Class 7

SCERT AP 7th Class Science Study Material Pdf 4th Lesson Respiration and Circulation Textbook Questions and Answers.

AP State Syllabus 7th Class Science 4th Lesson Questions and Answers Respiration and Circulation

7th Class Science 4th Lesson Respiration and Circulation Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. Respiration is the process essential for ______ of living things.
2. Inhaled air has ____ oxygen and _____ has carbon dioxide.
3. ______ can breathe both with lungs and skin.
4.The latest respiratory disorder is
1. survival
2. more, less
3. Frog
4. COVID- 19

II. Choose the correct answer.

1. Colour of Haemoglobin is
a) Colourless
b) Blue
c) Red
d) Green
Answer:
c) Red

2. Lime water turns milky when it reacts with
a) Oxygen
b) Nitrogen
c) Carbon
d) Carbon dioxide
Answer:
d) Carbon dioxide

3. The part of respiratory system which branches into
a) Nasal cavity
b) Bronchi
c) Lungs
d) Trachea
Answer:
b) Bronchi

4. Humans normally breathe times per minute.
a) 14 to 20 times
b) 20 to 30 times
c) 72 times
d) Upto 80 times
Answer:
a) 14 to 20 times

III. Matching

Answer:

IV. Answer the following questions.

Question 1.
What is respiration?
Answer:
The oxygen reacts with the glucose present in digested food and breaks it down into Carbon dioxide and water to release energy. This process is called Respiration.
Glucose + Oxygen → Carbon dioxide + water + energy

Question 2.
Name the two types of respiration. Write its word equation.
Answer:
Respiration is primarily of two types (a) Aerobic (b) Anaerobic
(a) Aerobic respiration: oxygen involves in this process
Glucose + Oxygen → Carbon dioxide + water + Energy

(b) Anaerobic respiration: oxygen does not involve in this process
Glucose → Carbon dioxide + alcohol + Energy

Question 3.
Write the differences in composition of inhaled and exhaled air.
Answer:

Question 4.
Explain in detail the pathway of respiration in humans with the help of a flowchart.
Answer:
Parts that form the pathway of air in human respiratory system are
1) Nostrils,
2) Nasal cavity,
3) Pharynx,
4) Wind pipe,
5) Bronchi,
6) Lungs.

Question 5.
What are the different respiratory organs seen in animals and their method of working?
Answer:
The organs of respiration are different in different organisms.
A. Tracheae :

1. Respiration that take place via tracheae is called Tracheal respiration.
2. This is present in all insects. In this system there are openings called spiracles which enter the body of the insect through a network of air tubes called Tracheae.
3. The tracheae reach all parts of the body and help in the exchange of gases.
4. Ex : Grasshopper, Cockroach, Honey bee etc.

B. Skin :
Cutaneous respiration :

1. In some animals the skin is moist and slimy with mucus which helps in breathing through the skin. Ex : Earthworm.
2. Frogs have lungs for breathing which they use when they are on land.
3. But can breathe through the skin which is moist and slippery while in water.

C. Gills:
Branchial respiration :

1. Gills are seen in fishes. Gills are present beneath membranous covers on either side of the head.
2. The gills have a rich supply of blood, for exchange of oxygen and carbon dioxide.
3. Fish take in water through their mouth which passes out over the gills, the specialized organs to absorb the dissolved oxygen present in the water.

D. Lungs:

1. Respiration through lungs is called Pulmonary respiration.
2. In all land animals and some water animals the organs for breathing are lungs.
3. They are meant for taking oxygen from the air. Ex : Cow, dog, whale, humans etc.

Question 6.
If the diaphragm and ribs do not expand and contract what will be the consequences?
Answer:

1. A large thin muscular sheet called diaphragm is attached to the lower side of the ribcage and forms the floor of the chest cavity.
2. The process of breathing involves the movement of the diaphragm and the ribcage, in men and women respectively.
3. If they can’t contract inhalation and exhalation is not possible
4. It means we are not able to breath and it Leads to death

Question 7.
Write a report on the lab activity done in your school to prove that carbon dioxide is released during exhalation.
Answer:
Aim : To prove that carbon dioxide is released during exhalation What you need: Two small beakers .straw, lime water

How to do:

1. Take two beakers. Label them as A and B.
2. Fill two beakers up to half with clear lime water.
3. Blow air into beaker ‘A’ using a straw..
4. Pass atmospheric air repeatedly into beaker ‘B’ using a dropper.
5. Observe the colour change in both the beakers.

What you see:
Lime water is clear and colourless in beaker B , but it turns milky white in beaker A when it reacts with carbon dioxide.

What you learn:
This concludes that there is more carbon dioxide in exhaled air when compared to inhaled air, it is proven that exhaled air contains carbon dioxide

8. Prepare slogans to create awareness about a) evil effects of smoking b) COVID-19 prevention.
Answer:
a) Evil effects of smoking :
Smoking is injurious to health.
Say YES to NO SMOKING
Be smart, Don’t start
Smoking? You must be joking
You don’t have to smoke to be hot!
Be Cool – Don’t Be a Smoking Fool
Don’t take smoke as a joke
Smokers are jokers
Don’t put your lips on fire

b) COVID-19 prevention
Follow the SMS always
SMS means social distance, mask
and sanitization
Don’t neglect be cautious
Wash the hand regularly
Break the chain to kill the corona
Minimum distance is maximum safe

Question 9.
Make use of the Stethoscope made by you to measure the number of heart beats of five of your friends and note down your findings in the table given.
Answer:

7th Class Science 4th Lesson Respiration and Circulation InText Questions and Answers

7th Class Science Textbook Page No. 53

Question 1.
How fish could breathe in water?
Answer:
Fish has gills, the specialized organs to absorb the dissolved oxygen present in the water. Hence, fish breathe through gills.

Question 2.
Why humans and other animals cannot breathe in water?
Answer:

1. Humans, and some other animals have only lungs for breathing.
2. But lungs cannot absorb the oxygen from the water.
3. ence, they cannot take breathe in water.

Question 3.
What is breathing?
Answer:
The process of inhalation and exhalation of air is called Breathing.

Question 4.
Inhale deeply. Do you feel air moving inwards?
Answer:
Yes.

Question 5.
Now hold a finger under your nose and slowly release the air. How do you feel now? Do you feel air moving outwards?
Answer:
Yes.

Question 6.
How does the air reach the lungs? What are the organs involved in this process?
Answer:
a) Air reaches the lungs from nostril through various organs.
b) The organs involved in this process are 1) nostril 2) nasal cavity 3) pharynx 4) wind pipe 5) bronchi 6) lungs.

7th Class Science Textbook Page No. 54

Question 7.
How are the lungs expanding and contracting?
Answer:

1. When we breathe in, the chest along with ribcage moves upwards and air enters into our lungs. So, lungs expand in inspiration.
2. When we breatheout, the chest along with ribcage moves downwards and air goes out. So, lungs contracts in expiration.

7th Class Science Textbook Page No. 56

Question 8.
Why does the amount of oxygen vary between inhaled and exhaled air?
Answer:
Because the oxygen in inhaled air is absorbed in the lungs.

Question 9.
Which gas has increased in quantity in the exhaled air? Why?
Answer:

1. The quantity of carbondioxide is more in the exhaled air.
2. Because carbondioxide from the blood is transferred to the exhaled air.

Question 10.
In which beaker does the lime, water turned milky white?
Answer:
Beaker ‘A’.

Question 11.
What does this change indicate?
Answer:

1. Carbondioxide turns lime water milky.
2. The change indicates that beaker ‘A’ contains more carbondioxide.
3. This concludes that there is more carbondioxide in exhaled air when compared to inhaled air.

Question 12.
What happens to the air in the lungs?
Answer:
Oxygen from the inhaled air is absorbed by the blood vessels present in the lungs. Carbon dioxide collected by the blood vessels from all parts of the body enters the Lungs. The absorbed oxygen is transported to every part (Cell) of our body.

7th Class Science Textbook Page No. 57

Question 13.
Checklist.
Tick (✓) the statements which you think are correct and (✗) for wrong me.
1) There is no harm in trying a cigarette once, because one can stop after that. [ ✗ ]
2) Smoking a cigarette once a day is not harmful. [ ✗ ]
3) One can stop smoking only with his will power. [ ✓ ]
4) Smoking helps you feel good and relaxed. [ ✗]
5) Smoking is not harmful to health. [ ✗ ]

Question 14.
What about the smoke inhaled by smokers?
Answer:

1. The smoke inhaled by smokers is being sent to all parts of the body.
2. Tobacco smoke contains a highly dangerous substance called Nicotine.
3. This poisonous substance is also carried to all cells of the body. . ‘
4. Smoking leads to lung cancer, tuberculosis and other respiratory disorders.

Question 15.
Do all animals have the same type of respiratory organs?
Answer:
No.

Question 16.
What are the respiratory organs in whales?
Answer:

1. Whales breathe air into their lungs,
2. They cannot breathe under water like fish can as they do not have gills.
3. They breathe through nostrils called a blowhole, located right on top of their heads.

Question 17.
Why is the skin of frog always wet and slimy?
Answer:

1. In some animals like frog the skin is moist and slimy with mucus.
2. It helps in breathing through the skin while in water.

7th Class Science Textbook Page No. 58

Question 18.
How do plants breathe?
Answer:

1. Plants take in oxygen and leave out carbondioxide in breathing.
2. They breathe through stomata, lenticels and root hairs.

Question 19.
What are their (plants) respiratory organs?
Answer:
The respiratory organs in plants are

1. Stomata in leaves
2. Lenticels in stems
3. Root hairs in roots.

7th Class Science Textbook PageNo. 59

Question 20.
What changes did you observe in the lime water?
Answer:
The lime water turns milky white.

Question 21.
Why was there a change in the lime water?
Answer:
The carbondioxide, which was released by the plants (sprouts)’is turned lime water milky white.

Question 22.
How does blood reach al! parts of the body?
Answer:

1. Our circulatory system consists of heart, blood vessels and blood.
2. There are three types of blood vessels – arteries, veins and blood capillaries.
3. These blood vessels are connected to the heart.
4. While heart is pumping the blood, the blood capillaries which are connected to the arteries with the veins distribute the blood to the body parts.

Question 23.
How is the oxygen and glucose absorbed by the blood transported to all the parts of our body?
Answer:

1. Oxygen and glucose absorbed by the blood transported to all parts of the body through circulatory system.
2. The circulatory system consists of heart, blood vessels and blood.
3. Blood absorbs the glucose (Digested food) from the digestive system, and oxygen from lungs.
4. This blood is pumped by heart to all parts of the body through a kind of blood vessels called arteries.
5. These arteries ends in another type of blood vessels called blood capillaries.
6. These capillaries are thin arid narrow.
7. So, through their thin walls, blood is distributed to all body parts.
8. Now oxygen and glucose absorbed by the blood transported to the body parts.

7th Class Science Textbook Page No. 61

Question 24.
Is the blood in all organisms same as In humans?
Answer:
No, the blood in all organisms is not same as in humans.
For example, the colour of human blood is red whereas blue in prawns.

7th Class Science Textbook Page No. 62

Question 25.
Which new disease created a global impact recently?
Answer:
COVID -19 was the new disease that created a global impact recently.

Question 26.
What is a Pandemic?
Answer:
A disease which infects most of the people in a country or the whole world at the same time is called a Pandemic.

Question 27.
How can we prevent it?
Answer:
We can prevent any disease or infection by improving our immunity.

7th Class Science Textbook Page No. 63

Question 28.
Why do we sneeze?
Answer:

1. Sneezing occurs when we inhale air with dust, smoke, pollen or strong smells.
2. Sneezing expels the unwanted, harmful substances from the lungs.

Question 29.
Why do we yawn?
Answer:

1. Yawning is caused when the respiratory rate gets slowed down resulting in insufficient supply of oxygen to the brain.
2. To overcome this situation, the body goes for the involuntary opening of the mouth to take in a long deep breathe of air.

7th Class Science Textbook Page No. 64

Question 30.
What is first aid?
Answer:
The simple assistance provided to a person suffering from injury or serious situations before a doctor attending is called First Aid.

Question 31.
What is the first aid tor drowning?
Answer:
When a person drowns we should first bring him out of the water and make him lay down on the back with the face turned to one side. Check for any sand or mud in the nose, mouth and ears and remove them.

Now, slowly press the abdomen to create pressure on the diaphragm and on the lungs to push out the water that has entered the lungs. Repeat the process by turning the person to lie on the abdomen until recovery. Give some warm clothing and hot tea on recovery.

Think & Respond

Page No. 54

Question 1.
Why is the right lung larger than the left lung?
Answer:

1. Heart is located in the centre of the chest cavity slightly bent towards the left.
2. So, as a result the left lung being smaller than the right lung.

Activities and Projects

Question 1.
Make use of a water bottle with water and two straws to test your lungs power.
Answer:

1. Take a bottle with a lid.
2. Make two holes in the lid and insert two straws through them.
   
3. Fill the bottle with water and close its mouth with the lid along with straws.
4. Adjust the straws such that one should be till the bottom of the bottle and other till the surface of the water.
5. Now blow air with force through the straw placed above the water surface of water.
6. As this blowing air put pressure on water, it escapes out through the second straw.
7. The hight of the water raises through the second straw or the empty space formed in the bottle indicates your lung power.

Question 2.
Make use of a mirror to know whether our breath contains water vapour as given in the table 2 showing what our breath contains.
Answer:

1. Take a mirror in to your hands during morning and blow air on to that.
2. You can find that your image in the mirror become blarred.
3. Now wipe the mirror with your hand.
4. You will find that the image become clear and some moisture touches to your hand.
5. This moisture is nothing but the water vapour present in your breath.
6. This phenomena can be observed more clearly during winter mornings than summer mornings.

Question 3.
Prepare a model of lungs using a water bottle, balloons and Y tube to show the importance of diaphragm in respiration.
Answer:
Materials:
Y-shaped tube, a large balloon, two small balloons, a one litre plastic bottle, cork.

Method of construction :
Cut the plastic bottle to half its size. Fix two small balloons at both ends of the Y-tube. Make a small hole in the cork and fix the Y-tube through the hole as shown in the picture. Cut the large balloon into two halves and fix one half tightly around the open part of the bottle.

Method of working :
Hold the balloon at the middle and pull it slowly downwards. The balloons become inflated (Expands), Now leave the balloon free. Then the balloons collapse (contracts).

The expansion and contraction of the lungs is almost like this. Similar to the balloon tied firmly at the base of the bottle, there is a part called ‘diaphragm’ in our body to control the expansion and contraction of the lungs.

Activities

Activity – 1

Question 1.
How do you perform breathing by an activity?
(OR)
Do an activity to show the stages in breathing?
Answer:

1. Inhale deeply.
2. You may feel air moving inwards.
3. The process of inhaling air is called inspiration.
4. Now hold a finger under your nose and slowly release the air.
5. You may feel air coming outside.
6. The process of exhalation of air is called expiration.
7. This activity shows us that there are two stages in breathing – Inhalation and Exhalation.

Activity – 2

Question 2.
How do you prove that there is a difference in the size of the chest cavity during inhalation and exhalation process?
(OR)
Fill in the table with an activity.

Answer:

1. Take a measuring tape.
2. Hold the tape around the chest of one of your friend and measure the width of her/ his chest.
3. Hold the tape lightly and ask your friend to breathe in and again measure the width of her/his chest.
4. Repeat it with more friends and record your observations in the given table.

5. We can conclude from this activity that there is a difference in the size of the chest cavity during inhalation and exhalation process.

Activity – 3

Question 3.
How do you prove that there is more carbon dioxide in exhaled air when compared to inhaled air?
Answer:

1. Take two beakers. Label them as A and B.
2. Fill two beakers upto half with clear lime water.
3. Blow air into beaker ‘A’ using a straw.
4. Pass atmospheric air repeatedly into beaker B’ using a dropper.
5. Observe the colour change in both the beakers.
6. Lime water is clear and colourless, but it turns milky white when it reacts with carbon dioxide.
7. This concludes that there is more carbon dioxide in exhaled air when compared to inhaled air.

Activity – 4

Question 4.
How do you prove that the sprouting seeds respire to take in oxygen and leave out carbon dioxide?
Answer:
Take a wide mouthed bottle and place a handful of sprouting seeds in it. Prepare some fresh lime water in a small container and place it carefully in one corner of the bottle. Close the cap of the bottle and apply vaseline on the edges to make it air-tight. Leave the apparatus undisturbed for a day or two. After two days open the cap and carefully take out the lime water container and observe the changes.

1. What changes did you observe in the lime water?
Answer:
The lime water changes into milky white colour.

2. Why was there a change in the lime water?
Answer:
The sprouting seeds respire to take in oxygen and -leave out carbon dioxide. The lime water in the small container reacts with the carbon dioxide released by sprouting seeds to change into milky white colour.

Activity – 5

Question 5.
How do you make a stethoscope? What are the material do you require?
(OR)
How do you know the rate of heart beat?
Answer:
Aim : To know the rate of heart beat.

What we need :
Rubber tube, Y – shaped attachment, small funnel, rubber sheet /balloon, steel tongue cleaner, beads or earphone buds, insulation tape.

Procedure:
1. Insert rubber tubes to the three ends of a Y shaped attachment.
2. Arrange a small funnel to the rubber tube attached to the lower single arm of the Y shaped attachment.
3. Tie a rubber sheet / balloon to the broad end of the funnel.

4. Arrange ear buds to the ends of rubber tubes attached to the upper two arms of the Y shaped attachment.
5. Arrange a steel tongue cleaner connecting the two upper tubes over the Y shaped attachment to give support.

Working:
Keep upper two rubber tube ends in your ears. Put the funnel on the chest of your friend and observe the sound of heart beat.

SCERT AP 7th Class Science Study Material Pdf 3rd Lesson Nutrition in Organisms Textbook Questions and Answers.

AP State Syllabus 7th Class Science 3rd Lesson Questions and Answers Nutrition in Organisms

7th Class Science 3rd Lesson Nutrition in Organisms Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. During photosynthesis ________ gas is released.
2. The tiny pores present on the surface of the leaves are __________
3. ____________ is the outermost layer of teeth.
Answer:
1. oxygen
2. stomata
3. Enamel

II. Choose the correct answer.

1. Muscular tubular structure that connects pharynx with Stomach
a) Duodenum
b) Buccal cavity
c) Oesophagus
d) Rectum
Answer:
c) Oesophagus

2. Non insectivorous plant,
a) Drosera
b) Nepenthes
c) Utricularia
d) Dodder
Answer:
d) Dodder

3. The green colour pigment in the leaf
a) chloroplast
b) stomata
c) chlorophyll
d) all the above
Answer:
c) chlorophyll

III. Matching

Answer:

IV. Answer the following questions.

Question 1.
Distinguish between autotrophic and heterotrophic nutrition.
Answer:

Question 2.
What is Photosynthesis? Write its word equation.
Answer:
The process by which green plants make their own food from carbon dioxide and water by using light energy in the presence of chlorophyll is called Photosynthesis.

Question 3.
Describe different types of teeth and their functions.
Answer:
There are four different types of teeth in our human beings.

1. Incisors :
The four front teeth in both the upper and lower jaws are called incisors. Thus there are eight incisors in the oral cavity. They have sharp incisal edge. Their primary function is to cut food.

2. Canines :
They are behind and adjacent to the lateral incisors on both the jaws. There are four canines in the oral cavity. They have single sharpcusp. Their main function is to tear food.

3. Premolars :
These teeth are located behind and adjacent to the canines. There are two premolars on the upper jaw and also two premolars on the lower jaw on either side of the mouth. Thus there are eight premolars in the oral cavity.These teeth can have 3-4 cusps. They are designed to crush food.

4. Molars :
The most posterior teeth in the mouth are the molars. There are 12 molars in the permanent dentition.They have broader and flatter surfaces with 4-5 cusps. They are designed to grind food.

Question 4.
What happens if leaves of a green plant are coated with green paint?
Answer:

1. If the leaves of a green plant are coated with green paint, then stomata present on the surface of the leaf will get blocked.
2. As a result exchange of gases will be affected.
3. So, plant may not get sufficient amount of carbon dioxide for photosynthesis and oxygen for respiration.
4. oreover, paint over the leaf prevents the reaching of sunlight to the leaf.
5. As a result plant will not be able to perform photosynthesis and respiration, gradually it leads to its death.

Question 5.
What questions will you ask a doctor to know about gastritis?
Answer:

1. What is gastritis?
2. What are the reasons for gastritis?
3. How can it be prevented?
4. Is there any relation between gastritis and life style?
5. Are there any home remedies to get relief from gastritis?
6. What is the treatment for gastritis?

Question 6.
How can you prove that the leaves other than green also carry out photo synthesis? (Activity – 2)
Answer:
Aim: To know whether leaves other than green perform photosynthesis or not.

What you need:
Red or brown coloured leaves, dropper; test tube, iodine solution, water.

What to do:
Take few Red or brown coloured leaves, add few drops of water and mash them to make a paste, collect the extract by squeezing the paste (pulp). Take 5-6 drops of this extract into a test tube. Also add two drops of iodine solution. Note your observations.

What do you see:
The colour of the leaf extract turns blue black.

What do you learn:
This shows the presence of starch in the leaves, confirming that red or brown coloured leaves also perform photosynthesis.

These non-green leaves also have chlorophyll. The large amount of red, brown and other pigments mask the green colour. So photosynthesis takes place in these leaves also.

Question 7.
Draw a neat labelled diagram of human digestive system.
Answer:

Question 8.
Draw a flow chart showing nutrition in amoeba.
Answer:

Question 9.
How will you appreciate the role of saprophytes in cleaning of the earth surface?
Answer:

1. Saprophytes grow on dead and decaying matter.
2. They secrete digestive juices on it, convert it into a solution and then absorb the nutrients from it.
3. During this process they decompose them and mix them with the soil.
4. In this Way they help in cleaning the earth surface by removing the dead and decaying matter.
5. This process help in recycling the nutrients too.
6. Thus they are doing a great service to the entire living world.

Question 10.
What precautions will you take to keep your digestive system healthy?
Answer:
To keep my digestive system healthy, I will

1. take simple and balanced diet
2. drink sufficient quantity of water.
3. brush my teeth regularly.
4. prefer natural foods.
5. avoid Junk foods and cool drinks.
6. never smoke.
7. never consume alcohol.
8. never chew tobacco and tobacco products.
9. do moderate exercises regularly.
10. take fibre rich food.

7th Class Science 3rd Lesson Nutrition in Organisms InText Questions and Answers

7th Class Science Textbook Page No. 37

Question 1.
My mother prepares food for me. Who prepares food for animals?
Answer:

1. Animals cannot make food.
2. They depend on other plants and animals for food.

Question 2.
How do animals get their food?
Answer:
Animals get their food from plants and animals.

Question 3.
Plant is a living thing. What is the food for it?
Answer:
Plants also need food, such as glucose, carbohydrates, minerals etc.

Question 4.
How do they get their food?
Answer:

1. Plants prepare their food by using photosynthesis.
2. Some plants get their food from other plants and animals.

Question 5.
What is the food for mushrooms?
Answer:
Mushrooms get their food from dead and decaying matter.

Question 6.
Even you might have got the same doubts?
Answer:
Yes.

Question 7.
Is the nutrition in plants same as in animals?
Answer:
No.

Question 8.
We eat food daily. But plants are not eating food then how are they living?
Answer:
Plants are able to live as they make their food themselves.

7th Class Science Textbook Page No. 38

Question 9.
How can the green plants prepare their own food?
Answer:
Green plants make their own food from carbon dioxide and water by using light energy in the presence of chlorophyll. This process is called photosynthesis.

7th Class Science Textbook Page No. 39

Question 10.
In which part of the plant, does photosynthesis takes place?
Answer:
The synthesis of food occurs in all green parts of plant body. For Ex : leaves.

Question 11.
How do the raw materials required for photosynthesis reach there?
Answer:

1. Plants get carbon dioxide from air through stomata in leaves.
2. Water obsorbed by the roots are transported to the leaves through the stem.
3. Chlorophyll captures the energy of the sunlight.

Question 12.
Do they perform photosynthesis?
Answer:
Yes, the plants with red and brown coloured leaves make their food in the process of photosynthesis.

7th Class Science Textbook Page No. 40

Question 13.
What is the importance of sunlight in photosynthesis?
Answer:

1. Sunlight is the source of energy in the process of photosynthesis.
2. The solar energy is captured by the leaves and stored in the plant in the form of food.
3. If there is no sunlight starch cannot be prepared in plants.

Question 14.
Plants get water from the soil through their roots. How does this water reach the leaves from the roots? What path does it follow?
Answer:

1. Water absorbed by the roots are transported to the leaves through the stem.
2. It follows the path from roots to leaves via stem.

7th Class Science Textbook Page No. 41

Question 15.
How do carbondioxide enters the leaves?
Answer:
Carbondioxide from air is entered the leaves through stomata which are present in the surface of the leaf.

Question 16.
How do oxygen comes out of the leaves?
Answer:
Through the stomata, oxygen produced in photosynthesis goes out of the leaves.

Question 17.
Where do these substances (starch, fats and proteins) come from?
Answer:
Plants produce sugar first, which is converted to starch and then other compounds like fats and proteins.

7th Class Science Textbook Page No. 42

Question 18.
There are some plants which do not have chlorophyll. How do they get their food?
Answer:

1. Organisms which do not have chlorophyll, cannot synthesis their food. They are heterotrophs.
2. Heterotrophic nutrition is of different types basing on how they obtain their food from other organisms.
3. Some plants depends on dead and decaying matter and some on other plants.

Question 19.
How do they (mushrooms) get their food?
Answer:

1. Mushrooms grow on dead and decaying matter.
2. They secrete digestive juices on it convert it into a solution and then absorb the nutrients from it.
3. This type of nutrition is called saprophytic nutrition.

Question 20.
A rat died at Pullaiah’s house. What happens to the body after few days? Do you find any traces of its body after a month? Where did that body go?
Answer:

1. Saprotrophs like fungi and bacteria will grow on the dead body of the rat, after a few days.
2. We cannot find any traces (except strong bones) of the dead body of rat, after a month.
3. The decomposed body of the rat is mixed with the soil.

Question 21.
How does Cuscuta survive? From where does it get nutrition?
Answer:

1. Cuscuta take readymade food from the plant on which it is climbing.
2. It develops special roots called haustoria, which penetrate into the tissues of the host plant and absorb food materials from them.

7th Class Science Textbook Page No. 43

Question 22.
How do animals take their food?
Answer:

1. Animals obtain their food from other organisms.
   Ex : Cow eats grass, cat drinks milk, dog eats meat.
2. They take the food in the form of solid or liquid.

Question 23.
Where do they digest it?
Answer:

1. Animals take food into the body for digestion.
2. Digestion occurs inside the body.

7th Class Science Textbook Page No. 44

Question 24.
What do you understand from Sabiha observations?
Answer:
Holozoic nutrition is the mode of heterotrophic nutrition in which the food is taken in solid or liquid form from the outside and is digested inside the body.

Question 25.
How do Amoeba takes its food?
Answer:

1. Amoeba has many small bubbles – like vacuoles in its cytoplasm.
2. Amoeba pushes out one or more finger – like projections called pseudopodia for movement and capture of food.
3. Food vacuole forms around the captured food.
4. Food get digested in it, absorbed into the cytoplasm and assimilates.
5. Finally, undigested food is sent out by opening this vacuole out at the body surface.

Question 26.
Is the nutrition in human beings also holozoic?
Answer:
Yes. The nutrition in human beings also holozoic.

7th Class Science Textbook Page No. 45

Question 27.
Have you ever wondered what happens to the food inside the body?
Answer:
Seeing the food that went into the body being digested, I was very surprised.

Question 28.
Do you have any structures like vacuole in our body?
Answer:
No.

Question 29.
Where do the food eaten by us go?
Answer:

1. We take food in the form of solids or liquids.
2. It goes to the digestive system, which is consisting of the alimentary canal and digestive glands.
3. Food get’s digested in the digestive system.
4. Digested food is absorbed by the blood.
5. Blood transports digestive food to different parts of the body for assimilation.
6. Undigested food will be ejected out of the body.

Question 30.
What change does the food undergo inside the mouth?
Answer:

1. We chew the food with the teeth and break it down mechanically into small pieces.
2. Carbohydrates are digested in the mouth.
3. Food also mixed with saliva and thrown into the stomach.

7th Class Science Textbook Page No. 46

Question 31.
How many kinds of teeth could you find?
Answer:
Four types of teeth.

Question 32.
Which teeth do you use for biting and cutting and which one for tearing?
Answer:

1. Teeth which are located at front side of the mouth are used for biting and cutting. (These are called incisors.)
2. Canines are used for tearing, which are located behind incisors.

Question 33.
Have you noticed formation of new teeth in the mouth of small children? Do all the teeth, wfiich sprouted first, remain forever?
Answer:

1. Yes. f have noticed that the formation of new teeth in the mouth of small children.
2. The teeth which sprouted first do not remain forever.

Question 34.
Do you know how does tooth decay happens?
Answer:
When food remains stuck in teeth, bacteria feed on them. As a result, lactic acid is produced, which causes the destruction of enamel and it leaves toothache.

7th Class Science Textbook Page No. 47

Question 35.
How does acid come into contact with teeth?
Answer:
When food remains stuck in teeth, bacteria feed on them. Bacteria produce lactic acid. This acid come to contact with teeth.

Question 36.
Do you now realise the importance of cleaning your teeth after food?
Answer:
Yes. It is very important. If we do not clean odr teeth properly after food, some food material will stuck in between teeth and causes tooth decay.

Think & Respond

7th Class Science Textbook Page No. 39

Question 1.
Testing tiie starch in the leaf directly with iodine had certain problems. To obtain results successfully think and discuss regarding this with your teacher.
Answer:

1. Leaves are green in colour.
2. When iodine solution is put on a leaf, it should turn blue if starch is present.
3. However, the green colour of the leaf disguises the blue colour.
4. More over cuticle present on the leaf hardly allows the iodine to reach he starch present in side the leaf.

7th Class Science Textbook Page No. 47

Question 2.
Which habit should be practised for the health of teeth? Why?
Answer:

1. Brush teeth twice a day with fluoride tooth paste.
2. Use dental floss to clean between your teeth.
3. Change your tooth brush every 3 months.
4. Cut down on how often you have sugary foods like chocolates, sweets and cool drinks. Visit dentist regularly.

The above habits should practise for healthy teeth. Because, if we do not clean our teeth and mouth after eating, many harmful bacteria begin to live and grow in mouth. These bacteria breakdown the sugars present from left over food and release acids. These acids gradually damage teeth and causes severe toothache.

Activities and Projects

Question 1.
Designer leaves making select any broad-leaved potte d plant. Cut a card board with a design of your choice and seal the selected leaf with the card board. Let the plant stand under the sun for a week then remove the card board. You will get designer leaves plant. Try to make more leaves with designs and display your plant but don’t forget to present your writeup.
Answer:

I have selected a potted plant with broad leaves.

I cut a design of my name first letter on a card board.

I sealed the selected leaf with the card board.

I kept this plant under the sun for a week.

Then I remove card board.

I surprised to see the design of my name first letter on the leaf.

Question 2.
Collect small plants, or the branches or other parts of big plants from your locality and classify them into autotrophs, parasites, saprophytes and symbionts. With the help of your teacher, preserve them in the form of specimens in the biology laboratory of your school.
Answer:
I have collected the following plants and classified them in to following categories.

Autotrophs:
Neem, mango, hibiscus, Paddy

Parasites:
Cuscuta, loranthus.

Saprophytes:
Mush rooms, Moulds Symbionts: Red gram, beans,

I preserved these specimens in the biology laboratory of our school.

Activities

Activity – 1

Question 1.
Fill the following table basing on your own observations and information collected from elders.

Answer:

→ Are all plants Autotrophs?
Answer:
No.

→ What kind of nutrition is seen in mushrooms?
Answer:
Heterotrophic nutrition (Saphrophytic).

→ What kind of nutrition is seen in animals?
Answer:
Heterotrophic (Parasitic, Holozoic).

7th Class Science Textbook Page No : 38 (Autotrophic Nutrition – Photosynthesis)

Question 2.
Anu said plants also prepare their food same as we prepare boiled rice.
Comparative table prepared by her was given below. Study the table and answer the questions.

→ What are the raw materials required for the preparation of food by green plants?
Answer:
Carbondioxide, water.

→ Which gas is taken by plants during this process?
Answer:
Carbondioxide.

→ Name the food material formed in plants?
Answer:
Glucose / Carbohydrates.

→ What do you conclude from the above information?
Answer:
Plants prepare their own food.

Activity – 3

Question 3.
How do you prove that the sunlight is essential for photosynthesis?
Answer:
Aim: To know the importance of light in photosynthesis.

What you need :
Two potted plants, dropper, test tube, iodine solution, water.

What to do:
Take two potted plants of the same kind. Keep one in the dark (or in a black box) for 72 hours and the other in the sunlight. Perform iodine test with the Leaf extracts of both the plants as you did in activity-2. Note your observations.

What do you see :
The colour of the leaf extract of first plant does not change. The colour of the leaf extract of second plant turns blue black.

What do you learn :
This shows the presence of starch in the leaves of plant kept in the sunlight indicating the occurrence of photosynthesis. There is no starch in the leaves of the plant kept in the dark. This confirms that sunlight is essential for photosynthesis.

Activity – 4

Question 4.
How do you conduct an activity to show the saprotrophic nutrition in bread moulds?
Answer:
Aim:
To observe the saprotrophic nutrition ih bread mould.

What you need:
Piece of bread, water, a container and a hand lens.

What to do:
Take a piece of Bread, in a container. Sprinkle some water and close the container. Open the container after few days and observe. (Use mask and gloves while doing this activity).

What do you see:
You will see cotton-like threads spread on the piece of bread and bread size is decreased.

What do you learn:
These thread-like structures are some sort of plants called Fungi. They don’t have chlorophyll so they obtain their food from dead and decaying matter.

Activity – 5

Question 5.
Wash your hands. Look into the mirror and count your teeth. Use your index finger to feel the teeth. How many kinds of teeth could you find? Take a piece of an apple, sugar cane or bread and eat it. Which teeth do you use for biting and cutting and which one for tearing?

Also find out the ones that are used for chewing and grinding? Compare your observations with the given picture and fill the table.

(Note: The picture of teeth on lower jaw was given here. The number and type of teeth on the upper jaw is same as on the lower jaw.)

→ What do you find?
Answer:
I found different types of teeth in my mouth.

→ How many teeth do you have? Is it equal to the number given in the picture?
Answer:
I have 28 teeth. No, in the figure there are 32 teeth. But 1 have only 28 teeth.

→ Which teeth are absent?
Answer:
I have no (wisdom teeth). 4 molars are absent in my mouth.

Activity – 6

Question 6.
How does acid damage our teeth? Prove with an activity.
Answer:
Aim : To know the process of tooth decay.

What you need :
Marbles, dilute hydrochloric acid and test tube.

What to do:
Put a few small pieces of marble in dilute hydrochloric acid. Examine after awhile.

What do you see :
Acid reacts with marble and dissolves it.

What do you learn :
The enamel, which is a calcium compound, reacts with acid and gets destroyed in the same way a marble reacts with hydrochloric acid.

When food remains stuck in teeth, bacteria feed on them. As a result, lactic acid is produced, which causes the destruction of enamel.

SCERT AP 7th Class Science Study Material Pdf 7th Lesson Reproduction in Plants Textbook Questions and Answers.

AP State Syllabus 7th Class Science 7th Lesson Questions and Answers Reproduction in Plants

7th Class Science 7th Lesson Reproduction in Plants Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. Hibiscus propagated by _______ .
2. The male reproductive part in a flower is _______ .
3. The lower swollen part in gynoecium is _______ .
Answer:
1. stem
2. Stamens dr Androecium
3. Ovary

II. Choose the correct answer.

1. The plant which reproduces through leaves is
a) Bryophyllum
b) Rose
c) Hydrilla
d) Balsam
Answer:
a) Bryophyllum

2. The reproductive part in a plant is
a) Root
b) Stem
c) Leaf
d) Flower
Answer:
d) Flower

3. The agents of pollination are
a) Air
b) Water
c) Insects
d) All the above
Answer:
d) All the above

III. Matching

Answer:

IV. Answer the following questions.

Question 1.
Identify whether the sentences below are true or not. Correct the statements which are not true
a) The flowers in pumpkin are unisexual.
b) Seeds are formed in asexual reproduction.
c) Generally, roses are propagated through seeds.
Answer:
a) The flowers in pumpkin are unisexual. This statement is true.

b) Seeds are formed in asexual reproduction. This statement is false. Seeds are formed in sexual reproduction.

c) Generally, roses are propagated through seeds. This statement also false. Generally, rose plants are propagated through a vegetative propagation method called stem cuttings.

Question 2.
What do you call the transfer of pollen grains to the stigma? Explain its types with the help of labelled diagrams.
Answer:

1. The process of transferring pollen grains from the anther to stigma is known as POLLINATION.
2. If the pollen grains are transferred from the anther of a flower to stigma of the same flower, is known as SELF POLLINATION.
3. If the transfer of pollen grains takes place from the anther of one flower to the stigma of another flower, it is called CROSS POLLINATION.

Question 3.
Can plants produce new plants without the seeds? Explain those methods with the help of examples.
Answer:

1. Yes, Some plants can produce new plants without seeds. This method of reproduction without formation of seeds is called asexual reproduction.
2. In some plants asexual reproduction occurs through some vegetative parts like stem, roots and leaf. Such a propagation is called vegetative propagation.
3. When the banana plants grow, a small new plant rises from the base of mother plant. Such small plants are called suckers (pilakulu). If we separate them and plant, they grow as new plants.
4. If we cut the sugarcane stem into pieces with at least one node and plant it in the soil. After few days new plants will develop from the nodes.
5. In Jasmine and chrysanthemum plants, small new plants arise from the underground stem of the mature parent plant. They are called layers. If these are separated from the parental plant and plant them in soil they grow into a new plants.
6. If we cut the mint twigs with nodes and sow them in soil, they develops roots and give the crop.
7. Potatoes have notches on them called eyes. If they are cut and plant in the suitable soil they gives rise to new potato plants.
8. In addition to these natural vegetative propagation methods, some artificial vegetative propagation methods such as stem cuttings, layering and grafting are also helpful to grow new plants with out seeds.

Question 4.
What happens if the pollen of mango flower reaches the stigma of guava flower?
Answer:

1. If the pollen of mango flower reaches the stigma of guava flower, fertilization does not occur.
2. Fertilization occurs freely between the individuals of the same specious (type).
3. But mango and guava plants belong to two different specious.
4. So, there will be no fertilization and formation of new plants.

Question 5.
If all the honeybees in nature become extinct. Imagine, what will be the consequences.
Answer:

1. Honeybees visit the flowers in search of nectar.
2. When insects come in contact with a flower, the pollen grains stick to their legs and wings.
3. Then insects visit another flower, the pollen fall on its slimy stigma.
4. Thus they play key role in pollination, intern helps in fertilization and formation of seeds.
5. If all the honey bees in nature become extinct, the fertilization in plants will reduce tremendously.
6. This results in the extinction of several specious of plants.
7. Agricultural productivity also will reduce to a great extent.
8. So we should stop indiscriminate use of chemical pesticides and protect honey bees.

Question 6.
What are the materials required, procedure and precautions taken by you in the lab activity conducted to study the parts of Datura flower?
Answer:
AIM: To observe the parts of a flower.

Material required:
Two datura flowers, new blade, magnifying glass, pencil. Procedure: Hold a datura flower by its stalk and observe its external features. Now draw the diagram of that flower in the box given below. Note down your observations. Now cut the second datura flower vertically into two equal parts from bottom to top. Check that all parts are cut into two equal halves. Observe and draw the internal view.

Precautions:
We should be careful while cutting the flower as the new blade is sharp and may cause injury. We should not inhale the pollen as it may cause allergy.

Question 7.
Draw the diagram of a complete flower and label the parts.
Answer:

Question 8.
Rahul goes to a field trip with his classmates, he tries to catch an insect on a flower. Do you support this?
Answer:

1. No, I never support Rahul.
2. Insects on the flowers helps in pollination and there by fertilization.
3. Catching and killing such insects effects the fertilization.
4. This leads to decrease the productivity of plants.
5. This hampers the propagation of that plants also.
6. All the organisms on this planet have the equal right live.
7. Killing of other organisms just for fun is a cruel act Which leads to imbalance in the nature.
8. “Live and let live” should be out motive.

Question 9.
Venkat, lives in a city. He maintains a roof garden at the top of his six storied building. The ridge gourd creeper bears plenty of flowers. But the flower do not grow into ridge gourds. Can you give him any suggestions to get ridge gourds?
Answer:

1. Flowers should pollinate and fertilize to form fruits.
2. Flowers on the ridge guard plant are unisexual flowers.
3. So, self-pollination is not possible.
4. For Cross pollination pollinating agents such as insects are required.
5. To allow the insects, if there is any net surrounding the plants, those should be removed.
6. Should not spray harmful chemicals on plants.
7. Along with ridge guard, other attractive flowers having nectar should be grown in the roof garden.
8. As a last trial do artificial pollination. (Transfer the pollen from male flower to the stigma of the female flower.)

Question 10.
Draw the various methods of Artificial propagations on a chart and exhibit in your classroom.
Answer:

7th Class Science 7th Lesson Reproduction in Plants InText Questions and Answers

7th Class Science Textbook Page No. 2

Question 1.
Can we grow all plants by planting their stems?
Answer:
No, we cannot grow all the plants by planting their stems.

Question 2.
How do new plants grow from the stems?
Answer:
New plants grow from the stems with some vegetative propagation methods.

7th Class Science Textbook Page No. 3

Question 3.
Have you ever seen seeds in Banana?
Answer:
In general we never found seeds in most of the hybrid banana varities. In certain varities of banana we can find small seeds. But in the wild varities we can find large sized seeds.

Question 4.
Have you ever seen seeds in Jasmine?
Answer:
Yes, I have seen seeds in Jasmine plants. Jasmine plants give long fruits from their flowers. These fruits contain seeds.

Question 5.
Did you observe how the Hibiscus plant propagate?
Answer:
Hibiscus plant propagate by means of stem cuttings.

7th Class Science Textbook Page No. 6

Question 6.
What would your mother have done to grow a basil / tulasi plant in your house?
Answer:
My mother sow the seeds of basil / tulasi in soil to grow the basil plant.

Question 7.
She might have put some seeds in the soil. Isn’t it?
Answer:
Yes. She would sow some seeds in the soil to get new plants.

7th Class Science Textbook Page No. 7

Question 8.
Which part of the plant form the fruit?
Answer:
Flower form the fruit.

7th Class Science Textbook Page No. 9

Question 9.
Do bitter gourd flower have whorls?
Answer:
No, bitter gourd flower has only three whorls. Either androecium or gynoecium are absent in them.

7th Class Science Textbook Page No. 11

Question 10.
Do you know which part of the flower will develop into a fruit?
Answer:

1. Ovary develop into fruits.
2. Ovules develop into seeds.

7th Class Science Textbook Page No. 12

Question 11.
Think, what changes will take place in the flower after fertilization?
Answer:

1. After fertilization, the ovary ripens and turns into fruit. ’
2. The remaining floral parts will fall off.
3. The ovules become the seeds.

7th Class Science Textbook Page No. 13

Question 12.
In which cup we observe healthy plants?
Answer:
The second cup contains healthy plants.

Question 13.
If all the seeds of a plant fall at the same place, how do the baby plants grow?
Answer:

1. Baby plants will suffer from deficiency of nutrients, water and place.
2. Their growth will be stunted.

Question 14.
Will they get sufficient place, nutrients, water to grow? What will happen if plants grow in such conditions?
Answer:

1. Plants will not have enough space to grow freely.
2. They will not get adequate amounts of nutrients and water.
3. They do not obtain healthy growth.

Question 15.
Do you have any idea that how plant overcome this situation?
Answer:

1. To overcome this problem, plants adopt different methods to disperse their seeds.
2. Seed dispersal takes place with the help of various agents.

Think & Respond

7th Class Science Textbook Page No. 12

Question 1.
Why do some plants produce small and numerous seeds?
Answer:

1. All the seeds of a fruit should be able to germinate to produce new plants. Actually, it does not happen in nature.
2. All the seeds don’t germinate, some will germinate but die before their maturity. Some of the seeds never germinate.
3. To over come all these problems, plants produce a large number of seeds.

Question 2.
Why do some seeds have wings?
Answer:

1. This type of winged seeds are dispersed by wind.
2. These wings help the seeds to move easily by winds to settle in a suitable place to germinate.

Question 3.
Why are some seeds with more fibre?
Answer:

1. Fibres make seeds light weight and help in floating.
2. These seeds dispersed through water for longer distance.
   Ex : Coconut.

Question 4.
Why do some pods of dry fruits explode?
Answer:

1. Some pods of dry fruits explode is a mechanical reaction.
2. In such way that they scatter the seeds all around.
3. In this way they dispersal the seeds.

Question 5.
Why do some seeds have hairs?
Answer:

1. Hairs help the seeds to travel in air.
2. Long haired seeds are travel long distances in air.
3. In that way seeds are settle at a suitable place to germinate.
   Ex : Calotropis.

Question 6.
Why are many fruits sweet and fleshy?
Answer:

1. Animals, birds and human beings eat these fruits and the seeds are thrown away.
2. Like that way, seeds will be dispersed to different places.
3. For the purpose of reproduction many fruits are sweet and fleshy,

Question 7.
Why do some seeds-have hooks?
Answer:

1. Seeds have hooks and thorns, help them to stick to the animals
2. So they travel one place to another place. Ex : Xanthium, urena, grass.

Question 8.
Why are some seeds heavy and round?
Answer:

1. Seeds that are heavy and round are dispersal by water.
2. They fall into the bottom of water
3. They carried by the flow of water for long distances.
   Ex : Seeds of Lotus.

Question 9.
Why are some seeds are delicate and small?
Answer:

1. This type of seeds are dispersed through winds
2. They are delicate, small, light in weight and hairy.
3. These qualities enable them to travel with the mind and settle at a suitable place to germinate

Activities and Projects

Question 1.
Collect information from elders, internet or your school library regarding the method of vegetative propagation in Onion, Garlic, Ginger, Gladioli, Sweet potato, Bryophyllum, Begonia.

Answer:

Question 2.
Grow a plant using seeds and exhibit in the classroom record its growth.
Answer:

Activities

Activity – 1

Question 1.
Discuss with your friends and teachers and fill the table given below with ‘Yes’ or ‘No’ options. Answer the questions given below.

i) Which plants are reproduce through seeds?
ii) Which plants are reproduce without seeds?
iii) Which plants are reproduce by both means?
Answer:
i) Tamarind, Coriander, Drumsticks plants repr oduce through seeds.
ii) Jasmine, Curry leaf, Banana plants reproduce without seeds.
iii) Jasmine, Curry leaf and some Banana plants reproduce by both means.

Activity – 2

Sheet For Lab Activity
Student Name : _____________________________________________
Date : _____________________________________________
AIM: To observe the parts of a flower.
Material required:
Two datura flowers, new blade, magnifying glass, pencil.

Procedure :
Hold a datura flower by its stalk and observe its external features. Now draw the diagram of that flower in the box given below. Note down your observations.
Parts of the flower :

External parts :
SEPALS:
Colour : Green Petal
Shape : Round tubular
Number : 5
Are they United/Separate ? United PETALS:
Colour : White
Shape v : Round tubular
Number : 5
Are they United/Separate? United

Procedure:
Cut the second datura flower vertically into two equal parts from bottom to top. Check that all parts are cut into two equal halves. Observe and draw the internal view.

Internal parts:
ANDROECIUM:
Colour : White
Shape : Thin filamentous
Number : 5
Are they United/Separate?
Separate
GYNOECIUM: Style
Colour : Light Yellow
Shape : Long tube like
. swollen at the base Ovary
Number : 1-Ovary – style – stigma
Are they United/Separate ? Separate Gynoecium

Activity – 3

Question 3.
Collect different varieties of flowers from your school garden. Take each flower and count the parts in it. Record the details in the given table.

i) Which flowers have all four whorls?
ii) Mention the flowers in which one or two whorls are missing.
iii) Write the whorl \ whorls which are missing.
Answer:
i) Lady’s fingers, Hibiscus and Datura flowers have all the four whorls.
ii) In Pumpkin, Bauhinia, Ipomea flowers one or two whorls are missing.
iii) Either androecium or gynoecium are missing.

Give examples for complete flowers …………………, ………………………, ………………….
Hibiscus, Datura etc.

Give examples for incomplete flowers …………………., …………………
Pumpkin, Ridge gourd, etc.

Activity – 4

Question 4.
Collect flowers of hibiscus, papaya, ridge gourd and some other flowers. Fill the table given below by observing the androecium and gynoecium.

i) In which plants either androecium or gynoecium is present?
ii) In which plants both androecium and gynoecium are present in the same flower?
iii) In which plant both androecium and gynoecium are present separately in different flowers of the same plant?
iv) In which plant both androecium and gynoecium are present in different flowers of different plants?
Answer:
i) Papaya, Ridge gourd, Bitter gourd
ii) Hibiscus, Datura.
iii) Ridge gourd, Bitter gourd
iv) Papaya

Activity – 5

Question 5.
How do you prove that seed dispersal is necessary for plants?
Answer:

1. Take two cups with soil.
2. Take handful of mustard seeds in the first cup and only four mustard seeds in the second cup.
3. Pour equal amount of water daily and observe them regularly for 15 days.

Observations:

1. Numerous plants germinated in the first cup. They have not enough space to grow freely. They are also not getting sufficient nutrients and water to grow. The plants present in the first cup are not healthy.
2. In the second cup, the four mustard seeds germinated into new plants. They have enough space, water and nutrients to grow. They are healthy.

Inference:
All the plants try to spread their seeds to distant places to increase the chances of survival and propagation. So, they adopt different methods to disperse their seeds. It occurs with the help of different agents.

PROJECT : Page No. 13

Question 6.
Now Collect information about various agents of seed dispersal from internet, your school library or by observing your surroundings. Make a Scrap book with pictures and your illustrations. And fill the below table with at least three examples each.

SCERT AP 7th Class Maths Solutions Pdf Chapter 78 Exponents and Powers InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers InText Questions

Check Your Progress [Page No. 28]

Question 1.
Write the following in exponential form by using 10 as the base number :
(i) 10,00,00,000
(ii) 100,00,00,000
Answer:

Let’s Explore [Page No. 29]

Question 1.
Observe and complete the following table. First one is done for you.

Answer:

Question 2.
Write the following numbers in exponential form. Also state the base, exponent and how to read. .
(i) 16
(ii) 49
(iii) 512
(iv) 243
Answer:

Question 3.
Compute the following and write the greater one.
(i) 4³ or 3⁴
Answer:
4³ = 4 × 4 × 4 = 64
3⁴ = 3 × 3 × 3 × 3 = 81 is greater.
64 < 81 (or) 81 > 64
So, 4³ < 3⁴ (or) 3⁴ > 4³

(ii) 5³ or 3⁵
Answer:
5³ = 5 × 5 × 5 = 125
3⁵ = 3 × 3 × 3 × 3 × 3 = 243 is greater.
125 < 243 (or) 243 > 125
So, 5³ < 3⁵ (or) 3⁵ > 5³

Question 4.
Is 32 equal to 23 ? Justify your answer.
Answer:
3² = 3 × 3 = 9
23 = 2 × 2 × 2 = 8
9 ≠ 8
3² ≠ 2³
So, a^b ≠ b^a unless a = b

Check Your Progress [Page No. 30]

Express the following number in exponential form using prime factorisation:
(i) 432
Answer:

432 = 2 × 216
= 2 × 2 × 108
= 2 × 2 × 2 × 54
= 2 × 2 × 2 × 2 × 27
= 2 × 2 × 2 × 2 × 3 × 9
= 2 × 2 × 2 × 2 × 3 × 3 × 3
∴ 432 = 2⁴ × 3³

(ii) 1296
Answer:
1296 = 2 × 648
= 2 × 2 × 324
= 2 × 2 × 2 × 162
= 2 × 2 × 2 × 2 × 81
= 2 × 2 × 2 × 2 × 3 × 27
= 2 × 2 × 2 × 2 × 3 × 3 × 9
= 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

∴ 1296 = 2⁴ × 3⁴

(iii) 729
Answer:

729 = 3 × 243
= 3 × 3 × 81
= 3 × 3 × 3 × 27
= 3 × 3 × 3 × 3 × 9
= 3 × 3 × 3 × 3 × 3 × 3
729 = 3⁶

(iv) 1600
Answer:

1600 = 2 × 800
= 2 × 2 × 400
= 2 × 2 × 2 × 200
= 2 × 2 × 2 × 2 × 100
= 2 × 2 × 2 × 2 × 2 × 50
= 2 × 2 × 2 × 2 × 2 × 2 × 25
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
1600 = 2⁶ × 5²

Let’s Explore [Page No. 32]

Question 1.
Write the appropriate number in place of ▢ in the following.
Let ‘b’ be any non-zero integer.

Answer:

= b² × b³
= b × b × b × b × b = b⁵

Answer:

then b¹⁰ × b¹ = b¹⁴
b¹⁰⁺¹ = b¹⁴
b¹¹ ≠ b¹⁴

then b¹⁰ × b² = b¹⁴
b¹⁰⁺² = b¹⁴
b¹² ≠ b¹⁴

then b¹⁰ × b³ = b¹⁴
b¹⁰⁺³ = b¹⁴
b¹³ ≠ b¹⁴

then b¹⁰ × b⁴ = b¹⁴
b¹⁰⁺⁴ = b¹⁴
∴ b¹⁴ = b¹⁴

Question 2.
Simplify the following using the formula a^m × aⁿ = a^m+n
(i) 5⁷ × 5⁴
Answer:
5⁷ × 5⁴
We know that a^m × aⁿ = a^m+n;
5⁷ × 5⁴ = 5⁷⁺⁴ = 5¹¹
∴ 5⁷ × 5⁴ = 5¹¹

(ii) p³ × p²
Answer:
p³ × p²
We know that a^m × aⁿ = a^m+n
p³ × p² = p³⁺² = p⁵
∴ p³ × p² = p⁵

(iii) (-4)¹⁰ × (-4)³ × (-4)²
Answer:
(-4)¹⁰ × (-4)³ × (-4)²
We know that a^m × aⁿ = a^m+n
(- 4)¹⁰⁺³⁺² = (- 4)¹⁵
∴ (-4)¹⁰ × (-4)³ × (-4)² = (- 4)¹⁵

Let’s Explore [Page No. 33]

Question 1.
Write the following in exponential form using the formula (a^m)ⁿ = a^mn.
(i) (6²)⁴
Answer:
(6²)⁴
We know (a^m)ⁿ = a^mn
(6²)⁴ = 6^2×4 = 6⁸
∴ (6²)⁴ = 6⁸

(ii) (2²)¹⁰⁰
Answer:
(2²)¹⁰⁰
We know (a^m)ⁿ = a^mn
(2²)¹⁰⁰ = 2^2×100 = 2²⁰⁰
∴ (2²)¹⁰⁰ = 2²⁰⁰

(iii) (20⁶)²
Answer:
(20⁶)²
We know (a^m)ⁿ = a^mn
(20⁶)² = 20^6×2 = 20¹²
∴ (20⁶)² = 20¹²

(iv) [(-10)³]⁵
Answer:
[(-10)³]⁵
We know (a^m)ⁿ = a^mn
[(-10)³]⁵ =(-10)^3×5 = (-10)¹⁵
∴ [(- 10)³]⁵ = (- 10)¹⁵

Check Your Progress [Page No. 34]

Simplify the following by using the law , a^m × b^m = (ab)^m.
(i) 7⁶ × 3⁶
Answer:
7⁶ × 3⁶
We know, a^m × b^m = (ab)^m
7⁶ × 3⁶ = (7 × 3)⁶ = (21)⁶
∴ 7⁶ × 3⁶ = 21⁶

(ii) (3 × 5)⁴
Answer:
(3 × 5)⁴
We know, (ab)^m = a^m × b^m
(3 × 5)⁴ = 3⁴ × 5⁴
∴ (3 × 5)⁴ = 3⁴ × 5⁴

(iii) a⁴ × b⁴
Answer:
a⁴ × b⁴
We know, a^m × b^m = (ab)^m
a⁴ × b⁴ = (a . b)⁴
∴ a⁴ × b⁴ = (a . b)⁴

(iv) 3² × a²
Answer:
3² × a²
We know, a^m × b^m = (ab)^m
3² × a² = (3 × a)² = (3a)²
∴ 3² × a² = (3a)²

Let’s Explore [Page No. 37]

Question 1.
Simplify and write In the form of a^m-n Or \(\frac{1}{\mathbf{a}^{\mathbf{n}-\mathbf{m}}}\)
(i) \(\frac{10^{8}}{10^{4}}\)
Answer:
We Know \(\frac{10^{8}}{10^{4}}\) = a^m-n
\(\frac{10^{8}}{10^{4}}\) = 10^8-4 = 10⁴
∴\(\frac{10^{8}}{10^{4}}\) = 10⁴

(ii) \(\frac{(-7)^{13}}{(-7)^{10}}\)
Answer:
\(\frac{(-7)^{13}}{(-7)^{10}}\)
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n (m > n)
\(\frac{(-7)^{13}}{(-7)^{10}}\) = (-7)^13-10
= (-7)³ = -7 × -7 × -7 = – 343
∴\(\frac{(-7)^{13}}{(-7)^{10}}\) = – 343

(iii) \(\frac{12^{5}}{12^{8}}\)
Answer:

(iv) \(\frac{3^{4}}{3^{7}}\)
Answer:

Question 2.
Fill the appropriate number In the box


Answer:

We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n (m > n)
\(\frac{7^{12}}{7^{7}}\) = 7^12-7 = 7⁵
∴ \(\frac{7^{12}}{7^{7}}\) = 7⁵

Answer:

Answer:

Answer:

Question 3.
Simplify the following :
(i) \(\frac{6^{8}}{6^{8}}\)
Answer:
\(\frac{6^{8}}{6^{8}}\)
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n
\(\frac{6^{8}}{6^{8}}\) = 6^8-8 = 6⁰ = 1 (∵ a⁰ = 1)

(ii) \(\frac{t^{10}}{t^{10}}\)
Answer:
\(\frac{t^{10}}{t^{10}}\)
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n
\(\frac{t^{10}}{t^{10}}\) = t^10-10 = t⁰ = 1 (∵ a⁰ = 1)

(iii) \(\frac{12^{7}}{12^{7}}\)
Answer:
\(\frac{12^{7}}{12^{7}}\)
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n
\(\frac{12^{7}}{12^{7}}\) = 12^7-7 = 12⁰ = 1 (∵ a⁰ = 1)

(iv) \(\frac{p^{5}}{p^{5}}\)
Answer:
\(\frac{p^{5}}{p^{5}}\)
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n
\(\frac{p^{5}}{p^{5}}\) = p^10-10 = p⁰ = 1 (∵ a⁰ = 1)

Check Your Progress [Page No. 38]

Question 1.
Complete the following boxes .

Answer:

Answer:

Answer:

Answer:

Check Your Progress [Page No. 39]

Question 1.
Express the following in exponential form.
(i) \(\frac{-27}{125}\)
Answer:

(ii) \(\frac{-32}{243}\)
Answer:

(iii) \(\frac{-125}{1000}\)
Answer:

(iv) \(\frac{-1}{625}\)
Answer:

Let’s Think [Page No. 39]

Question 1.
Deekshltha and Harsha computed 4(3)² in different ways.
Deeksbitha did it like this
4(3)² = (4 × 3)²
= 12²
= 144

Harsha did it like this
4(3)² = 4 × (3 × 3)
= 4 × 9
= 36
Who has done the problem Incorrectly?
Discuss the reason for the mistake with your friends.
Answer:
In 4(3)² square only belongs to 3, but not 4.
So, Deekshitha did wrong and Harsha did correct.

Let’s Do Activity [Page No. 40]

Finding the pair : Divide the classroom into two groups. Each group has a set of cards. Each student of group 1 has to pair with one suitable student of group 2 by stating in the reason.

Answer:

Note : This activity can be extended till all the children in the class are familiarised with the laws of exponents.

Project Work [Page No. 42]

Collect the annual income of 5 families in your location by observing their ration card and rounded into the nearest thousand / Lakh and express in the exponential form. One done for you.

Answer:

Examples:

Question 1.
Which one is greater 8² or 2⁸? Justify.
Answer:
8² = 8 × 8 = 64
2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
256 >64
Therefore, 2⁸ > 8².

Question 2.
Simplify the following using the formula a^m × aⁿ = a^m+n
(i) (- 5)⁷ × (- 5)⁴
Answer:
(- 5)⁷ × (- 5)⁴ = (- 5)⁷⁺⁴
(∵ a^m × aⁿ = a^m+n)
= (- 5)¹¹
∴ (- 5)⁷ × (- 5)⁴ = (- 5)¹¹

(ii) 3³ × 3² × 3⁴
Answer:
3³ × 3² × 3⁴ = 3³⁺²⁺⁴
(∵ a^m × aⁿ = a^m+n)
= 3⁹
∴ 3³ × 3² × 3⁴ = 3⁹

Question 3.
Simplify the following using the formula (a^m)ⁿ = a^mn
(i) (8³)⁴
Answer:
(8³)⁴ = 8^3×4
= 8¹²
∴ (8³)⁴ = 8¹²

(ii) [(-11)⁵]²
Answer:
[(-11)⁵]² = (- 11)^5×2
= (- 11)¹⁰
∴ [(-11)⁵]² = (- 11)¹⁰

(iii) (7⁵⁰)²
Answer:
(7⁵⁰)²= 7^50×2 = 7¹⁰⁰
∴ (7⁵⁰)² = 7¹⁰⁰

Question 4.
Simplify the following using the expo-nential law a^m × b^m = (ab)^m
(i) 5² × 3²
Answer:
5² × 3² = (5 × 3)² [∵ a^m × b^m = (ab)^m]

(ii) p³ × q³
Answer:
p³ × q³ = (p × q)³

(iii) (7 × 8)⁴
Answer:
(7 × 8)⁴ = 7⁴ × 8⁴ [∵ (ab)^m = a^m × b^m]

Question 5.
Simplify the following and write in the form of \(\frac{a^{m}}{a^{n}}\) = a^m-n or \(\frac{a^{m}}{a^{n}}=\frac{1}{a^{n-m}}\)
(i) \(\frac{2^{9}}{2^{3}}\)
Answer:
\(\frac{2^{9}}{2^{3}}\) = 2^9-3 [∵ \(\frac{a^{m}}{a^{n}}\) = a^m-n]
= 2⁶

(ii) \(\frac{(-9)^{11}}{(-9)^{7}}\)
Answer:
\(\frac{(-9)^{11}}{(-9)^{7}}\) = (- 9)^11-7 = (-9)⁴

(iii) \(\frac{7^{10}}{7^{13}}\)
Answer:
\(\frac{7^{10}}{7^{13}}=\frac{1}{7^{13-10}}\)
[∵ \(\frac{a^{m}}{a^{n}}=\frac{1}{a^{n-m}}\)]
= \(\frac{1}{7^{3}}\)

(iv) \(\frac{6^{2}}{6^{5}}\)
Answer:
\(\frac{6^{2}}{6^{5}}=\frac{1}{6^{5-2}}=\frac{1}{6^{3}}\)

Question 6.
Simplify the following by using formula \(\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}\).
(i) \(\frac{5^{3}}{2^{3}}\)
Answer:
\(\frac{5^{3}}{2^{3}}=\left(\frac{5}{2}\right)^{3}\) [∵ \(\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}\)]

(ii) \(\left(\frac{8}{5}\right)^{4}\)
Answer:
\(\left(\frac{8}{5}\right)^{4}=\frac{8^{4}}{5^{4}}\)
[∵ \(\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}\)]

Question 7.
Evaluate:
(1)⁴, (1)⁵, (1)⁷, (- 1)², (- 1)³, (- 1)⁴, (- 1)⁵
Answer:
(1)⁴ = 1 × 1 × 1 × 1 = 1
(1)⁵ = 1 × 1 × 1 × 1 × 1 = 1
(1)⁷ = 1 × 1 × 1 × 1 × 1 × 1 × 1 = 1
(- 1)²= (-1) × (-1) = 1
(- 1)³ = (-1) × (1) × (-1) = – 1
(- 1)⁴ =(-1) × (-1) × (-1) × (-1) = 1
(- 1)⁵ = (-1) × (-1) × (-1) × (-1) × (- 1) = – 1

From the above illustrations,
(i) It raised. to any power is 1.
(ii) (-1) raised to even power is (- 1) and
(-1) raised to an odd power is (-1).

Thus (- 1)^m = 1 if ’m’ is even
(- 1)^m 1 if ‘m’ is odd

Question 8.
Express \(\) In exponential form.
Answer:
-8 = (-2) × (-2) × (-2) = (-2)³
27 = 3 × 3 × 3 = (3)³
∴ \(\frac{-8}{27}: \frac{(-2)^{3}}{3^{3}}=\left(\frac{-2}{3}\right)^{3}\)

Question 9.
Abhllash computed a³. a² as a⁶. Is it correct?
Answer:
Abhilash has done it incorrectly.
Bcause a³. a² = a³⁺² = a⁵ [∵ a^m . aⁿ = a^mn]
Therefore, a³. a² = a⁵ is correct answer.

Question 10.
Riyaz computed \(\frac{a^{8}}{a^{2}}\) as a⁴. Has he done It correctly? Justify your answer.
Answer:
Riyaz has done it incorrectly.
Because \(\frac{a^{8}}{a^{2}}\) = a^8-2
= a⁶ [∵ \(\frac{a^{m}}{a^{n}}\) = a^m-n]
∴ \(\frac{a^{8}}{a^{2}}\) = a⁶ is correct answer.

Question 11.
Write the following into standard form.
(i) 7465
Answer:
7465 = 7.465 × 1000 (Decimal is shifted three places to the left)
= 7.465 × 10³

(ii) The height of Mount Everest is 8848 m.
Answer:
The height of Mount Everest
= 8848 m
= 8.848 × 1000 m (Decimal is shifted three places to the left)
= 8.848 × 10³m

(iii) The distance from the Sun to Earth is 149,600,000,000 m.
Answer:
The distance from the Sun to Earth
= 149,600,000,000 m
= 1.496 × 100000000000 m
= 1.496 × 10¹¹ m

Reasoning Corner [Page No. 45: Odd one out in numbers]

In each of the following questions, there are 4 numbers. Three of them are similar in a certain way but one is not like the other three. One has to identify the similarity and then strike the odd one out as answer option.
The number can be odd/ even /consecutive, prime numbers, multiple of some number, single, square or cubes of different numbers, plus/minus of some other number or combinations of any mathematical calculation.
Question 1.
(a) 12
(b) 25
(c) 37
(d) 49
Answer:
(c) 37

Hint:
Prime number

Question 2.
(a) 13
(b) 63
(c) 83
(d) 43
Answer:
(b) 63

Hint:
Not a prime number

Question 3.
(a) 21
(b) 49
(c) 56
(d) 36
Answer:
(d) 36

Hint:
Not divisible by 7

Question 4.
(a) 112
(b) 256
(c) 118
(d) 214
Answer:
(b) 256

Hint:
Square number

Question 5.
(a) 42
(b) 21
(c) 84
(d) 35
Answer:
(d) 35

Hint:
Not divisible by 3

Question 6.
(a) 11
(b) 13
(c) 15
(d) 17
Answer:
(c) 15

Hint:
Not a prime number

Question 7.
(a) 10
(b) 11
(c) 15
(d) 16
Answer:
(b) 11

Hint:
Prime number

Question 8.
(a) 49
(b) 63
(c) 77
(d) 81
Answer:
(d) 81

Hint:
Not divisible by 7

Question 9.
(a) 28
(b) 65
(c) 129
(d) 215
Answer:
(a) 28

Hint:
Even number

Question 10.
(a) 51
(b) 144
(c) 64
(d) 121
Answer:
(a) 51

Hint:
Not square number

Practice Questions [Page No. 46]

Question 1.
(a) 3
(b) 9
(c) 5
(d) 7
Answer:
(b) 9

Explanation:
9 is composite number. The remaining numbers are primes.

Question 2.
(a) 6450
(b) 1776
(c) 2392
(d) 3815
Answer:
(d) 3815

Explanation:
All others are even numbers.

Question 3.
(a) 24
(b) 48
(c) 42
(d) 12
Answer:
(c) 42

Explanation:
12, 24 and 48 are multiples of 12.

Question 4.
(a) 616
(b) 252
(c) 311
(d) 707
Answer:
(c) 311

Explanation:
616, 252 and 707 are Palindromes.

Question 5.
(a) 18
(b) 12
(c) 30
(d) 20
Answer:
(d) 20

Explanation:
12, 18 and 30 are 3 (or) 6 multiples. But, 20 is not 3 (or) 6 multiple.

Question 6.
Find the odd one from the given
(a) 3730
(b) 6820
(c) 5568
(d) 4604
Answer:
(a) 3730

Explanation:
3730, All others are divisible by 4.

Question 7.
(a) 2587
(b) 7628
(c) 8726
(d) 2867
Answer:
(a) 2587

Explanation:
All others are formed by 2, 6, 7 and 8.

Question 8.
(a) 63
(b) 29
(c) 27
(d) 25
Answer:
(d) 25

Explanation:
63, 29 and 27 are not squares. 25 only the square.

Question 9.
(a) 23
(b) 37
(c) 21
(d) 31
Answer:
(c) 21

Explanation:
23,37 and 31 are prime numbers.
21 only the composite number.

Question 10.
(a) 18
(b) 9
(c) 21
(d) 7
Answer:
(d) 7

Explanation:
18, 9, 21 are composite numbers.
7 only the prime number.

SCERT AP 7th Class Maths Solutions Pdf Chapter 8 Exponents and Powers Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers Unit Exercise

Question 1.
Answer the following.
(i) The exponential form 14⁹ should read as
Answer:
14 is raised to the power of 9.

(ii) When base is 12 and exponent is 17, it’s exponential form is _________
Answer:
12¹⁷.

(iii) The value of (14 × 21)⁰ is
Answer:
We know a⁰ = 1
So, (14 × 21)⁰ = 1

Question 2.
Express the following numbers as a product of powers of prime factors :
(i) 648
Answer:

Given 648 = 2 × 324
= 2 × 2 × 162
= 2 × 2 × 2 × 81
= 2 × 2 × 2 × 3 × 27
= 2 × 2 × 2 × 3 × 3 × 9
= 2 × 2 × 2 × 3 × 3 × 3 × 3
∴ 648 = 2³ × 3⁴

(ii) 1600
Answer:
Given 1600 = 2 × 800
= 2 × 2 × 400
= 2 × 2 × 2 × 200
= 2 × 2 × 2 × 2 × 100
= 2 × 2 × 2 × 2 × 2 × 2 × 25
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

∴ 1600 = 2⁶ × 5²

(iii) 3600
Answer:

Given 3600 = 2 × 1800
= 2 × 2 × 900
= 2 × 2 × 2 × 450
= 2 × 2 × 2 × 2 × 225
= 2 × 2 × 2 × 2 × 3 × 75
= 2 × 2 × 2 × 2 × 3 × 3 × 25
= 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
∴ 3600 = 2⁴ × 3² × 5²

Question 3.
Simplify the following using laws of exponents.
(i) a⁴ × a¹⁰
Answer:
a⁴ × a¹⁰
We know a^m × aⁿ = a^m+n
= a⁴⁺¹⁰
∴ a⁴ × a¹⁰ = a¹⁴

(ii) 18¹⁸ ÷ 18¹⁴
Answer:
18¹⁸ ÷ 18¹⁴
We Know a^m ÷ aⁿ = a^m-n
= 18^18-14
∴ 18¹⁸ ÷ 18¹⁴ = 18⁴

(iii) (x^m)⁰
Answer:
(x^m)⁰
We Know (a^m)ⁿ = a^m.n
= x^m×n = x⁰(∵ a⁰ = 1)
∴ (x^m)⁰ = 1

(iv) (6² × 6⁴) ÷ 6³
Answer:
(6² X 6⁴) ÷ 6³
We Know a^m × aⁿ = a^m+n
= (6²⁺⁴) ÷ 6³
= 6⁶ ÷ 6³
We Know a^m ÷ aⁿ = a^m-n
= 6^6-3
∴ (6² × 6⁴) ÷ 6³ = 6³

(v) \(\left(\frac{2}{3}\right)^{p}\)
Answer:
\(\left(\frac{2}{3}\right)^{p}\)
We Know \(\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}=\frac{2^{p}}{3^{p}}\)
∴ \(\left(\frac{2}{3}\right)^{\mathrm{p}}=\frac{2^{\mathrm{p}}}{3^{\mathrm{p}}}\)

Question 4.
Identify the greater number in each of the following andjustify your answer.
(i) 2¹⁰ or 10²
Answer:
2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
2¹⁰ = 1024
10² = 10 × 10
10² = 100
1024 > 100
So, 2¹⁰ > 10²
∴ 2¹⁰ is greater.

(ii) 5⁴ or 4⁵
Answer:
5⁴ = 5 × 5 × 5 × 5 = 625
4⁵ =4 × 4 × 4 × 4 × 4 = 1024
1024 > 625
So, 4⁵ > 5⁴
∴ 4⁵ is greater number.

Question 5.
If \(\left(\frac{4}{5}\right)^{2} \times\left(\frac{4}{5}\right)^{5}=\left(\frac{4}{5}\right)^{k}\), then find the value of ‘k’
Answer:
Given \(\left(\frac{4}{5}\right)^{2} \times\left(\frac{4}{5}\right)^{5}=\left(\frac{4}{5}\right)^{k}\)
We Know a^m × aⁿ = a^m+n
⇒ \(\left(\frac{4}{5}\right)^{2+5}=\left(\frac{4}{5}\right)^{\mathrm{k}}\)
⇒ \(\left(\frac{4}{5}\right)^{7}=\left(\frac{4}{5}\right)^{\mathrm{k}}\)
If the bases are equal powers should be equal
⇒ 7 = k
∴ k = 7

Question 6.
If 5^2p+1 ÷ 5² = 125, then find the value of ‘p’.
Answer:
Given 5^2p+1 ÷ 5² = 125
We Know a^m ÷ aⁿ = a^m-n
⇒ 5^2p+1-2 = 5 × 5 × 5
⇒ 5^2p-1 = 5³
If the bases are equal, powers should be equal.
⇒ 2p – 1 =3
⇒ 2p = 3 + 1
⇒ 2p = 4
⇒ \(\frac{2 \mathrm{p}}{2}=\frac{4}{2}\)
∴ p = 2

Question 7.
Prove that \(\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b} \times\left(\frac{x^{a}}{x^{b}}\right)^{c}\) = 1
Answer:
Given \(\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b} \times\left(\frac{x^{a}}{x^{b}}\right)^{c}\) = 1

Question 8.
Express the following numbers in the expanded form.
(i) 20068
Answer:
20068 = (2 × 10,000) + (0 × 1000) + (0 × 100) + (6 × 10) + (8 × 1)
∴ 20068 = (2 × 104) + (6 × 101) + (8 × 1)

(ii) 120718
Answer:
120718 = (1 × 1,00,000) + (2 × 10,000) + (0 × 1000) + (7 × 100) + (1 × 10) + (8 × 1)
∴ 120718 = (1 × 105) + (2 × 104) + (7 × 102) + (1 × 101) + (8 × 1)

Question 9.
Express the number appearing in the following statements in standard form :
(i) The Moon is 384467000 meters away from the Earth approximately.
Answer:
Distance of Moon from the Earth = 384467000 metres
= 3.84467000 × 100000000

Decimal is shifted eight places to the left.
= 3.84 4 67 × 10⁸ m

Distance of Moon from the earth = 3.84 4 67 × 10⁸ m

(ii) Mass of the Sun is 1 ,989,000,000,000,000,000,00000,000000 kg.
Answer:
Mass of Sun = 1 ,989,000,000,000,000,000,00000,000000 kg
= 1.989 × 1 ,000,000,000,000,000,00000,000000
= 1.989 × 10³⁰ kg
∴ Mass of Sun = 1.989 × 10³⁰ kg.

Question 10.
Lasya solved some problems of exponents and powers in the following way. Do you agree with the solution ? If not why? Justify your answer.
Answer:
(i) x³ × x² = x⁶
Answer:
No. I won’t agree with this solution.
Given, x³ × x²
We know a^m × aⁿ = a^m+n
= a³⁺²
= x⁵ which is ≠ x⁶
so, x³ × x² ≠ x⁶
We have to add to powers. But, Lasya multiplied the powers. That’s why Lasya’s solution is wrong.

(ii) (6³)¹⁰ = 6¹³
Answer:
No, (6³)¹⁰ is not equal to 6.
We know (a^m)ⁿ = a^mn
= 6^3×10
= 6³⁰ which is ≠ 6¹³
so, (6³)¹⁰ ≠ 6¹³
We have to multiply. the powers. But, Lasya added the powers. That’s why Lasya’s solution is wrong.

(iii) \(\frac{4 x^{6}}{2 x^{2}}\) = 2x³
Answer:
No, I won’t agree with this solution.
\(\frac{4 x^{6}}{2 x^{2}}=\frac{2^{2}}{2^{1}} \times \frac{x^{6}}{x^{2}}\)
We know a^m ÷ aⁿ = a^m-n
= 2^2-1 × x^6-2
= 2¹.x⁴
= 2x⁴ which is ≠ 2x³
so, \(\frac{4 x^{6}}{2 x^{2}}\) ≠ 2x³
We have to subtract the powers. But, Lasya divided the powers. That’s why Lasya’s solution is wrong.

(iv) \(\frac{3^{5}}{9^{5}}=\frac{1}{3}\)
Answer:
No. I won’t agree with this solution.
\(\frac{3^{5}}{9^{5}}=\frac{3^{5}}{\left(3^{2}\right)^{5}}\)

We Know (a^m)ⁿ = a^mn

We have to subtract the powers. But, Lasya divided the powers. That’s why Lasya’s solution is wrong.

Question 11.
Is – 2² is equal to 4? Justify your answer.
Answer:
– 2² = – (2 × 2) = – 4 ≠ 4
∴ – 2² = – 4

Question 12.
Beulah computed 2⁵ × 2¹⁰ = 2⁵⁰. Has she done it correctly? Give the reason.
Answer:
Given 2⁵ × 2¹⁰
We know a^m × aⁿ = a^m+n
= 2⁵⁺¹⁰ = 2¹⁵ ≠ 2⁵⁰
∴ 2⁵ × 2¹⁰ ≠ 2⁵⁰
Beulah did wrong.
Here we have to add the powers.
But, he multiplied the powers.

Question 13.
Rafi computed \(\frac{3^{9}}{3^{3}}\) as 3³. Has he done
Answer:
Given \(\frac{3^{9}}{3^{3}}\)
We know \(\frac{a^{m}}{a^{n}}\) = a^m-n
= 3^9-3 = 3⁶ ≠ 3³
So, \(\frac{3^{9}}{3^{3}}\) ≠ 3³
Rafi computed wrong.
Here we have to subtract the powers. But, he divided the powers.

Question 14.
Is (a²)³ equal to a⁸? Give the reason.
Answer:
Given (a²)³ equal to a⁸
We Know (a^m)ⁿ = a^mn
= (a²)³ = a^2×3 = a⁶ ≠ a⁸
∴ (a²)³ = a⁶

We have to do 2 × 3 = 6, But not
2³ = 2 × 2 × 2 = 8

SCERT AP 7th Class Maths Solutions Pdf Chapter 8 Exponents and Powers Ex 8.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers Ex 8.3

Question 1.
Write the following numbers in expanded form.
(i) 23468
Answer:
23468 = (2 × 10,000) + (3 × 1000) + (4 × 100) + (6 × 10) + (8 × 1)
= (2 × 10⁴) + (3 × 10³) + (4 × 10²) + (6 × 10¹) + (8 × 1)

(ii) 120718
Answer:
120718 = (1 × 1,00,000) + (2 × 10,000) + (0 × 1000) + (7 × 100) + (1 × 10) + (8 × 1)
= (1 × 10⁵) + (2 × 10⁴) + (7 × 10²) + (1 × 10¹) + (8 × 1)

(iii) 806190
Answer:
806190 = (8 × 1,00,000) + (0 × 10,000) + (6 × 1000) + (1 × 100) + (9 × 10) + (0 × 1)
= (8 × 10⁵) + (6 × 10³) + (1 × 10²) + (9 × 10¹)

(iv) 3006194
Answer:
3006194 = (3 × 10,00,000) + (0 × 1,00,000) + (0 × 10,000) + (6 × 1000) + (1 × 100) + (9 × 10) + (4 × 1)
= (3 × 10⁶) + (6 × 10³) + (1 × 10²) + (9 × 10¹) + (4 × 1)

Question 2.
Write the following numbers in standard fortti:
(i) 5,00,000
Answer:
5,00,000 = 5 × 1,00,000 = 5 × 10⁵

(ii) 48,30,000
Answer:
48,30,000 = 4.830000 × 10,00,000 (Decimal is shifted six places to the left).
= 4.830000 × 10⁶
∴ 48,30,000 = 4.83 × 10⁶

(iii) 3,94,00,00,00,000
Answer:
3,94,00,00,00,000 = 3.94000000000 × 100000000000 = 3.94 × 10¹¹

(iv) 30000000
Answer:
30000000 = 3 × 10000000 = 3 × 10⁷
∴ 30000000 = 3 × 10⁷

(v) 180000
Answer:
180000 = 1.80000 × 100000 = 1.8 × 10⁵

Question 3.
Express the number appearing in the following statements in standard form,
(i) The Universe is estimated to be about 12,000,000,000 years old.
Answer:
The Universe is 12,000,000,000 years old = 12 × 1,000,000,000
= 12 × 10⁹ years.
= 1.2 × 10¹⁰ years.

(ii) Earth circumference is about 402000000 km.
Answer:
Earth circumference = 402000000 km
= 4.02000000 × 100000000 (Decimal is shifted eight places to left)
= 4.02 × 10⁸ km.
∴ Earth circumference = 4.02 × 10⁸ km.

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion InText Questions

[Page No. 1]

There are several situations in our daily life where we use ratio and proportion. Let us look at the following pictures and answers to the given questions:
Question 1.
Can we say the ratio of speeds of Cheetah to Man?

Answer:
Speed of Cheetah = 120 kmph
Speed of a man = 20 kmph
∴ Ratio of speed of Cheetah to a man
= 120 : 20
= 6 : 1

Question 2.
What will be the ratio of heights of HematoAmir?

Answer:
Height of Hema = 150 cm
Height of Amir 75 cm
∴ Ratio of heights 150 : 75
= 2 : 1

Question 3.

Answer:
Cost of 2 Hands of banana = ₹ 80
∴ Cost of 3 Hands of banana
= \(\frac{3}{2}\) × 80 = ₹ 120

Question 4.
Do you know which aspect ratio makes constructions more attractive and beautiful?
Answer:
Golden ratio is more beautiful. It is the ratio of a line segment cut in to two pieces of different lengths such that the ratio of whole segment to that of the longer segment is equal to that of longer segment to the shorter segment.

Check your Progress [Page No. 4]

Question 1.
Write the compound ratio of the following given ratios.
(i) 3 : 5 and 4 : 3
Answer:
Given ratios are 3 : 5 and 4 : 3.
If a : b and c : d are ratios, then their compound ratio is product of antecedents : product of extremes.
that is a × c : b : d
3 × 4 : 5 × 3
12 : 15
∴ Compound ratio = 4 : 5

(ii) 8 : 3 and 6 : 5
Answer:
Given ratios are 8 : 3 and 6 : 5.
If a : b and c : d are ratios, then their compound ratio is product of antecedents : product of extremes, that is
a × c : b : d
8 × 6 : 3 × 5
∴ Compound ratio = 16 : 5

(iii) 2: 1 and 8: 7
Answer:
Given Ratios are 2 : 1 and 8 : 7.
If a : b and c: d are ratios, then their compound ratio is product of antecedents : product of extremes that is a × c : b : d
2 × 8 : 1 × 7
16 : 7
∴ Compound ratio= 16 : 7

Question 2.
Fill in the boxes with correct answers.

Answer:

[Page No. 6]

Discuss and answer the following questions related to real-life situations:
Question 1.
If the cost of 3 ball pens is ₹ 15, then what is the cost of 6 such pens?
Answer:
3 : 15 : : 6 : cost
Cost x 3 = 15 × 6
∴ Cost = \(\frac{15 \times 6}{3}\) = ₹ 30

Question 2.
In a school, for implementing ‘Mid day Meal’ scheme for class VII of 40 students they need 6 kgs of rice for one day.
(i) How much rice is needed for 6 days?
Answer:
Rice required for 40 students for one day = 6 kgs
Rice required for 6 days = 6 × 6 = 36 kg

(ii) How much rice is needed for 10 days?
Answer:
Rice required for 10 days = 10 × 6 = 60 kg

(iii) How much rice is needed for 5 days?
Answer:
Rice required for 5 days = 5 × 6 = 30 kg

Let’s Explore [Page No. 6]

Question 1.
Given below the number of workers working for construction of a house and their wages in total.

Answer:

Let’s Think [Page No. 7]

Question 1.
If the cost of 4 note books is 80. What would be the cost of7 note books? The above problem can be solved by using unitary method. Can you solve? How? Think.
Answer:
Given the cost of 4 note books = ₹ 80
Cost of 1 note book = 80 ÷ 4 = ₹ 20
Cost of 7 note books = ₹ 20 × 7
= ₹ 140

Check Progress [Page No 7]

Question 1.
Fill in the blanks, if the given quantities are in direct proportion.

Answer:

Let’s’ Do Activity [Page No. 8]

Take a clock and fix Its minute hand at a particular number on clock (if it is 12 It will be easy). Then find and note the angles made by minute hand in every 15 minutes interval of time:


Check whether they are in direct proportion or not?
Answer:

15 : 30 and 90 : 180
Here 1 :2 = 1 :2
If ratios are equal, then they are in direct proportion.
So, time passed and angle are in direàt proportion.

Let’s Think [Page No. 8]

Question 1.
What is the angle made by minutes hand in a minute?
Answer:
We know angle made by minutes hand in 15 m = 90°
Let angle made by minutes hand in 1 m = x°
Then 15 : 1 = 90°: x°
If the ratios are equal,
The product of means = The product of extremes
⇒ 15 × x = 1 × 90°
⇒ \(\frac{15 x}{15}=\frac{90^{\circ}}{15}\)
⇒ x = 6°
∴ Angle made by minutes hand in 1 minute = 6°

Question 2.
What is the angle made by hours hand in one minute?
Answer:
We know angle made by hours hand in 1 hour (60 minutes) = 30°
Let angle made by hours hand 1 minute = x°
Then 60 : 1 = 30°: x°
If the ratios are equal,
The product of means = The product of the extremes
⇒ 60 × x = 1 × 30
⇒ \(\frac{60 x}{60}=\frac{30}{60}=\frac{1}{2}\)
⇒ x = \(\frac{1}{2}\) (or) 0.5°
∴ Angle made, by hours hand in 1 minute = \(\frac{1}{2}\)° (or) 0.5°

Check your Progress [Page No. 9]

Question 1.
Fill in the blanks if the given quantities are in inverse proportion.

Answer:

Check your Progress [Page No. 13]

Question 1.
Analyse how the three quantities given below are related and find ‘x’.

Answer:
From the above,

• Ration and the number of days are in inversely proportional.
• Ration and number of men are in directly proportional.
• Number of men is depends on both Ration and number of days.

So we have to take compound ratio of 108 : 70 and 25 : 15
i. e., 108 × 25 : 70 × 15
∴ 18 : x = 108 × 25 : 70 × 15
If the ratios are equal.
The Product of means = The Product of extremes
⇒ 108 × 25 × x = 18 × 70 × 15

∴ x = 7

Let’s Think [Page No. 18]

Question 1.
A man buys chocolates 10 for 10 rupees and sells them at 10 for ₹ 12. Does this result profit or loss ? What percent?
Answer:
Given cost price of 10 chocolates = ₹ 10
Selling price of 10 chocolates = ₹ 12
SP > CP, then man gets profit.
Profit = S.P – C.P
= 12 – 10 = ₹ 2
We know profit percent

Profit percentage = 20%

Question 2.
If a shopkeeper bought sofa sets and increases their prices by 50% and sell at 50% less, is it a loss or gain?
Answer:
Let the cost price of sofa set = ₹ 100

= ₹ 5o
Total printed price = ₹ 100 + ₹ 50
= ₹ 150
Loss percent = 50%

Loss = ₹ 75
Selling price Printed Price – Loss
= 150 – 75 = ₹ 75
C.P= ₹ 100; S.P = ₹ 75
and C.P > S.P
So, the shopkeeper will get loss of 25%.

Check Your Progress [Page No. 20]

Answer:

Check Your Progress [Page No. 24]

Answer:

Let’s Think [Page No. 24]

Question 1.
At what rate per annum will the principle doubles in 10 years?
Answer:
Method 1:
Let the principle = 100
Interest rate = x% ; Time = 10 years

Amount = Principle + Interest
= Double the principle
= 100 + 10x = 200
⇒ 100 + 10x – 100 = 200 – 100
⇒ 10x = 100
⇒ \(\frac{10 x}{10}=\frac{100}{10}\)
⇒ x = 10%
∴ Rate of interest = 10%

Method 2:
Let the principle =?
Rate of Interest = R%
T = 10 years
Given that A = 2P = P + P [∵ A = P + I]
∴ I = P
∴ \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}\) = P
⇒ \(\frac{\mathrm{P} \times 10 \times \mathrm{R}}{100}\) = P
R = 10%
∴ The principle become double in 10 years at 10% rate of interest.

Question 2.
At what rate per annum will the principle be 4 times in 15 years?
Answer:
Method 1:
Let the principle = ₹ 100
Interest rate = x %
Tirne = 15 years
Interest = \(\frac{\mathrm{PTR}}{100}\)

Amount = Principle + Interest
= 4 times the principle
= 100 + 15x = 400
⇒ 100 + 15x – 100 = 400 – 100
⇒ 15x = 300
⇒ \(\frac{15 x}{15}=\frac{300}{15}\) = 20%
∴ x = 20%

Method 2:
Let the principle = P
Time = 15 years
A = 4timesP = 4P
⇒ A = P + 3P
But A = P + I
So I = 3P

We know that I = \(\frac{\text { PTR }}{100}\)
⇒ 3P = \(\frac{P \times 15 \times R}{100}\)
⇒ R = \(\frac{100 \times 3}{15}\) = 20%
∴ The principle become 4 times in 15 years at 20% rate of interest.

Examples:

Question 1.
Find the compound ratio of 8 : 7 and 9 : 13.
Answer:
Given ratio = 8 : 7 :: 9 : 13
Compound ratio = 8 × 9 : 7 × 13 = 72 : 91
(∵ Compound ratio product of antecedent product consequents).

Question 2.
Two friends Prabhu and Suresh started a business with ₹ 1,00,000 each. After 3 months Suresh left the business. At the end of the year, there was a profit of ₹ 20,000. Calculate the profits shared by Prabhu and Suresh?
Answer:
Here Prabhu and Suresh started a business with ₹ 1,00,000.
Prabhu continued till the end of the year.

Suresh continued only for 3 months.
The ratio of the contributions
= 100000: 100000 = 1 : 1

Ratio of their period of business
= 12 : 3 = 4 : 1
So, their profits sho’uld be divided on the basis of compound ratio
Compound ratio = 1 × 4 : 1 × 1
Profit = ₹ 20,000
Total parts= 4 + 1 = 5
Suresh’s profit = 20000 × \(\frac{1}{5}\)
= ₹ 4000
Prabhus profit = 20000 – 4000
= ₹ 16000

Question 3.
Rani started a beauty parlour with an amount ₹ 75,000. After 4 months, Vani also joined with Rani with an amount ₹ 50,000. After one year, they got a profit of ₹ 52,000. Calculate the profits shared by Rani and Vani.
Answer:
Rani’s investment = ₹ 75,000
Rani’s period in business = 1 year = 12 months
Vani’s investment = ₹ 56,000
Vani’s period in business = 8 months
The ratio of investments of Rani to Vani = 75,000 : 50,000 = 3:2
The ratio of periods of business of Rani to Vani =12:8 = 3:2

The profit should be distributed on the, basis of compound ratio.
Compound ratio = 3 ×. 3 : 2 × 2
= 9 : 4
Profit = ₹ 52,000

Total parts = 9 + 4 = 13
Rani’s profit = 52,000 × \(\frac{9}{13}\)
= ₹ 36000
Vani’s profit = 52,000 – 36,000 = ₹ 16,000

Question 4.
If 3 : 4 and 9 : x are in direct proportion, then what is the value of x?
Answer:
If 3 : 4 and 9 : x are in direct proportion, then their ratio is constant.
∴ \(\frac{3}{4}=\frac{9}{x}\)
⇒ 3 × x = 4 × 9
⇒ x = \(\frac{4 \times 9}{3}\) = 12

Question 5.
If the cost of 4 note books is ₹ 80. What would be the cost of 7 note books?
Answer:
We know that as number of note books increases, the cost also increases such that the ratio of number of note books and the ratio of their costs will remain the same. That means here number of note books and the cost are in direct proportion.

Let the cost of 7 note books be ‘x’.
Then, 4 : 80 = 7 : x
If the ratios are equal, the product of means = The product of the extremes
4 × x = 80 × 7
⇒ x = \(\frac{80 \times 7}{4}\) = ₹ 140
Thus, the cost of 7 note book is equal to ₹ 140.

Question 6.
The scale of a map is given as 1 : 30000. If two cities are 20 cm apart on the map, then o-j, . find the actual distance between them.
Answer:
Let the actual distance be ‘x’ cm. Since the distance on the map is directly proportional to the actual distance.
1 : 30000 = 20 : x
If the ratios are equal, the product of means = The product of the extremes
∴ 1 × x = 30,000 × 20
⇒ x = 6,00,000 cm = 6 km. [1,00,000 cm = 1000 m = 1 km.]
Thus, two cities which are 20cm apart on the map are actually 6 km away from each other.

Question 7.
If 4, 7 and 2, x are inverse proportion, then what is the value of x?
Answer:
4, 7 and 2, x are in inverse proportion . Therefore, 4 × 7 = 2 × x
⇒ x = \(\frac{4 \times 7}{2}\) = 14

Question 8.
If 18 workers can build a wall in 12 days, how many days will eight workers take to build the same wall?

Answer:
If the number of workers decreases, the time taken to build the wall increases in the same proportion.
So, number of workers and the number of days to complete work are in inverse proportion.

Let the number of days to complete the work be ‘x’.

By taking inverse proportion,
18 : 8 = x : 12
Then,
∴ 18 × 12 = 8 × x
⇒ 8 × x = 18 × 12
⇒ x = \(\frac{18 \times 12}{8}\) = 27days
∴ Eight workers can complete the will in 27 days.

Observe this:
In inverse Proportion, product is always constant.
∴ 18 × 12 = 8 × x
⇒ 8 × x = 18 × 12
⇒ x = \(\frac{18 \times 12}{8}\) = 27days

Question 9.
4 Pumps are required to fill a tank in 1 hr 30 min. How long will it take if only 3 pumps of the same type are used?

Answer:
Let the time be ‘x’.
1 hr 30min = 60 + 30 = 90 minutes ,
If the number of pumps are decrease, the time taken to fill the tank increases.
So, the number of pumps and time taken to fill the tank are in inverse proportion.

By taking inverse proportion,
4 : 3 = x : 90
Thus,
∴ 4 × 90 = 3 × x
⇒ 3 × x = 4 × 90
⇒ x = \(\frac{4 \times 90}{3}\) = 120 min
∴ 3 pumps will fill the tank in 120 min or 2 hrs.

Observe this:
In inverse proportion, product is always constant
∴ 4 × 90 = 3 × x
⇒ 3 × x = 4 × 90
⇒ x = \(\frac{4 \times 90}{3}\) = 120 min

Question 10.
If 30 persons use 40 kg. of sugar in 10 days, find in how many days 80 persons will use 320 kg. of sugar? How can we solve?
Answer:
Here, the 3 quantities are persons, weight and number of days.

From the above comparison mustbe in between unknown form to know form.

• Here days and persons are in inverse we denote it with Tmark.
• Here days and sugar are in direct proportion we denote it with 4r mark.

Here number of days is depends on both persons and weight of sugar. So we have to take compound ratio of 80 : 30 and 40 : 320.
∴ 10 : x = 80 × 40 : 30 × 320 = 3200 : 9600
Since the ratios are equal, the product of extremes is equal to product of means.
10 : x = 3200 : 9600 ⇒ 10 × 9600 = x × 3200
⇒ x × 3200 = 10 × 9600 ⇒ x = \(\frac{10 \times 9600}{3200}\) = 30

Question 11.
8 painters can paint a wall of 160m long in 5 days. How many painters are required to paint 240kn. wall in 10 days?
Answer:
Here we have three quantities – number of painters, length of wall and number of days.

Number of painters, is directly proportional to length of the wall.
Number of painters is inversely proportional to the number of days.

Since the number of painters depends both on length of wall and number of days. We will take the compound ratio of 160 : 240 and 10 : 5 1
∴ 8 : x = 160 × 10 : 240 × 5
Since the ratios are equal, product of means is equal to product of extremes
8 : x = 160 × 10 : 240 × 5
⇒ x × 160 × 10 = 8 × 240 × 5
⇒ x = \(\frac{8 \times 240 \times 5}{160 \times 10}\) = 6
Hence the required number of painters = 6

Question 12.
195 men working 10 hours a day can finish a job in 20 days. How many men are employed to finish the job in 15 days if they work 13 hours a day?
Answer:
Here we have three quantities – number of workers, number of hours and number of days.

Here the number of workers inversely proportional to number of hours per day.
Here the number of workers inversely proportional to number of days.
Since number of workers depends on both number of hours and number of days, we will take compound ratio of 13 : 10 and 15 : 20.
Since the ratios are equal, the product of means is equal to product of extremes.
195 : x = 13 × 15 : 20 × 10
⇒ x × 13 × 15 = 195 × 20 × 10
⇒ x = \(\frac{195 \times 20 \times 10}{13 \times 15}\) = 200
Hence the required number of workers = 200

Question 13.
Express the following percentages as fraction, decimal and In ratio.
(i) 45%
Answer:
45% = \(\frac{45}{100}=\frac{9}{20}\)(Fraction)
= 0.45 (decimal form)
= 9:20 (ratio)

(ii) 62%
Answer:
62% = \(\frac{62}{100}=\frac{31}{50}\) = (Fraction)
= 0.62 (decimal form)
= 31:50 (ratio)

Question 14.
Find 24% of 150 and àlso the remaining of that number.
Answer:
24% of 150 = \(\frac{24}{100}\) × 150 = 36
The remaining of that number
= 150 – 36 = 114

Question 15.
Raghu bought pens for 400 and he sold them for ₹ 480 what Is his profit or loss percent?
Answer:
Jyothi solved It this way:
Pen C.P of = ₹ 400, S.P = ₹ 480
S.P > C.P.
So, Raghu gets a profit P = 480 – 400 = ₹ 80
Profit Percentage = \(\frac{80}{400}\) × 100 = 20%

Anwar solved it this way :
Raghu’s Profit = S.P. – C.P = 480 – 400 = ₹ 80
The ratio of profit and cost or the fraction is \(\frac{80}{400}\)
∴ Profit Percentage = \(\frac{80}{400}\) × 100 = 20%

Suresh solved it using proportion :
When C.P is ₹ 400, the profit is ₹ 80 C.P. is ₹ 100, let the profit be ₹ x.
Here C.P and profit are in direct proportion.
x : 80 = 100 : 400 ⇒ \(\frac{x}{80}=\frac{100}{400}\)
⇒ x × 400 = 100 × 80
⇒ x = \(\frac{100 \times 80}{400}\) = 20%
Profit = 20 per 100
Profit Percentage = 20%

Question 16.
Ramana bought a cycle for ₹ 1200 and sold it to his friend Rehman for ₹ 900, then what is Ramanas profit or loss percentage?
Answer:
Here Ramana’s Cost price = ₹ 1200
Selling price = ₹ 900
S.P < C.P, then Ramana got loss
Loss = C.P – S.P = 1200 – 900 = ₹ 300
Loss percentage = \(\frac{300}{1200}\) × 100 = 25%

Question 17.
If John buys a ear for ₹ 1,50,000 and gains 10% on selling it. Then find the selling price.

Answer:
Cost Price = ₹ 1,50,000
Gain % = 10%
Profit = 10% of ₹ 1,50,000
= \(\frac{10}{100}\) × 150000 = ₹ 15000
Selling Price = C.P. + Profit
= ₹ 1,50,000 + ₹ 15,000
= ₹ 1,65,000
This can be solved using proportion:
Gain 10% means
If CP is ₹ 100,the gain is ₹ 10
Thus S.P = 100 + 10 = 110
Now, here C.P = ₹ 1,50,000

Let S.P = x
CP and SP are directly proportional.
\(\frac{110}{100}=\frac{x}{1,50,000}\)
\(\frac{x}{1,50,000}=\frac{110}{100}\)
x = \(\frac{1,50,000 \times 110}{100}\) = ₹ 1,65,000

Question 18.
Kiran sold a refrigerator for ₹ 16800 at a gain of 12%. Then what is the cost price of it?

Answer:
Roopa did the problem uslnj unitary method:
S.P = ₹ 16800 ; Gain% = 12%
If CP is ₹ 100, then Profit is ₹ 12 and thus S.P = ₹ 112
So when S.P is ₹ 112, then C.P is ₹ 100
When S.P is One Rùpee, C.P is \(\frac{110}{112}\) .
Here S.P is ₹ 16800
So, C.P = \(\frac{110}{112}\) × 16800 = ₹ 15000

Sneha solve did the prot1em using proportion as follows:
Gain % = 12% ;S.P = ₹ 16800
If C.P is ₹ 100 then profit is ₹ 12, thus SP = ₹ 12
here S.P = ₹ 16800
C.P be x
C.P and S.P are directly proportional.
∴ \(\frac{x}{16800}=\frac{100}{112}\)
⇒ x = \(\frac{100 \times 16800}{112}\) = ₹ 15000

Question 19.
The cost of an article goes down every year by 10% of its previous value. Find its original cost, if its cost after 2 years is ₹ 32400.
Answer:
Let the cost at the beginning of 1st year be ₹ 100. At the beginning of,2nd year i.e at the end of 1st year it will be decreased by 10% means cost will be ₹ 90.
At the end of 2nd year i.e at the beginning of 3rd year it will be reduced by 10%
i. e. ₹ 90 is reduced by 10%.
90 – 9 = ₹ 81
If the cost of object is ₹ 100 at the beginning, then after 2 years it’s cost will be ₹ 81.
Let the cost of the object ₹ ‘x’ at the beginning, after 2 years it’s cost is ₹ 32,400 .
Thus ratio of original costs = ratio of costs after 2 years
⇒ x : 100 = 32400 : 81
⇒ \(\frac{x}{100}=\frac{32400}{81}\)
⇒ x = \(\frac{32400 \times 100}{81}\) = ₹ 40000

Question 20.
Find discount if,
(i) marked price is ₹ 450, selling price = ₹ 415
Answer:
Discount = marked price – selling price ⇒ 450 – 415 = ₹ 35.

(ii) marked price = ₹ 810, selling price = ₹ 765.
Answer:
Discount = marked price – selling price ⇒ 810 – 765 = ₹ 45.

Question 21.
If discount is ₹ 40, marked price is ₹ 400, then find discount percentage.
Answer:
Discount = ₹ 40, marked price = ₹ 400,
then discount percentage = \(\frac{40}{400}\) × 100 = 10

Question 22.
A shopkeeper marks his goods 20% above the cost price and allows a discount of 10% on them. What percent does he gain?
Answer:
Let the cost price be ₹ 100.
Then the marked price = 100 + 20 = ₹ 120
Discount = 10%, so discount = x 120 = 12% /
SP = Marked price – Discount = ₹ 120 – ₹ 12 = ₹ 108
Gain = \(\frac{8}{100}\) × 100 = 8%
The shopkeeper gains 8% after discount.

Question 23.
Calculate simple interest and the total amount, if
(i) principal = ₹ 5000, time = 2 years, rate = 10%
Answer:
Simple interest
I = \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{5000 \times 2 \times 10}{100}\) = ₹ 1000
Total amount = principal + interest = 5000 + 1000 = ₹ 6000

(ii) principal = ₹ 25000, time = 3 years, rate = 12%
Answer:
Simple interest = \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{25000 \times 3 \times 12}{100}\) = ₹ 9000
Total amount = principal + interest = 25000 + 9000 = ₹ 34000

Question 24.
If Raheem borrowed a sum of ₹ 25000 at a rate 10% per annum, what is the simple interest and total amount he has to pay for 3 years?
Answer:
Rajesh did like this :
Principal = ₹ 25000 ; Time = 3 years ; Rate of interest = 10%
Simple interest = I = \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{25000 \times 3 \times 10}{100}\) = ₹ 7500
Total amount = ₹ 25000 + ₹ 7500 = ₹ 32500

Sangeetha did like this :
For 1 year we have to pay 10%,
for 3 years we have to pay 3 × 10 = 30% as interest.
Interest = \(\frac{30}{100}\) × 25000 = ₹ 7500
Total amount = ₹ 25000 + ₹ 7500 = ₹ 32500

Question 25.
What sum will yield an interest of ₹ 6000 at 9% per annum in 3 years 4 months?
Answer:
S.I = ₹ 6000 ; R = 9%

Question 26.
At what rate per annum will ₹ 70000 yield an interest of ₹ 14000 in 2\(\frac{1}{2}\) years?
Answer:
Principal = ₹ 70000 ; Time = 2\(\frac{1}{2}\)years = \(\frac{5}{2}\) years ; Simple interest = ₹ 14000

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Unit Exercise

Question 1.
If the cost of 7 toys is ₹ 1575, then what would be the cost of 6 such toys?
Answer:
We know that as number of toys decreases, the cost also decreases such that the ratio of number of toys and the ratio of their costs will remain the same. That means here number of toys and the cost are in direct proportion.
Let the cost of 6 toys be x.
Then 7 : 1575 = 6 : x
If the ratios are equal, the product of extremes = Product of means
⇒ 7 × x = 1575 x 6

⇒ x = ₹ 1350
∴ Cost of 6 toys is ₹ 1350.

Question 2.
A boy went to a hotel and wants to buy 5 plates of Idly worth ₹ 24 each. After, going to Hotel he observed that rates of Idly were increased to ₹ 30. Now if the boy wants to buy idlies, how many plates of idlies he can buy with the same amount?
Answer:
If the cost of idly increases number of plates decreases.
So, cost of idly and number of plates are in inverse proportion.
Let number of plates at ₹ 30 for the same amount is x.

By taking inverse proportion, 24 : 30 = x : 5
Then, Product of means = Product of extremes

Number of Plates at ₹ 30 for the same amount is 4.

Question 3.
Raju covers a distance of 28 kilometers in 2 hours. Find the time taken by him to cover a distance of 56 km with the same speed.
Answer:
We know that as number of kilometers (distance) increases the time taken is also increases with the same speed. That means distance and the time taken are in direct proportion.

Let the time taken to cover the 56 km distance is x hours.
Then 28 : 2 = 56 : x

If the ratios are equal, the product of extremes = the product of means

The time taken to cover the 56 km distance is 4 hours.

Question 4.
24 men working at 8 hours per day can do a piece of work in 15 days. In how many days 20 men working 9 hours per day do the same work?
Answer:
Here, we have three quantities number of workers, number of hours and number of days.

Here the number of worker’s inversely proportional to number of hours perday.
Here the number of workers inversely proportional to number of days.

Let the number of days be ‘x’.
Since number of workers depends on both number of hours and number of days, we will take compound ratio of
24 : 20 and 8 : 9 is 24 × 8 : 20 × 9

Since the ratios are equal,
The product of means = The product of extemes

∴ Number of days required to complete 20 men at 9 hours per day is 16.

Question 5.
Out of 15000 voters in a constituency, 60% of the voters voted. Find the number of people not voted in the constituency.
Answer:
Given number of voters in the constituency = 15000 Percentage of voters voted = 60 %
Number of people voted = 60% of 15000

Number of people voted in the constituency = 9000
Number of people not voted = 15000 – 9000
∴ Number of people not voted = 6000

Question 6.
A shopkeeper bought a suitcase for ₹ 950 and sold it for ₹ 1200. Find its profit or loss percentage.
Answer:
Given cost price of suitcase = ₹ 950
Selling price of suitcase = ₹ 1200
S.P > C.P
So, shopkeeper got profit.
Profit = S.P – C.P = 1200 – 950 = ₹ 250

∴ Profit percent = 26\(\frac{6}{19}\)%.

Question 7.
On selling a mobile for ₹ 4500, a shop-keeper losses 10%. For what amount should be sell it to gain of 15%?
Answer:
Method 1 :
Given selling of mobile = ₹ 4500
Loss = 10%

S.P = C.P – Loss

∴ x = 5000

Method 2:
Let the C.P of mobile = 100%
S.P at a loss of 10% = (100 – 10)%
= ₹ 4500
90% of C.P = ₹ 450O
S.P at a gain of 15% = (100 + 15)%.
= 115%
So, if 90% of CP = ₹ 4500
115% of CP = ?
s.p= \(\frac{115}{90}\) × 4500= ₹ 5750
∴ Cost price of mobile = ₹ 5000
If he gain 15%.

Gain = 15% of 5000

S.P = C.P + Gain = 5000 + 750
∴ Selling price of Mobile = ₹ 5750

Question 8.
A carpenter allows 15% discount on his goods. Find the marked. price of a chair which is sold by him for ₹ 680.
Answer:
Given selling price = ₹ 680
Discount percent = 15%
Let marked price = x

Note:
85% of M.P = ₹ 680
100% of M.P = \(\frac{100}{85}\) × 680
= ₹ 800.

Question 9.
What Is the simple Interest accrued on asum of ₹ 75000 at the rate of 11% for 3 years? Find the total amount.
Answer:
Given Principle = ₹ 7500
Rate of Interest = 11%
Time = 3 years

Total amount = Principle + Interest = ₹ 75000 + ₹ 24750
∴ Total amount = ₹ 99,150

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.7 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.7

Question 1.
Calculate the simple interest accrued on a sum oR 12600, at the rate of 9% per annum for 2 years.
Answer:
Given principle = ₹ 12600
Time = 2 years
Rate of Interest = 9%

∴ Simple Interest = ₹ 2268

Question 2.
Calculate the simple interest accrued for 3 years and the total amount on a sum of ₹ 85000, at the rate of 11% per annum.
Answer:
Given Principle = ₹ 85000
Time = 3 years
Rate of Interest = 11%

Simple Interest = ₹ 28,050
Total amount = Principle + Interest
= 85000 + 28050
∴ Total amount = ₹ 1,13,050

Question 3.
In what time will ₹ 45000 amounts to ₹ 63000, If the simple Interest Is calculated 10% per annum?
Answer:
Method 1:
Given principle = ₹ 45000
Rate of Interest = 10%
Amount = ₹ 63000
Let the time = x years

Interest = ₹ 4500 x
Amount = Principle + Interest
= 45000 + 4500x = 63000
= 4500x = 63000 – 45000 = 18000
⇒ x = \(\frac{18000}{4500}\) = 4
∴ Time = 4 years

Method 2:
Given A = ₹ 63000;
P = ₹ 45000; R = 10%; T=?
So I = A – P = 63000 – 45000 = ₹ 18000

But I = \(\frac{\text { PTR }}{100}\)
⇒ 18000 = \(\frac{45000 \times T \times 10}{100}\)
∴ T = \(\frac{18000 \times 100}{45000 \times 10}\) = 4 years

Question 4.
On a certain amount at the rate of 12% per annum for time 3 years the toti1 interest becomes ₹ 18000. What is the principal amount?
Answer:
Given Rate of Interest = 12%
Time.= 3 years
Interest = ₹ 18000
Let Principle = ₹ x

∴ Principle = ₹ 50,000

Question 5.
In what time the simple interest accrued on a sum of?35000 at the rate of 13% per annum becomes?27300?
Answer:
Given Principle = ₹ 35000 ;
Rate of Interest = 13% ;
Time = x years
Interest = ₹ 27300 .

Time = 6 years

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.6 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.6

Question 1.
A shop selling sewing machines offers a 3% discount on purchases. If the marked price is ₹ 6500. then what is the selling price?
Answer:
Method 1:
Given the marked price = ₹ 6500
bis count = 3%
Disount calculated on marked price.
Discount = 3% of ₹ 6500

∴ Discount = ₹ 195

Selling price
= Marked price – Discount = 6500 – 195
∴ Selling price of sewing machine
= ₹ 6305

Method 2 :
M.P = 100% = ₹ 6500
S.P at a discount of 3%
= (100 – 3)% =?
So, if 100% = 6500
97% = ?
S.P = \(\frac{97}{100}\) × 6500 = ₹ 6305

Question 2.
The marked price of a ceiling fan is ₹ 720 during off season, it is sold for ₹ 684. Determine the discount percentage.

Answer:
Given marked price = ₹ 720
Selling price = ₹ 684
Discount = M.P – S.P = 720 – 684
= ₹ 36
Discount percent = \(\frac{\text { Discount }}{\mathrm{MP}}\) × 100

= 5%
∴ Discount percent = 5%

Question 3.
A publisher gives 32% discount on the printed price of books to book sellers. If the printed price is ₹ 275, then what amount does the seller has to pay to publisher?
Sol. Given printed (marked) price = ₹ 275
Discount percent = 32%

Selling price = M.P – Discount = 275 – 88
∴ Selling price of books = ₹ 187

Question 4.
Rohit buys an item at 25% discount on the marked price. If he bought it for ₹ 660, what is the marked price?
Answer:
Method 1 :
Given selling price = ₹ 660
Discount percentage = 25%
Let the marked price = ₹ x

Marked price – Selling price = Discount
x – 660 = \(\frac{x}{4}\)

∴ Marked price = ₹ 880

Method 2 : M.P = 100% (say)
S.P at a discount of 25% means
S.P = (100 – 25) = 75%
∴ If 75% = ₹ 660 then 100% =?
⇒ SP = \(\frac{100}{75}\) × 660 = ₹ 880

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.5 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.5

Question 1.
Rekha bought a wristwatch for ₹ 2250 and sold it for ₹ 1890. Then find her loss or gain percentage?

Answer:
Given cost price of wristwatch = ₹ 2250
Selling price of wrist watch = ₹ 1890
C.P > S.P, then Rekha got loss.
Loss = C.P – S.P = 2250 – 1890 = ₹ 360

We know loss percent = \(\frac{\text { Loss }}{\text { C.P }}\) × 100

∴ Loss percentage 16%.

Question 2.
A shopkeeper buys a toy for ₹ 250 and sells it for ₹ 300. Find his gain or loss percentage?

Answer:
Given cost price of a toy = ₹ 250
Selling price of a toy = ₹ 300
C.P < S.P, then Shopkeeper got profit.
Profit = S.P – C.P
= 300 – 250 = ₹ 50

We know profit percent

∴ Profit percentage = 20%

Question 3.
The cost price of a chair ₹ 480. 1f he sold at a profit of 10%. what would be selling price of it’?
Answer:
Method: 1
Given cost price of a chair = ₹ 480
Profit percent = 10%
Profit = 10% of ₹ 480

Selling price = C.P + Profit
= ₹ 480 + ₹ 48
∴ Selling price = ₹ 528

Method: 2
C.P = 100% = ₹ 480
Selling it 10% gain.
= S.P = 100% + 10% =?
z 100% = ₹ 480
110% =?
S.P = \(\frac{110}{100}\) × 480= ₹ 528
∴ Selling price = ₹ 528

Question 4.
If Sharma purchased a car for ₹ 350000. After two years he sold at a loss of 12%. Find its selling price’?
Answer:
Given cost price of a car = ₹ 350000
Loss = 12 %
Loss 12% of ₹ 350000

Loss = ₹ 42000
Selling price = CP – Loss
= 350000 – 42000
∴ Selling price of a car = ₹ 3,08,000

Question 5.
A shopkeeper buys wooden tables each at ₹ 2800 and expends ₹ 400 on each table for painting. If he sells it at a cost of ₹ 4000. Find his profit percentage?

Answer:
Given cost price of wooden table = ₹ 2800
Amount spent on painting = ₹ 400
Effective cost price of wooden table
= ₹ 2800 + ₹ 400 = ₹ 3200

Selling price wooden table = 4000
C.P < SP, then shopkeeper got profit.
Profit = S.P – C.P
= ₹ 4000 – ₹ 3200 = ₹ 800

We know profit percentage

∴ Profit percentage = 25%

Question 6.
In a garment shop a saree is sold at a cost of ₹ 1800 after a profit of ₹ 600. Find the profit percentage and cost price?
Answer:
Given selling price of a saree = ₹ 1800
Profit = ₹ 600
Cost price of a saree = S.P – Profit
= 1800 – 600
= ₹ 1200.
We know that profit percentage

∴ Profit percentage = 50%

Question 7.
After incurring a loss of ₹ 258, Jean pant was sold at ₹ 1750. Then find the cost price and loss percentage?
Answer:
Given selling price of Jean pant = ₹ 1750
Loss = ₹ 258
Cost price of Jean pant = S.P + Loss
= 1750 + 258
= ₹ 2008
Loss percent = \(\frac{\text { Loss }}{\text { C.P }}\) × 100
⇒ \(\frac{258}{2008}\) × 100
∴ Loss percentage = 12.85%

Question 8.
The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percentage?
Answer:
Method 1:
Let Cost price of 10 articles be ₹ x.
Cost price of 10 articles = 10 x
Selling price of 9 articles = 10 x
Cost price of 9 articles 9x
Profit = S.P — C.P = 10x – 9x
= ₹ x
Profit percentage = \(\frac{\text { Profit }}{\text { C.P }}\) × 100

∴ Profit percentage = \(\frac{100}{9}\)% (or)
11.11% (or) 11\(\frac{1}{9}\)%

Method 2:
By selling 9 articles, there is a gain of 1 article.
∴ Profit Percentage = \(\frac{1}{9}\) × 100
= 11\(\frac{1}{9}\)%

Question 9.
By selling a book for ₹ 258, a bookseller gains 20%. For how much should he sell it to gain 30%?
Answer:
Method 1:
Given selling priceof a book = ₹ 258
Gain = 20%
Cost price of book = ₹ x

∴ Cost price of a book = ₹ 215

If profit percent = 30%

Profit = ₹ 64.50
∴ S.P = C.P + Profit
= 215 + 64.50 = 279.50
To get 30% profit he has to sold at ₹ 279.50.

Method 2:
Let the C.P of the book = 100%
Selling at a gain of 20% means (100 + 20)% = ₹ 258
120% = ₹ 258
Selling at a gain of 30% means (100+30)%=?
If 120% = ₹ 258
then 130% =?
∴ SP = \(\frac{130}{120}\) × 258 = ₹ 279.5

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.3

Question 1.
Find out whether the given quantities vary directly or inversely.
(i) Time taken to cover a distance, speed.
Answer:
Inversely proportional.

(ii) Area of land and its cost.
Answer:
Direct proportion.

(iii) Number of men for work, time taken to complete the work.
Answer:
Inverselv proportional.

(iv) Number of people, quantity of food grains each one gets (total quantity remains same).
Answer:
Inversely proportional.

(v) The length of a journey by bus and price of the ticket.
Answer:
Direct proportion.

Question 2.
If 24 men can construct a wall in. 10 days. In how many days will 15 men do it?
Answer:
If the number of men decreases working days will increase in the same proportion.
So, number of men and the working days are in inverse proportion.

Let the number of days to complete the work be ’x.

By taking inverse proportion,
24 : 15 = x : 10

Then, product of means = Product of the extremes
⇒ 15 × x = 24 × 10

⇒ x = 16
∴ 15 men will complete the wall in 16 days.

Question 3.
In a hostel there are food provisions for 50 girls for 40 days. If 30 more girls join the hostel, how long will the provisions last?
Answer:
If the number of girls increases, then number of days will decrease in the same proportion.
So, number of girls and the days for food provisions lost are in inverse proportion.
Let the number of food provisions days be ’x’.

By taking inverse proportion,
50 : 80 = x : 40
Then, product of means = Product of extremes
⇒ 80 × x = 50 × 40

⇒ x = 25
∴ Number of days required to com-plete the food provisions for 80 girls is 25 days.

Question 4.
Suman travels a distance for 5 hours with a speed of 48 kilometres per hour. If he wants to travel the same in 4 hours at what speed he should travel?
Answer:
If the time decreases, then the speed will increase in the same proportion. Let the speed to cover the distance in 4 hours is ’x’.

By taking inverse proportion,
5 : 4 = x : 48
Then, Product of means = Product of extremes
⇒ 4 × x = 48 × 5

⇒ x = 60 kmph.
∴ Distance covered in 4 hours with the speed of 60 kmph.

Question 5.
A person has money to buy 8 bicycles of worth ₹ 4500 each. If the cost of the bicycle is decreased by ₹ 500, then how many bicycles cap he buy with the amount he has?

Answer:
If the cost of the bicyble decreases, then the number of cycles will increase in the same proportion.
Let the number of bicycles he can buy each with ₹ 4000 (4500 – 500) be ‘x’.

By taking inverse proportion,
4500 : 4000 = x : 8
Then, Product of means = Product of extremes
⇒ 4000 × x = 4500 × 8
⇒ \(\frac{4000 x}{4000}=\frac{4500 \times 8}{4000}\) = 9
⇒ x = 9
∴ Number of bicycles bought with the same amount is 9.

Question 6.
2 pumps are required to fill a tank in 1 hour. How many pumps of the same type are used to fill the tank in 24 minutes?
Answer:
If the filling time decreases, then the number of pumps will increase in the same proportion.
Let the number of pumps to fill the tank be x.

By taking inverse proportion,
60 : 24 = x : 2

Then, Product of means = Product of extremes
⇒ 24 × x = 60 × 2

⇒ x = 5
∴ Number of pumps required to fill the tank in 24 minutes is 5.

Question 7.
18 men can reap a field in 10 days. For reaping the same field in 6 days, how many more men are required?

Answer:
If the reaping days decreases, then the number of men increases in the same proportion.
Let the number of men required to reap the field in 6 days is x.

By taking inverse proportion, 18 : x = 6 : 10
Then, Product of means = Product of extremes

∴ Number of men required to reap the field in 6 days = 30
Number of more men = 30 – 18 = 12
Therefore 12 more men required to reap the field in 6 days.

Question 8.
1200 soldiers in a checkpost had enough food for 28 days. After 4 days some soldiers were transferred to another checkpost and thus remaining food is sufficient for 32 more days. How many soldiers left the checkpost?

Answer:
If the number of soldiers decreases, then the number of days food lost will increases in the same proportion.
Let the number of soldiers after transfer is x.

After 4 days,

By taking inverse proportion, 1200 : x = 32 : 24
Then, Product of means = Product of extremes

∴ Number of Soldiers remained in the checkpost = 900
Number of Soldiers transferred = 1200 – 900 = 300
So, 300 soldiers left the checkpost.