NCERT Solutions For Class 8 Maths Chapter 9 Algebraic .

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NCERT solutions for class 8 maths chapter 9 algebraic expressions andidentitieswhat are expressions?utetopic 9.1Question: 1 Give five examples of expressions containing one variable and fivetitexamples of expressions containing two variables.Answer:InsFive examples of expressions containing one variable are:shFive examples of expressions containing two variables are:kaQuestion: 2(i) Show on the number line :AaAnswer:x on the number line:

utetitQuestion: 2(ii) Show on the number line :Aakashx-4 on the number line:InsAnswer:Question: 2(iii) Show on the number line :2x 1Answer:2x 1 on the number line:

Answer:uteAakash3x - 2 on the number lineInstitQuestion: 2(iv) Show on the number line:NCERT solutions for class 8 maths chapter 9 algebraic expressions andidentities topic 9.2 terms, factors and coefficientsQuestion:1 Identify the coefficient of each term in the expression.

Answer:utecoefficient of each term are given belowNCERT solutions for class 8 maths chapter 9 algebraic expressions andtitidentities topic 9.3 monomials, binomials and polynomialsInsQuestion: 1(i) Classify the following polynomials as monomials, binomials, trinomials.Answer:shBinomial since there are two terms with non zero coefficients.kaQuestion: 1(ii) Classify the following polynomials as monomials, binomials, trinomials.Answer:AaTrinomial since there are three terms with non zero coefficients.Question:1(iii) Classify the following polynomials as monomials, binomials, trinomials.Answer:

Trinomial since there are three terms with non zero coefficients.uteQuestion: 1(iv) Classify the following polynomials as monomials, binomials, trinomials.Answer:Binomial since there are two terms with non zero coefficients.InstitQuestion: 1(v) Classify the following polynomials as monomials, binomials, trinomials.Answer:Monomial since there is only one term.Answer:shQuestion: 2(a) Construct 3 binomials with onlyas a variable;AakaThree binomials with the only x as a variable are:Question: 2(b) Construct 3 binomials withandAnswer:Three binomials with x and y as variables are:as variables;

Question: 2(c) Construct 3 monomials withandas variables;Answer:uteThree monomials with x and y as variables areAnswer:InsTwo polynomials with 4 or more terms are:titQuestion: 2(d) Construct 2 polynomials with 4 or more terms .NCERT solutions for class 8 maths chapter 9 algebraic expressions andshidentities topic 9.4 like and unlike termskaQuestion:(i) Write two terms which are likeAaAnswer:Question:(ii) Write two terms which are likeAnswer:

we can write more like termsuteQuestion:(iii) Write two terms which are likeInstitAnswer:NCERT solutions for class 8 maths chapter 9 algebraic expressions andidentities-Exercise: 9.1kaAnswershQuestion:1(i) Identify the terms, their coefficients for each of the following expressions.Aafollowing are the terms and coefficientThe terms areand the coefficients are 5 and -3.Question: 1(ii) Identify the terms, their coefficients for each of the followingexpressions.

Answer:utethe following is the solutionQuestion:1(iii) Identify the terms, their coefficients for each of the followingtitexpressions.InsAnswer:Question: 1(iv) Identify the terms, their coefficients for each of the followingkaAnswer:shexpressions.The terms are 3, -pq, qr,and -rp and the coefficients are 3, -1, 1 and -1 respectively.AaQuestion:1(v) Identify the terms, their coefficients for each of the following expressions.Answer:Above are the terms and coefficients

Question: 1(vi) Identify the terms, their coefficients for each of the followingexpressions.uteAnswer:The terms are 0.3a, -0.6ab and 0.5b and the coefficients are 0.3, -0.6 and 0.5.titQuestion: 2(a) Classify the following polynomials as monomials, binomials, trinomials.InsWhich polynomials do not fit in any of these three categories?Answer:shBinomial.Question: 2(b) Classify the following polynomials as monomials, binomials, trinomials.kaWhich polynomials do not fit in any of these three categories?AaAnswer:Monomial.Question: 2(c) Classify the following polynomials as monomials, binomials, trinomials.Which polynomials do not fit in any of these three categories?

Answer:This polynomial does not fit in any of these three categories.Question: 2(d) Classify the following polynomials as monomials, binomials, trinomials.uteWhich polynomials do not fit in any of these three categories?titAnswer:InsTrinomial.Question: 2(e) Classify the following polynomials as monomials, binomials, trinomials.Answer:kaBinomial.shWhich polynomials do not fit in any of these three categories?Question: 2(f) Classify the following polynomials as monomials, binomials, trinomials.AaWhich polynomials do not fit in any of these three categories?Answer:Trinomial.

Question: 2(g) Classify the following polynomials as monomials, binomials, trinomials.Which polynomials do not fit in any of these three categories?uteAnswer:Trinomial.titQuestion: 2(h) Classify the following polynomials as monomials, binomials, trinomials.InsWhich polynomials do not fit in any of these three categories?Answer:shBinomial.Question: 2(i) Classify the following polynomials as monomials, binomials, trinomials.kaWhich polynomials do not fit in any of these three categories?AaAnswer:This polynomial does not fit in any of these three categories.Question:2(j) Classify the following polynomials as monomials, binomials, trinomials.Which polynomials do not fit in any of these three categories?

Answer:Monomial.Question: 2(k) Classify the following polynomials as monomials, binomials, trinomials.uteWhich polynomials do not fit in any of these three categories?titAnswer:InsBinomial.Question: 2(i) Classify the following polynomials as monomials, binomials, trinomials.Answer:kaBinomial.shWhich polynomials do not fit in any of these three categories?AaQuestion: 3(i) Add the following.Answer:ab-bc bc-ca ca-ab 0.Question:3 (ii) Add the following.

Answer:uteQuestion:3 (iii) Add the followingInstitAnswer:kaAnswer:shQuestion: 3(iv) Add the following.Question: 4(a) SubtractfromAaAnswer:12a-9ab 5b-3-(4a-7ab 3b 12) (12-4)a (-9 7)ab (5-3)b (-3-12) 8a-2ab 2b-15Question: 4(b) Subtractfrom

Answer:Question: 4(c) SubtractutefromtitAnswer:InsNCERT solutions for class 8 maths chapter 9 algebraic expressions andidentities topic 9.7.2 multiplying three or more monomialsQuestion:1 Find; or first.kaAnswer:and multiply it byand multiply it byshfind. First findWe observe that the result is same in both cases and the result does not depend onAathe order in which multiplication has been carried out.NCERT solutions for class 8 maths chapter 9 algebraic expressions andidentities-Exercise: 9.2Question: 1(i) Find the product of the following pairs of monomials.

uteAnswer:Question: 1(ii) Find the product of the following pairs of monomials.InstitAnswer:kaAnswer:shQuestion: 1(iii) Find the product of the following pairs of monomialsAaQuestion: 1(iv) Find the product of the following pairs of monomials.Answer:

Question:1(v) Find the product of the following pairs of monomials.uteAnswer:Question:2(A) Find the areas of rectangles with the following pairs of monomials asInstittheir lengths and breadths respectively.Answer:shThe question can be solved as followsQuestion:2(B) Find the areas of rectangles with the following pairs of monomials asAakatheir lengths and breadth respectively.Answer:the area is calculated as follows

Question:2(C) Find the areas of rectangles with the following pairs of monomials astheir lengths and breadths respectively.uteAnswer:titthe following is the solutionInsQuestion:2(D) Find the areas of rectangles with the following pairs of monomials asAnswer:shtheir lengths and breadths respectively.kaarea of rectangles isAaQuestion:2(E) Find the areas of rectangles with the following pairs of monomials astheir lengths and breadths respectively.Answer:The area is calculated as follows

uteQuestion:3 Complete the table of products.Second monomial.Ins.titFirst monomial.Aakash.Answer:Firstmonomial.

SecondshInstitutemonomialQuestion:4(i) Obtain the volume of rectangular boxes with the following length, breadthkaand height respectively.AaAnswer:Question:4(ii) Obtain the volume of rectangular boxes with the following length,breadth and height respectively.

Answer:utethe volume of rectangular boxes with the following length, breadth and height isInsbreadth and height respectively.titQuestion:4(iii) Obtain the volume of rectangular boxes with the following length,Answer:shthe volume of rectangular boxes with the following length, breadth and height iskaQuestion:4(iv) Obtain the volume of rectangular boxes with the following length,Aabreadth and height respectively.Answer:the volume of rectangular boxes with the following length, breadth and height is

Question:5(i) Obtain the product ofthe productInsQuestion:5(ii) Obtain the product ofAnswer:shthe productkaQuestion:5(iii) Obtain the product ofAaAnswer:the producttituteAnswer:

Question:5(iv) Obtain the product ofthe productInsQuestion:5(v) Obtain the product ofkashAnswer:the producttituteAnswer:NCERT solutions for class 8 maths chapter 9 algebraic expressions andAaidentities topic 9.8.1 multiplying a monomial by a binomialQuestion:(i) Find the productAnswer:Using distributive law,

uteQuestion:(ii) Find the productAnswer:Using distributive law,titWe have :InsNCERT solutions for class 8 maths chapter 9 algebraic expressions andidentities topic 9.8.2 multiplying a monomial by a trinomialAnswer:shQuestion:1 Find the product:AakaBy using distributive law,NCERT solutions for class 8 maths chapter 9 algebraic expressions andidentities-Exercise: 9.3Question:1(i) Carry out the multiplication of the expressions in each of the followingpairs.

Answer:uteMultiplication of the given expression gives :By distributive law,titQuestion:1(ii) Carry out the multiplication of the expressions in each of the followingInspairs.shAnswer:We have ab, (a-b).kaUsing distributive law we get,AaQuestion:1(iii) Carry out the multiplication of the expressions in each of the followingpairs.Answer:Using distributive law we can obtain multiplication of given expression:

Question:1(iv) Carry out the multiplication of the expressions in each of the followingutepairs.Answer:InstitWe will obtain multiplication of given expression by using distributive law :Question:1(v) Carry out the multiplication of the expressions in each of the followingAnswer:shpairs.kaUsing distributive law :AaQuestion:2 Complete the table(i)FirstSecondexpressionexpressionProduct.

.(iii).(iv).(v).titute(ii)InsAnswer:We will use distributive law to find product in each iii)(iv)(v)Question:3(i) Find the product.Product

Opening brackets :orInsQuestion:3(ii) Find the product.kashAnswer:We have,AaQuestion:3(iii) Find the product.Answer:We havetituteAnswer:

Question:3(iv) Find the product.uteAnswer:We haveQuestion:4(a) SimplifyAnswer:ka(a) We haveand find its values forsh(i)InstitorAaPut x 3,We get :Question:4(a) Simplify(ii)and find its values for

Answer:uteWe havePutQuestion:4(b) SimplifyAnswer:kaWe have :and find its value forsh(i)Instit. So We get,Put a 0 :AaQuestion:4(b) Simplify(ii)Answer:We havePut a 1 ,and find its value for

we get :Question:4(b) Simplifyand find its value forute(iii)Answer:We have.titorInsPut a (-1)Question:5(a) Add:shAnswer:and(a)First we will solve each brackets individually.;ka;AaAddind all we get :Question:5(b) Add:Answer:Firstly, open the brackets:and

andQuestion:5(c) Subtract:fromInsAnswer:titoruteAdding both, we get :andAaorkaSubtracting:shAt first we will solve each bracket individually,orQuestion:5(d) Subtract:Answer:Solving brackets :from

uteandInstitSubtracting :NCERT solutions for class 8 maths chapter 9 algebraic expressions andidentities-Exercise: 9.4andkaAnswer:shQuestion:1(i) Multiply the binomials.We have (2x 5) and (4x - 3)Aa(2x 5) X (4x - 3) (2x)(4x) (2x)(-3) (5)(4x) (5)(-3) 8- 6x 20x - 15 8 14x -15Question:1(ii) Multiply the binomials.and

Answer:We need to multiply (y - 8) and (3y - 4) 3- 4y - 24y 32 3- 28y 32ute(y - 8) X (3y - 4) (y)(3y) (y)(-4) (-8)(3y) (-8)(-4)Question:1(iii) Multiply the binomialstitandInsAnswer:We need to multiply (2.5l - 0.5m) and (2.5l 0.5m)(2.5l - 0.5m) X (2.5l 0.5m) 6.25- 0.25usingkaandshQuestion:1(iv) Multiply the binomials.Answer:Aa(a 3b) X (x 5) (a)(x) (a)(5) (3b)(x) (3b)(5) ax 5a 3bx 15bQuestion:1(v) Multiply the binomials.andAnswer:

(2pq 3q 2 ) X (3pq - 2q 2 ) (2pq)(3pq) (2pq)(-2q 2 ) ( 3q 2 )(3pq) (3q 2 )(-2q 2 ) 6p 2 q 2 - 4pq 3 9pq 3 - 6q 4 6p 2 q 2 5pq 3 - 6q 4uteQuestion:1(vi) Multiply the binomials.andX sh InsMultiplication can be done as followstitAnswer: kaQuestion:2(i) Find the product.AaAnswer:(5 - 2x) X (3 x) (5)(3) (5)(x) (-2x)(3) (-2x)(x) 15 5x - 6x - 2 15 - x - 2Question:2(ii) Find the product.

Answer: 7- xy 49xy - 7 7 48xy - 7ute(x 7y) X (7x - y) (x)(7x) (x)(-y) (7y)(7x) (7y)(-y)Answer:( b) X (a ) ( InstitQuestion:2(iii) Find the product.)(a) (kaAnswer:Aafollowing is the solution() (b)(a) (b)(shQuestion:2(iv) Find the product.)() X (2p q) Question:3(i) Simplify.)

Answer:this can be simplified as follows(-5) X (x 5) 25 ()(x) ()(5) (-5)(x) (-5)(5) 25ute Answer:This can be simplified as( 5) X ()() ()(3) (5)(sh 3) 5 (InstitQuestion:3(ii) Simplify . kaQuestion:3(iii) Simplify.AaAnswer:simplifications can be(t )(- s) (t)() (t)(-s) ( Question:3(iv) Simplify.)() ()(-s)) (5)(3) 5

Answer:(a b) X ( c -d) (a - b) X (c d) 2(ac bd ) ac - ad bc - bd ac ad -bc - bd 2(ac bd ) 2(ac - bd ) 2(ac bd ) 2ac - 2bd 2ac 2bdAnswer:InsQuestion:3(v) Simplify.tit 4acute (a)(c) (a)(-d) (b)(c) (b)(-d) (a)(c) (a)(d) (-b)(c) (-b)(d) 2(ac bd )sh(x y) X ( 2x y) (x 2y) X (x - y) (x)(2x) (x)(y) (y)(2x) (y)(y) (x)(x) (x)(-y) (2y)(x) (2y)(-y) xy 2xy - xy 2xy - 2ka 2 3 4xy -AaQuestion:3(vi) Simplify.Answer:simplification is done as follows

(x y) X () xX() y() uteQuestion:3(vii) Simplify.titAnswer:(1.5x - 4y) X (1.5x 4y 3) - 4.5x 12y (1.5x) X (1.5x 4y 3) -4y X (1.5x 4y 3)- 4.5x 12y 6xy 4.5x - 6xy - 16 2.25- 16shQuestion:3(viii) Simplify.- 12y -4.5x 12 yIns 2.25kaAnswer:(a b c) X (a b - c) a X (a b - c) b X (a b - c) c X (a b - c) --bc ac bc - 2abAa ab - ac ab NCERT solutions for class 8 maths chapter 9 algebraic expressions andidentities topic 9.11 standerd identitiesQuestion:1(i) Put -b in place of b in identity 1. Do you get identity 2?Answer:

Identity 1If we replace b with -b in identity 1We get,utewhich is equal towhich is identity 2So, we get identity 2 by replacing b with -b in identity 1titNCERT free solutions for class 8 maths chapter 9 algebraic expressionsand identities topic 9.11 standard identitiesAnswer:shIdentity IV(a x)(b x) .InsQuestion:1 Verify Identity (IV), forSo, it is given that a 2, b 3 and x 5Lets put these value in identity IV (2 3)5 2 X 3ka(2 5)(3 5) 7 X 8 25 5 X 5 6Aa56 25 25 6 56L.H.S. R.H.S.So, by this we can say that identity IV satisfy with given value of a,b and xQuestion:2 Consider, the special case of Identity (IV) with a b, what do you get? Is itrelated to Identity

Answer:Identity IV isIf a b thanute(a x)(a x) Which is identity IAnswer:Identity IV isIf a b -c than,WhatInsdo you get? Is it related to Identity ?andtitQuestion:3 Consider, the special case of Identity (IV) withsh(x - c)(x - c) kaWhich is identity IIQuestion:4 Consider the special case of Identity (IV) withIs it related to Identity?AaAnswer:Identity IV isIf b -a than,. What do you get?

(x a)(x - a) uteWhich is identity IIINCERT solutions for class 8 maths chapter 9 algebraic expressions andidentities-Exercise: 9.5Answer:(x 3) X (x 3) InstitQuestion:1(i) Use a suitable identity to get each of the following products.shSo, we use identity I for this which is kaIn this a x and b xAaQuestion:1(ii) Use a suitable identity to get each of the following products in bracket.Answer:(2y 5) X ( 2y 5) We use identity I for this which is

IN this a 2y and b 5 Answer:(2a -7) X (2a - 7) in this a 2a and b 7sh InsWe use identity II for this which istituteQuestion:1(iii) Use a suitable identity to get each of the following products in bracket.kaQuestion:1(iv) Use a suitable identity to get each of the following products in bracket.AaAnswer:We use identity II for this which isin this a 3a and b -1/2

Question:1(v) Use a suitable identity to get each of the following products in bracket.uteAnswer:We use identity III for this which is(a - b)(a b) titIn this a 1.1m and b 4 - 16Ins 1.21Question:1(vi) Use a suitable identity to get each of the following products in bracket.shAnswer:-katake the (-)ve sign common so our question becomesWe use identity III for this which is(a - b)(a b) and b AaIn this a Question:1(vii) Use a suitable identity to get each of the following.

Answer:(6x -7) X (6x - 7) We use identity III for this which is(a - b)(a b) uteIn this a 6x and b 7(6x -7) X (6x - 7) Answer:InstitQuestion:1(viii) Use a suitable identity to get each of the following product.take (-)ve sign common from both the brackets So, our question become(a -c) X (a -c) shWe use identity II for this which iskaIn this a a and b cAaQuestion:1(ix) Use a suitable identity to get each of the following product.Answer:We use identity I for this which is

and b uteIn this a titQuestion:1(x) Use a suitable identity to get each of the following products.shInsAnswer:We use identity II for this which is kaIn this a 7a and b 9bAaQuestion:2(i) Use the identityfollowing products.Answer:to find the

We use identityin this a 3 and b 7 Question:2(ii) Use the identityuteto find thetitfollowing products.We use identityIn this a 5 , b 1 and x 4x sh InsAnswer:Question:2(iii) Use the identitykafollowing products.AaAnswer:We use identityin this x 4x , a -5 and b -1 to find the

Question:1(iv) Use the identityto find thefollowing products.uteAnswer:We use identityIn this a 5 , b -1 and x 4xtit Question:2(v) Use the identityAnswer:to find theshfollowing products.Ins We use identitykaIn this a 5y , b 3y and x 2x Aa Question:2(vi) Use the identityfollowing products.Answer:to find the

We use identityIn this a 9 , b 5 and x titfollowing products.to find theuteQuestion:2(vii) Use the identityWe use identityInsAnswer:In this a -4 , b -2 and x xyz sh kaQuestion:3(i) Find the following squares by using the identities.Answer:AaWe use identityIn this a b and b 7 Question:3(ii) Find the following squares by using the identities.

Answer:uteWe useIn this a xy and b 3ztit InsQuestion:3(iii) Find the following squares by using the identities.We use and b kaIn this a shAnswer:AaQuestion:3(iv) Find the following squares by using the identities.Answer:we use the identity

In this a and b ute titQuestion:3(v) Find the following squares by using the identities.we useIn this a 0.4p and b 0.5qsh InsAnswer:kaQuestion:3(vi) Find the following squares by using the identities.AaAnswer:we use the identityIn this a 2xy and b 5y Question:4(i) Simplify:

Answer:In this a utewe useand b tit InsQuestion:4(ii) Simplify.Answer:shwe use kaIn this a (2x 5) and b (2x - 5) (4x)(10)Aa 40xorremember thathere a 2x, b 5

uteQuestion:4(iii) Simplify.Answer:we usetitandIn this a 7m and b 8nIns andSo,orka sh Aaremember thatQuestion: 4(iv) Simplify.Answer:

we use 2 ) in this a 5m and b 4n Answer:kawe useshQuestion: 4(v) Simplify.Aa1 ) In this a (2.5p- 1.5q) and b (1.5p - 2.5q) 4(p q ) (p - q) 4Question:4(vi) Simplify. InsSo,titute1 ) In this a 4m and b 5n

Answer:uteWe use identityIn this a ab and b bc -titNow,Ins Question:4(vii) Simplify.shAnswer:kaWe use identityIn this a Aa and b Now, Question:5(i) Show that

Answer:uteL.H.S. R.H.S.titHence it is proovedInsQuestion:5(ii) Show thatL.H.S. shAnswer:Aaka) R.H.S.Question:5(iii) Show that.Answer:(Using

First we will solve the LHS :uteor RHSAnswer:kashOpening both brackets we get,InstitQuestion:5(iv) Show that. R.H.S.AaQuestion:5(v) Show thatAnswer:Opening all brackets from the LHS, we get :

Question:6(i) Using identities, evaluate.titAnswer:InsWe will use the identity:shSo,kaQuestion:6(ii) Using identities, evaluate.AaAnswer:Here we will use the identity :So :ute RHS

oruteQuestion:6(iii) Using identities, evaluate.Answer:InsSo :shortitHere we will use the identity :AakaQuestion:6(iv) Using identities, evaluate.Answer:Here we will the identity :or

oruteQuestion:6(v) Using identities, evaluate.Answer:titHere we will use :InsThusshorkaQuestion:6(vi) Using identities, evaluate.AaAnswer:This can be written as :usingor

uteQuestion:6(vii) Using identities, evaluate.Answer:titThis can be written in form of :InsororAnswer:shQuestion:6(viii) Using identities, evaluate.AakaHere we will use the identity :Thus :oror

Question:6(ix) Using identities, evaluate.uteAnswer:This can be written as :titororAnswer:AakaWe know,shQuestion:7(i) UsingInsorUsing this formula, (51 49)(51 - 49) (100)(2) 200, find

Question:7(ii) Using, finduteAnswer:Using this formula, (2.00)(0.04) 0.08kaAnswer:shQuestion:7(iii) UsingIns (1.02 0.98)(1.02 - 0.98)AaWe know,Using this formula, (153 - 147)(153 147) (6) (300)titWe know,, find.

1800Question:7(iv) Usingute, findAnswer:titWe know,InsUsing this formula, (1.02 0.98)(1.02 - 0.98) 0.08sh (2.00)(0.04)Question:8(i) UsingkaAnswer:AaWe know,Using this formula, (100 3)(100 4)Here x 100, a 3, b 4

ute 11212Question:8(ii) Using, findtitAnswer:Using this formula,InsWe know,sh (5 0.1)(5 0.2)AakaHere x 5, a 0.1, b 0.2 26.52Question:8(iii) UsingAnswer:, find

We know, (100 3)(100 - 2) (100 3){100 (-2)} 10094Answer:AakaWe know,shQuestion: 8(iv) UsingInstitHere x 100, a 3, b -2uteUsing this formula,Using this formula, (10 - 0.3)(10 - 0.2) {10 (-0.3)}{10 (-0.2)}Here x 10, a -0.3, b -0.2, find

kaAashitutestIn 95.

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.1 what are expressions? Question: 1 Give five examples of expressions containing one variable a

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