NCERT Exemplar Problems Class 8 Mathematics Algebraic Expressions, Identities and Factorisation

NCERT  Exemplar Problems Class 8 Mathematics Chapter 7 Algebraic Expressions, Identities and Factorisation

Multiple Choice Questions
Question. 1 The product of a monomial and a binomial is a
(a) monomial (b) binomial
(c) trinomial (d) None of these
Solution. (b) Monomial consists of only single term and binomial contains two terms. So, the multiplication of a binomial by a monomial will always produce a binomial, whose first term is the product of monomial and the binomial’s first term and second term is the product of monomial and the binomial’s second term.

Question. 2 In a polynomial, the exponents of the variables are always (a)’integers (b) positive integers (c) non-negative integers (d) non-positive integers
Solution. (c) In a polynomial, the exponents of the variables are either positive integers or 0. Constant term C can be written as C x°. We do not consider the expressions as a polynomial which consist of the variables having negative/fractional exponent.

Question. 3 Which of the following is correct?
(a) \({{\left( a-b \right)}^{2}}={{a}^{2}}+2ab-{{b}^{2}}\) (b) \({{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\)
(c) \({{\left( a-b \right)}^{2}}={{a}^{2}}-{{b}^{2}}\) (d) \({{\left( a+b \right)}^{2}}={{a}^{2}}+2ab-{{b}^{2}}\)
Solution.

Question. 4 The sum of -7pq and 2pq is
(a) -9pq   (b) 9pq
(c) 5pq   (d) -5pq
Solution.

Question. 5 If we subtract \(-3{ x }^{ 2 }{ y }^{ 2 }\) from \({ x }^{ 2 }{ y }^{ 2 }\), then we get

Solution.

Question. 6 Like term as \(4{ m }^{ 3 }{ n }^{ 2 }\) is
(a)\(4{ m }^{ 2 }{ n }^{ 2 }\) (b) \(-6{ m }^{ 3 }{ n }^{ 2 }\)
(c) \(6p{ m }^{ 3 }{ n }^{ 2 }\) (d) \(4{ m }^{ 3 }{ n }\)
Solution. (b) We know that, the like terms contain the same literal factor. So, the like term as \(4{ m }^{ 3 }{ n }^{ 2 }\) , is \(-6{ m }^{ 3 }{ n }^{ 2 }\), as it contains the same literal factor \({ m }^{ 3 }{ n }^{ 2 }\).

Question. 7 Which of the following is a binomial?

Solution.

Question. 8 Sum of a – b + ab, b + c – bc and c – a – ac is

Solution.

Question. 9 Product of the monomials 4p, -7\({ q }^{ 3 }\), -7pq is

Solution.

Question. 10 Area of a rectangle with length 4ab and breadth 6\({ b }^{ 2 }\) is

Solution.

Question. 11 Volume of a rectangular box (cuboid) with length = 2ab, breadth = 3ac and height = 2ac is

Solution.

Question. 12 Product of 6\({ a }^{ 2 }\) -7b + 5ab and 2ab is

Solution.

Question. 13 Square of 3x – 4y is

Solution.

Question. 14 Which of the following are like terms?

Solution.

Question. 15 Coefficient of y in the term of \({ -y }^{ 3 }\) is
(a)-1 (b)-3 (c)\({ -1 }^{ 3 }\) (d)\({ 1 }^{ 3 }\)
Solution.

Question. 16 \({ a }^{ 2 }-{ b }^{ 2 }\) is equal to

Solution.

Question. 17 Common factor Of 17abc, 34a\({ b }^{ 2 }\), 51\({ a }^{ 2 }\)b is
(a)17abc (b)17ab (c)17ac (d)17\({ a }^{ 2 }\)\({ b }^{ 2 }\)c
Solution.

Question. 18 Square of 9x – 7xy is

Solution.

Question. 19 Factorised form of 23xy – 46x + 54y -108 is

Solution.

Question. 20 Factorised form of \({ r }^{ 2 }\)-10r + 21 is
(a)(r-1)(r-4) (b)(r-7)(r-3) (c)(r-7)(r+3) (d)(r+7)(r+3)
Solution.

Question. 21 Factorised form of \({ p }^{ 2 }\) – 17p – 38 is
(a) (p -19)(p + 2) (b) (p -19) (p – 2) (c) (p +19) (p + 2) (d) (p + 19) (p – 2)
Solution.

Question. 22 On dividing 57 \({ p }^{ 2 }\) qr by 114pq, we get

Solution.

Question. 23 On dividing p(4\({ p }^{ 2 }\) – 16) by 4p (p – 2), we get
(a) 2p + 4 (b) 2p – 4 (c) p + 2 (d) p – 2
Solution.

Question. 24 The common factor of 3ab and 2cd is
(a) 1 (b) -1 (c) a (d) c
Solution. (a) We have, monomials 3ab and 2cd Now, 3ab = 3xaxb 2cd =2 x c x d
Observing the monomials, we see that, there is no common factor (neither numerical nor literal) between them except 1.

Question. 25 An irreducible factor of24\({ x }^{ 2 }\)\({ y }^{ 2 }\) is
(a)\({ a }^{ 2 }\) (b)\({ y }^{ 2 }\) (c)x (d)24x
Solution. (c) A factor is said to be irreducible, if it cannot be factorised further.
We have, 24\({ x }^{ 2 }\)\({ y }^{ 2 }\) =2 x 2 x 2 x 3 x x x x x y x y Hence, an irreducible factor of 24\({ x }^{ 2 }\)\({ y }^{ 2 }\) is x.

Question. 26 Number of factors of \({{\left( a+b \right)}^{2}}\) is
(a) 4 (b) 3 (c) 2 (d) 1
Solution. (c) We can write \({{\left( a+b \right)}^{2}}\) as, (a + b) (a + b) and this cannot be factorised further.
Hence, number of factors of \({{\left( a+b \right)}^{2}}\) is 2.

Question. 27 The factorised form of 3x – 24 is
(a) 3x x 24 (b)3 (x – 8) (c)24(x – 3) (d)3(x-12)
Solution. (b) We have,
3x – 24 = 3 x x – 3 x 8= 3 (x – 8) [taking 3 as common]

Question. 28 The factors of \({ x }^{ 2 }\) – 4 are
(a) (x – 2), (x – 2) (b) (x + 2), (x – 2)
(c) (x + 2), (x + 2) (d) (x – 4), (x – 4)
Solution.

Question. 29 The value of \((-27{ x }^{ 2 }y)\div (-9xy)\) is
(a)3xy (b)-3xy (c)-3x (d)3x
Solution.

Question. 30 The value of \((2{ x }^{ 2 }+4)\div (2)\) is

Solution.

Question. 31 The value of \((3{ x }^{ 3 }+9{ x }^{ 2 }+27x)\div 3x\) is

Solution.

Question. 32 The value of \({{\left( a+b \right)}^{2}}+{{(a-b)}^{2}}\) is

Solution.

Question. 33 The value of \({{\left( a+b \right)}^{2}}-{{(a-b)}^{2}}\) is

Solution.

Fill in the Blanks
In questions 34 to 58, fill in the blanks to make the statements true.
Question. 34 The product of two terms with like signs is a term.
Solution. Positive
If both the like terms are either positive or negative, then the resultant term will always be positive.

Question. 35 The product of two terms with unlike signs is a term.
Solution. Negative
As the product of a positive term and a negative term is always negative.

Question. 36 a (b + c) = a x ——– + a x ———-
Solution. b,c
we have , a(b+c)=a x b + a x c [using left distributive law]

Question. 37 (a-b) ————- =\( { a }^{ 2 }-2ab+{ b }^{ 2 }\)
Solution.

Question. 38 \({ a }^{ 2 }-{ b }^{ 2 }\)=(a+b)—————-
Solution.

Question. 39 \({{(a-b)}^{2}}\)+—————-=\({ a }^{ 2 }-{ b }^{ 2 }\)
Solution.

Question. 40 \({{(a+b)}^{2}}\)-2ab=————- + ———–.
Solution.

Question. 41 (x+a)(x+b)=\({ x }^{ 2 }\) + (a+b) x + ———–.
Solution.

Question. 42 The product of two polynomials is a ————–.
Solution. Polynomial
As the product of two polynomials is again a polynomial.

Question. 43 Common factor of ax2 + bx is——————.
Solution.

Question. 44 Factorised form of 18mn + 10mnp is —————–.
Solution.

Question. 45 Factorised form of 4\({ y }^{ 2 }\) – 12y + 9 is———– .
Solution.

Question. 46 \(38{ x }^{ 2 }{ y }^{ 2 }z\div 19x{ y }^{ 2 }\) is equal to———–.
Solution.

Question. 47 Volume of a rectangular box with length 2x, breadth 3y and height 4z is ——.
Solution. 24 xyz
We know that, the volume of a rectangular box,
V = Length x Breadth x Height = 2x x 3y x 4z = (2 x 3 x 4) xyz = 24 xyz

Question. 48 \( 6{ 7 }^{ 2 }-3{ 7 }^{ 2 }\) =(67 -37) x ———–=————.
Solution.

Question. 49 \( { 103 }^{ 2 }-{ 102 }^{ 2 }\)=————- x (103-102)=————–.
Solution.

Question. 50 Area of a rectangular plot with sides 4\({ y }^{ 2 }\) and 3\({ y }^{ 2 }\) is————–.
Solution.

Question. 51 Volume of a rectangular box with l = b = h = 2x is ———-.
Solution.

Question. 52 The numerical coefficient in -37abc is————–.
Solution. -37
The constant term (with their sign) involved in term of an algebraic expression is called the numerical coefficient of that term.

Question. 53 Number of terms in the expression \({ a }^{ 2 }\) and + bc x d is –.
Solution.

Question. 54 The sum of areas of two squares with sides 4o and 4b is————-.
Solution.

Question. 55 The common factor method of factorisation for a polynomial is based on————-property.
Solution.Distributive
In this method, we regroup the terms in such a way, so that each term in the group contains a common literal or number or both.

Question. 56 The side of the square of area 9\({ y }^{ 2 }\) is————.
Solution.

Question. 57 On simplification, \(\frac { 3x+3 }{ 3 }\) =————.
Solution.

Question. 58 The factorisation of 2x + 4y is————-.
Solution. 2 (x + 2y)
We have, 2x + 4y = 2x + 2 x 2y = 2 (x + 2y)

True/False
In questions 59 to 80, state whether the statements are True or False
Question. 59 \({{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}\).
Solution.

Question. 60 \({{(a-b)}^{2}}={{a}^{2}}-{{b}^{2}}\).
Solution.

Question. 61 (a+b) (a-b)=\({{a}^{2}}-{{b}^{2}}\)
Solution.

Question. 62 The product of two negative terms is a negative term.
Solution.False
Since, the product of two negative terms is always a positive term, i.e. (-) x (-) = (+).

Question. 63 The product of one negative and one positive term is a negative term.
Solution.True
When we multiply a negative term by a positive term, the resultant will be a negative term, i-e. (-) x (+) = (-).

Question. 64 The numerical coefficient of the term -6\({ x }^{ 2 }{ y }^{ 2 }\) is -6.
Solution. True
Since, the constant term (i.e. a number) present in the expression -6\({ x }^{ 2 }{ y }^{ 2 }\) is -6.

Question. 65 \({ p }^{ 2 }\)q+\({ q }^{ 2 }\)r+\({ r }^{ 2 }\)q is a binomial.
Solution. False
Since, the given expression contains three unlike terms, so it is a trinomial.

Question. 66 The factors of \({ a }^{ 2 }\) – 2ab + \({ b }^{ 2 }\)are (a + b) and (a + b).
Solution.

Question. 67 h is a factor of \(2\pi (h+r)\).
Solution.

Question. 68 Some of the factors of \(\frac { { n }^{ 2 } }{ 2 } +\frac { n }{ 2 }\) are \(\frac { 1 }{ 2 } n\) and (n+1).
Solution.

Question. 69 An equation is true for all values of its variables.
Solution. False
As equation is true only for some values of its variables, e.g. 2x – 4= 0 is true, only for x =2.

Question. 70 \({ x }^{ 2 }\) + (a+b)x +ab =(a+b)(x +ab)
Solution.

Question. 71 Common factors of \(11p{ q }^{ 2 },121{ p }^{ 2 }{ q }^{ 3 },1331{ p }^{ 2 }q\) is \(11{ p }^{ 2 }{ q }^{ 2 }\)
Solution.

Question. 72 Common factors of 12 \(11{ a }^{ 2 }{ b }^{ 2 }\) +4a\({ b }^{ 2 }\) -32 is 4.
Solution.

Question. 73 Factorisation of -3\({ a }^{ 2 }\)+3ab+3ac is 3a (-a-b-c).
Solution.

Question. 74 Factorised form of \({ p }^{ 2 }\)+30p+216 is (p+18) (p-12).
Solution.

Question. 75 The difference of the squares of two consecutive numbers is their sum.
Solution.

Question. 76 abc + bca + cab is a monomial.
Solution. True
The given expression seems to be a trinomial but it is not as it contains three like terms which can be added to form a monomial, i.e. abc + abc + abc = 3abc

Question. 77 On dividing \(\frac { p }{ 3 }\) by \(\frac { 3 }{ p }\) ,the quotient is 9
Solution.

Question. 78 The value of p for 5\({ 1 }^{ 2 }\)-4\({ 9 }^{ 2 }\)=100 p is 2.
Solution.

Question. 79 \((9x-51)\div 9\) is x-51.
Solution.

Question. 80 The value of (a+1) (a-1)(\({ a }^{ 2 }\) +1) is \({ a }^{ 4 }\)-1.
Solution.

The post NCERT Exemplar Problems Class 8 Mathematics Algebraic Expressions, Identities and Factorisation appeared first on Learn CBSE.