1. How many periods and groups are present in the periodic table?

a) 7 periods and 18 groups

b) 8 periods and 7 groups

c) 7 periods and 7 groups

d) 8 periods and 8 groups

Answer: (a) 7 periods and 18 groups

Explanation: Modern periodic table consists of 7 horizontal rows known as periods and 18 vertical columns named as groups.

2. Which of the following forms the basis of the modern periodic table?

a) Atomic mass

b) Atomic number

c) Number of nucleons

d) All of these

Answer: (b) Atomic number

Explanation: Modern periodic table is based on the atomic numbers of elements as according to the modern periodic law the properties of elements are a periodic function of their atomic numbers.

3. What happens to the electropositive character of elements on moving from left to right in a periodic table?

a) Increase

b) Decreases

c) First increases than decreases

d) First decreases than increases

Answer: (b) Decreases

Explanation: Electropositive character of an element is its ability to lose electrons and form positive ions. Now, as on moving from left to right in a period of periodic table, the nuclear charge increases due to the gradual increase in number of protons, so the valence electrons are pulled more strongly by the nucleus. Thus, it becomes more and more difficult for the atoms to lose electrons causing a decrease in the electropositive character of elements on moving from left to right in a periodic table.

4. The electronic configuration of an element M is 2, 8, 4. In modern periodic table, the element M is placed in

a) 4^th group

b) 2^nd group

c) 14^th group

d) 18^th group

Answer: (c) 14^th group

Explanation: In the periodic table, elements having 4 valence electrons are placed in group 14.

5. Which of the following is the correct order of the atomic radii of the elements oxygen, fluorine and nitrogen?

a) O < F < N

b) N < F < O

c) O < N < F

d) F < O < N

Answer: (d) F < O < N

Explanation: Oxygen (8), fluorine (9) and nitrogen (7) belong to the same period of the periodic table, in the order nitrogen, oxygen and fluorine. Now in a period, on moving from left to right the atomic radius of the elements decreases. Therefore, the atomic radius of nitrogen is the largest.

6. What is the other name for group 18^th elements?

a) Noble gases

b) Alkali metals

c) Alkali earth metals

d) Halogens

Answer: (a) Noble gases

Explanation: Group 18^th elements are named as noble gases as they are very stable due to having the maximum number of valence electrons their outermost shell can hold, hence they rarely react with other elements. 

7. Which of the following is the most reactive element of the group 17?

a) Oxygen

b) Sodium

c) Fluorine

d) Magnesium

Answer: (c) Fluorine

Explanation: As we move down in a group, the size of the atoms of elements goes on increasing. So, fluorine being on the top position in the halogen’s group, is the smallest element and has the maximum tendency to gain an electron to complete its octet. Thus fluorine is the most reactive element of the group 17.

8. Element X forms a chloride with the formula XCl₂, which is a solid with a high melting point. X would most likely be in the same group of the Periodic Table as

a) Na

b) Mg

c) Al

d) Si

Answer: (b) Mg

Explanation: Group 2 alkaline earth metal atoms have two valence electrons each. They can donate their two valence electrons to two other chlorine atoms to form the solid compounds of the form XCl₂.

This XCl₂ compound being ionic in nature, has a very strong electrostatic forces of attraction between 2 chloride atoms and 1 metal atom. Thus a large amount of heat is required to break these strong bonds, causing the compound to have very high melting and boiling points.

9. Which group elements are called transition metals?

a) Group number 1 to 2

b) Group number 13 to 18

c) Group number 3 to 12

d) Group number 1 to 8

Answer: (c) Group number 3 to 12

Explanation: The elements occurring in the group 3 to 12 are named as transition metals because they are metallic elements that form a transition between the main group elements, which occur in groups 1 and 2 on the left side, and groups 13–18 on the right side of the periodic table.

10. Which of the following elements has 2 shells and both are completely filled?

a) Helium

b) Neon

c) Calcium

d) Boron

Answer: (b) Neon

Explanation: Neon with the atomic number 10, has the electronic configuration as:

Hence, both its K and L shells are completely filled.

11. Which of the following is the atomic number of an element that forms basic oxide?

a) 18

b) 17

c) 19

d) 15

Answer: (c) 19

Explanation: The elements which can donate their valence electrons to other atoms are the metallic elements which form basic oxides as they give hydroxides in their aqueous solutions.

12. The elements A, B and C belong to group 2, 14 and 16 respectively, of the periodic table. Which of the two elements will form covalent bonds?

a) A and B

b) B and C

c) C and A

d) None of these

Answer: (b) B and C

Explanation: The covalent bond is formed by the sharing of electrons between two atoms. As the element B (which belongs to group 14) has 4 valence electrons which it can share with two elements of C type (from group 16) electrons to complete the octet of each included atom:

13. Which of the following does not decrease while moving down the group of the periodic table?

a) Atomic radius

b) Metallic character

c) Number of shells in the atom

d) Valence electrons

Answer: (d) Valence electrons

Explanation: Number of valence electrons in a group remain the same.

14. An element X belongs to the 3^rd period and 1^st group of the periodic table. What is the number of valence electrons in its atom?

a) 1

b) 3

c) 6

d) 8

Answer: (a) 1

Explanation: As the element belongs to the 1^st group of the periodic table, so the number of valence electrons in its atom is one.

15. An element M is in group 13^th of the periodic table, the formula for its oxide is

a) MO

b) M₂O₃

c) M₃O₂

d) None of these

Answer: (b) M₂O₃

Explanation: As the element M belongs to group 13^th of the periodic table so it has 3 valence electrons, i.e., it can have +3 oxidation state while oxygen atom (with 2 valency) has −2 oxidation state. So the formula for the corresponding oxide is M₂O₃.

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1. Which of the following structures correctly represents the electron dot structure of a chlorine molecule?

Answer: (a)

Explanation: In an electron dot structure of a molecule there must be shown eight electrons (in the form of dots or crosses) around each element of the molecule, to represent the complete octet of the element.

2. While cooking, if the bottom of the vessel is getting blackened on the outside, it means that:

a) The food is not cooked completely

b) The fuel is not burning completely

c) The fuel is wet

d) The fuel is burning completely

Answer: (b) The fuel is not burning completely

Explanation: In case the fuel doesn’t burn completely, i.e., there is not enough oxygen to react with the carbon to produce carbon dioxide, then the unburnt carbon particles are left behind in the form of black particles known as soot. These soot particles stick to the bottom of the vessel making it black.

3. Cation is formed when:

a) Atom gains electrons

b) Atom loses electrons

c) Proton is lost by the atom

d) Atom shares electrons

Answer: (b) Atom loses electrons

Explanation: A cation is formed by loss electrons from the atom of an element which acquires positive charge due to the presence of greater number of protons as compared to that of electrons.

4. The I.U.P.A.C name of CH₃CH₂CH=CH₂ is?

a) 3-Butene

b) Prop-1-ene

c) But-1-ene

d) Butyne

Answer: (c) But-1-ene

Explanation: As the compound, CH₃CH₂CH=CH₂ contains four carbon atoms and a double bond attached to the first carbon, so the I.U.P.A.C name of CH₃CH₂CH=CH₂ is But-1-ene.

5. Which of the following compounds of carbon does not consist of ions?

a) CHCl₃

b) CaCO₃

c) NaHCO₃

d) Ca₂C

Answer: (a) CHCl₃

Explanation: Carbon always forms covalent compounds by sharing its electrons with other atoms. Now, in covalent bonding, the two electrons shared by the atoms are attracted to the nucleus of both atoms and neither atom completely loses or gains electrons as in ionic bonding. So the compounds in which all the atoms are directly attached to C-atom, contain covalent bonding and no ionic bond.

In CHCl₃, all the three chlorine atoms are bonded covalently to the carbon atom, not to the hydrogen atom. So CHCl₃ is a covalent compound and does not consist of ions.

6. The property of self-linkage among identical atoms to form long chain compounds is known as:

a) Catenation

b) Isomerisation

c) Superposition

d) Halogenation

Answer: (a) Catenation

Explanation: Catenation is the property of self-linking of an element by which an atom combines with the other atoms of the same element to form long chains.

7. Which of the following is the molecular formula of cyclobutane?

a) C₄H₁₀

b) C₄H₆

c) C₄H₈

d) C₄H₄

Answer: (c) C₄H₈

Explanation: Cyclobutane is a cyclic hydrocarbon consisting of four carbon atoms where each carbon atom is attached to the two other carbon atoms and two hydrogen atoms, as shown below:

8. Which of the following statements about graphite and diamond is true?

a) They have the same crystal structure

b) They have the same degree of hardness

c) They have the same electrical conductivity

d) They can undergo the same chemical reactions

Answer: (d) They can undergo the same chemical reactions

Explanation: Both Graphite and diamond being the allotropes of the same element , carbon, have similar chemical properties. So they undergo the same chemical reactions.

9. How many number of carbon atoms are joined in a spherical molecule of buckminsterfullerene?

a) 30

b) 60

c) 90

d) 120

Answer: (b) 60

Explanation: Buckminsterfullerene is a molecule of carbon in the form of a hollow sphere consisting of 60 C-atoms and is having the formula C₆₀.

10. Which of the followings is the major constituent of the liquefied petroleum gas?

a) Methane

b) Ethane

c) Propane

d) Butane

Answer: (d) Butane

Explanation: The major constituent of the liquefied petroleum gas is butane.

11. The organic compounds having functional group are known as:

a) Aldehyde

b) Ketone

c) Carboxylic acids

d) Alcohol

Answer: (c) Carboxylic acids

Explanation: Carboxylic acids are compounds which contain a group also known as carboxyl group. 

12. From which of the following substance pencil lead is formed?

a) Charcoal

b) Wood

c) Lead

d) Graphite

Answer: (d) Graphite

Explanation: Pencil lead is formed of graphite. Graphite is an allotropic form of carbon in which each carbon atom is joined to three others, forming layers:

These layers are put together by weak van der Waals forces which enable the layers to slide over each other, making graphite soft and slippery. So graphite is used as pencil ‘lead’. As the pencil moves across the paper, layers of graphite rub off leaving the dark marks on paper.

13. Ester is formed by the reaction between:

a) An acid and an alcohol

b) An acid and a base

c) A base and an alcohol

d) An acid and an alkene

Answer: (a) An acid and an alcohol

Explanation: Reaction between an acid and an alcohol results in the formation of ester, and the reaction is named as estrification.

For example: Acetic acid reacts with ethyl alcohol in the presence of concentrated sulphuric acid to form Ethyl acetate:

14. What is denatured alcohol?

a) Ethyl alcohol which has been made unfit for drinking purpose by adding small amount of poisonous substance

b) Methyl alcohol which has been made unfit for drinking purpose by adding small amount of poisonous substance

c) Alcohol having properties of an acid

d) Ethyl alcohol containing 60% of water by weight

Answer: (a) Ethyl alcohol which has been made unfit for drinking purpose by adding small amount of poisonous substance

Explanation: Denatured alcohol is the ethyl alcohol which has been made unfit for drinking purpose by adding small amount of poisonous substance like methanol, pyridine, etc. This is mainly done to prevent the misuse of industrial alcohol for drinking purposes.

15. Which of the following substance produces brisk effervescence with baking soda solution?

a) Ethanoic acid

b) Table salt

c) Vinegar

d) Sunflower oil

Answer: (a) Ethanoic acid

Explanation: Ethanoic acid when treated with baking soda (Sodium hydrogencarbonate) gives brisk effervescence of Carbon dioxide gas.

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1. The non-metal which is liquid at room temperature is:

a) Mercury

b) Bromine

c) Carbon

d) Helium

Answer: (b) Bromine

Explanation: Bomine is the only non-metal which exists as a liquid at room temperature.

2. The number of protons in an atom of an element A is 19 then, the number of electron in its ion A⁺ is:

a) 18

b) 19

c) 20

d) 21

Answer: (a) 18

Explanation: In the neutral atom of an element,

3. Bauxite is an ore of

a) Iron

b) Aluminium

c) Mercury

d) Copper

Answer: (b) Aluminium

Explanation: Bauxite is a commercial ore of aluminium which consists largely of hydrated aluminium oxide, Al₂O₃.2H₂O. 

4. The metal which is liquid at room temperature is

a) Bromine

b) Mercury

c) Iodine

d) Potassium

Answer: (b) Mercury

Explanation: Mercury is the only metal which exists as a liquid at room temperature.

5. The sulphide ores are converted into oxides by heating strongly in the presence of excess air. This process is known as

a) Roasting

b) Smelting

c) Calcination

d) Refining

Answer: (a) Roasting

Explanation: The process of heating the sulphide ore strongly in the presence of air to convert it into metal oxide, is known as roasting.

6. In electrolytic refining, the cathode is made up of

a) Pure metal

b) Impure metal

c) Alloy

d) Metallic salt

Answer: (a) Pure metal

Explanation: In electrolytic refining of a metal, the cathode is made up of pure metal whereas the anode is made up of impure metal.

7. In the given reaction, Al₂O₃ + NaOH   →  ……X……   +  H₂O

What is element X?

a) NaAlO₂

b) Na₃Al

c) Na₂O₃

d) NaAl₂O₃

Answer: (a) NaAlO₂

Explanation: Aluminium oxide is amphoteric in nature, i.e., it reacts with acids as well as bases to form salt and water.

Here, aluminium oxide behaves as an acid as it reacts with NaOH, a base and forms sodium aluminate (NaAlO₂) and water:

                                   Al₂O₃ + NaOH   →  2NaAlO₂+  H₂O

8. Which of the following represent the correct order of decreasing reactivity?

a) Mg > Al > Zn > Fe

b) Mg > Zn > Al > Fe

c) Al > Zn > Fe > Mg

d) Mg > Fe > Zn > Al

Answer: (a) Mg > Al > Zn > Fe

Explanation: The decreasing order of the reactivity of the common metals is given below:

            Li, K, Na, Ba, Ca, Mg , Al, Mn, Zn, Fe, Ni, Sn, Pb, [H], Cu,Hg, Ag, Au,Pt 

9. An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be

(a) Ca

(b) C

(c) Si

(d) Fe

Answer: (a) Ca

Explanation:                         

Calcium reacts with oxygen to give calcium oxide (CaO) which is having a high melting point and dissolves in water to  form calcium hydroxide (Ca(OH)₂)along with the release of large amount of thermal energy.

10. Which of the following pairs will give displacement reactions?

(a) NaCl solution and copper metal

(b) MgCl₂ solution and aluminium metal

(c) FeSO₄ solution and silver metal

(d) AgNO₃ solution and copper metal

Answer: (d) AgNO₃ solution and copper metal

Explanation: Copper (Cu) being more reactive than silver (Ag), displaces silver from silver nitrate (AgNO₃) to form copper nitrate

                                                                 2AgNO₃ + Cu  →  Cu(NO₃)₂+  2Ag

11. Which among the following is the most abundant metal found in the earth’s crust?

(a) Magnesium

(b) Aluminium

(c) Oxygen

(d) Iron

Answer: (b) Aluminium

Explanation: Aluminium is the most abundant metal found in the earth’s crust.

12. Which of the following pairs of reactants will go undergo a displacement reaction?

(a) CuSO₄ + Fe 

(b) ZnSO₄ + Fe

(c) MgSO₄ + Fe

(d) Ca(SO₄)₂ + Fe

Answer: (a) CuSO₄ + Fe

Explanation: As per the reactivity series of metals, iron is more reactive than copper metal so it can displace copper from copper sulphate solution and form iron (II) sulphate and copper:                     

13. Galvanisation is a method of protecting steel and iron from rusting by coating them with a thin layer of

(a) Copper

(b) Aluminum

(c) Zinc

(d) Bauxite

Answer: (c) Zinc

Explanation: In this method a thin layer of zinc metal is deposited over the surface of steel or iron objects, which does not corrode on exposure to damp air and prevents the coated metals from rusting.

14. Which of the following alloys contains a non-metal as one of its constituents?

(a) Steel

(b) Brass

(c) Amalgam

(d) Bronze

Answer: (a) Steel

Explanation: Stainless steel is an alloy of iron (a metal) and carbon (a non metal).

15. An element X is soft and can be cut with the help of a knife. It is very reactive to air and cannot be kept open in the air. It reacts vigorously with water. Identify the element from the following:

(a) Mg

(b) Na

(c) P

(d) Ca

Answer: (b) Na

Explanation: Na is a metal which is soft enough to be cut with a knife. It is so reactive that it reacts vigorously with air or moisture and catches fire when kept in open. So to prevent it from coming in contact with oxygen and moisture, it is kept in kerosene.

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1.Sodium carbonate is a basic salt because it is a salt of
(a) weak acid and weak base
(b) strong acid and weak base
(c) weak acid and strong base

Answer: (c) weak acid and strong base.
Explation= Sodium carbonate is a basic salt which is formed by the combination of sodium hydroxide and carbonic acid.

Sodium hydroxide is a strong base, while cabonic acid is a weak acid resulting in the formation of sodium carbonate salt.

2NaOH+H2CO3→Na2CO3+2H2O

2. Which gas is evolved when acids react with metals?
   (a) O2
   (b) CO2
   (c) H2
   (d) N2
   Answer: (c) H2
   Metals readily react with acid and produce hydrogen gas which burns with a pop sound.
3. Dilute acid does not produce carbon dioxide on being treated with: a. Marble
   b. Lime
   c. Baking soda
   d. Limestone
   Answer. b. Lime
   In the given options, marble, limestone, and baking soda are all either metal carbonates or bicarbonates. All these when react with a dilute acid then they produce carbon dioxide. Whereas lime which is calcium oxide when react with a dilute acid then they do not produce carbon dioxide.
4. Which of the following salt will give acidic solution when dissolved in water?
   a. NH4Cl
   b. NaCl
   c. Na2CO3
   d. CH3COONa
   Answer. a. NH4Cl
   NaCl is a neutral salt hence, it will give a neutral solution. NH4Cl is an acidic salt hence, it will give an acidic solution. CH3COONa is a basic salt hence, it will give a basic solution.
5. Bleaching powder is used as a disinfectant for water to: a. Make water tastier
   b. Remove all the dirt from water
   c. Make water germ-free
   d. Make water clear
   Answer. Make water germ-free

Bleaching powder, CaOCl2 is used to kill germs and bacteria of water because it is a good disinfecting agent

5. Which one of the following salts will dissolve in water to form an alkaline solution?
   a. Potassium carbonate
   b. Sodium chloride
   c. Sodium carbonate
   d. Potassium sulphate
   Answer. a. Potassium carbonate
   Potassium carbonate will dissolve in water to give alkaline
   solution. This is because they are formed by strong base and weak acid and thus are basic in nature
6. Which among the following represents the chemical formula for ‘Plaster of Paris’?

Explanition= The chemical formula for the plaster of Paris is (CaSO4) H2O and is better known as calcium sulfate hemihydrate.

7. The nature of calcium phosphate is present in tooth enamel is
   (a) Basic
   (b) Amphoteric
   (c) Acidic
   (d) NeutralRead
   ANSWER .Basic
   Calcium phosphate is a family of substances and minerals which consist of calcium ions ( Ca2 + Ca2 + ) with inorganic phosphate anions. In some calcium phosphates, oxides and hydroxides are also present. They are white solids of nutritious value. Calcium phosphates are found in many living organisms.
   Calcium phosphate is basic salt since it is a source of weak phosphoric acid and a slightly stronger base of calcium hydroxide. Calcium phosphate is a mineral comprising calcium ion and phosphate ion which is inorganic in nature. It is present in the crown area of the tooth and also in bones.
   Although calcium phosphate is a very hard substance, it can be dissolved in acids. Hence, we can state that the nature of calcium phosphate is basic and hence, the correct option is option (A).
8. Which one of the following is acidic?
   (a) Lemon juice
   (b) Tomatoes
   (c) Milk
   (d) All
   Answer . All
9. Acids react with metals to liberate _________gas
   (a) Carbon dioxide
   (b) Carbon monoxide
   (c) Hydrogen
   (d) Water
   Answer . Hydrogen
   Explation= acid reacts with metal, hydrogen gas is released.
10. . Generally, when certain metals react with an acid they release _ gas.
    A. Nitrogen
    B. Oxygen
    C. Hydrogen
    D. Argon
    Answer. Hydrogen
11. Which one of the given is commonly known as blue vitriol and is used as a fungicide?
    A. Potassium nitrate

B. Copper sulphate

C. Sodium carbonate

D. Sodium chloride
Answer Copper sulphate
Copper(II) sulfate, also known as cupric sulfate, or copper sulphate, is the inorganic compound with the chemical formula CuSO4(H2O)x, where x can range from 0 to 5. The pentahydrate (x = 5) is the most common form. Older names for this compound include blue vitriol, bluestone, the vitriol of copper, and Roman vitriol.

12. Vinegar is used in pickling as it

A. Is an acid

B. Prevents the growth of microbes

C. Prevents drying of a pickle

D. Increases taste
Answer= Prevents the growth of microbes

13. PH scale of a neutral solution is
    A. 14
    B. 7
    C. 10
    D. 12
    Answer. 7
    The pH scale ranges from 0 to 14. A pH of 7 is neutral. A pH less than 7 is acidic.
14. Butyric acid is found in
    A. Rancid butter
    B. Rancid cake
    C. Stings of bees
    D. All of these
    Answer. Rancid cake
15. An indicator is one kind of the following compound
    A. Strong acid only
    B. Reducing agent
    C. Weak base or acid only
    D. Complex salt
    Answer= Weak base or acid only
    indicators are chemical based compounds used to identify the acidity/alkalinity of a given compound so they should be a very weak acid /base

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1.In which mode of nutrition an organism de-rives its food from the body of another living organism without killing it?
(a) Saprotrophic nutrition
(b) Parasitic nutrition
(c) Holozoic nutrition
(d) Autotrophic nutrition

Answer: b

2. The mode of nutrition found in fungi is:
(a) Parasitic nutrition
(b) Holozoic nutrition
(c) Autotrophic nutrition
(d) Saprotrophic nutrition

Answer (c )

the mode of nutrition found in fungi is saprotrophic nutrition.

3. The site of photosynthesis in the cells of a leaf is
(a) chloroplast
(b) mitochondria
(c) cytoplasm
(d) protoplasm

Answer. (a) chloroplastThe process of photosynthesis takes place in the chloroplasts, using

chlorophyll, the green pigment involved in photosynthesis.

 4.In amoeba, food is digested in the:
(a) food vacuole
(b) mitochondria
(c) pseudopodia
(d) chloroplast

Answer. (a) food vacuole

 Food vacuole is formed when food is engulfed through phagocytosis.

5.The contraction and expansion movement of the walls of the food pipe is called:
(a) translocation
(b) transpiration
(c) peristaltic movement
(d) digestion

Answer. (c)

Peristalsis Is the Contraction of Muscle Tissue That Helps Move and Break Down Foodstuffs

6. . When a few drops of iodine solution are added to rice water, the solution turns blue- black in colour. This indicates that rice water contains:
(a) fats
(b) complex proteins
(c) starch
(d) simple proteins

Amylose in starch is the main factor for colour formation as iodine comes in contact with beta coils structure of amylose and gives blue – black colour with iodine. Thus, the reaction of iodine with rice water indicates that rice water contains starch.

7. The exit of unabsorbed food material is regu-lated by
(a) liver
(b) anus
(c) small intestine
(d) anal sphincter

(d) anal sphincter

Anal sphincter regulates the unabsorbed food , which releases it in small amounts into the small intestine.

8. The breakdown of pyruvate to give carbon di-oxide, water and energy takes place in
(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) nucleus

Answer.

(b) mitochondriaThe break down of pyruvate to give carbon dioxide, energy and water takes place in the presence of oxygen and is termed as aerobic respiration. This process takes place in the mitochondria of a cell

9.The kidneys in human beings are a part of the system for

 (a)  nutrition.

(b)  respiration.

(c)  excretion.

(d)  transportation.

Ans. (c) excretion.

Excretion refers to removal of metabolic waste from body. This makes kidneys part of excretory system.

10.  The xylem in plants are responsible for

     (a)  transport of water.

     (b)  transport of food.

     (c)  transport of amino acids.

     (d) transport of oxygen.

    Ans. (a) transport of water.

The role of xylem tissue in plants is to transport water and minerals.

11.  The autotrophic mode of nutrition requires

         (a)  carbon dioxide and water.

          (b)  chlorophyll.

          (c)  sunlight.

          (d)  all of the above.

Ans. (d) All of the above.

12.  The breakdown of pyruvate to give carbon dioxide, water and energy takes place in

          (a)  cytoplasm.

          (b)  chloroplast.

          (c) mitochondria.

(d)  nucleus.

Ans. (c) mitochondria.

The break down of pyruvate to give carbon dioxide, energy and water takes place in the presence of oxygen and is termed as aerobic respiration. This process takes place in the mitochondria of a cell.

 13.Name the pores in a leaf through which respi-ratory exchange of gases takes place.

(a) Lenticels
(b) Vacuoles
(c) Xylem
(d) Stomata

Answer.

(d) Stomata

The pore through which the exchange of respiratory gases takes place is known as stomata. These are the tiny pores that are mainly found under the surfaces of the leaves, epidermis, stems etc.

14.Which plant tissue transports water and min-erals from the roots to the leaf?
(a) Xylem
(b) Phloem
(c) Parenchyma
(d) Collenchyma

Answer. XylemXylem tissue transports water and nutrients from the roots to different parts of the plant, and also plays a

role in structural support in the stem.

15. The movement of food in phloem is called:
(a) transpiration
(b) translocation
(c) respiration
(d) evaporation

Answer. (b) translocation

The transport of food in plants is called translocation. It takes place with the help of a conducting tissue called phloem. Phloem transports glucose, amino acids and other substances from leaves to root, shoot, fruits and seeds.

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1. For some integer m, every odd integer is of the form

(A) m                                      

(B) m + 1

(C) 2m                                    

(D) 2m + 1

Answer:  D

Explanation: As the number 2m will always be even, so if we add 1 to it then, the number will always be odd.

2. If two positive integers a and b are written as a = p³q² and b = pq³; p, q are prime numbers, then HCF (a, b) is:

(A) pq                                                 

(B) pq²

(C) p³q³                                               

(D) p²q²

Answer:  B

Explanation:  Since a = p × p × p × q × q,

                                           b = p × q × q × q

Therefore H.C.F of a and b = pq²

3. The product of a non-zero number and an irrational number is:

(A) always irrational                          

(B) always rational

(C) rational or irrational                     

(D) one

Answer:  A

Explanation: Product of a non-zero rational and an irrational number is always irrational i.e.,

4. If the HCF of 65 and 117 is expressible in the form 65 m – 117, then the value of m is

(A) 4                                                   

(B) 2

(C) 1                                                   

(D) 3

Answer: B

Explanation: By Euclid’s division algorithm,

5 The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is

(A) 13                                                             

(B) 65

(C) 875                                                           

(D) 1750

Answer: A

Explanation: Since 5 and 8 are the remainders of 70 and 125, respectively. Thus after subtracting these remainders from the numbers, we have the numbers

65 = (70 − 5), 117 = (125 − 8) which is divisible by the required number.

Now required number = H.C.F of (65,117)

6. If two positive integers p and q can be expressed as p = ab² and q = a³b; a, b being prime numbers, then LCM (p, q) is

(A) ab                                     

(B) a²b²

(C) a³b²                                   

(D) a³b³

Answer: C

Explanation:

p = a × b × b

q = a × a × a × b

Since L.C.M is the product of the greatest power of each prime factor involved in the numbers

Therefore, L.C.M of p and q = a³b²

7. The values of the remainder r, when a positive integer a is divided by 3 are:

(A) 0, 1, 2, 3                           

(B) 0, 1

(C) 0, 1, 2                               

(D) 2, 3, 4

Answer: C

Explanation:

According to Euclid’s division lemma,

a = 3q + r, where 0  r < 3

As the number is divided by 3.So the remainder cannot be greater than divisor 3 also r is an integer. Therefore, the values of r can be 0, 1 or 2.

(A) one decimal place       

(A) Terminating decimal expansion   

(B) Non–Terminating Non repeating decimal expansion       

(C) Non–Terminating repeating decimal expansion   

(D) None of these

Answer: A

Explanation: After simplification,

As the denominator has factor 5³ × 2² and which is of the type 5^m × 2ⁿ, So this is a terminating decimal expansion.

10. A rational number in its decimal expansion is 327.7081. What would be the prime factors of q when the number is expressed in the p/q form?

(A) 2 and 3                                         

(B) 3 and 5

(C) 2, 3 and 5                                     

(D) 2 and 5

Answer: D

Explanation: This can be explained as,

11. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(A) 10                                                             

(B) 100

(C) 2060                                                         

(D) 2520

Answer: D

Explanation: Factors of 1 to 10 numbers

L.C.M of numbers from 1 to 10 is =

12. n² – 1 is divisible by 8, if n is

(A) an integer                               

(B) a natural number

(C) an odd integer greater than 1

(D) an even integer

Answer: C

Explanation: n can be even or odd

Case 1: If n is even

Case 2: If n is odd

Which is divisible by 8.

Similarly we can check for any integer.

13. If n is a rational number, then 5²ⁿ − 2²ⁿ is divisible by

(A) 3                                                                           

(B) 7

(C) Both 3 and 7                                                        

(D) None of these

Answer: C

Explanation:

5²ⁿ −2²ⁿ is of the form a²ⁿ − b²ⁿ which is divisible by both (a + b) and (a – b).

So, 5²ⁿ − 2²ⁿ is divisible by both 7, 3.

14. The H.C.F of 441, 567 and 693 is

(A) 1                                                                           

(B) 441

(C) 126                                                                       

(D) 63

Answer: D

Explanation:

693 = 3×3×7×7

567 = 3×3×3×3×7

441 = 3×3×7×11

Therefore H.C.F of 693, 567 and 441 is 63.

15. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

(A) 2520cm                                                                

(B) 2525cm

(C) 2555cm                                                                

(D) 2528cm

Answer: A

Explanation: We need to find the L.C.M of 40, 42 and 45 cm to get the required minimum distance.

40 = 2×2×2×5

42 = 2×3×7

45 = 3×3×5

L.C.M. = 2×3×5×2×2×3×7 = 2520

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MCQs: Real Numbers

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Question 1.
In the given figure, AD = BC and BD = AC, prove that ∠DAB = ∠CBA.
Solution:

In ∆DAB and ∆CBA, we have
AD = BC [given]
BD = AC [given]
AB = AB [common]
∴ ∆DAB ≅ ∆CBA [by SSS congruence axiom]
Thus, ∠DAB =∠CBA [c.p.c.t.]

Question 2.
In the given figure, ∆ABD and ABCD are isosceles triangles on the same base BD. Prove that ∠ABC = ∠ADC.
Solution:

In ∆ABD, we have
AB = AD (given)
∠ABD = ∠ADB [angles opposite to equal sides are equal] …(i)
In ∆BCD, we have
CB = CD
⇒ ∠CBD = ∠CDB [angles opposite to equal sides are equal] … (ii)
Adding (i) and (ii), we have
∠ABD + ∠CBD = ∠ADB + ∠CDB
⇒ ∠ABC = ∠ADC

Question 3.
In the given figure, if ∠1 = ∠2 and ∠3 = ∠4, then prove that BC = CD.
Solution:

In ∆ABC and ACDA, we have
∠1 = ∠2 (given)
AC = AC [common]
∠3 = ∠4 [given]
So, by using ASA congruence axiom
∆ABC ≅ ∆CDA
Since corresponding parts of congruent triangles are equal
∴ BC = CD

Question 4.
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:
Here, ∠B < ∠A
⇒ AO < BO …..(i)
and ∠C < ∠D
⇒ OD < CO …..(ii)
[∴ side opposite to greater angle is longer]
Adding (i) and (ii), we obtain
AO + OD < BO + CO
AD < BC

Question 5.
In the given figure, AC > AB and D is a point on AC such that AB = AD. Show that BC > CD.
Solution:

Here, in ∆ABD, AB = AD
∠ABD = ∠ADB
[∠s opp. to equal sides of a ∆]
In ∆BAD
ext. ∠BDC = ∠BAD + ∠ABD
⇒ ∠BDC > ∠ABD ….(ii)
Also, in ∆BDC .
ext. ∠ADB > ∠CBD …(iii)
From (ii) and (iii), we have
∠BDC > CD [∵ sides opp. to greater angle is larger]

Question 6.
In a triangle ABC, D is the mid-point of side AC such that BD = 12 AC. Show that ∠ABC is a right angle.
Solution:

Here, in ∆ABC, D is the mid-point of AC.
⇒ AD = CD = 12AC …(i)
Also, BD = 12AC… (ii) [given]
From (i) and (ii), we obtain
AD = BD and CD = BD
⇒ ∠2 = ∠4 and ∠1 = ∠3 …..(iii)
In ∆ABC, we have
∠ABC + ∠ACB + ∠CAB = 180°
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠1 + ∠2 + ∠1 + ∠2 = 180° [using (iii)]
⇒ 2(∠1 + ∠2) = 180°
⇒ ∠1 + ∠2 = 90°
Hence, ∠ABC = 90°

Question 7.
ABC is an isosceles triangle with AB = AC. P and Q are points on AB and AC respectively such that AP = AQ. Prove that CP = BQ.
Solution:

In ∆ABQ and ∆ACP, we have
AB = AC (given)
∠BAQ = ∠CAP [common]
AQ = AP (given)
∴ By SAS congruence criteria, we have
∆ABQ ≅ ∆ACP
CP = BQ

Question 8.
In the given figure, ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC, AD is extended to intersect BC at P. Show that : (i) ∆ABD ≅ ∆ACD (ii) ∆ABP ≅ ∆ACP

Solution:
(i) In ∆ABD and ∆ACD
AB = AC [given]
BD = CD [given]
AD = AD [common)]
∴ By SSS congruence axiom, we have
∆ABD ≅ ∆ACD
(ii) In ∆ABP and ∆ACP
AB = AC [given]
∠BAP = ∠CAP [c.p.cit. as ∆ABD ≅ ∆ACD]
AP = AP [common]
∴ By SAS congruence axiom, we have
∆ABP ≅ ∆ACP

Question 9.
In the given figure, it is given that AE = AD and BD = CE. Prove that ∆AEB ≅ ∆ADC.

Solution:
We have AE = AD … (i)
and CE = BD … (ii)
On adding (i) and (ii),
we have AE + CE = AD + BD
⇒ AC = AB
Now, in ∆AEB and ∆ADC,
we have AE = AD [given]
AB = AC [proved above]
∠A = ∠A [common]
∴ By SAS congruence axiom, we have
∆AEB = ∆ADC

Question 10
In the given figure, in ∆ABC, ∠B = 30°, ∠C = 65° and the bisector of ∠A meets BC in X. Arrange AX, BX and CX in ascending order of magnitude.

Solution:
Here, AX bisects ∠BAC.
∴ ∠BAX = ∠CAX = x (say)
Now, ∠A + ∠B + C = 180° [angle sum property of a triangle]
⇒ 2x + 30° + 65° = 180°
⇒ 2x + 95 = 180°
⇒ 2x = 180° – 95°
⇒ 2x = 85°
⇒ x = 85∘2 = 42.59
In ∆ABX, we have x > 30°
BAX > ∠ABX
⇒ BX > AX (side opp. to larger angle is greater)
⇒ AX < BX
Also, in ∆ACX, we have 65° > x
⇒ ∠ACX > ∠CAX
⇒ AX > CX [side opp. to larger angle is greater]
⇒ CX > AX … (ii)
Hence, from (i) and (ii), we have
CX < AX < BX

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Question 1.
Define :
(a) a square (b) perpendicular lines.
Solution:
(a) A square : A square is a rectangle having same length and breadth. Here, undefined terms are length, breadth and rectangle.
(b) Perpendicular lines : Two coplanar (in a plane) lines are perpendicular, if the angle between them at the point of intersection is one right angle. Here, the term one right angle is undefined.

Question 2.
In the given figure, name the following :
(i) Four collinear points
(ii) Five rays
(iii) Five line segments
(iv) Two-pairs of non-intersecting line segments.

Solution:
(i) Four collinear points are D, E, F, G and H, I, J, K
(ii) Five rays are DG, EG, FG, HK, IK.
(iii) Five line segments are DH, EI, FJ; DG, HK.
(iv) Two-pairs of non-intersecting line segments are (DH, EI) and (DG, HK).

Question 3.
In the given figure, AC = DC and CB = CE. Show that AB = DE. Write the Euclid’s axiom to support this.

Solution:
We have
AC = DC
CB = CE
By using Euclid’s axiom 2, if equals are added to equals, then wholes are equal.
⇒ AC + CB = DC + CE
⇒ AB = DE.

Question 4.
In figure, it is given that AD=BC. By which Euclid’s axiom it can be proved that AC = BD?

Solution:
We can prove it by Euclid’s axiom 3. “If equals are subtracted from equals, the remainders are equal.”
We have AD = BC
⇒ AD – CD = BC – CD
⇒ AC = BD

Question 5.
In the given figure, AB = BC, BX = BY, show that
AX = CY.

Solution:
Given that AB = BC
and BX = BY
By using Euclid’s axiom 3, equals subtracted from equals, then the remainders are equal, we have
AB – BX = BC – BY
AX = CY

Question 6.

In the above figure, if AB = PQ, PQ = XY, then AB = XY. State True or False. Justify your answer.
Solution:
True. ∵ By Euclid’s first axiom “Things which are equal to the same thing are equal to one another”.
∴ AB = PQ and XY = PQ ⇒ AB = XY

Question 7.
In the given figure, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4, write the relation between ∠1 and ∠2, using an Euclid’s axiom.

Solution:
Here, ∠3 = ∠4, ∠1 = ∠3 and ∠2 = ∠4. Euclid’s first axiom says, the things which are equal to equal thing are equal to one another. So ∠1 = ∠2.,

Question 8.
In the given figure, we have ∠1 = ∠2, ∠3 = ∠4. Show that ∠ABC = ∠DBC. State the Euclid’s Axiom used.
Solution:
Here, we have 1 = ∠2 and ∠3 = ∠4. By using Euclid’s Axiom 2. If equals are added to
equals, then the wholes are equal..
∠1 + ∠3 = ∠2 + ∠4
∠ABC = ∠DBC.

Question 9.
In the figure, we have BX and 12 AB =12 BC. Show that BX = BY.

Solution:
Here, BX = 12 AB and BY = 12 BC …(i) [given]
Also, AB = BC [given]
⇒ 12AB = 12BC …(ii)
[∵ Euclid’s seventh axiom says, things which are halves of the same thing are equal to one another]
From (i) and (ii), we have BX = BY

Question 10.
In the given figure, AC = XD, C is mid-point of AB and D is mid-point of XY. Using an Euclid’s axiom, show that AB = XY.

Solution:
∵ C is the mid-point of AB
AB = 2AC
Also, D is the mid-point of XY
XY = 2XD
By Euclid’s sixth axiom “Things which are double of same things are equal to one another.”
∴ AC = XD = 2AC = 2XD ⇒ AB = XY

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Question 1.
Represent √3 on the number line.
Solution:


On the number line, take OA = 1 unit. Draw AB = 1 unit perpendicular to OA. Join OB.
Again, on OB, draw BC = 1 unit perpendicular to OB. Join OC.
By Pythagoras Theorem, we obtain OC = √3. Using
compasses, with centre O and radius OC, draw an arc, which intersects the number line at point
D. Thus, OD = √3 and D corresponds to √3.

Question 2.
Represent √3.2 on the number line.
Solution:
First of all draw a line of length 3.2 units such that AB = 3.2 units. Now, from point B, mark a distance of 1 unit. Let this point be ‘C’. Let ‘O’ be the mid-point of the distance AC. Now, draw a semicircle with centre ‘O’ and radius OC. Let us draw a line perpendicular to AC passing through the point ‘B’ and intersecting the semicircle at point ‘D’.
∴ The distance BD = √3.2

Now, to represent √3.2 on the number line. Let us take the line BC as number line and point ‘B’ as zero, point ‘C’ as ‘1’ and so on. Draw an arc with centre B and radius BD, which intersects the number line at point ‘E’.
Then, the point ‘E’ represents √3.2.

Question 3.
Express 1.32 + 0.35 as a fraction in the simplest form.
Solution:
Let . x = 1.32 = 1.3222…..(i)

Multiplying eq. (i) by 10, we have
10x = 13.222…
Again, multiplying eq. (i) by 100, we have
100x = 132.222… …(iii)
Subtracting eq. (ii) from (iii), we have
100x – 10x = (132.222…) – (13.222…)
90x = 119
⇒ x = 11990
Again, y = 0.35 = 0.353535……
Multiply (iv) by 100, we have …(iv)
100y = 35.353535… (v)
Subtracting (iv) from (u), we have
100y – y = (35.353535…) – (0.353535…)
99y = 35
y = 3599

Question 4.
Find the square root of 10 + √24 + √60 + √40.
Solution:

Question 5.
If x = 9 + 4√5, find the value of √x – 1x√.
Solution:
Here,
x = 9 + 4√5
x = 5 + 4 + 2 x 2√5
x = (√5² + (2² + 2 x 2x √5).
x = (√5 + 2)²
√x = √5 + 2
Now, 1x√ = 15√+2

Question 6.
If x = 15√−2 , find the value of x³ – 3² – 5x + 3
Solution:

∴ x – 2 = √5
Squaring both sides, we have
x² – 4x + 4 = 5
x² – 4x – 1 = 0 …(i)
Now, x³ – 3² – 5x + 3 = (x² – 4x – 1) (x + 1) + 4
= 0 (x + 1) + 4 = 4 [using (i)]

Question 7.
Find ‘x’, if 2^x-7 × 5^x-4 = 1250.
Solution:
We have 2^x-7 × 5^x-4 = 1250
⇒ 2^x-7 × 5^x-4 = 2 5 × 5 × 5 × 5
⇒ 2^x-7 × 5^x-4 = 21 × 54
Equating the powers of 2 and 5 from both sides, we have
⇒ x – 7 = 1 and x – 4 = 4
⇒ x = 8 and x = 8
Hence, x = 8 is the required value.

Question 8.
Evaluate:

Solution:

`Question 9.
If x = p+q√+p−q√p+q√−p−q√, then prove that q² – 2px + 9 = 0.
Solution:


Squaring both sides, we have
⇒ q²x² + p² – 2pqx = p² – q²
⇒ q²x² – 2pqx + q² = 0
⇒ q(q² – 2px + q) = 0
⇒ qx² – 2px + q = 0 (∵ q ≠ 0)

Question 10.
If a = 13−11√ and b = 1a, then find a² – b²
Solution:

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Question 1.
In the given figure, AB || CD, ∠FAE = 90°, ∠AFE = 40°, find ∠ECD.

Solution:
In AFAE,
ext. ∠FEB = ∠A + F
= 90° + 40° = 130°
Since AB || CD
∴ ∠ECD = FEB = 130°
Hence, ∠ECD = 130°.

Question 2.
In the fig., AD and CE are the angle bisectors of ∠A and ∠C respectively. If ∠ABC = 90°, then find ∠AOC.

Solution:
∵ AD and CE are the bisector of ∠A and ∠C

In ∆AOC,
∠AOC + ∠OAC + ∠OCA = 180°
⇒ ∠AOC + 45o = 180°
⇒ ∠AOC = 180° – 45° = 135°.

Question 3.
In the given figure, prove that m || n.

Solution:
In ∆BCD,
ext. ∠BDM = ∠C + ∠B
= 38° + 25° = 63°
Now, ∠LAD = ∠MDB = 63°
But, these are corresponding angles. Hence,
m || n

Question 4.
In the given figure, two straight lines PQ and RS intersect each other at O. If ∠POT = 75°, find the values of a, b, c.

Solution:
Here, 4b + 75° + b = 180° [a straight angle]
5b = 180° – 75° = 105°
b – 105∘5 = 21°
∴ a = 4b = 4 × 21° = 84° (vertically opp. ∠s]
Again, 2c + a = 180° [a linear pair]
⇒ 2c + 84° = 180°
⇒ 2c = 96°
⇒ c = 96∘2 = 48°
Hence, the values of a, b and c are a = 84°, b = 21° and c = 48°.

Question 5.
In figure, if AB || CD. If ∠ABR = 45° and ∠ROD = 105°, then find ∠ODC.

Solution:

Through O, draw a line ‘l’ parallel to AB.
⇒ line I will also parallel to CD, then
∠1 = 45°[alternate int. angles]
∠1 + ∠2 + 105° = 180° [straight angle]
∠2 = 180° – 105° – 45°
⇒ ∠2 = 30°
Now, ∠ODC = ∠2 [alternate int. angles]
= ∠ODC = 30°

Question 6.
In the figure, ∠X = 72°, ∠XZY = 46°. If YO and ZO are bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OYZ and ∠YOZ.

Solution:
In ∆XYZ, we have
∠X + XY + ∠Z = 180°
⇒ ∠Y + ∠Z = 180° – ∠X
⇒ ∠Y + ∠Z = 180° – 72°
⇒ Y + ∠Z = 108°
⇒ 12 ∠Y + 12∠Z = 12 × 108°
∠OYZ + ∠OZY = 54°
[∵ YO and ZO are the bisector of ∠XYZ and ∠XZY]
⇒ ∠OYZ + 12 × 46° = 54°
∠OYZ + 23° = 54°
⇒ ∠OYZ = 549 – 23° = 31°
In ∆YOZ, we have
∠YOZ = 180° – (∠OYZ + ∠OZY)
= 180° – (31° + 23°) 180° – 54° = 126°

Question 7.
In the given figure, m and n are two plane mirrors perpendicular to each other. Show that incident rays CA is parallel to reflected ray BD.

Solution:
Let normals at A and B meet at P.
As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.
So, BP ⊥ PA i.e., ∠BPA = 90°
Therefore, ∠3 + ∠2 = 90° [angle sum property] …(i)
Also, ∠1 = ∠2 and ∠4 = ∠3 [Angle of incidence = Angle of reflection]
Therefore, ∠1 + ∠4 = 90° [from (i)) …(ii]
Adding (i) and (ii), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
i.e., ∠CAB + ∠DBA = 180°
Hence, CA || BD

Question 8
If in ∆ABC, the bisectors of ∠B and ∠C intersect each other at O. Prove that ∠BOC = 90° + 12∠ A.

Solution:
Let ∠B = 2x and ∠C = 2y
∵OB and OC bisect ∠B and ∠C respectively.
∠OBC = 12∠B = 12 × 2x = x
and ∠OCB = 12∠C = 12 × 2y = y
Now, in ∆BOC, we have
∠BOC + ∠OBC + ∠OCB = 180°
⇒ ∠BOC + x + y = 180°
⇒ ∠BOC = 180° – (x + y)
Now, in ∆ABC, we have
∠A + 2B + C = 180°
⇒ ∠A + 2x + 2y = 180°
⇒ 2(x + y) = 12(180° – ∠A)
⇒ x + y = 90° – 12∠A …..(ii)
From (i) and (ii), we have
∠BOC = 180° – (90° – 12∠A) = 90° + 12 ∠A

Question 9.
In figure, if I || m and ∠1 = (2x + y)°, ∠4 = (x + 2y)° and ∠6 = (3y + 20)°. Find ∠7 and ∠8.

Solution:
Here, ∠1 and ∠4 are forming a linear pair
∠1 + ∠4 = 180°
(2x + y)° + (x + 2y)° = 180°
3(x + y)° = 180°
x + y = 60
Since I || m and n is a transversal
∠4 = ∠6
(x + 2y)° = (3y + 20)°
x – y = 20
Adding (i) and (ii), we have
2x = 80 = x = 40
From (i), we have
40 + y = 60 ⇒ y = 20
Now, ∠1 = (2 x 40 + 20)° = 100°
∠4 = (40 + 2 x 20)° = 80°
∠8 = ∠4 = 80° [corresponding ∠s]
∠1 = ∠3 = 100° [vertically opp. ∠s]
∠7 = ∠3 = 100° [corresponding ∠s]
Hence, ∠7 = 100° and ∠8 = 80°

Question 10.
In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28o and ∠QRT = 65°. Find the values of x, y and z.

Solution:
Here, PQ || SR .
⇒ ∠PQR = ∠QRT
⇒ x + 28° = 65°
⇒ x = 65° – 28° = 37°
Now, in it. ∆SPQ, ∠P = 90°
∴ ∠P + x + y = 180° [angle sum property]
∴ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
Now, ∠SRQ + ∠QRT = 180° [linear pair]
z + 65° = 180°
z = 180° – 65° = 115°

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MCQs :Lines and Angles

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