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Selina Concise Physics Class 7 ICSE Solutions – Physical Quantities and Measurement

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Selina Concise ICSE Solutions for Class 7 Physics Chapter 1 Physical Quantities and Measurement

• Points to Remember
• The mass of a body is the quantity of matter contained in a body regardless of its volume or any force acting on it.
•  The weight of a body is the force with which every body is attracted towards its centre.
•  The unit of mass in S.I. system is kilogram (kg). Higher units of mass are quintal and metric tonne.
•  The weight of body changes with acceleration due to gravity.
•  Weight is zero at the centre of the earth.
•  Mass per unit volume of a substance is called density of the body.
•  The unit of density in S.I. system is kg m^-3 and gcm^-3 in C.G.S. system.
•  The density in S.I. system is 1000 × numerical value in C.G.S. system.
•  The density of liquids and gases decreases or increases with the rise or fall in temperature.
• The cycle of upward and downward movements of the fluid form currents in the medium which are known as convectional currents.

Test Yourself

A. Objective Questions 

1. Write true or false for each statement

(a) The S.I. unit of volume is litre.
Answer. False.
The S.I. unit of volume is cubic metre.

(b) A measuring beaker of capacity 200 ml can measure only the volume. 200 ml of a liquid.
Answer. True.

(c) cm2 is a smaller unit of area than m².
Answer. True.

(d) Equal volumes of two different substances have equal masses.
Answer. False.
Equalvolumes of two different substances have different masses.

(e) The S.I. unit of density is g cm^-3.
Answer. False.
The S.I. unit of density is Kg m^-3.

(f) 1 g cm^-3 = 1000 kg m^-3.
Answer. True.

(g) The density of water is maximum at 4°C.
Answer. True.

(h) The speed 5 ms^-1 is less than 25 km h^-1.
Answer. True.

(i) The S.I. unit of speed is ms^-1.
Answer. True

2. Fill in the blanks

(a) l m³ = 10⁶ cm³
(b) The volume of an irregular solid is determined by the method of displacement of liquid.
(c) Volume of a cube = (one side)
(d) The area of an irregular lamina is measured by using a grapl paper.
(e) Mass = density × volume.
(f) The S.I. unit of density is kg m^-3.
(g) 1 g cm^-3 = 1000 kg m^-3.
(h) 36 km h^-1 = 10 ms^-1.
(i) Distance travelled d = speed v × time t.

3. Match the following

4. Select the correct alternative 

(a) One litre is equal to :

1.  1 cm^-3
2. 1 m³
3.  10^-3  cm³
4. 10^-3 m³

(b) A metallic piece displaces water of volume 15 ml. The volume of piece is :

1.  15 cm³
2.  15 m³
3.  15 × 10³ cm³
4.  15 × 10³ cm³

(c) A piece of paper of dimensions 1.5 m x 20 cm has area :

1.  30 m2
2.  300 cm2
3.  0.3 m2
4.  3000 m3

(d) The correct relation is :

1.  d = M × V
2.  M = d × Y  
3.  V = d × M
4.  d = M + V

(e) The density of alcohol is 0.8 g cm-3. In S.I. unit, it will be :

1.  0.8 kg m^-3
2.  0.0008 kg m^-3
3.  800 kg m^-3
4.  8 x 10³ kg m^-3

(f) The density of aluminium is 2.7 g cm-3 and of brass is 8.4 g cm^-3. For the same mass, the volume of:

1.  both will be same
2.  aluminium will be less than that of brass
3.  aluminium will be more than that of brass
4.  nothing can be said.

(g) A block of wood of density 0.8 g cm^-3 has a volume of 60 cm³. The mass of block will be :

1.  60.8 g
2.  75 g
3.  48 g
4.  0.013 g

(h) The correct relation for speed is

1.  Speed = distance x time
2.  speed = distance / time
3.  speed = time / distance
4.  speed = 1 / distance x time

(i) A boy travels a distance 150 m in 1 minute. His speed is

1.  150 m s^-1
2.  2.5 m s^-1
3.  25 m s^-1
4.  9 m s^-1

B. Short/Long Answer Questions

Question 1.
Define the term volume of an object.
Answer:
The space occupied by an object is called its volume.

Question 2.
State and define the S.I. unit of volume.
Answer:
S.I. unit of volume – The S.I. unit of volume is cubic metre. In short form, it is written as m³.
One cubic metre is the volume of a cube of each side 1 metre as shown in figure below i.e., 1 m³ = 1 m × 1 m × 1 m.

Question 3.
State two smaller units of volume. How are they related to the S.I. unit?
Answer:
A smaller unit of volume is cubic centimetre (symbol cm³) and cubic decimetre (symbol 1 dm³). One cubic centimetre is the volume of a cube of each side 1 centimetre, i.e.,
1 cm³ = 1 cm × 1 cm × 1 cm.
Relationship between m³ and cm³
l m³ = lm × lm × lm
= 100 cm × 100 cm × 100 cm
= 10,00,000 cm³ = 10⁶ cm³.
Relationship between m³ and dm³
l m³ = lm × 1 m × 1 m .
= 10 dm × 10 dm × 10 dm
= 1000 dm
= 10³ dm
Note 1 m = 10 dm

Question 4.
How will you determine the volume of a cuboid ? Write the formula you will use.
Answer:
Volume of a cuboid = length × breadth × height.

Question 5.
Name two devices which are used to measure the volume of an object. Draw their neat diagrams.
Answer:
Two devices that are used to measure the volume of an object are :
(i) Measuring cylinder and
(ii) Measuring beaker.

Question 6.
How can you determine the volume of an irregular solid (say a piece of brass) ? Describe in steps with neat diagrams.
Answer:
To measure the volume of a piece of stone.
Take a piece of brass, a measuring cylinder, fine thread of sufficient length and some water.
Place a measuring cylinder on a flat horizontal surface and fill it partially with water. Note the reading of the water level very carefully. Now tie the piece of brass with a thread and dip it completely into water. We see that the level of water rises. Note the reading of the new water level.

The difference in the two levels of water gives the volume of the piece of brass
Initial level of water = 60 ml
Level of water when brass is immersed = 80 ml
∴ Volume of water displaced = 80 ml – 60 ml = 20 ml
∴ Volume of the piece of brass = 20 cm³
Note : 1 ml = 1 cm³

Question 7.
You are required to take out 200 ml of milk from a bucket full of milk. How will you do it ?
Answer:
By using the measuring beaker A measuring beaker is used to measure a fixed volume of liquid from a large volume. Suppose it is required to measure 200 ml of milk from the milk contained in a bucket. For this, take the measuring beaker of capacity 200 ml. Wash it and dry it. Then, immerse the measuring beaker well inside the milk contained in the bucket so that the beaker gets completely filled with the milk.
Take out the measuring beaker from the bucket gently so that no milk splashes out and then pour the milk from the measuring beaker into the another empty vessel.

Question 8.
Describe the method in steps to find the area of an irregular lamina using a graph paper.
Answer:
Method to find the area of an irregular lamina using a graph paper : First, place the lamina over a graph paper and draw its boundary line on the graph paper with a pencil. Then remove the lamina and count and note the number of complete squares as well as the number of squares more than half within the boundary line (only the squares less than half, are left while counting). The area of lamina is equal to the sum of the area of complete squares and the area of squares more than half. Let n be the total number of complete and more than half or half squares within the boundary of lamina. Since area of one big square is 1 cm × 1cm = 1 cm², so the area of lamina will be n x

Question 9.
Define the term density of a substance.
Answer:
The density of a substance is defined as the mass of a unit volumx of that substance.

Question 10.
State the S.I. and C.G.S. units of density. How are they inte related ?
Answer:
The S.I. unit of mass is kilogram (symbol kg) and of volume is
cubic metre (symbol m³). Therefore S.I. unit of density is kg/m³
or kg m^-3.
The C.G.S. unit of mass is gram (symbol g) and of volume is cubic centimetre (symbol cm3). Therefore the C.G.S. unit of
density is g/cm³ or g cm^-3.

Question 11.
‘The density of brass is 8.4 g cm’3’. What do you mean by the statement ?
Answer:
Density of brass is 8.4 g cm^-3. This means that unit volume of brass contain 8.4 g mass.

Question 12.
Arrange the following substances in order of their increasing density:
(a) iron
(b) cork
(c) brass
(d) water
(e) mercury
Answer:
b<a<c<d<e

Question 13.
How does the density of water changes when :
(a) it is heated from 0°C to 4°C,
(b) it is heated from 4°C to 10°C ?
Answer:
(a) Water contracts on heating from 0°C to 4°C and expands on heating above 4°C.
(b) The density of water is maximum at 4°C. It decreases when it is cooled from 4°C to 0°C or it is heated above 4°C.

Question 14.
Write the density of water at 4°C.
Answer:
The density of water at 4°C is 1.0 g cm^-3, or 1,000 kg m^-3

Question 15.
Explain the meaning of the term speed.
Answer:
The distance covered or travelled by a body in one second is called the speed of the body, i.e.

Question 16.
Write the S.I. unit of speed.
Answer:
The S.I. unit of speed is metre/second or metre per second. Its symbol is m s^-1.

Question 17.
A car travels with a speed 12 m s”1, while a scooter travels with a speed 36 km h-1. Which of the two travels faster ?
Answer:
Speed of car = 12 m s^-1
Speed of scooter = 36 km h^-1
here, 1 km = 1000 m
1 hr = 3600 sec

∴ Speed of car is more. Car travels faster than scooter.

C. Numericals

Question 1.
The length, breadth and height of a water tank are 5 m, 2.5 m and 1.25 m respectively. Calculate the capacity of the water tank in (a) m³ (b) litre.
Answer:
Given,
Length (1) = 5m
Breadth (b) = 2.5 m
and Height (h) = 1.25 m

Question 2.
A solid silver piece is immersed in water contained in a measuring cylinder. The level of water rises from 50 ml to 62 ml. Find the volume of silver piece.
Answer:
Given, initial level of water .v₁ = 50 ml
Final level of water v₂ = 62 ml
Volume of silver piece V = v₂ – v₁
= 62 ml – 50 ml
= 12 ml or 12 cm³

Question 3.
Find the volume of a liquid present in a dish of dimensions 10 cm x 10 cm x 5 cm.
Answer:
Volume of water = Length × breadth × height
= 10 cm × 10 cm × 5 cm
= 500 cm³ or 500 ml.

Question 4.
A rectangular field is of length 60 m and breadth 35 m. Find the area of the field.
Answer:
Length of a rectangular field = 60 m
Breadth of rectangular field = 35 m
∴ Area = 60 m × 35 m
= 2100 m²

Question 5.
Find the approximate area of an irregular lamina of which boundary line is drawn on the graph paper shown in fig. 1.16. below.

Answer:
From figure, the number of complete squares = 11
The number of squares more than half = 9
∴ Total number of squares = 11 + 9 = 20
∴ Area of the 1 square = 1 cm × 1cm = 1 cm²
∴ Area of 20 squares = 20 × 1 cm² = 20 cm²
∴ Approximate area of irregular lamina = 20 cm²

Question 6.
A piece of brass of volume 30 cm³ has a mass of 252 g. Find the density of brass in (i) g cm^-3, (ii) kg m^-3.
Answer:

Question 7.
The mass of an iron ball is 312 g. The density of iron is 7.8 g cm^-3. Find the volume of the ball.
Answer:

Question 8.
A cork has a volume 25 cm³. The density of cork is 0.25 g cm^-3. Find the mass of cork.
Answer:

Question 9.
The mass of 5 litre of water is 5 kg. Find the density of water in g cm^-3.
Answer:

Question 10.
A cubical tank of side 1 m is filled with 800 kg of a liquid. Find: (i) the volume of tank, (ii) the density of liquid in kg m^-3.
Answer:

Question 11.
A block of iron has dimensions 2 m × 0.5 m × 0.25 m. The density of iron is 7.8 g cm^-3. Find the mass of block.
Answer:
Given, l = 2m
b = 0.5 m

Question 12.
The mass of a lead piece is 115 g. When it is immersed into a measuring cylinder, the water level rises from 20 ml mark to 30 ml mark.
Find:
(i) the volume of the lead piece,
(ii) the density of the lead in kg m^-3.
Answer:

Question 13.
The density of copper is 8.9 g cm^-3. What will be its density in kg m^-3 ?
Answer:

Question 14.
A car travels a distance of 15 km in 20 minute. Find the speed of the car in (i) km h^-1, (ii) m s^-1.
Answer:
Distance travelled by car =15 km
Time taken = 20 minutes
(i) Speed of car in km h^-1
Convert 20 minutes to hour

Question 15.
How long a train will take to travel a distance of 200 km with a speed of 60 km h^-1 ?
Answer:
Distance covered by train = 200 km
Speed of train = 60 km h^-1

Question 16.
A boy travels with a speed of 10 m s^-1 for 30 minute. How much distance does he travel ?
Answer:
Speed of boy = 10 m s^-1
Time taken = 30 minutes
speed = distance travelled / time taken
Distance travelled = Speed × Time taken
Convert 30 minutes to seconds
1 minute = 60 sec
30 minute 60 × 30 = 1800 seconds
Putting the value of speed and time we get
Distance travelled = 10 ms^-1 × (1800 sec) = 18000 m
= 18000 metre or 18 km Ans.

Question 17.
Express 36 km h^-1 in m s^-1
Answer:

Question 18.
Express 15 m s^-1 in km h^-1.
Answer:

Selina Concise Chemistry Class 7 ICSE Solutions – Metals and Non-metals

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

ICSESolutions.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.

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Selina Concise ICSE Solutions for Class 7 Chemistry Chapter 6 Metals and Non-metals

Points to Remember :

1. Knowledge of chemistry plays a vital role in the development of human society and civilization.
2. Metals are known to man from ancient times. Metals are used to make our life comfortable.
3. Non-metals form another class of elements, e.g. hydrogen, oxygen, carbon, etc. They are used for various purposes.
4. Alloys are homogeneous solid mixtures containing two or more metals e.g. steel, brass, bronze.
5. Common salt, hydrocloric acid, carbohydrates, fats, proteins, vitamins, occur naturally and can also be prepared artificially.
6. Fertilizers are artificially prepared substances, which are necessary for the proper growth of crops.
7. There are a number of man-made materials that are used in our daily life for various purposes, e.g., cement, plaster of pairs, plastics.
8. Medicines are used to cure diseases.
9. Solution is a homogenous mixture of solute and solvent.
10. Soda water is prepared by dissolving carbon dioxide in water under high pressure.
11. Syrup is a highly concentrated sugar solutions. It contains a specific flaviour.

EXERCISE

1. Name a metal

1. that is most malleable : Pure gold
2. that is brittle : Zinc
3. as precious as gold : Platinum
4. that can be cut with knife : Sodium
5. used in making electric cables : Copper
6. used as a thermometric liquid : Mercury
7. that is the best conductor of electricity : Silver

2. Name a non-metal that is :

1. a good conductor of heat and electricity : Graphite (Carbon)
2. hardest naturally occurring substance : Diamond (Carbon)
3. used to kill germs in water : Chlorine
4. lustrous : Iodine
5. used for filling into electric bulbs : Argon
6. used for cancer therapy : Radon
7. liquid at room temperature : Bromine

3. Mention two uses of the following metals and non-metals

(a) Iron :
It is used to make pipes, tanks, railing, etc.
It is used in the construction of power transmission towers.

(b) Aluminium :
It is used to make electric wires.
It is used to make utensils, cans, window fram’es, etc.

(c) Gold :
It is used for making ornaments and coins.
It is used in the manufacture of electronic devices like computers, telephones, home appliances, etc.

(d) Oxygen :
It is used by all living beings for breathing.
It is important for combustion.

(e) Iodine :
It is used in photographic films in the form of potassium iodide.
It is added to salt to make it iodized salt which is necessary for the growth of human body.

4. Give reasons :

(a) Magnesium is used in fire works.
Ans : Magnesium is used in fire works because it bums with a dazzling light.

(b) Aluminium is used in making aircrafts.
Ans : Aluminium is used in making aircrafts because it is light and strong. It is mixed with other metals to make it stronger.

(c) Copper is used in making electric cables.
Ans : Copper is ductile and a very good conductor of heat and electricity. This is the reason that copper is used in making electric cables.

(d) Graphite is used in the leads of pencils 
Ans : Graphite turns paper black that is why it is used in the leads of pencils.

(e) Impure diamond is used to cut glass
Ans : Impure diamond is used to cut glass becuase it is the hardest substance and can easily exert force required for cutting.

(f) Gold is mixed with copper and nickel.
Ans :  Pure gold is a very soft metal. It cannot be moulded into ornaments so it is mixed with copper and nickel so that it becomes harder and bit cheaper also.

(g) Tungsten is used in electric bulbs.
Ans : It is a shiny grey metal, in solid state at room temperature. It can withstand high temperature because it has highest melting point among metals. Hence, it is used in electric bulbs.

5. Name the metals present in the following alloys

1. Brass— Copper and zinc
2. Bronze— Copper and tin
3. Duralumin— Aluminium and copper
4. Stainless steel— Iron, chromium, nickel

6. Give four differences between metals and non-metals with reference to their
(a) Melting point and boiling point,
(b) Conductivity of heat and electricity,
(c) Malleability
(d) Solubility

Metals		Non-metals
Melting point and boiling point	Metals have both high high melting point and boiling point.	Non-metals have both low melting and low boiling point.
Conducti­vity of heat and electricity	They are good conductors of heat and electricity.	Nofi-metals are bad conductors of heat and electricity.
Malleabi­ lity	Metals are ususally malleable.	All non-metal are non- malleable.
Solubility	Metals are generally insoluble in water and other organic solvents.	They are both soluble and insoluble

7. What are metalloids?
Ans : Metalloids are the elements which show some properties of metals and some properties of non-metals. They all are solids. They are silicon, boron, arsenic, antimony, germanium, tellurium and polonium.

8. Give two uses of

(a) Silicon :

• Highly pure silicon is used in making microchips for computers, transistors, solar cells, rectifiers and other solid state devices that are used extensively in the electronic and present space age industries.
• It is used in the manufacture of a waterproof material called “silicone”. Silicone is used to make bags, umbrellas, raincoats, etc.
• It is an important substance present in steel, an alloy of carbon.

(b) Antimony :

• Antimony is used in electric industry to make semiconductor devices.
• It is alloyed with lead to improve its hardness and strength and is used in batteries.
• It is also used in printing presses as type metal.

(c) Tungsten :

• It is used in making electrodes.
• It is used in heating elements.
• It is used as filaments in electric bulbs and cathode ray tubes.

(d) Germanium:

• Germanium is used as a semiconductor.
• It is used as a transistor in many electronic applications when mixed with arsenic, gallium, antiomony, etc.
• Germanium is also used to form alloys and as a phosphor in fluorescent lamps

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks :

(a) The most ductile metal is silver.
(b) A metal stored in kerosene oil is sodium.
(c) Tungsten metal is a poor conductor of heat.
(d) Pure gold is a soft metal.
(e) Silicon carbide is the hardest compound known to us.
(f) A non-metal used to purify water is phosphorus.
(g) A metal that gives dazzling effect to crackers when they explode is magnesium.
(h) A chemical compound that makes up the striking heads of match sticks is sulphur.

2. Match the following :

3. Write ‘true’ or ‘false’ for the following statements :
(a) Silver is used to make electric cables : False
(b) Iodine acts as an antiseptic in the form of tincture of iodine : True
(c) Sodium can be cut with a knife : True
(d) Antimony is a metal : False
(e) Sand is an oxide of silicon : True

MULTIPLE CHOICE QUESTIONS

1. The noble gas used in advertising signboards is
(a) Helium
(b) Neon
(c) Argon
(d) Krypton

2. A metal with melting point less than 50°C is
(a) Gallium
(b) Iron
(c) Gold
(d) Aluminium

3. A metal which is neither ductile nor malleable is
(a) Copper
(b) Silver
(c) Zinc
(d) Aluminium

4. Rust is a hydrated oxide of iron which is
(a) Reddish brown
(b) Green
(c) White
(d) Black

5. Aluminium is not used to make :
(a) Foils
(b) Wires
(c) Fireworks
(d) Utensils

6. A metalloid used in the manufacture of microchips used in computer is :
(a) Antimony
(b) Germanium
(c) Silicon
(d) Arsenic

7. A metalloid used to make glass :
(a) Sulphur
(b) Germanium
(c) Silicon
(d) Antimony

 

Selina Concise Physics Class 7 ICSE Solutions – Electricity and Magnetism

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

Selina Class 7 Physics ICSE SolutionsChemistryBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 7 Physics Chapter 7 Electricity and Magnetism

• Points to Remember
•  Some materials behave in a particular manner showing magnetic properties.
•  A freely suspended bar magnet always point in North-South direction.
•  Like poles repel each other while unlike poles attract each other.
•  Magnet is neutral at its centre and has maximum magnetic effects at its ends called the poles.
•  Poles always exist in pairs with opposite polarities and can never be isolated.
•  Examples of magnetic substances are:- iron, cobalt, nickel.
•  Examples of non-magnetic substances are:- wood, plastic, aluminium, copper.
•  The poles are not exactly the ends of the magnet but thev are slightly inside.
•  For a given magnet, both the poles are of equal strength.
•  Earth’s magnet has its south pole situated near the geographic north.
•  Electromagnets are the temporary magnets which are made up of soft iron core with a winding of insulated copper wire.
•  Freely suspended current carrying solenoid always rests in north- south direction.
•  A direct current source has the fixed positive and negative terminals, e.g. a battery.
•  Hans Christian Oersted discovered that if an electric current is passed through a conductor, a magnetic field is developed around it.
•  The study of magnetic effects produced due to electric current is known as electromagnetism.
•  The direction of magnetic field due to a straight current carrying conductor is obtained by any of the following rules:
  (a) Right Hand Thumb Rule.
  (b) Right Hand Cork Screw Rule.
•  The property due to which a changing magnetic field within a closed conducting coil induces electric current in the coil is called electromagnetic induction.
•  The current produced in a closed coil when magnetic lines of force rapidly change within it is called the induced current.
•  The symbol for alternating current is 0.
•  The strength of induced current can be increased by increasing the
  (a) the number of turns in the coil
  (b) strength of the magnet used
  (c) relative speed between the magnet and closed coil.
•  We cannot think of modem life without electricity. We light our homes and other places of work with electricity.
•  It is used to run electric fans, televisions, geyser, electric irons, room heaters, refrigerators, music system etc.
•  Cell is a primary source of electricity. A combination of two or more cells is a battery.
•  Some other sources of electricity are generator and solar cells.
•  The path along which an electric current flows is called a circuit.
•  Electricity has the following effects
  (a) Heating effect
  (b) Magnetic effect
  (c) Chemical effect
  (d) Mechanical effect
•  When an electric path is complete is called closed circuit and the path with a break is called open circuit.
•  The substances which allow the electricity to flow through them are called conductors, e.g. metals, human body etc.
•  The substances which do not allow electricity to flow through them are called insulators, e.g. wood, paper, glass etc.
•  The consumption of electricity is calculated from the meter in kWh.
•  Electric fuse is a device which limits the current in an electric circuit.
•  All electrical appliances are connected in parallel in household circuits.
•  We should be cautious in using electricity. Certain precautions should be taken before working on an electrical gadget or circuit.

Activity 6

List five such electrical gadgets in your house in which electromagnet is used.

1. ……………………
2. ……………………
3. ……………………
4. ……………………
5. ……………………

Answer.

1. Computer
2. Electric motor
3. Fan, Toaster
4. Refrigerator
5. Television
6. Electric Bell

Activity 9

Answer.

 

Test Yourself

A. Objective Questions 

1. Write true or false for each statement

(a) A current carrying coil when suspended freely can rest in any direction.
Answer. False.

(b) A coil carrying current behaves like a magnet.
Answer. True.

(c) In an electromagnet, the core is made up of copper.
Answer. False.

(d) An electric bell uses an electromagnet.
Answer. True.

(e) An electromagnet with soft iron core is a temporary magnet.
Answer. True.

(f) We use cell as the source of electricity to run an electric immersion rod.
Answer. False.

(g) A torch bulb glows if the terminals of the bulb are connected to the terminals of a cell by the metallic wire.
Answer. True.

(h) Wool is a conductor of electricity.
Answer. False.
Wool is a insulator of electricity.

(i) Silver is an insulator of electricity.
Answer. False.
Silver is good conductor of electricity.

(j) Our body is a conductor of electricity.
Answer. True.

(k) For a circuit to be complete, every part of it must be made up of conductors.
Answer. True.

(l) All metals are conductors of electricity.
Answer. True.

(m) The switch should not be touched with wet hands.
Answer. True.

(n) A switch is an on-off device in an electric circuit.
Answer. True.

2. Fill in the blanks

(a) A magnet has two poles.
(b) Like poles repel each other and unlike poles attract.
(c) An electromagnet is used to separate large mass of iron scrap.
(d) The strength of magnetic field of an electromagnet is increased by inserting a core of soft iron.
(e) In a torch we use dry cell as the source of electricity.
(f) To light a table lamp and to run a refrigerator, we use mains as the source of electricity.
(g) A group of two or more cells is called a battery.
(h) Conductors pass electricity through them.
(i) Insulators do not pass electricity through them.

3. Match the following

4. Select the correct alternative

(a) A freely suspended magnet rests in

1.  east-west direction
2.  north-south direction
3.  north-east direction
4.  north-west direction.

(b) Electromagnets are made up of

1.  steel
2.  copper
3.  brass
4.  soft iron.

(c) An electromagnet is used in

1.  electric oven
2.  ammeter
3.  electric bell
4.  radio set.

(d) The purpose of armature in an electric bell is

1.  to make and break the circuit
2.  to produce sound
3.  to produce magnetic field
4.  to provide spring action.

(e) In a torch, the source of electricity is

1.  the bulb
2.  the switch
3.  the cell
4.  the mains.

(f) Electricity can flow through

1. wood
2.  rubber
3.  plastic
4.  copper wire.

(g) Electricity does not flow through

1.  human body
2.  animals body
3.  rubber
4.  silver.

(h) We should not touch the switch with wet hands otherwise

1.  electricity may pass through our body
2.  electricity may not pass through the appliance
3.  circuit may break
4.  the switch may get off.

B. Short/Long Answer Questions

Question 1.
State two properties of a bar magnet.
Answer:
Properties of a bar magnet

1.  Attractive property: A magnet attracts small pieces of iron, cobalt or nickel.
2.  Directive property: A magnet when suspended freely, always point towards North and South direction.
3.  Like poles, i.e. North and North or South and South poles repel each other.
4.  Unlike poles i.e. North and South attract each other.
5.  Poles always exist in pairs, i.e. poles of a magnet cannot be separated.

Question 2.
How will you test whether a given rod is a magnet or not?
Answer:
Aim— To test whether a given rod is a magnet or not.
Apparatus —

1.  rod to be tested
2.  a bar magnet
3.  a stand
4.  a thread

Procedure —

1.  Suspend the rod to be tested with a thread on the stand.
2.  Bring the bar magnet near the rod with its north pole towards the rod’s end.
3.  Observe.

Inference — In every case, the magnet comes to rest in a north- south direction which shows its directive property.
Observation and Inference —
— If the rod is attracted towards the bar magnet, the rod may be magnet or not.
— But if the rod remains in the same direction of rest as in the start of the experiment, it show it is not a magnet.

Question 3.
How will you test whether a given rod is made of iron or not?
Answer:
Bring a magnet near the rod if the rod is attracted by the magnet then it will be made of iron else not.

Question 4.
You are given two similar bars. One is a magnet and the other is of soft iron. How will you distinguish and identify them ?
Answer:
Take first bar and suspend it in a stand with the help of a thread. So that it is free to rotate in horizontal plane. Note the direction in which it sets itself. If the direction is North and South it may be magnet. Again rotate it, if this time again it sets itself in north and south direction, it is a magnet, otherwise, it is iron.
Now repeat above experiment with second bar and in the same way find it if it sets always in north and south direction then it will be a magnet.

Question 5.
You are given a magnet. How will you use it to find north-south direction at a place?
Answer:
The earth is a huge magnet in itself with its North and South poles. The North and South of the place can easily be detected with the help of a bar magnet. A freely suspended bar magnet always rest in the geographic N-S direction. Since the South pole of the earth’s magnet is closer to the geographic North, the North pole of the suspended magnet will always rest in the geographic North, and the South pole of the freely suspended magnet will point towards geographic South.

Question 6.
Describe a simple experiment to illustrate that like poles of two magnets repel each other while the unlike poles attract.
Answer:
Like poles repel and unlike poles attract.
Two like poles (both North poles or both South poles) repel each other. Two unlike poles (one North pole and the other South pole) attract each other. This can be demonstrated by the following simple experiment.
Take two bar magnets A and B. Suspend one magnet A with a silk thread from a support so that it is free to swing. The magnet A will come to rest in the North-South direction. The North pole of the magnet A is in the North direction and its South pole is in the South direction. Now holding the other magnet B in your hand if you bring its North pole near the North pole of the suspended magnet A as shown in figure you will observe that the two poles repel each other. Care is taken that the two magnets do not touch each other.

Now if you bring the South pole of the magnet B near the north pole of the suspended magnet A as shown in figure without touching it, you will observe that the two poles attract each other.

The above experiment shows that the like poles repels each other while the unlike poles attract each other.

Question 7.
“Poles exist in pair”. Comment on this statement.
Answer:
The magnetic poles always exist in pairs. It is not possible to separate the two poles of a magnet.
If a bar magnet is broken at the middle in two parts each part is found to be a magnet. Each part has the property to attract the small iron pieces. Each part rests in the North-South direction when suspended such as to swing freely. This shows that the new poles are formed at the broken ends.
If these pieces are broken again and again, each part will still found to be a complete magnet. Each part contains both the poles (N-pole and S-pole). Thus, the two poles of a magnet exist simultaneously.

Question 8.
What is a magnetic compass ? State its use.
Answer:
Magnetic compass is a device which is used to locate the direction of a place. It always rests in a North-South direction. It is used in the navigators in ships, submarines, aeroplanes etc.

Question 9.
Explain the meaning of the term magnetic field.
Answer:
The space around the magnet where its influence can be experienced is known as magnetic field. This field is formed by the magnetic lines of force which run from the North pole to the South pole. These lines can be found to be maximum crowded at the two ends of the magnet which are the poles i.e. the North pole and the South pole.

Question 10.
What is an electromagnet ?
Answer:
An electromagnet — An electromagnet is a temporary magnet which behaves as a magnet when electric current is passed through the insulated copper wire and loses its magnetism when current is stopped. It has a soft iron piece called the core with an insulated copper wire wound on it.

Question 11.
Name the material of an electromagnet.
Answer:
Iron bar, insulated copper wire, battery.

Question 12.
Draw a labelled diagram to make a soft iron bar as an electromagnet. Describe in steps the procedure.
Answer:
Usually, the electromagnets are made in two shapes :
(1) bar or I shaped magnet and (2) horse shoe or U shaped magnet.

1.  To make a fiar or I shaped electromagnet: Take a soft iron bar PQ and wind a thin insulated copper wire around the bar. Connect a cell or a battery B, and a key K in series between the ends of the coil. The circuit diagram is shown in figure.
   
   When key K is closed, current passes through the winding of the coil and the bar becomes a magnet. As the key K is opened, the current stops flowing in the coil and the bar loses its magnetism. Thus, the bar behaves like an electromagnet.
2.  To make a horse shoe or U shaped electromagnet: Take a U shaped soft iron piece. Wind a thin insulated copper wire on its arms such that the winding in the two arms is in opposite direction. In figure winding in the arm A starts from the front and is in clockwise direction (when seen from the bottom).
   On reaching the upper end of the arm A, winding starts from the back at the top of the arm B and is in anticlockwise direction. Connect a battery B and a key K between the two ends of the wire.
   

Question 13.
You are given a U shaped soft iron piece, insulated copper wire and a battery. Draw a circuit diagram to make a horse shoe electromagnet.
Answer:
End A becomes S-pole and B becomes N-pole.

Question 14.
Name two factors on which the strength of magnetic field of an electromagnet depends.
Answer:
The magnetic field of an electromagnet (I or U-shaped) can be increased by the following two ways :

1.  By increasing the number of turns of winding in the solenoid.
2.  By increasing the current through the solenoid.

Question 15.
State two ways by which the strength of magnetic field of an electromagnet can be increased.
Answer:
The magnetic field of the electromagnet can be increase in the following two ways :

1.  By inserting a rod of soft iron or steel inside the cylindrical tube. This rod is called the core.
2.  By increasing the total number of turns of the coil.

Question 16.
State two common uses of electromagnets.
Answer:
Uses of electromagnet —

1.  In electrical appliances such as electric bell, fan etc.
2.  In lifting heavy loads of iron scrap.
3.  To remove tiny particles of iron from the wound.
4.  In loading furnaces with iron.
5.  In separation of magnetic substances from the non-magnetic substances.

Question 17.
Name a domestic device in which an electromagnet is used.
Answer:
Electromagnet is used in ELECTRICAL APPLIANCES like ELECTRIC BELL, RADIO, T.V., FAN and MOTORS etc.

Question 18.
Draw a neat and labelled diagram of an electric bell and describe its working.
Answer:
Working of the bell— Principle of working of electromagnetism When the switch is pushed on, the circuit gets completed and current stalls flowing through the U-shaped electromagnet which creates magnetic field in the core. This attracts the iron armature. Now when the armature moves towards the electro-magnet, the hammer strikes the gong and the bell rings. But as the armature
moves towards the electromagnet, the contact with the adjustment screw breaks which breaks the closed circuit and stops the current. Now when there is no current there is no electromagnetism and the armature returns to the original position. This making and breaking of the circuit of the electromagnet continues as long as the button remains pressed.

Question 19.
The incomplete diagram of an electric bell is given in fig. Complete the diagram and label its different parts.

Answer:

Question 20.
What is declination ? Draw a diagram to show the angle between the declination and true direction of geographic north.
Answer:

Magnetic declination is the angle of the horizontal plane between the magnetic North and the geographic North (or true North). This angle is shown in figure by symbol θ.
The angle of declination is different at different places on the earth surface and it also changes at a place with time. The declination is taken positive if the magnetic North is towards the east of the true North as in figure and is negative if the magnetic north if towards the west of the true North.

Question 21.
Define the term current.
Answer:
An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire.
The S.I. unit of electric current is the ampere.

Question 22.
Name four appliances which work using electricity.
Answer:

1.  an electric iron
2.  an electric heater
3.  an electric kettle
4.  an immersion rod

Question 23.
Name two sources of electricity.
Answer:

1.  dry cell and battery
2.  generator and solar cell

Question 24.
What is a battery?
Answer:
If we use a group of two or more cells, it is called a battery. A battery is used where we require more electricity.

Question 25.
What is an electric circuit?
Answer:
For a smooth flow of electric current, a complete circuit is needed. This is known as electric circuit.

Question 26.
Describe an experiment to show that electricity flows only if the circuit is complete and it does not flow if the circuit is incomplete.
Answer:
Take two torch bulbs A and B. Connect them to a cell through a switch as shown in fig. The bulbs are said to be in series. Close the switch (i.e., the circuit it completed), you will see that both the bulbs glow.

Now take out the connection of the bulb B as shown in fig. Now close the switch, you will observe that the bulb A does not glow because the circuit is now incomplete.

Now replace the bulb B by a fused bulb fig. and close the switch. Again you will see that the bulb A does not glow. This is because the circuit being in series, is still incomplete.

Question 27.
You are provided with a torch bulb, a cell and two plastic coated f metal wires. Draw a diagram to show a complete circuit to light the bulb.
Answer:
Take two bulbs A and B. Connect them through switches S₁ and S₂ in parallel as shown in fig. Close both the switches. You will see that both the bulbs glow.

Question 28.
In which of the following case the bulb will glow :

1.  Only one terminal of a cell is joined with a metal wire to one terminal of the bulb.
2.  Both terminals of the bulb are joined with two metal wires to one terminal of the cell.
3.  One terminal of the cell is joined to one terminal of the bulb and other terminal of the cell to the other terminal of the bulb.
   Answer:
   The bulb will glow in (3) case i.e.

Question 29.
Distinguish between conductors and insulators of electricity. Give two examples of each.
Answer:
Conductors

1.  Conductors are those substances which allow electricity to flow through them.
2.  e.g. all metals, human body.
   Insulators
   Insulators are those which do not allow electricity to pass through them.
   e.g. wood, paper, glass.

Question 30.
Select conductors and insulators from the following :
Glass, silver, copper, wood, paper, pure water, impure water, aluminium, iron, leather, plastic, steel, human body and ebonite.
Answer:
Conductors — Silver, copper, impure water, aluminium, iron, steel, human body.
Insulators — Glass, wood, paper, pure water, leather, plastic and ebonite.

Question 31.
The following diagram shows four circuits A, B, C and D. Each circuit has a cell and a torch bulb. Name the circuits in which the bulb will glow ? Give a reason to your answer

Answer:
The bulb will glow in circuit (D).
This is because copper is the best conductor of electricity as compared to aluminium. Silk is a non-conductor of electricity.

Question 32.
The diagram given below shows a bulb connected with a cell having terminals A and B. Mark the direction of current in the bulb.

Answer:

Current always flows from +ve terminal to the -ve terminal of a cell.

Question 33.
State the function of each of the following in an electric circuit and draw its symbolic representation: (1) Switch and (2) Cell.
Answer:

1.  Switch – A switch or key is used to put the circuit on and off. fig. shows the symbol of a switch or key when it is open (to put the circuit off and when it is closed (to put the circuit on)
   
2.  Cell – A cell or a group of cells is generally used as a source of electricity. A positive (+) and a negative (-). It is represented by the two vertical lines of unequal lengths. The long vertical line represents the positive terminal and the short line represents the negative terminal as shown in fig.
   

Question 34.
Draw a circuit diagram for a bulb connected to a cell with a switch. Mark arrow in the diagram to indicate the direction of flow of current.
Answer:

Question 35.
In which arrangement are the appliances connected in the electric circuit of our homes, Series or Parallel ? Give one reason for your answer.
Answer:
Parallel circuit: When the circuit is in parallel, the appliances work independently. This is the reason that in our household wiring system, all the circuits are in parallel. Every appliance when put on, works on its own without the interruption of the other appliance.

Question 36.
State two precautions that you must take when switching on an electric circuit.
Answer:
Precautions to be taken before the circuit is switched on
Before the circuit is switched on, following precautions must be taken :

1.  See that all the components of the circuit are properly connected.
2.  See that the connection wire is tightly connected to each appliance or component.
3.  Do not touch the switch or any component with wet hands.

 

Selina Concise Chemistry Class 7 ICSE Solutions – Air and Atmosphere

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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

Oxygen constitutes about 21% of air by volume. It is the active part of air.

Nitrogen dioxide then reacts with the water vapour present in air to form nitrous and nitric acids.

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.

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.

(b) Manganese dioxides acts as a catalyst.

(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.

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

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

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

 

Selina Concise Chemistry Class 7 ICSE Solutions – Language of Chemistry

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ICSESolutions.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.

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Selina Concise ICSE Solutions for Class 7 Chemistry Chapter 5 Language of Chemistry

Points to Remember :

1. A chemical reaction involves the transformation of original substance into an altogether new substance(s).
2. A chemical reaction can be represented with the help of the symbols or the formulae of the elements and the compounds taking part in that reaction. This gives a chemical equation.
3. Certain necessary conditions for a chemical reaction to happen are  close contact, solution form, heat, light and catalyst.
4. Characteristics of chemical reactions are change of colour, evolution of a gas, formation of a precipitate, change of state, change of smell and evolution/absorption of heat.
5. A complete chemical equation symbolically represents the reactants, products and their physical states.
6. The substances that react with each other are called reactants and they are represented on the left hand side of the equation. The substances that are formed as a result of the reaction are called products. They are represented on the right hand side of the equation.
7. A chemical equation needs to be balanced to make it follow the law of the conservation of mass.
8. The law of conservation of mass states that mass can be neither created nor destroyed, it can only be transformed from one form to another.
9. A chemical equation gives both qualitative and quantitative information about the reactants and products.

EXERCISE

Question 1.
(a) Define chemical reaction.
(b) What is a chemical equation?
(c) Why do we need to balance chemical equations?
Answer:
(a) Chemical reaction : Any chemical change in matter which involves its transformation into one or more new substances is called a chemical reaction.
(b) Chemical equation : A chemical equation is the symbolic representation of a chemical reaction using the symbols and the formula of the substances involved in the reaction.
(c) A chemical equation needs to be balanced so as to make the number of the atoms of the reactants equal to the number of the atoms of the products.

Question 2.
State four conditions necessary for chemical reactions to take place.
Answer:
Conditions necessary for chemical reactions :

1. Close contact
2. Solution form
3. Heat
4. Light
5. Catalyst

3. Differentiate between :
(a) Reactants and products.

Reactants	Products
The substances that react with one another are called reactants. Reactants are written on the left hand side of equation.	The new substances formed are called products. Products are written on the right hand side of equation.

(b) Chemical reaction and chemical equation.

Chemical reaction	Chemical Equation
Any chemical change in matter which involves its transformation into one or more new substances is called a chemical reaction.	A chemical equation is the symbolic representation of a chemical reaction using the symbols and the formula of the substances involved in the reaction.

(c) A balanced and a skeletal chemical equation.

Balanced Equation	Skeletal Equation
A balanced chemical equation is one in which the number of atoms each element on the reactant side is equal to the number of atoms of that element on the product side.	In a skeletal equation the number of atoms on reactant side are not equal to number of atoms of product side.

Question 4.
Write word equations for the following skeletal equations:
(a) KClO₃ →  KCl + O₂
(b) Zn + HCl → ZnCl₂ + H₂
(c)FeCl₂ + Cl₂ → FeCl₃
(d) CO + O₂ → CO₂
(e) Ca + O₂ → CaO
(f) Na + O_{2 →}  Na₂O
(g) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
(h) AgBr →  Ag + Br₂
(i) KNO₂ →   KNO₂ + O₂

Answer:
(a) 2KClO₃  →  2KCl+ 3O₂
(b) Zn + 2HCl    →   ZnCl₂ + H₂
(c) 2FeCl₂ + Cl₂  →  2FeCl₃
(d) 2CO + O₂   →   2CO₂
(e) 2Ca + O₂ →  2CaO
(f) 4Na + O₂   → 2Na₂O
(g) 2NaOH + H₂SO₄   → Na₂SO₄ + 2H₂O
(h) 2AgBr     → 2Ag + Br₂
(i) 2KNO₃   → 2KN0₂ + O₂

Question 5.
Balance the following chemical equations :
(a) FeS + HCl →  FeCl₂ + H₂S
(b) Na₂CO₃ + HCl  →  NaCI + H₂O + CO₂
(c) H₂ + O₂   →  H₂O
(d) Na₂0 + H₂0  →  NaOH
Answer:
(a) FeS + 2HCl   →   FeCl₂ + H₂S
(b) Na₂CO₃ + 2HCl  →  2NaCl + H₂O + CO₂
(c) 2H₂+ O₂    →    2H₂O
(d) Na₂O + H₂O   →   2NaOH

Question 6.
What information do you get from the equation H₂+ Cl₂  →  2HCl ?
Answer:
(a)Hydrogen and chlorine molecules are the reactants.
(b)They are in gaseous form.
(c)The product is hydrogen chloride gas.
(d)Two molecules of hydrogen chloride are formed.

Question 7.
Write your observations for the following chemical reactions and name the products formed :
(a) When sugar is heated.
(b) When manganese dioxide is added to potassium chlorate and heated.
(c) When dilute acetic acid is poured on baking soda.
(d) When an aqueous solution of sodium chloride is mixed with an aqueous solution of silver nitrate.
(e) When ammonium chloride is heated with sodium hydroxide.
(f) When water is added to quick lime?

Answer:
(a) Black solid mass (charcoal) is formed along with water vapours.
(b) Manganese dioxide acts as a catalyst for the decomposition of potassium chlorate into potassium chloride and oxygen at a lower temperature.
(c) Sodium acetate, CO₂ and water is formed.
(d) A white insoluble solid precipitate of silver chloride is formed along with Sodium nitrate.
(e) When solid ammonium chloride is heated with sodium hydroxide solution, a gas ammonia is evolved which is recognised by its strong pungent smell.

(f) When water is added to quick lime, a large amount of heat energy is evolved.

Question 8.
Write symbolic representation for the following word equations and balance them :
(a) Calcium carbonate → Calcium oxide + Carbon dioxide
(b) Carbon + Oxygen → Carbon dioxide
(c) Calcium oxide + Water → Calcium hydroxide
(d) Aluminium + Chlorine → Aluminium chloride
(e) Iron + Sulphur → Iron sulphide

Answer:

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks:
(a) The substances which undergo chemical change are called reactants.
(b) The substances formed as a result of a chemical reaction are called products.
(c) During a chemical reaction transfer of energy takes place.
(d) The basic conditions necessary for a chemical reaction is close contact.
(e) In some chemical reactions an insoluble precipitate is formed when two solutions are mixed.

2. Write ‘true’ or ‘false’ for the following statements :
(a) No new substance is formed during a chemical reaction : True
(b) Hydrogen sulphide has rotten egg smell : True
(c) When potassium iodide solution is added to lead acetate solution a red precipitate is formed : False
(d) A black residue is formed when sugar is heated : True
(e) When iron and sulphur are heated together a grey mass is formed which is attracted by a magnet : False
(f) A chemical equation gives only qualitative information of a chemical reaction : False

MULTIPLE CHOICE QUESTIONS

1. A chemical equation is a statement that describes a chemical change in terms of
(a) symbols and formulae
(b) energy
(c) number of atoms
(d) colours

2. Balancing a chemical equation is based on
(a) Law of conservation of mass
(b) Mass of reactants and products
(c) Symbols and formulae
(d) None of the above

3. Copper carbonate when heated, it turns :
(a) Blue
(b) Green
(c) Black
(d) Yellow

 

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 6 Sets. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

 

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Sets Exercise 6A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Write the following sets in roster (Tabular) form :

Solution:

Question 2.
Write the following sets in set-builder (Rule Method) form :

Solution:

Question 3.
(i) Is {1, 2, 4, 16, 64} = {x : x is a factor of 32} ? Give reason.
(ii) Is {x : x is a factor of 27} ≠ {3, 9, 27, 54} ? Give reason.
(iii) Write the set of even factors of 124.
(iv) Write the set of odd factors of 72.
(v) Write the set of prime factors of 3234.
(vi) Is {x : x² – 7x + 12 = 0} = {3, 4} ?
(vii) Is {x : x² – 5x – 6 = 0} = {2, 3} ?
Solution:
(i) No, {1, 2, 4, 16, 64} ≠ {x : x is factor of 32}
Because 64 is not a factor of 32
(ii) Yes, {x : x is a factor of 27} + {3, 9, 27, 54}
Because 54 is not a factor of 27
(iii) 1 x 124 = 124
2 x 62 = 124
4 x 31 = 124
Factors of 124 = 1, 2, 4, 31, 62, 124
Set of even factors of 124 = {2, 4, 62, 124}
(iv) 1 x 72 = 72
2 x 36 = 72
3 x 24 = 72
4 x 18 = 72
6 x 12 = 72
8 x 9 = 72
Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Set of odd factors of 72 = {1, 3, 9}

Question 4.
Write the following sets in Roster form :
(i) The set of letters in the word ‘MEERUT’.
(ii) The set of letters in the word ‘UNIVERSAL’.
(iii) A = {x : x = y + 3, y ∈N and y > 3}
(iv) B = {p : p ∈ W and p² < 20}
(v) C = {x : x is composite number and 5 < x < 21}
Solution:
(i) Roster form of the set of letters in the word “MEERUT” = {m, e, r, u, t}
(ii) Roster form of the set of letters in the word “UNIVERSAL” = {u, n, i, v, e, r, s, a, l}
(iii) A = {x : x = y + 3, y ∈ N and y > 3}
x = y + 3

Question 5.
List the elements of the following sets :
(i) {x : x² – 2x – 3 = 0}
(ii) {x : x = 2y + 5; y ∈ N and 2 ≤ y < 6}
(iii) {x : x is a factor of 24}
(iv) {x : x ∈ Z and x² ≤ 4}
(v) {x : 3x – 2 ≤ 10, x ∈ N}
(vi) {x : 4 – 2x > -6, x ∈ Z}
Solution:

Sets Exercise 6B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the cardinal number of the following sets :

Solution:

Question 2.
If P = {P : P is a letter in the word “PERMANENT”}. Find n (P).
Solution:
P = (P : P is a letter in the word “PERMANENT”}
or P = {p, e, r, m, a, n, t)
n (P) = 7

Question 3.
State, which of the following sets are finite and which are infinite :

Solution:

Question 4.
Find, which of the following sets are singleton sets :
(i) The set of points of intersection of two non-parallel st. lines in the same plane
(ii) A = {x : 7x – 3 = 11}
(iii) B = {y : 2y + 1 < 3 and y ∈ W}
Note : A set, which has only one element in it, is called a SINGLETON or unit set.
Solution:

Question 5.
Find, which of the following sets are empty :
(i) The set of points of intersection of two parallel lines.
(ii) A = {x : x ∈ N and 5 < x < 6}
(iii) B = {x : x² + 4 = 0, x ∈ N}
(iv) C = {even numbers between 6 & 10}
(v) D = {prime numbers between 7 & 11}
Note : The set, which has no element in it, is called the empty or null set.
Solution:

Question 6.
(i) Are the sets A = {4, 5, 6} and B = {x : x² – 5x – 6 = 0} disjoint ?
(ii) Are the sets A = {b, c, d, e} and B = {x : x is a letter in the word ‘MASTER’} joint ?
Note :
(i) Two sets are said to be joint sets, if they have atleast one element in common.
(ii) Two sets are said to be disjoint, if they have no element in common.
Solution:

Question 7.
State, whether the following pairs of sets are equivalent or not :
(i) A = {x : x ∈ N and 11 ≥ 2x – 1} and B = {y : y ∈ W and 3 ≤ y ≤ 9}
(ii) Set of integers and set of natural numbers.
(iii) Set of whole numbers and set of multiples of 3.
(iv) P = {5, 6, 7, 8} and M = {x : x ∈ W and x < 4}
Note : Two sets are said to be equivalent, if they contain the same number of elements.
Solution:

B = {3,4,5,6,7,8,9}
n (B) = 7
Cardinal number of set A = 6 and cardinal number of set B = 7
Set A and set B are not equivalent.
(ii) Set of integers has infinite number of elements. Set of natural numbers has infinite number of elements.
Set of integers and set of natural numbers are equivalent because both these sets have infinite number of elements.
(iii) Set of whole numbers, has infinite number of elements. Set of multiples of 3, has infinite number of element.
Set of whole numbers and set of multiples of 3 are equivalent because both these sets have infinite number of elements.
(iv) P = {5,6,7,8}
n (P) = 4
M = {x : x ∈ W and x ≤ 4}
M = {0, 1, 2, 3, 4}
n (M) = 5
Now Cardinal number of set P = 4 and
Cardinal number of set M = 5
These sets are not equivalent.

Question 8.
State, whether the following pairs of sets are equal or not :

Solution:

Question 9.
State whether each of the following sets is a finite set or an infinite set:
(i) The set of multiples of 8.
(ii) The set of integers less than 10.
(iii) The set of whole numbers less than 12.
(iv) {x : x = 3n – 2, n ∈ W, n ≤ 8}
(v) {x : x = 3n – 2,n ∈ Z, n ≤ 8}
(vi) {x : x = \(\frac { n-2 }{ n+1 }\) , n ∈ w)
Solution:

Question 10.
Answer, whether the following statements are true or false. Give reasons.
(i) The set of even natural numbers less than 21 and the set of odd natural numbers less than 21 are equivalent sets.
(ii) If E = {factors of 16} and F = {factors of 20}, then E = F.
(iii) The set A = {integers less than 20} is a finite set.
(iv) If A = {x : x is an even prime number}, then set A is empty.
(v) The set of odd prime numbers is the empty set.
(vi) The set of squares of integers and the set of whole numbers are equal sets.
(vii) In n(P) = n(M), then P → M.
(viii) If set P = set M, then n(P) = n(M).
(ix) n(A) = n(B) => A = B.
Solution:

Sets Exercise 6C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find all the subsets of each of the following sets :
(i) A = {5, 7}
(ii) B = {a, b, c}
(iii) C = {x : x ∈ W, x ≤ 2}
(iv) {p : p is a letter in the word ‘poor’}
Solution:
(i) A = {5,7}

Question 2.
If C is the set of letters in the word “cooler”, find :
(i) Set C
(ii) n(C)
(iii) Number of its subsets
(iv) Number of its proper subsets.
Note : (i) If a set has n elements, the number of its subsets = 2ⁿ
(ii) If a set has n elements, the number of its proper subsets = 2ⁿ – 1
Solution:
(i) C = {c, o, l, e, r}
(ii) n(C) = 5
(iii) Number of its subsets : 2⁵ = 2 x 2 x 2 x 2 x 2 = 32
(iv) Number of its proper subsets = 2⁵ – 1 = 32 – 1 = 31

Question 3.
If T = {x : x is a letter in the word ‘TEETH’}, find all its subsets.
Solution:
T = {t,e,h}
Subsets of set T = φ, {r}, {e}, {h}, {t,e}, {t,h}, {e,h}, {t,e,h}

Question 4.
Given the universal set = {-7,-3, -1, 0, 5, 6, 8, 9}, find :
(i) A = {x : x < 2}
(ii) B = {x : -4 < x < 6}
Solution:
Universal set = {-7, -3, -1, 0, 5, 6, 8, 9},
(i) A = {x : x < 2} = {-7, -3, -1, 0}
(ii) B = {x : -4 < x < 6} = {-3, -1, 0, 5}

Question 5.
Given the universal set = {x : x ∈ N and x < 20}, find :
(i) A = {x : x = 3p ; p ∈ N}
(ii) B = {y : y – 2n + 3, n ∈ N}
(iii) C = {x : x is divisible by 4}
Solution:

Question 6.
Find the proper subsets of {x : x² – 9x – 10 = 0}
Solution:

x = 10
=> x = -1
Given set = {-1, 10}
Proper subsets of this set = φ, {-1}, {10}

Question 7.
Given, A = {Triangles}, B = {Isosceles triangles}, C = {Equilateral triangles}. State whether the following are true or false. Give reasons.

Solution:

Question 8.
Given, A = {Quadrilaterals}, B = {Rectangles}, C = {Squares}, D= {Rhombuses}. State, giving reasons, whether the following are true or false.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Sets Exercise 6D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.

Solution:

Question 2.

Solution:

Question 3.


Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Sets Exercise 6E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.

Solution:

Question 2.
From the given diagram, find :
(i) A’
(ii) B’

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Selina Concise Chemistry Class 8 ICSE Solutions – Hydrogen

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 7 Hydrogen. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 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 8 Chemistry Chapter 7 Hydrogen

Points to Remember:

1. Hydrogen is the most abundant element found in the universe.
2. Hydrogen is much more common in the form of compounds. The most important compound of hydrogen is water.
3. The chief sources of hydrogen are water, acids and alkalies.
4. Hydrogen is prepared by the action of water, acids or alkalies on active metals.
5. Electrolysis of water results in the formation of hydrogen and oxygen.
6. Hydrogen is lighter than air.
7. Hydrogen bums in air with pop sound.
8. Hydrogen acts as a reducing agent.
9. Hydrogen is used to produce oxyhydrogen flame and in weather forecast balloons.

Exercise

1. Fill in the blanks:

(a) Hydrogen is lighter than air.
(b) Hydrogen is sparingly soluble in water.
(c) Hydrogen bums with a oxyhydrogen pale blue flame and pop sound is heard.
(d) A metal sodium hydrogen in the reactivity series gives hydrogen with water.
(e) Hydrogen reacts with metal oxides to form metal and water.
(f) Oxidation is the removal of hydrogen and addition of oxygen.
(g) In redox reaction oxidation and reduction occur simultaneously.

2. Indicate which of the following statements are true and which are false:

(a) Hydrogen molecule is monovalent.
(b) The removal of hydrogen from a substance is called reduction.
(c) Nitric acid can not be used to prepare hydrogen by its action on active metals ?
(d) The reaction between hydrogen and nitrogen to form ammonia is reversible.
(e) Zinc can liberate hydrogen from water, acid and alkali solution.
(f) Hydrogen is combustible as well as a supporter of combustion.
(g) Hydrogen gas is easily liquefiable.
Answer:
(a) False
(b) True
(c) False. Hydrogen cannot be prepared by the action of nitric acid on metals because it also releases nitrous oxide and nitric oxide.
(d) True
(e) True
(f) False
(g) False

3. Complete and balance the following equations:


Answer:

4. Give reasons for the following:

(a) Hydrogen be used as a fuel?
(b) Though hydrogen is lighter than air it cannot be collected by downward displacement of air.
(c) A pop sound produced when hydrogen is burnt?
(d) Helium replaced hydrogen in weather observation balloons?
(e) Nitric acid not used for the preparation of hydrogen gas?
(a) Because of its high heat of combustion, it is used as a fuel.
Answer:
(a) Coal gas, water gas and liquid hydrogen are some significant fuel.
(b) Since hydrogen is lighter than air. it is possible to collect the gas by downward displacement of air. But it is not safe to do so since a mixture of hydrogen and air can lead to an explosion.
(c) Impure hydrogen gas bums in air with a pop sound. This is because of the presence of impurities in it.
(d) If there is small leakage of hydrogen in a balloon, it forms a mixture with air that can explode. So helium has replaced hydrogen.
(e) Hydrogen cannot be prepared by the action of nitric acid on metals because it also releases nitrous oxide and nitric oxide and oxides the hydrogen to form water.

5. Name the following:
(a) Two metals which give hydrogen with cold water.
(b) A metal which liberates hydrogen only when steam is passed over red hot metal.
(c) The process in which oxygen is added or hydrogen is removed.
(d) A metallic oxide which can be reduced into metal by hydrogen.
Answer:
(a) Sodium (Na) and Potassium (K) give hydrogen with cold water.
(b) Iron
(c) Oxidation
(d) Copper oxide (CuO)

6. (a) Name the chemicals required to prepare hydrogen
gas in the laboratory.
(b) Give a balanced chemical equation for the reaction.
(c) Draw a neat and well-labelled diagram for the laboratory preparation ofhydrogen.
(d) How is hydrogen gas collected?
Answer:
(a) Granulated Zinc and dil. Hydrochloric acid.
(b) Zn + 2 HCl → ZnCl₂ + H₂ (g)
(c)

(d) Hydrogen gas is collected by the down-ward displacement of water.

7. How would you show that hydrogen:
(a) is a non-supporter of combustion?
(b) is lighter than air?
Answer:
(a) Hold a hydrogen gas-filled jar with its mouth downwards.
Place a lighted candle inside the jar. The candle gets extinguished but the gas bums with a pop sound. This shows that hydrogen is a non-supporter of combustion.

(b) Take a delivery tube and place one of its ends in a soap solution kept in a trough and the other one in a flat bottom jar as shown in the figure. The soap bubbles containing hydrogen rise upward the air. The rising soap bubbles prove that hydrogen is lighter than air.

Hydrogen-filled soap bubbles rising upward in the soap solution and into the air shows that hydrogen is lighter than air.

8. Hydrogen is a good reducing agent: What do you understand by the above statement? Explain with the help of copper oxide as an example.
Answer:
Hydrogen acts as a good reducing agent means, when hydrogen gas is passed over hot metallic oxides of copper, lead, iron, etc. it removes oxygen from them and thus reduces them to their corresponding metal.
Let us consider the following example, in each of which metallic oxide react with hydrogen. Metallic oxide act as oxidising agents and hydrogen acts as a reducing agent.

9. (a) Name a process by which hydrogen gas is manufactured.
(b) Give equations for the reactions.
(c) How is hydrogen separated from carbon dioxide and carbon monoxide?
Answer:
(a) Commercially, hydrogen is prepared by Bosch process.

(b) (i) Steam is passed over hot coke at 1000°C in a furnace called converters. As a result water gas is produced which is a mixture of carbon monoxide and hydrogen gases.

This reaction is endothermic in nature.
(ii) Water gas is mixed with excess of steam and passed over a catalyst ferric oxide (Fe₂O₃) and a promotor chromium trioxide (Cr₂O₃).

This reaction is exothermic in nature

(c) (i) The products are hydrogen, carbon dioxide and some unreacted carbon monoxide. Hydrogen is separated from carbon dioxide by passing the mixture through water under pressure, in which carbon dioxide gets dissolved leaving behind hydrogen. Carbon dioxide can also be separated by passing it through caustic potash (KOH) solution.
2KOH + CO₂ → K₂CO₃ + H₂O
(ii) To separate carbon monoxide the gaseous mixture is passed through ammoniacal cuprous chloride in which carbon monoxide dissolves leaving behind hydrogen.

Thus hydrogen gas is obtained.

10. Match the statements in Column A with those in Column B.

Answer:

11. State four uses of hydrogen:
Answer:

1. Hydrogen with oxygen produces an oxy-hydrogen flame which is used for cutting and welding.
2. Hydrogen gas is used as a fuel.
3. Hydrogen is used for the hydrogenation of vegetable oil.
4. Hydrogen gas is used extensively in the manufacture of ammonia gas, which is used to produce fertilizers.

12. Define:
(a) catalytic hydrogenation (b) oxidation
(c) reduction (d) redox reaction
Answer:
(a) Catalytic hydrogenation: catalytic hydrogenation is a process by which hydrogen gas is passed through vegetable oils in the presence of catalysts like Ni, Pt or Pd to convert them into solid vanaspati ghee.

(b) Oxidation: A reaction in which a substance combines with oxygen or in which hydrogen is removed is called an oxidation reaction.
Example: H₂S + Cl → 2HCl + S

(c) Reduction: Those reactions in which hydrogen combines with a substance or oxygen is removed from a substance, are known as reduction reactions.

(d) Redox reaction: Redox reactions are those in which reduction and oxidation both takes place simultaneously i.e. one substance is reduced while the other gets oxidised.

13. Multiple Choice Questions

(a) Equal volumes of hydrogen and chlorine are exposed to diffused sunlight to prepare

1. hydrogen chloride
2. water
3. sodium hydroxide
4. hydrochloric acid

(b) The metal which reacts with cold water to produce hydrogen is

1. magnesium
2. aluminium
3. calcium
4. iron

(c) In metal activity series the more reactive metals are at

1. top
2. bottom
3. middle
4. none

(d) Hydrogen is responsible for producing

1. heat and light
2. hydrogenated oil
3. fertilizers
4. all of the above

(e) Hydrogen is

1. combustible
2. non-combustible
3. supporter of combustion
4. neither supporter nor combustible

(f) Water gas is a mixture of

1. carbon monoxide and oxygen
2. carbon monoxide and hydrogen
3. hydrogen and oxygen
4. hydrogen and nitrogen.

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 10 Direct and Inverse Variations. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Direct and Inverse Variations Exercise 10A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
In which ofthe following tables, x and y vary directly:

Solution:

Question 2.
If x and y vary directly, find the values of x, y and z:

Solution:

Question 3.
A truck consumes 28 litres of diesel for moving through a distance of 448 km. How much distance will it cover in 64 litres of diesel?
Solution:

Question 4.
For 100 km, a taxi charges ₹ 1,800. How much will it charge for a journey of 120 km?
Solution:

Question 5.
If 27 identical articles cost ₹ 1,890, how many articles can be bought for ₹ 1,750?
Solution:

Question 6.
7 kg of rice costs ₹ 1,120. How much rice can be bought for ₹ 3,680?
Solution:

Question 7.
6 note-books cost ₹ 156, find the cost of 54 such note-books.
Solution:

Question 8.
22 men can dig a 27 m long trench in one day. How many men should be employed for digging 135 m long trench of the same type in one day?
Solution:

Question 9.
If the total weight of 11 identical articles is 77 kg, how many articles of the same type would weigh 224 kg?
Solution:

Question 10.
A train is moving with uniform speed of 120 km per hour.
(i) How far will it travel in 36 minutes?
(ii) In how much time will it cover 210 km?
Solution:

Direct and Inverse Variations Exercise 10B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Check whether x and y vary inversely or not.

Solution:
x and y are inversely proportional.
Then xy are equal.
(i) xy = 4 x 6 = 24
xy = 3 x 8 = 24
xy = 12 x 2 = 24
xy = 1 x 24 = 24
xy in each case is equal.
x and y are inversely proportional
(ii) xy = 30 x 60 = 1800
xy= 120 x 30 = 3600
xy = 60 x 30= 1800
xy = 24 x 75 = 1800
xy in each case is not equal.
x and y are not inversely proportional.
(iii) xy = 10 x 90 = 900
= 30 x 30 = 900
= 60 x 20= 1200
= 10 x 90 = 900
xy in each case is not equal.
x and y are not inversely proportional.

Question 2.
If x and y vary inversely, find the values of l, m and n :

Solution:

Question 3.
36 men can do a piece of work in 7 days. How many men will do the same work in 42 days?
Solution:

Question 4.
12 pipes, all of the same size, fill a tank in 42 minutes. How long will it take to fill the same tank, if 21 pipes of the same size are used?
Solution:

Question 5.
In a fort 150 men had provisions for 45 days. After 10 days, 25 men left the fort. How long would the food last at the same rate?
Solution:

Question 6.
72 men do a piece of work in 25 days. In how many days will 30 men do the same work?
Solution:

Question 7.
If 56 workers can build a wall in 180 hours, how many workers will be required to do the same work in 70 hours?
Solution:

Question 8.
A car takes 6 hours to reach a destination by travelling at the speed of 50 km per hour. How long will it take when the car travels at the speed of 75 km per hour?
Solution:

Direct and Inverse Variations Exercise 10C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Cost of 24 identical articles is Rs. 108, Find the cost of 40 similar articles.
Solution:

Question 2.
If 15 men can complete a piece of work in 30 days, in how many days will 18 men complete it?
Solution:

Question 3.
In order to complete a work in 28 days, 60 men are required. How many men will be required if the same work is to be completed in 40 days ?
Solution:
Let x be number of men required 60 men can do the work in = 28 days
1 man can do the work in = 28 x 60 days
x man can do the work in = \(\frac { 28\times 60 }{ x }\) day

Question 4.
A fort had provisions for 450 soldiers for 40 days. After 10 days, 90 more soldiers come to the fort. Find in how many days will the remaining provisions last at the same rate ?
Solution:
After 10 days :
For 450 soldiers, provision are sufficient for (40 – 10) days = 30 days
For 1 soldier, provision are sufficient for 30 x 450 days
For 540 soldiers, the provision are sufficient for

Question 5.
A garrison has sufficient provisions for 480 men for 12 days. If the number of men is reduced by 160; find how long will the provisions last.
Solution:

Question 6.
\(\frac { 3 }{ 5 }\) quintal of wheat costs Rs.210. Find the cost of :
(i) 1 quintal of wheat
(ii) 0.4 quintal of wheat
Solution:
(i) \(\frac { 3 }{ 5 }\) quintal of wheat costs = Rs.210
1 quintal of wheat costs = 210 x \(\frac { 3 }{ 5 }\) = 70 x 5 = Rs.350
(ii) 1 quintal of wheat costs = Rs.350
0.4 quintal of wheat costs = 350 x 0.4 = Rs. 140.0 = Rs.140

Question 7.
If \(\frac { 2 }{ 9 }\) of a property costs Rs.2,52,000; find the cost of \(\frac { 4 }{ 7 }\) of it.
Solution:

Question 8.
4 men or 6 women earn Rs. 360 in one day. Find, how much will:
(i) a man earn in one day ?
(ii) a woman earn in one day ?
(iii) 6 men and 4 women earn in one day ?
Solution:
4 men earn Rs. 360 in one day

Question 9.
16 boys went to canteen to have tea and snacks together. The bill amounted to Rs. 114.40. What will be the contribution of a boy who pays for himself and 5 others ?
Solution:

Question 10.
50 labourers can dig a pond in 16 days. How many labourers will be required to dig an another pond, double in size in 20 days ?
Solution:

Question 11.
If 12 men or 18 women can complete a piece of work in 7 days, in how many days can 4 men and 8 women complete the same work?
Solution:

Question 12.
If 3 men or 6 boys can finish a work in 20 days, how long will 4 men and 12 boys take to finish the same work ?
Solution:
3 men = 6 boys
4 men = \(\frac { 6 }{ 3 }\) x 4 = 8
Total boys in second case :
= 4 men + 12 boys = 8 + 12 = 20 boys
6 boys can do a piece of work in 20 days
Then let 20 boys will do the same work in x days

Question 13.
A particular work can be completed by 6 men and 6 women in 24 days; whereas the same work can be completed by 8 men and 12 women in 15 days. Find :
(i) according to the amount of work done, one man is equivalent to how many women.
(ii) the time taken by 4 men and 6 women to complete the same work.
Solution:
6 men + 6 women can finish the work in = 24 days
144 men + 144 women can finish it in = 1 day
8 m + 12 women can finish the work in = 15 days
120 men + 180 women can finish it in = 1 day
(i) 144 men + 144 women = 120 men + 180 women
=> 144 men – 120 men
= 180 – women – 144 women
=> 24 men = 36 women
1 man = \(\frac { 36 }{ 24 }\)

Question 14.
If 12 men and 16 boys can do a piece of work in 5 days and, 13 men and 24 boys can do it in 4 days, how long will 7 men and 10 boys take to do it ?
Solution:
12 men + 16 boys can do a piece of work in = 5 days
60 men + 80 boys can do a piece of work in = 1 day ………. (i)
and 13 men + 24 boys can do the same work in = 4 days
52 men + 96 boys can do the same work in = 1 day ………….(ii)
From (i) and (ii)
60 men + 80 boys = 52 men + 96 boys
=> 60 men – 52 men = 96 boys – 80 boys
=> 8 men = 16 boys
1 men = \(\frac { 16 }{ 8 }\) = 2 boys
Now, in first case,
12 men + 16 boys = 12 x 2 + 16 = 24+16 = 40 boys
In the second case,
7 men + 10 boys = 7 x 2 + 10 = 14 + 10 = 24 boys
Now 40 boys can do a piece of work in = 5 days
1 boy can do the same work in = 5 x 40 days
and 24 boys will do the same work in

Direct and Inverse Variations Exercise 10D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Eight oranges can be bought for Rs. 10.40. How many more can be bought for Rs. 16.90?
Solution:
Number of oranges bought for Rs. 10.40 = 8

Question 2.
Fifteen men can build a wall in 60 days. How many more men are required to build another wall of same size in 45 days ?
Solution:
In 60 days a wall can be built by = 15 men
In 1 day a wall can be built by = 15 x 60 men
In 45 days a wall can be built by = \(\frac { 15\times 60 }{ 45 } =\frac { 900 }{ 45 }\) = 20 men
No. of more men required to build the wall in 45 days = 20 – 15 = 5 men

Question 3.
Six taps can fill an empty cistern in 8 hours. How much more time will be taken, if two taps go out of order ? Assume, all the taps supply water at the same rate.
Solution:
Total no. of taps = 6
Out of order taps = 2
Taps in working condition =6 – 2 = 4
6 taps can fill an empty cistern in = 8 hours
1 tap can fill an empty cistern in = 6 x 8 hours
4 taps can fill an empty cistern in = \(\frac { 48 }{ 4 }\) = 12 hours
More time taken when 2 taps are out of order = 12 – 8 = 4 hour

Question 4.
A contractor undertakes to dig a canal, 6 kilometres long, in 35 days and employed 90 men. He finds that after 20 days only 2 km of canal have been completed. How many more men must be employed to finish the work in time ?
Solution:
Length of canal = 6 km
In 20 days canal made = 2 km
Remaining length of canal = 6 – 2 = 4 km
Remaining time = 35 – 20 = 15 days
In 20 days 2 km canal is made by = 90 men
In 1 day 2 km canal is made by = 90 x 20 men
In 15 days 2 km canal is made by = \(\frac { 90\times 20 }{ 15 }\) men
In 15 days 1 km canal is made by = \(\frac { 90\times 20 }{ 15\times 2 }\) men
In 15 days 4 km canal is made by = \(\frac { 90\times 20\times 4 }{ 15\times 2 }\) men = 6 x 10 x 4 men = 240 men
Number of more men to be employed to finish the work in time = 240-90 = 150 men

Question 5.
If 10 horses consume 18 bushels in 36 days. How long will 24 bushels last for 30 horses ?
Solution:

Question 6.
A family of 5 persons can be main¬tained for 20 days with Rs.2,480. Find, how long Rs.6944 maintain a family of 8 persons
Solution:
A family of 5 persons can be maintained with Rs.2480 for = 20 days

Question 7.
90 men can complete a work in 24 days working 8 hours a day. How many men are required to complete the same work in 18 days working 7\(\frac { 1 }{ 2 }\) hours a day ?
Solution:

Question 8.
Twelve typists, all working with same speed, type a certain number of pages in 18 days working 8 hours a day. Find, how many hours per day must sixteen typists work in order to type the same number of pages in 9 days ?
Solution:
12 typists can type in 18 days with number of working hours in day = 8 hours
1 typist can type in 18 days = 8 x 12 hour
1 typist can type in 9 days = 2 (8 x 12) hour
16 typist can type in a day = \(\frac { 2(8\times 12) }{ 16 }\) = 12 hours

Question 9.
If 25 horses consume 18 quintal in 36 days, how long will 28 quintal last for 30 horses ?
Solution:

Question 10.
If 70 men dig 15,000 sq. m of a field in 5 days, how many men will dig 22,500 sq. m field in 25 days ?
Solution:

Question 11.
A contractor undertakes to build a wall 1000 m long in 50 days. He employs 56 men, but at the end of 27 days, he finds that only 448 m of wall is built. How many extra men must the contractor employ so that the wall is completed in time ?
Solution:
Number of men employed in the beginning = 56
Length of wall = 1000 m No. of days = 50
In the time of 27 days, only 448 m of wall was completed
Remaining period = 50 – 27 = 23 days
and length of wall to be completed = 1000 – 448 = 552
Now in 27 days, 448 m long wall was completed by = 1000 m
in 1 day, 448 m long was completed by = 56 x 27
inf 1 day, 1 m long wall will be completed by

Question 12.
A group of labourers promises to do a piece of work in 10 days, but five of them become absent. If the remaining labourers complete the work in 12 days, find their original number in the group.
Solution:
Total period = 10 days
But work completed in = 12 days
No. of men were absent = 5
Let the number of men in the beginning = x
Now x men can do a piece work in = 10 days
1 man will do it in = 10x x days

Question 13.
Ten men, working for 6 days of 10 hours each, finish \(\frac { 5 }{ 21 }\) of a piece of work. How many men working at the same rate and for the same number of hours each day, will be required to complete the remaining work in 8 days ?
Solution:

Direct and Inverse Variations Exercise 10E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A can do a piece of work in 10 days and B in 15 days. How long will they take together to finish it ?
Solution:

Question 2.
A and B together can do a piece of work in 6\(\frac { 2 }{ 3 }\) days ; but B alone can do it in 10 days. How long will A take to do it alone ?
Solution:

Question 3.
A can do a work in 15 days and B in 20 days. If they together work on it for 4 days ; what fraction of the work will be left ?
Solution:

Question 4.
A, B and C can do a piece of work in 6 days, 12 days and 24 days respectively. In what time will they all together do it ?
Solution:

Question 5.
A and B working together can mow a field in 56 days and with the help of C, they could have mowed it in 42 days. How long would C take by himself ?
Solution:

Question 6.
A can do a piece of work in 24 days, A and B can do it in 16 days and A, B and C in 10\(\frac { 2 }{ 3 }\) days. In how many days can A and C do it ?
Solution:

Question 7.
A can do a piece of work in 20 days and B in 15 days. They worked together on it for 6 days and then A left. How long will B take to finish the remaining work ?
Solution:

Question 8.
A can finish a piece of work in 15 days and B can do it in 10 days. They worked together for 2 days and then B goes away. In how many days will A finish the remaining work?
Solution:

Question 9.
A can do a piece of work in 10 days ; B in 18 days; and A, B and C together in 4 days. In what time would C alone do it ?
Solution:

Question 10.
A can do \(\frac { 1 }{ 4 }\) of a work in 5 days and B can do \(\frac { 1 }{ 3 }\) of the same work in 10 days. Find the number of days in which both working together will complete the work.
Solution:

Question 11.
One tap can fill a cistern in 3 hours and the waste pipe can empty the full cistern in 5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together ?
Solution:

Question 12.
A and B can do a work in 8 days; B and C in 12 days, and A and C in 16 days. In what time could they do it, all working together ?
Solution:
A and B can do a work in = 8 days
B and C can do a work in = 12 days
A and C can do a work in = 16 days
(A+B)’s 1 day work = \(\frac { 1 }{ 8 }\)
(B+C)’s 1 day work = \(\frac { 1 }{ 12 }\)

Question 13.
A and B complete a piece of work in 24 days. B and C do the same work in 36 days ; and A, B and C together finish it in 18 days. In how many days will (i) A alone,
(ii) C alone,
(iii) A and C together, complete the work ?
Solution:
A and B complete a piece of work in = 24 days
B and C complete a piece of work in = 36 days
(A+B+C) complete a piece of work in = 18 days
(A+B)’s 1 day work = \(\frac { 1 }{ 24 }\)
(B+C)’s 1 day work = \(\frac { 1 }{ 36 }\)
(A+B+C)’s 1 day work = \(\frac { 1 }{ 18 }\)

Question 14.
A and B can do a piece of work in 40 days; B and C in 30 days; and C and A in 24 days.
(i) How long will it take them to do the work together ?
(ii) In what time can each finish it working alone ?
Solution:
A and B can do a piece of work in = 40 days
B and C can do a piece of work in = 30 days
C and A can do a piece of work in = 24 days
(A+B)’s 1 day work = \(\frac { 1 }{ 40 }\)
(B+C)’s 1 day work = \(\frac { 1 }{ 30 }\)
(C+A)’s 1 day work = \(\frac { 1 }{ 24 }\)
(i) [(A+B)+(B+C)+(C+A)]’s 1 day work = \(\frac { 1 }{ 40 }\) + \(\frac { 1 }{ 30 }\) + \(\frac { 1 }{ 24 }\)


C can do the work in 40 days
Hence A can do the work in = 60 days
B can do the work in = 120 days
C can do the work in = 40 days

Question 15.
A can do a piece of work in 10 days, B in 12 days and C in 15 days. All begin together but A leaves the work after 2 days and B leaves 3 days before the work is finished. How long did the work last ?
Solution:

Question 16.
Two pipes P and Q would fill an empty cistern in 24 minutes and 32 minutes respectively. Both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern may be just filled in 16 minutes.
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

ICSESolutions.com Provides Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 5 Playing with APIusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 5 Playing with Number. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Playing with Number Exercise 5A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Write the quotient when the sum of 73 and 37 is divided by
(i) 11
(ii) 10
Solution:

Question 2.
Write the quotient when the sum of 94 and 49 is divided by
(i) 11
(ii) 13
Solution:

Question 3.
Find the quotient when 73 – 37 is divided by
(i) 9
(ii) 4
Solution:

Question 4.
Find the quotient when 94 – 49 is divided by
(i) 9
(ii) 5
Solution:

Question 5.
Show that 527 + 752 + 275 is exactly divisible by 14.
Solution:
Property :
abc = 100a + 106 + c ………(i)
bca = 1006 + 10c + a ……..(ii)
and cab = 100c + 10a + b ……….(iii)
Adding, (i), (ii) and (iii), we get abc + bca + cab = 111a + 111b + 111c = 111(a + b + c) = 3 x 37(a + b + c)
Now, let us try this method on
527 + 752 + 275 to check is it exactly divisible by 14
Here, a = 5, 6 = 2, c = 7
527 + 752 + 275 = 3 x 37(5 + 2 + 7) = 3 x 37 x 14
Hence, it shows that 527 + 752 + 275 is exactly divisible by 14

Question 6.
If a = 6, show that abc = bac.
Solution:
Given : a = 6
To show : abc = bac
Proof: abc = 100a + 106 + c …….(i)
(By using property 3)
bac = 1006 + 10a + c —(ii)
(By using property 3)
Since, a = 6
Substitute the value of a = 6 in equation (i) and (ii), we get
abc = 1006 + 106 + c ………(iii)
bac = 1006 + 106 + c ………(iv)
Subtracting (iv) from (iii) abc – bac = 0
abc = bac
Hence proved.

Question 7.
If a > c; show that abc – cba = 99(a – c).
Solution:
Given, a > c
To show : abc – cba = 99(a – c)
Proof: abc = 100a + 10b + c ……….(i)
(By using property 3)
cba = 100c + 10b + a ………..(ii)
(By using property 3)
Subtracting, equation (ii) from (i), we get
abc – cba = 100a + c – 100c – a
abc – cba = 99a – 99c
abc – cba = 99(a – c)
Hence proved.

Question 8.
If c > a; show that cba – abc = 99(c – a).
Solution:
Given : c > a
To show : cba – abc = 99(c – a)
Proof:
cba = 100c + 106 + a ……….(i)
(By using property 3)
abc = 100a + 106 + c ………(ii)
(By using property 3)
Subtracting (ii) from (i)
cba – abc= 100c+ 106 + a- 100a- 106-c
=> cba – abc = 99c – 99a
=> cba – abc = 99(c – a)
Hence proved.

Question 9.
If a = c, show that cba – abc = 0.
Solution:
Given : a = c
To show : cba – abc = 0
Proof:
cba = 100c + 106 + a …………(i)
(By using property 3)
abc = 100a + 106 + c …………(ii)
(By using property 3)
Since, a = c,
Substitute the value of a = c in equation (i) and (ii), we get
cba = 100c + 10b + c ……….(iii)
abc = 100c + 10b + c …………(iv)
Subtracting (iv) from (iii), we get
cba – abc – 100c + 106 + c – 100c – 106 – c
=> cba – abc = 0
=> cba = abc
Hence proved.

Question 10.
Show that 954 – 459 is exactly divisible by 99.
Solution:
To show : 954 – 459 is exactly divisible by 3 99, where a = 9, b = 5, c = 4
abc = 100a + 10b + c
=> 954 = 100 x 9 + 10 x 5 + 4
=> 954 = 900 + 50 + 4 ………(i)
and 459 = 100 x 4+ 10 x 5 + 9
=> 459 = 400 + 50 + 9 ……..(ii)
Subtracting (ii) from (i), we get
954 – 459 = 900 + 50 + 4 – 400 – 50 – 9
=> 954 – 459 = 500 – 5
=> 954 – 459 = 495
=> 954 – 459 = 99 x 5
Hence, 954 – 459 is exactly divisible by 99
Hence proved.

Playing with Number Exercise 5B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:
As we need A at unit place and 9 at ten’s place,
A = 6 as 6 x 6 = 36

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:
As we need B at unit place and A at ten’s place,
B = 0 as 5 x 0 = 0
Now we want to find A, 5 x A = A (at unit’s place)
A = 5, as 5 x 5 = 25
C = 2

Question 10.

Solution:

Question 11.

Solution:

Playing with Number Exercise 5C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find which of the following nutpbers are divisible by 2:
(i) 192
(ii) 1660
(iii) 1101
(iv) 2079
Solution:
A number having its unit digit 2,4,6,8 or 0 is divisible by 2,
So, Number 192, 1660 are divisible by 2.

Question 2.
Find which of the following numbers are divisible by 3:
(i) 261
(ii) 111
(iii) 6657
(iv) 2574
Solution:
A number is divisible by 3 if the sum of its digits is divisible by 3,
So, 261, 111 are divisible by 3.

Question 3.
Find which of the following numbers are divisible by 4:
(i) 360
(ii) 3180
(iii) 5348
(iv) 7756
Solution:
A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
So, Number 360, 5348, 7756 are divisible by 4.

Question 4.
Find which of the following numbers are divisible by 5 :
(i) 3250
(ii) 5557
(iii) 39255
(iv) 8204
Solution:
A number having its unit digit 5 or 0, is divisible by 5.
So, 3250, 39255 are all divisible by 5.

Question 5.
Find which of the following numbers are divisible by 10:
(i) 5100
(ii) 4612
(iii) 3400
(iv) 8399
Solution:
A number having its unit digit 0, is divisible by 10.
So, 5100, 3400 are all divisible by 10.

Question 6.
Which of the following numbers are divisible by 11 :
(i) 2563
(ii) 8307
(iii) 95635
Solution:
A number is divisible by 11 if the difference of the sum of digits at the odd places and sum of the digits at even places is zero or divisible by 11.
So, 2563 is divisible by 11.

Playing with Number Exercise 5D – Selina Concise Mathematics Class 8 ICSE Solutions

For what value of digit x, is :
Question 1.
1×5 divisible by 3 ?
Solution:
1×5 is divisible by 3
=> 1 + x + 5 is a multiple of 3
=> 6 + x = 0, 3, 6, 9,
=> x = -6, -3, 0, 3, 6, 9
Since, x is a digit
x = 0, 3, 6 or 9

Question 2.
31×5 divisible by 3 ?
Solution:
31×5 is divisible by 3
=> 3 + 1 + x + 5 is a multiple of 3
=> 9 + x = 0, 3, 6, 9,
=> x = -9, -6, -3, 0, 3, 6, 9,
Since, x is a digit
x = 0, 3, 6 or 9

Question 3.
28×6 a multiple of 3 ?
Solution:
28×6 is a multiple of 3
2 + 8+ x + 6 is a multiple of 3
=> 16 + x = 0, 3, 6, 9, 12, 15, 18
=> x = -18, -5, -2, 0, 2, 5, 8
Since, x is a digit = 2, 5, 8

Question 4.
24x divisible by 6 ?
Solution:
24x is divisible by 6
=> 2 + 4+ x is a multiple of 6
=> 6 + x = 0, 6, 12
=> x = -6, 0, 6
Since, x is a digit
x = 0, 6

Question 5.
3×26 a multiple of 6 ?
Solution:
3×26 is a multiple of 6
3 + x + 2 + 6 is a multiple of 3
=> 11 + x = 0, 3, 6, 9, 12, 15, 18,21,
=> x = -11, -8, -5, -2, 1, 4, 7, 10, ….
Since, x is a digit
x = 1, 4 or 7

Question 6.
42×8 divisible by 4 ?
Solution:
42×8 is divisible by 4
=> 4 + 2 + x + 8 is a multiple of 2
=> 14 + x = 0, 2, 4, 6, 8,
=> x = -8, -6, -4, -2, 2, 4, 6, 8,
Since, x is a digit 2, 4, 6, 8

Question 7.
9142x a multiple of 4 ?
Solution:
9142x is multiple of 4
=> 9 + 1 + 4 + 2 + x is a multiple of 4
=> 16 + x = 0, 4, 8, ………
x = -8, -4, 0, 4, 8
Since, x is a digit
4, 8

Question 8.
7×34 divisible by 9 ?
Solution:
7×34 is multiple of 9
=> 7 + x + 3+ 4 is a multiple of 9
=> 14 + x = 0, 9, 18, 27,
=> x = -1, 4, 13,
Since, x is a digit
x = 4

Question 9.
5×555 a multiple of 9 ?
Solution:
Sum of the digits of 5×555
=5 + x + 5 + 5 + 5 = 20 + x
It is multiple by 9
The sum should be divisible by 9
Value of x will be 7

Question 10.
3×2 divisible by 11 ?
Solution:
Sum of the digit in even place = x
and sum of the digits in odd place = 3 + 2 = 5
Difference of the sum of the digits in even places and in odd places = x – 5
3×2 is a multiple of 11
=> x – 5 = 0, 11, 22,
=> x = 5, 16, 27,
Since, x is a digit x = 5

Question 11.
5×2 a multiple of 11 ?
Solution:
Sum of a digit in even place = x
and sum of the digits in odd place = 5 + 2 = 7
Difference of the sum of the digits in even places and in odd places = x – 7
5×2 is a multiple of 11
=> x – 7 = 0, 11, 22,
=> x = 7, 18, 29,
Since, x is a digit
x = 7

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage

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Percent and Percentage Exercise 7A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate :
(i) 55% of 160 + 24% of 50 – 36% of 150
(ii) 9.3% of 500 – 4.8% of 250 – 2.5% of 240
Solution:

Question 2.
(i) A number is increased from 125 to 150 ; find the percentage increase.
(ii) A number is decreased from 125 to 100 ; find the percentage decrease.
Solution:

Question 3.
Find :
(i) 45 is what percent of 54 ?
(ii) 2.7 is what percent of 18 ?
Solution:

Question 4.
(i) 252 is 35% of a certain number, find the number.
(ii) If 14% of a number is 315 ; find the number.
Solution:

Question 5.
Find the percentage change, when a number is changed from :
(i) 80 to 100
(ii) 100 to 80
(iii) 6.25 to 7.50
Solution:

Question 6.
An auctioneer charges 8% for selling a house. If a house is sold for Rs.2, 30, 500; find the charges of the auctioneer.
Solution:

Question 7.
Out of 800 oranges, 50 are rotten. Find the percentage of good oranges.
Solution:
Total number of oranges = 800

Question 8.
A cistern contains 5 thousand litres of water. If 6% water is leaked. Find how many litres of water are left in the cistern.
Solution:

Question 9.
A man spends 87% of his salary. If he saves Rs.325 ; find his salary.
Solution:

Question 10.
(i) A number 3.625 is wrongly read as 3.265; find the percentage error.
(ii) A number 5.78 x 10³ is wrongly written as 5.87 x 10³; find the percentage error
Solution:

Question 11.
In an election between two candidates, one candidate secured 58% of the votes polled and won the election by 18, 336 votes. Find the total number of votes polled and the votes secured by each candidate.
Solution:

Question 12.
In an election between two candidates, one candidate secured 47% of votes polled and lost the election by 12, 366 votes. Find the total votes polled and die votes secured by the winning candidate.
Solution:

Question 13.
The cost of a scooter depreciates every year by 15% of its value at the beginning of the year. If the present cost of the scooter is
₹ 8,000; find its cost:
(i) after one year
(ii) after 2 years
Solution:

Question 14.
In an examination, the pass mark is 40%. If a candidate gets 65 marks and fails by 3 marks ; find the maximum marks.
Solution:

Question 15.
In an examination, a candidate secured 125 marks and failed by 15 marks. If the pass percentage was 35% ; find the maximum marks.
Solution:

Question 16.
In an objective type paper of 150 questions; John got 80% correct answers and Mohan got 64% correct answers.
(i) How many correct answers did each get?
(ii) What percent is Mohan’s correct answers to John’s correct answers ?
Solution:

Question 17.
The number 8,000 is first increased by 20% and then decreased by 20%. Find the resulting number.
Solution:

Question 18.
The number 12,000 is first decreased by 25% and then increased by 25%. Find the resulting number.
Solution:

Question 19.
The cost of an article is first increased by 20% and then decreased by 30%, find the percentage change in the cost of the article.
Solution:

Question 20.
The cost of an article is first decreased by 25% and then further decreased by 40%. Find the percentage change in the cost of the article.
Solution:

Percent and Percentage Exercise 7B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A man bought a certain number of oranges ; out of which 13 percent were found rotten. He gave 75% of the remaining in charity and still has 522 oranges left. Find how many had he bought?
Solution:

Question 2.
5% pupil in a town died due to some diseases and 3% of the remaining left the town. If 2, 76, 450 pupil are still in the town; find the original number of pupil in the town.
Solution:

Question 3.
In a combined test in English and Physics ; 36% candidates failed in English ; 28% failed in Physics and 12% in both ; find:
(i) the percentage of passed candidates
(ii) the total number of candidates appeared, if 208 candidates have failed.
Solution:

Question 4.
In a combined test in Maths and Chemistry; 84% candidates passsed in Maths; 76% in Chemistry and 8% failed in both. Find :
(i) the percentage of failed candidates ;
(ii) if 340 candidates passed in the test ; then how many appeared ?
Solution:

Question 5.
A’s income is 25% more than B’s. Find, B’s income is how much percent less than A’s.
Solution:

Question 6.
Mona is 20% younger than Neetu. How much percent is Neetu older than Mona ?
Solution:

Question 7.
If the price of sugar is increased by 25% today; by what percent should it be decreased tomorrow to bring the price back to the original ?
Solution:

Question 8.
A number increased by 15% becomes 391. Find the number.
Solution:

Question 9.
A number decreased by 23 % becomes 539. Find the number.
Solution:

Question 10.
Two numbers are respectively 20 percent and 50 percent more than a third number. What percent is the second of the first ?
Solution:

Question 11.
Two numbers are respectively 20 percent and 50 percent of a third number. What percent is the second of the first ?
Solution:

Question 12.
Two numbers are respectively 30 percent and 40 percent less than a third number. What percent is the second of the first ?
Solution:

Percent and Percentage Exercise 7C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A bag contains 8 red balls, 11 blue balls and 6 green balls. Find the percentage of blue balls in the bag.
Solution:

Question 2.
Mohan gets Rs. 1, 350 from Geeta and Rs. 650 from Rohit. Out of the total money that Mohan gets from Geeta and Rohit. what percent does he get from Rohit ?
Solution:

Question 3.
The monthly income of a man is Rs. 16, 000. 15 percent of it is paid as income-tax and 75% of the remainder is spent on rent, food, clothing, etc. How much money is still left with the man?
Solution:

Question 4.
A number is first increased by 20% and the resulting number is then decreased by 10%. Find the overall change in the number as percent.
Solution:

Question 5.
A number is increased by 10% and the resulting number is again increased by 20%. What is the overall percentage increase in the number ?
Solution:

Question 6.
During 2003, the production of a factory decreased by 25%. But, during 2004, it (production) increased by 40% of what it was at the beginning of2004. Calculate the resulting change (increase or decrease) in production during these two years.
Solution:

Question 7.
Last year, oranges were available at Rs. 24 per dozen ; but this year, they are available at Rs. 50 per score. Find the percentage change in the price of oranges.
Solution:

Question 8.
In an examination, Kavita scored 120 out of 150 in Maths, 136 out of 200 in English and 108 out of 150 in Science. Find her percentage score in each subject and also on the whole (aggregate).
Solution:

Question 9.
A is 25% older than B. By what percent is B younger than A ?
Solution:

Question 10.
(i) Increase 180 by 25%.
(ii) Decrease 140 by 18%.
Solution:

Question 11.
In an election, three candidates contested and secured 29200, 58800 and 72000 votes. Find the percentage of votes scored by winning candidate.
Solution:

Question 12.
(i) A number when increased by 23% becomes 861 ; find the number.
(ii) A number when decreased by 16% becomes 798 ; find the number.
Solution:

Question 13.
The price of sugar is increased by 20%. By what percent must the consumption of sugar be decreased so that the expenditure on sugar may remain the same ?
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers

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Rational Numbers Exercise 1A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Add, each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:

Solution:

Question 2.
Evaluate:

Solution:

Question 3.
Evaluate:

Solution:

Question 4.
For each pair of rational numbers, verify commutative property of addition of rational numbers:

Solution:

This verifies the commutative property for the addition of rational numbers.


This verifies the commutative property for the addition of rational numbers.


This verifies the commutative property for the addition of rational numbers.

This verifies the commutative property for the addition of rational numbers.

This verifies the commutative property for the addition of rational numbers.


This verifies the commutative property for the addition of rational numbers.

Question 5.
For each set of rational numbers, given below, verify the associative property of addition of rational numbers:

Solution:


This verifies associative property of the addition of rational numbers.


This verifies associative property of the addition of rational numbers.



This verifies associative property of the addition of rational numbers.

Question 6.
Write the additive inverse (negative) of:

Solution:

Question 7.
Fill in the blanks:

Solution:

Question 8.
State, true or false:

Solution:
(i) False
(ii) False
(iii) True
(iv) True
(v) False
(vi) False

Rational Numbers Exercise 1B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:

Solution:

Question 2.
Subtract:

Solution:

Question 3.
The sum of two rational numbers is \(\frac { 9 }{ 20 }\). If one of them is \(\frac { 2 }{ 5 }\), find the other.
Solution:
The sum of two rational numbers = \(\frac { 9 }{ 20 }\)
And, one of the numbers = \(\frac { 2 }{ 5 }\)
The other rational number

Question 4.
The sum of the two rational numbers is \(\frac { -2 }{ 3 }\). If one of them is \(\frac { -8 }{ 5 }\), find the other.
Solution:

Question 5.
The sum of the two rational numbers is -6. If one of them is \(\frac { -8 }{ 5 }\), find the other.
Solution:

Question 6.
Which rational number should be added to \(\frac { -7 }{ 8 }\) to get \(\frac { 5 }{ 9 }\) ?
Solution:

Question 7.
Which rational number should be added to \(\frac { -5 }{ 9 }\) to get \(\frac { -2 }{ 3 }\) ?
Solution:

Question 8.
Which rational number should be subtracted from \(\frac { -5 }{ 6 }\) to get \(\frac { 4 }{ 9 }\) ?
Solution:

Question 9.
(i) What should be subtracted from -2 to get \(\frac { 3 }{ 8 }\)
(ii) What should be added to -2 to get \(\frac { 3 }{ 8 }\)
Solution:

Question 10.
Evaluate:


Solution:

Rational Numbers Exercise 1C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:

Solution:

Question 2.
Multiply:

Solution:

Question 3.
Evaluate:

Solution:

Question 4.
Multiply each rational number, given below, by one (1):

Solution:

Question 5.
For each pair of rational numbers, given below, verify that the multiplication is commutative:

Solution:

Question 6.
Write the reciprocal (multiplicative inverse) of each rational number, given below :

Solution:

Question 7.
Find the reciprocal (multiplicative inverse) of:


Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
Name the multiplication property of rational numbers shown below :

Solution:
(i) Commutativity property.
(ii) Associativity property.
(iii) Distributivity property.
(iv) Existence of inverse.
(v) Existence of identity.
(vi) Existence of inverse.

Question 11.
Fill in the blanks:
(i) The product of two positive rational numbers is always ……………
(ii) The product of two negative rational numbers is always ……………
(iii) If two rational numbers have opposite signs then their product is always …………..
(iv) The reciprocal of a positive rational number is ………. and the reciprocal of a negative raitonal number is ……………
(v) Rational number 0 has ………….. reciprocal.
(vi) The product of a rational number and its reciprocal is ………..
(vii) The numbers ……….. and ……….. are their own reciprocals.
(viii) If m is reciprocal of n, then the reciprocal of n is ………….
Solution:
(i) The product of two positive rational numbers is always positive.
(ii) The product of two negative rational numbers is always positive.
(iii) If two rational numbers have opposite signs then their product is always negative.
(iv) The reciprocal of a positive rational number is positive and the reciprocal of a negative raitonal number is negative.
(v) Rational number 0 has no reciprocal.
(vi) The product of a rational number and its reciprocal is 1.
(vii) The numbers 1 and -1 are their own reciprocals.
(viii)If m is reciprocal of n, then the reciprocal of n is m.

Rational Numbers Exercise 1D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:


Solution:

Question 2.
Divide:

Solution:

Question 3.
The product of two rational numbers is -2. If one of them is \(\frac { 4 }{ 7 }\), find the other.
Solution:

Question 4.
The product of two numbers is \(\frac { -4 }{ 9 }\). If one of them is \(\frac { -2 }{ 27 }\), find the other.
Solution:

Question 5.
m and n are two rational numbers such that

Solution:

Question 6.
By what number must \(\frac { -3 }{ 4 }\) be multiplied so that the product is \(\frac { -9 }{ 16 }\) ?
Solution:

Question 7.
By what number should \(\frac { -8 }{ 13 }\) be multiplied to get 16?
Solution:

Question 8.
If 3\(\frac { 1 }{ 2 }\) litres of milk costs ₹49, find the cost of one litre of milk?
Solution:

Question 9.
Cost of 3\(\frac { 2 }{ 5 }\) metre of cloth is ₹88\(\frac { 1 }{ 2 }\). What is the cost of 1 metre of cloth?
Solution:

Question 10.
Divide the sum of \(\frac { 3 }{ 7 }\) and \(\frac { -5 }{ 14 }\) by \(\frac { -1 }{ 2 }\).
Solution:

Question 11.

Solution:

Question 12.
The product of two rational numbers is -5. If one of these numbers is \(\frac { -7 }{ 15 }\), find the other.
Solution:

Question 13.

Solution:

Rational Numbers Exercise 1E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.

Solution:

Question 2.

Solution:

Question 3.
Insert one rational number between (0 7 and 8 (ii) 3.5 and 5
(i) 2 and 3.2
(ii) 3.5 and 5
(iii) 2 and 3.2
(iv) 4.2 and 3.6
(v) \(\frac { 1 }{ 2 }\) and 2
Solution:

Question 4.
Insert two rational numbers between
(i) 6 and 7
(ii) 4.8 and 6
(iii) 2.7 and 6.3
Solution:

Question 5.
Insert three rational numbers between
(i) 3 and 4
(ii) 10 and 12
Solution:

Question 6.
Insert five rational numbers between \(\frac { 3 }{ 5 }\) and \(\frac { 2 }{ 5 }\)
Solution:
LCM of denominators 5 and 3 is 15

Question 7.
Insert six rational numbers between \(\frac { 5 }{ 6 }\) and \(\frac { 8 }{ 9 }\)
Solution:

Question 8.
Insert seven rational numbers between 2 and 3.
Solution:

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

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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:

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:

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:

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:

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:

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:

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.

Answer:

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:

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

Answer:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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: