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CHAPTER -3 Fibre to Fabric | CLASS 7TH |NCERT SCIENCE IMPORTANT QUESTIONS & MCQS | EDUGROWN

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Get Chapter Wise MCQ Questions for Class 7 Science with Answers prepared here according to the latest CBSE syllabus and NCERT curriculum. Students can practice CBSE Class 7 Science MCQs Multiple Choice Questions with Answers to score good marks in the examination. Students can also visit the most accurate and elaborate NCERT Solutions for Class 7Science. Every question of the textbook has been answered here.

Chapter - 3 Fibre to Fabric

MCQs

Question 1.
Which one of the following is not a breed of sheep?
(a) Murrah
(b) Marwari
(c) Lohi
(d) Nali

Answer

Answer: (a) Murrah

Question 2.
What is the scientific name of mulberry tree?
(a) Magnifera indica
(b) Morus alba
(c) Desmodium girence
(d) None of these

Answer

Answer: (b) Morus alba

Question 3.
Silk is derived from
(a) cocoon
(b) pupa
(c) egg
(d) moth

Answer

Answer: (a) cocoon

Question 4.
Selective breeding is a process of
(a) selecting the offspring with desired properties.
(b) selecting the parents with desired properties.
(c) selecting an area for breeding.
(d) selecting fine hair for good quality wool.

Answer

Answer: (b) selecting the parents with desired properties.

Question 5.
The general process that takes place at a sheep shearing shed is:
(a) removal of fleece.
(b) separating hair of different textures.
(c) washing of sheep fibre to remove grease.
(d) rolling of sheep fibre into yam.

Answer

Answer: (a) removal of fleece.

Question 6.
The rearing of silkworms for obtaining silk is called:
(a) cocoon
(b) silk
(c) sericulture
(d) silviculture

Answer

Question 7.
Which of the following is not a type of silk?
(a) Mulberry silk
(b) Tassar silk
(c) Mooga silk
(d) Moth silk

Answer

Answer: (d) Moth silk

Question 8.
Paheli wanted to buy a gift made of animal fibre obtained without killing the animal. Which of the following would be the right gift for her to buy?
(a) Woollen shawl
(b) Silk scarf
(c) Animal fur cap
(d) Leather jacket

Answer

Answer: (a) Woollen shawl

Question 9.
Silk fibre is obtained from:
(a) fleece of sheep
(b) cotton ball
(c) cocoon
(d) shiny jute stalk

Answer

Answer: (c) cocoon

Question 10.
Wool fibre cannot be obtained from which of the following?
(a) Goat
(b) Llama
(c) Alpaca
(d) Moth

Answer

Answer: (d) Moth

Match the following:

Column A	Column B
(i) Selective breeding	(a) Separating different textures
(ii) Shearing	(b) Washing of hair removed from sheep
(iii) Scouring	(c) Removing fleece from sheep
(iv) Reeling	(d) Process of taking out threads from cocoon
(v) Sorting	(e) Selecting parents for obtaining special characters in their offspring

Answer

Answer:

Column A	Column B
(i) Selective breeding	(e) Selecting parents for obtaining special characters in their offspring
(ii) Shearing	(c) Removing fleece from sheep
(iii) Scouring	(b) Washing of hair removed from sheep
(iv) Reeling	(d) Process of taking out threads from cocoon
(v) Sorting	(a) Separating different textures

Fill in the blanks:

1. A pile of ……………… is used for obtaining silk fibres.

Answer

Answer: cocoon

2. In India mostly ……………… are reared for getting wool.

Answer

Answer: sheep

3. Silk fibres are made of a ………………

Answer

Answer: protein

4. Silkworms are the ……………… of silk moth.

Answer

Answer: caterpillar

5. The hair of sheep are termed as ………………

Answer

Answer: fleece

6. Silk and ……………… are animal fibres.

Answer

Answer: wool

7. ……………… is a very old occupation in India.

Answer

Answer: Sericulture

8. ……………… and ……………… fibres are obtained from animals.

Answer

Answer: Silk, wool

9. Wool yielding animals bear ……………… on their body.

Answer

Answer: hair

10. Hair trap a lot of ………………, which is a poor ……………… of heat.

Answer

Answer: air, conductor

Choose the true and false statements from the following:

1. Wool and silk are the animal fibres.

Answer

Answer: True

2. Silk is obtained from the fleece of sheep.

Answer

Answer: False

3. Pashmina shawls woven from the wool obtained from the under fur of Kashmiri goats.

Answer

Answer: True

4. Silk moths spin the silk fibres.

Answer

Answer: False

5. Silk fibre is chemically a protein.

Answer

Answer: True

6. The most common silk moth is the mulberry silk moth.

Answer

Answer: True

7. For reeling of silk, cocoon are boiled.

Answer

Answer: True

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

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

Question 3.
in which part of India, yak wool is common?
Answer:
Yak wool is common in the hilly regions of Ladakh.

Question 4.
Rampur bushair and Bakharwal are the breeds of which animal-sheep or goat?
Answer:
Rampur bushair and Bakharwal are the breeds of sheep.

Question 5.
How do the hair of certain animals help in keeping their bodies warm? [NCERT Exemplar]
Answer:
Hair traps a lot of air which is a poor conductor of heat resulting their body warm.

Question 6.
Write the name of the animal from which wool for pashmina shawls is obtained.
Answer:
Kashmiri goats.

Question 7.
Name the goats which are found in Tibet and Ladakh.
Answer:
In Tibet and Ladakh, angora goats are found.

Question 8.
wool is obtained from which body part of sheep?
Answer:
Wool is obtained from the fleece or hair of sheep.

Question 9.
Name the natural fibre obtained from insects.
Answer:
Silk.

Question 10.
In which part, the wool yielding animals liama and alpaca are found?
Answer:
Liama and alpaca are found in South America.

Question 11.
For what purpose, rearing of sheep is done?
Answer:
The main aim to rear the sheep is to obtain wool.

Question 12.
What term is used for the following process? Washing of sheep’s fleece to remove dust, dirt, dried sweat and grease.
Answer:
Scouring.

Question 13.
Out of scouring or shearing, which process comes first in the production of wool from sheep?
Answer:
The first step is shearing then scouring.

Question 14.
To which class, animal fibre or plant fibre, artificial silk belongs to?
Answer:
Artificial silk is a plant fibre.

Question 15.
Two fibres are made up of proteins. Name them.
Answer:
Silk and wool are two fibres which are made up of proteins.

Question 16.
Name the process of taking out silk fibres from the cocoon for use as silk.
Answer:
The process of taking out threads from the cocoons for use as silk is called reeling the silk.

Question 17.
Name the stage next to caterpillar in the life cycle of silk moth.
Answer:
The stage next to caterpillar in the life cycle of silk moth is called pupa.

Question 18.
which stage in the life cycle of silk moth is called cocoon?
Answer:
The covering of silk fibres inside which the caterpillar covers itself is called cocoon.

Question 19.
Write the maximum length of continuous silk thread that can be obtained from a cocoon.
Answer:
The maximum length of continuous silk thread that can be obtained from a cocoon is more than that of 1000 feet (1000 to 1500 feet) continuous in length.

Question 20.
Define the term ‘selective breeding’.
Answer:
The process of breeding of selective parents for obtaining special characters in their offspring such as soft under hair in sheep, is termed as selective breeding.

Question 21.
Silk fibre belongs to which class of organic substance?
Answer:
Protein.

Question 22.
In the life history of silk moth as
Egg ➝ Larva (or caterpillar) ➝ Pupa ➝ Silk moth which actually makes the silk fibre? [HOTS]
Answer:
The larva (or caterpillar) of a silk moth makes the silk fibre.

Fibre to Fabric Class 7 Science Extra Questions Short Answer Type

Question 1.
Sheep are herbivores, so which type of feed (or food) is provided by shepherds? [HOTS]
Answer:
Sheep are herbivores, and prefer grass and leaves of various trees. Their food also contain a mixture of pulses, corn, jowar, oil cakes (material left after taking out oil from seeds) and minerals.

Question 2.
Write a caption for each of the figures given as figure (a-d).

Answer:
(a) Eggs of silk moth on mulberry leaves
(b) Silkworm
(c) Cocoon
(d) Cocoon with developing moth

Question 3.
Silk yarn of different textures can be prepared. Define the statement.
Answer:
The silk yarn is obtained from the cocoon of the silk moth. There is a variety of silk moths which look very different from one another and the silk yarn they yield is different in textures (coarse, smooth, shiny, etc.) Thus, tassar silk, mooga silk, kosa silk, etc are obtained from cocoons spun by different types of moths.

Question 4.
Write the difference between natural silk and artificial silk.
Answer:
Natural silk is obtained from the cocoons of silkworms and it is made up of a protein. Natural silk is an animal fibre. Artificial silk is obtained from wood pulp and it is made up of modified plant material ‘cellulose’.
If we perform the burning test, then natural silk fibre burns giving a smell of burning hair while artificial fibre burns giving a smell of burning paper.

Question 5.
Sorter’s disease is an occupational hazard. Explain.
Answer:
Wool industry is an important source of livelihood for many people in our country. The people who do the job of sorting (separating) the fleece of sheep into fibres of different qualities are called sorters. The sorter’s job is very risky because sometimes, they get infected by the bacteria called ‘anthrax’ which cause a deadly blood disease called sorter’s disease.

Question 6.
Write a short note on how first silk industry began in China.
Answer:
According to an old Chinese legend, the empress Si-lung-Chi was asked by the emperor Huang-ti to find the cause of the damaged leaves of mulberry trees growing in their garden. The empress found white worms eating up mulberry leaves. She also noticed that they were spinning shiny cocoons around them.
Accidentally, a cocoon dropped into her cup of tea and a tangle of delicate threads separated from the cocoon. This is how silk was discovered by chance. In this way, the first silk industry began in China.

Question 7.
Various steps involved to obtain wool from fleece are given here.

Picking out the burrs
Dyeing in various colours
Shearing
Scouring
Sorting

Write the above steps in the correct sequence in which they are carried out. [NCERT Exemplar]
Answer:

Question 8.
Steps for the production of silk are given below in a jumbled order. Arrange them in their proper sequence. [NCERT Exemplar]
(a) Eggs are warmed to a suitable temperature for the larvae to hatch from eggs.
(b) Fibres are taken out from the cocoon.
(c) After 25 to 30 days, the caterpillars stop eating and start spinning cocoons.
(d) The larvae/caterpillars or silkworms are kept in clean trays along with freshly chopped mulberry leaves.
(e) Female silk moths lay eggs.
(f) Cocoons are kept under the sun or boiled in water.
Answer:
The correct order is
(e) Female silk moths lay eggs.
(a) Eggs are warmed to a suitable temperature for the larvae to hatch from eggs.
(d) The larvae/caterpillars or silkworms are kept in clean trays along with freshly chopped mulberry leaves.
(c) After 25 to 30 days, the caterpillers stop N eating and start spinning cocoons.
(f) Cocoons are kept under the sun or boiled in water.
(g) Fibres are taken out from the cocoon.

Question 9.
Cocoon is used to obtain silk thread. How?
Answer:
A pile of cocoon is used for obtaining silk fibre. The cocoon are kept under the sun, boiled water or exposed to steam.
The silk fibres separate out. The process of taking out threads from the cocoon for use as silk is called reeling the silk.

Question 10.
Name different types of silk. Which variety of silk is most common and how it is obtained?
Answer:
Different types of silk are mulberry silk, tassar silk, mooga silk and kosa silk. The most common variety of silk is mulberry silk. It is obtained from the cocoons of mulberry silkworm and is made up of protein. It is a natural silk and is an animal fibre.

Question 11.
Four different types of fibres are given to us. Out of these, two fibres (1 and 2) are obtained from plants and other two (3 and 4) fibres are obtained from animals. Fibre (1) is used in filling quilts and the yarn made of fibre (2) is used in making gunny bags, The yarn made from fibre (3) is used for knitting sweaters and yarn of fibre (4) is used for weaving saries. Name the four given fibres marked 1,2, 3 and 4. [HOTS]
Answer:
Fibre (marked 1) which is used in filling quilts is cotton and fibre (marked 2) is jute. (cotton and jute are plant fibres).
Fibre (marked 3) is wool and fibre (marked 4) which is used for weaving saris is silk. (wool and silk are animal fibres).

Question 12.
Why caterpillars need to shed their skin when they grow bigger but humans do not? Do you have any idea? [HOTS]
Answer:
The caterpillar eats day and night the leaves of mulberry tree and grows big in size whereas its skin does not increase in size, shape or length. During the feeding period, a silkworm sleeps four times (24 h each time) at intervals of six days.

While sleeping, its skin cracks and on awakening, the worm leaves the old skin and comes out in a new one. So, it sheds skin and this phenomenon is called moulting. After the final moulting begins, the last feeding period (of about ten days) after which the worm grows to its full size.

Question 13.
Radhika wanted to buy a silk frock and went to the market with her mother. There they found that the artificial (synthetic) silk was much cheaper and wanted to know why? Do you know why? Find out. [HOTS]
Answer:
Artificial (synthetic) silk called rayon is obtained from wood pulp and it is made of modified plant material cellulose. Synthetic silk can be prepared at a large scale in factories/mills. So, it is cheap. For obtaining pure silk, we have to rear silk moth, their larvae to get pupa.

To get silk thread, we have to dip cocoon in hot water to get silk thread which is wrapped over the cocoon. The pupa inside the cocoon dies. To obtain silk for commercial purposes, a large number of cocoons (containing living pupa) are killed. As a result, natural silk is costly.

Question 14.
From what type of health problems the workers suffered while working in a wool industry? [HOTS]
Answer:
People working in the wool industry sometimes get infected by a bacterium called anthrax, during the sorting (separating) the fleece of sheep into fibres of different qualities.
It leads to a fatal blood disease called sorter’s disease.

Question 15.
Write the different types of fibres that form the hair of sheep.
Answer:
There are following two types of fibres that form the hair to sheep

The coarse beard hair.
Fine soft under hair close to the skin.

Question 16.
Name the most common silk moth. What are the characteristics of silk fibres obtained from the cocoons of this silk moth?
Answer:
The most common silk moth is the mulberry silk moth. The silk obtained from the cocoons of mulberry silk moth is called mulberry silk. Mulberry silk is soft, lustrous (shiny) and elastic and can be dyed in different colours.

Fibre to Fabric Class 7 Science Extra Questions Long Answer Type

Question 1.
Paheli went to the market to buy sarees for her mother. She took out a thread from the edge of the two sarees shown by the shopkeeper and burnt them. One thread burnt with a smell of burning hair and the other burnt with the smell of burning paper. Which thread is from a pure cotton saree and which one from a pure silk saree? Give reason for your answer. [NCERT Exemplar; HOTS]
Answer:
In first saree, one thread which burnt with a smell of burning hair is from pure silk, silk and hair are protein fibres. So, on burning these threads, a smell of burning hair comes out. In second saree, second thread which burnt with the smell of burning cotton and paper because cotton and paper both are carbohydrates and on burning they give similar smell.

Question 2.
Describe the life history of silk moth with the help of figures of various stages.
Answer:
Life History of Silk Moth Formation of Silkworm
The female silk moth lays eggs on mulberry leaves. The eggs are hatched into very small larvae within a week. The larvae of silk moth are called caterpillar or silkworm. The silkworms feed on the leaves of mulberry tree and grow bigger in size.

Development of Cocoon

When the silkworm (or caterpillar) is ready to enter the next stage of its development called pupa, it first weaves a net to hold itself. Then, it swings its head from side to side. During these movement of head, the silkworm secrets fibre made of protein which hardens on exposure to air and becomes silk fibre (or silk thread). Soon the silkworm (or caterpillar) covers itself by silk fibres and turns into pupa. This covering is known as cocoon. The silkworm continues to develop in the form of pupa inside the cocoon to form the silk moth.

Production of Silk

In order to produce silk, the silkworm developing inside the cocoon (as pupa) is not allowed to mature into an adult silk moth. So, as soon as the cocoon is formed, it is used to obtain silk fibres and the developing silkworm (as pupa) gets killed. Some of the silkworms (as pupae) are however, allowed to live and mature into silk moths so that they can lay eggs to produce more silkworms.

There is a variety of silk moths which look very different from one another and the silk yarn they yield is different in texture (coarse, smooth, shiny, etc). Thus, tassar silk, kosa silk, mooga silk, etc are obtained from cocoons spun by different types of moths. The most common silk moth is the mulberry silk moth. The silk obtained from the cocoons of mulberry silk moth is called mulberry silk. Mulberry silk is soft, lustrous (shiny) and elastic and can be dyed in beautiful colours.

Take a thread of pure silk and another thread of an artificial (synthetic) silk. Burn them separately and observe the smell produced. The thread which burns giving a smell of burning hair will be pure silk thread. The thread which burns giving a smell of burning paper will be artificial silk thread.

Question 3.
How silkworms are reared? Explain in brief.
Answer:
Pure and Artificial Silk
Pure silk is obtained from the cocoons of silkworm and it is made up of protein. Artificial silk is obtained from wood pulp and it is made of modified plant material ‘cellulose’. Just like silk, wool is also made up of proteins. So, a piece of woollen fabric also burns giving the smell of burning hair. The thread which burns giving a smell of burning paper will be cotton fibres. Cotton and paper both are carbohydrates. Paper is made of cellulose obtained from wood pulp. So, on burning cotton and paper both give similar smell.

From Cocoon to Silk
For obtaining silk, silk moths are reared and their cocoons are collected to get silk thread.

Rearing Silkworms
A female silk moth lays hundred of eggs at a time. The eggs are stored carefully on strips of paper or cloth and sold to silkworm farmers. The farmers keep eggs under hygienic condition. They warm them to a suitable temperature for the larvae to hatch from egg.

The larvae are kept in clean bamboo trays along with young and freshly chopped mulberry leaves. After 25-30 days, the silkworms stop eating and start spinning the cocoons. Small racks or twigs may be provided in the trays to which cocoons get attached.

Processing Silk
The cocoons are collected and boiled in water to kill the insect inside them. The resulting fibre is known as raw silk. The silk fibres separate out.

Reeling the Silk
The process of taking out fibres from the cocoon for use as silk is known as reeling the silk. Reeling is done in special machines. Silk fibres are spun into silk threads which are woven into silk cloth by weavers.

Question 4.
Explain the phrase – ‘Unity is Strength’ on the basis of the making of fabric from fibre. [NCERT Exemplar; HOTS]
Answer:
Fibres and fabric play a large role in everyday applications. A fibre is a hair-like strand of material. They are the smallest visible unit of a fabric and denoted by being extremely long in relation to their width. Fibres can be spun into yarn and made into fabric. A single fibre is too weak to break but when it once made a fabric it is difficult to tear. Fabric needs more energy to tear apart as compared to a single fibre.

Question 5.
Write various steps for processing fibres into wool. [NCERT Exemplar]
Answer:
Processing of Fibres into Wool
The wool which is used for knitting sweaters or for weaving shawls is the finished product of a long process. Processing of fibres into wool involves the following steps:

Step I
The fleece of the sheep along with thin layer of skin is removed from the body. This process is called shearing. The hair of the sheep are shaved off by using a saving machine similar to that used by barbers.
Shearing does not hurt the sheep because the uppermost layer of the skin of sheep is ‘dead’. The shearing (cutting the hair) of sheep is done in hot weather of summer so that sheep may survive without their protective coat of hair. The hair of sheep grow again before the onset of winter and protect them in cold weather. The fleece (or hair) of sheep provides woollen fibres. Woollen fibres are then processed to obtained woollen yarn.

Step II
The fleece of sheep (or cut hair of sheep) contains dust, dirt, dried sweat and grease, etc. So, the sheared hair of sheep are thoroughly cleaned by washing with soap (or detergent) and a lot of water in tanks. This process of washing of sheared hair is called scouring. Scouring makes the fleece of sheep clean. The scoured fleece is then dried. Now-a-days scouring is done by machines.

Step III
After scouring, sorting is done. The process of separating the fleece of a sheep into sections according to the quality of woollen fibres (such as fine, coarse, long, short, etc) is called sorting. In sorting, the hairy skin is sent to a factory where hair of different textures are separated or sorted. Every section of wool obtained after sorting contains the same quality wool. The same quality wool obtained are then mixed together.

Step IV
The small fluffy fibres, called burrs, are picked out from the hair (burrs are soft, fluffy fibres in wool).
(After this, the fibres are scoured again and dried. The wool obtained after this is ready to be drawn into fibres).

Step V
The natural fleece or hair of sheep (or goat) is white, brown or black in colour. The white woollen fibre obtained by sorting can be dyed in different colours.

Step VI
The fibres are straightened, combed and rolled into yarn. The long woollen fibres are spun (or twisted) into thick yarn called wool which is used for knitting sweaters, etc.
The short woollen fibres are spun into fine yarn and then woven on a loom to make woollen clothes (like shawls, etc).
Finally, we conclude that the sheep’s hair is sheared off from the body, scoured, sorted, dyed, combed and spun to obtain wool (for knitting sweaters) and woollen yarn (for weaving cloth). The quality of woollen cloth depends on the breed of sheep from which wool is obtained.

Question 6.
Neha went to the market with her mother to buy a silk saree for her grandmother. The shopkeeper was showing sarees of different varieties of silk but her mother wanted to take only a pure silk saree and not an artificial one. Neha was confused that how to distinguish between a pure silk and an artificial silk saree. But her mother helped her in selecting a pure silk saree for her grandmother and she was quite elated about that.
Now, answer the following questions:
(a) How her mother distinguished between pure silk saree and an artificial silk saree?
(b) Is the pure silk saree more costly than an artificial one?
(c) What values are shown by Neha’s mother? [Value Based Question]
Answer:
(a) Neha’s mother took a thread of natural silk fabric and another thread of artificial silk fabric and burn them separately. The thread which burns giving a smell of burning hair will be natural silk (or pure silk). The thread which burns giving a smell a burning paper will be an artificial silk.
(b) Yes, pure silk saree is more costly than an artificial one.
(c) Neha’s mother is intelligent and knowledgeable.

Question 7.
Ajay’s father works in a wool industry. In his line of work, he does the job of sorting the fleece of sheep into fibres of different qualities. Ajay noticed that his father and other workers who were doing the similar kind of jobs faced a number of health problems.
Ajay discussed this problem with his friends and he came to know that his father is getting health problems due to the nature of his work.
Ajay is worried about his father’s health. He advised his father to take care of their health and told him about the occupational hazards.
Now answer the following questions.
(a) What do you mean by occupational hazard?
(b) Name the disease that workers may at risk.
(c) What values are shown by Ajay? [Value Based Question]
Answer:
(a) The risks faced by people working in any industry due to the nature of their work are called occupational hazards.
(b) The people who do the job of sorting, sometimes, get infected by the bacteria called anthrax which cause a deadly blood disease called sorter’s disease.
(c) Ajay is very caring, intelligent and sensible boy.

Question 8.
Name some breeds of sheep reared in our country. Explain the quality of wool they provide and the state where they are found.
Answer:

Name of breed of sheep	Quality of wool	Name of the state where found
Lohi	Good quality wool	Rajasthan, Punjab
Rampur bushair	Brown fleece	Uttar Pradesh, Himachal Pradesh
Nali	Carpet wool	Rajasthan, Haryana, Punjab
Bakharwal	For woollen shawls	Jammu and Kashmir
Marwari	Coarse wool	Gujarat
Patanwadi	For hosiery	Gujarat

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RD SHARMA SOLUTION CHAPTER –17 Understanding Shapes-III (Special Types of Quadrilaterals)| CLASS 8TH MATHEMATICS-EDUGROWN

Exercise 17.1

Question 1.
Given below is a parallelogram ABCD. Complete each statement along with the definition or property used:
(i) AD = ………
(ii) ∠DCB = ……….
(iii) OC’ = …….
(iv) ∠DAB + ∠CDA = …….
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 1
Solution:
(i) AD = BC
(ii) ∠DCB = ∠ADC

(iii) OC = OA
(iv) ∠DAB + ∠CDA = 180°

Question 2.
The following figures are parallelograms. Find the degree values of the unknowns x,y, z.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 3

Solution:
In a parallelogram, opposite angles are equal and sum of adjacent angle is 180°.
(i) In parallelogram ABCD,
∠B = 100°
∠A = ∠C = 180° (Co-interior angles)

⇒ x + 100° = 180°
⇒ x = 180° – 100°
⇒ x = 80°
But ∠A = ∠C and ∠B = ∠D (Opposite angles)
A = x
⇒ z = x
⇒ z = 80°
and ∠D = ∠B
⇒ y = 100°
x = 80°, y = 100° and z = 80°
(ii) In parallelogram PQRS, side PQ is produced to T.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 6
∠S = 50°
∠PQR = ∠S (Opposite angles)
w = 50°
But ∠P + ∠PQR = 180° (Sum of adjacent angles)
⇒ x + 50° = 180°
⇒ x = 180° – 50° = 130°
⇒ But ∠P = ∠R (Opposite angles)
x = y
⇒ y = 130°
But w + z = 180° (A linear pair)
⇒ 50° + z = 180°
⇒ z = 180° – 50° = 130°
⇒ x = 130°, y = 130°, ∠ = 130
(iii) In parallelogram LMNP, PM is its diagonal ∠NPM = 30°, ∠PMN = 90°
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 7
PN || LM (Opposite sides of a parallelogram) and PM is its transversal
∠NPM = ∠PML
⇒ 30° = x
x = 30°
In ∆PMN,
∠P = 30°, ∠M = 90°
But ∠P + ∠M + ∠N = 180° (Sum of angles of a triangle)
⇒ 30° + 90° + z = 180°
⇒ 120° + z = 180°
⇒ z = 180° – 120° = 60°
But ∠L = ∠N (Opposite angles of a parallelogram)
y = z
⇒ y = 60°
Hence x = 30°, y = 60° and ∠ = 60°
(iv) In rhombus ABCD, diagonals AC and BD bisect each other at right angles.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 8
x = 90°
In ∆OCD,
∠O + ∠C + ∠D = 180°
⇒ 90° + 30° + y = 180°
⇒ 120° + y = 180°
⇒ y = 180° – 120° = 60°
y = 60°
CD || AB, BD is the transversal
y = z (Alternate angles)
z = 60°
Hence x = 90°, y = 60° and z = 60°
(v) In parallelogram PQRS, side QR is produced to T
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 9
∠Q = 80°
∠P + ∠Q = 180° (Sum of adjacent angles)
⇒ x + 80° = 180°
⇒ x = 180° – 80° = 100°
∠Q = ∠S (Opposite angles of a parallelogram)
⇒ 80° = y
⇒ y = 80°
PQ || SR and QRT is transversal
∠TRS = ∠RQP (Corresponding angles)
⇒ ∠ = 80°
Hence x = 100°, y = 80° and z = 80°
(vi) In parallelogram TUVW, UW is its diagonal
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 10
∠TUW = 40° and ∠V = 112°
∠T = ∠V (Opposite angles)
y = 112°
In ∆TUW,
∠T + ∠V + ∠W = 180° (Sum of angles of a parallelogram)
⇒ y + 40° + x = 180°
⇒ 112° + 40° + x = 180°
⇒ 152° + x = 180°
⇒ x = 180° – 152° = 28°
UV || TW and UW is its transversal
∠WUV = ∠TWU (Alternate angles)
⇒ z = x
⇒ z = 28°
Hence x = 28°, y = 112°, z = 28°

Question 3.
Can the following figures be parallelograms. Justify your answer.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 11
Solution:
(i) In quadrilateral PLEH
∠H = 100°, ∠L = 80°
But there are opposite angles
∠H ≠ ∠L
PLEH is not a parallelogram.
(ii) In quadrilateral GNIR,
RI = 8 cm, GN = 8 cm, RG = 5 cm and IN = 5 cm
PI = GH and RG = IN
But there are opposite sides of the quadrilateral.
GNIR is a parallelogram.
(iii) In quadrilateral BEST,
BS and ET are its diagonals
But these diagonal do not bisect each other.
BEST is not a parallelogram

Question 4.
In the adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the geometrical truths you use to find them.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 12
Solution:
In the figure, HOPE is a parallelogram in which HG is produced and HP is the diagonal

∠EHP = 40° and ∠POQ = 70°
But ∠POQ + ∠POH = 180° (Linear pair)
⇒ 70° + w = 180°
⇒ w = 180° – 70° = 110°
But ∠E = ∠POE (Opposite angles of a parallelogram)
x = 110°
HE || OP and HP is its transversal.
∠EHP = ∠HPO (Alternate angles)
⇒40° = y
⇒ y = 40°
In ∆PHO,
Ext. ∠POQ = ∠PHO + ∠HPO
⇒ 70° = z + y
⇒ 70° = z + 40°
⇒ z = 70° – 40° = 30°
Hence x = 110°, y = 40°, z = 30°

Question 5.
In the following figures GUNS and RUNS are parallelograms. Find x and y.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 14
Solution:
(i) In parallelogram GUNS,
Opposite sides are parallel and equal
3x = 18
⇒ x = 6
and 3y – 1 = 26
⇒ 3y = 26 + 1 = 27
⇒ y = 9
x = 6, y = 9
(ii) Diagonals of a parallelogram bisect each other.
y – 7 = 20
⇒ y = 20 + 7 = 27
and x – 27 = 16
⇒ x – 27 = 16
⇒ x = 16 + 27 = 43
x = 43, y = 27

Question 6.
In the following figure RISK and CLUE are parallelograms. Find the measure of x.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 15
Solution:
In the figure, RISK and CLUE are parallelograms
∠K = 120° and ∠L = 70°
In parallelogram RISK

RK || IS
∠RKS = ∠ISU (Corresponding angles)
⇒ ∠ISU = 120°
In parallelogram CLUE,
∠E = ∠L (Opposite angles of a parallelogram)
∠E = 70° (∠L = 70°)
Now in ∆EOS,
Ext. ∠ISU = x + ∠E
⇒ 120° = x + 70°
⇒ x = 120° – 70° = 50°
x = 50°

Question 7.
Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each angle of the parallelogram.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 17
∠A = (3x – 2)° and ∠C = (50 – x)°
∠A = ∠C (Opposite angles of a parallelogram)
⇒ 3x – 2° = 50° – x
⇒ 3x + x = 50° + 2°
⇒ 4x = 52°
x = 13°

Question 8.
If an angle of a parallelogram is two- third of its adjacent angle, find the angles of the parallelogram.
Solution:
Let the parallelogram is ABCD and ∠A of a parallelogram ABCD be x
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 18
Hence other angles will be
∠C = ∠A = 108° (Opposite angles)
and ∠D = ∠B = 72° (Opposite angles)
Hence ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°

Question 9.
The measure of one angle of a parallelogram is 70°. What are the measures of the remaining angles ?
Solution:
Let in parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 19
∠A = 70°
But ∠A + ∠B= 180° (Sum of adjacent angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70°
⇒ ∠B = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

Question 10.
Two adjacent angles of a parallelogram are as 1 : 2. Find the measures of all the angles of the parallelogram.
Solution:
Let in parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 20
∠A : ∠B = 1 : 2
Let ∠A = x, then ∠B = 2x
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ x + 2x = 180°
⇒ 3x = 180°
∠A = x = 60°
and ∠B = 2x = 2 x 60°= 120°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 60° and ∠D = 120°
Hence ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°

Question 11.
In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 21
∠D = 135°
But ∠A + ∠D= 180° (Sum of adjacent angles)
∠A + 135° = 180°
∠A = 180° – 135° = 45°
But ∠B = ∠D (Opposite angles)
∠B = 135°
Hence ∠A = 45° and ∠B = 135°

Question 12.
ABCD is a parallelogram in which ∠A = 70°, compute ∠B, ∠C and ∠D.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 22
∠A = 70°
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70° = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

Question 13.
The sum of two opposite angles of a parallelogram is 130°. Find all the angles of the parallelogram.
Solution:
Let in parallelogram
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 23
∠A + ∠C = 130°
But ∠A = ∠C (Opposite angles)
⇒ ∠C = 1302 = 65°
∠B + ∠D = 180° (Sum of adjacent angles)
⇒ 65° + ∠B = 180°
⇒ ∠B = 180° – 65° = 115°
⇒ ∠B = 115°
But ∠D = ∠B (Opposite angles)
∠D = 115°
Hence ∠A = 65°, ∠B = 115°, ∠C = 65° and ∠D = 115°

Question 14.
All the angles of a quadrilateral are equal to each other. Find the measure of each. Is the quadrilateral a parallelogram ? What special type of parallelogram is it ?
Solution:
All the angles of a quadrilateral are equal and sum of the four angles = 360°
Each angle will be = 3604 = 90°
Let in quadrilateral ABCD,
∠A = ∠B = ∠C = ∠D = 90°
It is a parallelogram as opposite angles are equal
i.e., ∠A = ∠C and ∠B = ∠D.
Each angle is of 90°
This parallelogram is a rectangle.

Question 15.
Two adjacent sides of a parallelogram are 4 cm and 3 cm respectively. Find its perimeter.
Solution:
Length of two adjacent sides = 4 cm and 3 cm
i.e., l = 4 cm and b = 3 cm
Perimeter = 2 (l + b) = 2 (4 + 3) cm = 2 x 7 = 14 cm

Question 16.
The perimeter of a parallelogram is 150 cm. One of its sides is greater ii.au the other by 25 cm. Find the length of the sides of the parallelogram.
Solution:
Perimeter of a parallelogram = 150 cm
Let l he the longer side and b be the shorter side
l = b + 25 cm.
⇒ 2 (l + b) = 150
⇒ l + b = 75
⇒ b + 25 + b = 75
⇒ 2b = 75 – 25 = 50
⇒ b = 25
l = b + 25 = 25 + 25 = 50
Sides are 50 cm, 25 cm

Question 17.
The shorter side of a parallelogram is 4.8 cm and the longer side is half as much again as the shorter side. Find the perimeter of the parallelogram.
Solution:
In a parallelogram shorter side (b) = 4.8 cm.
longer side (l) = 4.8 + 12 x 4.8
= 4.8 + 2.4 = 7.2 cm
Perimeter = 2 (l + b) = 2 (7.2 + 4.8) cm = 2 x 12.0 = 24 cm

Question 18.
Two adjacent angles of a parallelogram are (3x – 4)° and (3x + 10)°. Find the angles of the parallelogram.
Solution:
Let in parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 24
∠A = (3x + 4)° and ∠B = (3x + 10)°
But ∠A + ∠B = 180° (Sum of adj adjacent angles)
⇒ 3x – 4 + 3x + 10 = 180°
⇒ 6x + 6° = 180°
⇒ 6x = 180° – 6° = 174°
⇒ x = 29°
∠A = 3x – 4 = 3 x 29 – 4 = 87° – 4° = 83°
∠B = 3x + 10 = 3 x 29° + 10° = 87° + 10° = 97°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 83° and ∠D = 97°
Hence ∠A = 83°, ∠B = 97°, ∠C = 83° and ∠D = 97°

Question 19.
In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC = 30°, ∠BDC = 10° and ∠CAB = 70°.
Find : ∠DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC and ∠DBA.
Solution:
In parallelogram ABCD, diagonal AC and ED bisect each other at O.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 25
∠ABC = 30°, ∠CAB = 70° and ∠BDC = 10°
∠ADC = ∠ABC = 30° (Opposite angles)
and ∠ADB = ∠ADC – ∠BDC = 30° – 10° = 20°
AB || DC and AC is the transversal
∠ACD = ∠CAB = 70°(Altemate angles)
AB || DC and BD is transversal
∠CDB = ∠ABD = 10°(Altemate angles)
In ∆ABC
∠CAB + ∠ABC + ∠BCA = 180° (Sum of anlges of a triangle)
⇒ 70° + 30° + ∠BCA = 180°
⇒ 100° + ∠BCA = 180°
∠BCA = 180° – 100° = 80°
∠BCD = ∠BCA + ∠ACD = 80° + 70° = 150°
∠BCD = ∠DAB (Opposite angles)
⇒ ∠DAB = 150° and ∠CAD = 150° – 70° = 80°
In ∆OCD,
∠ODC + ∠OCD + ∠COD = 180° (Angles of a triangle)
⇒ ∠ACD + ∠ACD + ∠COD = 180°
⇒ 70° + 10° + ∠COD = 180°
⇒ 80° + ∠COD = 180°
⇒ ∠COD = 180° – 80° = 100°
∠COD = 100°
But ∠AOD + ∠COD = 180° (Linear pair)
⇒ ∠AOD + 100° = 180°
⇒ ∠AOD = 180° – 100° = 80°
⇒ ∠AOD = 80°
But ∠AOB = ∠COD
and ∠BOC = ∠AOD (Vertically opposite angles)
∠AOB = 100° and ∠BOC = 80°
Hence ∠DAB = 150°, ∠ADC = 30°, ∠BCD = 150°
∠AOD = 80° ∠DOC = 100°
∠BOC = 80° ∠AOB = 100°
∠ACD = 70°, ∠CAB = 70°, ∠AD = 20°, ∠ACB = 80°, ∠DBC = 20°, ∠DBA = 10

Question 20.
Find the angles marked with a question mark shown in the figure.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 26
Solution:
In parallelogram ABCD,

CE ⊥ AB and CF ⊥ AD
∠BCE = 40°
In ∆BCE,
∠BCE + ∠CEB + ∠EBC = 180° (Sum of angles of a triangle)
⇒ 40° + 90° + ∠EBC = 180°
⇒ 130° + ∠EBC = 180°
⇒ ∠EBC = 180° – 130° = 50°
or ∠B = 50°
But ∠D = ∠B (Opposite angles)
∠D = 50° or ∠ADC = 50°
Similarly in ∆DCF,
∠DCF + ∠CFD + ∠FDC = 180°
⇒ ∠DCF + 90° + 50° = 180°
⇒ ∠DCF + 140° = 180°
⇒ ∠DCF = 180° – 140° = 40°
But ∠C + ∠B= 180° (Sum of adjacent angles)
⇒ ∠BCE + ∠ECF + ∠DCF + ∠B = 180°
⇒ 40° + ∠ECF + 40° + 50° = 180°
⇒ ∠ECF + 130° = 180°
⇒ 40° + ∠ECF + 40° + 50° = 180°
⇒ ∠ECF + 130° = 180°
⇒ ∠ECF = 180° – 130° = 50°
∠ECF = 50°

Question 21.
The angle between the altitudes of a parallelogram through the same vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 28
∠A is an obtused angle
AE ⊥ BC and AF ⊥ DC
∠EAF = 60°
In quadrilateral AECF,
∠EAF + ∠F + ∠C + ∠E = 360° (Sum of angles of a quadrilateral)
⇒ 60° + 90° + ∠C + 90° = 360°
⇒ 240° + ∠C = 360°
⇒ ∠C = 360° – 240° = 120°
∠C = 120°
But ∠A = ∠C (Opposite angles)
∠A = 120°
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ 120° + ∠B = 180°
⇒ ∠B = 180° – 120° = 60°
∠B = 60°
But ∠D = ∠B (Opposite angles)
∠D = 60°
Hence ∠A = 120°, ∠B = 60°, ∠C = 120° and ∠D = 60°

Question 22.
In the figure, ABCD and AEFG are parallelograms. If ∠C = 55°, what is the measure of ∠F ?
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 29
Solution:
In the parallelogram ABCD,
∠A = ∠C …..(i) (Opposite angles of a parallelogram)
Similarly in parallelogram AEFG,
∠A = ∠F …(ii)
From (i) and (ii),
∠C = ∠F = 55° (∠C = 55°)
Hence ∠F = 55°

Question 23.
In the figure, BDEF and DCEF are each a parallelogram. It is true that BD = DC ? Why or why not ?
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 30
Solution:
In parallelogram BDEF,
BD = EF ……(i) (Opposite sides of a parallelogram)
Similarly, in parallelogram DCEF
DC = EF
From (i) and (ii),
BD = DC
Hence it is true that BD = DC.

Question 24.
In the figure, suppose it is known that DE = DF. Then is ∆ABC isosceles ? Why or why not ?
Solution:
In parallelogram BDEF,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 31
∠B = ∠E ……(i) (Opposite angles of a parallelogram)
Similarly, in parallelogram DCEF,
∠C = ∠F ……(ii)
But DE = DF (Given)
In ∆DEF
∠E = ∠F
From (i) and (ii),
∠B = ∠C
AC = AB (Sides opposite to equal angles)
∆ABC is an isosceles triangle.

Question 25.
Diagonals of parallelogram ABCD intersect at O as shown in the figure. XY contains O, and X, Y are points on opposite sides of the parallelogram. Give reasons for each of the following:
(i) OB = OD
(ii) ∠OBY = ∠ODX
(iii) ∠BOY = ∠DOX
(iv) ∆BOY = ∆DOX
Now, state if XY is bisected at O.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 32
diagonals AC and BP intersect each other at O
O is the mid-point of AC and BD.
Through O, XY is draw such that X lies on AD and Y, on BC.
(i) OB = OD (O is mid-point of BD)
(ii) AD || BC and BD is transversal
∠OBY = ∠ODX (Alternate angles)
(iii) ∠BOY = ∠DOX (Vertically opposite angles)
(iv) Now in ∆BOY and ∆DOX,
OB = OD
∠OBY = ∠ODX
∠BOY = ∠DOX
∆BOY = ∆DOX (ASA axiom)
OY = OX (c.p.c.t.)
Hence XY is bisected at O.

Question 26.
In the figure ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A. In each of the following, if the statement is true, give a reason for the same.
(i) ∠A = ∠C
(ii) ∠FAB = 12 ∠A
(iii) ∠DCE = 12 ∠C
(iv) ∠CEB = ∠FAB
(v) CE || AF.
Solution:
In parallelogram ABCD,
CE is the bisector of ∠C and and AF is the bisector of ∠A.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 33
(i) ∠A = ∠C (Opposite angles of a parallelogram)
(ii) AF is the bisector of ∠A
∠FAB = 12 ∠A
(iii) CE is the bisector of ∠C
∠DCE = 12 ∠C
(iv) From (i), (ii) and (iii)
∠FAB = ∠DCE
(v) ∠FAB = ∠DCE
But these are opposite angles of quadrilateral AECF
AB or AE || DC or FC
AECF is a parallelogram
CE || AF
Hence proved

Question 27.
Diagonals of a parallelogram ABCD intersect at O. AL and CM are drawn perpendiculars to BD such that L and M lie on BD. Is AL = CM ? Why or why not ?
Solution:
In parallelogram ABCD, diagonals AC and BD intersect each other at O.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 34
O is the mid-point of AC and BD.
AL ⊥ BD and CM ⊥ BD.
In ∆ALO and ∆CMO
∠L = ∠M (Each 90°)
∠AOL = ∠COM (Vertically opposite angles)
AO = CO (O is mid-point of AC)
∆ALO = ∆CMO (AAS axiom)
AL = CM (c.p.c.t.)
Hence proved

Question 28.
Points E and F lie on diagonal AC of a parallelogram ABCD such that AE = CF. What type of quadrilateral is BFDE ?
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 35
AC is its diagonal. E and F are points on AC such that AE = CF
Join EB, BF, FD and DE .
Join also diagonal BD which intersects AC at O
O is the mid-point of AC and BD
AO = OC
But AE = CF
⇒ AO – AE = CO – CF
⇒ EO = OF
But BO = OD (O is mid-point of BD)
Diagonals EF and BD of quadrilateral bisect each other at O.
BFDE is a parallelogram.

Question 29.
In a parallelogram ABCD, AB = 10 cm, AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC produced meet at F. Find the length of CF.
Solution:
AB || DC
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 36
∠DEA = ∠EAB (Alternate angles)
= ∠DAE (EA is bisector of ∠A)
In ∆DAE,
∠DEA = ∠DAE
AD = DE = 6 cm
But DE = AB = 10 cm.
EC = DC – DE = 10 – 6 = 4 cm
AD || BC or BF and AF is transversal
∠DAE = ∠EFC (Alternate angle)
But ∠DAE = ∠DEA Prove
= ∠FEC (DEA = FEC vertically opposite angles)
In ∆ECF,
CE = CF = 4 cm (CE = 4 cm)

Exercise 17.2

Question 1.
Which of the following statements are true for a rhombus ?
(i) It has two pairs of parallel sides.
(ii) It has two pairs of equal sides.
(iii) It has only two pairs of equal sides.
(iv) Two of its angles are at right angles.
(v) Its diagonals bisect each other at right angles.
(vi) Its diagonals are equal and perpendicular.
(vii) it has all its sides of equal lengths.
(viii) It is a parallelogram.
(ix) It is a quadrilateral.
(x) It can be a square.
(xi) It is a square.
Solution:
(i) True
(ii) True
(iii) False (Its all sides are equal)
(iv) False (Opposite angles are equal)
(v) True
(vi) False (Diagonals are not equal)
(vii) True
(viii) True
(ix) True
(x) True (It is a rhombus)
(xi) False

Question 2.
Fill in the blanks, in each of the following, so as to make the statement true:
(i) A rhombus is a parallelogram in which ………..
(ii) A square is a rhombus in which ………..
(iii) A rhombus has all its sides of …….. length.
(iv) The diagonals of a rhombus each ………. other at ………. angles.
(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a ………..
Solution:
(i) A rhombus is a parallelogram in which adjacent sides are equal.
(ii) A square is a rhombus in which one angle is right angle.
(iii) A rhombus has all its sides of equal length.
(iv) The diagonals of a rhombus bisect each other at right angles.
(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a rhombus.

Question 3.
The diagonals of a parallelogram are not perpendicular. Is it a rhombus ? Why or why not ?
Solution:
By definition of a rhombus, its diagonals bisect each other at right angle.
So, the given parallelogram is not a rhombus.

Question 4.
The diagonals of a quadrilateral are perpendicular to each other. Is such a quadrilateral always a rhombus. If your answer is ‘No’, draw a figure to justify your answer.
Solution:
The diagonals of a quadrilateral are perpendicular to each.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 1
It is not always possible to be a rhombus. It can be of the diagonals bisect each other at right angles and if not, then it is not rhombus as shown in the figure given above:

Question 5.
ABCD is a rhombus. If ∠ACB = 40°, find ∠ADB.
Solution:
In rhombus ABCD, diagonals AC and BD intersect each other at O.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 2
∠ACB = 40°, we have to find ∠ADB.
BD || AD and AC.is its transversal..
∠ACB = ∠CAD (Alternate angles)
Now in ∆AOD
∠OAD + ∠AOD + ∠ADO = 180° (Sum of angles of a triangle)
⇒ 40° + 90° + ∠ADO = 180°
⇒ 130° + ∠ADO = 180°
⇒ ∠ADO = 180° – 130° = 50°
∠ADB = 50°

Question 6.
If the diagonals of a rhombus are 12 cm and 16 cm, find the length of each side.
Solution:
In rhombus ABCD, diagonals AC and BD bisect each other at O at right angles.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 3
AC = 16 cm, BD = 12 cm
AO = OC = 162 = 8 cm
BO = OD = 122 = 6 cm.
Now, in right angled ∆AOB,
AB² = AO² + BO² = (8)² + (6)² = 64 + 36 = 100 = (10)²
AB = 10 cm
Each side of rhombus = 10 cm

Question 7.
Construct a rhombus whose diagonals are of length 10 cm and 6 cm.
Solution:
(i) Draw a line segment AC =10 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 4
(ii) Draw its perpendicular bisector and cut off OB = OD = 3 cm (12 of 6 cm).
(iii) Join AB, BC, CD and DA.
Then ABCD is the required rhombus.

Question 8.
Draw a rhombus, having each side of length 3.5 cm and one of the angles as 40°.
Solution:
Steps of construction
(i) Draw a line segment AB = 3.5 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 5
(ii) Draw a ray BX making an angle of 40° at B and cut off BC = 3.5 cm.
(iii) With centres C and A, and radius 3.5 cm. Draw arcs intersecting each other at D.
(iv) Join AD and CD.
Then ABCD is a required rhombus.

Question 9.
One side of a rhombus is of length 4 cm and the length of an altitude is 3.2 cm. Draw the rhombus.
Solution:
(i) Draw a line segment AB = 4 cm.
(ii) At B, draw a perpendicular BX and cut off BL = 3.2 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 6
(iii) At L, draw a line LY parallel to AB.
(iv) With centres B and radius 4 cm, draw an arc intersecting the line LY at C.
(v) With centre C cut off CD = 4 cm.
(vi) Join BC and AD.
Then ABCD is the required rhombus.

Question 10.
Draw a rhombus ABCD if AB = 6 cm and AC = 5 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 6 cm.
(ii) With centre A and radius 5 cm and with centre B and radius 6 cm, draw arcs intersecting each other at C.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 7
(iii) Join AC and BC.
(iv) Again with centres C and A and radius 6 cm, draw arcs intersecting each other at D.
(v) Join AD and CD.
Then ABCD is the required rhombus.

Question 11.
ABCD is a rhombus and its diagonals intersect at O.
(i) Is ∆BOC = ∆DOC ? State the congruence condition used ?
(ii) Also state, if ∠BCO = ∠DCO.
Solution:
In rhombus ABCD, diagonals AC and BD bisect each other at O.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 8
(i) Now in ∆BOC and ∆DOC
OC = OC (Common)
BC = CD (Sides of rhombus)
OB = OC (Diagonals bisect each other at O)
∆BOC = ∆DOC (SSS. condition)
(ii) ∠BCO = ∠DCO

Question 12.
Show that each diagonal of a rhombus bisects the angle through which it passes:
Solution:
In rhombus ABCD, AC is its diagonal we have to prove that AC bisects ∠A and ∠C.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 9
Now, in ∆ABC and ∆ADC
AC = AC (Common)
AB = CD (Sides of a rhombus)
BC = AD
∆ABC = ∆ADC (SSS condition)
∠BAC = ∠DAC and ∠BCA = ∠DCA (c.p.c.t.)
Hence AC bisects the angle A.
Similarly, by joining BD, we can prove that BD bisects ∠B and ∠D.
Hence each diagonal of a rhombus bisects the angle through which it passes.

Question 13.
ABCD is a rhombus whose diagonals intersect at O. If AB = 10 cm, diagonal BD = 16 cm, find the length of diagonal AC.
Solution:
In rhombus ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 10
AB = 10 cm, diagonal BD = 16 cm.
Draw diagonal AC which bisects BD at O at right angle.
BO = OD = 8 cm and AO = OC.
Now in ∆AOB,
AB² = AO² + BO²
⇒ (10)² = AO² + (8)²
⇒ 100 = AO² – 64
⇒ AO² = 100 – 64 = 36 = (6)²
AO = 6.
AC = 2AO = 2 x 6 = 12 cm

Question 14.
The diagonals of a quadrilateral are of lengths 6 cm and 8 cm. If the diagonals bisect each other at right angles, what is the length of each side of the quadrilateral ?
Solution:
In a quadrilateral ABCD, diagonals AC and BD bisect each other at right angles.
AC = 8 cm and BD = 6 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.2 11
AO = OC = 4 cm and BO = OD = 3 cm
Now, in right ∆AOB,
AB² = AO² + BO² = (4)² + (3)² = 16 + 9 = 25 = (5)²
AB = 5 cm
Hence each side of quadrilateral will be 5 cm.

Exercise 17.3

Question 1.
Which of the following statements are true for a rectangle ?
(i) It has two pairs of equal sides.
(ii) It has all its sides of equal length.
(iii) Its diagonals are equal.
(iv) Its diagonals bisect each other.
(v) Its diagonals are perpendicular.
(vi) Its diagonals are perpendicular and bisect each other.
(vii) Its diagonals are equal and bisect each other.
(viii) Its diagonals are equal and perpendicular, and bisect each other.
(ix) All rectangles are squares.
(x) All rhombuses are parallelograms.
(xi) All squares are rhombuses and also rectangles.
(xii) All squares are not parallelograms.
Solution:
(i) True.
(ii) False. (Only pair of opposite sides is equal)
(iii) True
(iv) True
(v) False (Diagonals are not perpendicular)
(vi) False (Diagonals are not perpendicular to each other)
(vii) True
(viii) False (Diagonals are equal but not perpendicular)
(ix) False (All rectangles are not square but a special type can be a square)
(x) True
(xi) True
(xii) False (All squares are parallelograms because their opposite sides are parallel and equal)

Question 2.
Which of the following statements are true for a square ?
(i) It is a rectangle.
(ii) It has all its sides of equal length.
(iii) Its diagonals bisect each other at right angle.
(iv) Its diagonals are equal to its sides.
Solution:
(i) True
(ii) True
(iii) True
(iv) False (Each diagonal of a square is greater than its side)

Question 3.
Fill in the blanks in each of the following so as to make the statement true :
(i) A rectangle is a parallelogram in which ……..
(ii) A square is a rhombus in which ……….
(iii) A square is a rectangle in which ………
Solution:
(i) A rectangle is a parallelogram in which one angle is right angle.
(ii) A square is a rhombus in which one angle is right angle.
(iii) A square is a rectangle in which adjacent sides are equal.

Question 4.
A window frame has one diagonal longer than the other. Is the window frame a rectangle ? Why or why not ?
Solution:
No, it is not a rectangle as rectangle has diagonals of equal length.

Question 5.
In a rectangle ABCD, prove that ∆ACB = ∆CAD.
Solution:
In rectangle ABCD, AC is its diagonal.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 1
Now in ∆ACB and ∆CAD
AB = CD (Opposite sides of a rectangle)
BC = AD
AC = AC (Common)
∆ACB = ∆CAD (SSS condition)

Question 6.
The sides of a rectangle are in the ratio 2 : 3 and its perimeter is 20 cm. Draw the rectangle.
Solution:
Perimeter of a rectangle = 20 cm
Ratio in the sides = 2 : 3
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 2
Let breadth (l) = 2x
Then length (b) = 3x
Perimeter = 2 (l + b)
⇒ 20 = 2 (2x + 3x)
⇒ 4x + 6x = 20
⇒ 10x = 20
⇒ x = 2010 = 2
Length = 3x = 3 x 2 = 6
and breadth = 2x = 2 x 2 = 4 cm
Steps of construction:
(i) Draw a line segment AB = 6 cm.
(ii) At A and B draw perpendicular AX and BY.
(iii) Cut off from AX and BY,
AD = BC = 4 cm.
(iv) Join CD.
Then ABCD is the required rectangle.

Question 7.
The sides of a rectangle are the ratio 4 : 5. Find its sides if the perimeter is 90 cm.
Solution:
Perimeter of a rectangle = 90 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 3
Ratio in sides = 4 : 5
Let first side = 4x
Then second side = 5x
Perimeter = 2 (l + b)
⇒ 2 (4x + 5x) = 90
⇒ 2 x 9x = 90
⇒ 18x = 90
⇒ x = 5
First side = 4x = 4 x 5 = 20 cm
and second side = 5x = 5 x 5 = 25 cm

Question 8.
Find the length of the diagonal of a rectangle whose sides are 12 cm and 5 cm.
Solution:
In rectangle ABCD, AB = 12 cm and AD = 5 cm
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 4
BD is its diagonal.
Now, in right angled ∆ABD,
BD² = AB² + AD² (Pythagoras theorem)
= (12)² + (5)² = 144 + 25 = 169 = (13)²
BD = 13 cm
Length of diagonal = 13 cm

Question 9.
Draw a rectangle whose one side measures 8 cm and the length of each of whose diagonals is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8 cm
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 5
(ii) At B, draw a perpendicular BX
(iii) With centre A and radius 10 cm, draw an arc which intersects BX at C.
(iv) With centre C and radius equal to AB and with centre A and radius equal to BC, draw arcs which intersect at D.
(v) Join AD, AC, CD and BD.
Then ABCD is the required rectangle.

Question 10.
Draw a square whose each side measures 4.8 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4.8 cm.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 6
(ii) At A and B, draw perpendiculars AX and BY.
(iii) Cut off AD = BC = 4.8 cm
(iv) Join CD.
Then ABCD is the required square.

Question 11.
Identify all the quadrilaterals that have:
(i) Four sides of equal length.
(ii) Four right angles.
Solution:
(i) A quadrilateral whose four sides are equal can be a square or a rhombus.
(ii) A quadrilateral whose four angle are right angle each can be a square or a rectangle.

Question 12.
Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle ?
Solution:
(i) A square is a quadrilateral as it has four sides and four angles.
(ii) A square is a parallelogram, because its opposite sides are parallel and equal.
(iii) A square is a rhombus because it has all sides equal and opposite sides are parallel.
(iv) A square is a rectangle as its opposite sides are equal and each angle is of 90°.

Question 13.
Name the quadrilaterals whose diagonals:
(i) bisect each other
(ii) are perpendicular bisector of each other
(iii) are equal.
Solution:
(i) A quadrilateral whose diagonals bisect each other can be a square, rectangle, rhombus or a parallelogram.
(ii) A quadrilateral whose diagonals are perpendicular bisector of each other can be a square or a rhombus.
(iii) A quadrilateral whose diagonals are equal can be a square or a rectangle.

Question 14.
ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B and C.
Solution:
In ∆ABC, ∠B = 90°.
O is the mid-point of AC i.e. OA = OC.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.3 7
BO is joined.
Now, we have to prove that OA = OB = OC
Produce BO to D such that OD = OB.
Join DC and DA.
In ∆AOB and ∆COD
OA = OC (O is the mid point of AC)
OB = OD (Construction)
∠AOB = ∠COD (Vertically opposite angles)
∆AOB = ∆COD (SAS condition)
AB = CD (c.p.c.t.) …..(i)
Similarly, we can prove that
∆BOC = ∆AOD
BC = AD …….(ii)
From (i) and (ii)
ABCD is a rectangle.
But diagonals of a rectangle bisect each other and are equal in length.
AC and BD bisect each other at O.
OA = OC = OB.
O is equidistant from A, B and C.

Question 15.
A mason has made a concrete slap. He needs it to be rectangular. In what different ways can he make sure that it is a rectangular ?
Solution:
By definition, a rectangle has each angle of 90° and their diagonals are equal.
The mason will check the slab whether it is a rectangular in shape by measuring that
(i) its each angle is 90°
(ii) its both diagonals are equal.

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CHAPTER -2 Nutrition in Animals | CLASS 7TH |NCERT SCIENCE IMPORTANT QUESTIONS & MCQS | EDUGROWN

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 2 Nutrition in Animals

MCQs

Choose the correct answer:

Question 1.
Bile is produced in
(a) Gall bladder
(b) Blood
(c) Liver
(d) Spleen

Answer

Answer: (c) Liver

Question 2.
Cud is the name given to the food of ruminants which is:
(а) swallowed and undigested.
(b) swallowed and partially digested.
(c) properly chewed and partially digested.
(d) properly chewed and completely digested.

Answer

Answer: (b) swallowed and partially digested.

Question 3.
The false feet of Amoeba are used for:
(a) movement only
(b) capture of food only
(c) capture of food and movement
(d) exchange of gases only

Answer

Answer: (c) capture of food and movement

Question 4.
Read the following statements with reference to the villi of small intestine.
(i) They have very thin walls.
(ii) They have a network of thin and small blood vessels close to the surface.
(iii) They have small pores through which food can easily pass.
(iv) They are finger-like projections.
Identify those statements which enable the villi to absorb digested food.
(a) (i), (ii) and (iv)
(b) (ii), (iii) and (iv)
(c) (iii) and (iv)
(d) (i) and (iv)

Answer

Answer: (a) (i), (ii) and (iv)

Question 5.
The finger-like outgrowths of Amoeba helps to ingest food. However, the finger-like outgrowths of human intestine helps to:
(a) digest the fatty food substances
(b) make the food soluble
(c) absorb the digested food
(d) absorb the undigested food

Answer

Answer: (c) absorb the digested food

Question 6.
The main function of the lacteals of intestine is the absorption of
(a) amino acids
(b) glucose and vitamins
(c) lactic acid
(d) fatty acids and glycerol

Answer

Answer: (d) fatty acids and glycerol

Question 7.
Gastric digestion takes place efficiently in
(a) acidic medium
(b) alkaline medium
(c) neutral medium
(d) highly alkaline medium

Answer

Answer: (a) acidic medium

Question 8.
Which is not digested by human?
(a) Protein
(b) Fats
(c) Glucose
(d) Cellulose

Answer

Answer: (d) Cellulose

Question 9.
How many premolars teeth found in mouth?
(a) 2
(b) 4
(c) 6
(d) 8

Answer

Answer: (b) 4

Question 10.
Given below from (i) to (iv) are some food items.
(i) Boiled and mashed potato
(ii) Glucose solution
(iii) A slice of bread
(iv) Mustard oil
Which of the above will give blue-black colour when tested with iodine?
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (iii) and (iv)

Answer

Answer: (b) (i) and (iii)

Question 11.
Which of the following pair of teeth differ in structure but are similar in function?
(a) canines and incisors
(b) molars and premolars
(c) incisors and molars
(d) premolars and canines

Answer

Answer: (a) canines and incisors

Question 12.
Read carefully the terms given below. Which of the following set is the correct combination of organs that do not carry out any digestive functions?
(a) Oesophagus, Large Intestine, Rectum
(b) Buccal Cavity, Oesophagus, Rectum
(c) Buccal Cavity, Oesophagus, Large Intestine
(d) Small Intestine, Large Intestine, Rectum

Answer

Answer: (a) Oesophagus, Large Intestine, Rectum

Question 13.
The swallowed food moves downwards in the alimentary canal because of:
(a) force provided by the muscular tongue.
(b) the flow of water taken with the food.
(c) gravitational pull.
(d) the contraction of muscles in the wall of food pipe.

Answer

Answer: (a) force provided by the muscular tongue.

Question 14.
The acid present in the stomach:
(a) kills the harmful bacterial that may enter along with the food.
(b) protects the stomach lining from harmful substances.
(c) digests starch into simpler sugars.
(d) makes the medium alkaline.

Answer

Answer: (c) digests starch into simpler sugars.

Question 15.
The enzymes predsent in the saliva convert:
(a) fats into fatty acids and glycerol
(b) starch into simple sugars
(c) proteins into amino acids
(d) complex sugars into simple sugars

Answer

Answer: (b) starch into simple sugars

Match the following:

Column A	Column B
(i) Autotrophs	(a) Carbohydrate
(ii) Sugar	(b) Sucking
(iii) Photosynthesis	(c) Chlorophyll
(iv) Hydra	(d) Green plants

Answer

Answer:

Column A	Column B
(i) Autotrophs	(d) Green plants
(ii) Sugar	(a) Carbohydrate
(iii) Photosynthesis	(c) Chlorophyll
(iv) Hydra	(b) Sucking

Fill in the blanks:

1. Ingestion is the first step in the process of ……………..

Answer

Answer: holozoic nutrition

2. Digestion is the breakdown of complex food materials into ……………..

Answer

Answer: simple materials

3. …………….. plant is the common example of insectivorous plants.

Answer

Answer: Pitcher

4. The water is present even in the dry soil in the form of a …………….. layer on the sand particles.

Answer

Answer: hygroscopic

5. The source of energy for living organism is ……………..

Answer

Answer: food

Choose the true and false statements from the following:

1. The breakdown of complex component of food into simpler substance is called digestion.

Answer

Answer: True

2. Food is not synthesised in green plant.

Answer

Answer: False

3. Animals are heterotrophic.

Answer

Answer: True

4. Amoeba is a many celled animal.

Answer

Answer: False

5. Salivary glands secrete saliva.

Answer

Answer: True

6. Ruminants are cud chewing animals.

Answer

Answer: True

7. There are four kinds of permanent teeth in our mouth.

Answer
False

Question 1.
What is the total number of teeth in an adult human?
Answer:
In an adult human, there are total 32 teeth.

Question 2.
Name the parts of the alimentary canal where
(a) water gets absorbed from undigested food
(b) digested food gets absorbed
(c) taste of the food is perceived
(d) bile juice is produced [NCERT Exemplar]
Answer:
(a) large intestine
(b) small intestine
(c) tongue
(d) liver

Question 3.
Identify the location of salivary gland.
Answer:
Salivary gland is present in the buccal cavity and it secretes saliva.

Question 4.
You were blindfolded and asked to identify the drinks provided in two different glasses. You could identify drink A as lime juice and B as bitter gourd juice. How could you do it inspite of being blindfolded? [NCERT Exemplar; HOTS]
Answer:
Inspite of being blindfolded, one could identify two different drinks with the help of taste buds present in the tongue.

Question 5.
We should not eat hurriedly. Give reason.
Answer:
We should not eat hurriedly because if we ingest food in hurry or we talk or laugh while eating we experience hiccups, coughing or choking sensation.

Question 6.
Name the secretions of stomach which digest food.
Answer:
The inner lining of stomach secretes mucous hydrochloric acid and digestive juices.

Question 7.
Explain the role of mucus secreted by stomach.
Answer:
The function of mucus is to protect the lining of stomach from the action of hydrochloric acid secreted by stomach lining.

Question 8.
The long structure of small intestine is accommodated in small space within our body. Comment. [HOTS]
Answer:
The small intestine is about 7.5 metre long It is accommodated in a coiled form inside our body.

Question 9.
Suggest the organ of digestive system where the digestive juices from liver and pancreas is poured.
Answer:
The digestive juices from liver and pancreas is poured into small intestine which helps in complete digestion and absorption of food.

Question 10.
From which organ of digestive system, the undigested faecal matter is removed?
Answer:
The undigested faecal matter is removed through the anus by the process called egestion.

Question 11.
Mention the position of the rumen in ruminants.
Answer:
The rumen is the sac-like structure which is present between small intestine and large intestine in ruminants.

Question 12.
What is assimilation?
Answer:
The process by which absorbed food is taken by body cells and is used for energy, growth and repair is called assimilation.

Question 13.
Describe alimentary canal briefly.
Answer:
Alimentary canal is the tube running from mouth to anus of human and animals here digestion and absorption of food take place.

Question 14.
Name the simple forms of carbohydrates, fats and proteins.
Answer:
The food components and their simple forms are carbohydrates (glucose), fatty (fats) acids and glycerol, proteins (amino acids).

Question 15.
Write the shape of stomach.
Answer:
The shape of stomach is like flattened U-shaped.

Question 16.
Discuss the role of hydrochloric acid secreted by gastric glands.
Answer:
It helps in the breakdown of food particles. It creates an acidic environment which facilitates the action

Nutrition in Animals Class 7 Science Extra Questions Short Answer Type

Question 1.
With the help of labelled diagram show the gradual decay of tooth.
Answer:
Sweets and Tooth Decay The tooth is covered by white, hard outer covering of tooth called Enamel enamel below which dentine is present. It is similar to bone which Pulp cavity (with nerves and blood vessels) protects the pulp cavity having nerves and blood vessels. Bacteria are Gum present in our mouth but they are not harmful to us. However, if we do not clean our teeth and mouth after eating, many harmful bacteria also begin to live and grow in it. These bacteria breakdown the sugars present from the leftover food and release acids. The acids gradually damage the tooth. This is called tooth decay.

Therefore, tooth decay is defined as the process of rotting of tooth and formation of cavity or holes in it which leads to the toothache.
When the holes or cavity reaches to the pulp cavity, it causes pain. If these cavities are not treated on time it causes severe toothache and may result in tooth loss.
Nutrition in Animals Class 7 Extra Questions Science Chapter 2 sh Q 7
Tooth decay can be prevented by adapting following measures.
(i) One should rinse and clean its teeth thoroughly after every Pulp cavity meal.
(ii) We should clean our teeth with the help of datun or brush Gradual decaying of human tooth and toothpaste, twice a day.
(iii) We should use dental floss which is a special strong thread. It is moved between two teeth to take out
trapped food particles.
(iv) Dirty fingers or unwashed objects must be avoided to put in the mouth.
(v) We should avoid the use of sweets, chocolates, toffees, ice-cream, etc. Much use of cold drink should also be avoided.”

Question 2.
Name the various components of food and their simpler forms.
Answer:
The various components of food and their simpler forms are

Components of food	Simpler form
Carbohydrate	Glucose
Fats	Fatty acids and glycerol
Proteins	Amino acids
Vitamins	Vitamins
Minerals and water	Minerals and water

Question 3.
‘A’ got her gall bladder removed surgically as she was diagnosed with stones in her gall bladder. After the surgery, she faced problems in digestion of certain food items when consumed in bulk. Can you tell which kind of food items would they be and why? [NCERT Exemplar; HOTS]
Answer:
After surgical removal of gall bladder, ‘A’ would face problems in digestion of fat and fatty substances when consumed in bulk. This is because the bile juice stored in the gall bladder helps in digestion of fats.

Question 4.
Discuss the various associated glands of digestive system and their role in digestion.
Answer:
The various associated glands of digestive system and their role in digestion are as follows

Salivary gland digestion of starch in mouth.
Liver secretes bile juices which help in the digestion of fats.
Pancreas secretes pancreatic juices which act on carbohydrate, fats and proteins and change them into simpler compounds.

Question 5.
Boojho took some grains of boiled rice in test tube ‘A’ and Paheli took boiled and chewed rice in test tube ‘S’. Both of them poured 1-2 drops of iodine solution into the test tube and observed the colour change. What colour change would they have observed? Give reasons for your Answer: [NCERT Exemplar; HOTS]
Answer:
In test tube A, blue black colour appeared because of presence of starch.
In test tube 6, colour of iodine will not change because of digestion of starch into sugars by the action of saliva in our mouth.

Question 6.
Mention the different steps of nutrition in animals.

Question 7.
List the preventive measures that one should adopt for avoiding tooth decay.
Answer:
Sweets and Tooth Decay The tooth is covered by white, hard outer covering of tooth called Enamel enamel below which dentine is present. It is similar to bone which Pulp cavity (with nerves and blood vessels) protects the pulp cavity having nerves and blood vessels. Bacteria are Gum present in our mouth but they are not harmful to us. However, if we do not clean our teeth and mouth after eating, many harmful bacteria also begin to live and grow in it. These bacteria breakdown the sugars present from the leftover food and release acids. The acids gradually damage the tooth. This is called tooth decay.

Question 8.
Write the difference between milk teeth and permanent teeth.
Answer:
The difference between milk teeth and permanent teeth are

Milk teeth	Permanent teeth
They grow during infancy, i.e. when one is small baby.	They grow at the age of 6-8 years.
They are also called as temporary teeth because these are lost at the age of 6-8 years.	They do not fall till the old age.
They can be replaced by permanent teeth.	If these teeth fall down no new teeth arises on its place.

Question 9.
Complete the following table, from the options given below:
(Scraping, chewing, siphoning, capturing and swallowing, sponging, sucking, etc.)

Name of animal	Kind of animal	Mode of feeding
Snail
Ant
Eagle
Humming bird
Lice
Mosquito
Butterfly
Housefly

Answer:
The complete table is

Name of animal	Kind of food	Mode of feeding
Snail	Leaves and insects	Scraping
Ant	Food particles	Chewing
Eagle	Small animals	Capturing and swallowing
Humming bird	Nectar of flower	Sucking
Lice	Blood	Sucking
Mosquito	Blood	Sucking
Butterfly	Nactar of flower	Siphoning
Housefly	All most everything	Sponging

Question 10.
Briefly describe the mechanism of producing hiccups while we take food in hurry.

Question 11.
Boojho and Paheli were eating their food hurriedly so that they could go out and play during the recess. Suddenly, Boojho started coughing violently. Think of the reasons, why he was coughing and discuss with your friends?
[NCERT Exemplar; HOTS]
Answer:
Sometimes when we eat hurriedly, talks or laughs while eating, the flap-like valve (called epiglottis) which closes the passage of windpipe remains open. Therefore, the food may enter into the windpipe. Coughing helps to clear the passage and returns the food particle back to the foodpipe.

Question 12.
Gastric glands in stomach release hydrochloric acid, enzyme pepsin and mucus. What will happen if mucus is not secreted by the gastric glands?
Answer:
Mucus protects the inner lining of stomach form the action of hydrochloric acid and enzyme pepsin. If mucus is not released, it will lead to erosion of inner lining of stomach leading to acidity and ulcers.

Question 13.
Choose the odd one out from each group and give reasons.
(a) liver, salivary gland, starch, gall bladder
(b) stomach, liver, pancreas, salivary gland
(c) tongue, absorption, taste, swallow
(d) oesophagus, small intestine, large intestine, rectum [NCERT Exemplar]
Answer:
(a) Starch, because rest all are glands and starch is a type of carbohydrate.
(b) Stomach, because rest all are digestive glands and stomach is a digestive organ.
(c) Tongue, because rest all are digestive processes and tongue is a part of digestive system.
(d) Small intestine, because it carriers the process of digestion and rest are not involved in digestion.

Question 14.
Following statements describe the five steps in animal nutrition. Read each statement and give one word for each statement. Write the terms that describe each process.
(a) Transportation of absorbed food to different parts of body and their utilisation.
(b) Breaking of complex food substances into simpler and soluble substances.
(c) Removal of undigested and unabsorbed solid residues of food from the body.
(d) Taking food into the body.
(e) Transport of digested and soluble food from the intestine to blood vessels.[NCERT Exemplar]
Answer:
(a) Assimilation
(b) Digestion
(c) Egestion
(d) Ingestion
(e) Absorption

Question 15.
Small intestine in herbivores is longer than in carnivores. Do you agree? Support your Answer:
Answer:
Yes, carnivores animals cannot digest cellulose, hence they have a shorter small intestine. In herbivores, digestion of cellulose takes a longer time. Hence, herbivores need a longer small intestine to allow complete digestion of cellulose.

Question 16.
Draw a neat and clean diagram of Amoeba showing the correct location of the following components : nucleus, vacuole, pseudopodia.
Answer:
Nutrition in Animals Class 7 Extra Questions Science Chapter 2 sh Q 16

Question 17.
Draw the labelled diagram of tongue showing different region for taste buds.
Answer:
Refer to figure on page 17.
Nutrition in Animals Class 7 Extra Questions Science Chapter 2 sh Q 17

Question 18.
Ruminants such as cows and buffaloes swallow their food hurriedly and then sit restfully and chew their food. Give reason. [NCERT Examplar; HOTS]
Answer:
Ruminants such as cows and buffaloes swallow their food hurriedly and store it in a part of the stomach called rumen. The cellulose of the food is digested here by the action of certain bacteria which are not present in humans. Later, this partially digested food is returned to the buccal cavity of the animals in small lumps and animal chews it to complete the process of digestion. This process is called rumination.

Question 19.
Discuss the position and number of molars in buccal cavity.
Answer:
Molars are very large teeth which are present behind the premolar, towards the back of our mouth. They are only present in the permanent set of teeth and are 6 in each jaw.

Question 20.
Name the three digestive glands in our body.
Answer:
The three digestive glands are

liver
Pancreas
Salivary glands

Question 21.
The swallowed food moves downwards in the alimentary canal. Explain.
Answer:
The swallowed food moves downwards in the alimentary canal because of the atternate relaxation contraction movement of muscles in the wall of foodpipe called peristalsis.

Question 22.
Explain how assimilation is different from absorption.
Answer:
The process by which nutrients from the digested food are absorbed by the body is called absorption whereas the process by which the absorbed nutrients are utilised by the body for providing energy is called assimilation.

Question 23.
Food moves in the opposite direction during vomiting. How?
Answer:
The intense pressure is formed in the stomach when the food is not accepted by the stomach. The content in the stomach is then pushed back. This returned content is expelled out from the mouth in the form of vomiting.

Question 24.
Briefly explain, why animals like cow cannot chew their food properly at the time they take it in.
Answer:
Animals like cow cannot chew their food properly due to the presence of cellulose in their diet. At the time they take in food, the food is moistend and is sent for cellulose digestion and softening in rumen.

Question 25.
Is there any role of liver in digestion of fats? Explain.
Answer:
Yes, liver produces bile which has bile salts. These salts break large fat molecules to fine droplets. These fine droplets are further converted into fatty acids and glycerol.

Question 26.
Cellulose rich food substances are good source of roughage in human beings. Justify. [HOTS]
Answer:
Cellulose rich food substances are good source of roughage in human beings. It is because the cellulose digesting bacteria are not present in the body of human beings due to which human beings cannot digest cellulose (present in plant foods).

Question 27.
Recall and name the main organs of the digestive system in our body.
Answer:
The different organs of the alimentary canal are as follows :

Mouth and mouth cavity
Oesophagus
Stomach
Small intestine
Large intestine
Anus

Question 28.
Alimentary canal is different from digestive system. Comment.
Answer:
Alimentary canal is a long, muscular coiled tube. It is also known as digestive tract. The alimentary canal with its associated glands constitute the digestive system. These glands are salivary glands, liver and pancreas.

Question 29.
Windpipe runs adjacent to the foodpipe. What will happen if food particles enter the windpipe? Explain.
Answer:
The windpipe carries air from the nostrils to the lungs. It runs adjacent to the foodpipe. If, by chance, food particles enter the windpipe, we feel choked, get hiccups or cough.

Question 30.
Explain how is small intestine designed to absorb digested food.
Answer:
The finger-like projections called villi are present in the inner walls of the small intestine. The villi increase the surface area. The large surface area of small intestine helps in the rapid absorption of the digested food.

Nutrition in Animals Class 7 Science Extra Questions Long Answer Type

Question 1.
Label the given figure as directed below in A to D and give the name of each type of teeth.
(a) The cutting and biting teeth as A
(b) The piercing and tearing teeth as B
(c) The grinding and chewing teeth as C
(d) The grinding teeth present only in adult as D
Nutrition in Animals Class 7 Extra Questions Science Chapter 2 lo Q 1
Answer:

Question 2.
Briefly describe the process of digestion in Amoeba with the help of labelled diagram. [NCERT Examplar]
Answer:
Feeding and Digestion in Amoeba

Amoeba is a microscopic single celled organism, which is found in pond water. It is a very simple animal and cannot be seen by naked eyes. Amoeba has a cell membrane, a rounded dense nucleus and many small bubble-like vacuoles in its cytoplasm. These vacuoles are of two types, i.e. food vacuole and contractile vacuole. Food vacuole contains food surrounded by water while contractile vacuole contains liquid or water and controls water regulation activity in Amoeba. Its shape is not fixed, i.e. it constantly changes its shape and position. The body of Amoeba has finger-like projections, called pseudopodia or false feet. It captures food and helps in locomotion of Amoeba.
Nutrition in Animals Class 7 Extra Questions Science Chapter 2 lo Q 2
The food of Amoeba are microscopic organisms like tiny plants and animals present in pond water. When Amoeba senses its food, it pushes out pseudopodia around the food particle and engulfs it. The two pseudopodia join around the food particle and trap the food particle with a little water forming vacuole around food, thus the food gets trapped. Digestive juices present inside the vacuole, acts on the food and break it into simpler substances. This digested food is then absorbed and is used for growth, maintenance and multiplication of Amoeba. The undigested food residue is expelled outside by the vacuole. The basic process of digestion of food and release of energy is as similar to the other organisms.

Question 3.
Label the following parts of above figure and name them.
(a) The largest gland in our body.
(b) The organ where protein digestion starts.
(c) The organ that releases digestive juice into the small intestine.
(d) The organ where bile juice gets stored. [NCERT Examplar]
Nutrition in Animals Class 7 Extra Questions Science Chapter 2 lo Q 3
Answer:

Question 4.
Little Rishi (student of class VI) was watching his favourite cartoon serial on television. Suddenly he got hiccups. His elder brother Shubham who was sitting near by him gave him a glass of water and suggested not to eat too fast in a hurry. Little Rishi got confused as he had heard that ‘hicki’ comes when someone remembers. He asked his father. His father smiled and explained him that it is only a myth. He also explained him the proper scientific reason behind it.
(a) What is hiccup?
(b) Why do we get hiccup?
(c) What are the values shown by Rishi? [Value Based Question]
Answer:
(a) Hiccup is a choking sensation that produces a characteristic gulping sound repeatedly. It is called ‘hicki’ in our local language.
(b) Sometimes, when we eat too fast in a hurry or talk too much or laugh while eating, then a little of windpipe remains open due to which food particles may enter the windpipe. It may result in a choking sensation called hiccups.
(c) He is sincere and curious to know about the things at an early age.

Question 5.
Jaya returned from school and found that grandmother was scolding her maid, Rani as she did not come yesterday. The maid told that his son was passing watery stools frequently that’s why she didn’t come. Jaya’s mother who was listening the discussion came to them and told Rani not to come for coming 3-4 days. She also suggested her to give his son a solution of sugar and salt in clean water, many times a day for fast recovery. Jaya was surprised. She rushed to her mother and asked the scientific reason for it. Her mother smiled and explained her the importance of this solution.
(a) Name the term used to describe the condition in which a person passes out watery stools.
(b) Name the solution of sugar and salt in water. Why is it given to a person suffering from diarrhoea?
(c) What are the values shown by Jaya? [Value Based Question]
Answer:
(a) Diarrhoea.
(b) Oral Rehydration Solution (ORS). It is given to a person suffering from diarrhoea to prevent the dehydration.
(c) She is sincere, curious and has interest in science.

Question 6.
Read the following passage carefully and answer the questions that follow it.
Bile juice is stored in a sac called, gall bladder, located near its organ of secretion, liver. The gall bladder releases the bile juice into the small intestine whenever food reaches there. Though bile juice is devoid of any digestive enzymes, it is required for the digestion of fats. The fats cannot be digested easily because they are insoluble in water and are present as large globules. Bile juice breaks down big fat droplets into smaller droplets. These are then easily digested by the enzymes released from the pancreas
(a) Which organ secretes the bile juice?
(b) Why is digestion of fats difficult as compared to that of other nutrients?
(c) How does bile juice help in digestion of that of other nutrients?
(d) Where is the digestion of fat completed?
(e) Does bile juice digest fat completely? [NCERT Exemplar]
Answer:
(a) Bile juice is secreted by liver.
(b) Digestion of fats is difficult as compared to that of other nutrients because of insolubility of fat in water.
(c) Bile juice helps in digestion of fat by breaking down big fat droplets into smaller droplet.
(d) Digestion of fat is completed in small intestine.
(e) No, fat is not completely digested by bile juice.

Question 7.
Define oral rehydration solution and when it is given to the patient? How can you prepare ORS at home?
Answer:
Oral rehydration solution is the solution of sugar and salt in a particular ratio in the clean water. When a person passes out watery stools frequently, the disease is called diarrhoea. In this condition there is a loss of water and salt from the body of a person.

This is called dehydration which may be fatal if not cured at proper time. In order to prevent dehydration, the person or patient should be given ORS. ORS makes up the loss of water and salts in the body and sugar provides energy which helps in the recovery of disease. It should be given to a patient suffering from diarrhoea at a regular interval.

At home the ORS can be prepared by dissolving a teaspoonful of sugar and pinch of salt in a glass of clean water. The water used for preparing ORS should be first boiled and then cooled so that all the microorganisms or harmful bacteria may be killed.

Question 8.
Open your mouth, look into a mirror and try to count the different types of teeth of teeth in your mouth. Compare them with figure 2.3 on page 13 of your NCERT textbook. Record your observations in the table below: [NCERT Exemplar]

Type of teeth	Number of teeth
	In my mouth	In the figure
Incisors
Canines
Premolars
Molars

(a) Did you observe any difference in the number of teeth? If yes, could you identify which type of teeth showed the difference?
(b) Compare the number and type of teeth in an adult (say your parents or cousins who have reached the age of 25-30 or more). Note your observation.
Answer:

Type of teeth	Number of teeth
	In my mouth	In the figure
Incisors	4	4
Canines	8	8
Premolars	8	8
Molars	8	12

(a) Yes, the difference has been observed in the number of molars.
(b) The number and type of teeth varries in an adult as compared to the child. Children have 28 teeth in their mouth.There are only four molars in each jaw.
While, adults have 32 teeth in their mouth which means six molars in each jaw.

Question 9.
Explain how the digestion of cellulose occurs in grass eating animals.
Answer:
Digestion in Grass-Eating Animals

The herbivorous animals such as cow, buffaloes, etc eat grass. These animals quickly swallow the grass and store it in a part of stomach called rumen. The food is not chewed completely. Rumen possess cellulose digesting bacteria which breakdown the food by fermentation. This partially digested food or grass present in the rumen of cow is called cud.

This cud is brought back into the mouth of the cow from the rumen into small lumps and animal chews it again. This process is called rumination and animals are called ruminants.
When this cud is thoroughly chewed in the mouth of the cow, it is swallowed again. This time the chewed cud does not go back to rumen but enter into the other compartments of cow’s stomach and then into the small intestine for complete digestion and absorption of food. The cellulose digesting bacteria are not present in the body of human being, therefore human beings and other carnivore cannot digest cellulose present in plant food items.

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RD SHARMA SOLUTION CHAPTER –14 Compound Interest | CLASS 8TH MATHEMATICS-EDUGROWN

Exercise 14.1

Question 1.
Find the compound interest when principal = Rs 3,000, rate = 5% per annum and time = 2 years.
Solution:
Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 1
Amount after one year = Rs 3,000 + Rs 150 = 3,150
and principal for the second year = Rs 3,150
and interest for the second year

Compound interest for two years = Rs 150 + Rs 157.50 = Rs 307.50

Question 2.
What will be the compound interest on Rs 4,000 in two years when rate of interest is 5% per annum ?
Solution:
Principal (P) = Rs 4,000
Rate (R) = 5% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 3
Amount after one year = Rs 4,000 + Rs 200 = Rs 4,200
Principal for the second year = Rs 4,200
Interest for the second year

Compound interest for 2 years = Rs 200 + Rs 210 = Rs 410

Question 3.
Rohit deposited Rs 8,000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years ?
Solution:
Principal (P) = Rs 8,000
Rate (R) = 15% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 5
Amount after first year = Rs 8,000 + RS 1,200 = Rs 9,200
or Principal for the second year = Rs 9,200
Interest for the second year

Amount after 2 years = Rs 9,200 + Rs 1,380 = Rs 10,580
or Principal for the third year = Rs 10,580
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 7
Compound for the 3 years = Rs 1,200 + Rs 1,380 + Rs 1,587 = Rs 4,167

Question 4.
Find the compound interest on Rs 1,000 at the rate of 8% per annum for 112 years when interest is compounded half-yearly ?
Solution:
Principal (P) = Rs 1,000
Rate (R) = 8% p.a.
Period (T) = 112 years = 3 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 8
Amount after one half-year = Rs 1,000 + Rs 40 = 1,040
Or principal for the second half-year = Rs 1,040
Interest for the second half-year

Amount after second half-year = Rs 1,040 + 41.60 = Rs 1,081.60
Or principal for the third half-year = Rs 1081.60
Interest for the third half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 10
Compound interest for the third half-years or 112 years
= Rs 40 + Rs 41.60 + Rs 43.264 = Rs 124.864

Question 5.
Find the compound interest on Rs 1,60,000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.
Solution:
Principal (P) = Rs 1,60,000
Rate (R) = 20% p.a. or 5% quarterly
Period (T) = 1 year or 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 11
Amount after first quarter = Rs 1,60,000 + 8,000 = 1,68,000
Or principal for the second quarter = Rs 1,68,000
Interest for the second quarter

Amount after the second quarter = Rs 1,68,000 + Rs 8,400 = 1,76,400
Or principal for the third quarter = Rs 1,76,400
Interest for the third quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 13
Amount after third quarter = Rs 1,76,400 + 8,820 = Rs 1,85,220
or Principal for the fourth quarter = Rs. 1,85,220
Interest for the fourth quarter

Total compound interest for the 4 quarters = Rs 8,000 + Rs 8,400 + Rs 8,820 + 9,261 = Rs 34,481

Question 6.
Swati took a loan of Rs 16,000 against her insurance policy at the rate of 1212 % per annum. Calculate the total compound interest payable by Swati after 3 years.
Solution:
Amount of loan or principal (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 15
Amount after first year = Rs 16,000 + Rs 2,000 = Rs 18,000
Principal for the second year = Rs 18,000
Interest for the second year

Amount after second year = Rs 18,000 + 2,250 = Rs 20,250
Principal for the third year = Rs 20,250

Compound for 3 years = Rs 2,000 + Rs 2,250 + 2531.25 = Rs 6,781.25

Question 7.
Roma borrowed Rs 64,000 from a bank for 112 years at the rate of 10% per annum. Compute the total compound interest payable by Roma after 112 years, if the interest is compounded half-yearly.
Solution:
Principal (sum borrowed) (P) = Rs 64,000
Rate (R) = 10% p.a. or 5% half-yearly
Period (T) = 112 years or 3 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 18
Amount after first half-year = Rs 64,000 + Rs 3,200 = Rs 67,200
Or principal for the second half-year = Rs 67,200
Interest for the second half-year

Amount after second half-year = Rs 6,7200 + 3,360 = Rs 70,560
Or principal for the third half-year = Rs 70,560
Interest for the third half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 20
Total compound interest for 3 half-years
or 112 years = Rs 3,200 + Rs 3,360 + Rs 3,528 = Rs 10,088

Question 8.
Mewa Lai borrowed Rs 20,000 from his friend RoopLal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.
Solution:
Principal (P) = Rs 20,000
Rate (R) = 18% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 21
In second case
Interest for the first year

Amount after one year = Rs 20,000 + Rs 3,600 = Rs 23,600
Or principal for the second year = Rs 23,600
Interest for the second year

Interest for two years = Rs 3,600 + 4,248 = Rs 7,848
Gain = Rs 7,848 – Rs 7,200 = Rs 648

Question 9.
Find the compound interest on Rs 8,000 for 9 months at 20% per annum compounded quarterly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 20% p.a. or 5% p.a. quarterly
Period (T) = 9 months or 3 quarters
Interest for the first quarterly
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 24
Amount after first quarter = Rs 8,000 + Rs 400 = Rs 8,400
Or principal for second quarter = Rs 8,400
Interest for the second quarter

Amount after second quarter = Rs 8,400 + Rs 420 = Rs 8,820
Or principal for the third quarter = Rs 8,820
Interest for the third quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 26
Compound interest for 9 months or 3 quarters = Rs 400 + Rs 420 + Rs 441 = Rs 1,261

Question 10.
Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum gives Rs. 200 as simple interest.
Solution:
Simple interest = Rs 200
Rate (R) = 10% p.a.
Period (T) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 27
Now in second case,
Principal CP) = Rs 1,000
Rate (R) = 10% p.a.
Period (T) = 2 years.

Amount after one year = Rs 1,000+ Rs 100 = Rs 1,100
Or principal for the second year = Rs 1,100

Total interest for two years = Rs 100 + Rs 110 = Rs 210

Question 11.
Find the compound interest on Rs 64,000 for 1 year at the rate of 10% per annum compounded quarterly.
Solution:
Principal (P) = Rs 64,000
Rate (R) = 10% p.a. or 52 % quarterly
Period (T) = 1 year = 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 30
Amount after first quarter = Rs 64,000 + Rs 1,600 = Rs 65,600
Or principal for the second quarter = Rs 65,600
Interest for the second quarter

Amount after second quarter = Rs 65,600 + Rs 1,640 = Rs 67,240
Or principal for the third year = Rs 67,240
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 32
= Rs 1,681
Amount after third quarter = Rs 67,240 + Rs 1,681 = Rs 68,921
Or principal for the fourth quarter

Total compound interest for 4 quarters or one year
= Rs 1,600 + Rs 1,640 + Rs 1,681 + Rs 1723.025 = Rs 6644.025

Question 12.
Ramesh deposited Rs 7,500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months ?
Solution:
Principal (P) = Rs 7,500
Rate (R) = 12% p.a. or 3% quarterly
Time (T) = 9 months or 3 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 34
Amount after one quarter = Rs 7,500 + Rs 225 = Rs 7,725
Or Principal for second quarter = Rs 7,725
Interest for the second quarter

Amount after second quarter = Rs 7,725 + Rs 231.75 = Rs 7956.75
Or principal for the third quarter

Total amount he received after 9 months = Rs 7956.75 + Rs 238.70 = Rs 8195.45

Question 13.
Anil borrowed a sum of Rs 9,600 to install a hand pump in his dairy. If the rate of interest is 512 % .per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years.
Solution:
Principal (P) = Rs 9,600
Rate of interest (R) = 512 % = 112 % p.a.
Period (T) = 3 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 37
Amount after one year = Rs 9,600 + Rs 528 = Rs 10,128
Or principal for second year = Rs 10,128
Interest for the second year

Amount after second year = Rs 10,128 + Rs 557.04 = Rs 10685.04
or Principal for the third year = Rs 10685.04
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 39
Total compound interest = Rs 528 + Rs 557.04 + Rs 587.68 = Rs 1672.72

Question 14.
Surabhi borrowed a sum of Rs 12,000 from a finance company to purchase a refrigerator. If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after 3 years.
Solution:
Sum of money borrowed (P) = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 40
Amount after one year = Rs 12,000 + Rs 600 = Rs 12,600
Or principal for the second year = Rs 12,600
Interest for the second year

Amount after second year = Rs 12,600 + Rs 630 = Rs 13,230
Or Principal for the third year = Rs 13,230
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 42
Total compound interest for 3 years = Rs 600 + Rs 630 + Rs 661.50 = Rs 1891.50

Question 15.
Daljit received a sum of Rs 40,000 as a loan from a finance company. If the rate of interest is 7% per annum compounded annually, calculate the compound interest that Daljit pays after 2 years.
Solution:
Amount of loan (P) = Rs 40,000
Rate (R) = 7% p.a.
Period = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 43
Amount after one year = Rs 40,000 + Rs 2,800 = Rs 42,800
Or principal for the second year = Rs 42,800
Interest for the second year

Total interest paid after two years = Rs 2,800 + 2,996 = Rs 5,796

Exercise 14.2

Question 1.
Compute the amount and the compound interest in each of the following by using the formula when :
(i) Principal = Rs 3,000, Rate = 5%, Time = 2 years
(ii) Principal = Rs 3,000 Rate = 18%, Time = 2 years
(iii) Principal = Rs 5,000 Rate = 10 paise per rupee per annum, Time = 2 years
(iv) Principal = Rs 2,000, Rate = 4 paise per rupee per annum, Time = 3 years.
(v) Principal = Rs 12,800, Rate = 712 %, Time = 3 years
(vi) Principal = Rs 10,000, Rate = 20% per annum compounded half-yearly, time = 2 years
(vii) Principal = Rs 1,60,000, Rate = 10 paise per rupee per annum compounded half- yearly, Time = 2 years.
Solution:
(i) Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Time (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 1
and compound interest (C.I) = A – P = Rs 3307.50 – Rs 3,000 = Rs 307.50
(ii) Principal (P) = Rs 3,000
Rate (R) = 18% p.a.
Time (n) = 2 years

and compound interest (C.I.) = A – P = Rs 4177.20 – Rs 3,000 = Rs 1177.20
(iii) Principal (P) = Rs 5,000
Rate (R) =10 paise per rupee or 10% p.a.
Time (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 3
C.I. = A – P = Rs 6,050 – Rs 5,000 = Rs 1,050
(iv) Principal (P) = Rs 2,000
Rate (R) = 4 paise per rupee or 4% p.a.
Time (n) = 3 years

C.I. = A – P = Rs 2249.73 – Rs 2,000 = Rs 249.73
(v) Principal (P) = Rs 12,800

C.I. = A – P = Rs 15901.40 – Rs 12,800 = Rs 3101.40
(vi) Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Time = 2 years or 4 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 6
C.I. = A – P = Rs 14,641 – Rs 10,000 = Rs 4,641
(vii) Principal (P) = Rs 1,60,000
Rate (R) = 10 paise per rupee or 10% p.a. or 5% half-yearly
Time (n) = 2 years or 4 half-years.

C.I. = A – P = Rs 1,94,481 – Rs 1,60,000 = Rs 34,481

Question 2.
Find the amount of Rs 2,400 after 3 years, when the interest is compounded annually at the rate of 20% per annum.
Solution:
Principal (P) = Rs 2,400
Rate (R) = 20%
Time (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 8

Question 3.
Rahman lent Rs 16,000 to Rasheed at the rate of 1212 % per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years.
Solution:
Principal (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 9

Question 4.
Meera borrowed a sum of Rs 1,000 from Sita for two years. If the rate of interest is 10% compounded annually find the amount that Meera has to pay back.
Solution:
Amount of loan (P) = Rs 1,000
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 10

Question 5.
Find the difference between the compound interest and simple interest. On a sum of Rs 50,000 at 10% per annum for 2 years.
Solution:
Principal (P) = Rs 50,000
Rate (R) = 10% p.a.
Period (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 11
Difference between C.I. and S.I. = Rs 10,500 – Rs 10,000 = Rs 500

Question 6.
Amit borrowed Rs 16,000 at 1712 % per annum simple interest on the same day, he lent it to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years ?
Solution:
Amount of loan (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 12

C.I. = A – P = Rs 22,090 – Rs 16,000 = Rs 6,090
Now gain = C.I. – S.I = Rs 6,090 – 5,600 = Rs 490

Question 7.
Find the amount of Rs 4,096 for 18 months at 1212 % per annum, interest being compounded semi-annually ?
Solution:
Principal (P) = Rs 4,096
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 14

Question 8.
Find the amount and the compound interest on Rs 8,000 for 112 years at 10% per annum, compounded half-yearly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 10% p.a. or 5% half yearly
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 15
and C.I. = A – P = Rs 9,261 – Rs 8,000 = Rs 1,261

Question 9.
Kamal borrowed Rs 57,600 from LIC against her policy at 1212 % per annum to build a house. Find the amount that she pays LIC after 112 years if the interest is calculated half-yearly.
Solution:
Amount of loan (P) = Rs 57,600
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 16

Question 10.
Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs 64,000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.
Solution:
Price of house (P) = Rs 64,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 17
Compound interest (C.I.) = A – P = Rs 68,921 – Rs 64,000 = Rs 4,921

Question 11.
Rakesh lent out Rs 10,000 for 2 years at 20% per annum compounded annually. How much more he could earn if the interest be compounded half-yearly ?
Solution:
Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 18

C.I. = Rs 14,400 – Rs 10,000 = Rs 4,400
Now difference in C.I. = Rs 4,641 – Rs 4,400 = Rs 241

Question 12.
Romesh borrowed a sum of Rs 2,45,760 at 12.5% per annum compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest but compounded semi-annually. Find his gain after 2 years.
Solution:
In first case,
Principal (P) = Rs 2,45,760
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 20

C.I. = A – P = Rs 313203.75 – Rs 2,45,760 = Rs 67443.75
Gain = 67443.75 – Rs 65,280 = Rs2163.75

Question 13.
Find the amount that David would receive if he invests Rs 8,192 for 18 months at 1212 % per annum, the interest being compounded half-yearly.
Solution:
Principal (P) = Rs 8,192
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 22

Question 14.
Find the compound interest on Rs 15,625 for 9 months at 16% per annum, compounded quarterly.
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16% p.a. or 4% quarterly
Period (n) = 9 months or 3 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 24
Compound interest = A – P = Rs 17,576 – Rs 15,625 = Rs 1,951

Question 15.
Rekha deposited Rs 16,000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year.
Solution:
Principal (P) = Rs 16,000
Rate (R) = 20% p.a. or 5% quarterly
Period (n) = one year or 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 25
C.I. = A – P = Rs 19448.10 – Rs 16,000 = Rs 3448.10

Question 16.
Find the amount of Rs 12,500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.
Solution:
Principal (P) = Rs 12,500
Rate (R1) = 15% p.a. for first year
R2 = 16% p.a. for second year
Period = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 26

Question 17.
Ramu borrowed Rs 15,625 from a finance company to buy a scooter. If the rate of interest be 16% per annum compounded annually, what payment he will have to make after 214 years ?
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16%
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 27

Question 18.
What will Rs 1,25,000 amount to at the rate of 6% if interest is calculated after every 4 months for one year ?
Solution:
Principal (P) = Rs 1,25,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 29

Question 19.
Find the compound interest at the rate of 5% for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 12,000 as simple interest.
Solution:
In first case,
S.I. = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 30
= Rs 80,000
In second case,
Principal (P) = Rs 80,000
Rate (R) = 5% p.a.
Period (n) = 3 years

C.I. = A – P = Rs 92,610 – 80,000 = Rs 12,610

Question 20.
A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs 482 more. Find the sum.
Solution:
Let Sum (P) = Rs x
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half years
In first case,
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 32
Interests = A – P = Rs 146.41 – Rs 100 = Rs 46.41
Now difference in interests = Rs 46.41 – Rs 44.00 = Rs 2.41
If difference is 2.41 then sum is 100 If difference is Rs 482, then sum

Question 21.
Simple interest on a sum of money for 2 years at 612 % per annum is Rs 5,200. What will be the compound interest on the sum at the same rate for the same period ?
Solution:
In first case,
S.I. = Rs 5,200
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 34

Compound interest = A – P = Rs 45,369 – Rs 40,000 = Rs 5,369

Question 22.
Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 1,200 as simple interest.
Solution:
In first case,
S.I. = Rs 1,200
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 36
In second case,
Principal (P) = Rs 8,000
Rate (R) = 5% p.a.
Period (n) = 3 years

= Rs 9,261
C.I. = A – P = Rs 9,261 – Rs 8000 = Rs 1,261

Exercise 14.3

Question 1.
On what sum will the compound interest at 5% p.a. annum for 2 years compounded annually be Rs 164 ?
Solution:
Let Principal (P) = Rs 100
Rate (R) = 5% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 1

Question 2.
Find the principal of the interest compounded annually at the rate of 10% for two years is Rs 210.
Solution:
Let principal (P) = Rs 100
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 3

Question 3.
A sum amounts to Rs 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.
Solution:
Amount (A) = Rs 756.25
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 5

Question 4.
What sum will amount to Rs 4913 in 18 months, if the rate of interest is 1212 % per annum, compounded half-yearly.
Solution:
Amount (A) = Rs 4,913
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 6

Question 5.
The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum.
Solution:
Let sum (P) = Rs 100
Rate (R) = 15% p.a.
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 7

Question 6.
Rachna borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1,290 as interest compounded annually, find the sum she borrowed.
Solution:
C.I. = Rs 1,290
Rate (R) = 15% p.a.
Period (n) = 2 years
Let sum (P) = Rs 100
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 10

Question 7.
The interest on a sum of Rs 2,000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.
Solution:
Sum (P) = Rs 2,000
C.I. = Rs 163.20
Amount (A) = P + C.I. = Rs 2000 + Rs 163.20 = Rs 2163.20
Rate (R) = 4%
Let period = n years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 11

Question 8.
In how much time would Rs 5,000 amount to Rs 6,655 at 10% per annum compound interest ?
Solution:
Principal (P) = Rs 5,000
Amount (A) = Rs 6,655
Rate (R) = 10%
Let period = n years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 13

Question 9.
In what time will Rs 4,400 becomes Rs 4,576 at 8% per annum interest compounded half-yearly ?
Solution:
Principal (P) = Rs 4,400
Amount (A) = Rs 4,576
Rate (R) = 8% or 4% half-yearly
Let period = n half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 14
n = 1
Period = 1 half year

Question 10.
The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum.
Solution:
Difference between C.I. and S.I. = Rs 20
Rate (R) = 4% p.a.
Period (n) = 2 years
Let principal (P) = Rs 100
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 15

Question 11.
In what time will Rs 1,000 amount to Rs 1,331 at 10% per annum compound interest.
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1,331
Rate (R) = 10% p.a.
Let period = n year
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 17

Question 12.
At what rate percent compound interest per annum will Rs 640 amount to Rs 774.40 in 2 years ?
Solution:
Principal (P) = Rs 640
Amount (A) = Rs 774.40
Period (n) = 2 years.
Let R be the rate of interest p.a.
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 18

Question 13.
Find the rate percent per annum if Rs 2000 amount to Rs 2,662 in 112 years, interest being compounded half-yearly ?
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2,662
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 20

Question 14.
Kamala borrowed from Ratan a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years, she received Rs 210 as compound interest, but paid Rs 200 only as simple interest. Find the sum and the rate of interest.
Solution:
Simple interest = Rs 200
and compound interest = Rs 210.
Period = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 21

Question 15.
Find the rate percent per annum, if Rs 2,000 amount to Rs 2,315.25, in an year and a half, interest being compounded six monthly.
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2315.25
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 22

Question 16.
Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.
Solution:
Let Principal (P) = Rs 100
then Amount (A) = Rs 200
Period (n) = 3 years
Let R be the rate % p.a.
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 24

Question 17.
Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half- yearly.
Solution:
Let Principal (P) = Rs 100
Then Amount (A) = Rs 400
Period (n) = 2 years or 4 half years
Let R be the rate % half-yearly, then
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 25
Rate % = 41.42% half yearly and 82.84% p.a.

Question 18.
A certain sum amounts to Rs 5,832 in 2 years at 8% compounded interest. Find the sum.
Solution:
Amount (A) = Rs 5,832
Let P be the sum
Rate (R) = 8% p.a.
Period (n) = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 26

Question 19.
The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum.
Solution:
Let sum (P) = Rs 100
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 27

Question 20.
The difference in simple interest and compound interest on a certain sum of money at 623 % per annum for 3 years is Rs 46. Determine the sum.
Solution:
Let sum (P) = Rs 100
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 29

Question 21.
Ishita invested a sum of Rs 12,000 at 5% per annum compound interest. She received an amount of Rs 13,230 after n years, Find the value of n.
Solution:
Principal (P) = Rs 12,000
Amount (A) = Rs 13,230
Rate (R) = 5% p.a.
Period = n years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 32

Question 22.
At what rate percent per annum will a sum of Rs 4,000 yield compound interest of Rs 410 in 2 years ?
Solution:
Principal (P) = Rs 4,060
C.I. = Rs 410
Amount (A) = Rs 4,000 + 410 = Rs 4,410
Let rate = R % p.a.
Period (n) = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 33

Question 23.
A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years. Find the sum deposited.
Solution:
Amount (A) = Rs 10,404
Rate (R) = 2% p.a.
Period (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 18

Question 24.
In how much time will a sum of Rs 1,600 amount to Rs 1852.20 at 5% per annum compound interest ?
Solution:
Principal (P) = Rs 1,600
Amount (A) = Rs 1852.20
Rate (R) = 5% p.a.
Let n be the time
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 20

Question 25.
At what rate percent will a sum of Rs 1,000 amount to Rs 1102.50 in 2 years at compound interest ?
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1102.50 .
Period (n) = 2 years
Let R be the rate of interest p.a.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 21

Question 26.
The compound interest on Rs 1,800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.
Solution:
Principal (P) = Rs 1,800
C.I. = Rs 378
Amount (A) = P + C.I. = Rs 1,800 + 378 = Rs 2,178
Rate (R) = 10% p.a.
Let n be the period in years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 22
Comparing, we get:
n = 2
Period = 2 years

Question 27.
What sum of money will amount to Rs 45582.25 at 634 % per annum in two years, interest being compounded annually ?
Solution:
Amount (A) = Rs 45582.25
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 23

Question 28.
Sum of money amounts to Rs 4,53,690 in 2 years at 6.5% per annum compounded annually. Find the sum.
Solution:
Amount (A) = Rs 4,53,690
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 24

Exercise 14.4

Question 1.
The present population of a town is 28,000. If it increases at the rate of 5% per annum, what will be its population after 2 years ?
Solution:
Present population = 28000
Rate of increase (R) = 5% p.a.
Period (n) = 2 years
Population after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 1

Question 2.
The population of a city is 125000. If the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of city after 3 years.
Solution:
Present population = 125000
Rate of birth = 5.5%
and rate of death = 3.5%
Increase = 5.5 – 3.5 = 2% p.a.
Period = 3 years.
Population after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 2

Question 3.
The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.
Solution:
Present population = 25000
Increase in first year = 4%
in second year= 5% and
in third year = 8%
Population after 3 years =
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 4

Question 4.
Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.
Solution:
Three years ago,
Population of a town = 50000
Annual increase in population in first year = 4%
in second year = 5%
and in third year = 3%
Present population
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 5

Question 5.
There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9261, what it was 3 years ago ?
Solution:
Let 3 years ago, population = P
Present population = 9261
Rate of increase (R) = 5% p.a.
Period (n) = 3 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 6

Question 6.
In a factory, the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.
Solution:
Production of scooters 3 years ago (P) = 40000
Present production (A) = 46305
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 7

Question 7.
The annual rate of growth in population of a certain city is 8%. If its present population is 196830, what it was 3 years ago ?
Solution:
Let 3 years ago, the population of a city = P
Rate of growth (R) = 8% p.a.
Present population = 196830
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 9

Question 8.
The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.
Solution:
Population after 2 years = 22050
Rate of increase = 50 per thousand
Period (n) = 2 years
Let present population = P, then
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 10

Question 9.
The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours ?
Solution:
Present count of bacteria = 13125000
In first hour increase = 10%
decrease in second hour = 8%
increase in third hour = 12%
Count of bacteria after 3 hours
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 11

Question 10.
The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000 ?
Solution:
On the last day of 1998,
Population of a town = 72000
In the first year, increase = 7%
In the second year, decrease = 10%
Population in the last day of 2000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 12

Question 11.
6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year ?
Solution:
Number of workers at the beginning = 6400
Period = 4 years.
At the end of 1st year, workers retrenched = 25%
At the end of second year, workers retrenched = 25%
At the end of third year, workers increased = 25%
Total number of workers during the 4 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 13

Question 12.
Aman started a factory with an initial investment of Rs 1,00,000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.
Solution:
Initial investment = Rs 100000
Loss in the first year = 5%
Profit in the second year = 10%
Profit in the third year = 12%
Investment at the end of 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 14

Question 13.
The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago ?
Solution:
Present population (A) = 175760
Increase rate = 40 per 1000
Period = 3 years
Let 3 years ago,
population was = P
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 15

Question 14.
The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years ?
Solution:
Production of Mixi in 1996 = 8000
Increase in next 2 years = 15%
Decrease in the third year = 5%
Production after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 16

Question 15.
The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997.
Solution:
Population of a city in 1999 = 6760000
Increase = 4%
(i) Population in 2001 is after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 17

Question 16.
Jitendra set up a factory by investing Rs 25,00,000. During the first two successive years his profits were 5% and f 10% respectively. If each year the profit was on previous year’s capital, compute his total profit.
Solution:
Investment in the beginning = Rs 25,00,000
Profit during the first 2 years = 5% and 10% respectively
Investment after 2 years will be
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 19
= Rs 28,87,500
Amount of profit = Rs 28,87,500 – Rs 25,00,000 = Rs 3,87,500

Exercise 14.5

Question 1.
Mr. Cherian purchased a boat for Rs 16,000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.
Solution:
Cost of boat = Rs 16,000
Rate of depreciating = 5% p.a.
Period = 2 years
Value of boat after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 1

Question 2.
The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 1,0,000 ? Also, find the total depreciation during this period.
Solution:
Present value of machine = Rs 1,00,000
Rate of depreciation = 10% p.a.
Period (n) = 2 years
Value of machine after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 2

Question 3.
Pritam bought a plot of land for Rs 6,40,000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years ?
Solution:
Present value of plot = Rs 6,40,000
Increase = 5% per half year
Period (n) = 2 years or 4 half years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 3

Question 4.
Mohan purchased a house for Rs 30,000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.
Solution:
Present value of the house (P) = Rs 30,000
Rate of depreciation = 25% p.a.
Period (n) = 3 years
Value of house after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 5

Question 5.
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43,740, find its purchased price.
Solution:
Let the purchase price of machine = Rs P
Rate of depreciation = 10% p.a.
Period (n) = 3 years.
and present value = Rs 43,740
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 6

Question 6.
The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9,680, for how much was it purchased ?
Solution:
Let the refrigerator was purchased for = Rs P
Rate of depreciation (R) = 12% p.a.
Period (n) = 2 years
and present value (A) = Rs 9,680
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 8

Question 7.
The cost of a TV set was quoted Rs 17,000 at the beginning of 1999. In the beginning of2000, the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What is the cost of the TV set in 2001 ?
Solution:
List price of TV set in 1999 = Rs 17,000
Rate of hike in 2000 = 5%
Rate of decrease in 2001 = 4%
Price of TV set in 2001
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 9

Question 8.
Ashish started the business with an initial investment of Rs 5,00,000. In the first year, he incurred a loss of 4%. However, during the second year he earned a profit of 5% which in third year, rose to 10%. Calculate the net profit for the entire period of 3 years.
Solution:
Initial investment = Rs 5,00,000
In the first year, rate of loss = 4%
In the second year, rate of gain = 5%
and in the third year, rate of gain = 10%
Investment after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 10

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RD SHARMA SOLUTION CHAPTER –13 Profit, Loss, Discount and Value Added Tax (VAT) | CLASS 8TH MATHEMATICS-EDUGROWN

Exercise 13.1

Question 1.
A student buys a pen for Rs. 90 and sells it for Rs. 100. Find his gain and gain percent ?
Solution:
C.P. of a pen = Rs. 90
and S.P. = Rs. 100
Gain = S.P. – C.P. = Rs. 100 – 90 (S.P > C.P.)
= Rs. 10
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 1

Question 2.
Rekha bought a saree for Rs. 1240 and sold it for Rs. 1147. Find her loss and loss percent.
Solution:
C.P. of saree.= Rs. 1240 and
S.P. = Rs. 1147
Loss = C.P – S.P. = Rs. 1240 – Rs. 1147 (C.P. > S.P.)
= Rs. 93
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 2

Question 3.
A boy buys 9 apples for Rs. 9.60 and sells them at 11 for Rs. 12. Find his gain or loss percent.
Solution:
L.C.M. of 9 and 11 = 99
Let 99 apples were purchased.
C.P. of 99 apples at the rate of 9 apples for Rs. 9.60
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 3

Question 4.
The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percent.
Solution:
C.P. of 10 articles = S.P. of 9 articles = 90 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 4

Question 5.
A retailer buys a radio for Rs. 225. His overhead expenses are Rs. 15. If he sells the radio for Rs. 300, determine his profit percent.
Solution:
Cost of radio = Rs. 225
Over head expenses = Rs. 15
Total C.P. of the ratio = Rs. 225 + 15 = Rs. 240
S.P. of radio = Rs. 300
Gain = S.P. – C.P. = Rs. 300 – Rs. 240 = Rs. 60
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 5

Question 6.
A retailer buys a cooler for Rs. 1200 and overhead expenses on it are Rs. 40. If he sells the cooler for Rs. 1550, determine his profit percent.
Solution:
Cost of cooler = Rs. 1200
Overhead expenses = Rs. 40
Total cost price of cooler = Rs. 1200 + Rs. 40 = Rs. 1240
Selling price = Rs. 1550
Gain = S.P. – C.P. = Rs. 1550 – Rs. 1240 = Rs. 310
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 6

Question 7.
A dealer buys a wristwatch for Rs. 225 and spends Rs. 15 on its repairs. If he sells the same for Rs. 300, find his profit percent.
Solution:
Cost of wristwatch = Rs. 225
Cost on repairs = Rs. 15
Total cost price = Rs. 225 + Rs. 15 = Rs. 240
Selling price = Rs. 300
Gain = S.P. – C.P. = Rs. 300 – Rs. 240 = Rs. 60
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 7

Question 8.
Ramesh bought two boxes for Rs. 1300. He sold one box at a profit of 20% and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.
Solution:
Total cost price of two boxes = Rs. 1300
S.P. of each box is same.
Let S.P of each box = Rs. 100
S.P. of first box = Rs. 100
Gain = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 8

Question 9.
If the selling price of 10 pens is equal to cost price of 14 pens, find the gain percent.
Solution:
S.P. of 10 pens = C.P. of 14 pens = Rs. 100 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 10

Question 10.
If the cost price of 18 chairs be equal to selling price of 16 chairs, And the gain or loss percent.
Solution:
C.P. of 18 chairs = S.P. of 16 chairs = Rs. 100 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 11

Question 11.
If the selling price of 18 oranges is equal to the cost price of 16 oranges, And the loss percent.
Solution:
S.P. of 18 oranges = C.P. of 16 oranges = Rs. 100 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 13

Question 12.
Ravish sold his motorcycle to Vineet at a loss of 28%. Vineet spent Rs. 1680 on its repairs and sold the motor cycle to Rahul for Rs. 35910, thereby making a profit of 12.5%, find the cost price of the motor cycle for Ravish.
Solution:
Cost price of cycle for Rahul or Selling price for Vineet = Rs. 35910
Gain = 12.5%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 14

Question 13.
By selling a book for Rs. 258, a book-seller gains 20%. For how much should he sell it to gain 30% ?
Solution:
S.P. of a book = Rs. 258
Gain = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 16

Question 14.
A defective briefcase costing Rs. 800 is being sold at a loss of 8%. If its price is further reduced by 5%, find its selling price.
Solution:
C.P. of a briefcase = Rs. 800
Loss = 8%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 17

Question 15.
By selling 90 ball pens for Rs. 160, a person losses 20%. How many ball pens should be sold for Rs. 96 so as to have a profit of 20%.
Solution:
S.P of 90 ball pens = Rs. 160
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 19

Question 16.
A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs. 36.75 less, he would have gained 30%. Find the cost price of the article.
Solution:
Let C.P. of the article = Rs. 100
In first case gain = 25%
S.P. = 100 + 25 = Rs. 125
In second case,
C.P. = 20% less of Rs. 100 = 100 – 20 = Rs. 80
Gain = 30%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 20

Question 17.
A dishonest shopkeeper professes to sell pulses at his cost price but uses a false weight of 950 for each kilogram. Find his gain percent.
Solution:
Let C.P. of 1 kg of pulses = Rs. 100
S.P. of 950 gm = Rs. 100
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 21

Question 18.
A dealer bought two tables for Rs. 3120. He sold one of them at a loss of 15% and other at a gain of 36%. Then he found that each table was sold for the same price, find the cost price of each table.
Solution:
Cost price of two tables = Rs. 3120
Let S.P of each table = Rs. 100
Now S.P of one table = Rs. 100
Loss = 15%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 23

Question 19.
Mariam bought two fans for Rs. 3605. She sold one at a profit of 15% and the other at a loss of 9%. If Mariam obtained the same amount for each fan, find the cost price of each fan.
Solution:
Total cost price of two fans = Rs. 3605
Let selling price of each fan = Rs. 100
Now S.P. of first fan = Rs. 100
Profit = 15%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 25

Question 20.
Some toffees are bought at the rate of 11 for Rs. 10 and the same number at the rate of 9 for Rs. 10. If the whole lot is sold at one rupee per toffee, find the gain or loss percent on the whole transaction.
Solution:
L.C.M. of 11 and 9 = 99
Let each time 99 toffees are bought.
In first case, the C.P. of 99 toffees at the rate of 11 for Rs. 10
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 26

Question 21.
A tricycle is sold at a gain of 16%. Had it been sold for Rs. 100 more, the gain would have been 20%. Find the C.P. of the tricycle.
Solution:
Let the C.P. of tricycle = Rs. 100
In first case, gain = 16%
S.P. = Rs. 100 + 16 = Rs. 116
In second case, gain = 20%
S.P. = Rs. 100 + 20 = Rs. 120
Difference in S.P.’s = Rs. 120 – Rs. 116 = Rs. 4
If difference is Rs. 4,
then C.P. of the tricycle = Rs. 100
and if difference is Re. 1, then C.P.
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 27

Question 22.
Shabana bought 16 dozen ball pens and sold then at a loss equal to S.P. of 8 ball pens. Find
(i) her loss percent
(ii) S.P. of 1 dozen ball pens, if she purchased these 16 dozen ball pens for Rs. 576.
Solution:
C.P. of 16 dozen ball pens = S.P. of 16 dozen pens – loss
C.P. of 16 x 12 pens = S.P. of 16 dozen pens x S.P. of 8 pens
C.P. of 192 pens = S.P. of 16 x 12 pens x S.P. of 8 pens
S.P. of 192 pens + S.P. of 8 pens = S.P. of 200 pens
(i) Let C.P. of 1 pen = Re. 1
Then C.P. of 192 pens = Rs. 192
and S.P. of 200 pens = Rs. 192
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 28

Question 23.
The difference between two selling prices of a shirt at profits of 4% and 5% is Rs. 6. Find
(i) C.P. of the shirt
(ii) The two selling prices of the shirt
Solution:
The difference of two selling prices of shirt = Rs. 6
Difference in profits of 4% and 5% = 5 – 4 = 1%
(i) C.P. = 1 x 6 x 100 = Rs. 600
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 29

Question 24.
Toshiba bought 100 hens for Rs. 8000 and sold 20 of these at a gain of 5%. At what gain percent she must sell the remaining hens so as to gain 20% on the whole ?
Solution:
Total number of hens bought = 100
C.P. of 100 hens = Rs. 8000
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 30

Exercise 13.2

Question 1.
Find the S.P. If
(i) M.P. = Rs. 1300 and Discount = 10%
(ii) M.P. = Rs. 500 and Discount = 15%
Solution:
(i) M.P. = Rs. 1300, Discount = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 1

Question 2.
Find the M.P. If
(i) S.P. = Rs. 1222 and Discount = 6%
(ii) S.P. = Rs. 495 and Discount = 1%
Solution:
(i) S.P. = Rs. 1222, discount = 6%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 4

Question 3.
Find discount in percent when
(i) M.P. = Rs. 900 and S.P. = Rs. 873
(ii) M.P. = Rs. 500 and S.P. = Rs. 425
Solution:
(i) M.P. = Rs. 900
S.P. = Rs. 873
Discount = M.P. – S.P. = Rs. 900 – Rs. 873 = Rs. 27
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 5
(ii) M.P. = Rs. 500
S.P. = Rs. 425
Discount M.P. – S.P. = Rs. 500 – Rs. 425 = = Rs. 75

Question 4.
A shop selling sewing machines offers 3% discount on ail cash purchases. What cash amount does a customer pay for a sewing machine, the price of which is marked as Rs. 650.
Solution:
Marked price (M.P.) of one sewing machine = Rs. 650
Rate of discount = 3%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 7

Question 5.
The marked price of a ceiling fan is Rs. 720. During off season, it is sold for Rs. 684. Determine the discount percent.
Solution:
Marked price (M.P.) of fan = Rs. 720
Sale price (S.P.) = Rs. 684
Amount of discount = M.P. – S.P. = Rs. 720 – Rs. 684 = Rs. 36
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 8

Question 6.
On the eve of Gandhi Jayanti, a saree is sold for Rs. 720 after allowing 20% discount. What is the marked price ?
Solution:
S.P. of saree = Rs. 720
Rate of discount = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 9

Question 7.
After allowing a discount of 712 % on the marked price, an article is sold for Rs. 555. Find its marked price.
Solution:
Selling price (S.P.) = Rs. 555
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 10

Question 8.
A shopkeeper allows his customers 10% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs. 250?
Solution:
Marked price = Rs. 250
Discount allowed = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 11

Question 9.
A shopkeeper allows 20% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs. 500 ?
Solution:
Marked price (M.P.) of an article = Rs. 500
Discount allowed = 20%
Selling price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 13

Question 10.
A tradesman marks his goods at such a price that after allowing a discount of 15%, he makes a profit of 20%. What is the marked price of an article whose cost price is Rs 170 ?
Solution:
Rate of discount = 15% gain = 20%
Cost price (C.P.) of an article = Rs 170
Selling price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 14

Question 11.
A shopkeeper marks his goods in such a way that after allowing a discount of 25% on the marked price, he still makes a profit of 50%. Find the ratio of the C.P. to the M.P.
Solution:
Let cost price (C.P.) = Rs 100
Profit = 50%
Selling Price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 15

Question 12.
A cycle dealer offers a discount of 10% and still makes a profit of 26%. What is the actual cost to him of a cycle whose marked price is Rs 840 ?
Solution:
Rate of discount = 10%
Gain = 26%
Marked Price (M.P.) = Rs 840
Selling Price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 16

Question 13.
A shopkeeper allows 23% commission on his advertised price and still makes a profit of 10%. If he gains Rs 56 on one item, find his advertised price.
Solution:
Rate of commission = 23%
Profit = 10%
Total gain = Rs 56
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 17

Question 14.
A shopkeeper marks his goods at 40% above the cost price but allows a discount of 5% for cash payment to his customers. What actual profit does he make, if he receives Rs 1064 after paying the discount ?
Solution:
Let cost price (C.P.) = Rs 100
Marked price = Rs 100 + 40 = Rs 140
Rate of discount = 5%
Selling price (S.P)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 18

Question 15.
By selling a pair of ear rings at a discount of 25% on the marked price, a jeweller makes a profit of 16%. If the profit is Rs 48, what is the cost price ? What is the marked price and the price at which the pair was eventually bought ?
Solution:
Total profit = Rs 48
Profit percent = 16%
Cost price = 48×10016 = Rs 300
Selling Price = C.P. + profit = Rs 300 + Rs 48 = Rs 348
Rate of discount = 25%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 19

Question 16.
A publisher gives 32% discount on the printed price of a book to booksellers. What does a bookseller pay for a book whose printed price is Rs 275 ?
Solution:
Printed price of a book = Rs 275
Rate of discount = 32%
Selling price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 20

Question 17.
After allowing a discount of 20% on the marked price of a lamp, a trader loses 10%. By what percentage is the marked price above the cost price ?
Solution:
Rate of discount = 20%
Loss = 10%
Let the cost price of the lamp = Rs 100
Loss = 10%
Selling price = Rs 100 – 10 = Rs 90
Rate of discount = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 21

Question 18.
The list price of a table fan is Rs 480 and it is available to a retailer at 25% discount. For how much should a retailer sell it to gain 15% ?
Solution:
List price of table fan (M.P.) = Rs 480
Rate of discount = 25%
Selling price (S.P)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 22

Question 19.
Rohit buys an item at 25% discount on the marked price. He sells it for Rs 660, making a profit of 10%. What is the marked price of the item ?
Solution:
Rate of discount = 25%
Selling price (S.P.) for Rohit = Rs 660
Profit = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 23

Question 20.
A cycle merchant allows 20% discount on the marked price of the cycles and still makes a profit of 20%. If he gains Rs 360 over the sale of one cycle, find the marked price of the cycle.
Solution:
Rate of discount = 20%
Profit = 20%
Total gain = Rs 360
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 25

Question 21.
Jyoti and Meena run a ready-made garment shop. They mark the garments at such a price that even after allowing a discount of 12.5%, they make a profit of 10%. Find the marked price of a suit which costs them Rs 1470.
Solution:
Rate of discount = 12.5%
Profit = 10%
Cost price = Rs 1470
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 26

Question 22.
What price should Aslam mark on a pair of shoes ? Which costs him Rs 1200 so as to gain 12% after allowing a discount of 16% ?
Solution:
Cost price of shoes = Rs 1200
Gain = 12%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 28

Question 23.
Jasmine allows 4% discount on the marked price of her goods and still earns a profit of 20%. What is the cost price of a shirt for her marked at Rs 850 ?
Solution:
Marked price of a shirt = Rs 850
Discount = 4%
Selling price
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 29

Question 24.
A shopkeeper offers 10% off-season discount to the customers and still makes a profit of 26%. What is the cost price for the shopkeeper on a pair of shoes marked at Rs 1120 ?
Solution:
Marked price = Rs 1120
Rate of discount = 10%
S.P. of shoes
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 31

Question 25.
A lady shopkeeper allows her customers 10% discount on the marked price of the goods and still gets a profit of 25%. What is the cost price of a fan for her marked at Rs 1250 ?
Solution:
Marked price (M.P.) of fan = Rs 1250
Discount = 10%
S.P. of the fan
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 32

Exercise 13.3

Question 1.
The list price of a refrigerator is Rs 9700. If a value added tax of 6% is to be charged on it, how much one has to pay to buy the refrigerator ?
Solution:
List price of refrigerator = Rs 9700
Rate of VAT = 6%
Amount of VAT = Rs. 9700×6100 = Rs 582
Total price to be paid = Rs 9700 + 582 = Rs 10282

Question 2.
Vikram bought a watch for Rs 825. If this amount includes 10% VAT on the list price, what was the list price of the watch ?
Solution:
Price of watch including VAT = Rs 825
Rate of VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 1

Question 3.
Aman bought a shirt for Rs 374.50 which includes 7% VAT. Find the list price of the shirt.
Solution:
Cost price of the shirt = Rs 374.50
Rate of VAT = 7%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 2

Question 4.
Rani purchases a pair of shoes whose sale price is Rs 175. If he pays a VAT at the rate of 7%, how much amount does he pays as VAT. Find the net value of the pair of shoes.
Solution:
Sale price of a pair of shoes = Rs 175
Rate of VAT = 7%
Total VAT paid = Rs. 175×7100 = Rs 12.25
and total amount paid = Rs 175 + Rs 12.25 = Rs 187.25https://googleads.g.doubleclick.net/pagead/ads?client=ca-pub-7601472013083661&output=html&h=280&adk=2523109437&adf=980553861&pi=t.aa~a.4089255474~i.16~rp.4&w=750&fwrn=4&fwrnh=100&lmt=1642576099&num_ads=1&rafmt=1&armr=3&sem=mc&pwprc=1894297687&psa=1&ad_type=text_image&format=750×280&url=https%3A%2F%2Fwww.learninsta.com%2Frd-sharma-class-8-solutions-chapter-13-profits-loss-discount-and-value-added-tax-vat-ex-13-3%2F&flash=0&fwr=0&pra=3&rh=188&rw=750&rpe=1&resp_fmts=3&wgl=1&fa=27&adsid=ChEIgN6ykAYQlqfqreX-0tatARI5AOa2GRQuFTRuP5gxgF63-8ooeukU8cZ7QCQZcMs0Bo08KBiUSCO_wUJ4DWFWZv_YD9YpnCl_xW2w&uach=WyJXaW5kb3dzIiwiMTAuMC4wIiwieDg2IiwiIiwiOTguMC40NzU4LjEwMiIsW10sbnVsbCxudWxsLCI2NCIsW1siIE5vdCBBO0JyYW5kIiwiOTkuMC4wLjAiXSxbIkNocm9taXVtIiwiOTguMC40NzU4LjEwMiJdLFsiR29vZ2xlIENocm9tZSIsIjk4LjAuNDc1OC4xMDIiXV1d&dt=1645025146248&bpp=2&bdt=6721&idt=2&shv=r20220214&mjsv=m202202090101&ptt=9&saldr=aa&abxe=1&cookie=ID%3D1be37a61a5e7cca6-2207ff21a8d0005c%3AT%3D1643370310%3ART%3D1645025060%3AS%3DALNI_MZlVwsQQNfk-mglMlxYK4Mv7pUwgg&prev_fmts=0x0%2C750x280%2C750x280&nras=3&correlator=3657959876739&frm=20&pv=1&ga_vid=528868594.1643370309&ga_sid=1645025145&ga_hid=193832627&ga_fc=1&u_tz=330&u_his=4&u_h=715&u_w=1270&u_ah=681&u_aw=1270&u_cd=24&u_sd=1.513&dmc=8&adx=57&ady=2221&biw=1254&bih=610&scr_x=0&scr_y=667&eid=42531397%2C44750773%2C21067496&oid=2&pvsid=1097847813229336&pem=972&tmod=1825076183&uas=3&nvt=1&ref=https%3A%2F%2Fwww.learninsta.com%2Frd-sharma-class-8-solutions%2F&eae=0&fc=1408&brdim=0%2C0%2C0%2C0%2C1270%2C0%2C1270%2C681%2C1270%2C610&vis=1&rsz=%7C%7Cs%7C&abl=NS&fu=128&bc=31&jar=2022-02-16-15&ifi=4&uci=a!4&btvi=3&fsb=1&xpc=BHm11P10xT&p=https%3A//www.learninsta.com&dtd=65

Question 5.
Swarna paid Rs 20 as VAT on a pair of shoes worth Rs 250. Find the rate of VAT.
Solution:
Amount of VAT paid = Rs 20
Price of the pair of shoes = Rs 250
Rate of VAT = 20×100250 = 8%

Question 6.
Sarita buys goods worth Rs 5,500. She gets a rebate of 5% on it. After getting the rebate if VAT at the rate of 5% is charged, find the amount she will have to pay for the goods.
Solution:
Price of goods = Rs 5,500
Rate of rebate = 5%
Sales price after rebate
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 3
= Rs 5,225
Rate of VAT = 5%
Amount of VAT

Amount to be paid after paying VAT = Rs 5,225 + 261.25 = Rs 5486.25

Question 7.
The cost of furniture inclusive of VAT is Rs 7,150. If the rate of VAT is 10%, find the original cost of the furniture.
Solution:
Cost of furniture including VAT = Rs 7,150
Rate of VAT = 10%
Original cost of furniture
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 5

Question 8.
A refrigerator is available for Rs 13,750 including VAT. If the rate of VAT is 10%, find the original cost of refrigerator.
Solution:
Cost of refrigerator including VAT = Rs 13,750
Rate of VAT = 10%
Actual price of refrigerator
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 6

Question 9.
A colour TV is available for Rs 13,440 inclusive of VAT. If the original cost of TV is Rs 12,000, find the rate of VAT.
Solution:
Cost of TV including VAT = Rs 13,440
Actual cost = Rs 12,000
Amount of VAT = Rs 13,440 – Rs 12,000 = Rs 1,440
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 7

Question 10.
Reena goes to a shop to buy a radio, costing Rs 2,568. The rate of VAT is 7%. She tells the shopkeeper to reduce the price of the radio such that she has to pay Rs 2,568 inclusive of VAT. Find the reduction needed in the price of radio.
Solution:
Price of radio inclusive of VAT = Rs 2,568
Rate of VAT = 7%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 8
But the cost of radio in the beginning = Rs 2,568
Reduction = Rs 2568 – Rs 2,400 = Rs 168

Question 11.
Rajat goes to a departmental store and buys the following articles :
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 9
Calculate the total amount he has to pay to the store.
Solution:
Price of 2 pairs of shoes @ Rs 800 = Rs800 x 2 = Rs 1,600
Rate of VAT = 5%

Total amount to be paid = Rs 1,680 + Rs 1,590 + Rs 1,352 = Rs 4,622

Question 12.
Ajit buys a motorcycle for Rs 17,600 including value added tax. If the rate of VAT is 10%, what is the sale price of the motorcycle ?
Solution:
Cost price of motorcycle (including VAT) = Rs 17,600
Rate of VAT = 10%
Sale price of motorcycle
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 12

Question 13.
Manoj buys a leather coat costing Rs 900 at Rs 990 after paying the VAT. Calculate the VAT charged on the cost.
Solution:
Cost price of coat = Rs 900
And sale price including VAT = Rs 990
Amount of VAT = Rs 990 – Rs 900 = Rs 90
Rate of VAT = 90×100900 = 10%

Question 14.
Rakesh goes to a departmental store and purchases the following articles:
(i) biscuits and bakery products costing Rs 50, VAT @ 5%.
(ii) medicines costing Rs 90. VAT @ 10%,
(iii) clothes costing Rs 400, VAT @ 1% and
(iv) cosmetics costing Rs 150, VAT @ 10% Calculate the total amount to be paid by Rakesh to the store.
Solution:
(i) Cost of biscuits and bakery product = Rs 50
VAT = 5%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 13
Total amount of the bill = Rs 52.50 + Rs 99 + Rs 404 + Rs 165 = Rs 720.50

Question 15.
Rajeeta purchased a set of cosmetics. She paid Rs 165 for it including VAT. If the rate of VAT is 10%, find the sale price of the set.
Solution:
Total price of set including VAT = Rs 165
Rate of VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 14

Question 16.
Sunita purchases a bicycle for Rs 660. She has paid a VAT of 10%. Find the list price of bicycle.
Solution:
Cost price of bicycle including VAT = Rs. 660
Rate of VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 15

Question 17.
The sales price of a television inclusive of VAT is Rs 13,500. If VAT is charged at the rate of 8% of the list price, find the list price of the television.
Solution:
Sale price of television including VAT = Rs 13,500
Rate of VAT = 8%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 16

Question 18.
Shikha purchased a car with a marked price of Rs 2,10,000 at a discount of 5%. If VAT is charged at the rate of 10%, find the amount Shikha had paid for purchasing the car.
Solution:
Marked price of car = Rs 2,10,000
Rate of discount = 5%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 18

Question 19.
Sharuti bought a set of cosmetic items for Rs 345 including 15% value added tax and a purse for Rs 110 including 10% VAT. What percent is the VAT charged on the whole transactions ?
Solution:
Cost price of set = Rs 345
VAT = 15%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 19

Question 20.
List price of a cooler is Rs 2,563. The rate of VAT is 10%. The customer requests the shopkeeper to allow a discount in the price of the cooler to such an extent that the price remains Rs 2,563 inclusive of VAT. Find the discount in the price of the cooler.
Solution:
List price of cooler = Rs. 2,563
On request the price of cooler is paid Rs 2,563 including VAT
VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 21
Amount of discount = Rs 2,563 – Rs 2,330 = Rs 233

Question 21.
List price of a washing machine is Rs 9,000. If the dealer allows a discount of 5% on the cash payment, how much money will a customer pay to the dealer in cash, if the rate of VAT is 10%.
Solution:
List price of washing machine = Rs 9,000
Rate of discount = 5%
Amount after giving discount
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 22

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CHAPTER – 1 Nutrition in Plants | CLASS 7TH |NCERT SCIENCE IMPORTANT QUESTIONS & MCQS | EDUGROWN

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 1 Nutrition in Plants

MCQs

Question 1.
Which of the following is a nutrient?
(a) Protein
(b) Fat
(c) Vitamin
(d) All of these

Answer

Answer: (d) All of these

Question 2.
Human beings can be categorised as
(a) heterotrophs
(b) autotrophs
(c) parasites
(d) saprotrophs

Answer

Answer: (a) heterotrophs

Question 3.
The food making process in plants is called as
(a) glycolysis
(b) photosynthesis
(c) photolysis
(d) chemosynthesis

Answer

Answer: (b) photosynthesis

Question 4.
Which part of the plant is called its food factory ?
(a) Fruits
(b) Seeds
(c) Leaves
(d) Flowers

Answer

Answer: (c) Leaves

Question 5.
Tiny pores present on the surface of leaves through which gaseous exchange occurs are called
(a) stomata
(b) guard cells
(c) food holes
(d) gas holes

Answer

Answer: (a) stomata

Question 6.
What is the ultimate source of energy for all living organisms?
(a) Water energy
(b) Wind energy
(c) Solar energy
(d) Chemical energy

Answer

Answer: (c) Solar energy

Question 7.
Green pigment present in the leaves is called
(a) haemoglobin
(b) globulin
(c) albumin
(d) chlorophyll

Answer

Answer: (d) chlorophyll

Question 8.
During photosynthesis plants
(a) take oxygen and release carbon dioxide
(b) take carbon dioxide and release oxygen
(c) take carbon dioxide but do not release oxygen
(d) take oxygen but do not release carbon dioxide

Answer

Answer: (b) take carbon dioxide and release oxygen

Question 9.
During photosynthesis
(а) solar energy is converted into chemical energy
(b) solar energy is converted into mechanical energy
(c) chemical energy is converted into mechanical energy
(d) bioenergy is converted into chemical energy

Answer

Answer: (а) solar energy is converted into chemical energy

Question 10.
The raw materials for photosynthesis are
(а) CO₂
(b) CO₂, O₂, H₂
(c) N₂ water
(d) O₂ water

Answer

Answer: (а) CO₂

Question 11.
The end products of photosynthesis are
(a) carbohydrates, oxygen
(b) carbohydrates, hydrogen
(c) carbohydrates, water vapours
(d) carbohydrates, oxygen and water vapours

Answer

Answer: (a) carbohydrates, oxygen

Question 12.
Which one of the following is a parasite?
(a) Lichen
(b) Cuscuta
(c) Pitcher plant
(d) Rhizobium

Answer

Answer: (b) Cuscuta

Question 13.
Which of the following class of organisms belongs to saprotrophs?
(a) Fungi
(b) Algae
(c) Lichens
(d) Bryophytes

Answer

Answer: (a) Fungi

Question 14.
Which one of the following is a pair of symbiotic organisms?
(a) Lichens
(b) Rhizobium and a legume
(c) None of these
(d) Both (a) and (b)

Answer

Answer: (d) Both (a) and (b)

Question 15.
Which of the following is an insectivorous plant?
(a) Pitcher plant
(b) Indian telegraph plant
(c) 4 ‘O’clock plant
(d) Cuscuta

Answer

Answer: (a) Pitcher plant

Match the following:

Column A	Column B
(i) Autotrophs	(a) Fungi
(ii) Heterotroph	(b) Lichen
(iii) Parasite	(c) Pitcher plant
(iv) Saproptroph	(d) Algae
(v) Symbiont	(e) Man
(vi) Insectivorous	(f) Cuscuta

Answer

Answer:

Column A	Column B
(i) Autotrophs	(d) Algae
(ii) Heterotroph	(e) Man
(iii) Parasite	(f) Cuscuta
(iv) Saproptroph	(a) Fungi
(v) Symbiont	(b) Lichen
(vi) Insectivorous	(c) Pitcher plant

Fill in the blanks:

1. All organisms take ……………… and utilize it to get energy for the growth and the maintenance of their bodies.

Answer

Answer: food

2. Green plants synthesise their food themselves by the process of ………….. they are called ……………..

Answer

Answer: photosynthesis, autotrophs

3. …………………. energy is stored by the leaves with the help of chlorophyll.

Answer

Answer: Solar

4. …………………. derive nutrition from, dead, decaying matter.

Answer

Answer: Fungi/s aprotrophs

5. Plants like cuscuta take food from ………………… plant.

Answer

Answer: host

6. All animals are categorised as ……………..

Answer

Answer: heterotrophs

7. …………….. is produced and …………….. is utilized during photosynthesis.

Answer

Answer: Oxygen, carbon dioxide

8. …………….. is the site of reception of light energy in leaves.

Answer

Answer: Chlorophyll

Choose the true and false statements from the following:

1. Food is essential for all living organisms.

Answer

Answer: True

2. Leaves are the food factories of plant.

Answer

Answer: True

3. Water comes into leaves through stomata in the form of vapours.

Answer

Answer: False

4. Plants utilize the carbon dioxide dissolved in the water absorbed by the roots for photosynthesis.

Answer

Answer: False

Important Questions

Question 1.
Potato and ginger are both underground parts that store food. Where is the food prepared in these plants? [NCERT Exemplar]
Anwer:
In both the plants, shoot system and leaves are above ground. They prepare food through photosynthesis and transport it to the underground part for storage.

Question 2.
Plants prepare their food using a different mode of nutrition than us. What is it?
Answer:
The mode of nutrition in plant is autotrophic, i.e. they synthesise their own food.

Question 3.
Photosynthesis requires chlorophyll and a few other raw materials. Add the missing raw materials to the list given below:
Water, minerals, (a) …… (b) …….
Answer:
(a) Sunlight
(b) Carbon dioxide

Question 4.
The tiny openings present on the leaf surface. What are they called?
Answer:
Stomata are the tiny pores present on the surface of leaves through which gaseous exchange takes place in plants.

Question 5.
What is the function of guard cells of stomata?
Answer:
Guard cells help in controlling the opening and closing of stomata for gaseous exchange.

Question 6.
Which parts of the plant are called food factories of the plant?
Answer:
Leaves are referred to as food factories of plants. This is because, leaves synthesise food by the process of photosynthesis.

Question 7.
A carbohydrate is produced by plants as food source. It is constituted from which molecules?
Answer:
Carbohydrates are composed of carbon, hydrogen and oxygen.

Question 8.
Why do some plants feed on insects?
Answer:
Insectivorous plants grow in soil which lack nitrogen, therefore they eat insects to fulfill their need of nitrogen.

Question 9.
Define parasites.
Answer:
Parasites they are those organisms which grow on other plants or animals for their food, e.g. Cuscuta.

Question 10.
Name the bacteria that can fix atmospheric nitrogen.
Answer:
Rhizobium is the bacterium which can fix atmospheric nitrogen.

Question 11.
Except plants, why can’t other living organisms prepare their food using CO₂, water and minerals? [HOTS]
Answer:
Our body does not contain chlorophyll for absorbing solar energy which is necessary for preparing food using air, water, etc.

Question 12.
A leguminous plant can restore the soil’s concentration of mineral nutrients. Can you give examples of some such plants?
Answer:
Plants such as gram, pulses and beans are leguminous.

Question 13.
Algae are green in colour. Why?
Answer:
Algae contain chlorophyll which imparts green colour to them.

Question 14.
what do you understand by nutrition?
Answer:
The process of utilising nutrients like carbohydrates, proteins, fats, etc., to generate energy is called nutrition.

Question 15.
Fungus can be harmful and useful. Give an example showing both of these traits of fungus.
Answer:
Fungus produces antibiotics like penicillin used to treat diseases and fungus can also harm us by causing fungal infections on skin and hair.

Question 16.
A unique feature in leaves allows them to prepare the food while other parts of plants cannot. Write the possible reason for this. [HOTS]
Answer:
Leaves contain chlorophyll which is essential for food preparation and is absent in other parts of plant.

Question 17.
Algae and fungi form a unique association sharing benefits from each other. What is the name of association between them?
Answer:
Lichens.

Question 18.
In a plant, photosynthesis occurs in a part other than leaf. Name that plant and the part where photosynthesis occurs.
Answer:
Cactus, the part where photosynthesis occurs are stem and branches which are green.

Question 19.
Why is Cuscuta, categorised as a parasite?
Answer:
Cuscuta derives its nutrition using an association where it deprives its host of all valuable nutrients and absorbs them itself. Hence, it is called a parasitic plant.

Question 20.
Plant cannot use the nitrogen present in the soil directly. Why?
Answer:
Plants can use nitrogen only in soluble form while in soil nitrogen is present in inorganic form.

Question 21.
Why are insectivorous plants called partial heterotrophs?
Answer:
Insectivorous plants are autotrophs, i.e. they prepare their own food. They are partial heterotrophs as they eat insects for obtaining nitrogen.

Question 22.
What is the stored food form in sunflower seeds?
Answer:
In sunflower seeds, glucose is stored in the form of oils (fats).

Question 23.
What do you understand by saprotrophic mode of nutrition?
Answer:
The mode of nutrition in which organisms take their nutrients from dead and decaying matter is called saprotrophic mode of nutrition.

Question 24.
A mutually beneficial relationship that occurs between two plants. It is known by what name? Give an example.
Answer:
Symbiosis is the mutually benefitting association between two plants, e.g. lichens.

Question 25.
For testing the presence of starch in leaves, a boiled leaf is used. Why?
Answer:
Boiling the leaf remove chlorophyll/green colour from the leaves.

Question 26.
Mosquitoes, bed bugs, lice and leeches suck our blood. Can they be called as parasites? [HOTS]
Answer:
Yes, these animals/insects are parasites as they harm the hosts while they suck blood.

Question 27.
Insectivorous plants have one or the other specialised organs to catch their prey. What is that organ?
Answer:
Leaves of insectivorous plants catches the prey.

Question 28.
Farmers spread manure of fertilisers in the field or in gardens, etc. Why are these added to the soil?
Answer:
Plants absorb mineral nutrients from soil. Thus, declining their concentration in soil fertilisers and manures enhance or add these essential nutrients back in soil.

Question 29.
A cell is formed of many sub-components. Identify different constituents of the cell. Are animal and plant cells similar?
Answer:
A cell contains nucleus, cytoplasm, vacuole, cell organelles like chloroplast, mitochondria, etc. No, animal cells are different from plant cells.

Question 30.
A goat eats away all the leaves of a small plant (balsam). However, in a few days, new leaves could be seen sprouting in the plant again. How did the plant survive without leaves? [NCERT Exemplar; HOTS]
Answer:
The plant of balsam survived on the food stored in the stem and roots.

Nutrition in Plants Class 7 Science Extra Questions Short Answer Type

Question 1.
Different modes of nutrition has been observed in plants. What are they? Give example of each.
Answer:
Plants show two major modes of nutrition, i.e.
(i) Autotrophs are those which can synthesise their own food.
(ii) Heterotrophs are those which are dependent on other plants and animals for their food. They are of following types:
(a) Parasites, e.g. Cuscuta
(b) Saprotrophs, e.g. fungi.

Question 2.
Sunlight, chlorophyll, carbon dioxide, water and minerals are raw materials essential for photosynthesis. Do you know where they are available? Fill in the blanks with the appropriate raw materials.
(a) Available in the plant: ………
(b) Available in the soil: ………
(c) Available in the air: ………
(d) Available during day : ……… [NCERT Exemplar]
Answer:
(a) Available in the plant: chlorophyll
(b) Available in the soil : water, minerals
(c) Available in the air : carbon dioxide
(d) Available during day : sunlight

Question 3.
Plants are considered an essential part of earth as they keep a check on lot of process occurring all over. What would happen if all the green plants are wiped from earth? [HOTS]
Answer:
Green plants are the source of energy for all the living organisms so that they can perform their normal functions. If all green plants and trees disappear, all the organism depending on them for food and shelter will also die.

The lack of gaseous exchange will lead to increase in amount of CO₂, causing death in humans and other animals also. The cycle of life will gradually disappear.

Question 4.
Autotrophs and heterotrophs are two different organisms with distinct modes of nutrition state. How are they different from each other?
Answer:
The difference between autotrophs and heterotrophs are as follows:

Autotrophs	Heterotrophs
They can prepare their own food.	They cannot prepare their own food.
Autotrophs take simple inorganic substances and change it into complex organic food, e.g. green plants.	They take in complex food and breakdown it into simple compounds, e.g. all animals, fungi and non-green plants.

Question 5.
Wheat dough if left in the open, after a few days, starts to emit a foul smell and becomes unfit for use. Give reason. [NCERT Exemplar; HOTS]
Answer:
Carbohydrates in wheat dough encourage the growth of yeast and other saprophytic fungi which breakdown carbohydrates into simpler compounds like CO₂ and alcohol and emit a foul smell.

Question 6.
What are the various raw materials for photosynthesis?
Answer:
Plants utilise carbon dioxide from air and water and minerals are derived from soil (through roots) as raw material for photosynthesis. Besides these chlorophyll present in green leaf is necessary for the process and sunlight is the source of energy which is converted into chemical energy during the process of photosynthesis.

Question 7.
Observe the given figure and label the following terms given in the box. Stomatal opening, guard cell

Answer:
Labelled figure is given below:

Question 8.
Nitrogen is an essential nutrient for plants growth. But farmers who cultivate pulses as crops like green gram, bengal gram, black gram, etc., do not apply nitrogenous fertilisers during t cultivation. Why? [NCERT Exemplar; HOTS]
Answer:
Roots of pulses (leguminous plants) have a symbiotic association with a bacterium called Rhizobium. This bacteria convert gaseous nitrogen of air into water soluble nitrogen compounds and give them to the leguminous plants for their growth. Hence, farmers need not use nitrogenous fertilisers.

Question 9.
Pooja is worried about her new shoes which she wore on special occassions that they were spoiled by fungus during rainy season. Is she right to worry, if yes, then tell why does fungi suddenly appears during the rainy season? [HOTS]
Answer:
Yes, the fungi reproduces by spores which are generally present in the air and grow on any article that are left in hot and humid weather for a long time. During rainy season they land on wet and warm things and begin to germinate and grow.

Question 10.
In what unique manner does a pitcher plant derive its nutrition?
Answer:
Nepenthes or pitcher plant modifies its leaf axis into a long tubular pitcher to form a pitfall trap. Inside the pitcher sticky liquid is present. When any insect comes in contact with the leaf, the lid present on it is closed and insect is trapped. The liquid contains digestive enzymes which slowly digest the trapped insects.

Question 11.
Water and minerals are absorbed by the roots and then transported to leaves. How?
Answer:
Water and minerals are transported to the leaves by the vessels which run like pipes throughout the root, stem, branches and the leaves. These vessels are xylem and phloem, forming a continuous path or passage for the nutrients to make them reach the leaf.

Question 12.
Some plants have deep red, violet or brown coloured leaves. Can these leaves perform the photosynthesis process? [HOTS]
Answer:
Yes, plants having deep red, violet or brown coloured leaves can also carry out photosynthesis because they contain chlorophyll. But their green colour of chlorophyll is masked by the large amount of all other coloured pigments.

Question 13.
If plant has a requirement for nitrogen, then from where will they obtain it?
Answer:
Soil contains nitrogen in the form that is not usable by plants. Bacteria like Rhizobium converts nitrogen into soluble form that can be easily used by plants. So, if plant has a requirement for nitrogen, then it will obtain that which the help of bacteria.

Question 14.
In the absence of photosynthesis, life would be impossible on earth. Is it true or false?
Answer:
True, because photosynthesis is important for the existence of life on the earth. Photosynthesis is important process as it is provides food to all living organisms and maintains CO₂ – O₂ balance of nature.

Nutrition in Plants Class 7 Science Extra Questions Long Answer Type

Question 1.
Describe the process by which plants prepare their food using different raw materials.
Answer:
The process by which green plants can prepare their own food is called photosynthesis. Green plants possess chlorophyll in their leaf and utilises carbon dioxide (from air) water, minerals (from soil, through root) as raw material and sunlight as source of energy and convert light energy into chemical energy. The food thus synthesised is in the form of starch (carbohydrate). The overall reaction for photosynthesis can be given as follows:

Question 2.
Describe the method for replinishing the soils with minerals and other essential constituents used by plants growing in those soil by farmers.
Answer:
Replenishment of Nutrients in Soil

Crops require a lot of nitrogen to make proteins. After the harvest, the soil becomes deficient in nitrogen. Plants cannot use the nitrogen gas available in atmosphere directly. Action of certain bacteria can convert this nitrogen into form readily used by plants. Rhizobium bacteria live in the root nodules of leguminous plants. These bacteria take nitrogen gas from the atmosphere and convert it into water soluble nitrogen compounds making it available to the leguminous plants for their growth.

In return, leguminous plants provide food and shelter to the bacteria as Rhizobium cannot prepare its food. They, thus have a symbiotic relationship. This association is very important for the farmers, as they do not need to add nitrogen fertilisers to the soil in which leguminous plants are grown.

Question 3.
Harish went to visit his grandfather in village where he saw that his grandfather’s field of wheat are infected with fungus but no one is aware of this. Harish rushed to his grandfather’s side and told him that the field have been infected with fungi. He should use an antifungal agent in his fields to stop this infection.
(a) What is fungus?
(b) Can fungus only cause diseases or can it be helpful also?
(c) What values are shown by Harish? [Value Based Question]
Answer:
(a) Fungus are saprophytic organisms usually present as spores in atmosphere which can germinate on any substrate in optimal conditions.
(b) Fungus are also useful in that they produce many antibiotics which can cure different types of infections like penicillin.
(c) Harish is sincere, curious and knowledgeable with a keen sense of applying it where necessary.

Question 4.
Wild animals like tiger, wolf, lion and leopard do not eat plants. Does this mean that they can survive without plants? Can you provide a suitable explanation? [HOTS]
Answer:
Animals like tiger, wolf, lion and leopard are carnivores and do not eat plants. They hunt and eat herbivorous animals like deer, gaur, bison, zebra, giraffe, etc., which are dependent on plants for food.

If there are no plants, herbivorous animals will not survive and ultimately animals like tiger, wolf, lion and leopard will have nothing to eat.

Question 5.
Asha went to visit her grandfather in his village. He was having a serious discussion with his fellow members regarding the productivity level of crops for present year. They all were worried about how to increase the productivity of crop Asha listened to this and then suggested to the group that the reason may be decreased level of minerals in soil.
She told her grandfather to plant crops like pulses, gram, beans, etc., for a year then follow with regular crops. This will increase the crop productivity?
(a) What will you name the process suggested by Asha? Why is there decrease in crop productivity?
(b) What are noted benefits of this process? Will the results be as what Asha expressed?
(c) What values are shown by Asha? [Value Based Question]
Answer:
(a) This process is known as crop rotation. All the plants/crops grown in soil use the minerals present in soil for their own use. This continuous usage depletes the concentration of mineral in soil.
(b) After growing leguminous plants, the mineral content of soil is restored and enriched to new level. Yes, the benefit of leguminous plant is the re-enrichment of soil minerals.
(c) Asha is observant, sincere and interested in applying her knowledge to situation

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CHAPTER – 15 Visualising Solid Shapes | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Get Chapter Wise MCQ Questions for Class 7 Maths with Answers prepared here according to the latest CBSE syllabus and NCERT curriculum. Students can practice CBSE Class 7 Maths MCQs Multiple Choice Questions with Answers to score good marks in the examination. Students can also visit the most accurate and elaborate NCERT Solutions for Class 7 Maths. Every question of the textbook has been answered here.

Chapter - 15 Visualising Solid Shapes

MCQs

Question 1.
The name of the solid shape is

(a) cone
(b) cylinder
(c) sphere
(d) cube

Answer

Answer: (b) cylinder

Question 2.
The name of the solid shape is

(a) cuboid
(b) cube
(c) pyramid
(d) cone

Answer

Answer: (a) cuboid

Question 3.
The name of the solid shape is

(a) cylinder
(b) cone
(c) sphere
(d) cube

Answer

Answer: (c) sphere

Question 4.
The name of the solid shape is

(a) cube
(b) cylinder
(c) cone
(d) sphere

Answer

Answer: (a) cube

Question 5.
The name of the solid shape is

(a) cylinder
(b) cone
(c) cuboid
(d) sphere

Answer

Answer: (b) cone

Question 6.
The name of the solid shape is

(a) cylinder
(b) cone
(c) sphere
(d) pyramid

Answer

Answer: (d) pyramid

Question 7.
The number of vertices of a cube is
(a) 8
(b) 12
(c) 6
(d) 3

Answer

Answer: (a) 8

Question 8.
The number of edges of a cube is
(a) 8
(b) 12
(c) 6
(d) 3

Answer

Answer: (b) 12

Question 9.
The number of faces of a cube is
(a) 8
(b) 12
(c) 6
(d) 3

Answer

Answer: (c) 6

Question 10.
The number of vertices of the solid shape is

(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 11.
The number of faces of the solid shape is

(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 12.
The number of edges of the solid shape is

(a) 1
(b) 2
(c) 3
(d) 6

Answer

Answer: (d) 6

Question 13.
The number of vertices of the solid shape is

(a) 9
(b) 4
(c) 6
(d) 8

Answer

Answer: (a) 9

Question 14.
The number of faces of the solid shape is

(a) 4
(b) 6
(c) 9
(d) 8

Answer

Answer: (c) 9

Question 15.
The number of edges of the solid shape is

(a) 16
(b) 9
(c) 6
(d) 4

Answer

Answer: (a) 16

Question 16.
Two cubes of edge length 2 cm are placed side by side. The length of the resulting cuboid is
(a) 2 cm
(b) 4 cm
(c) 1 cm
(d) none of these

Answer

Answer: (b) 4 cm

Important Questions

Question 1.
If three cubes of dimensions 2 cm × 2 cm × 2 cm are placed end to end, what would be the dimension of the resulting cuboid?
Solution:
Length of the resulting cuboid = 2 cm + 2 cm + 2 cm = 6 cm
Breadth = 2 cm
Height = 2 cm

Hence the required dimensions = 6 cm × 2 cm × 2 cm.

Question 2.
Answer the following:
(i) Why a cone is not a pyramid?
(ii) How many dimension a solid have?
(iii) Name the solid having one curved and two flat faces but no vertex.
Solution:
(i) Cone is not a pyramid because its base is not a polygon.
(ii) Three.
(iii) Cylinder

Question 3.
Write down the number of edges on each of the following solid figures:
(i) Cube
(ii) Tetrahedron
(iii) Sphere
(iv) Triangular prism
Solution:
(i) 12
(ii) 6
(iii) 0
(iv) 9

Question 4.
What cross-section do you get when you give a horizontal cut to an ice cream cone?
Solution:
Circle.

Question 5.
Determine the number of edges, vertices and faces in the given figure.

Solution:
Edges = 8
Vertices = 5
Faces = 5

Question 6.
Draw the sketch of two figure that has no edge.
Solution:

Question 7.
Draw the sketches of two figures that have no vertex.
Solution:

Question 8.
Name any three objects which resemble a sphere and cone.
Solution:
Sphere: Football, Earth, Round table
Cone: Conical funnel, ice cream cone, conical cracker

Question 9.
What shape would we get from the given figure?

Solution:
From the given net, we get a rectangular pyramid.

Visualising Solid Shapes Class 7 Extra Questions Short Answer Type

Question 10.
For the solids given below sketch the front, side and top view

Solution:

Question 11.
Match the following:

Solution:
(i) → (e)
(ii) → (a)
(iii) → (b)
(iv) → (c)
(v) → (d)

Question 12.
Complete the following table:

Solution:

Question 13.
Draw a plan, front and side elevations of the following solids.

Solution:

Question 14.
Name the solid that would be formed by each net:

Solution:
(i) Triangular pyramid
(ii) Square pyramid
(iii) Hexagonal pyramid

Question 15.
Name the solids that have:
(i) 1 curved surface
(ii) 4 faces
(iii) 6 faces
(iv) 5 faces and 5 vertices
(v) 8 triangular faces
(vi) 6 triangular faces and 2 hexagonal faces.
Solution:
(i) Cylinder
(ii) Tetrahedron
(iii) Cube and cuboid
(iv) Square pyramid or rectangular pyramid
(v) Regular octahedron
(vi) Hexagonal prism.

Question 16.
Draw the top, side and front views of the given solids:

Solution:

Question 17.
Draw the net of a cuboid having same breadth and height, but length double the breadth.
Solution:

Question 18.
Draw the nets of the following:
(i) Triangular prisms
(ii) Tetrahedron
(iii) Cuboid.
Solution:

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CHAPTER – 14 Symmetry | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 14 Symmetry

MCQs

Question 1.
How many lines of symmetry are there in an equilateral triangle?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3

Question 2.
How many lines of symmetry are there in a square?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 3.
How many lines of symmetry are there in a rectangle?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (b) 2

Question 4.
How many lines of symmetry are there in a regular pentagon?
(a) 1
(b) 2
(c) 3
(d) 5

Answer

Answer: (d) 5

Question 5.
How many lines of symmetry are there in a regular hexagon?
(a) 2
(b) 4
(c) 6
(d) 3

Answer

Answer: (c) 6

Question 6.
How many lines of symmetry are there in the following figure?

(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1

Question 7.
How many lines of symmetry are there in the following figure?

(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1

Question 8.
How many lines of symmetry are there in the following figure?

(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1

Question 9.
How many lines of symmetry are there in the following figure?

(a) 4
(b) 3
(c) 2
(d) 1

Answer

Answer: (d) 1

Question 10.
How many lines of symmetry are there in the following figure?

(a) 2
(b) 1
(c) 4
(d) 3

Answer

Answer: (b) 1

Question 11.
How many lines of symmetry are there in following figure?

(a) 1
(b) 2
(c) 3
(d) None of these

Answer

Answer: (a) 1

Question 12.
How many lines of symmetry are there in the following figure?

(a) 1
(b) 2
(c) 3
(d) Infinitely many

Answer

Answer: (d) Infinitely many

Question 13.
How many lines of symmetry are there in an isosceles triangle?
(a) 4
(b) 3
(c) 1
(d) 2

Answer

Answer: (c) 1

Question 14.
How many lines of symmetry are there in a scalene triangle?
(a) 1
(b) 0
(c) 2
(d) 4

Answer

Answer: (b) 0

Question 15.
How many lines of symmetry are there in a rhombus?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (b) 2

Question 16.
How many lines of symmetry are there in a parallelogram?
(a) 0
(b) 1
(c) 2
(d) None of these

Answer

Answer: (a) 0

Question 17.
How many lines of symmetry are there in a quadrilateral?
(a) 0
(b) 2
(c) 4
(d) None of these

Answer

Answer: (a) 0

Question 18.
The order of rotational symmetry of an equilateral triangle is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3

Question 19.
The order of rotational symmetry of a square is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 20.
What is the order of the rotational symmetry of the following figure?

(a) 4
(b) 3
(c) 2
(d) 1

Answer

Answer: (a) 4

Important Questions

Question 1.
Draw any two English alphabets having an only a vertical line of symmetry.
Solution:

Question 2.
Draw any two English alphabets having a horizontal line of symmetry.
Solution:

Question 3.
Draw any two English alphabets having both horizontal and vertical line of symmetry.
Solution:

Question 4.
Dray any two English alphabets which has no line of symmetry.
Solution:

Question 5.
Draw any two figures which have the order of rotational symmetry 4.
Solution:

Question 6.
Draw any two figures which have the order of rotational symmetry 2.
Solution:

Question 7.
State the order of rotational symmetry of the following figures.

Solution:
(i) Order of equilateral triangle = 3
(ii) Order of regular pentagon = 5.

Question 8.
Draw a figure having an infinite number of lines of symmetry.
Solution:

A circle has the infinite number of lines of symmetry.

Question 9.
Draw any two figure having no lines of symmetry.
Solution:
English alphabet R and P have no lines of symmetry.

Question 10.
State the English alphabet which has only the horizontal line of symmetry.
Solution:
B, C, D, E, H, I, O, X and K are the English alphabets having an only horizontal line of symmetry.

Symmetry Class 7 Extra Questions Short Answer Type

Question 11.
Give the order of rotational symmetry of each of the following figures:

Solution:
(a) Order of rotational symmetry = 4
(b) Order of rotational symmetry = 5
(c) Order of rotational symmetry = 3
(d) Order of rotational symmetry = 6
(e) Order of rotational symmetry = 3
(f) Order of rotational symmetry = 4

Question 12.
How many lines of symmetry do the following have:
(a) a parallelogram
(b) an equilateral triangle
(c) a right angle with equal legs
(d) an angle with equal arms
(e) a semicircle
(f) a rhombus
(g) a square
(h) scalene triangle
Solution:

Question 13.
What letters of the English alphabet have reflectional symmetry about
(а) a vertical mirror
(b) a horizontal mirror
(c) both horizontal and vertical mirrors
Solution:
(a) The letters of English alphabet having reflectional symmetry about a vertical mirror are:
A, H, I, M, O, T, U, V, W, X, Y
(b) Letters of the English alphabets having reflectional symmetry about a horizontal mirror are:
B, C, D, E, H, I, O, X
(c) Letters of the English alphabet having reflectional of symmetry about vertical and horizontal mirror are:
O, X, I, H.

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CHAPTER – 13 Exponents and Powers | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 13 Exponents and Powers

MCQs

Question 1.
The exponential form of 10000 is
(a) 10³
(b) 10⁴
(c) 10⁵
(d) none of these

Answer

Answer: (b) 10⁴
Hint:
10000 = 10 × 10 × 10 × 10 = 10⁴

Question 2.
The exponential form of 100000 is
(a) 10³
(b) 10⁴
(c) 10⁵
(d) none of these

Answer

Answer: (c) 10⁵
Hint:
100000 = 10 × 10 × 10 × 10 × 10 = 10⁵

Question 3.
The exponential form of 81 is
(a) 3⁴
(b) 3³
(c) 3²
(d) none of these

Answer

Answer: (a) 3⁴
Hint:
81 = 3 × 3 × 3 × 3 = 3⁴

Question 4.
The exponential form of 125 is
(a) 5⁴
(b) 5³
(c) 5²
(d) none of these

Answer

Answer: (b) 5³
Hint:
125 = 5 × 5 × 5 = 5³

Question 5.
The exponential form of 32 is
(a) 2³
(b) 2⁴
(c) 2⁵
(d) none of these

Answer

Answer: (c) 2⁵
Hint:
32 = 2 × 2 × 2 × 2 × 2 = 2⁵

Question 6.
The exponential form of 243 is
(a) 3⁵
(b) 3⁴
(c) 3³
(d) 3²

Answer

Answer: (a) 3⁵
Hint:
243 = 3 × 3 × 3 × 3 × 3 = 3⁵

Question 7.
The exponential form of 64 is
(a) 2⁵
(b) 2⁶
(c) 2⁷
(d) 2⁸

Answer

Answer: (b) 2⁶
Hint:
64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶

Question 8.
The exponential form of 625 is
(a) 5²
(b) 5³
(c) 5⁴
(d) 5⁵

Answer

Answer: (c) 5⁴
Hint:
625 = 5 × 5 × 5 × 5 = 5⁴

Question 9.
The exponential form of 1000 is
(a) 10¹
(b) 10²
(c) 10³
(d) 10⁴

Answer

Answer: (c) 10³
Hint:
1000 = 10 × 10 × 10 = 10³

Question 10.
The value of (- 2)³ is
(a) 8
(b) -8
(c) 16
(d) -16

Answer

Answer: (b) -8
Hint:
(-2)³ = (-2) × (-2) × (-2) = -8

Question 11.
The value of (- 2)⁴ is
(a) 8
(b) -8
(c) 16
(d) -16

Answer

Answer: (c) 16
Hint:
(-2)⁴ = (-2) × (-2) × (-2) × (-2) = 16

Question 12.
What is the base in 8²?
(a) 8
(b) 2
(c) 6
(d) 10

Answer

Answer: (a) 8

Question 13.
What is the exponent in 8²?
(a) 8
(b) 2
(c) 16
(d) 6

Answer

Answer: (b) 2

Question 14.
(-1)^{even number} =
(a) -1
(b) 1
(c) 0
(d) none of these

Answer

Answer: (b) 1

Question 15.
(-1)^{odd number} =
(a) -1
(b) 1
(c) 0
(d) none of these

Answer

Answer: (a) -1

Question 16.
0 × 10⁴ =
(a) 0
(b) 10⁴
(c) 1
(d) none of these

Answer

Answer: (a) 0

Question 17.
If 2³ × 2⁴ = 2^?, then ? =
(a) 3
(b) 4
(c) 1
(d) 7

Answer

Answer: (d) 7
Hint:
2³ × 2⁴ = 2³⁺⁴ = 2⁷

Important Questions

Question 1.
Express 343 as a power of 7.
Solution:
We have 343 = 7 × 7 × 7 = 7³
Thus, 343 = 7³

Question 2.
Which is greater 3² or 2³?
Solution:
We have 3² = 3 × 3 = 9
2³ = 2 × 2 × 2 = 8
Since 9 > 8
Thus, 3² > 2³

Question 3.
Express the following number as a powers of prime factors:
(i) 144
(ii) 225
Solution:
(i) We have
144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3²
Thus, 144 = 2⁴ × 3²

(ii) We have
225 = 3 × 3 × 5 × 5 = 3² × 5²
Thus, 225 = 3² × 5²

Question 4.
Find the value of:
(i) (-1)¹⁰⁰⁰
(ii) (1)²⁵⁰
(iii) (-1)¹²¹
(iv) (10000)⁰
Solution:
(i) (-1)¹⁰⁰⁰ = 1 [∵ (-1)^{even number} = 1]
(ii) (1)²⁵⁰ = 1 [∵ (1)^{even number} = 1]
(iii) (-1)¹²¹ = -1 [∵ (-1)^{odd number} = -1]
(iv) (10000)⁰ = 1 [∵ a⁰ = 1]

Question 5.
Express the following in exponential form:
(i) 5 × 5 × 5 × 5 × 5
(ii) 4 × 4 × 4 × 5 × 5 × 5
(iii) (-1) × (-1) × (-1) × (-1) × (-1)
(iv) a × a × a × b × c × c × c × d × d
Solution:
(i) 5 × 5 × 5 × 5 × 5 = (5)⁵
(ii) 4 × 4 × 4 × 5 × 5 × 5 = 4³ × 5³
(iii) (-1) × (-1) × (-1) × (-1) × (-1) = (-1)⁵
(iv) a × a × a × b × c × c × c × d × d = a³b¹c³d²

Question 6.
Express each of the following as product of powers of their prime factors:
(i) 405
(ii) 504
(iii) 500
Solution:
(i) We have
405 = 3 × 3 × 3 × 3 × 5 = 3⁴ × 5¹
Thus, 405 = 3⁴ × 5¹

(ii) We have
504 = 2 × 2 × 2 × 3 × 3 × 7 = 2³ × 3² × 7¹
Thus, 504 = 2³ × 3² × 7¹

(iii) We have
500 = 2 × 2 × 5 × 5 × 5 = 2² × 5³
Thus, 500 = 2² × 5³

Question 7.
Simplify the following and write in exponential form:
(i) (5²)³
(ii) (2³)³
(iii) (a^b)^c
(iv) [(5)²]²
Solution:
(i) (5²)³ = 5^2×3 = 5⁶
(ii) (2³)³ = 2^3×3 = 2⁹
(iii) (a^b)^c = a^b×c = a^bc
(iv) [(5)²]² = 5^2×2 = 5⁴

Question 8.
Verify the following:

Solution:

Question 9.
Simplify:

Solution:

Question 10.
Simplify and write in exponential form:

Solution:

Exponents and Powers Class 7 Extra Questions Short Answer Type

Question 11.
Express each of the following as a product of prime factors is the exponential form:
(i) 729 × 125
(ii) 384 × 147
Solution:
(i) 729 × 125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 3⁶ × 5³
Thus, 729 × 125 = 3⁶ × 5³

(ii) 384 × 147 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 = 2⁷ × 3² × 7²
Thus, 384 × 147 = 2⁷ × 3² × 7²

Question 12.
Simplify the following:
(i) 10³ × 9⁰ + 3³ × 2 + 7⁰
(ii) 6³ × 7⁰ + (-3)⁴ – 9⁰
Solution:
(i) 10³ × 9⁰ + 3³ × 2 + 7⁰
= 1000 + 54 + 1
= 1055
(ii) 6³ × 7⁰ + (-3)⁴ – 9⁰
= 216 × 1 + 81 – 1
= 216 + 80
= 296

Question 13.
Write the following in expanded form:
(i) 70,824
(ii) 1,69,835
Solution:
(i) 70,824
= 7 × 10000 + 0 × 1000 + 8 × 100 + 2 × 10 + 4 × 10⁰
= 7 × 10⁴ + 8 × 10² + 2 × 10¹ + 4 × 10⁰
(ii) 1,69,835
= 1 × 100000 + 6 × 10000 + 9 × 1000 + 8 × 100 + 3 × 10 + 5 × 10⁰
= 1 × 10⁵ + 6 × 10⁴ + 9 × 10³ + 8 × 10² + 3 × 10¹ + 5 × 10⁰

Question 14.
Find the number from each of the expanded form:
(i) 7 × 10⁸ + 3 × 10⁵ + 7 × 10² + 6 × 10¹ + 9
(ii) 4 × 10⁷ + 6 × 10³ + 5
Solution:
(i) 7 × 10⁸ + 3 × 10⁵ + 7 × 10² + 6 × 10¹ + 9
= 7 × 100000000 + 3 × 100000 + 7 × 100 + 6 × 10 + 9
= 700000000 + 300000 + 700 + 60 + 9
= 700300769
(ii) 4 × 10⁷ + 6 × 10³ + 5
= 4 × 10000000 + 6 × 1000 + 5
= 40000000 + 6000 + 5
= 40006005

Question 15.
Find the value of k in each of the following:

Solution:

Question 16.
Find the value of
(a) 3⁰ ÷ 4⁰
(b) (8⁰ – 2⁰) ÷ (8⁰ + 2⁰)
(c) (2⁰ + 3⁰ + 4⁰) – (4⁰ – 3⁰ – 2⁰)
Solution:
(a) We have 3⁰ ÷ 4⁰ = 1 ÷ 1 = 1 [∵ a⁰ = 1]
(6) (8⁰ – 2⁰) ÷ (8⁰ + 2⁰) = (1 – 1) ÷ (1 + 1) = 0 ÷ 2 = 0
(c) (2⁰ + 3⁰ + 4⁰) – (4⁰ – 3⁰ – 2⁰)
= (1 + 1 + 1) – (1 – 1 – 1) [∵ a⁰ = 1]
= 3 – 1
= 2

Question 17.
Express the following in standard form:
(i) 8,19,00,000
(ii) 5,94,00,00,00,000
(iii) 6892.25
Solution:
(i) 8,19,00,000 = 8.19 × 10⁷
(ii) 5,94,00,00,00,000 = 5.94 × 10¹¹
(iii) 6892.25 = 6.89225 × 10³

Question 18.
Evaluate:

Solution:

Exponents and Powers Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 19.
Find the value of x, if

Solution:

Question 20.

Solution:

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CHAPTER – 12 Algebraic Expressions | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 12 Algebraic Expressions

MCQs

Question 1.
How many terms are there in the expression 2x²y?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1

Question 2.
How many terms are there in the expression 2y + 5?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (b) 2

Question 3.
How many terms are there in the expression 1.2ab – 2.4b + 3.6a?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3

Question 4.
How many terms are there in the expression – 2p³ – 3p² + 4p + 7?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 5.
What is the coefficient of x in the expression 4x + 3y?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 6.
What is the coefficient of x in the expression 2 – x + y?
(a) 2
(b) 1
(c) -1
(d) None of these

Answer

Answer: (c) -1

Question 7.
What is the coefficient of x in the expression y²x + y?
(a) y²
(b) y
(c) 1
(d) 0

Answer

Answer: (a) y²

Question 8.
What is the coefficient of x in the expression 2z – 3xz?
(a) 3
(b) z
(c) 3z
(d) -3z

Answer

Answer: (d) -3z

Question 9.
What is the coefficient of x in the expression 1 + x + xz?
(a) z
(b) 1 + z
(c) 1
(d) 1 + x

Answer

Answer: (b) 1 + z

Question 10.
What is the coefficient of x in the expression 2x + xy²?
(a) 2 + y²
(b) 2
(c) y²
(d) None of these

Answer

Answer: (a) 2 + y²

Question 11.
What is the coefficient of y² in the expression 4 – xy²?
(a) 4
(b) x
(c) -x
(d) None of these

Answer

Answer: (c) -x

Question 12.
What is the coefficient of y² in the expression 3y² + 4x?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3

Question 13.
What is the coefficient of y² in the expression 2x²y – 10xy² + 5y²?
(a) 5 – 10x
(b) 5
(c) -10 x
(d) None of these

Answer

Answer: (a) 5 – 10x

Question 14.
What is the coefficient of x in the expression ax³ + bx² + d?
(a) a
(b) b
(c) d
(d) 0

Answer

Answer: (d) 0

Question 15.
What is the coefficient of x² in the expression ax + b?
(a) a
(b) b
(c) a + b
(d) 0

Answer

Answer: (d) 0

Imporatant Questions

Question 1.
Identify in the given expressions, terms which are not constants. Give their numerical coefficients.
(i) 5x – 3
(ii) 11 – 2y²
(iii) 2x – 1
(iv) 4x²y + 3xy² – 5
Solution:

Question 2.
Group the like terms together from the following expressions:
-8x²y, 3x, 4y, −32x , 2x²y, -y
Solution:
Group of like terms are:
(i) -8x²y, 2x²y
(ii) 3x, −32x
(iii) 4y, -y

Question 3.
Identify the pairs of like and unlike terms:
(i) −32x, y
(ii) -x, 3x
(iii) −12y2x, 32xy²
(iv) 1000, -2
Solution:
(i) −32x, y → Unlike Terms
(ii) -x, 3x → Like Terms
(iii) −12y2x, 32xy² → Like Terms
(iv) 1000, -2 → Like Terms

Question 4.
Classify the following into monomials, binomial and trinomials.
(i) -6
(ii) -5 + x
(iii) 32x – y
(iv) 6x² + 5x – 3
(v) z² + 2
Solution:
(i) -6 is monomial
(ii) -5 + x is binomial
(iii) 32x – y is binomial
(iv) 6x² + 5x – 3 is trinomial
(v) z² + z is binomial

Question 5.
Draw the tree diagram for the given expressions:
(i) -3xy + 10
(ii) x² + y²
Solution:

Question 6.
Identify the constant terms in the following expressions:
(i) -3 + 32x
(ii) 32 – 5y + y²
(iii) 3x² + 2y – 1
Solution:
(i) Constant term = -3
(ii) Constant term = 32
(iii) Constant term = -1

Question 7.
Add:
(i) 3x²y, -5x²y, -x²y
(ii) a + b – 3, b + 2a – 1
Solution:
(i) 3x²y, -5x²y, -x²y
= 3x²y + (-5x²y) + (-x²y)
= 3x²y – 5x²y – x²y
= (3 – 5 – 1 )x²y
= -3x²y
(ii) a + b – 3, b + 2a – 1
= (a + b – 3) + (b + 2a – 1)
= a + b – 3 + b + 2a – 1
= a + 2a + b + b – 3 – 1
= 3a + 2b – 4

Question 8.
Subtract 3x² – x from 5x – x².
Solution:
(5x – x²) – (3x² – x)
= 5x – x² – 3x² + x
= 5x + x – x² – 3x²
= 6x – 4x²

Question 9.
Simplify combining the like terms:
(i) a – (a – b) – b – (b – a)
(ii) x² – 3x + y² – x – 2y²
Solution:
(i) a – (a – b) – b – (b – a)
= a – a + b – b – b + a
= (a – a + a) + (b – b – b)
= a – b
(ii) x² – 3x + y² – x – 2y²
= x² + y² – 2y² – 3x – x
= x² – y² – 4x

Algebraic Expressions Class 7 Extra Questions Short Answer Type

Question 10.
Subtract 24xy – 10y – 18x from 30xy + 12y – 14x.
Solution:
(30xy + 12y – 14x) – (24xy – 10y – 18x)
= 30xy + 12y – 14x – 24xy + 10y + 18x
= 30xy – 24xy + 12y + 10y – 14x + 18x
= 6xy + 22y + 4x

Question 11.
From the sum of 2x² + 3xy – 5 and 7 + 2xy – x² subtract 3xy + x² – 2.
Solution:
Sum of the given term is (2x² + 3xy – 5) + (7 + 2xy – x²)
= 2x² + 3xy – 5 + 7 + 2xy – x²
= 2x² – x² + 3xy + 2xy – 5 + 7
= x² + 5xy + 2
Now (x² + 5xy + 2) – (3xy + x² – 2)
= x² + 5xy + 2 – 3xy – x² + 2
= x² – x² + 5xy – 3xy + 2 + 2
= 0 + 2xy + 4
= 2xy + 4

Question 12.
Subtract 3x² – 5y – 2 from 5y – 3x² + xy and find the value of the result if x = 2, y = -1.
Solution:
(5y – 3x² + xy) – (3x² – 5y – 2)
= 5y – 3x² + xy – 3x² + 5y + 2
= -3x² – 3x² + 5y + 5y + xy + 2
= -6x² + 10y + xy + 2
Putting x = 2 and y = -1, we get
-6(2)² + 10(-1) + (2)(-1) + 2
= -6 × 4 – 10 – 2 + 2
= -24 – 10 – 2 + 2
= -34

Question 13.
Simplify the following expressions and then find the numerical values for x = -2.
(i) 3(2x – 4) + x² + 5
(ii) -2(-3x + 5) – 2(x + 4)
Solution:
(i) 3(2x – 4) + x² + 5
= 6x – 12 + x² + 5
= x² + 6x – 7
Putting x = -2, we get
= (-2)² + 6(-2) – 7
= 4 – 12 – 7
= 4 – 19
= -15
(ii) -2(-3x + 5) – 2(x + 4)
= 6x – 10 – 2x – 8
= 6x – 2x – 10 – 8
= 4x – 18
Putting x = -2, we get
= 4(-2) – 18
= -8 – 18
= -26

Question 14.
Find the value of t if the value of 3x² + 5x – 2t equals to 8, when x = -1.
Solution:
3x² + 5x – 2t = 8 at x = -1
⇒ 3(-1)² + 5(-1) – 2t = 8
⇒ 3(1) – 5 – 2t = 8
⇒ 3 – 5 – 2t = 8
⇒ -2 – 2t = 8
⇒ 2t = 8 + 2
⇒ -2t = 10
⇒ t = -5
Hence, the required value of t = -5.

Question 15.
Subtract the sum of -3x³y² + 2x²y³ and -3x²y³ – 5y⁴ from x⁴ + x³y² + x²y³ + y⁴.
Solution:
Sum of the given terms:

Required expression

Question 16.
What should be subtracted from 2x³ – 3x²y + 2xy² + 3y² to get x³ – 2x²y + 3xy² + 4y²? [NCERT Exemplar]
Solution:
We have

Required expression

Question 17.
To what expression must 99x³ – 33x² – 13x – 41 be added to make the sum zero? [NCERT Exemplar]
Solution:
Given expression:
99x³ – 33x² – 13x – 41
Negative of the above expression is
-99x³ + 33x² + 13x + 41
(99x³ – 33x² – 13x – 41) + (-99x³ + 33x² + 13x + 41)
= 99x³ – 33x² – 13x – 41 – 99x³ + 33x² + 13x + 41
= 0
Hence, the required expression is -99x³ + 33x² + 13x + 41

Algebraic Expressions Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 18.
If P = 2x² – 5x + 2, Q = 5x² + 6x – 3 and R = 3x² – x – 1. Find the value of 2P – Q + 3R.
Solution:
2P – Q + 3R = 2(2x² – 5x + 2) – (5x² + 6x – 3) + 3(3x² – x – 1)
= 4x² – 10x + 4 – 5x² – 6x + 3 + 9x² – 3x – 3
= 4x² – 5x² + 9x² – 10x – 6x – 3x + 4 + 3 – 3
= 8x² – 19x + 4
Required expression.

Question 19.
If A = -(2x + 3), B = -3(x – 2) and C = -2x + 7. Find the value of k if (A + B + C) = kx.
Solution:
A + B + C = -(2x + 13) – 3(x – 2) + (-2x + 7)
= -2x – 13 – 3x + 6 – 2x + 7
= -2x – 3x – 2x – 13 + 6 + 7
= -7x
Since A + B + C = kx
-7x = kx
Thus, k = -7

Question 20.
Find the perimeter of the given figure ABCDEF.

Solution:
Required perimeter of the figure
ABCDEF = AB + BC + CD + DE + EF + FA
= (3x – 2y) + (x + 2y) + (x + 2y) + (3x – 2y) + (x + 2y) + (x + 2y)
= 2(3x – 2y) + 4(x + 2y)
= 6x – 4y + 4x + 8y
= 6x + 4x-4y + 8y
= 10x + 4y
Required expression.

Question 21.
Rohan’s mother gave him ₹ 3xy² and his father gave him ₹ 5(xy² + 2). Out of this total money he spent ₹ (10 – 3xy²) on his birthday party. How much money is left with him? [NCERT Exemplar]
Solution:
Money give by Rohan’s mother = ₹ 3xy²
Money given by his father = ₹ 5(xy² + 2)
Total money given to him = ₹ 3xy² + ₹ 5 (xy² + 2)
= ₹ [3xy² + 5(xy² + 2)]
= ₹ (3xy² + 5xy² + 10)
= ₹ (8xy² + 10).
Money spent by him = ₹ (10 – 3xy)²
Money left with him = ₹ (8xy² + 10) – ₹ (10 – 3xy²)
= ₹ (8xy² + 10 – 10 + 3x²y)
= ₹ (11xy²)
Hence, the required money = ₹ 11xy²

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 3 Fibre to Fabric

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Fibre to Fabric Class 7 Science Extra Questions Long Answer Type

Exercise 17.1

Exercise 17.2

Exercise 17.3

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Chapter - 2 Nutrition in Animals

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Nutrition in Animals Class 7 Science Extra Questions Long Answer Type

Exercise 14.1

Exercise 14.2

Exercise 14.3

Exercise 14.4

Exercise 14.5

Exercise 13.1

Exercise 13.2

Exercise 13.3

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Chapter - 1 Nutrition in Plants

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Nutrition in Plants Class 7 Science Extra Questions Long Answer Type

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Chapter - 14 Symmetry

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Chapter - 12 Algebraic Expressions

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Algebraic Expressions Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

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