Class 9

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 9th Standard Biology Notes Textbook Solutions Pdf Download English Medium and Malayalam Medium of 9th Class Biology Questions and Answers Pdf are part of Kerala Syllabus 9th Standard Textbooks Solutions. Here we have given SCERT Class 9 Biology Solutions of 9th Std Biology Textbook Answers Solutions Part 1 and Part 2.

Kerala SCERT Class 9 Biology Solutions

Kerala Syllabus 9th Standard Biology Notes Textbook Solutions Pdf English Medium

9th Std Biology Textbook Answers Part 1

• Class 9 Biology Chapter 1 Kerala Syllabus – To Life Processes
• Class 9 Biology Chapter 2 Kerala Syllabus – Digestion and Transport of Nutrients
• Class 9 Biology Chapter 3 Kerala Syllabus – Respiration and Excretion

9th Class Biology Textbook Answers Part 2

• Class 9 Biology Chapter 4 Kerala Syllabus – Behind Movements
• Class 9 Biology Chapter 5 Kerala Syllabus – Reproductive Health
• Class 9 Biology Chapter 6 Kerala Syllabus – Classification

9th Class Biology Questions and Answers Pdf English Medium

SCERT Class 9 Biology Textbook Solutions (Extra Questions)

1. To Life Processes Class 9 Questions and Answers
2. Digestion and Transport of Nutrients Class 9 Questions and Answers
3. Respiration and Excretion Class 9 Questions and Answers
4. Behind Movements Class 9 Questions and Answers
5. Reproductive Health Class 9 Questions and Answers
6. Classification Class 9 Questions and Answers

Kerala Syllabus 9th Standard Biology Notes Textbook Solutions Pdf Malayalam Medium

9th Class Biology Notes Pdf Malayalam Medium Part 1

• Class 9 Biology Chapter 1 Malayalam Medium – ജീവൽപ്രക്രിയകളിലേക്ക്
• Class 9 Biology Chapter 2 Malayalam Medium – ദഹനവും പോഷകസംവഹനവും
• Class 9 Biology Chapter 3 Malayalam Medium – ശ്വസനവും വിസർജനവും

SCERT Class 9 Biology Solutions Malayalam Medium Part 2

• Class 9 Biology Chapter 4 Malayalam Medium – ചലനത്തിനുപിന്നിൽ
• Class 9 Biology Chapter 5 Malayalam Medium – പ്രത്യുൽപാദന ആരോഗ്യം
• Class 9 Biology Chapter 6 Malayalam Medium – വർഗീകരണം

9th Class Biology Questions and Answers Pdf Malayalam Medium

9th Standard Biology Notes Malayalam Medium (Extra Questions)

1. ജീവൽപ്രക്രിയകളിലേക്ക് Class 9 Questions and Answers
2. ദഹനവും പോഷകസംവഹനവും Class 9 Questions and Answers
3. ശ്വസനവും വിസർജനവും Class 9 Questions and Answers
4. ചലനത്തിനുപിന്നിൽ Class 9 Questions and Answers
5. പ്രത്യുൽപാദന ആരോഗ്യം Class 9 Questions and Answers
6. വർഗീകരണം Class 9 Questions and Answers

We hope the given Class 9 Biology Notes Kerala Syllabus English Medium and Malayalam Medium of SCERT Class 9 Biology Textbook Solutions will help you. If you have any queries regarding Biology Class 9 Notes Kerala Syllabus Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Students rely on Kerala Syllabus 9th Standard Chemistry Textbook Solutions Chapter 3 Chemical Bonding Notes Questions and Answers English Medium to help self-study at home.

Kerala SCERT Class 9 Chemistry Chapter 3 Solutions Chemical Bonding

Kerala Syllabus Std 9 Chemistry Chapter 3 Chemical Bonding Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 3 Let Us Assess Answers Chemical Bonding

Question 1.
Draw the electron dot diagram of hydrogen (H), helium (He), lithium (Li), beryllium (Be) and fluorine (F).
Answer:

Question 2.
Illustrate the formation of the chemical bond in chlorine (Cl₂) using an electron dot diagram as illustrated in the fluorine (F₂) molecule.
Answer:

Question 3.
Represent the covalent bond in chlorine molecules using symbols.
Answer:
Cl – Cl

Question 4.
Represent the formation of ionic bonds in the following ionic compounds using the electron dot diagram and orbit model.
a) Sodium fluoride (NaF)
b) Sodium oxide (Na₂O)
c) Magnesium fluoride (MgF₂)
d) Calcium oxide (CaO)
Answer:

Question 5.
Assume that calcium (Ca) and fluorine (F) combine.
a) Complete the following table accordingly.

Element	Atomic number	Electronic configuration	Number of electrons received or donated
Ca	20	……………………………	………………………………….
F	9	……………………………..	……………………………..

b) Write the chemical formula of calcium fluoride.
c) Similarly, write the chemical formula of magnesium chloride and aluminium chloride.
Answer:

Element	Atomic number	Electronic configuration	Number of electrons received or donated
Ca	20	2, 8, 8, 2	2
F	9	2,7	1

b) CaF₂

c)

Element	Atomic number	Electronic configuration	Number of electrons received or donated
Mg	12	2, 8, 2	2
Al	13	2, 8, 3	3
Cl	17	2, 8, 7	1

Magnesium chloride – MgCl₂
Aluminium chloride – AlCl₃

Question 6.
Some cations and anions are given in the table. Fill in the blanks.

Answer:

Question 7.
Complete the following chemical equations and answer the questions given below.
(Hint: Atomic number Mg – 12, Cl – 17)
Mg → Mg²⁺ + ________
Cl + 1 e^– → ________
________ + ________ → MgCl²
(a) Identify the cation and anion in these compounds.
(b) What is the nature of the chemical bond in MgCh?
Answer:
Mg → Mg²⁺ + 2e^–
Cl + l e^– → Cl^–
Mg²⁺ + 2Cl^– → MgCl₂

(a) Cation – Mg²⁺
Anion – Cl^–
(b) Ionic bonding

Question 8.
Complete the following table. (Hint: Atomic number F – 9, H – 1, O – 8, N – 7)

Answer:

Question 9.
Complete the following table. (Symbols are not real)

Element	Atomic number	Electron configuration
P	12	………………..
Q	………………..	2, 7
R	10	………………..
S	17	………………..

a) Which among these is the most stable element?
b) Which element donates electrons during chemical reactions?
c) Write the chemical formula of the compound formed when the elements P and S combine.
Answer:

Element	Atomic number	Electron configuration
P	12	2, 8, 2
Q	9	2, 7
R	10	2, 8
S	17	2, 8, 7

a) R is the most stable element.
b) P is the element that donates electrons.

Formula – PS₂

Question 10.
Atom models of two elements are represented below.

a) Draw the electron dot diagram of the formation of sodium fluoride.
b) What is the nature of the chemical bond in sodium fluoride?
c) Write any two characteristics of compounds having this type of bond.
Answer:
a)
b) Ionic bond
c)

• Dissolves in polar solvents like water.
• Exhibits high melting and boiling points.
• Conducts electricity in a molten state or solutions.

Question 11.
The electron configuration of the elements P, Q, and R are given below. (Symbols are not real)
P – 2, 8, 6
Q – 2, 8, 1
R – 2, 8, 8
a) Which is the most stable element among these? What is the reason?
b) What is the atomic number of Q?
c) Draw the atom model of Q.
d) What are the valencies of the elements P and Q?
e) Write the chemical formula of the compound formed when P and Q combine.
Answer:
a) R is the most stable element among these. Because it has a stable octet electron configuration in
the outermost shell.
b) Atomic number of Q = 11
c)
d) Valency of P = 2
Valency of Q = 2

e)
Formula – Q₂P

Question 12.
A, B, C and D are four elements (Symbols are not real). Information about them is given in the following table.

Element	Atomic number	Electronegativity
A	6	2.55
B	8	3.44
C	12	1.31
D	17	3.16

Based on these, find the type of bond in the compounds formed by the combination of the
following pairs of elements.
1. C, B
2. C, D
3. A, B
Answer:
1. C, B
Difference in electronegativity = 3.44 – 1.31 = 2 : 13
Greater than 1.7, so it is ionic bonding.

2. C, D
The difference in electronegativity = 3.16 – 1.31 = 1.85
Greater than 1.7, so it is ionic bonding.

3. A, B
Difference in electronegativity = 3.44 – 2.55 = 0.89
Less than 1.7, so it is covalent bonding.

Extended Activities

Question 1.
Magnesium nitride is obtained when nitrogen is passed over heated magnesium. Write the chemical equation of this reaction. Find out whether the formed compound is ionic or covalent using the electronegativity scale given in this unit. (Hint – Valency: Nitrogen – 3, Magnesium – 2)
Answer:
The chemical equation for the reaction is 3Mg + N₂ → Mg₃N₂
The difference in electronegativity = 3.04 – 1.31 = 1.73
Greater than 1.7, so it is an ionic compound.

Question 2.
Draw the electron dot diagram of the chemical bonds in ethane (C₂H₆), ethene (C₂H₄) and ethyne (C₂H₂). Find out whether these compounds are ionic or covalent. Calculate the total number of bonds in each compound.
Answer:
a) Ethane (C₂H₆)

Total number of bonds = 7

b) Ethane (C₂H₄)

Total number of bonds = 5

c) Ethyne (C₂H₂)

Total number of bonds = 3

Question 3.
Conduct the experiment arranging the apparatus as shown in the figure.

Record your observations and identify what types of compounds are sodium chloride and glucose.
Answer:
The galvanometer in the first experiment shows deflection, indicating that electricity passes through a common salt solution. This observation also notes that gases evolve from metal rods. Thus, common salt is confirmed as an ionic compound, as ionic compounds conduct electricity in their molten state or solutions.

No deflection is observed in the second experiment, signifying that electricity does not pass through the glucose solution. Therefore, glucose is identified as a covalent compound. Typically, covalent compounds do not conduct electricity.

Question 4.
Draw the chemical bonds in different compounds and exhibit them on the bulletin board.
Answer:

Cl – Cl


a)
b) Ionic bond
c)

• Dissolves in polar solvents like water.
• Exhibits high melting and boiling points.
• Conducts electricity in a molten state or solutions.

Chemical Bonding Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Some substances are given below. Differentiate them into elements and compounds and list them.
Potassium, oxygen, water, common salt, nitrogen, helium, hydrogen, and sugar.

Answer:

You know that there are two atoms in one molecule of hydrogen. If so, how many atoms are there in each substance given.

Question 2.
How many atoms are there in each substance given below?

Answer:

Molecule	Number of atoms
Oxygen (O2)	2
Water (H2O)	3
Nitrogen (N2)	2
Helium (He)	1
Methane (CH4)	5
Sugar (C12H22O11)	45

Some molecules have more than one atom.

Question 3.
Why do atoms in a molecule stay together?
Answer:
The atoms of a molecule stay together because of chemical bonds.

Question 4.
Why do atoms combine to form molecules?
Answer:
An atom combines in order to attain stability. Also, different compounds are formed by the combination of different atoms.

Question 5.
How do atoms combine?
Answer:
Atoms can combine either by sharing electrons or by completely transferring the electrons.

Question 6.
Do all atoms combine in the same way?
Answer:
No, not all atoms combine in the same way.

Question 7.
Do all atoms combine with other atoms?
Answer:
Not all atoms combine with other atoms.

Question 8.
How many atoms are there in a molecule of noble gases?
Answer:
One atom only.
Generally, noble gases do not combine with other atoms due to their stability. They are able to exist independently. Examples of noble gases, along with their atomic number and electronic configuration, are given in the table below.

Element (Symbol)	Atomic number	Electronic Configuration
Helium (He)	2	2
Neon (Ne)	10	2, 8
Argon (Ar)	18	2, 8, 8
Krypton (Kr)	36	2, 8, 18, 8
Xenon (Xe)	54	2, 8, 18, 18, 8
Radon (Rn)	86	2, 8, 18, 32, 18, 8

Question 9.
How many electrons are there in the outermost shell of noble gases except helium?
Answer:
Eight electrons.
The atomic number of helium is 2; hence, it contains a maximum of two electrons in its outermost shell. All the other noble gases have a total of eight electrons in their outermost shell.

The arrangement of eight electrons in the outermost shell is called octet configuration.

Atoms having octet configuration are more stable. Such atoms are generally reluctant to take part in chemical reactions. So noble gases are also called inert gases. In the case of helium, the configuration is called duplet configuration, which is stable like that of the other noble gases.

Question 10.
Look at the electron configuration of magnesium and oxygen given in the table.

Element	Atomic number	Electronic configuration
Magnesium	12	2, 8, 2
Oxygen	8	2, 6

(i) Are these atoms stable?
(ii) How can they attain stability?
(iii) What is the name of the compound formed when these atoms combine?
Answer:
(i) No
(ii) Stability can be attained by gaining octet electron configuration in the outermost shell through chemical bonding. (In here, magnesium must lose two electrons, and oxygen must gain two- electrons).
(iii) Magnesium oxide (MgO).
The force that binds together the component particles in a compound is called chemical bonding,

Question 11.
The chemical name of table salt is sodium chloride. Answer the following questions.
(i) What are the constituent elements of sodium chloride?
(ii) Write the electron configuration of the sodium atom (atomic number – 11).
(iii) How many electrons are there in the outermost shell of a sodium atom?
(iv) How does the sodium atom attain octet electron configuration?
Answer:
(i) Sodium and Chlorine
(ii) ₁₁Na – 2, 8, 1
(iii) One atom
(iv) By losing one electron in the outermost shell.
In the case of a sodium atom, it loses one electron and changes into a sodium ion.
Na → Na⁺ + le^–

The removal of the outermost electron from the sodium atom can only be achieved by overcoming the force of attraction exerted by the nucleus.

This energy required, or in other terms, the amount of energy required to remove the most loosely bound electron from the outermost shell of an isolated gaseous atom of an element is called its ionisation energy or ionisation enthalpy.

The amount of energy required to remove the most loosely bound electron from the outermost shell of an isolated gaseous atom of an element is called its ionisation energy.

Question 12.
Write the electron configuration of a chlorine atom (atomic number 17).
Answer:
₁₇Cl – 2, 8, 7

Question 13.
How many electrons are needed for the chlorine atom to attain octet electron configuration?
Answer:
One.
I Non-metals like chlorine accept an electron to become a chloride ion.

Energy is released when atoms become negative ions by accepting electrons. This energy released when an electron is added to a neutral gaseous atom to form a negative ion is called electron gain enthalpy.

Illustrate the electron exchange in the formation of sodium chloride through shell electron configuration.

Question 14.
Represent the electron dot diagram of a chlorine atom.
Answer:


The electron dot diagram of the formation of sodium chloride can be represented as

Question 15.
Observe the electron dot diagram and shell electron configuration diagram of sodium chloride formation and complete the table.

Answer:

The equations of the electron transfer during the formation of sodium-chloride are:
Na → Na⁺ + le^–
Cl + le- → Cl^–
In, the reaction between sodium and chlorine to form sodium chloride, the sodium atom loses one electron and gets converted into sodium ion (Na⁺). The chlorine atom accepts an electron to form a chloride ion (Cl^–).

The positive ions formed by losing electrons during chemical reactions are called cations, and the negative ions formed by accepting electrons are called anions.

The particles Na⁺ and Cl^–, which possess opposite charges, show mutual attraction and are bound together by an extremely strong electrostatic force of attraction, resulting in the formation of NaCl.

The electrostatic force of attraction that holds together the oppositely charged ions in an ionic compound is called an ionic bond. An ionic bond is also known as an electrovalent bond. An ionic bond is always formed between a metal and a non-

Question 16.
What is the compound formed when magnesium burns in the air?
Answer:
Magnesium oxide (MgO)
The equation for the chemical reaction between magnesium and oxygen can be represented as
2Mg + O₂ → 2MgO

Question 17.
The electron dot diagram of magnesium oxide formation is given. Analyse the diagram and complete the table.

Answer:

Question 18.
Which are the ions present in magnesium oxide?
Answer:
Magnesium ion (Mg²⁺) and oxygen ion (O^2-).

Question 19.
How many electrons are transferred from magnesium to oxygen during the formation of magnesium oxide?
Answer:
Two electrons.
By the transfer of two electrons from magnesium to oxygen, an ionic bond is formed between them. The
compounds that are formed by ionic bonding are known as ionic compounds or electrovalent compounds.
The distribution of electrons of fluorine is given.

Question 20.
How many electrons are there in the outermost shell of fluorine?

Answer:
Seven electrons.

Question 21.
How many more electrons are required for one fluorine atom to attain octet configuration? 63caj
Answer:
One electron.

Question 22.
Is it possible to transfer electrons from one fluorine atom to another? If so, what, type of arrangement might have taken place, between the atoms in order to attain octet configuration?
Answer:
No, it is not possible to move electrons from one fluorine atom to another. In order to attain stability, atoms will share electrons.
In the case of some molecules like fluorine, the octet configuration is attained by the sharing of electrons.

Question 23.
Analyse the electron dot diagram of fluorine molecule formation through chemical bonding and answer the following questions.

(i) How many electrons are donated by each fluorine atom for sharing?
(ii) How many pairs of electrons are shared in the chemical bonding of fluorine molecule?
Answer:
(i) One each.
(ii) One pair.
The chemical bond formed as a result of the sharing of electrons between the combining atoms is called a covalent bond. The covalent bond formed by the sharing of one pair of electrons is a single bond.
A single bond is represented by a small line (-) between the symbols of the combining elements in molecules. The single bond in fluorine molecule can be represented using symbols such as F – F.

Question 24.
Oxygen is a diatomic molecule. Now answer the following questions.
(i) What is the atomic number of oxygen?
(ii) Write the electronic configuration of oxygen.
(iii) How many more electrons are required for one oxygen atom to attain the octet configuration?
Answer:
(i) Atomic number of oxygen is eight.
(ii) ₈O – 2, 6.
(iii) 2 more electrons.

Question 25.
The illustration of a chemical bond in an oxygen molecule is given below

How many pairs of electrons are shared in the oxygen molecule?
Answer:
Two pairs
The covalent bond formed by the sharing of two electron pairs or four electrons is called a double covalent bond.
The double bond in oxygen molecules can be represented by using symbols such as 0 = 0.

Question 26.
The illustration of a chemical bond in a nitrogen molecule is given below

How many pairs of electrons are shared here to complete the octet configuration?
Answer:
3 pairs
The covalent bond formed by the sharing of three electron pairs or six electrons is called a triple covalent bond.
The triple bond in nitrogen molecules can be represented by using symbols such as N ≡ N

Question 27.
Illustrate the chemical bond in a hydrogen molecule using an electron dot diagram.
Answer:

In this case, two hydrogen atoms share one pair of electrons, which forms a single bond. Stability is achieved by adopting the electronic configuration of helium, which is the nearest noble gas.

Formation of hydrogen chloride molecule:
The atomic number of hydrogen is 1.
Electronic configuration = 1.
The atomic number of chlorine is 17.
Electronic configuration = 2, 8, 7.
Chlorine needs one more electron to complete the octet, which the hydrogen atom will share, thereby forming the hydrogen chloride molecule. Hence, a single bond is present in hydrogen chloride.

Question 28.
Represent the covalent bond in hydrogen chloride using symbols.
Answer:
H – Cl

Question 29.
Depict the chemical bonding in hydrogen fluoride molecules.
Answer:

A water molecule consists of two hydrogen atoms and one oxygen atom. The chemical bonding present in water molecules can be represented as:

Question 30.
How many covalent bonds are formed here?
Answer:
2 covalent bonds.
Compounds formed by covalent bonding are called covalent compounds. When non-metals combine, usually covalent compounds are formed.

Question 31.
Is the shared pair of electrons in the HF molecule attracted equally by both atoms?
Answer:
No, the electrons are more attracted by fluorine.

Question 32.
Find out the electronegativity difference of the constituent elements and complete the table.

Compounds	Difference in electronegativity of constituent elements	Nature of the compound
Sodium chloride (NaCl)	3.16 – 0.93 = …………………	Ionic
Hydrogen Chloride (HCl)	3.16 – 2.20 = …………………	Covalent
Sodium Oxide (Na2O)	………………..	…………………
Calcium Chloride (CaCl2)	…………………	…………………
Methane (CH4)	…………………	…………………
Magnesium Fluoride (MgF2)	…………………	…………………

Answer:

Compounds	Difference in electronegativity of constituent elements	Nature of the compound
Sodium chloride (NaCl)	3.16 – 0.93 = 2.23	Ionic
Hydrogen Chloride (HCl)	3.16 – 2.20 = 0.96	Covalent
Sodium Oxide (Na2O)	3.44 – 0.93 = 2.51	Ionic
Calcium Chloride (CaCl2)	3.16 – 1.00 = 2.16	Ionic
Methane (CH4)	2.55 – 2.20 = 0.35	Covalent
Magnesium Fluoride (MgF2)	3.98 – 1.31 = 2.67	Ionic

Question 33.
Answer the questions given below
(i) What is the electronegativity value of hydrogen?
(ii) What is the electronegativity value of chlorine?
(iii) The nucleus of which of these atoms has a greater tendency to attract the shared pair of electrons involved in covalent bonding?
Answer:
(i) 2.20
(ii) 3.16
(iii) Chlorine will attract the shared pair of electrons.

Question 34.
Analyse the change in the electron arrangement in atoms during the formation of each compound and
Complete the table given below.

Answer:

Question 35.
Complete the following table regarding the combination of magnesium (Mg) and fluorine (F).

Element	Atomic number	Electron configuration	Number of electrons donated or accepted
Mg	12	–	–
F	9	–	–

Answer:

Element	Atomic number	Electron configuration	Number of electrons donated or accepted
Mg	12	2, 8, 2	2
F	9	2, 7	1

Question 36.
How many fluorine atoms are required to receive the electrons donated by magnesium?
Answer:
2 fluorine atoms.
During the formation of magnesium fluoride, one magnesium atom combines with two fluorine atoms. Hence, the chemical formula of magnesium fluoride will be MgF₂.

Question 37.
What are the constituent elements of aluminium oxide?
Answer:
Aluminium and Oxygen

Question 38.
What is the valency of aluminium? (Atomic number – 13)
Answer:
Electron configuration – 2, 8, 3
Therefore, the valency of aluminium is three.

Question 39.
What is the valency of oxygen? (Atomic number – 8)
Answer:
Electron configuration – 2, 6
Therefore, the valency of oxygen is two.

Question 40.
What are the constituent elements of carbon dioxide?
Answer:
Carbon and oxygen

Question 41.
Write the symbols of elements together, considering their electronegativity.
Answer:
C and O.

Question 42.
The valency of carbon is 4, and that of oxygen is 2. Interchange the valencies and write them as base indices.
Answer:

Chemical formula – C₂O₄
Divide the base indices by common factor, C_2/2O_4/2 = C₁O₂.
If the base index is 1, then there is no need to write. Therefore, the chemical formula of carbon dioxide is CO₂.

Question 43.
The constituent elements of some compounds and the valencies of their constituent elements are given in the following table. Find out the chemical formula.

Answer:

Question 44.
Which are the ions derived from hydrochloric acid? Why is it a monobasic acid?
Answer:
Hydrochloric acid contains H⁺ and Cl^–. The ionisation of one molecule of HC1 releases one H⁺ ion; hence, it is a monobasic acid.
In the case of sulphuric acid, two H⁺ and one SO₄^2- ions are released. Therefore, it is a dibasic acid. Hence, the chemical formula of sulphuric acid is H₂SO₄.

Question 45.
The basicity and the negative ions of certain acids are given in the table. Find their chemical formula and complete the table.

Answer:

Bases that are soluble in water are called alkalies. The number of OH” ions in an alkali will be equal to the number of positive ions.

Question 46.
Which is the positive ion present in sodium hydroxide?
Answer:
Sodium ion (Na⁺)

Question 47.
How many OH” ions, equal to the positive charge on sodium ion, will be present in sodium hydroxide?
Answer:
One

Question 48.
If so, what is the chemical formula of sodium hydroxide?
Answer:
NaOH

Question 49.
The positive ions of some bases are given in the table below. Find the chemical formulae and complete the table.

Answer:

Question 50.
Which is the positive ion in magnesium hydroxide, Mg(OH)₂?
Answer:
Magnesium ion (Mg²⁺).

Question 51.
Which is the negative ion in phosphoric acid, H₃PO₄?
Answer:
Phosphate ion (PO₄^3-).
Let us write the chemical formula of magnesium phosphate, which is formed from magnesium hydroxide and phosphoric acid.
Step 1 – Write the symbols of the ions Mg²⁺ + PO₄^3-
Step 2 – Interchange the number indicating charge and write as base indices. Mg₄(PO₄)₂
Therefore, the chemical formula of magnesium phosphate is Mg₃(PO₄)₂.

Question 52.
The reaction between sulphuric acid and calcium hydroxide forms salt calcium sulphate. To find the chemical formula of calcium sulphate, answer the following questions.
(i) Which is the positive ion in calcium hydroxide, Ca(OH)₂?
(ii) Which is the negative ion in sulphuric acid, H₂SO₄?
(iii) Write the symbol of the positive ion and then the symbol of the negative ion.
(iv) Write the number indicating the charge of each ion/radical as the base index after interchanging them.
Answer:
(i) Calcium ion, Ca²⁺
(ii) Sulphate ion, SO₄^2-
(iii) Ca²⁺ SO₄^2-
(iv) Ca₂(SO₄)₂
Simplify the base indices into simple whole number ratio
Ca_2/2(SO₄)_2/2 = CaSO₄
Therefore, the chemical formula of calcium sulphate is CaSO₄.

Question 53.
Certain positive ions and negative ions are given in the following’ table. Complete the table by writing the chemical formula and the name of the salt formed from these ions.

Answer:

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 15 Malayalam Medium സ്ഥിതിവിവരക്കണക്ക് can save valuable time.

Kerala SCERT Class 9 Maths Chapter 15 Solutions Malayalam Medium സ്ഥിതിവിവരക്കണക്ക്

Class 9 Maths Chapter 15 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 15 Malayalam Medium Textual Questions and Answers

Question 1.
ട്വന്റി-ട്വന്റി ക്രിക്കറ്റിൽ ഒരു ടീം ആദ്യത്തെ 5 ഓവറിൽ 51 റൺ നേടി.
i) അപ്പോഴത്തെ റൺ നിരക്ക് എത്രയാണ്?
Answer:

ii) ഇതേ റൺ നിരക്ക് തുടരുകയാണെങ്കിൽ 20 ഓവറിൽ എത്ര റൺ പ്രതീക്ഷിക്കാം?
Answer:
ഇതേ റൺ നിരക്ക് തുടർന്നാൽ 20 ഓവറിൽ എടുക്കാവുന്ന റൺസ് = 10.2 × 20 = 204

Question 2.
ക്ലാസിൽ ഒരു കണക്കു പരീക്ഷ നടത്തി, മാർക്കിന്റെ അടിസ്ഥാനത്തിൽ കുട്ടികളെ തരംതിരിച്ച പട്ടികയാണ് ചുവടെ കാണിച്ചിരിക്കുന്നത്:

i) കുട്ടികൾക്ക് കിട്ടിയ മാർക്കുകളുടെ മാധ്യം കണക്കാക്കുക.
ii) മാധ്യത്തെക്കാൾ കുറവ് മാർക്ക് കിട്ടിയവർ എത്ര പേരാണ്?
iii) മാധ്യത്തെക്കാൾ കൂടുതൽ മാർക്ക് കിട്ടിയവർ എത്ര പേരാണ് ?
Answer:

മാർക്ക്	കുട്ടികൾ	ആകെ മാർക്ക്
2	1	2 × 1 = 2
3	2	3 × 2 = 6
4	5	4 × 5 = 20
5	4	5 × 4 = 20
6	6	6 × 6 = 36
7	11	7 × 11 = 77
8 ‘	10	8 × 10 = 80
9	4	9 × 4 = 36
10	2	10 × 2 = 20
ആകെ	45	297

കുട്ടികൾക്ക് കിട്ടിയ മാർക്കുകളുടെ മാധ്യം = \(\frac{297}{45}\) = 6.6
ii) മാധ്യത്തെക്കാൾ കുറവ് മാർക്ക് കിട്ടിയവർ = 1 + 2 + 5 + 4 + 6 = 18
iii) മാധ്യത്തെക്കാൾ കൂടുതൽ മാർക്ക് കിട്ടിയവർ = 11 + 10 + 4 + 2 = 27

Question 3.
ഒരു കർഷകന് ഒരു മാസം കിട്ടിയ റബ്ബർഷീറ്റിന്റെ വിവരങ്ങൾ ചുവടെയുള്ള പട്ടികയിലുണ്ട്:

i) ഈ മാസത്തിൽ കിട്ടിയ റബ്ബർഷീറ്റിന്റെ മാധ്യ ഭാരം എത്രയാണ്?
Answer:

റബ്ബർ (കിഗ്രാം)	ദിവസങ്ങൾ	ആകെ റബ്ബർ (കിഗ്രാം)
09	3	9 × 3 = 27
10	4	10 × 4 = 40
11	3	11 × 3 = 33
12	3	12 × 3 = 36
13	5	13 × 5 = 65
14	6	14 × 6 = 84
16	6	16 × 6 = 96
ആകെ	30	381

ഈ മാസത്തിൽ കിട്ടിയ റബ്ബർഷീറ്റിന്റെ മാധ്യ ഭാരം = \(\frac{381}{30}\) = 12.7 കിഗ്രാം

ii) റബ്ബറിന്റെ വില കിലോഗ്രാമിന് 175 രൂപയാണ്. ഈ മാസത്തിൽ റബ്ബറിൽ നിന്നു കിട്ടിയ മാധ്യ വരുമാനം എത്ര രൂപയാണ്?
Answer:
റബ്ബറിന്റെ വില കിലോഗ്രാമിന് 175 രൂപ ആയതിനാൽ ഈ മാസത്തിൽ റബ്ബറിൽ നിന്നു കിട്ടിയ മാധ്യ വരുമാനം = 12.7 × 175
= 2222.5 രൂപ

Question 4.
ഒരു പ്രദേശത്തു പെയ്ത മഴയുടെ അളവനുസരിച്ച് ഒരു മാസത്തിലെ ദിവസങ്ങളെ തരം തിരിച്ച പട്ടികയാണിത്.

ആ മാസം അവിടെ ഒരു ദിവസം പെയ്ത മഴയുടെ മാധ്യ അളവെന്താണ്?
Answer:

മഴ (മിമീ)	ദിവസങ്ങൾ	ആകെ മഴ (മിമീ)
54	3	54 × 3 = 162
56	5	56 × 5 = 280
58	6	58 × 6 = 348
55	3	55 × 3 = 165
50	2	50 × 2 = 100
47	4	47 × 4 = 188
44	5	44 × 5 = 220
41	2	41 × 2 = 82
ആകെ	30	1545

ആ മാസം അവിടെ ഒരു ദിവസം പെയ്ത മഴയുടെ മാധ്യം

= \(\frac{1545}{30}\)
= 51.5 മിമീ

Question 5.
ഒരു ക്ലാസിലെ കുട്ടികളെ ഉയരത്തിന്റെ അടിസ്ഥാനത്തിൽ തരംതിരിച്ച പട്ടികയാണ് ചുവടെ കാണുന്നത്.

ഈ ക്ലാസിലെ കുട്ടികളുടെ മാധ്യഉയരം എത്രയാണ് ?
Answer:

ഉയരം (സെമീ)	കുട്ടികളുടെ എണ്ണം	വിഭാഗമാധ്യം	ആകെ ഉയരം
148- 152	8	150	1200
152- 156	10	154	1540
156 – 160	15	158	2370
160- 164	10	162	1620
164- 168	7	166	1162
ആകെ	50		7892

= \(\frac{7892}{50}\)
= 157.84 സെമീ

Question 6.
ഒരു സർവകലാശാലയിലെ അധ്യാപകരുടെ എണ്ണം പ്രായമനുസരിച്ച് തരംതിരിച്ചെഴുതിയ പട്ടികയാണ് ചുവടെയുള്ളത്.

അധ്യാപകരുടെ മാധ്യ പ്രായം കണക്കാക്കുക.
Answer:

പ്രായം	അധ്യാപകരുടെ എണ്ണം	വിഭാഗമാധ്യം	ആകെ പ്രായം
25-30	06	27.5	165
30-35	14	32.5	455
35-40	18	37.5	675
40-45	20	42.5	850
45-50	05	47.5	237.5
50-55	04	52.5	210
55-60	03	57.5	172.5
ആകെ	70		2765

= \(\frac{2765}{70}\)
= 39.5

Question 7.
ഒരു ക്ലാസിലെ കുട്ടികളെ ഭാരമനുസരിച്ചു തരംതിരിച്ച പട്ടികയാണിത്.

മാധ്യ ഭാരം കണക്കാക്കുക.
Answer:

ഭാരം (കിഗ്രാം)	കുട്ടികളുടെ എണ്ണം	വിഭാഗമാധ്യം	ആകെ ഭാരം
21-23	4	22	88
23-25	7	24	168
25-27	8	26	208
27-29	6	28	168
29-31	3	30	90
31-33	1	32	32
ആകെ	29		754

= \(\frac{754}{29}\)
= 26 കിഗ്രാം

Class 9 Maths Chapter 15 Malayalam Medium Intext Questions and Answers

Question 1.
ഏതെങ്കിലും കുറേ സംഖ്യകൾ എടുത്തു മാധ്യം കണക്കാക്കുക; ഓരോ സംഖ്യയും മാധ്യത്തേക്കാൾ എത്ര കൂടുതൽ, അല്ലെങ്കിൽ എത്ര കുറവ് എന്നു കണക്കാക്കി, കൂടുതലും കുറവും വെവ്വേറെ കൂട്ടി നോക്കുക. ഒരേ തുകയാണോ?
Answer:
ഇത് എന്തുകൊണ്ട് എന്നു വിശദീകരിക്കാമോ?
10, 15, 20, 25, 30 എന്നീ സംഖ്യകൾ പരിഗണിക്കുക.
മാധ്യം = \(\frac{10+15+20+25+30}{5}=\frac{100}{5}\) = 20
ഓരോ സംഖ്യയും മാധ്യത്തേക്കാൾ എത്ര കൂടുതൽ, അല്ലെങ്കിൽ എത്ര കുറവ് എന്ന് കണക്കാക്കാം.

സംഖ്യകൾ	മാധ്യത്തിൽ നിന്ന് എത്ര കൂടുതൽ	മാധ്യത്തിൽ നിന്ന് എത്ര കുറവ്
10		20 – 10 = 10
15		20 – 15 = 5
20
25	25 – 20 = 5
30	30 – 20 = 10

മാധ്യത്തിൽ നിന്ന് കൂടുതൽ വന്ന സംഖ്യകളുടെ തുക = 10 + 5 = 15
മാധ്യത്തിൽ നിന്ന് കുറവ് വന്ന സംഖ്യകളുടെ തുക = 10 + 5 = 15
ഇവിടെ, മാധ്യത്തിൽ നിന്ന് കൂടുതൽ വന്ന സംഖ്യകളുടെ തുകയും മാധ്യത്തിൽ നിന്ന് കുറവ് വന്ന സംഖ്യകളുടെ തുകയും തുല്യമാണ്. അതായത്, മാധ്യം അഥവാ ശരാശരി ഒരു സംഖ്യാസമൂഹത്തിന്റെ തുലനബിന്ദുവായാണ് പ്രവർത്തിക്കുന്നത്.

Statistics Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
ഒരു തൊഴിൽ ശാലയിലെ തൊഴിലാളികളുടെ ദിവസക്കൂലി പട്ടികപ്പെടുത്തിയിരിക്കുന്നു. ശരാശരി കൂലി എത്ര?

കൂലി	എണ്ണം
500	3
600	7
700	10
900	8
1000	2

Answer:

കൂലി	എണ്ണം	ആകെ കൂലി
500	3	1500
600	7	4200
700	10	7000
900	8	7200
1000	2	2000
ആകെ	30	21900

ശരാശരി കൂലി = \(\frac{21900}{30}\) = 730

Question 2.
ഒരു ക്ലാസിലെ കുട്ടികളെ പരീക്ഷയ്ക്ക് കിട്ടിയ മാർക്കിന്റെ അടിസ്ഥാനത്തിൽ തരംതിരിച്ച പട്ടികയാണിത്

i) ശരാശരി മാർക്ക് 6 ആണ്. എത്ര കുട്ടികൾക്കാണ് 8 മാർക്ക് കിട്ടിയത്?
ii) ക്ലാസിൽ അകെ എത്ര കുട്ടികളുണ്ട്?
Answer:

മാർക്ക്	കുട്ടികൾ	ആകെ മാർക്ക്
3	2	6
4	4	16
5	5	25
6	6	36
7	7	49
8	x	8x
9	2	18
10	1	10
ആകെ	27 + x	160 + 8x

i) ശരാശരി മാർക്ക്

6 = \(\frac{160+8 x}{27+x}\)
6(27 + x) = 160 + 8x
162 + 6x = 160 + 8x
2x = 2
x = 1
8 മാർക്ക് കിട്ടിയ കുട്ടികളുടെ എണ്ണം = 1
ii) ക്ലാസ്സിലെ ആകെ കുട്ടികളുടെ എണ്ണം = 27 + x = 27 + 1 = 28

Question 3.
ഒരു സ്ഥാപനത്തിലെ തൊഴിലാളികളുടെ ദിവസവേതനം പട്ടികയിൽ ചുവടെ കൊടുത്തി രിക്കുന്നു. മാധ്യവേതനം കണക്കാക്കുക.

ദിവസ വേതനം (രൂപ)	തൊഴിലാളികളുടെ എണ്ണം
15000 – 18000	1
18000 – 21000	3
21000 – 24000	5
24000 – 27000	4
27000 – 30000	1
30000 – 33000	1

Answer:

ദിവസ വേതനം (രൂപ)	എണ്ണം തൊഴിലാളികളുടെ	വിഭാഗമാധ്യം	ആകെ വേതനം
15000 – 18000	1	16500	16500
18000 – 21000	3′	19500	58500
21000 – 24000	5	22500	112500
24000 – 27000	4	25500	102000
27000 – 30000	1	28500	28500
30000 – 33000	1	31500	31500
ആകെ	15		349500

= \(\frac{3,49,500}{15}\)
= 23,300 രൂപ

Question 4.
ഒരു സ്ഥാപനത്തിലെ തൊഴിലാളികളുടെ ദിവസവേതനം പട്ടികയിൽ ചുവടെ കൊടുത്തിരി ക്കുന്നു. മാധ്യവേതനം കണക്കാക്കുക.

ദിവസ വേതനം (രൂപയിൽ)	തൊഴിലാളികളുടെ എണ്ണം
450 – 550	7
550 – 650	8
650 – 750	10
750 – 850	10
850 – 950	9
950 – 1050	6

Answer:

ദിവസ വേതനം (രൂപയിൽ)	തൊഴിലാളികളുടെ എണ്ണം	വിഭാഗമാധ്യം	ആകെ വേതനം
450 – 550	7	500	3500
550 – 650	8	600	4800
650 – 750	10	700	7000
750 – 850	10	800	8000
850 – 950	9	900	8100
950 – 1050	6	1000	6000
ആകെ	50		37,400

= \(\frac{37,400}{50}\)
= 748 രൂപ

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 15 Statistics Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 15 Solutions Statistics

Statistics Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 15 Statistics Solutions Questions and Answers

Class 9 Maths Chapter 15 Kerala Syllabus – Average

Intext Questions And Answers

Question 1.
Take some numbers and calculate the arithmetic means. Calculate the excess or deficit of each of the numbers from the mean and add them up separately. Are the sums equal? Can you explain the reason for this?
Answer:
Let’s consider some numbers 10, 15, 20, 25, 30
Arithmetic mean = \(\frac{10+15+20+25+30}{5}=\frac{100}{5}\) = 20
Now calculate the excess or deficit of each of the number from the mean.

Numbers	Excess of the number from 20	Deficit of the number from 20
10		20 – 10 = 10
15		20 – 15 = 5
20
25	25 – 20 = 5
30	30 – 20 = 10

Sum of excess of the number from 20 = 10 + 5 = 15
Sum of deficit of the number from 20 = 10 + 5 = 15
Here, the sum of excess of the numbers from the mean and deficit of the numbers from the mean are equal. Because the mean act as a balance between the values of the numbers.

Class 9 Maths Kerala Syllabus Chapter 15 Solutions – Tables

Textual Questions And Answers

Question 1.
In a T20 match, 51 runs were scored in the first 5 overs
i) What is the mean run rate then?
ii) If this run rate is maintained, what is the total they can expect?
Answer:
i) To find the mean run rate, we divide the total runs scored by the number of overs:
Mean Run Rate = \(\frac{\text { Total Runs Scored }}{\text { Overs Bowled }}\) = \(\frac{51}{5}\) = 10.2 runs per over

ii) Expected Total = Mean Run Rate × Total Overs
= 10.2 × 20
= 204 runs

Question 2.
The table below shows the children in a class grouped according to their marks in a math test:

i) What is the mean marks of the class?
ii) How many got less marks than the mean?
iii) How many got more marks than the mean?
Answer:

Marks	Children	Total Mark
2	1	2 × 1 = 2
3	2	3 × 2 = 6
4	5	4 × 5 = 20
5	4	5 × 4 = 20
6	6	6 × 6 = 36
7	11	7 × 11 = 77
8 ‘	10	8 × 10 = 80
9	4	9 × 4 = 36
10	2	10 × 2 = 20
Total	45	297

Mean mark = \(\frac{297}{45}\) = 6.6
i) Children who got mark less than mean is = 1 + 2 + 5 + 4 + 6 = 18
ii) Children who got mark more than mean is = 11 + 10 + 4 + 2 = 27

Question 3.
The details of rubber sheets a farmer got during a month are shown below:

i) How many kilograms of rubber did he get a day on average in this month?
Answer:

Rubber(kg)	Days	Total Rubber (kg)
09	3	9 × 3 = 27
10	4	10 × 4 = 40
11	3	11 × 3 = 33
12	3	12 × 3 = 36
13	5	13 × 5 = 65
14	6	14 × 6 = 84
16	6	16 × 6 = 96
Total	30	381

Average Quantity of rubber per day = \(\frac{381}{30}\) = 12.77 kg

ii) The price of a kilogram of rubber is 175 rupees. How much did he get a day on average this month from rubber?
Answer:
If the price is Rs. 175 per kg, then average income per day = 12.77 × 175
= Rs.2234.75

Question 4.
The table below shows the days in a month sorted according to the amount of rainfall in a locality:

What is the mean rainfall per day during this month?
Answer:

Rainfall (mm)	Days	Total Rainfall (mm)
54	3	54 × 3 = 162
56	5	56 × 5 = 280
58	6	58 × 6 = 348
55	3	55 × 3 = 165
50	2	50 × 2 = 100
47	4	47 × 4 = 188
44	5	44 × 5 = 220
41	2	41 × 2 = 82
Total	30	1545

The average of the rain fall per day during that month = \(\frac{\text { Total rain fall }}{\text { Number of days }}=\frac{1545}{30}\) = 51.5 m

Textual Questions And Answers

Question 1.
The table below shows the children in a class, grouped according to their heights:

What is the mean height?
Answer:

Height (cm)	Number of children	Class Mark	Total Height
148 – 152	8	150	1200
152 – 156	10	154	1540
156 – 160	15	158	2370
160 – 164	10	162	1620
164 – 168	7	166	1162
Total	50		7892

Mean height = \(frac{\text { Total height }}{\text { Number of children }}\)
= \(\frac{7892}{50}\)
= 157.84 cm

Question 2.
The table below shows the classification of teachers in a university based on their ages:

Calculate the mean age of the teachers
Answer:

Age	Number of teachers	Class Mark	Total Age
25-30	06	27.5	165
30-35	14	32.5	455
35-40	18	37.5	675
40-45	20	42.5	850
45-50	05	47.5	237.5
50-55	04	52.5	210
55-60	03	57.5	172.5
Total	70		2765

Mean age = \(\frac{\text { Total age }}{\text { Number of persons }}\)
= \(\frac{2765}{70}\)
= 39.5

Question 3.
The classification of a group of children according to their weights is given in the table below:

Calculate the mean weight.
Answer:

Weight (kg)	Number of children	Class Mark	Total Weight
21-23	4	22	88
23-25	7	24	168
25-27	8	26	208
27-29	6	28	168
29-31	3	30	90
31-33	1	32	32
Total	29		754

Mean Weight = \(=\frac{\text { Total Weight }}{\text { Number of children }}\)
= \(\frac{754}{29}\)
= 26

Statistics Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
The daily wages of workers of a factory are given below. Find the average daily wage.

Daily wages	Number
500	3
600	7
700	10
900	8
1000	2

Answer:

Daily wages	Number	Total Wages
500	3	1500
600	7	4200
700	10	7000
900	8	7200
1000	2	2000
Total	30	21900

Average daily wage = \(\frac{21900}{30}\) = 730

Question 2.
The table shows the students in a class sorted according to their marks in an exam

i) The average marks is 6 .How many students got 8 marks?
ii) How many students are there in the class?
Answer:

Mark	Students	Total mark
3	2	6
4	4	16
5	5	25
6	6	36
7	7	49
8	X	8x
9	2	18
10	1	10
Total	27 + x	160 + 8x

i) Average mark = \(\frac{\text { Total Mark }}{\text { Number of Students }}\)
6 = \(\frac{160+8 x}{27+x}\)
6(27 + x)= 160 + 8x
162 + 6x = 160 + 8x
2x = 2
x = 1
Therefore only one student got 8 marks ii) Total Number of students in the class = 27 + x = 27 + 1 = 28

Question 3.
A table categorizing the workers in an office on the basis of their salary is given below.

Salary (Rs)	Number of workers
15000 -18000	1
18000 – 21000	3
21000 – 24000	5
24000 – 27000	4
27000 – 30000	1
30000-33000	1

Find the mean of salary.
Answer:

Salary (Rs)	Number of workers	Class interval	Total salary
15000- 18000	1	16500	16500
18000-21000	3	19500	58500
21000-24000	5	22500	112500
24000 – 27000	4	25500	102000
27000 – 30000	1	28500	28500
30000 – 33000	1	31500	31500
Total	15		349500

Mean income = \(=\frac{\text { Total salary }}{\text { Number of workers }}\)
= \(\frac{349500}{15}\)
= 23,300
Therefore the mean salary of the workers = 23,300

Question 4.
The table below shows the daily wages of the workers in a firm. Calculate the mean daily wage.

Daily wage (Rs)	Number of workers
450 – 550	7
550 – 650	8
650 – 750	10
750 – 850	10
850 – 950	9
950 -1050	6

Answer:

Daily wages (Rs)	Number of workers	Class interval	Total salary
450-550	7	500	3500
550-650	8	600	4800
650 – 750	10	700	7000
750-850	10	800	8000
850-950	9	900	8100
950- 1050	6	1000	6000
Total	50		37,400

Mean value = \(\frac{\text { Total salary }}{\text { Number of workers }}\)
= \(\frac{37,400}{50}\)
= 748
Therefore the mean daily wage = 748

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 13 Malayalam Medium ബഹുപദചിത്രങ്ങൾ can save valuable time.

Kerala SCERT Class 9 Maths Chapter 13 Solutions Malayalam Medium ബഹുപദചിത്രങ്ങൾ

Class 9 Maths Chapter 13 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 13 Malayalam Medium Textual Questions and Answers

Question 1.
ചുവടെയുള്ള ബഹുപദങ്ങളുടെ ചിത്രരൂപം വരയ്ക്കുക
i) p(x) = 2x – 1
ii) p(x) = x – 1
iii) p(x) = 1 – x
iv) p(x) = x
v) p(x) = −x
Answer:
i) p(x) = 2x – 1 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	-1	1	3

ii) p(x) = x – 1 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	-1	0	1

iii) p(x) = 1 – x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	1	0	-1

iv) p(x) = x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	0	1	2

v) p(x) = -x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	0	-1	-2

Question 2.
ചുവടെയുള്ള വരകളുടെ ബഹുപദരൂപം കണ്ടുപിടിക്കുക

Answer:
i) ചിത്രത്തിൽ നിന്നും
p(0) = 1
p(\(\frac{-1}{2}\)) = 0 എന്ന് കിട്ടും
ഇവിടെ P(x) ഒന്നാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax + b
p(0) = a × 0 + b = 1
b = 1
p(\(\frac{-1}{2}\)) = a × (\(\frac{-1}{2}\)) + b = 0
\(\frac{-a}{2}\) + 1 = 0
a = 2
ബഹുപദ രൂപം = 2x + 1

ii) ചിത്രത്തിൽ നിന്നും
p(0) = 0
p(\(\frac{1}{2}\)) = 1 എന്ന് കിട്ടും
ഇവിടെ P(x) ഒന്നാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax + b
p(0) = a × 0 + b = 0
b = 0
p(\(\frac{1}{2}\)) = a × (\(\frac{1}{2}\)) + b = 1
\(\frac{a}{2}\) = 1
a = 2
ബഹുപദ രൂപം = 2x

iii) ചിത്രത്തിൽ നിന്നും
p(0) = -2
p(2) = 0 എന്ന് കിട്ടും
ഇവിടെ P(x) ഒന്നാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax + b
p(0) = a × 0 + b = -2
b = -2
p(2) = a × (2) + b = 0
2a – 2 = 0
a = 1
ബഹുപദ രൂപം = x – 2

Question 3.
ചില രണ്ടാംകൃതി ബഹുപദങ്ങളുടെ ചിത്രങ്ങളാണ് ചുവടെ കൊടുത്തിരിക്കുന്നത്.

ഓരോന്നിന്റെയും ബഹുപദരൂപം കണക്കാക്കുക.
Answer:
i) ചിത്രത്തിൽ നിന്നും
P(0) = 3
p(1) = 0
p(3) = 0 എന്ന് കിട്ടും
p(x) ഒരു രണ്ടാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax² + bx + c
ഇനി, p(0) = a × 0² + b × 0 + c = 3
c = 3
p(1) = a × 1² + b × 1 + c = 0
= a + b + 3 = 0 … (1)
p(3) = a × 3² + b × 3 + c = 0
= 9a + 3b + 3 = 0
= 3a + b + 1 = 0 … (2)

സമവാക്യം (2) ൽ നിന്നും സമവാക്യം (1) കുറച്ചാൽ
2a + (1 – 3) = 0
2a = 2
a = 1

a = 1 എന്നത് സമവാക്യം (1) ൽ ആരോപിച്ചാൽ
1 + b + 3 = 0
b + 4 = 0
b = -4
ആയതിനാൽ, ബഹുപദരൂപം= p(x) = x² – 4x + 3.

ii) ചിത്രത്തിൽ നിന്നും
2a + 2 = 0
a = -1
a = – 1 എന്നത് സമവാക്യം (1) ൽ ആരോപിച്ചാൽ
-1 + b – 3 = 0
b – 4 = 0
b = 4
ആയതിനാൽ, ബഹുപദരൂപം = p(x) = -x² + 4x – 3.

p(0) = 4
P(1) = 0
p(2) = 0 എന്ന് കിട്ടും
p(x) ഒരു രണ്ടാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax² + bx + c
ഇനി, p(0) = a × 0² + b x 0 + c = 4.
c = 4
p(1) = a x 1² + b x 1 + c = 0
= a + b + 4 = 0… (1)

p(2) = a x 2² + b x 2 + c = 0
= 4a + 2b + 4 = 0
= 2a + b + 2 = 0… (2)

സമവാക്യം (2) ൽ നിന്നും സമവാക്യം (1) കുറച്ചാൽ
a + (2 – 4) = 0
a = 2

2 എന്നത് സമവാക്യം (1) ൽ ആരോപിച്ചാൽ
2 + b + 4 = 0
b + 6 = 0
b = -6

ആയതിനാൽ, ബഹുപദരൂപം= p(x) = 2x² – 6x + 4.

iii) ചിത്രത്തിൽ നിന്നും
p(0) = – 3
p(1) = 0
p(3) = 0 എന്ന് കിട്ടും
p(x) ഒരു രണ്ടാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax² + bx + c
ഇനി, p(0) = a × 0² + b × 0 + c = -3
c = -3
p(1) = a × 1² + b × 1 + c = 0
= a + b + -3 = 0… (1)

p(3) = a × 3² + b × 3 + c = 0
= 9a + 3b – 3 = 0
= 3a + b – 10… (2)

സമവാക്യം (2) ൽ നിന്നും സമവാക്യം (1) കുറച്ചാൽ
2a + (-1 – -3) = 0

Class 9 Maths Chapter 13 Malayalam Medium Intext Questions and Answers

Question 1.
ചുവടെപ്പറയുന്ന ബഹുപദങ്ങളുടെ ചിത്രം വരയ്ക്കുക:
i) p(x) = x
ii) p(x) = 2x
iii) p(x) = x
iv) p(x) = -x 1
v) p(x) = -2x
ഈ വരകൾക്കെല്ലാം പൊതുവായ എന്തെങ്കിലും സവിശേഷതയുണ്ടോ?
Answer:
i) p(x) = x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	0	1	2

ii) p(x) = 2x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	0	2	4

iii) p(x) = \(\frac{1}{2}\)x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	2	4
p(x)	0	1	2

iv) p(x) = – x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	0	-1	-2

v) p(x) = −2x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	l	2
P(x)	0	-2	-4

ഈ വരകൾ എല്ലാം ആധാര ബിന്ദുവിലൂടെ കടന്നു പോകുന്നു.

Question 2.
ഇതുപോലെ ചുവടെയുള്ള ബഹുപദങ്ങളുടെയും ചിത്രം വരയ്ക്കുക:
i) p(x) = x + 1
ii) p(x) = x + 2
iii) p(x) = x + \(\frac{1}{2}\)
iv) p(x) = x – 1
v) p(x) = x – 2
Answer:
i) p(x) = x + 1 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	1	2	3

ii) p(x) = x + 2 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	2	3	4

iii) p(x) = x + \(\frac{1}{2}\) ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	\(\frac{1}{2}\)	1
p(x)	\(\frac{1}{2}\)	1	1\(\frac{1}{2}\)

iv) p(x) = x – 1 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	-1	0	1

v) p(x) = x – 2 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	-2	-1	0

ഈ ചിത്രങ്ങളുടെ എല്ലാം ചരിവ് തുല്യമാണ്

Polynomial Pictures Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
ചുവടെയുള്ള ബഹുപദങ്ങളുടെ ചിത്രരൂപം വരയ്ക്കുക
i) p(x) = 3-x
ii) p(x) = 3x + 1
iii) p(x) = 2x – 3
Answer:
i) p(x) = 3 – x ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
P(x)	3	2	1

ii) p(x) = 3x + 1 ൽ x ആയി -1, 0, 1 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	-1	0	1
p(x)	-2	1	4

iii) p(x) = p(x) = 2x – 3 ൽ x ആയി 0, 1, 2 എന്നീ സംഖ്യകളായി എടുത്താൽ

X	0	1	2
p(x)	-3	-1	1

Question 2.
ചുവടെയുള്ള വരകളുടെ ബഹുപദരൂപം കണ്ടുപിടിക്കുക

Answer:
ചിത്രത്തിൽ നിന്നും
p(0) = 8
p(-4) = 0 എന്ന് കിട്ടും
ഇവിടെ p(x) ഒന്നാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax + b
p(0) = a × 0 + b = 8
b = 8
p(-4) = a × (-4)+ b = 0
= 4a + 8 = 0
a = 2
ബഹുപദ രൂപം = 2x + 8

Answer:
ചിത്രത്തിൽ നിന്നും
p(0) = -4
p(4) = 0 എന്ന് കിട്ടും
ഇവിടെ p(x) ഒന്നാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax + b
p(0) = a × 0 + b = -4
b = -4
p(4) = a × (4) + b = 0
= 4a – 4 = 0
a = 1
ബഹുപദ രൂപം = x – 4

Question 3.
ചില രണ്ടാംകൃതി ബഹുപദങ്ങളുടെ ചിത്രങ്ങളാണ് ചുവടെ കൊടുത്തിരിക്കുന്നത്.

Answer:
i) ചിത്രത്തിൽ നിന്നും
P(0) = 1
p(-2) = 5
p(2) = 5 എന്ന് കിട്ടും
p(x) ഒരു രണ്ടാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax² + bx + c
ഇനി, p(0) = a × 0² + b × 0 + c = 1
c = 1

p(-2) = a × (-2)² + b × -2 + c = 5
= 4a – 2b + 1 = 5
= 4a- 2b = 4
= 2a – b = 2…(1)

p(2) = a × 2² + b × 2 + c = 0
= 4a + 2b + 1 = 5
= 4a + 2b = 4
= 2a + b = 2… (2)

സമവാക്യം (1) ഉം സമവാക്യം (2) ഉം കൂട്ടിയാൽ
4a= 4
a = 1

a = 1 എന്നത് സമവാക്യം (1) ൽ ആരോപിച്ചാൽ
2 × 1 – b = 2
b = 0
ആയതിനാൽ, ബഹുപദരൂപം= p(x) = x² + 1

Answer:
ചിത്രത്തിൽ നിന്നും
p(0) = -2
p(-2) = 2
p(2) = 2 എന്ന് കിട്ടും
p(x) ഒരു രണ്ടാം കൃതി ബഹുപദമായതിനാൽ
p(x) = ax² + bx + c
ഇനി, p(0) = a × 0² + b × 0 + c = -2
c = -2

p(-2) = a × (-2)² + b x -2 + c = 2
= 4a – 2b – 2 = 2
= 4a- 2b = 4
= 2a – b = 2…(i)

p(2) = a × 2² + bx² + c = 0
= 4a + 2b – 2 = 2
= 4a + 2b = 4
= 2a + b = 2… (2)

സമവാക്യം (1) ഉം സമവാക്യം (2) ഉം കൂട്ടിയാൽ
4a = 4
a = 1

a = 1 എന്നത് സമവാക്യം (1) ൽ ആരോപിച്ചാൽ
2 × 1 – b = 2
b = 0
ആയതിനാൽ, ബഹുപദരൂപം= p(x) = x – 2

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 14 Proportion Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 14 Solutions Proportion

Proportion Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 14 Proportion Solutions Questions and Answers

Class 9 Maths Chapter 14 Kerala Syllabus – Proportional Changes

Textual Questions And Answers

Question 1.
In each of the instances below, show that the second quantity changes proportionally with respect to the first. Also find the proportionally constant in each:
i) The length of sides of squares and their perimeters.
ii) The lengths of wires bent into squares and the length of the sides of the squares.
iii) The number of rotations of a circle rolling along a line, and the distance travelled along the line.
Answer:
i) Let s be the length of a side of a square. The perimeter P of a square is given by:
P = 4s
Now, we can express the ratio of the perimeter to the side length:
\(\frac{\mathrm{P}}{\mathrm{~s}}=\frac{4 \mathrm{~s}}{\mathrm{~s}}\) = 4
Since this ratio is constant (4), we conclude that the perimeter changes proportionally with respect to the length of the sides of the squares. The proportionality constant is 4.

ii) Let L be the total length of the wire, and let s be the length of a side of the square formed by bending the wire. The total length of the wire is given by:
L = 4s
Now, we express the ratio of the total length of the wire to the side length:
\(\frac{L}{s}=\frac{4 s}{s}\) = 4
Since the ratio is constant (4), we find that the lengths of wires bent into squares change proportionally with respect to the length of the sides of the squares. The proportionality constant is 4.

iii) Let r be the radius of the circle. The circumference C of the circle is given by:
C = 2πr
When a circle rolls along a line, the distance d travelled along the line is equal to the number of rotations n times the circumference of the circle:
d = nC
Substituting the circumference:
d = n (2πr)
Now, we express the ratio of the ratio of the distance travelled to the number of rotations:
\(\frac{\mathrm{d}}{\mathrm{n}}\) = 2πr
Since this ratio is constant (specifically, 27rr), we find that the distance travelled along the line changes proportionally with respect to the number of rotations of the circle. The proportionality constant is 2πr.

Question 2.
We have seen that in the picture, the height of a point on the slanted line from the horizontal line changes proportionally with respect to its distance from the corner. Calculate the proportionality constants 30°, 45° and 60°.

Answer:

When the angle is 30°
A base triangle ΔABC is drawn such that AC = 2cm, BC = 1cm and AB = √3 cm
Here triangle ABC and APQ are similar, we get
\(\frac{A C}{A Q}=\frac{B C}{P Q}=\frac{A B}{A P}\)

BC and PQ are heights and AC and AQ are distance.
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{2}{1}\)
AQ = 2 PQ
Therefore, the proportionality constant is 2
Similarly, when the angle is 60°

A base triangle ΔABC is drawn such that AC = 2cm, BC = √3 cm and AB = 1cm
Here triangle ABC and APQ are similar, we get
\(\frac{A C}{A Q}=\frac{B C}{P Q}=\frac{A B}{A P}\)

BC and PQ are heights and AC and AQ are distance.
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{2}{\sqrt{3}}\)
AQ = \(\frac{2}{\sqrt{3}}\)PQ

Therefore, the proportionality constant is \(\frac{2}{\sqrt{3}}\)
Similarly, when the angle is 45°
A base triangle A ABC is drawn such that AC = √2 cm, BC = 1 cm and AB = 1 cm
Here triangle ABC and APQ are similar, we get
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}=\frac{\mathrm{AB}}{\mathrm{AP}}\)
BC and PQ are heights and AC and AQ are distance.
\(\frac{\mathrm{AC}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}\)
AQ = PQ × \(\frac{A C}{B C}\) = PQ × \(\frac{\sqrt{2}}{1}\)
AQ = \(\frac{\sqrt{2}}{1}\) PQ
Therefore, the proportionality constant is √2.

Question 3.
Prove that in equilateral triangles, the perimeter changes proportionally with respect to the length of the sides. What is the proportionality constant? What can we say about other regular polygons?
Answer:
Let’s consider an equilateral triangle with side length’s’.
Perimeter of an equilateral triangle, P = 3s
The perimeter of an equilateral triangle changes proportionally with respect to the length of its sides, with a proportionality constant of 3.

Let’s consider a regular polygon with ‘n’ sides, each of length’s’.
Perimeter of a regular polygon, P = ns
The perimeter of any regular polygon changes proportionally with respect to the length of its sides,
with a proportionality constant equal to the number of sides.

Question 4.
Prove that the lengths of arcs of a fixed circle change proportionally with respect to their central angles. What is the proportionality constant? What about the relation between the area of a sector and its central angle?
Answer:
Let’s consider a circle with radius ‘r’ and central angle ‘θ’
Arc Length (L) = θ × r
The lengths of arcs of a fixed circle change proportionally with respect to their central angles, with a proportionality constant of r.
Area of a sector (A) = \(\frac{1}{2}\) r² θ
The area of a sector of a fixed circle changes proportionally with respect to its central angle with a proportionality constant of \(\frac{1}{2}\) r².

Class 9 Maths Kerala Syllabus Chapter 14 Solutions – Scale And Proportion

Textual Questions And Answers

Question 1.
i) What is the sum of the angles of a triangle? And the sum of the angles of a hexagon?
ii) Does the sum of the angles of polygons change proportionally with respect to the number of sides? Explain the reason.
Answer:
i) Sum of the interior angles of a triangle = (3 – 2) × 180° = 180°
Sum of the interior angles of a hexagon = (6 – 2) × 180° = 720°

ii) Yes, the sum of the angles of polygons changes proportionally with respect to the number of sides.
The sum of interior angles (S) of a polygon with ‘n’ sides is given by:
S = (n- 2) × 180°
The sum of the angles of polygons changes proportionally with respect to the number of sides with proportionality constant 180°

Question 2.
Inside a triangle of base 6 centimeters and height 3 centimeters, lines are drawn parallel to the base. Prove that the lengths of these lines change proportionally with respect to the distance from the top vertex. Find the proportionality constant.

Answer:
Given, Base = 6 cm
Height = 3 cm
Consider parallel lines drawn from the top vertex.
Distance from top vertex = x
Length of parallel line = y
Using similar triangle property,
\(\frac{y}{x}=\frac{6}{3}\)
y = 2x
Thus we can say that lengths of these lines change proportionally with respect to the distance from the top vertex with proportionality constant 2

Question 3.
Within a semicircle of diameter 10 centimeters, lines are drawn parallel to the diameter:

(i) In each of the pictures below, calculate the length of the line parallel to the diameter:

ii) Does the length of the parallel line change proportionally with respect to the distance from the top of the semicircle. Explain the reason.
Answer:
i) Figure 1

Length of the parallel line = 2\(\sqrt{(5)^2-(4)^2}\) = 2 × 3 = 6 cm

Figure 2

Length of the parallel line = 2\(\sqrt{(5)^2-(3)^2}\) = 2 × 4 = 8 cm
ii) The length of the parallel line does not change proportionally with respect to the distance from the top of the semicircle.

SCERT Class 9 Maths Chapter 14 Solutions – Different Proportions

Textual Questions And Answers

Question 1.
i) Prove that the areas of equilateral triangles change proportionally with respect to the squares of the lengths of sides. What is the proportionality constant?
Answer:

Areas of equilateral triangles change proportionally with respect to the squares of the lengths of sides with proportionality constant of \(\frac{\sqrt{3}}{4}\).

ii) Are the areas of squares proportional to the squares of the lengths of sides? If so, what is the proportionality constant?
Answer:

Area of the squares are proportional to the squares of the lengths of side with proportionality constant of 1.

Question 2.
Consider all rectangles of area 1 square meter. The length of one side of such a rectangle depends on the length of the other side. Write this relation as an algebraic equation. How do we state it in terms of proportion?
Answer:
In rectangles of area one square meter
Let x be the length and y be the breadth of the rectangle
Then, area = length × breadth
1 = x × y
1 = x y
y = \(\frac{1}{x}\)
y is inversely proportional to x

Question 3.
Consider all triangles of a fixed area. How do we state in terms of proportion, the relation between the length of the longest side and the length of the perpendicular to it from the opposite vertex? What if we use the shortest side instead of the longest?
Answer:
Let ‘a’ be the longest side, ‘h’ be the length of perpendicular from opposite vertices, ‘A’ be the area then
A = \(\frac{1}{2}\)ah,
a = \(\frac{2A}{h}\)
Length of larger side is inversely proportional to the length of perpendicular from the opposite vertex.
That is, the length of small side is inversely proportional to the length of perpendicular line from the vertex of small side.

Question 4.
In regular polygons, can we say the relation between the number of sides and the measure of an outer angle, in terms of proportion? What is the proportionality constant?
Answer:
The sum of the exterior angles of all polygon is 360°.
If ‘n’ is the number of sides.
Measure of an exterior angle = \(\frac{\text { sum of exterior angle }}{\text { number of sides }}\)
If the measure of an outer angle is ‘x’
x = \(\frac{360^{\circ}}{n}\)
One outer angle and number of sides are inversely proportional.
The constant of proportionality is \(\frac{1}{n}\)

Proportion Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
A fixed volume of water is to flow into a rectangular water tank. The rate of flow can be changed by using different pipes. Write the relations between the following quantities as an algebraic equation and in terms of proportions.
i) The rate of water flow and the height of the water level.
ii) The rate of water flow and the time taken to fill the tank.
Answer:
i) Let x be the rate of water flowing, y be the height of water in the tank and A be the base area of the tank, then x = Ay
Height of the water level in the tank is proportional to the rate of the water flowing.

ii) If C is the volume of the tank, V be the volume of water flowing per second, the volume of water in’t’ second is given by C = V t
V = C × \(\frac{1}{t}\)
That is the rate of water flow and the time taken for filling the tank are inversely proportional.
C is the constant of proportionality.

Question 2.
Raghu invested Rs. 60000 and Nazar Rs. 100000 and started a business. Within one month a profit of Rs. 4800 was obtained. Raghu took 1800 and Nazar took Rs. 3000 out of the profit obtained. What is the ratio of the investment? Is the investment and the profit divided proportionally?
Answer:
Ratio of investments = 60000: 100000 = 6:10 = 3:5
Ratio of profit divided 1800: 3000 = 18:30 = 3:5
Ratio of investments and Ratio of profit divided are equal.
Hence they are proportional.

Question 3.
Are the length and breadth of a square having same perimeter inversely proportional?
Answer:
For a square the perimeter is 20 cm. so, length + breadth = 10.
Let x be the length and y be the breadth then possible values of x and y are
When x = 9 then y = 1
When x = 8 then y = 2
Here xy is not a constant term i.e. the changes in the x and y is not in the firm of xy = kept
Hence length and breadth are not in inversely proportional.

Question 4.
A car with 5L of petrol travels a distance of 75 km. What is the proportionality constant between the distance travelled and the quantity of petrol? How much petrol is needed for travelling 180 km?
Answer:
Taking the distance travelled as x and quantity of petrol as y, then the constant of proportionality 15
k = \(\frac{y}{x}=\frac{75}{5}\) = 15
k = \(\frac{y}{x}\) = 15 = \(\frac{100}{x}\)
x = \(\frac{180}{15}\) = 12
The quantity of petrol needed to travel 180 km = 12 litre

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 13 Polynomial Pictures Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 13 Solutions Polynomial Pictures

Polynomial Pictures Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 13 Polynomial Pictures Solutions Questions and Answers

Class 9 Maths Chapter 13 Kerala Syllabus – First Degree Polynomials

Textual Questions And Answers

Question 1.
Draw the graphs of these polynomials:
i) p(x) = 2x – 1
ii) p(x) = x – 1
iii) p(x) = 1 – x
iv) p(x) = x
v) p(x) = -x
Answer:
i) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 2x – 1

x	0	1	2
P(x)	-1	1	3

ii) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x – 1

x	0	1	2
P(x)	-1	0	1

iii) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 1 – x

x	0	1	2
P(x)	1	0	-1

iv) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x

x	0	1	2
P(x)	0	1	2

v) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = -x

x	0	1	2
P(x)	0	-1	-2

Question 2.
Find the polynomials which has these lines as their graphs:

Answer:
i) From the figure, we get
P(0) = 1
P(\(\frac{-1}{2}\)) = 0
Since p(x) is first degree polynomial
p(x) = ax + b
p(0) = a x 0 + b = 1
b = 1
p(\(\frac{-1}{2}\)) = a x (\(\frac{-1}{2}\)) + b = 0
\(\frac{-a}{2}\) + 1 = 0
a = 2
Therefore the polynomial is 2x + 1

ii) from the figure, we get
P(0) = 0
P(\(\frac{1}{2}\)) = 1
since p(x) is first degree polynomial
p(x) = ax + b
p(0) = a × 0 + b = 0
b = 0
P(\(\frac{1}{2}\)) = a × (\(\frac{1}{2}\)) + b = 1
= \(\frac{a}{2}\) = 1
= a = 2
Therefore the polynomial is 2x

iii) from the figure, we get
P(0) = -2
P(2) = 0
since p(x) is first degree polynomial
p(x) = ax + b
p(0) = a × 0 + b = -2
b = -2
p(2) = a x (2) + b = 0
= 2a – 2 = 0
= a = 1
Therefore the polynomial is x – 2

Class 9 Maths Kerala Syllabus Chapter 13 Solutions – Second Degree Polynomials

Intext Questions And Answers

Question 1.
Draw the graphs of these polynomials:
i) p(x) = x
ii) p(x) = 2x
iii) PO) = \(\frac{1}{2}\)x
iv) p(x) = -x
v) p(x) = -2x
Do you see any common feature of these graphs?
Answer:
i) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x

x	0	1	2
P(x)	0	1	-2

ii) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 2x

x	0	1	2
P(x)	0	1	4

iii) Let’s take x = 0, x = 2, and x = 4 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = \(\frac{1}{2}\)x

x	0	2	4
P(x)	0	1	2

iv) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = -x

x	0	1	2
P(x)	0	-1	-2

v) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = -2x

x	0	1	2
P(x)	0	-2	-4

All these lines pass through the origin

Question 2.
Now draw the graphs of these polynomials:
i) p(x) = x +1
ii) p(x) = x + 2
iii) P(x) = x + \(\frac{1}{2}\)
iv) p(x) = x – 1
v) p(x) = x – 2
Do you see any common feature of these graphs?
Answer:
i) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x + 1

x	0	1	2
P(x)	1	2	3

ii) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x + 2

x	0	1	2
P(x)	2	3	4

iii) Let’s take x = 0, x = \(\frac{1}{2}\), and x = 1 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x + \(\frac{1}{2}\)

x	0	\(\frac{1}{2}\)	1
P(x)	\(\frac{1}{2}\)	1	1\(\frac{1}{2}\)

iv) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x – 1

x	0	1	2
P(x)	-1	0	1

v) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x – 2

x	0	1	2
P(x)	-2	-1	0

All these lines have same slope

Polynomial Pictures Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
Draw the graphs of these polynomials:
i) P(x) = 3 – x
ii) p(x) = 3x + 1
iii) p(x) = 2x – 3
Answer:
i) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 3 – x

x	0	1	2
P(x)	3	2	1

ii) Lets take x = -1, x = 0, and x = 1 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 3x + 1

x	-1	0	1
P(x)	-2	1	4

iii) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 2x – 3

x	0	1	2
P(x)	-3	-1	1

Question 2.
Find the polynomials which has these lines as their graphs:

Answer:
i) from figure we get,
P(0) = 8
p(-4) = 0
since p(x) is first degree polynomial
now, p(x) = ax + b
p(0) = a × 0 + b = 8
b = 8

p(-4) = a × (-4) + b = 0
-4a + 8 = 0
a = 2
Therefore, the polynomial is p(x) = 2x + 8

Answer:
from figure we get,
p(0) = -4
p(4) = 0
since p(x) is first degree polynomial now,
p(x) = ax + b
p(0) = a × 0 + b = -4
b = -4
p(4) = a × (4) + b = 0
= 4a – 4 = 0
= a = 1
Therefore, the polynomial is p(x) = x – 4

Question 3.
The graphs of some second degree polynomials are given below:

Answer:
from figure we get, p(0) = 1
p(-2) = 5
P(2) = 5
since p(x) is first degree polynomial
p(x) = ax² + bx + c
now, p(0) = a × 0² + b × 0 + c = 1
= c = 1

p(-2) = a × (-2)² + bx – 2 + c = 5
= 4a – 2b + 1 = 5
= 4a- 2b = 4
= 2a-b = 2 …(1)

p(2) = a × 2² + bx² + c = 0
= 4a + 2b + 1 = 5
= 4a + 2b = 4
= 2a + b = 2 … (2)

Adding equation (1) and equation (2)
4a = 4
a = 1

Put a = 1 in equation (1) we get,
2 × 1 – b = 2
b = 0
Therefore, the polynomial is p(x) = x² + 1

Answer:
from the figure we get,
P(0) = -2
p(-2) = 2
p(2) = 2
since p(x) is second degree polynomial p(x) = ax² + bx + c
now, p(0) = a × 0² + b x 0 + c = -2
= c = -2

p(-2) = a × (-2)² + bx – 2 + c = 2
= 4a – 2b – 2 = 2
= 4a – 2b = 4
= 2a – b = 2 …(1)

p(2) = a × 2² + bx² + c = 0
= 4a + 2b – 2
= 2

Adding equation (1) and equation (2)
4a = 4
a = 5 = 1

Put a = 1 in equation (1) we get,
2 × 1 – b = 2
b = 0
Therefore, the polynomial is p(x) = x² – 2

You can Download पुल बनी थी माँ Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 1 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Hindi Solutions Unit 1 Chapter 1 पुल बनी थी माँ (कविता)

पुल बनी थी माँ Textual Questions and Answers

पुल बनी थी माँ आशयग्रहण के प्रश्न

प्रश्ना 1.
‘पुल बनी थी माँ’ से क्या तात्पर्य है?

उत्तर:
पुल दो किनारों को आपस में जोड़ता है। माँ परिवार के हर सदस्य को आपस में जोड़नेवाली कड़ी थी। इसलिए माँ को पुल कहा गया है।

प्रश्ना 2.
‘बुढ़ा रही है माँ’ इसका आशय क्या है?

उत्तर:
प्रस्तुत पंक्ति का आशय यह है कि माँ के शरीर पर बुढ़ापे का असर दिखने लगा। वह शारीरिक और मानसिक रूप से कमज़ोर होने लगी।

प्रश्ना 3.
‘माँ आख़िर माँ ही तो है’ इससे आपने क्या समझा?

उत्तर:
इसका मतलब है कि माँ का मातृत्व बच्चों की कठिनाइयों को अच्छी तरह जानता है। अर्थात् माँ अपने बच्चों के बारे में सबकुछ जानती है।

पुल बनी थी माँ Text Book Activities

पुल बनी थी माँ अभ्यास के प्रश्न

प्रश्ना 1.
बेटों का जीवन बेरोकटोक चलती गाड़ी के समान रहा।

उत्तर:
दौड़ती रहती थी बेधड़क
बिना किसी हरी लाल बत्ती के
हम लोगों की छुक छुक छक छक

प्रश्ना 2.
माँ की देख-भाल की ज़िम्मेदारी बेटों पर आ गई।

उत्तर:
हाथों हाथ रहती माँ
एक दिन हमारे कंधों में आ गई

प्रश्ना 3.
बेटे अपने दायित्व बदलते रहे।

उत्तर:
जब तक जीवित रही माँ।
हम बदलते रहे अपने कंधे

प्रश्ना 4.
माँ के चले जाने से बेटे बेसहारे बने।

उत्तर:
और माँ के कंधों से उतरते ही
उतर गए हमारे कंधे

पुल बनी थी माँ विधात्मक प्रश्न

प्रश्ना 1.
कविता का परिचय देते हुए टिप्पणी लिखें।

उत्तर:
पुल बनी थी माँ : बदलते पारिवारिक संबंधों की आलोचना
कविता ‘पुल बनी थी माँ’ बूढ़े-बुजुर्गों के प्रति उत्तरदायित्वों से विमुख होती जा रही नई पीढ़ी के व्यवहार को दर्शाता है। कविता में माँ के पुल होने और पुल से बोझ बनने की हालत पर चर्चा की गई है।

माँ भाइयों के बीच पुल बनी थी। पुल दो किनारों को आपस में जोड़ता है। माँ परिवार के हर सदस्य को आपस में जोड़नेवाली कड़ी रही। इस माँ रूपी पुल से बच्चों की जिंदगी रूपी रेल गाड़ी बेरोकटोक चलती रही। पिता के चल बसने के बाद भी भाइयों के बीच माँ पुल बनी रही। माँ धीरे-धीरे टूटने लगी। यानी मानसिक रूप से वह धीरे-धीरे कमज़ोर होती गई। उसके शरीर पर बुढ़ापे का असर दिखने लगा। वह शारीरिक रूप से भी
कमज़ोर होने लगी थी। एक ही बात को माँ बार-बार कहने लगी। बच्चे इस आदत को उनके बढ़ते हुए बुढापे की निशानी मानकर जीने लगे। उसकी आवाज़ कमज़ोर होती रही। वह धीरे-धीरे दुर्बल होती रही।

बच्चों के प्रति प्यार और दुलार से रहनेवाली माँ एक दिन बच्चों के आश्रय में आ गईं। धीरे-धीरे बच्चों के सशक्त कंधों में माँ बोझ बन गईं। जब तक बूढ़ी माँ जीवित रही, बच्चे माँ की देखरेख की ज़िम्मेदारी एक दूसरे के कंधों पर डालते रहे। सारी जिंदगी बच्चों के लिए जीनेवाली माँ बुढ़ापे में बच्चों के लिए भार बन गई। पर माँ का मातृत्व बच्चों की इस कठिनाई को सह नहीं पाया। वह स्वयं उनके कंधों से उतर गई मतलब उसका अंतिम प्रयाण हो गया। माँ के अभाव में बच्चे बेसहारे बन गए। कविता में प्रयुक्त शब्द, कथन और मुहावरे- वृषभ कंधा, कंधा बदलना, उतर गए कंधे आदि कविता को और सशक्त बनाया है।

पुल बनी थी माँ Summary in Malayalam and Translation

पुल बनी थी माँ शब्दार्थ

Students rely on Kerala Syllabus 9th Standard Chemistry Textbook Solutions Chapter 1 Structure of Atom Notes Questions and Answers English Medium to help self-study at home.

Kerala SCERT Class 9 Chemistry Chapter 1 Solutions Structure of Atom

Kerala Syllabus Std 9 Chemistry Chapter 1 Structure of Atom Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 1 Let Us Assess Answers Structure of Atom

Question 1.
Some observations related to experiments on cathode rays are given. Write the inference based on each observation.
a. A paddle wheel placed in the path of cathode rays rotates.
b. A shadow is formed if an object is placed in the path of cathode rays.
c. When an electric field is applied perpendicular to the path of cathode rays, the rays deflect towards the positive plate.
Answer:
a. A paddle wheel placed in the path of cathode rays rotates shows that the particles of cathode rays have mass

b. A shadow is formed if an object is placed in the path of cathode rays, indicating that the cathode
rays travel in a straight line.

c. When an electric field is applied perpendicular to the path of cathode rays, the rays deflect towards the positive plate because the cathode rays are composed of negatively charged particles (electrons)

Question 2.
The atomic number of an atom is 16 and mass number is 32.
Answer:
a. How many electrons, protons and neutrons are present in this atom?
b. Write the electron configuration of this atom.
c. Draw the orbit electron configuration of this atom.
Answer:
a) Atomic number of atom = 16
Mass number of atom = 32
Number of protons = Atomic number = 16
No: of electrons = 16
No: of neutrons = Mass number – Atomic number = 32 – 16 = 16

b. Electronic configuration of atom = 2, 8, 6

c. Orbit electronic configuration of the atom

Question 3.
Electrons are present in the K, L and M shells of an atom.
a. Which of these shells has the highest energy?
b. If M shell contains only 3 electrons, write the atomic number of this atom.
c. What is the number of electrons in this atom?
d. If the nucleus of this atom contains 16 neutrons, what is its mass number?
Answer:
a. M shell has maximum energy in K, L, and M shells

b. M shell contains 3 elecrons, indicates that inner shells are filled. i.e., K and L shells contain maximum of 2 and 8 electrons respectively.
Total number of electrons = 2 + 8 + 3 = 13
Atomic number = no: of protons = no: of electrons in a neutral atom = 13

c. No: of electrons in the atom = sum of electrons in each shell = 2 + 8 + 3= 13

d. No: of neutrons = 16
Mass number = No: of protons + No: of neutrons = 13 + 16 = 29

Question 4.
The orbit electron configuartion of an atom is given below.

a. What is the mass number of this atom?
b. Write its electron configuration.
Answer:
a. From the figure
No: of protons = 13
No: of neutrons = 14
Mass number of atom = No: of protons + No: of neutrons = 13 + 14 = 27

b. From the figure
No: of electrons = 13
Electron configuration = 2, 8, 3

Question 5.
The symbols of some elements are given.
\({ }_12^{24} \mathrm{Mg}\), \({ }_6^{12} \mathrm{C}\), \({ }_7^{15} \mathrm{N}\), \({ }_6^{14} \mathrm{C}\), \({ }_11^{24} \mathrm{Na}\)
a. Select a pair of isotopes from the given elements. Write the reason for selecting it.
b. Select a pair of isobars from the given elements.
Answer:
a. \({ }_6^{12} \mathrm{C}\) and \({ }_6^{14} \mathrm{C}\)C are the pair of isotopes. Isotopes have same atomic number and different mass number. They are atoms of same element with different mass number.

b. \({ }_12^{24} \mathrm{C}\)Mg and \({ }_11^{24} \mathrm{C}\)Na are the pair of isobars. Isobars are the atoms of different elements with same mass number and different atomic number.

Question 6.
Match the items in column A & B suitably.

A	B
Plum pudding model	James Chadwick
Planetary model of atom	Goldstein
Canal rays	J.J. Thomson
Neutron	Rutherford

Answer:

A	B
Plum pudding model	J.J. Thomson
Planetary model of atom	Rutherford
Canal rays	Goldstein
Neutron	James Chadwick

Question 7.
The atomic number and mass number of an element are 15 and 31 respectively.
a. What is the number of valence electrons in this atom?
b. How many neutrons are present in this atom?
c. Draw the orbit electron configuration of this atom.
Answer:
a. Atomic number of an atom = No: of protons = No: of electrons = 15
Electron configuration = 2,8,5
Valence electrons No: of electrons in outermost shell = 5

b. Mass number = 31
Mass number = atomic number + No: of nuetrons
No: of neutrons = Mass number – atomic number = 31 – 15=16
c. Orbit electron configuration of the atom

Question 8.
Isotope of an element is used to determine the age of fossils.
a. Which is this isotope?
b. Which are the other two main isotopes of this element?
c. Write the number of neutrons in each isotope.
Answer:
a. \({ }_6^{14} \mathrm{C}\) is the radioactive isotope of carbon used to determine the age of fossils.

b. \({ }_6^{12} \mathrm{C}\) and \({ }_6^{13} \mathrm{C}\) are the other two isotopes of carbon.

c. No: of neutrons in \({ }_6^{14} \mathrm{C}\) isotope = 8
No: of neutrons in \({ }_6^{13} \mathrm{C}\) isotope = 7
No: of neutrons in \({ }_6^{12} \mathrm{C}\) isotope = 6

Extended Activities

Question 1.
Prepare a presentation on scientists connected to the history of atom and their contributions and present it in the classroom.
Answer:
Here are some points for your reference:

• John Dalton – In 1809, found out all substances are made up of very minute particles called atoms.
• Heinrich Geissler – In 1854, developed discharge tubes and vaccum pumps.
• William Crookes – In 1875, passed high voltage electricity through gaes at low pressure in test tube.
• Eugen Goldstein – In 1886, found out canal rays.
• J.J Thomson – In 1906 found out cathode rays. Put forward the plumpudding model of atom
• Henry Becquerel – in 1896 discovered radioactivity.
• Ernest Rutherford – In 1911 conducted gold foil experiment. Found out the nucleus. Put forward the planetary model of atom.
• Neils Bohr – In 1913 put forward the Bohr model of atom.
• James Chadwick – In 1932 discovered neutron.

Question 2.
Prepare a timeline chart on the main events that led to the discovery of different subatomic particles.
Answer:
In 1886 Eugen Goldstein discovered canal rays. From the study of canal rays discovered protons.
Rutherford named the particle proton. In 1906 J.J.Thomson discovered electrons through discharge tube experiments. In 1932 James Chadwick discovered neutrons.

Question 3.
You have learned about isotopes. Find more examples for radio isotopes. Prepare an article on the uses of each radio isotope and publish it in the science magazine. Use word processor for this work.
Answer:

Radioactive isotopes	Uses
Iodine-131	For the study of thyroid glands and treatment
Uranium-235	Used as nuclear fuel
Cobalt-60	For the treatment of cancer
Sodium-24	To find out leakage in industrial pipe lines
Iron-59	To detect anaemia
Carbon-14	To detect the age of fossils
Dueterium	To produce heavy water
Phosphorus-31	As a tracer to find out material transportation in plants

Question 4.
If you get a chance to conduct an interview with Rutherford, what questions would you ask him? Prepare a questionnaire.
Answer:
Here are some questions for your reference:

• Can you remember the teachers who influenced you the most?
• What was your basic subject of interest?
• Which topic did you selected for your research?
• How long did you spend in research?
• Which do you think is your important discovery?

Structure of Atom Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Analyse table

How do molecules of different substances differ?
Answer:

• Differ in the component elements
• Differ in the ratio of the component elements

Question 2.
What are the important particles in an atom?
Answer:

• Electron
• Proton
• Neutron
  They are called sub-atomic particles.

Question 3.
How was it proved that electrons have mass?
Answer:
When a paddle wheel was placed in the path of cathode rays it rotates. This shows that electrons, the particles in cathode rays have mass.

Question 4.
Cathode rays, cast shadows of opaque objects placed in their path. What can be inferred from this?
Answer:
Cathode rays cast shadows of opaque objects placed in their path, indicating that the cathode rays travel in a straight line.

Question 5.
Some properties of the sub-atomic particles like electron, proton and neutron are given in the table. Complete the following table.

Answer:

Question 6.
Some statements are given. Which of them are related to J.J. Thomson?
a) Proposed the idea of the orbit
b) Conducted discharge tube experiments
c) Discovered neutron
d) Discovered electron
e) Proposed the plum pudding model
Answer:
b) Conducted discharge tube experiments
d) Discovered electron

Question 7.
Prepare a questionnaire about the scientists who conducted research on atomic structure and their contributions. Conduct a quiz programme in your classroom based on this.
Answer:
Here are some questions for your reference

• Who developed discharge tubes and vaccum tubes?
  – Heinrich Geissler
• Who discovered electrons?
  – J.J. Thomson
• Name the scientist who discovered the charge to mass ratio of electrons.
  – J.J. Thomson
• Who discovered canal rays?
  – Eugen Goldstein
• What is the charge of the particles of cathode rays?
  – Negative
• Who found the charge and mass of electron?
  – Robert Millikan
• Scientist who discovered radioactivity?
  – Henry Becquerel
• The Particles that do not deviate in electric and magnetic fields is
  – Neutron

Question 8.
What are the particles in the nucleus of an atom?
Answer:
Protons and neutrons are present in nucleus of an atom.

Question 9.
What is the mass number of an atom having 2 protons and 2 neutrons?
Answer:
No: of neutrons = 2
No: of protons = 2
Mass number of the atom = No: of protons + No: of neutrons = 2 + 2 = 4

Question 10.
Find the number of protons, electrons and neutrons in chlorine and calcium atoms

Answer:
1. Chlorine \({ }_17^{35} \mathrm{C}\)Cl
Atomic number of Cl, Z = 17
Mass number of Cl, A = 35
No: of protons = 17
No: of electrons = 17
Mass number = No: protons + No: of neutrons
∴ No: of neutrons = Mass number – No: of protons A – Z = 35 – 17 18

2. Calcium \({ }_20^{40} \mathrm{C}\)Ca
Atomic number of Ca. Z = 20
Mass number of Ca, A =40.
No: of protons = 20
No: of electrons = 20
Mass number = No: protons + No: of neutrons
∴ No: of neutrons Mass number – No: of protons = A – Z = 40 – 20 = 20

Question 11.
Complete the table given below.

Answer:
Atomic number, Z = No: of protons = No: of electrons
Mass Number, A = No: of protons (Z) + No: of neutrons
No: of neutrons = Mass number – Atomic number = A – Z

1. Hydrogen \({ }_1^{1} \mathrm{H}\)
Atomic number of H, Z = 1
Mass number of H, A =1
‘No: of protons = 1
No: of electrons = 1
Mass number = No: protons + No: of neutrons
∴ No: of neutrons = Mass number – No: of protons = A – Z = 1-1=0

2. Lithium \({ }_3^{7} \mathrm{Li}\)
Atomic number of Li, Z = 3
Mass number of Li. A = 7
No: of protons = 3
No: of electrons = 3
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 7 – 3 = 4

3. Oxygen \({ }_8^{16} \mathrm{O}\)
Atomic number of O, Z = 8
Mass number of O, A = 16
No: of protons = 8
No: of electrons = 8
Mass number = No: protons + No; of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 16 – 8 = 8

4. Sodium Na \({ }_11^{23} \mathrm{Na}\)
Atomic number of Na, Z = 11
Mass number of Na, A = 23
No: of protors = 11
No: of electrons = 11
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 23 – 11 = 12

5. Neon \({ }_10^{20} \mathrm{Ne}\)
Atomic number of Ne, Z = 10
Mass number of Ne, A = 20
No: of protons = 10
No: of electrons = 10
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 20 – 10 = 10

6. Titanium \({ }_22^{48} \mathrm{Ne}\)
Atomic number of Ti, Z = 22
Mass number of Ti, A =48
No: of protons = 22
No: of electrons = 22
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 48 – 22 = 26

7. Uranium \({ }_92^{235} \mathrm{U}\)
Atomic number of U, Z = 235
Mass number of U, A = 92
No: of protons = 92
No: of electrons = 92
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 235 – 92 = 143

8. Thorium \({ }_90^{232} \mathrm{Th}\)
Atomic number of Th, Z = 90
Mass number of Th, A = 232
No: of protons = 90
No: of electrons = 90
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 232 – 90 = 142

9. Zinc \({ }_30^{65} \mathrm{Th}\)
Atomic number of Zn, Z = 30
Mass number of Zn, A = 65
No: of protons = 30
No: of electrons = 30
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 65 – 30 = 35

Question 12.
According to the Bohr atom model, where is the electron situated in an atom?
Answer:
Electrons are situated in orbits of an atom.

Question 13.
What are the symbols given to the energy levels 1,2,3, and 4?
Answer:

n	Energy level
1	K
2	M
3	L
4	N

The arrangement of electrons in an atom is done in accordance with certain laws.

Orbit number (n)	Name	Maximum number of electrons that can be accommodated (2n2)
1	K	2 × l2 = 2
2	L	2 × 22 = 8
3	M	2 × 32 = 18
4	N	2 × 42 = 32
5	O	2 × 52 = 50

1. The maximum number of electrons that can be accomodated in an orbit is 2n2, where n is the Table
2. Normally, filling up of electrons in higher energy orbits will take place only after the lower energy orbits are filled.
3. The maximum number of electrons that can be accomodated in the outermost orbit of an atom is 8.

Question 14.
Let us write the electron configuration of some elements. Complete the following table

Answer:

Orbit Electron Configuration-Diagrammatic Representation
The orbit electron configuration of Hydrogen \({ }_{1}^{1} \mathrm{H}\)
No: of electrons in hydrogen = 1

The orbit electron configuration of Boron \({ }_{5}^{10} \mathrm{H}\)
No: of electrons in Boron = 5

Question 15.
Diagrammatically represent the orbit electron configuration of \({ }_{13}^{27} \mathrm{Al}\)
Answer:
The atomic number of aluminium, Z = 13
Mass number of aluminium, A = 27
Neutrons of aluminium = A – Z = 27 – 13 = 14
The orbit electronic’ configuration of aluminium is

Question 16.
The orbit electron configuration of an atom is given.

Analyse the figure and find the following.
(i) Atomic number
(ii) Number of protons
(iii) Number of neutrons
(iv) Mass number
Answer:
(i) Atomic number is the number of protons in an atom.
Atomic number of given atom, Z = 18
(ii) Number of protons (from the figure) = 18
(iii) Number of neutrons (from the figure) = 22
(iv) Mass number of the atom, A = sum of protons and neutron = 18 +22 = 40
(v) Electron configuration = 2,8,8

Question 17.
Write the electron configuration of elements from atomic number 1 to 18 and represent their shell electron configuration.
Answer:

Question 18.
Number of which sub-atomic particle determines the element? (proton/neutron)
Answer:
Proton

Question 19.
See figure given below.

Complete table regarding these atoms.

Answer:

Question 20.
What is the atomic number of these atoms?
Answer:
Atomic number in these atoms is 1.

Question 21.
Which is the element having atomic number 1?
Answer:
Hydrogen is the element with atomic number 1.
Then, all these three are hydrogen atoms.

Question 22.
In the number of which particle do these atoms differ?
Answer:
Neutrons

Question 23.
Are the mass number of these atoms same?
Answer:
Mass number of these atoms are not the same.

Question 24.
Which of them has no neutron in the nucleus?
Answer:
Protium has no nuetron in it.

Question 25.
These are the isotopes of hydrogen. Can you define an isotope?
Answer:
Isotopes are the atoms of the same element with same atomic number and different mass number. Isotopes exhibits the same chemical properties. But they show slight variations in the physical properties. Heavy water is the oxide of deuterium (an isotope of hydrogen) is used in nuclear reactors.

Question 26.
Let us see whether hydrogen alone has isotopes. See the figure given below.

Answer:

\({ }_6^{12} \mathrm{C}\), \({ }_6^{13} \mathrm{C}\), \({ }_6^{14} \mathrm{C}\)C are the three natural isotopes of carbon. Among which the most stable and most abundant is the \({ }_6^{12} \mathrm{C}\) isotope. The amount of ¹³C among the isotopes of carbon is approximately 1.1 %. This isotope is used to study the metabolic processes in plants and animals. ¹⁴C is a radioactive isotope. This is used to determine the age of fossils.

ISOTOPES	USES
Iodine-131	To study the functioning of thyroid gland and its treatment
Uranium-235	Fuel in nuclear reactors
Cobalt-60	Cancer treatment
Sodium-24	To detect the leakage in industrial pipelines
Iron-59	To diagnose Anaemia

Question 27.
Orbit electron configuration of argon (Ar), potassium (K) and calcium (Ca) is given below. Analyse the figure and complete table.


Answer:
From the figure:

Question 28.
What is the pecularity of the mass numbers of these elements?
Answer:
Mass number of the elements Ar, K and Ca are the same.i.e., A = 40

Question 29.
Are the atomic numbers the same?
Answer:
Atomic numbers are not the same for these elements.
Isobars are atoms having the same mass number and different atomic numbers. They are
atoms of different elements in which the number of nucleons (proton  neutron) is equal.
Atoihs having the same number of neutrons are called isotones. Examples: \({ }_6^{14} \mathrm{C}\) and \({ }_7^{15} \mathrm{C}\)

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 9 Real Numbers Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 9 Solutions Real Numbers

Real Numbers Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 9 Real Numbers Solutions Questions and Answers

Class 9 Maths Chapter 9 Kerala Syllabus – Absolute Value

Textual Questions And Answers

Question 1.
Complete the table below:

i) Expand the table by taking some more pairs x, y of numbers. Do you see any relation between \xy\ and |x||y|?
ii) Prove that |xy| = |x||y| for any two numbers x and y.
Answer:

i) If the table is expanded,

From the table, we get that the values of |xy| and |x||y| are equal,

ii) If both the numbers are positive.
Considering the values as, x = 6 and y = 4
Then, |xy| = xy = |x||y|
|(6) (4)| = 24 = |6||4|

If both the numbers are negative.
Considering the values as, x = -6 and y = -4
Then, |xy| – xy = |x||y|
|(-6)(-4)| = 24= |-6||-4|

If one of the number is positive and the another one is negative.
Considering the values as, x = 6 and y = -4
Then, |xy| = xy = |x||y|
|(6)(-4)| = 24= |6||-4|
Hence it is clear that |xy| = |x||y| for any two numbers x and y

Question 2.
Complete the table below:

Expand the table by taking some more pairs x, y of numbers. Do you see any relation between |x| + |y| and |x + y|?
Answer:

If the table is expanded,

From the table it is clear that the values of |x| + |y| is not equal to |x + y|

Class 9 Maths Kerala Syllabus Chapter 9 Solutions – Distances

Textual Questions And Answers

Question 1.
Find x satisfying each of the equations below
i) |x| = 5
ii) |x – 3 | = 2
iii) |x – 2 | = 3
iv) |x + 2 | = 3
Answer:
i) If | x | = 5
Then the values of x will be 5 and -5
That is, x = 5 or x = -5

ii) |x – 3| = 2
If x > 3
Then |x – 3| = x – 3
That means, x – 3 = 2
x = 2 + 3 = 5
x = 5

If x < 3
Then |x – 3| = 3 – x
That means, 3 – x = 2
-x = 2 – 3 = -1
x = 1
If |x – 3| = 2 then the values of x will be x = 1 or x = 5

iii) |x – 2| = 3
If x > 2
Then, |x — 2 | = x – 2
That means, x – 2 = 3
x = 3 + 2 = 5
x = 5

If x < 2
Then |x – 2| = 2 – x
That means, 2 – x = 3
-x = 3 – 2 = 1
x = – 1
If |x – 2 | = 3 then the values of x will be x

iv) |x + 2 | = 3
If x > 2
Then, |x + 2 | = x – (-2)
That means, x – (-2) = 3
x + 2 = 3
x = 3 – 2 = 1
x = 1

If x < 2
Then, |x + 2 | = |x – (-2)| = (-2) – x
That means, (-2) – x = 3
– x = 3 + 2= 5
x = -5
If |x + 2| = 3 then the values of x will be x = 1 or x = -5

Question 2.
Find between which numbers x should lie to satisfy each of the equations below:
i) |x| < 3
ii) |x – 2 | < 1
iii) |x – 11 < 2
iv) |x + 11 < 2
Answer:
i) If |x| = 3
Then the possible values of x will be 3 and -3
But if |x| < 3, then the possible values of x will be in between 3 and -3
That means, —3 < x < 3

ii) |x – 2| < 1
The point x can either be in the left or right of the point 2, (distance from 2 is less than 1).
If x is in the right side of the point 2, whose distance is 1.
Then x – 2 = 1
x = 1 + 2 = 3
x = 3
If x is in the left side of the point 2, whose distance is 1.
Then, 2 – x = 1
– x = 1 – 2 = – 1 .
x = 1
Therefore, |x – 2| < 1, then the possible values of x will be in between 1 and 3
That is, 1 < x < 3.

iii) |x – 1| < 2
The point x can either be in the left or right of the point 1, (distance from 1 is less than 2).
If x is in the right side of the point 1, whose distance is 2.
Then, x – 1 = 2
x = 2 + 1 = 3
x = 3
If x is in the left side of the point 1, whose distance is 2.
Then, 1 – x — 2
-x = 2 – 1 = 1
x = – 1
Therefore, |x – 1| < 2, then the possible values of x will be in between -1 and 3
That is, – 1 < x < 3.

iv) |x + 1| < 2
The point x can either be in the left or right of the point (-1), (distance from (-1) is less than 2).
If x is in the right side of the point (-1), whose distance is 2.
Then, x – (- 1) = 2 – x + 1 = 2
x = 2 – 1 = 1
x = 1
If x is in the left side of the point (-1), whose distance is 2.
Then, – 1 – x = 2
– x = 2 + 1 = 3
x = – 3
Therefore, |x + 1| < 2, then the possible values of x will be in between -3 and 1
That is, – 3 < x < 1.

Question 3.
Find the integers satisfying each of the equations in problem (2).
Answer:
i) |x| < 3
The possible values of x will be in between 3 and -3
Therefore the integer values of x = -2, -1, 0, 1, 2

ii) |x – 2| < 1
The possible values of x will be in between 1 and 3
Therefore the integer values of x – 2

iii) |x – 1| < 2
The possible values of x will be in between -1 and 3
Therefore the integer values of x = 0, 1, 2

iv) |x + 1| < 2
The possible values of x will be in between -3 and 1
Therefore the integer values of x = -2, -1, 0

SCERT Class 9 Maths Chapter 9 Solutions – Midpoint

Intext Questions and Answers

Question 1.
What number x satisfies the equation |x – 2| = |x – 5|
Answer:
Geometrical meaning of the equation |x – 2| = |x – 5| is that the point marked by x on the number line is at the same distance from the points marked by the numbers 1 and 4
That means, the point marked by x is the midpoint of the points marked by 2 and 5
So, x = \(\frac{1}{2}\) × (2 + 5) = \(\frac{7}{2}\)

Textual Questions And Answers

Question 1.
Find the number which mark the midpoint of the points marked by each pair of numbers given below on the number line:
i) 1, -5
ii) -1, -5
iii) –\(\frac{1}{2}\), –\(\frac{1}{3}\)
iv) –\(\frac{1}{3}\), \(\frac{3}{4}\)
v) -2.5, 3.5
vi) 1.3, 8.7
vii) -√2, -√3
viii) -√3, √7
Answer:

Question 2.
Find the number which mark the points dividing the distance between the points marked by 1 and 2 into four equal parts, on the number line.
Answer:
There should be three points which divides the distance between the points marked by 1 and 2 into four equal parts.
To get the three points we have to find the midpoint of
Midpoint of 1 and 2 = \(\frac{1}{2}\)(1 + 2) = \(\frac{3}{2}\)
Midpoint of 1 and \(\frac{3}{2}\) = \(\frac{1}{2}\)(1 + \(\frac{3}{2}\)) = \(\frac{5}{4}\)
Midpoint of \(\frac{3}{2}\) and 2 = \(\frac{1}{2}\)(\(\frac{3}{2}\) + 2) = \(\frac{7}{4}\)

Question 3.
Find x satisfying each of the equations below:
i) |x – 1| = |x – 3|
ii) |x – 3| = |x – 4|
iii) |x + 2| = |x – 5|
iv) |x| = |x + 1|
Answer:
i) x = \(\frac{1}{2}\)(1 + 3) = \(\frac{4}{2}\) = 2
ii) x = \(\frac{1}{2}\)(3 + 4) = \(\frac{7}{2}\)
iii) x = \(\frac{1}{2}\)(-2 + 5) = \(\frac{3}{2}\)
iv) x = \(\frac{1}{2}\)(0 + -1) = \(\frac{-1}{2}\)

Real Numbers Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
Find the absolute value of the following.
i) 7
ii) -11
iii) π
iv) \(\frac{-1}{2}\)
Answer:
i) |7| = 7
ii) |-11| = 11
iii) |π| = π
iv) |\(\frac{-1}{2}\)| = \(\frac{1}{2}\)

Question 2.
Complete the table below:

Expand the table by taking some more pairs x, y of numbers. Do you see any relation between |x – y| and |x| – |y|?
Answer:

If the table is expanded,

From the table it is clear that the values of |x| – |y| is not equal to|x – y|

Question 3.
Complete the missing columns of the given table.

Find a value for |x| = – x
Answer:

From the above table, the negative numbers are used.
Therefore, x = -2

Question 4.
What is the number x for which
i) |x – 2| = 5
ii) |x + 3| = 2
Answer:
i) |x – 2| = 5
If x < 2,
Then |x – 2 | = 2 – x
That means, 2 – x = 5
– x = 5 – 2 = 3
x = – 3

If x > 2,
Then |x – 2| = x – 2
That means, x – 2 = 5
x = 5 + 2 = 7
x = 7
Therefore|x – 2| = 5, then x = – 3 or x – 7

ii) |x + 3| = 2
If x < 3 ,
Then |x + 3| = |x – (-3) | = – 3 – x
That means, -3 – x = 2
– x = 2 + 3 = 5
x = – 5
Therefore |x + 3| = 2 , then x = – 5 or x = -1

Question 5.
Find between which numbers x should lie to satisfy each of the equations |x – 3| < 5 . Also find the integer values of the equation.
Answer:
|x – 3| < 5
The point x can either be in the left or right of the point 3, (distance from 3 is less than 5).
If x is in the right side of the point 3, whose distance is 5.
Then x — 3 = 5
x = 5 + 3 = 8
x = 8
If x is in the left side of the point 3, whose distance is 5.
Then, 3 – x = 5
– x = 5 – 3 = -2
x = – 2
Therefore, |x – 3| < 5, then the possible values of x will be in between -2 and 8
That is, —2 < x <8.
Therefore the possible integer values of x is = -1, 0, 1, 2, 3, 4, 5, 6, 7

Question 6.
Write down the numbers located 3 units apart from 0 on a number line.
Answer:
The number can either be in the left or right of the point 0,
If the number is in the right side of the point 0, whose distance is 3 units.
Then ,0 + 3 = 3
If the number is in the left side of the point 0, whose distance is 3 units.
Then, 0 – 3 = -3
That means, the numbers located 3 units apart from 0 on a number line is -3 and 3.

Question 7.
On a number line, calculate the number that represents the midpoint of each pair of numbers given below,
i) -3,-7
ii) √5, √6
iii) 4\(\frac{1}{9}\), -2\(\frac{3}{4}\)
iv) -1.8, 0.25
Answer:
i) \(\frac{1}{2}\)(-3 + (-7)) = \(\frac{-10}{2}\) = -5
ii) \(\frac{1}{2}\)(√5 + √6)
iii) \(\frac{1}{2}\left(\frac{37}{2}+\frac{-11}{4}\right)=\frac{1}{2} \times \frac{63}{4}=\frac{63}{8}\)
iv) \(\frac{1}{2}\)(-1.8 + 0.25) = 0.775

Question 8.
i) Find x,if |x – 2| = |x – 6|
ii) Find y, if |y – 3| = |y + 1|,
iii) Then find the value of |x – y|?
Answer:
i) x = \(\frac{2+6}{2}=\frac{8}{2}\) = 4
ii) y = \(\frac{3-1}{2}=\frac{2}{2}\) = 1
iii) |x – y| = |4 – 1| = 3

Question 9.
i) What is the distance between the points representing the numbers 2 and 7 on the number line?
ii) Find the value of x that satisfies the equation |x – 2| = |x – 7|.
Answer:
i) Distance = 7 – 2 = 5
ii) x = \(\frac{7+2}{2}=\frac{9}{2}\) = 4.5