radiation in the form of electromagnetic waves.

Differefnbe between radiation, conduction and convection:

Heat Class 7  HBSE Notes

• Our sense of touch is not always a reliable guide to the degree of hotness of an object.
• Heat: Heat is a form of energy which produces in us sensation of warmth. It causes hotness or coldness sensation in our body.
• Temperature: Temperature tells how hot a body is? Temperature of a body is the degree of hotness or coldness of the body.
• Heat flows in the direction of fall of temperature. Heat can be converted to other forms of energy like sound, light, mechanical energy etc. Other forms of energy can also be converted to heat energy.
• Effects of heat:
  (i) The object becomes hot.
  (ii) The object may expand in size.
  (iii) The object may change in state.
  (iv) Heat can speed up chemical reaction.
• Temperature is measured by a device called thermometer.
• The thermometer that measure our body temperature is called a clinical thermometer. The range of this thermometer is from 35°C to 42°C. For other purposes, we use the laboratory thermometers. The range of these thermometers is usually from 10°C to 110°c.
• The laboratory thermometer used in laboratories. It is called Celsius thermometer. It has one hundred markings on it.
• The melting point of pure ice is called lower standard point. Its value is 0°C.
• The boiling point of pure water is called upper standard point. Its value is 100°C.
• Transfer of heat: Heat flows from one body to another body by temperature difference. The flow of heat is from higher temperature to lower temperature. There are three different ways of heat transfer: conduction, convention and radiations.
• Conduction of heat: The process in which heat is handed over from one particle to another in the direction of fall of temperature without the actual movement of the particles of the medium is called conduction.
  
• Good conductor of heat: The materials which allow heat to pass through them easily, are called good conductor of heat. For example, silver, copper, aluminium, mercury etc. are good conductors of heat.
• Bad conductor of heat: The materials which do not easily allow heat to pass through them are called bad conductors of heat or insulators. Wood, plastics, mud, cork, cotton, ebonite, asbestos, most gases and liquids are bed conductor of heat.
• Convention of heat: Convention is a process by which heat is transferred from one part of a substance to another part through the actual movement of its constituents particles (atoms or molecules). Only liquids and gases convert heat. A solid cannot do so because solids do not flow like liquids or gases do.
• Radiation of heat: The process by which heat travels from one body to another without the intervention of any material medium is known as radiation.
  All hot objects transfer heat by radiation. Radiation does not require either a conducting medium or a convecting fluid. The sun transfers heat in all directions through the process of radiation. The radiations absorbed or emitted by a body depends upon the colour of the body.
  
• Thermos Flask (Vacuum Flask): Thermos flask is used to keep things warm and cold. It consists of a double walled glass flask, having vacuum between the walls, kept inside metallic case with the cork supporters.
  The outer and inner surfaces of the glass flask are highly silvered. The mouth of flask is fitted with air tight cork.
  
  (i) The vacuum between the two walls of the flask prevent heat transfer by conduction or convection.
  (ii) The silver coating of the walls of the vessels reflects back heat radiations. Thus heat by radiation can neither enter inside the flask nor go out of the flask.
  (iii) The air-tight cork, which is bad conductor of heat prevents any loss of heat by conduction or convection.
  Some heat may be transferred by conduction through thin glass walls at the neck and the poorly conducting cork. Total transfer of heat is very-very small, thus hot or cold liquid kept inside will keep its temperature for a long time.

Haryana State Board HBSE 7th Class Science Solutions Chapter 3 Fibre to Fabric Textbook Exercise Questions and Answers.

Haryana Board 7th Class Science Solutions Chapter 3 Fibre to Fabric

HBSE 7th Class Science Fibre to Fabric Textbook Questions and Answers

Question 1.
You must be familiar with the following nursery rhymes:
(i) Baa baa black sheep, have you any wool
(ii) Mary had a little lamb, whose fleece was white as snow
Answer the following:
(a) Which parts of the black sheep have wool?
(b) What is meant by the white fleece of the lamb?
Answer:
(a) The Hair (fleece) of the black sheep have wool.
(b) The white fleece is the v/hite hair of the lamb.

Question 2.
The silkworm is (a) a caterpiller, (b) a Larva. Choose the correct option.
(i) a
(ii) b
(iii) both a and b
(iv) neither a nor b.
Answer:
(iii) both a and b.

Question 3.
Which of the following does not yield wool?
(i) Yak
(ii) Camel
(iii) Goat
(iv) Woolly dog.
Answer:
(iv) Woolly Dog.

Question 4.
What is meant by the following terms?
(i) Reasing
(ii) Shearing
(iii) Sericulture
Answer:
(i) The fleece of the sheep alongwith a thin layer of skin is removed from its body. This process is called shearing.
(ii) Silk fibres are also animal fibres. Silkworms spin the ‘silk fibres’.
(iii) The rearing of silkworms for obtaining silk is called sericulture.

Question 5.
Given below is a sequence of steps in the processing of wool. Which are the missing steps? Add them.
Shearing, ………………. , sorting, ………………. , ………………. , ………………. , ………………. .
Answer:
Shearing, scouring, sorting, drying, dying, spinning, weaving.

Question 6.
Make sketches of the two stages in the life history of the silk moth which are directly related to the production of silk.
Answer:

(a) Female silkworm moth with eggs

Question 7.
Out of the following, which are the two terms related to silk production? Sericulture, Floriculture, Moriculture, Apiculture, Silviculture
Hints:
(i) Silk production involves cultivation of mulberry leaves and rearing silkworms.
(ii) Scientific name of mulberry is Morus alba.
Answer:
Sericulture, Moriculture.

Question 8.
Match the words of column I with those given in column II.

Answer:

Question 9.
Given below is a crossword puzzle based on this lesson. Use hints to fill in the blank spaces with letters that complete the words.

Answer:

Extended Learning-Activities And Projects

Question 1.
Paheli wants to know the maximum length of continuous silk thread that can be obtained from a cocoon.
Answer:
Do yourself. Take help to your teacher.

Question 2.
Boojho wants to know why caterpillars need to shed their skin when they grow bigger but we humans do not.
Answer:
Do yourself.

Question 3.
Boojho wants to know why caterpillars should not be collected with bare hands.
Answer:
Do yourself.

Question 4.
Paheli wanted to buy a silk frock and went to the market with her mother. There they found that the artificial (synthetic) silk was much cheaper and wanted to know why Do you know why? Find out.
Answer:
Do yourself.

Question 5.
Someone told Paheli that an animal called ‘Vicuna’ also gives wooljl Can you tell her where this animal is found? Look for this in a dictionary or an encyclopaedia.
Answer:
Do yourself.

Question 6.
When handloom and textile exhibitions are held, certain stalls display real moths of various varieties of silk and their life histories. Try and visit these stalls with elders or teachers and see these moths and stages of their life history.
Answer:
Do yourself.

Question 7.
Look for eggs of any moth or butterfly in your garden or park or any other place full of plants. They look like tiny specks (dots) laid in a cluster on the leaves. Pull out the leaves containing eggs and-place them in a cardboard box. Take some leaves of the same plant or another plant of the same variety, chop them and put them in the box. Eggs will hatch into caterpillars, which are busy eating day and night. Add leaves everyday for them to feed upon. Sometimes you may be able to collect the caterpillars. But be careful. Use a paper napkin or a paper to hold a caterpillar.
Observe everyday. Note the
(i) number of days taken for eggs to hatch
(ii) number of days taken to reach the cocoon stage and
(iii) to complete life cycle. Record your observations in your notebook.
Answer:
Do yourself.

HBSE 7th Class Science Fibre to Fabric Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Name the sources of obtaining fibres.
Answer:
The sources of obtaining fibres are plants and animals.

Question 2.
Name some wool yielding animals.
Answer:
Some wool yielding animals are: sheep, angora goat, goat, yak, camel, llama, alpaca.

Question 3.
What is shearing?
Answer:
The fleece (hair) of the sheep alongwith a thin layer of skin is removed from its body. This process is called shearing.

Question 4.
From where do we get wool and silk fibres?
Answer:
Wool and silk fibres are obtained from animals. Wool is obtained from hair of animals ; as sheep hairs and silk fibtes are obtained from silk worm.

Question 5.
What is wool?
Answer:
Wool is the common name of applied to soft curly fibres obtained chiefly from the fleece of domesticated sheep.

Question 6.
What determines the fineness of wool?
Answer:
The number of crimps per centimetre determines the fineness of wool fibre.

Question 7.
Name two breeds of sheep which has finest fleece?
Answer:
Two breeds of sheep which has finest fleece are: merino and Karakul.

Question 8.
Name two systems used in the processing of wool.
Answer:
Two system used in the processing of wool are:
1. Woollen system
2. Worsted system.

Question 9.
Name two substance removed from row wool, before manufacturing yam.
Answer:
Two substance removed from raw wool, before manufacturing yarn are:
1. Yolk
2. Suint.

Question 10.
What is woolmark?
Answer:
Woolmark is the mark of standardisation given to woollen yarns and woollen products. This mark has been instituted by International Wool Secretariat (I.W.S.) located in North Yorkshire (U.K.).

Question 11.
What are the uses of wool?
Answer:
Wool is used for making fabrics, shawls, blankets, carpets, felt and upholstery.

Question 12.
Write the properties of wool fibre.
Answer:
Wool fibre is considerably resilient, has high tensile strength, light weight and is heat insulator.

Question 13.
Name the domesticated moth whose larva produces silk.
Answer:
Silk is obtained from the cocoon of the pupa of mulberry silk moth.

Question 14.
Name the plant on which silk larva feeds.
Answer:
Silk larva feeds on the leaves of mulberry tree.

Question 15.
Name the person responsible for the discovery of silk.
Answer:
Silk was discovered by Xi-Ling-Shi, the bride of Chinese emperor Huang Di, in about 3000 B.C.

Question 16.
Name the strongest variety of silk.State whether the variety you have named is wild silk or domesticated silk.
Answer:
The strongest variety of silk is “Muga”. It is the variety of wild silk.

Question 17.
Name the different varieties of processed silk.
Answer:
Organize, Gepe, Tram and Thrown Singles are different, varieties of processed silk.

Question 18.
Give uses of silk, other than for making dresses.
Answer:
Other than for making dresses, silk is used for items such as parachutes, bicycle tyres, bullet proof vests and non-absorbable sutures in surgery.

Question 19.
What is sericulture?
Answer:
The science of raising silkworms, so as to obtain silk cocoons, is called sericulture.

Question 20.
Where are sheep reared in India?
Answer:
In India sheep are reared in Sub-Himalyan region which has a cooler climate.

Question 21.
Differentiate between staple fibres and filament fibres.
Answer:
The fibres like cotton, jute, and wool are short in length. Silk however, is the longest natural fibre. The short fibres are known as the stable fibres, where as long fibres are known as the filament fibres.

Question 22.
Name some Indian breeds of sheep.
Answer:
Some Indian breeds of sheep are Lohi, Rampur bushair, Nali, Bakharwal, Marwari, and Patanwadi.

Question 23.
What is reeling of silk?
Answer:
A pile of cocoons is used for obtaining silk fibres. This process is called reeling of silk.

Short Answer Type Questions

Question 1.
Differentiate between woollen yarn and worsted yam.
Answer:
The woolen yarn is made from the woolen fibres of mixed lenghts. These fibres are, carded and spun to form a woolen yarn. The articles made from this yarn do not have smooth finished surface.

In the worsted yarn, only the long fibres are used. They are formed into smooth compact strands and are then spun to form woolen yarn. The articles made from these have smooth finished surface.

Question 2.
What do you mean by the term ‘weighting of sillf’?
Answer:
Silk when unwound from the cocoon is covered with a sticky substance. This has to be removed by washing process. As a result, silk loses about 20% of its weight. To make up for this loss some manufactures dip silk in metallic salt solution. This is known as weighting of the silk. This silk is weaker than pure silk. It than not be cleaned properly. This silk is of substandard quality and hence, deceives the buyers.

Question 3.
What is raw silk? How is it produced?
Answer:
After brushing, filaments from four to eight cocoons are joined and twisted. They are then combined with a number of other similarly twisted filaments, to make a thread that is wound on a reel. The thread is called raw silk.

It usually consists of 48 individual silk fibres. When each cocoon is unwound, it is replaced by another cocoon.Unlike the threads spun from other natural fibres, such as cotton or wool, the silk thread is made of extremely long fibres.

Question 4.
Name and describe any two types of silk threads used in the making of silk cloth.
Answer:
Two types of silk threads used in the making of silk cloth are:
1. Organzine:
Organzine thread is made by giving raw silk thread a preliminary twist in one direction and then twisting two or more of these threads in the opposite direction at rate of about 4 turns / cm.
2. Crepe: Crepe is similar to organzine, but it twisted to much greater extent, usually, 16 to 32 turns / cm.

Question 5.
Why do the wool fibres have greater bulk as compared to other fibres?
Answer:
The scales and crimps in the wool fibre make it possible to spin and felt the fleece. They help the individual wool fibres “grab” each other so that they stay together. Because of the crimp, wool fabrics have a greater bulk; than other textiles. They can retain lot of air and hence wool fabrics are good insulators of heat. The heat insulation also works boths ways.

Question 6.
Name the some breeds of sheep reared in our country. Also, indicate the quality and texture of the fibres obtained from them.
Answer:
Some Indian breeds of sheep:
Answer:

Question 7.
Write short note on ‘wool production’.
Answer:
Australia is the world’s largest producer of raw wool and contributes 29% of total world supply. Other leading producers of wool are former Soviet Republics, New Zealand, China, Argentina, South Africa and Uruguay. In India sheep are reared in Sub-Himalayan regions which has a cooler climate. The sheep which are reared in Rajasthan have poor quality of wool and are mainly reared for meat.

Long Answer Type Questions

Question 1.
What is raw silk? How is it produced?
Answer:
After brushing, filaments from four to eight cocoons are joined and twisted. They are then combined with a number of other similarly twisted filaments, to make a thread that is wound oft a reel. The thread is called raw silk. It usually consists of 48 individual silk fibres. When each cocoon is unwound, it is replaced by another cocoon. Unlike the threads spun from other natural fibres, such as cotton or wool, the silk thread is made of extremely long fibres. About 5,500 cocoons are required to produced 1 kg of raw silk.

Question 2.
What is sericulture? Explain in brief.
Answer:
The silk is obtained from cocoon. When the complete cocoon is formed, they are collected. The pupa inside the cocoon is killed by treating the cocoons with boiling water. This hot water not only kills the cocoons but also dissolves the sticky substance that holds the cocoon filaments in place. The cocoons are, thereafter, dried and brushed. This removes the outer portion that is made of coarse filaments.

The thread is then unwound from the cocoons and is wound on a reel. The thread is called the raw silk. It generally consists of 48 individual silk fibres. The silk filaments are unwound from cocoon in a manufacturing plant known as filature. The silk thread is made up of extremely long fibres. About 5,500 silkworms are needed to produce 1 kg of raw silk. The breeding and management of silk moths for producing silk is known as sericulture.

Question 3.
Explain briefly the process of wool manufacturing.
Answer:
In order to make different articles, the raw wool is processed which starts with sorting of the fibres. The fibres are sorted on the basis of length, fineness, etc. The sorting is followed by cleaning 6f the fibres. Using a suitable detergent, the fibres are cleaned from lanolin (an oily substance) and suint (the dried perspiration of sheep). The cleaned wool fibres are then dried and disentangled. Thereafter, they are drawn into a straight continuous thread. This process is called cerding. The kind of yarns that are produced may be woollen yarn or worsted yarn.

For the production oi wooiien yarn, the web is split into fine thin strands, by a process called roving. It is then spun into woollen yarn. Woollen fabric such as tweed is woven from woollen yarn. It is a bulky fabric in which short woollen fibres are arranged randomly. It does not have smooth surface.

For the production of worsted yam, the web is processed by the machines that (i) straighten the fibres (ii) make them parallel (iii) remove all the short fibres. The resultant woolen strand is then condensed by passing it through several machines, until a very thin strand of worsted roving is obtained. It is then spun to a smooth yarn.

Question 4.
What are the uses of wool? What is woolmark?
Answer:
Wool is used for making fabrics, shawls, blankets, carpets, felt (compressed wool) and upholstery. Wool felt is used to cover piano hammers. It is also used to absorb noise in heavy machinery and stereo speakers.

Shoddy is made from the used wool. To make shoddy, the existing wool fabric is cut into small pieces and then carddd. The carded wool is then respun into yarn. Such a yarn is inferior to the fresh wool and is used for making cheap woollen garments and blankets.

Woolmark:
Wool is marketed worldwide by the International Wool Secretariat (IWS), which is based in Ilkley, North Yorkshise (U.K.). The “Woolmark” instituted by IWS indicates that garments bearing this sign are made from pure new wool, that has not been used previously in any process.

Fibre to Fabric Class 7  HBSE Notes

• Wool comes from sheep, goat, yak and some other animals. These wool- yielding animals bear hair (fleece) on their body.
• The hairy skin of the sheep has two types of fibres that form its fleece : (i) the coarse beard hair, and (ii) the fine soft under hair close to the skin.
• The wool which is used for knitting sweaters or for weaving shawls is the finished product of a long process.
• Sheep hair is sheared off from the body, scoured, sorted, dried, dyed, spun and woven to yield wool.
• The hairs of camel, llama and alpaca are also processed to yield wool.
• The wool may be classified as Virgin, Hogg or pulled depending upon its quality.
• In wool production, Australia and New Zealand produce two-thirds of the world’s supply.
• Woolmark is the mark of standardisation given to pure and new wool.
• Silk fibres are also animal fibres. Silkworms spin the ‘silk fibres’. The rearing of silkworms for obtaining silk is called sericulture.
• The female silk moth lays eggs, from which hatch larvae which are called caterpillars or silkworms.
• During their life cycle, the worms spin cocoons of silk fibres.
• Silk fibres are made of a protein.
• Tassar silk, mooga silk, kosa silk, etc., are obtained from cocoons spun by different types of moths.
• The most common silk moth is the mulberry silk moth. The silk fibre from the cocoon of this moth is soft, lustrous and elastic and can be dyed in beautiful colours.
• Sericulture is a very old occupation in India. India produces plenty of silk on a commercial scale.
• A pile of cocoons is used for obtaining silk fibres. This process is called reeling of silk. The cocoons are kept under the sun or boiled or exposed to steam. The silk fibres separate out. The process of taking out threads from the cocoon for use as silk is called reeling the silk.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 13 घातांक और घात InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 13 घातांक और घात InText Questions

प्रयास कीजिए (पृष्ठ सं. 266)

प्रश्न 1.
ऐसे पाँच और उदाहरण दीजिए, जहाँ एक संख्या को घातांकीय रूप में व्यक्त किया जाता है। प्रत्येक स्थिति में, घातांक व आधार की पहचान भी कीजिए।
हल :
पाँच उदाहरण हैं :
(i) 2⁵ = 2 × 2 × 2 × 2 × 2 = 32
2⁵ में 2 आधार और 5 घातांक है।

(ii) 3⁵ = 3 × 3 × 3 × 3 × 3 = 243
3⁵ में 3 आधार और 5 घातांक है।

(iiii) 4³ = 4 × 4 × 4 = 64
4³ में 4 आधार और 3 घातांक है।

(iv) 2⁶ = 2 × 2 × 2 × 2 × 2 × 2
= 64
2⁶ में 2 आधार और 6 घातांक है।

(v) 7⁵ = 7 × 7 × 7 × 7 × 7
= 16807
75 में 7 आधार और 5 घातांक है।

पृष्ठ सं. 267

प्रश्न 1.
व्यक्त कीजिए :
(i) 729 को 3 की घात के रूप में
(ii) 128 को 2 की घात के रूप में
(iii) 343 को 7 की घात के रूप में
हल :
(i)

∴ 729 = 3 × 3 × 3 × 3 × 3 × 3
= 3⁶

(ii)

∴ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁷

(iii)

343 = 7 × 7 × 7 = 7³

पृष्ठ सं. 270

प्रश्न 1.
सरल करके घातांकीय रूप में लिखिए :
(i) 2⁵ × 2³
(ii) p³ × p2
(iii) 4³ × 4²
(iv) a³ × a² × a⁷
(v) 5³ × 5⁷ × 5¹²
(vi) (-4)¹⁰⁰ × (-4)²⁰
हल :
(i) 2⁵ × 2³ = 2^{5 + 3} = 2⁸
(ii) p³ × p² = p^{3 + 2} = p⁵
(iii) 4³ × 4² = 4^{3 + 2} = 4⁵
(iv) a³ × a² × a⁷ = a3 + 2 + 7 = a¹²
(v) 5³ × 5⁷ × 5¹² = 5^{3 + 7 + 12} = 5²²
(vi) (- 4)¹⁰⁰ × (- 4)²⁰ = (- 4)^{100 + 20}
= (-4)¹²⁰

पृष्ठ सं. 271

प्रश्न 1.
सरल करके घातांकीय रूप में लिखिए:
(उदाहरण के लिए, 11⁶ ÷ 11² = 11⁴)
(i) 2⁹ ÷ 2³
(ii) 10⁸ ÷ 10⁴
(iii) 9¹¹ ÷ 9⁷
(iv) 20¹⁵ ÷ 20¹³
(v) 7¹³ ÷ 7¹⁰
हल :
(i) 2⁹ ÷ 2³ = 2^{9 – 3} = 2⁶
(ii) 10⁸ ÷ 10⁴ = 10^{8 – 4} = 10⁴
(iii) 9¹¹ ÷ 9⁷ = 9^{11 – 7} = 9⁴
(iv) 20¹⁵ ÷ 20¹³ = 20^{15 – 13} = 20²
(v) 7¹³ ÷ 7¹⁰ = 7^{13 – 10} = 7³

पृष्ठ सं. 271

प्रश्न 1.
सरल करके, उत्तर को घातांकीय रूप में व्यक्त कीजिए-
(i) (6²)⁴
(ii) (2²)¹⁰⁰
(iii) (7⁵⁰)²
(iv) (5³)⁷
हल :
(i) (6²)⁴ = 6^{2 × 4}  = 68
(ii) (2²)¹⁰⁰ = 2^{2 × 100} = 2²⁰⁰
(iii) (7⁵⁰)² = 7^{50 × 2} = 7¹⁰⁰
(iv) (5³)⁷ = 5^{3 × 7} = 5²¹

पृष्ठ सं. 273

प्रश्न 1.
a^m × b^m = (ab)^m का प्रयोग करके, अन्य रूप में बदलिए :
(i) 4³ × 2³
(ii) 2⁵ × 5⁵
(iii) a² × t²
(iv) 5⁶ × (-2)⁶
(v) (-2)⁴ × (-3)⁴
हल :
(i) 4³ × 2³ = (4 × 4 × 4) × (2 × 2 × 2)
= (4 × 2) × (4 × 2) (4 × 2)
= (4 × 2)³

(ii) 2⁵ × b⁵ = (2 × 2 × 2 × 2 × 2) × (b × b × b × b × b)
= (2 × b) × (2 × b) × (2 × b) × (2 × b) × (2 × b)
= (2 × b)⁵ = (2b)⁵

(iii) a² × t² = (a × a) × (1 × 1)
= (a × t) × (a × t)
= (at) × (at) = (at)²

(iv) 5⁶ × (-2)⁶ = 5 × 5 × 5 × 5 × 5 × 5 × (-2 × – 2 × – 2 × – 2 × – 2 × – 2)
= (5 × -2) (5 × – 2) × (5 × – 2) × (5 × – 2) (5 × – 2) × (5 × – 2)
= (5 × – 2)⁶

(v) (-2)⁴ × (-3)⁴ = (- 2 × – 2 × – 2 × – 2) × (- 3 × – 3 × – 3 × 3)
= (- 2 × – 3) × (- 2 × – 3) × (- 2 × – 3) × (- 2 × – 3)
= (- 2 × – 3)⁴

पृष्ठ सं. 273

प्रश्न 1.
a^m ÷ b^m = \(\left(\frac{a}{b}\right)^{m}\) का प्रयोग करके, अन्य रूप में बदलिए:
(i) 4⁵ ÷ 3⁵
(ii) 2⁵ ÷ b⁵
(iii) (-2)³ ÷ b³
(iv) p⁴ + q⁴
(v) 5⁶ + (-2)⁶
हल :
a^m ÷ a^m = \(\left(\frac{a}{b}\right)^{m}\) का प्रयोग करने पर,

पृष्ठ सं. 277

प्रश्न 1.
10 की घातों का प्रयोग करते हुए, घातांकीय रूप में प्रसारित कीजिए :
(i) 172
(ii) 5643
(ii) 56439
(iv) 176428
हल :
(i) 172 = 1 × 100 + 7 × 10 + 2 × 1
= 1 × 10² + 7 × 10¹ + 2 × 10⁰

(ii) 5643 = 5 × 1000 + 6 × 100 + 4 × 10 + 3 × 1
= 5 × 10³ + 6 × 10² + 4 × 10¹ + 3 × 10⁰

(iii) 56439 = 5 × 10000 + 6 × 1000 + 4 × 100 + 3 × 10 + 9 × 1
= 5 × 104 + 6 × 10³ + 4 × 10² + 3 × 10¹ + 9 × 10⁰

(iv) 176428 = 1 × 100000 + 7 × 10000 + 6 × 1000 + 4 × 100 + 2 × 10 +8 × 1
= 1 × 10⁵ + 7 × 10⁴ + 6 × 10³ + 4 × 10² + 2 × 10¹ + 8 × 10⁰

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 सरल समीकरण Ex 4.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 सरल समीकरण Ex 4.2

प्रश्न 1.
पहले चर को पृथक् करने वाला चरण बताइए और फिर समीकरण को हल कीजिए :
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4
हल :
(a) x – 1 = 0
दोनों पक्षों में 1 जोड़ने पर,
x – 1 + 1 = 0 + 1
⇒ x + 0 = 1
⇒ x = 1
∴ x = 1 दिये गये समीकरण का हल है। उत्तर

(b) x + 1 = 0
दोनों तरफ से 1 घटाने पर,
x + 1 – 1 = 0 – 1
⇒ x + 0 = – 1
⇒ x = – 1
∴ x = – 1 दिये गये समीकरण का हल है। उत्तर

(c) x – 1 = 5
दोनों तरफ 1 जोड़ने पर,
x – 1 + 1 = 5 + 1
⇒ x + 0 = 6
⇒ x = 6
∴ x = 6 दिये गये समीकरण का हल है। उत्तर

(d) x + 6 = 2
इसके लिए दोनों तरफ से 6 घटाने पर,
⇒ x + 6 – 6 = 2 – 6
⇒ x + 0 = – 4
⇒ x = – 4
∴ x = – 4 दिये गये समीकरण का हल है। उत्तर

(e) y – 4 = – 7
तब दोनों तरफ 4 जोड़ने पर,
⇒ y – 4 + 4 = – 7 + 4
⇒ y + 0 = – 3
⇒ y = – 3 दिये गये समीकरण का हल है। उत्तर

(f) y – 4 = 4
दोनों ओर 4 जोड़ने पर,
y – 4 + 4 = 4 + 4
⇒ y + 0 = 8
⇒ y = 8
∴ y = 8 दिये गये समीकरण का हल है। उत्तर

(g) y + 4 = 4
दोनों ओर से 4 घटाने पर,
y + 4 – 4 = 4 – 4
⇒ y + 0 = 0
⇒ y = 0
∴ y = 0 दिये गये समीकरण का हल है। उत्तर

(h) y + 4 = – 4
दोनों ओर से 4 घटाने पर,
y + 4 – 4 = – 4 – 4
⇒ y + 0 = – 8
⇒ y = – 8
∴ y = – 8 दिये गये समीकरण का हल है। उत्तर

प्रश्न 2.
पहले चर को पृथक् करने के लिए प्रयोग किए जाने वाले चरण को बताइए और फिर समीकरण को हल कीजिए :
(a) 3l = 42
(b) \(\frac {b}{2}\) = 6
(c) \(\frac {p}{7}\) = 4
(d) 4x = 25
(e) 8y = 36
(f) \(\frac{z}{3}=\frac{5}{4}\)
(g) \(\frac{a}{5}=\frac{7}{15}\)
(h) 20t = – 10
हल :
(a) 3l = 42
समीकरण को हल करने के लिए l को बाँयी ओर रखते हैं तथा 3 को बायीं तरफ से हटाते है। इसलिए लिए 3 का दोनों ओर भाग देने पर,
3l = 42
\(\frac{3 l}{3}=\frac{42}{3}\)
⇒ l = 14
∴ l = 14 दिये गये समीकरण का हल है। उत्तर

(b) \(\frac {b}{2}\) = 6
समीकरण को हल करने के लिए b को बाँयौं तरफ रखते हैं तथा बाँयी ओर से 2 को हटाते हैं। इसके लिये 2 का दोनों ओर गुणा करने पर,
= \(\frac {b}{2}\) × 2 = 6 × 2,
⇒ b = 12
[∵ \(\frac {b}{2}\) × 2 = b और 6 × 2 = 12]
∴ b = 12 समीकरण का हल है। उत्तर

(c) \(\frac {p}{7}\) = 4
समीकरण को हल करने के लिए p बायीं तरफ रखते हैं और 7 को बायीं ओर से हटाते हैं। इसके लिए 7 से दोनों ओर गुणा करने पर,
\(\frac {p}{7}\) × 7 = 4 × 7
⇒ p = 28
∴ p = 28 दिये गये समीकरण का हल है। उत्तर

(d) 4x = 25
समीकरण से 4 को बाँयीं तरफ से हटाने के लिए, समीकरण के दोनों ओर 4 का भाग देने पर,
\(\frac{4 x}{4}=\frac{25}{4}=\frac{25}{4}\)
∴ x = \(\frac {25}{4}\) दिये गये समीकरण का हल है। उत्तर

(e) 8y = 36
समीकरण से 8 को बायीं तरफ से हटाने के लिए, समीकरण के दोनों तरफ 8 से भाग देने पर,
\(\frac{8 y}{8}=\frac{36}{8}\)
⇒ y = \(\frac{36}{8}=\frac{9}{2}\)
∴ y = \(\frac {9}{2}\) दिये गये समीकरण का हल है। उत्तर

(f) \(\frac{z}{3}=\frac{5}{4}\)
इसको हल करने के लिए 2 को बायीं तरफ रखते हैं। और 3 को बायीं तरफ से हटाते हैं। इसके लिए इसके लिए दोनों ओर 3 से गुणा करने पर।
\(\frac{z}{3}=\frac{5}{4}\)
\(\frac {z}{3}\) × 3 = \(\frac {5}{4}\) × 3
⇒ z = \(\frac {15}{4}\)
∵ \(\frac {z}{3}\) × 3 = z और \(\frac {5}{4}\) × 3 = \(\frac {15}{4}\)]
∴ z = \(\frac {15}{4}\) दिये गये समीकरण का हल है। उत्तर

प्रश्न 3.
चर को पृथक् करने के लिए, जो आप चरण प्रयोग करेंगे, उसे बताइए और फिर समीकरण को हल कीजिए :
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) \(\frac {20p}{3}\) = 40
(d) \(\frac {3p}{10}\) = 6
हल :
(a) 3n – 2 = 46
दोनों ओर 2 जोड़ने पर,
3n – 2 + 2 = 46 + 2
⇒ 3n = 48
दोनों ओर 3 का भाग देने पर,
\(\frac{3 n}{3}=\frac{48}{3}\)
⇒ n = 16, समीकरण का हल है। उत्तर

(b) 5m + 7 = 17
दोनों ओर से 7 घटाने पर,
5m + 7 – 7 = 17 – 7
⇒ 5m + 0 = 10
⇒ 5m = 10
दोनों ओर 5 का भाग देने पर,
\(\frac{5 m}{5}=\frac{10}{5}\)
⇒ m = 2
∴ m = 2, समीकरण का हल है। उत्तर

प्रश्न 4.
निम्नलिखित समीकरणों को हल कीजिए :
(a) 10p = 100
(b) 10p + 10 = 100
(c) \(\frac {p}{4}\) = 5
(d) \(\frac {-p}{3}\) = 5
(e) \(\frac {3p}{4}\) = 6
(f) 3s = -9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
हल :
(a) 10p = 100
दोनों ओर 10 का भाग देने पर,
\(\frac{10 p}{10}=\frac{100}{10}\)
⇒ p = 10
∴ p = 10, समीकरण का हल है।

(b) 10p + 10 = 100
दोनों तरफ से 10 घटाने पर,
10p + 10 – 10 = 100 – 10
⇒ 10p + 0 = 90
⇒ 10p = 90
दोनों ओर 10 का भाग देने पर,
⇒ \(\frac{10 p}{10}=\frac{90}{10}\)
⇒ p = 9
∴ p = 9. समीकरण का हल है।

(c) \(\frac {p}{4}\) = 5
दोनों ओर 4 का गुणा करने पर,
\(\frac {p}{4}\) × 4 = 5 × 4
⇒ p = 20
∴ p = 20, समीकरण का हल है। उत्तर

(d) \(\frac {-p}{3}\) = 5
दोनों ओर – 3 का गुणा करने पर,
\(\frac {-p}{3}\) × (-3) = 5 × (-3)
⇒ \(\frac {3p}{3}\) = – 15
∴ p = – 15
∴ p = – 15, समीकरण का हल है।

(e) \(\frac {3p}{4}\) = 6
दोनों ओर \(\frac {4}{3}\) से भाग देने पर,
\(\frac{3 p}{4} \times \frac{4}{3}=6 \times \frac{4}{3}\)
⇒ p = \(\frac {24}{3}\)
⇒ p = 8
∴ p = 8 समीकरण का हल है। उत्तर

(f) 3s = – 9
⇒ \(\frac{3 s}{3}=\frac{-9}{3}\) [दोनों ओर 3 से भाग देने पर]
⇒ s = – 3
∴ s = – 3 समीकरण का हल है। उत्तर

(g) 3s + 12 = 0
⇒ 3s + 12 – 12 = 0 – 12
[दोनों ओर से 12 घटाने पर]
⇒ 3s = – 12
⇒ \(\frac{3 s}{3}=\frac{-12}{3}\) [दोनों ओर 3 से भाग देने पर]
⇒ s = -4
∴ s = – 4 समीकरण का हल है। उत्तर

(h) 3s = 0
⇒ \(\frac{3 s}{3}=\frac{0}{3}\) [दोनों ओर 3 से भाग देने पर]
⇒ s = 0
∴ s = 0 ही समीकरण का हल है। उत्तर

(i) 2q = 6
⇒ \(\frac{2 q}{2}=\frac{6}{2}\) [दोनों ओर 2 से भाग देने पर]
⇒ q = 3
∴ q = 3 समीकरण का हल है। उत्तर

(j) 2q – 6 = 0
दोनों ओर 6 जोड़ने पर,
⇒ 2q – 6 + 6 = 0 + 6
⇒ 2q + 0 = 6
⇒ 2q = 6
दोनों ओर 2 का भाग देने पर,
\(\frac{2 q}{2}=\frac{6}{2}\)
⇒ q = 3
∴ q = – 3, समीकरण का हल है। उत्तर

(k) 2q + 6 = 0
दोनों ओर से 6 घटाने पर,
2q + 6 – 6 = 0 – 6
⇒ 2q + 0 = – 6
⇒ 2q = – 6
दोनों ओर 2 का भाग देने पर,
\(\frac{2 q}{2}=\frac{-6}{2}\)
⇒ q = – 3
∴ q = – 3, समीकरण का हल है। उत्तर

(l) 2q + 6 = 12
दोनों ओर से 6 घटाने पर,
2q + 6 – 6 = 12 – 6
⇒ 2q + 0 = 6
⇒ 2q = 6
दोनों ओर 2 का भाग देने पर,
\(\frac{2 q}{2}=\frac{6}{2}\)
⇒ q = 3
∴ q = 3 समीकरण का हल है। उत्तर

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 सरल समीकरण InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 सरल समीकरण InText Questions

पृष्ठ सं. 89 (प्रयास कीजिए) 

प्रश्न 1.
व्यंजक (10y – 20) का मान y के मान पर निर्भर करता है। को पाँच भिन्न-भिन्न मान देकर सथा। के प्रत्येक मान के लिए (10y – 20) का मान ज्ञात करके इसकी पुष्टि कीजिए। (10y – 20) के प्राप्त किए गए विभिन्न मानों से, क्या आप 10y – 20 = 50 का कोई हल देख रहे हैं? यदि कोई हल प्राप्त नहीं हुआ है, तोy को कुछ अन्य मान देकर, ज्ञात कीजिए कि प्रतिबंध 10y – 20 = 50 संतुष्ट होता है या नहीं। .
हल :
y के विभिन्न मानों के लिए (10y – 20) निकालने पर, जब y = 1, तब
10y – 20= 10 × 1 – 20 = 10 – 20 = – 10
जब y = 2, तब
10y – 20 = 10 × 2 – 20 = 20 – 20 = 0
जब y = 3, तब
10y – 20 = 10 × 3 – 20 = 30 – 20 = 10
जब y = 4, तब
10y – 20 – 10 × 4 – 20 = 40 – 20 = 20
और जब y = 5, तब
10y – 20 = 10 × 5 – 20 = 50 – 20 = 30
स्पष्ट है कि (10y – 20) का मान के मानों पर निर्भर करता है।
लेकिन y के मान 1, 2, 3, 4 और 5 रखने पर हमें 10y – 20 = 50 प्राप्त नहीं हुआ।
इसलिए हम y का मान 5 से अधिक रखकर कोशिश करते हैं।
जब y = 6, तब
10y – 20 = 10 x 6 – 20 = 60 – 20 = 40
जब y = 7, तब
10y – 20 = 10 x 7 – 20 = 70 – 20 = 50
अत: y = 7 रखने पर 10y – 20 का मान 50 आता है।
∴ y = 7 इसका हल है।

पृष्ठ सं. 89

प्रश्न 2.
निम्न समीकरणों में से प्रत्येक के लिए कम-से-कम एक अन्य कथन के रूप में लिखिए :
समीकरण : 5p = 20; 3n +7 = 1; \(\frac {m}{5}\) – 2 = 6
हल :
स्थिति इस प्रकार है:
(i) p का गुणा 5 में करने से 20 प्राप्त होता हैं।
(ii) 3 और n के गुणा में 7 जोड़ा जाता है, तो 1 प्राप्त होता है।
(iii) m का \(\frac {1}{5}\) से 2 कम 6 होता है।

पृष्ठ सं. 99

प्रश्न 1.
x = 5 से प्रारम्भ करते हुए दो भिन्न समीकरण बनाइये।
हल :
प्रथम समीकरण :
x = 5 से प्रारम्भ करने पर, दोनों ओर 7 से गुणा करने पर,
7x = 35
दोनों ओर 2 घटाने पर,
7x – 2 = 35 – 2
⇒ 7x – 2 = 33

द्वितीय समीकरण
x – 5 से प्रारम्भ करने पर, दोनों ओर 3 से भाग देने पर,
\(\frac{x}{3}=\frac{5}{3}\)
दोनों ओर 1 घटाने पर,
\(\frac {x}{3}\) – 1 = \(\frac {5}{3}\) – 1
\(\frac {x}{3}\) – 1 = \(\frac {2}{3}\)

प्रश्न 2.
दो संख्या पहेलियों को बनाने का प्रयास कीजिए। एक हल 11 लेकर तथा दूसरा हल 100 लेकर।
हल :
प्रथम पहेली 11 लेकर : अंक सोचिए, उसमें 3 से गुणा करो और गुणनफल में 2 जोड़िए। परिणाम 35 मिलता है। मुझे अंक बताइए।
दूसरी पहेली 100 लेकर : अंक सोचिए, इसमें 10 से भाग दीजिए। भागफल से 5 घटाओ। परिणाम 5 मिलता है। मुझे अंक बताइए।

पृष्ठ सं. 101

प्रश्न 1.
(i) जब आप एक संख्या को 6 से गुणा करते हैं और फिर गुणनफल में से 5 घटाते हैं, तो आपको 7 प्राप्त होता है। आप बता सकते हैं कि वह संख्या क्या है ?
हल :
माना वह संख्या x है।
प्रश्नानुसार,
6x – 5 = 7
⇒ 6x = 7 + 5 = 12
⇒ \(\frac{6 x}{6}=\frac{12}{6}\)
⇒ x = 2
अत: संख्या 2 है। उत्तर

(ii) वह कौन-सी संख्या है, जिसके एक-तिहाई में 5 जोड़ने पर 8 प्राप्त होता है?
हल :
माना वह संख्या y है।
प्रश्न के अनुसार,
\(\frac {y}{3}\) + 5 = 8
⇒ \(\frac {y}{3}\) = 8 – 5 = 3
⇒ y = 3 × 3 = 9
⇒ y = 9
अत: संख्या 9 है। उत्तर

प्रश्न 2.
मापों के अनुसार दो पेटियाँ हैं, जिनमें आम रखे हुए हैं। प्रत्येक बड़ी पेटी में रखे आमों की संख्या 8 छोटी पेटियों में रखे आमों की संख्या से 4 अधिक है। प्रत्येक बड़ी पेटी में 100 आम हैं। प्रत्येक छोटी पेटी में कितने आम हैं ?
हल :
माना छोटी पेटी में आमों की संख्या x है।
∴ 8 पेटियों में आमों की संख्या = 8x
तथा बड़ी पेटी में आर्मों की संख्या = 8x + 4
प्रश्नानुसार, 8x + 4 = 100
⇒ 8x + 4 – 4 = 100 – 4
⇒ 8x = 96
⇒ \(\frac{8 x}{8}=\frac{96}{8}\)
⇒ x = 12
छोटी पेटी में आमों की संख्या 12 है। उत्तर

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 सरल समीकरण Ex 4.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 सरल समीकरण Ex 4.4

प्रश्न 1.
निम्नलिखित स्थितियों के लिए समीकरण बनाइए और फिर उन्हें हल करके अज्ञात संख्याएँ ज्ञात कीजिए:
(a) एक संख्या के आठ गुने में 4 जोडिए; आपको 60 प्राप्त होगा।
(b) एक संख्या का \(\frac {1}{5}\) घटा 4, संख्या 3 देता है।
(c) यदि मैं किसी संख्या का तीन-चौथाई लेकर इसमें 3 जोड़ दूं, तो मुझे 21 प्राप्त होते हैं।
(d) जब मैंने किसी संख्या के दुगुने में से 11 को घटाया, तो परिणाम 15 प्राप्त हुआ।
(e) मुना ने 50 में से अपनी अभ्यास-पुस्तिकाओं की संख्या के तिगुने को घटाया, तो उसे परिणाम 8 प्राप्त होता है।
(f) इबेनहल एक संख्या सोचती है। वह इसमें 19 जोड़कर योग को 5 से भाग देती है, उसे 8 प्राप्त होता है।
(g) अनवर एक संख्या सोचता है। यदि वह इस संख्या के \(\frac {5}{2}\) में से 7 निकाल दे, तो परिणाम 23 है।
हल :
(a) माना वह संख्या x है।
∴ समीकरण है:
8x + 4 = 60
⇒ 8x + 4 – 4 = 60 – 4
⇒ 8x = 56 है :
⇒ \(\frac{8 x}{8}=\frac{56}{8}\)
⇒ x = 7

(b) माना वह संख्या x है।
∴ समीकरण : \(\frac {x}{5}\) – 4 = 3
⇒ \(\frac {x}{5}\) × 5 – 4 = 3 × 5
⇒ x – 20 = 15
⇒ x – 20 + 20 = 15 + 20
⇒ x = 35

(c) माना अभीष्ट संख्या y है।
∴ समीकरण है:
\(\frac {3y}{4}\) + 3 = 21
⇒ 4 × \(\frac {3y}{4}\) + 4 × 3 = 4 × 21
⇒ 3y + 12 = 84
⇒ 3y + 12 – 12 = 84 – 12
⇒ 3y = 72
⇒ \(\frac{3 y}{3}=\frac{72}{3}\)
⇒ y = 24

(d) माना वह संख्या m है।
∴ समीकरण है: 2m – 11 = 15
⇒ 2m = 15 + 11
⇒ 2m = 26
⇒ \(\frac{2 m}{2}=\frac{26}{2}\)
⇒ m = 13 उत्तर

(e) माना मुन्ना के पास x अभ्यास-पुस्तिकाएँ हैं।
∴ समीकरण है: 50 – 3x = 8
⇒ 50 – 3x – 50 = 8 – 50
⇒ -3x = – 42
⇒ \(\frac{-3 x}{-3}=\frac{-42}{-3}\)
⇒ x = 14 उत्तर

(f) माना संख्या x है।
∴ समीकरण है : \(\frac{x+19}{5}\) = 8
⇒ 5 × \(\frac{x+19}{5}\) = 8 × 5
⇒ x + 19 = 40
⇒ x + 19 – 19 = 40 – 19
⇒ x = 21 उत्तर

(g) माना वह संख्या x है।
∴ समीकरण है: \(\frac{5x}{2}\) – 7 = 23
प्रत्येक पद में 2 से गुणा करने पर
2 × \(\frac{5x}{2}\) – 7 × 2 = 2 × 23
5x – 14 = 46
⇒ 5x – 14 + 14 = 46 + 14
⇒ 5x = 60
⇒ \(\frac{5 x}{5}=\frac{60}{5}\)
⇒ x = 12 उत्तर

प्रश्न 2.
निम्नलिखित को हल कीजिए :
(a) अध्यापिका बताती है कि उनकी कक्षा में एक विद्यार्थी द्वारा प्राप्त किए गए अधिकतम अंक, प्राप्त किए
न्यूनतम अंक का दुगुना जमा 7 है। प्राप्त किए गए अधिकतम अंक 87 हैं। प्राप्त किए गए न्यूनतम अंक क्या है? .
(b) किसी समद्विबाहु त्रिभुज में आधार कोण बराबर होते हैं। शीर्ष कोण 40° है। इस त्रिभुज के आधार कोण क्या हैं ? (याद कीजिए कि त्रिभुज के तीनों कोणों का योग 180° होता है।)
(c) सचिन द्वारा बनाए गए रनों की संख्या राहुल द्वारा बनाए गए रनों की संख्या की दुगुनी है। उन दोनों द्वारा मिलाकर बनाए गए कुल रन एक दोहरे शतक से 2 रन कम हैं। प्रत्येक ने कितने रन बनाए थे ?
हल :
(a) माना न्यूनतम अंक हैं, तब अधिकतम अंक = 2x + 7
लेकिन अधिकतम अंक 87 दिए हुए हैं।
∴ 2x + 7 = 87
⇒ 2x + 7 – 7 = 87 – 7
⇒ 2x = 80
⇒ \(\frac{2 x}{2}=\frac{80}{2}\)
⇒ x = 40
न्यूनतम अंक = 40 उत्तर

(b) माना आधार कोण = x°
और शीर्ष कोण = 40°
क्योंकि त्रिभुज के तीनों कोणों का योग 180° होता है।
∴ x° + x° + 40° = 180°
⇒ 2x° + 40° – 40° = 180° – 40°
⇒ 2x° = 140°
⇒ \(\frac{2 x^{\circ}}{2}=\frac{140^{\circ}}{2}\)
⇒ x° = 70°,
प्रत्येक आधार कोण = 70° उत्तर

(c) माना राहुल के रनों की संख्या x है। तब सचिन के रनों की संख्या 2x होगी।
प्रश्नानुसार,
x + 2x = 200 – 2
⇒ 3x = 198
⇒ \(\frac{3 x}{3}=\frac{198}{3}\)
⇒ x = 66.
∴ राहुल के रन = 66 और सचिन के रन = 2 × 66 = 132 उत्तर

प्रश्न 3.
निम्नलिखित को हल कीजिए:
(i) इरफान कहता है कि उसके पास परमीत के पास जितने कंचे हैं उनके पाँच गुने से 7 अधिक कंचे हैं। इरफान के पास 37 कैंचे हैं। परमीत के पास कितने कैचे हैं?
(ii) लक्ष्मी के पिता की आयु 49 वर्ष है। उनकी आयु लक्ष्मी की आयु के तीन गुने से 4 वर्ष अधिक है। लक्ष्मी की आयु क्या है?
(iii) सुंदरग्राम के निवासियों ने अपने गाँव के एक बाग में कुछ पेड़ लगाए। इनमें से कुछ पेड़ फलों के थे। उन पेड़ों की संख्या, जो फलों के नहीं थे, फलों वाले पेड़ों की संख्या के तिगुने से 2 अधिक थी। यदि ऐसे पेड़ों की संख्या, जो फलों के नहीं थे, 77 है, तो लगाए गए फलों के पेड़ों की संख्या क्या थी?
हल :
(i) माना परमीत के पास x कंचे हैं, तब
इरफान के पास कैचे = 5x + 7
परन्तु इरफान के पास 37 कैंचे हैं।
∴ 5x + 7 = 37
⇒ 5x + 7 – 7 = 37 – 7
⇒ 5x + 0 = 30
⇒ 5x = 30
⇒ \(\frac{5 x}{5}=\frac{30}{5}\)
⇒ x = 6
अतः परमीत के पास 6 कैचे हैं। उत्तर

(ii) माना लक्ष्मी को आयु y वर्ष है, तब
उसके पिता की आयु = 3y + 4 वर्ष
प्रश्नानुसार, पिता की आयु 49 वर्ष है।
∴ 3y + 4 = 49
⇒ 3y + 4 – 4 = 49 – 4
⇒ 3y + 0 = 45
⇒ 3y = 45
⇒ \(\frac{3 y}{3}=\frac{45}{3}\)
⇒ y = 15
अत: लक्ष्मी की आयु = 15 वर्ष। उत्तर

(iii) माना फलों वाले पेड़ों की संख्या = x
इसलिए बिना फलों पेड़ों की संख्या = 3x + 2
प्रश्नानुसार,
3x + 2 = 77
⇒ 3x = 77 – 2
⇒ x = \(\frac {75}{3}\)
⇒ x = 25
∴ फलों के पेड़ों की संख्या 25 है।

प्रश्न 4.
निम्नलिखित पहेली को हल कीजिए :
मैं एक संख्या हूँ,
मेरी पहचान बताओ!
मुझे सात बार लो,
और एक पचास जोड़ो !
एक तिहरे शतक तक पहुँचने के लिए
आपको अभी भी चालीस चाहिए !
हल :
माना वह संख्या x है।
प्रश्नानुसार, 7x + 50 = 3 × 100 – 40
⇒ 7x + 50 = 300 – 40
⇒ 7x + 50 – 50 = 300 – 40 – 50
⇒ 7x = 210
⇒ \(\frac{7 x}{7}=\frac{210}{7}\)
⇒ x = 30
अत: वह अभीष्ट संख्या 30 है। उत्तर

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 सरल समीकरण Ex 4.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 सरल समीकरण Ex 4.3

प्रश्न 1.
निम्नलिखित समीकरणों को हल कीजिए :

हल :

प्रश्न 2.
निम्नलिखित समीकरणों को हल कीजिए :
(a) 2(x + 4) = 12
(b) 3(n – 5) = 21
(c) 3(n – 5) = -21
(d) – 4(2 + x) = 8
(e) 4(2 – x) = 8
हल :
(a) 2(x + 4) = 12
⇒ 2x + 8 = 12, [कोष्ठक खोलने पर]
⇒ 2x = 12 – 8
[8 को दायीं ओर ले जाने पर]
⇒ 2x = 4
⇒ \(\frac{2 x}{2}=\frac{4}{2}\)
[दोनों ओर 2 का भाग देने पर] उत्तर
⇒ x = 2.

(b) 3(n – 5) = 21
⇒ \(\frac{3(n-5)}{3}=\frac{21}{3}\),
[दोनों ओर 3 का भाग देने पर]
⇒ n – 5 = 7
⇒ n = 7 + 5,
[- 5 को दाँयी ओर ले जाने पर]
⇒ n = 12. उत्तर

(c) (3n – 5) = – 21
⇒ \(\frac{3(n-5)}{3}=\frac{-21}{3}\)
[3 का दोनों ओर भाग देने पर]
⇒ n – 5 = – 7
[- 5 को दारों ओर लाने पर]
⇒ n = – 7 + 5 ,
⇒ n = – 2. उत्तर

(d) – 4(2+x) = 8
⇒ 8 – 4x = 8, [कोष्ठक खोलने पर]
⇒ – 4x = 8 + 8,
[- 8 को दायीं ओर लाने पर]
⇒ – 4x = 16
⇒ \(\frac{-4 x}{4}=\frac{16}{4}\)
[दोनों ओर 4 से भाग देने पर]
⇒ x = – 4. उत्तर

(e) 4(2 – x) = 8
⇒ 8 – 4x = 8. [कोष्ठक खोलने पर]
⇒ – 4x = 8 – 8
[8 को दायौं ओर ले जाने पर]
⇒ -4x = 0
⇒ \(\frac{-4 x}{-4}=\frac{0}{-4}\) [- 4 से भाग देने पर]
⇒ x = 0. उत्तर

प्रश्न 3.
निम्नलिखित समीकरणों को हल कीजिए :
(a) 4 = 5(p – 2)
(b) – 4 = 5(p – 2)
(c) 16 = 4 + 3(t + 2)
(d) 4 + 5(p – 1) = 34
(e) 0 = 16 + 4(m – 6)
हल :
(a) 4 = 5(p – 2)
⇒ 4 = 5p – 10, [कोष्ठक खोलने पर]
⇒ – 5p = -10 – 4 [4 को दायीं ओर तथा 5p को बायीं ओर लाने पर]
⇒ – 5p = – 14 [- 5 से दोनों ओर भाग देने पर]
⇒ \(\frac{-5 p}{-5}=\frac{-14}{-5}\)
⇒ p = \(\frac {14}{5}\) उत्तर

(b) – 4 = 5(p – 2)
⇒ -4 = 5p – 10, [कोष्ठक खोलने पर]
⇒ -5p = – 10 + 4, [-4 को दायीं ओर तथा 5p को बार्थी ओर लाने पर]
⇒ -5p = – 6
⇒ \(\frac{-5 p}{-5}=\frac{-6}{-5}\),
[- 5 से दोनों ओर भाग देने पर]
⇒ p = \(\frac {6}{5}\) उत्तर

(c) 16 = 4 + 3(t + 2)
⇒ 16 = 4 + 3t + 6,
[कोष्ठक खोलने पर]
⇒ – 3t = 4 + 6 – 16, [16 को दायीं ओर तथा 3t को बायीं ओर ले जाने पर]
⇒ – 3t = – 6
⇒ \(\frac{-3 t}{-3}=\frac{-6}{-3}\), [- 3 से दोनों ओर भाग देने पर]
⇒ t = 2. उत्तर

(d) 4 + 5(p – 1) = 34
⇒ 4 + 5p – 5 = 34 (कोष्ठक को खोलने पर)
⇒ 4 + 5p = 34 + 5
(5 को दायीं ओर ले जाने पर)
⇒ 4 + 5p = 39
⇒ 5p = 39 – 4
(4 को दायीं ओर ले जाने पर)
⇒ 5p = 35
⇒ \(\frac{5 p}{5}=\frac{35}{5}\)
(5 से दोनों ओर भाग देने पर)
⇒ p = 7 उत्तर

(e) 0 = 16 + 4(m – 6)
⇒ 0 = 16 + 4m – 24,
[कोष्ठक को खोलने पर]
⇒ 0 = 4m – 8,
⇒ -4m = – 8,
[4m को बायीं ओर ले जाने पर]
⇒ \(\frac{-4 m}{-4}=\frac{-8}{-4},\)
[दोनों ओर – 4 से भाग देने पर]
⇒ m = 2 उत्तर

प्रश्न 4.
(a) x = 2 से प्रारम्भ करते हुए, 3 समीकरण बनाइए।
(b) x = – 2 से प्रारम्भ करते हुए, 3 समीकरण बनाइए।
हल :
(a) प्रथम समीकरण:
x = 2 से शुरू करने पर
दोनों ओर 2 से गुणा करने पर,
2x = 4
दोनों ओर 3 जोड़ने पर,
2x + 3 = 4 + 3
⇒ 2x + 3 = 7.

द्वितीय समीकरण :
x = 2 से शुरू करने पर,
दोनों ओर – 5 से गुणा करने पर,
= 5x = – 10
दोनों ओर 5 जोड़ने पर,
– 5x + 5 = – 10 + 5
⇒ -5x + 5 = -5

तृतीय समीकरण:
x = 2 से शुरू करने पर, दोनों ओर से 3 से भाग देने पर,
\(\frac{x}{3}=\frac{2}{3}\)
दोनों ओर 5 घटाने पर,
\(\frac {x}{3}\) – 5 = \(\frac {2}{3}\)

(b) प्रथम समीकरण :
x = -2 से शुरू करने पर,
दोनों ओर 5 से गुणा करने पर,
5x = – 10
दोनों ओर से 3 घटाने पर,
5x – 3 = – 10 – 3
⇒ 5x – 3 = – 13

द्वितीय समीकरण :
x = – 2 से शुरू करने पर,
– 4 से दोनों ओर गुणा करने पर,
– 4x = 8
दोनों ओर 8 जोड़ने पर,
-4x + 8 = 8 + 8
⇒ – 4x + 8 = 16

तृतीय समीकरण :
x = -2 से शुरू करने पर, दोनों ओर 2 से भाग देने पर,
\(\frac {x}{2}\) = -1
दोनों ओर 3 जोड़ने पर,
\(\frac {x}{2}\) + 3 = – 1 + 3
⇒ \(\frac {x}{2}\) + 3 = 2