Selina Class 7 ICSE Solutions

Selina Concise Biology Class 7 ICSE Solutions – Plant And Animal Tissues

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Selina Concise Biology Class 7 ICSE Solutions – Plant And Animal Tissues

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A PlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Biology. You can download the Selina Concise Biology ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Biology for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 7 Biology Chapter 1 Plant And Animal Tissues

Synopsis

  • The unit of level of organisation is independent in its mode of existence and activity.
  • All multi cellular organisms start their life as a single cell.
  • Plant tissues are basically of two type
    1.  meristematic
    2.  permanent or non-dividing
  • The permanent – plant tissues are further of three types
    1. protective
    2. supportive: parenchyma, collenchyma, sclerenchyma
    3. conductive: xylem, phloem
  • Parenchymal cells have thin-walled cells and usually with a vacuole.
  • Potatoes mainly are composed of parenchymal cells.
  • Collenchyma are parenchymatous cells which are elongated and are thick at the comers. This helps to support the parts of the plant.
  • Sclerenchyma tissue is formed of long, narrow and thick cells. This provides strength to the plant parts.
  • Xylem is formed of thick-walled, tubular and often dead cells. They transport water and minerals absorbed by roots.
  • Old xylem forms the wood.
  • Phloem is formed of living tubular cells which provide a passage for the downward transport of food.
  • The four major groups of animal tissues

    1. epithelial tissue
    2. connective tissue
    3. muscular tissue
    4. nervous tissue
  • The epithelial tissue is further of four types:

    1. squamous epithelium (protective)
    2. cuboidal epithelium (absorption)
    3. columnar epithelium (secretory)
    4. ciliated epithelium (movement of substances)
  • Supportive connective tissue consists of
    1. Cartilage
    2.  Bone
  • Fibrous connective tissue:
    It packs and binds most of the organs. It is of the following types.

    1. areolar tissue: binds skin to underlying tissue.
    2. adipose tissue: filled with fat.
    3. tendon: connect muscles to bones.
    4. ligaments: connect bone to another bone.
  • Fluid connective tissue consists of
    1. Blood
    2. Lymph
  • The liquid part of the blood is called plasma and the cellular part includes:
    1. red blood cells
    2.  white blood cells
    3.  platelets.
  • Three distinct kinds of muscles are
    1. striated or skeletal
    2. unstriated or smooth
    3. cardiac or heart.
  • A nerve cell is formed of a cell body called cyton and one or more elongated hair-like extensions called dendrites. The longest dendrite is called axon.
  • Systems of the body with their primary vital function.Skeletal system: support and protection
    1. Muscular system: movement
    2. Digestive system: nutrition
    3. Respiratory system: exchange of gases
    4. Circulatory system: transport of materials
    5. Excretory system: waste removal
    6. Nervous system: sensation and co-ordination
    7.  Reproductive system: continuation of race.

Review Questions

MULTIPLE CHOICE QUESTIONS

1. Put a tick (✓) against the most appropriate alternative in the following statements.

(i) A group of similar cells to perform a specific function forms a
(a) organ
(b) species
(c) organ system
(d) tissue

(ii) The fine branches given out from the cell body of a nerve cell are
(a) dendrites
(b) cyton
(c) axon
(d) neurons

(iii) Fluid connective tissue of humans is
(a) blood and cartilage
(b) lymph and plasma
(c) blood and lymph
(d) stroma and matrix

Short Answer Questions

Question 1.
1. Define the following terms:

  1. Tissue
  2. Organ

Answer:

  1. Tissue: A group of similar cells which perform a specific function.
    example: Muscular tissue in animals.
  2. Organ: The different type of tissues which group together to function in a co-ordinated manner.
    example: liver

2. Answer the following:

Question 2(i).
What is a meristematic tissue ? How is it different from permanent tissues ?
Answer:
Plant tissues are classified into two types:

  1. Meristematic tissue
  2.  Permanent or non-diving tissue

Meristematic tissues are the plant tissues which are made up of actively dividing cells. These tissues actively divide and lead to the growth of the plant body. They are found at the growth points of the plant like tips of root, stem and branches etc.

  1. Cells are small with thin cell walls.
  2. Cells have large and conspicuous nuclei.
  3. Cells have no vacuoles.
  4. Cells are actively dividing type cells.

Difference between Meristematic and permanent
Meristematic tissue :

  1. Meristematic tissue is present at the tip of the root and stem and in between the xylem and phloem. Form apical meristematic tissue when present at the tips. It is in the form of cambium in between the xylem and phloem.
  2. Meristematic cells divide and form other types of tissues. The cells are thin walled.
  3. Meristematic cells may be intercalary as in case of monocots.
  4. The cells are small and isodiametric, vacuoles are small or absent.
  5. Respiratory and biosynthetic activities maximum.
  6. The cells are immature and mitochondria simple.
  7. Proplastids act as plastids.

Permanent tissue

  1. Permanent tissue may be simple as parenchyma, collenchyma or sclerenchyma and it may be complex as xylem and phloem.
  2. These are made up of more than one kind of cells. These perform a common function Xylem and phloem form vascular system of the plant. These cells do not have the power to divide.
  3. These cells may act as epidermis cortex or grit cells. Sclerenchyma gives strength.
  4. Living cells of permanent tissue have vacuoles. The cells are large and of different shapes.
  5. Both these activities are low.
  6. The cells fully mature, mitochondria fully developed.
  7. Living cells have plastids.

Question 2(ii).
Which living material would you take to demonstrate meristematic tissue ?
Answer:
Green gram seeds can be used to demonstrate meristematic tissue which when soaked in a petridish stuffed with wet cotton and left for 3-4 days would sprout out. These sprouted seeds have roots developing whose root tips have meristematic tissue.

Question 2(iii).
What is the function of meristematic tissue ?
Answer:
The meristematic tissue have the primary role in the growth of the plant tissue as it consists of active dividing cells

Question 3.
State whether the following statements are True or False. 

(i) A tissue is formed of only one type of cells.
Ans. True

(ii) Only one type of tissue forms an organ.
Ans. False.
Correct: Two or more types of tissue form an organ.

(iii) Permanent tissue is made up of undifferentiated and dividing Cells.
Ans. False.
Correct: Meristematic tissue is made up of undifferentiated and dividing cells.

(iv) Meristematic tissue is found at growing tips of a plant.
Ans. True

(v) Phloem is formed of dead tubular cells.
Ans. False.
Correct: Phloem is formed of living tubular cells.

Question 4.
Fill in the blanks by selecting suitable words from the list given below:
“Thin – walled, collenchyma, vascular, tissues, conducting”

  1. A group of different tissues working together to perform a function is called an organ.
  2.  Xylem and phloem form the conducting tissue.
  3.  Conducting tissue is also called vascular tissue.
  4. Cells are elongated and thick at the comers in collenchyma tissue.
  5. Parenchyma is composed of large thin-walled cell

Question 5.
Match the items given is column A with those given in column B:

Column A
(i) Fibrous connective tissue
(ii) Fluid connective tissue
(iii) Supportive connective tissue
(iv) Ligament
(v) Tendon
Column B
(a) blood
(b) cartilage
(iii) Supportive connective tissue
another bone.
(d) areolar tissue
(e) connects a muscle
with a bone.
Selina Concise Biology Class 7 ICSE Solutions - Plant And Animal Tissues 1

Question 6.
How do you rank the following among cells, tissues, organs, or organism ?

  1. Amoeba : organism
  2.  Euglena: organism
  3. Skin : organ
  4. Lungs : organ
  5. Neuron : tissue
  6. Cardiac muscles: Ti1ue

Question 7.
Each of the tissues listed in Column A is related to one of the functions given in Column B. Match the lines correct pairs by drawing
Selina Concise Biology Class 7 ICSE Solutions - Plant And Animal Tissues 2

Question 8.
Name the kind of tissue that

  1.  Carries oxygen around your body — Blood tissue.
  2.  Brings about movements in animals — muscular tissue.
  3.  Transports food to different parts of plant— phloem.
  4.  Transports water in plants — xylem.
  5.  Supports an animal’s body — connective tissue (supportive)
  6.  Binds different tissues together — Fibrous connective tissue.
  7.  Conducts messages from one part of the body to another — nervous tissue.

Question 9.
Based on the following information, identify the three types of epithelial tissue in the figures given below :
Selina Concise Biology Class 7 ICSE Solutions - Plant And Animal Tissues 3

(i) Cuboidal epithelium : It consists of a single layer of cuboidal cells.

(ii) Columnar epithelium: It is composed of tall, cylindrical cells with oval nuclei usually placed at the base of the cells.

(iii) Ciliated epithelium : It consists of cells being hair-like cilia on their free surface.
Answer:
(i) fig. b (ii) fig. a (iii) fig. c

Question 10.
Write three differences between the two principal vascular tissues found in plants.
Answer:
Xylem

  1. Transports water and minerals absorbed by the roots to other plant parts.
  2. Consists mainly of dead cells.
  3. Conduction is unidirectional i.e. only upwards from the roots.

Phloem

  1. Conducts food manufactured in the leaves to other plant parts.
  2. Consists mainly of living cells.
  3. Bidirectional conduction i.e. both upwards and downwards from the leaves.

 

Selina Concise Biology Class 7 ICSE Solutions – Classification of Animals

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Selina Concise Biology Class 7 ICSE Solutions – Classification of Animals

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Biology. You can download the Selina Concise Biology ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Biology for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 7 Biology Chapter 3 Classification of Animals

Synopsis

  •  The animals can move from one place to another in search of food and shelter and this is called locomotion.
  •  The vertebrates can be classified into five classes:
    1.  Pisces (Fishes)
    2.  Amphibia (Frogs)
    3.  Reptilia (Lizards and Snakes)
    4.  Aves (Birds)
    5. (Mammalia (Milk – nourishing animals)
  • Pisces / Fishes
    1.  have streamlined body shape
    2.  Locomotion with the help of fins
    3.  Body covered with scales.
    4.  Breathe through gills.
    5. Example: Dogfish, Catla
  •  Amphibia / Frogs
    1.  can live in water as well as on land.
    2. always lay their eggs in water.
    3. body covered by a slimy and slippery skin
    4. breathe through lungs and skin.
    5. Example: Frog and toads.
  •  Reptilia
    1.  Mostly live on land
    2.  Skin is dry and scaly
    3.  Breathe through lungs
    4. Females lay eggs on land
    5. Example: Lizards, snakes, crocodiles
  • Aves / Birds
    1. Body covered with feathers.
    2. Have wings to fly.
    3. Scales only on legs.
    4. Have jaws with homy beak and have no teeth.
    5. Example: Pigeon, hen
  • Mammalia / Milk – nourishing animals.
    1. Body covered with hairs.
    2. Posses projecting external ears.
    3. Give birth to young ones.
    4. Mothers suckle their young ones.
    5. Have a tail and four limbs. (Tail may become vestigeal)
    6. Example: dog, tiger, man.
  • Invertebrates can be further divided into nine groups.
    1. Protozoans
    2. Porifera
    3. Coelenterates
    4. platyhelminths
    5. Nemathelminths
    6. Annelids
    7. Molluscs
    8. Arthropoda
    9. Echinoderms
  • Coelenterates
    1. Now called cnidarians
    2. Body is tube like with only one opening called the mouth.
    3. Mouth is surrounded by finger like processes called tentacles for catching food.
    4. Body radially symmetrical
    5. Example: Hydra, Sea-anemone, jelly fish
  • Flatworms / Platyhelminths:
    are usually found as parasites in the bodies of other animals.
    Example: Tapeworm, liver fluke.
  •  Ascaris: The round worm is found in the small intestine of especially those who eat with the unwashed hands.
  •  Annelids:
    1.  are also called segmented worms
    2.  body is composed of rings or segments
    3.  have a body cavity.
    4. have special organs of excretion called nephridia.
      Example: earthworm, leech.
  •  Arthropods can be further divided into
    1.  Crustacea : head and thorax are fused and have many jointed legs.
      Example: crab, lobsters etc.
    2.  Myriagoda: Body is divided into many segments and has one or two pairs of legs on each segment.
      Example: Centipede, millipede.
  •  Insecta: Body is divided into three regions – head, thorax and abdomen.
    — Has three pairs of legs.
    — Have two pairs of wings.
    Example: ant, housefly, butterfly.
  •  Arachnida: Head and thorax fused
    — Have four pairs of legs.
    — Have no wings.
    Example: Spider, Scorpion
  •  Echinoderms
    — also called spiny-skinned animals.
    — Body is star – like or ball – like
    — Have no head or tail.
    — Have no left or right side.
    Example: Starfish, sea urchin.
  • A species can be defined as a group of individuals having common characteristics and which come together to pro¬duce young ones.
  •  Scientific name consists of two parts. The first part is the genus name while the second part is the species name.
    This type of naming is called Binomial nomenclature.
  • The animals can be classified also on the basis of their food habits into as follows.
    (a) Herbivorous:Feed on plants e.g. cow, goat.
    (b) Carnivorous:Feed on the flesh of other animals  e.g. lion, tiger etc.
    (c) Omnivorous:Feed on both plants as well as flesh of other animals, e.g. man, bear etc.
    (d) Parasites:Live either inside or on the outside of the body of other animals and plants and take food from them.
    Example: Leech, mosquitoes etc.

Activity 3
Look at the four animals shown alongside.
Which four classes of vertebrates are represented by them ? Name these classes.
Selina Concise Biology Class 7 ICSE Solutions - Classification of Animals 1

Answer:
1. Class Mammalia
2. Class Mammalia
3. Class Reptilia
4. Class Pisces

Review Questions

MULTIPLE CHOICE QUESTIONS

1. Tick (✓) the appropriate answer:

(i) Identify the aquatic animal with scaly skin which breathe with gills –
(a) Rohu
(b) Tortoise
(c) Sparrow
(d) Rat

(ii) The unicellular organism causing malaria –
(a) Amoeba
(b) Paramecium
(c) Euglena
(d) Plasmodium

(iii) Identify the animal which is not an Arthropoda —
(a) Prawn
(b) Butterfly
(c) Earthwonn
(d) Spider

(iv) Scientist who introduced binomial nomenclature is —
(a) Charles Darwin
(b) Carolus Linnaeus
(c) Robert Hooke
(d) Gregor Mendel

Short Answer Questions
1. Give two examples of each of the following:
(i) Amphibians:
Ans. Amphibians: 1. Frog 2. Toad
(ii) Segmented worms:
Ans. Segmented worms: 1. Earthworm 2. Leech
(iii) Reptiles:
Ans. Reptiles: 1. Snake 2. Lizard
(iv) Coelenterates:
Ans. Coelenterates : 1. Hydra 2. Jellyfish
(v) Arthropods:
Ans. Arthropods: 1. Crab 2. Centipede
(vi) Flatworms:
Ans. Flatworms: 1. Tapeworm 2. Liverfluke

2. Give names of two animals which are found as parasites inside the human intestine.
Ans. (a) Tapeworm (b)Ascaris ’

3. Name one example each of an animal which shows the following characteristics:
(i) Fixed animals with a pore-bearing body:
Ans. Fixed animals with a pore-bearing body: sponge
(ii) Star-shaped body:
Ans. Star-shaped body: Star-fish
(iii) Can live in water as well as on land:
Ans. Can live in water as well as on land: Frog
(iv) Has a flattened ribbon-like body:
Ans. Has a flattened ribbon-like body: Tapeworm

4. Write one difference each between the following pairs:
(i) Porifera and Coelenterata.
(ii) Arthropoda and mollusca.
(iii) Invertebrates and Vertebrates
(iv) Platyheminthes and Nematoda
Answer:
(i) Porifera and Coelenterata.
 Porifera

  1.  Body is porous i. e. bears many tiny pores to draw water into the body cavity.
  2.  e.g. Sponge

Coelenterata

  1. Sac-like body with only one opening i.e. mouth.
  2. e.g. Jelly fish, hydra,sea-anemone.

(ii) Arthropoda and mollusca.
Arthropoda 

  1.  These are animals with
  2.  They have segmented body.
  3.  They may or may not have wings
    Example: Crab.

Mollusca

  1.  Move with the help of a muscular foot.
  2.  Soft body which is not segmented.
  3.  Body enclosed in a hard shell Example: Octopus

(iii) Invertebrates and Vertebrates

Invertebrates

  1.  The animals which do not have a back bone.
  2.  They are further classified into nine groups.
    Example: Octopus, Starfish.

Vertebrates

  1.  The animals which have a back bone or a vertebral column.
  2.  They are further classified in to five groups.
    Example: Human Being, Lizard.

(iv) Platyheminthes and Nematoda
Platyheminthes

  1.  Body thin and flattened.
  2.  Mostly live as parasites in the bodies of other animals (hosts)
    e.g. Tapeworm.

Nematoda

  1. Body is rounded and unsegmented.
  2. Mostly live as parasites in the body of animals including humans.
    e.g. Roundworm commonly called Ascaris.

5. Match the animals given under column A with their respective classification group given under column B –

Column A                                          Column B
Selina Concise Biology Class 7 ICSE Solutions - Classification of Animals 2
Selina Concise Biology Class 7 ICSE Solutions - Classification of Animals 3
Answer:

Selina Concise Biology Class 7 ICSE Solutions - Classification of Animals 4
  

6. Write the characteristics of class Aves with reference to their body covering and jaws.
Answer:
The characteristics of class Aves are:

  1.  Body is covered with feathers.
  2.  They have wings to aid flying
  3.  They have scales on legs.
  4.  They have no teeth.
  5. They have jaws provided with homy beaks

7. Categorise the following animals under their appropriate columns of classification.
Selina Concise Biology Class 7 ICSE Solutions - Classification of Animals 5

Answer:
Worms – Arthropods, Butterfly, Ascaris, Scorpion, Honey bee, Liverfluke, Leech, grasshopper, Eathworm
Molluscs – Snail
Fishes – Rohu
Amphibians – Toad, Frog
Reptiles – Snake, Lizard, Turtle
Birds – Parrot, Pigeon
Mammals – Rat, Bat, Dog, Cattle, Cow, Rabbit, Monkey, Elephant

Selina Concise Mathematics class 7 ICSE Solutions – Data Handling

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Selina Concise Mathematics Class 7 ICSE Solutions Chapter 21 Data Handling

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 21 Data Handling

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Data Handling Exercise 21A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Consider the following numbers :
68, 76, 63, 75, 93, 83, 70, 115, 82, 105, 90, 103, 92, 52, 99, 73, 75, 63, 77 and 71.
(i) Arrange these numbers in ascending order.
(ii) What the range of these numbers?
Solution:
(i) When the above data are written in ascending order. We get,
52, 63, 63, 68, 70, 71, 73, 75, 75, 76, 77, 82, 83, 90, 92, 93, 99, 103, 105, 115
(ii) Range of given numbers = Largest number – Smallest number
= 115-52 = 48

Question 2.
Represent the following data in the form of a frequency distribution table :
16, 17, 21, 20, 16, 20, 16, 18, 17, 21, 17, 18, 19, 17, 15, 15, 19, 19, 18, 17, 17, 15, 15, 16, 17, 17, 19, 18, 17, 16, 15, 20, 16, 17, 19, 18, 19, 16, 21 and 17.
Solution:
The frequency distribution for these data will be as shown below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 2

Question 3.
A die was thrown 20 times and following scores were recorded.
2, 1, 5, 2, 4, 3, 6, 1, 4, 2, 5, 1, 6, 2, 6, 3, 5, 4, 1 and 3.
Prepare a frequency table for the scores.
Solution:
The frequency table for the scores will be as shown below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 3

Question 4.
Following data shows the weekly wages (in ₹) of 10 workers in a factory.
3500, 4250, 4000, 4250, 4000, 3750, 4750, 4000, 4250 and 4000
(i) Prepare a frequency distribution table.
(ii) What is the range of wages (in ₹)?
(iii) How many workers are getting the maximum wages?
Solution:
(i) The frequency table for the wages of 10 workers will be as shown below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 4
(ii) Range of wages (₹) = ₹4750 – ₹3500 = ₹1250
(iii) One

Question 5.
The marks obtained by 40 students of a class are given below :
80, 10, 30, 70, 60, 50, 50, 40, 40, 20, 40, 90, 50, 30, 70, 10, 60, 50, 20, 70, 70, 30, 80, 40,20, 80, 90, 50, 80, 60, 70, 40, 50, 60, 90, 60, 40, 40, 60 and 60
(i) Construct a frequency distribution table.
(ii) Find how many students have marks equal to or more than 70?
(iii) How many students obtained marks below 40?
Solution:
(i) The frequency distribution table will be shown as below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 5
(ii)Students have marks equal to or more than 70 = 5 + 4 + 3 = 12
(iii) Students obtained marks below 40 = 2 + 3 + 3 = 8 students

Question 6.
Arrange the following data in descending order:
3.3, 3.2, 3.1, 3.7, 3.6, 4.0, 3.5, 3.9, 3.8, 4.1, 3.5, 3.8, 3.7, 3.9 and 3.4.
(i) Determine the range.
(ii) How many numbers are less than 3.5?
(iii) How many numbers are 3.8 or above?
Solution:
Descending order : 4.1, 4.0, 3.9, 3.9, 3.8, 3.8, 3.7, 3.7, 3.6, 3.5, 3.5, 3.4, 3.3, 3.2, 3.1
(i) Range = 4.1 – 3.1 = 1
(ii) Number less than 3.5 = 4
i.e., 3.4, 3.3, 3.2, 3.1
(iii) Number are 3-8 or above = 6
i.e., 3.8, 3.8, 3.9, 3.9, 4.0, 4.1

Data Handling Exercise 21B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find the mean of 53, 61, 60, 67 and 64.
Solution:
Mean of 53, 6i, 60, 67 and 64
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 6

Question 2.
Find the mean of first six natural numbers.
Solution:
First six natural numbers are : 1, 2, 3, 4, 5, 6
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 7

Question 3.
Find the mean of first ten odd natural numbers.
Solution:
First ten odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 8

Question 4.
Find the mean of all factors of 10.
Solution:
The factor of 10 are 2 and 5
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 9

Question 5.
Find the mean of x + 3, x + 5, x + 7, x + 9 and x + 11.
Solution:
Mean of x + 3, x + 5, x + 7, x + 9 and x + 11
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 10

Question 6.
If different values of variable x are 19.8,15.4,13.7,11.71,11.8, 12.6,12.8,18.6,20.5 and 2.1, find the mean.
Solution:
19. +15.4 +13.7 +11.71 +11.8 +12.6 + 12.8 +18.6 + 20.5 +21.1
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 11

Question 7.
The mean of a certain number of observations is 32. Find the resulting mean, if each observation is,
(i) increased by 3
(ii) decreased by 7
(iii) multiplied by 2
(iv) divided by 0.5
(v) increased by 60%
(vi) decreased by 20%
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 12

Question 8.
The pocket expenses (per day) of Anuj, during a certain week, from monday to Saturday were ₹85.40, ₹88.00, ₹86.50, ₹84.75, ₹82.60 and ₹87.25. Find the mean pocket expenses per day.
Solution:
The pocket expenses (per day) during a certain week are : ₹85.40, ₹88.00, ₹86.50, ₹84.75, ₹82.60 and ₹87.25
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 13

∴Anuj expenses per day = ₹85.75

Question 9.
If the mean of 8, 10, 7, x + 2 and 6 is 9, find the value of x.
Solution:
The mean 8, 10, 7, x + 2 and 6 is 9
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 14

Question 10.
Find the mean of first six multiples of 3.
Solution:
The six multiples of 3 are 3, 6, 9, 12, 15, 18
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 15

Question 11.
Find the mean of first five prime numbers.
Solution:
The first five prime numbers are 2, 3, 5, 7, 11
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 16

Question 12.
The mean of six numbers :x-5,x- 1, x, x + 2, x + 4 and x + 12 is 15. Find the mean of first four numbers.
Solution:
The mean of six numbers are x – 5, x – 1,x,x + 2,x + 4 and x + 12 is 15
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 17
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 18

Question 13.
Find the mean of squares of first five whole numbers.
Solution:
First five whole numbers are 0, 1, 2, 3, 4
Then square the whole prime numbers
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 19

Question 14.
If the mean of 6, 4, 7, p and 10 is 8, find the value of p.
Solution:
The mean of 6, 4, 7, p and 10 is 8
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 20

Question 15.
Find the mean of first six multiples of 5.
Solution:
Six multiples of 5 are :
5, 10, 15, 20, 25 and 30
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 21

Question 16.
The rainfall (in mm) in a city on 7 days of a certain week is recorded as follows
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 22
Find the total and average (mean) rainfall for the week.
Solution:
The rainfall in a city on 7 days are 0.5, 2.7, 2.6, 0.5, 2, 5.8, 1.5
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 23

Question 17.
The mean of marks scored by 100 students was found to be 40, later on it was discovered that a score of 53 was misread as 83. Find the correct mean.
Solution:
Mean of 40 observations = 100
Total sum of 40 observations = 100 × 40 = 4000
Incorrect total of 40 observation is = 4000
Correct total of 40 observations = 4000 – 83 + 53 = 3970
∴ Correct mean = \(\frac { 3970 }{ 100 }\) = 39.70

Question 18.
The mean of five numbers is 27. If one number is excluded, the mean of remaining numbers is 25. Find the excluded number.
Solution:
Mean of 5 observations = 27
Total sum of 5 observations = 27 × 5 = 135
On excluding an observation, the mean of remaining 6 observations = 25
⇒ Total of remaining 4 observations = 25 x 4 = 100
⇒ Included observation = Total mean of 5 observations – Total mean of 4 observations
= 135- 100 = 35

Question 19.
The mean of 5 numbers is 27. If one new number is included, the new mean is 25. Find the included number.
Solution:
Mean of 5 observations = 27
Total sum of 5 observations = 27 x 5 = 135
On including an observation the mean of 6 observation = 25 x 6 = 150
⇒ Included observations = Total Mean of 6 observations – Total mean of 5 observations = 150- 135 = 15

Question 20.
Mean of 5 numbers is 20 and mean of other 5 numbers is 30. Find the mean of all the 10 numbers taken together.
Solution:
The mean of 5 number = 20
Then, mean of other 5 number = 30
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 24

Question 21.
Find the median of:
(i) 5,7, 9, 11, 15, 17,2, 23 and 19
(ii) 9, 3, 20, 13, 0, 7 and 10
(iii) 18, 19, 20, 23, 22, 20, 17, 19, 25 and 21
(iv) 3.6, 9.4, 3.8, 5.6, 6.5, 8.9, 2.7, 10.8, 15.6, 1.9 and 7.6.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 25
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 27
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 28

Question 22.
Find the mean and the mode for the following data :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 29
Solution:
We prepare the table given below :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 30

Question 23.
Find the mode of:
(i) 5, 6, 9, 13, 6, 5, 6, 7, 6, 6, 3
(ii) 7, 7, 8, 10, 10, 11, 10, 13, 14
Solution:
(i) Arranging the Numbers in ascending order : 3, 5, 5, 6, 6, 6, 6, 6, 7, 9, 13
Mostly repeated term = 6
∴ Mode = 6
(ii) Arranging the Numbers in ascending order = 7, 7, 8, 10, 10, 10, 11, 13, 14
Mostly repeated term =10
∴ Mode = 10

Question 24.
Find the mode of :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 31

Solution:
(i) Since, the frequency of number 18 is maximum
∴Mode = 18
(ii) Since, the frequency of number 41 is maximum
∴ Mode = 41

Question 25.
The heights (in cm) of 8 girls of a class are 140,142,135,133,137,150,148 and 138 respectively. Find the mean height of these girls and their median height.
Solution:
Arranging in ascending order : 133, 135, 137, 138, 140, 142, 148, 150
Here, number of girls = 8 which is even
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 32

Question 26.
Find the mean, the median and the mode of:
(i) 12, 24, 24, 12, 30 and 12
(ii) 21, 24, 21, 6, 15, 18, 21, 45, 9, 6, 27 and 15.
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 33.
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 34

Question 27.
The following table shows the market positions of some brands of soap.
Draw a suitable bar graph :
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 35
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 36

Question 28.
The birth rate per thousand of different countries over a particular period of time is shown below.
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 37
Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Data Handling imae - 38

Selina Concise Mathematics class 7 ICSE Solutions – Unitary Method (Including Time and Work)

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Selina Concise Mathematics class 7 ICSE Solutions – Unitary Method (Including Time and Work)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 7 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

The method in which the value of a unit (one) quantity is first calculated to get the value of any other quantity is called the unitary method.
In unitary method, we come across two types of variations :
(i) Direct-variation
(ii) Inverse-variation.

(i) Direct variation : Increase in one quantity causes increase in the other and decrease in one quantity causes decrease in the other.
(ii) Inverse variation : Increase in one quantity causes decrease in the other and decrease in one quantity causes increase in the other.
This is found in the sums of speed, work done etc.

EXERCISE 7 (A)

Question 1.
Weight of 8 identical articles is 4.8 kg. What is the weight of 11 such articles ?

Answer:
Weight of 8 articles = 4.8 kg
Weight of 1 article = \(\frac { 4.8 }{ 8 }\) kg
and weight of 11 articles =\(\frac { 4.8 }{ 8 }\) x 11 kg
= 0.6 x 11 = 6.6 kg

Question 2.
6 books weigh 1 .260 kg. How many books will weigh 3.150 kg ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 1

Question 3.
8 men complete a work in 6 hours. In how many hours will 12 men complete the same work ?

Answer:
8 men can complete a work in = 6 hours
1 man can complete the work in = 6×8 hours
12 men can complete the work in = \(\frac { 6 x 8 }{ 12 }\) = 4 hours

Question 4.
If a 25 cm long candle burns for 45 minutes, how long will another candle of the same material and same thickness but 5 cm longer than the previous one, burn ?

Answer:
25 cm long candle burn in = 45 minutes
1 cm long candle will burn in = \(\frac { 45 }{ 25 }\) mintues
25 + 5 = 30 cm long candle will burn in
= \(\frac { 45 x 30 }{ 25 }\)minutes = 54 minutes

Question 5.
A typist takes 80 minutes to type 24 pages. How long will he take to type 87 pages ?

Answer:
For typing 24 pages, time is required = 80 minutes
For typing 1 page, time is required =\(\frac { 80 }{ 24 }\) minutes
and for typing 87 pages, time is required
= \(\frac { 80 x 87 }{ 24 }\) minutes = 290 minutes

Question 6.
Rs. 750 support a family for 15 days. For how many days will Rs. 2,500 support the same family ?

Answer:
Rs. 750, can support a family for = 15 days
Re. 1 will support for = \(\frac { 15 }{ 750 }\)days
and Rs. 2,500 will support for = \(\frac { 15 }{ 750 }\)x 2500 days = 50 days

Question 7.
400 men have provisions for 23 weeks. They are joined by 60 men. How long will the provisions last ?

Answer:
400 men have provisions for = 23 weeks
1 man will have provisions for = 23 x 400 weeks
and 400 + 60 = 460 men will have provisions for = \(\frac { 23 x 400 }{ 460 }\) weeks = 20weeks

Question 8.
200 men have provisions for 30 days. If 50 men left, the same provisions would last for the remaining men, in how many days?

Answer:
200 men have provisions for = 30 days
1 man will have provisions for = 30 x 200 days
200 – 50 = 150 men will have provisions
for = \(\frac { 30 x 200 }{ 150 }\) days = 40 days

Question 9.
8 men can finish a certain amount of provisions in 40 days. If 2 more men join with them, find for how many days the same amount of provisions be sufficient ?

Answer:
8 men can finish a provision in = 40 days
1 man will finish in = 40 x 8 days
8+2=10 men will finish in =\(\frac { 40 x8 }{ 10 }\)
= 32 days

Question 10.
If interest on Rs. 200 be Rs. 25 in a certain time, what will be the interest on Rs 750 for the same time ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 2
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 3

Question 11.
If 3 dozen eggs cost Rs. 90, find the cost of 3 scores of eggs. (1 score = 20)

Answer:
3 dozen = 3 x 12 = 36 eggs,
3 scores = 3 x 20 = 60
The cost of 36 eggs is = Rs. 90
The cost of 1 egg will be = Rs. \(\frac { 90 }{ 36 }\)
∴ Cost of 60 eggs will be = Rs. \(\frac { 90 x60 }{ 36 }\)
= Rs. 150

Question 12.
If the fare for 48 km is Rs. 288, what will be the fare for 36 km ?

Answer:
Fare for 48 km = Rs. 288
fare for 1 km = Rs. \(\frac { 288 x 36 }{ 48 }\) = Rs. 216

Question 13.
What will be the cost of 3.20 kg of an item, if 3 kg of it costs Rs. 360 ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 4

Question 14.
If 9 lines of a print, in a column of a book contains 36 words. How many words will a column of 51 lines cqntain ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 5

Question 15.
125 pupil have food sufficient for 18 days. If 25 more pupil join them, how long will the food last now ? What assumption have you made to come to your answer ?

Answer:
Pupils in the beginning = 125
More pupils joined = 25
Total pupils = 125 + 25 = 150
Food is sufficient for 125 pupils for = 18 days
Food will be sufficient for 1 pupil for = 18 x 125 days (less pupil more days)
and food will be sufficient for 150 pupils = \(\frac { 18 x 125 }{ 150 }\) days (more pupil more days)
= \(\frac { 18 x 5 }{ 6 }\) 15 days

Question 16.
A carpenter prepares a new chair in 3 days, working 8 hours a day. Atleast how many hours per day must he work in order to make the same chair in 4 days ?

Answer:
A chair is completed in 3 days working per day = 8 hours
Then their will be completed in 1 day working for = 8 x 3 hours per day (less days more hours)
and it will be completed in 4 days working for = \(\frac { 8 x 3 }{ 4 }\)= 6 hours per day.

Question 17.
A man earns ₹5,800 in 10 days. How much will he earn in the month of February of a leap year?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 6

Question 18.
A machine is used for making rubber balls and makes 500 balls in 30 minutes. How many balls will it make in 3\(\frac { 1 }{ 2 }\) hours?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 7

Question 19.
In a school’s hostel mess, 20 children consume a certain quantity of ration in 6 days. However, 5 children did not return to the hostel after holidays. How long will the same amount of ration last now?

Answer:
Total number of children = 20
20 children consume a certain quantity of ration in = 6 days
1 children consume a certain quantity of ration in = 6 x 20 days
As 5 children did not return to the hostel after holidays.
Then number of children in hostel = 20-5 = 15
Hence, 15 children consume certain quantity 6×20
of ration in = \(\frac { 6 x 20 }{ 15 }\) days = 8 days

EXERCISE 7 (B)

Question 1.
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 8

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 9

Question 2.
3\(\frac { 1 }{ 2 }\)m of cloth costs Rs. 168 ; find the cost of 4\(\frac { 1 }{ 3 }\)m of the same cloth.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 10

Question 3.
A wrist watch loses 10 sec in every 8 hours; in how much time will it lose 15 sec. ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 12

Question 4.
In 2 days and 20 hours, a watch gains 20 sec ; find how much time will the watch take to gain 35 sec. ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 13

Question 5.
50 men mow 32 hectares of land in 3 days. How many days will 15 men take to mow it?

Answer:
Land is same in both the cases.
Now 50 men can mow land in = 3 days
∴ 1 man will mow it in = 3 x 50 days
and 15 men will mow it in = \(\frac { 3 x 50 }{ 15 }\) = 10 days

Question 6.
The wages of 10 workers for a six days week are Rs, 1,200. What are the one day wages: (i) of one worker ? (ii) of 4 workers?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 14

Question 7.
If 32 apples weigh 2 kg 800 g. How many apples will there be in a box, containing 35 kg of apples ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 15

Question 8.
A truck uses 20 litres of diesel for 240 km. How many litres will be needed for 1200 km?

Answer:
For 240 km, diesel is needed = 20 litres
∴ for 1 km, diesel will be needed 20

Question 9.
A garrison of 1200 men has provisions for 15 days. How long will the provisions last if the garrison be increased by 600 men ?

Answer:
1200 men has provision for = 15 days
1 man will have that provision for = 15 x 1200 days
∴1200 + 600 = 1800 men will has that provisions for =\(\frac { 15 x 1200 }{ 1800 }\)days
= 10 days

Question 10.
A camp has provisions for 60 pupil for 18 days. In how many days, the same provisions will finish off if the strength of the camp is increased to 72 pupil ?

Answer:
60 pupil have provision for = 18 days 1 pupil will have provision for = 18 x 60 days (less pupils more days)
and 72 pupils will have provision for = \(\frac { 18 x 60 }{ 72 }\) days
= 15 days.

EXERCISE 7 (C)

Question 1.
A can do a piece of work in 6 days and B can do it in 8 days. How long will they take to complete it together ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 16

Question 2.
A and B working together can do a piece of work in 10 days B alone can do the same work in 15 days. How long will A alone take to do the same work ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 17

Question 3.
A can do a piece of work in 4 days and B can do the same work in 5 days. Find, how much work can be done by them working together in : (i) one day (ii) 2 days.
What part of work will be left, after they have worked together for 2 days ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 18

Question 4.
A and B take 6 hours and 9 hours respectively to complete a work. A works for 1 hour and then B works for two hours.
(i) How much work is done in these 3 hours ?
(ii) How much work is still left ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 19

Question 5.
A, B and C can do a piece of work in 12, 15 and 20 days respectively. How long will they take to do it working together ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 20

Question 6.
Two taps can fill a cistern in 10 hours and 8 hours respectively. A third tap can empty it in 15 hours. How long will it take to fill the empty cistern, if all of them are opened together ?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 21

Question 7.
Mohit can complete a work in 50 days, whereas Anuj can complete the same work in 40 days.
Find:
(i) work done by Mohit in 20 days.
(ii) work left after Mohit has worked on it for 20 days.
(iii) time taken by Anuj to complete the remaining work.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 22
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 23

Question 8.
Joseph and Peter can complete a work in 20 hours and 25 hours respectively.
Find :
(i) work done by both together in 4 hrs.
(ii) work left after both worked together for 4 hrs.
(iii) time taken by Peter to complete the remaining work.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 24

Question 9.
A is able to complete \(\frac { 1 }{ 3 }\) of a certain work in 10 hrs and B is able to complete\(\frac { 2 }{ 5 }\) of the same work in 12 hrs.
Find:
(i) how much work can A do in 1 hour ?
(ii) how much work can B do in 1 hour ?
(iii) in how much time will the work be completed, if both work together.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 25

Question 10.
Shaheed can prepare one wooden chair in 3 days and Shaif can prepare the same chair in 4 days. If they work together, in how many days will they prepare :
(i) one chair ?
(ii)14 chairs of the same kind?

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 26

Question 11.
A, B and C together finish a work in 4 days. If A alone can finish the same work in 8 days and B in 12 days, find how long will C take to finish the work.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Unitary Method (Including Time and Work) image - 27

Selina Concise Chemistry Class 7 ICSE Solutions – Air and Atmosphere

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Selina Concise Chemistry Class 7 ICSE Solutions – Air and Atmosphere

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Chemistry. You can download the Selina Concise Chemistry ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Chemistry for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 7 Chemistry ICSE SolutionsPhysicsBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 7 Chemistry Chapter 7 Air and Atmosphere

Points to Remember :

  1. Air is a mixture of many gases, mainly Nitrogen = 78.1%, Oxygen = 20.9%, Carbon dioxide = 0.03 – 0.04%, Inert gases = 0.9%, [Water vapours, Dust particles and Impurities = Variable].
  2. Nitrogen is a colourless, an odourless and a tasteless gas. It is slightly lighter than air.
  3. The process of conversion of free atmospheric nitrogen into its compounds is called nitrogen-fixation.
  4. Oxygen constitutes about 21% of air by volume. It supports life on earth.
  5. Carbon dioxide is present in air in a very small quantity, e. 0.03 – 0.04%. It is essential for the process of photosynthesis.
  6. Inert gases like neon, argon do not react with any substance.
  7. The harmful substances added to air are called pollutants.
  8. Some pollutants are suspended particles like pollen grains, oxides of sulphur and nitrogen, oxides of carbon, chlorofluorocarbons etc.
  9. Symbol of oxygen = O ; atomic number = 8, relative mass = 16, molecule formula = O2.
  10. Oxygen is available in free and combined state.
  11. A catalyst is a substance that increases or decreases the rate of a chemical reaction without itself undergoing any chemical change.
  12. Oxides are binary compounds formed by the chemical combination of substance with oxygen.
  13. Rusting is the process in which iron slowly reacts with oxygen in the air and produces a flaky brown substance.
  14. Photosynthesis is a process by which C02 and water are used up by green plants in the presence of sunlight to produce glucose and oxygen gases.

A. AIR : A MIXTURE OF GASES

EXERCISE — I

Question 1.
Give one use for each of the following inert gases :
(a) argon
(b) helium
(c) neon
(d) radon
(e) krypton
(f) xenon

Answer:
(a) Argon— Argon is filled into electric bulbs to prevent the oxidation of their filaments.
(b) Helium— It is used in filling up weather observation balloons.
(c) Neon— Neon is used for making advertisement sign boards.
(d) Radon— It is used for treatment of Cancer.
(e) Krypton— It is used in photography.
(f) Xenon— It is also used in photography.

Question 2.
Answer the questions put against each of the following constituents of air :
(a) Nitrogen : Explain its significance for plants and animals.
(b) Oxygen : What is the percentage proportion of oxygen in air ? Why is oxygen called active air.
(c) Carbon dioxide : “Although carbon dioxide plays no role in respiration, all life would come to an end if there is no carbon dioxide in air.” Support this statement with relevant facts.
(d) Water vapours : Explain their role in modifying the earth’s climate.
Answer:
(a) Plants convert nitrogen into protein. It is an important constituent of proteins, which are necessary for the growth of animals, plants and human beings. Plants convert nitrogen into proteins.
(b) 20.9%, oxygen is called active air because it supports life on earth. It is essential for the process of combustion.
(c) Carbon dioxide is essential for photosynthesis by which green plants prepare their food. It minimises heat loss by radiation. Thus, it balances the temperature on earth.
(d) Water vapour determine the earth’s climate conditions. It causes rain. It controls the rate of evaporation from the bodies of plants and animals.

Question 3.
Define the following terms :
(a) pollutants
(b) acid rain
(c) Global warming
(d) smog

Answer:
(a) Pollutants : Air contains substances which are harmful to plants and animals. These harmful substances are called pollutants.
(b) Acid rain : When sulphur trioxide and nitrogen oxide present in the air mix with rainwater they form sulphuric acid and nitric acid respectively. Rainwater containing these acids is called acid rain.
(c) Global warming : An increase in the percentage of carbon dioxide, methane, nitrous oxide and chlorofluorocarbon traps the heat causing the temperature of the earth and its surroundings to rise. This is known as global warming.
(d) Smog : Oxides of nitrogen form a mixture of smoke and fog known as smog which affects our eyes too.

Question 4.
“Air is a mixture”. Support this statement citing at least three evidences.
Answer:
“Air is a mixture” The following are in evidences which prove that air is a mixture.

  1. The composition of air varies from place to place and from time to time.
  2. The components of air retain their individual properties.
  3. Liquid air has no definite boiling point.
  4. No energy exchange occurs when the components of air are mixed with each other.

Question 5.
What is air pollution ? What are the harmful effects of sulphur dioxide, nitrogen dioxide and hydrogen sulphide present in the air ?
Answer:
Air Pollution : Air is polluted by natural processes like volcanic eruption, crop pollination, etc. mostly it is polluted by human activities like burning of coal, wood, diesel oil, kerosene, petrol etc.
Fossil fuels contain sulphur and nitrogen as impurities. When fuels bum these substances combine with air to produce gasses like sulphur dioxide, nitrogen oxide and hydrogen sulphide. They cause many serious respiratory problems. They can destroy the ozone layer, which protects us from the ultra violet radiations of the Sun. They also cause acid rain, which damages crops and buildings.

Question 6.
(a) What are the causes of air pollution ?
(b) Suggest five measures to prevent air pollution.
Answer:
(a) When fuels bum they produce sulphur dioxide, sulphur trioxide, nitrogen dioxide, hydrogen sulphide when these gases mix with rain water. They produce sulphuric and nitric acid. These acids mix with rain water to form acid rain.
(b) Five measures for the prevention of air pollution are:

  1. By using smokeless sources of energy, like solar energy and electrical energy, in place of conventional fossil fuels.
  2.  By using filters for the. smoke coming out of the chimneys of factories and power plants.
  3. By using internal combustion engines in vehicles for complete and efficient burning of fuel.
  4. By locating industries away from residential areas.
  5. By growing more trees.

Question7.
(a) What is nitrogen-fixation ?
(b) What are the two ways in which nitrogen fixation occurs?
(c) Explain the conversion of nitrogen into nitrates during lightning.
Answer:
(a) Nitrogen fixation : Symbiotic bacteria living in the root nodules of leguminous plants like peas, beans, absorb nitrogen directly from air and convert into nitrates. Thereafter, the plants convert it into proteins. Nitrogen is returned to the soil when plant and animal matter decays. This decomposition work is done by organisms called denitrifying bacteria which reconvert dead organic tissue into its constituent nitrogen.
(b) 1. Natural process.
2. Non-biological fixation.
(c) During lightning, temperatures often reach as high as 3000°C. At such high temperatures, nitrogen and oxygen present in the air combine to form nitric oxide, which further react with oxygen to form nitrogen dioxide
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-7a
Oxygen constitutes about 21% of air by volume. It is the active part of air.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-7b
Nitrogen dioxide then reacts with the water vapour present in air to form nitrous and nitric acids.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-7c
Oxygen constitutes about“21% of air by volume. It is the active part of air.
Nitric acid, so formed, reaches the earth along with rain-water, and reacts with metal carbonates to form metal nitrates.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-7d
Oxygen constitutes about 21% of air by volume. It is the active part of air.

B. OXYGEN

EXERCISE — II

Question 1.
Name :
(a) The most abundant element in the earth’s crust.
Ans. Oxygen.

(b) A chemical called oxygenated water.
Ans. H2O2 (Hydrogen peroxide)

(c) A metal highly resistant to rusting.
Ans. Tin.

(d) A mixture of oxygen and carbon dioxide used for artificial respiration.
Ans. Carbogen

(e) Two substances from which oxygen can be obtained at a large scale.
Ans. Air, water.

(f) An oxide and a carbonate containing oxygen.
Ans. Mercuric oxide and potassium chlorate.

(g) Two substances which undergo rapid oxidation.
Ans. Sodium, carbon.

Question 2.
(a) Taking hydrogen peroxide, how would you prepare oxygen in the laboratory ?
(b) What is the role of manganese dioxide in the preparation of oxygen ?
(c) Write the balanced chemical equation for the above chemical reaction.
(d) Why is hydrogen peroxide preferred in the preparation of oxygen gas ?
(e) Why is oxygen collected by downward displacement of water ?
(f) What happens when a glowing splinter is introduced in a jar containing oxygen ?
(g) What happens when oxygen gas is passed through alkaline pyrogallol solution ?

Answer:
(a) Take manganese dioxide in a round bottom flask and add hydrogen peroxide drop by drop to it, which acts ; a catalyst as shown in the figure. Collect oxygen by downward displacement of water.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-e2

(b) Manganese dioxides acts as a catalyst.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-e3
(d) H2O2 is preferred for lab preparation of oxygen because of following reasons.

  1. No heating is required.
  2. The rate of evolution of oxygen (O2) is moderate and under control.
    H2O2 is a safe chemical.

(e) Since the water is displaced downward by the gas collecting in the jar, the process is called downward displacement of water. The reasons are :

  1. Oxygen is only slightly soluble in water. Therefpre it can be collected over water without fear of excessive dilution.
  2. Oxygen is slightly heavier than air, so it cannot collected over air.

(f) Introduction of glowing splinter in the jar. The glowing splinter rekindles, but the gas does not catch fire.
(g) Alkaline pyrogallol solution turns brown when oxygen is passed through it.

Question 3.
(a) What happens when

  1. mercuric oxide and
  2. potassium nitrate are heated ?

(b) Why is potassium chlorate not used for laboratory preparation of oxygen ?

Answer: (a) 

  1. When mercuric oxide is heated, it decomposes to give mercury and oxygen.
  2. Potassium nitrate on heating gets converted into molten potassium nitrite with the release of oxygen.

(b) Potassium chlorate needs heating for quite sometime (to a high temperature) before it decomposes.

Question 4.
What are oxides ? Give two examples for each of me – tallic and non-metallic oxides.

Answer:
Oxides are binary compounds formed by the chemical combination of a substance metal or a non-metal with oxygen.
Examples :
Metal:

  1. Sodium oxide (Na2O).
  2. Calcium oxide (CaO).

Non-metal:

  1. Sulphur dioxide (SO2).
  2. Carbon dioxide (CO2).

Question 5.
Name the three types of oxidation processes. In which of these large amount of heat and light energy are produced?

Answer:
Oxidation can be categorised into three types :

  1. Spontaneous oxidation
  2. Rapid oxidation
  3. Slow oxidation

Out of the above said three types, rapid oxidation produces large amount of heat and light energy.

Question 6.
What do you observe when the following substances are heated and then tested with moist blue and red litmus – paper?
(a) Sulphur
(b) Phosphorus
(b) Calcium
(d) Magnesium

Answer:
(a) Sulphur : Blue litmus turns red.
(b) Phosphorus : Blue litmus turns red.
(c) Calcium : Red litmus turns blue.
(d) Magnesium : Red litmus turns blue.

Question 7.
Complete and balance the following chemical equations.
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-L

Question 8.
(a) Give four uses of oxygen.
(b) How is oxygen naturally renewed in air ?

Answer:
(a) Uses of oxygen

  1. Oxygen is used by firemen, miners, aviators, sea divers and even by every living being.
  2. Oxygen is necessary for burning of fuels.
  3.  Oxygen mixed with hydrogen as fuel produces.a flame with a very high temperature about 2800°C.
  4. As a fuel in spacecraft.

(b) All living beings use atmospheric oxygen in breathing and burning of fuels and in the formation of oxides of nitrogen. Yet amount of oxygen in the air remains more or less constant. This is because green plants return oxygen to the atmosphere by the process of photosynthesis.

Question 9.
(a) What is rust ?
(b) State at least two ways of prevent rusting.

Answer:
(a) Rust: Rust is hydrated ferric oxide (Fe2O3 . x* H2O), which forms a brownish red coating over iron. (* x can be any number.)

(b) Two ways of prevention of rusting :

  1. Painting with red lead.
  2. Oil paint is applied on doors and windows.
  3. Enamel coating. Enamel is a mixture of iron, and steel with silicates.
  4. Coal tar it is used for coating the lower parts of ships and bridges.

Question 10.
State two differences between : Rusting and burning.

Answer:
Difference between rusting and burning

RustingBurning
  1. Rusting is the process in which iron slowly reacts with oxygen in the air and produces a flaky substance called rust.
  2. Air and moisture are necessary for rusting.
  1. Burning is fast oxidation process in which large amount of energy is produced.
  2. Only air is necessary for burning.

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks :

(a) Argon is the most abundant inert gas present in air.
(b) Oxides of sulphur and nitrogen combine with rain water to form sulphuric acid and nitric acid which cause acid rain.
(c) NO2 and CO are the most common air pollutants.
(d) Joseph Priestly discovered the oxygen gas.
(e) Oxygen occupies about 21% of air by volume.

2. Match the following :
Selina Concise Chemistry Class 7 ICSE Solutions - Air and Atmosphere-o2

MULTIPLE CHOICE QUESTIONS

1. A fuel when used releases least amount of pollutants in the air.
(a) sulphur dioxide
(b) chlorofluorocarbon
(c) smoke
(d) CNG

2. The natural way of adding oxygen to air which involves green plants is called
(a) photosynthesis
(b) respiration
(c) burning
(d) dissolution

3. Which one of the following is most likely to be corroded?
(a) a stainless steel cup-board
(b) a galvanised iron bucket
(c) an iron hammer
(d) a tin plated iron box

4. The process by which oxidation of food in our body takes place is
(a) photosynthesis
(b) respiration
(c) decomposition
(d) combustion