Exercise Page: 224
In questions 1 to 33, there are four options out of which one is correct. Write the correct answer.
Q.1. The product of a monomial and a binomial is a
(a) monomial
(b) binomial
(c) trinomial
(d) none of these
Ans: b
Solution: Let monomial = 2x, binomial = x + y
Then, product of a monomial and a binomial = (2x) × (x + y)
= 2x2 + 2xy
Q.2. In a polynomial, the exponents of the variables are always
(a) integers
(b) positive integers
(c) non-negative integers
(d) non-positive integers
Ans: b
Q.3. Which of the following is correct?
(a) (a – b)2 = a2 + 2ab – b2
(b) (a – b)2 = a2 – 2ab + b2
(c) (a – b)2 = a2 – b2
(d) (a + b)2 = a2 + 2ab – b2
Ans: b
Solution: We have, = (a – b) × (a – b)
= a × (a – b) – b × (a – b)
= a2 – ab – ba + b2
= a2 – 2ab + b2
Q.4. The sum of –7pq and 2pq is
(a) –9pq
(b) 9pq
(c) 5pq
(d) – 5pq
Ans: d
Solution: The given two monomials are like terms.
Then sum of -7pq and 2pg = – 7pq + 2pq
= (-7 + 2) pq
= -5pq
Q.5. If we subtract –3x2y2 from x2y2, then we get
(a) – 4x2y2
(b) – 2x2y2
(c) 2x2y2
(d) 4x2y2
Ans: d
Solution: We have,
The given two monomials are like terms.
Subtract –3x2y2 from x2y2 = x2y2 – (- 3x2y2)
= x2y2 + 3x2y2
= x2y2 (1 + 3)
= 4x2y2
Q.6. Like term as 4m3n2 is
(a) 4m2n2
(b) – 6m3n2
(c) 6pm3n2
(d) 4m3n
Ans: b
Solution: Like terms are formed from the same variables and the powers of these variables are also the same. But coefficients of like terms need not be the same.
Q.7. Which of the following is a binomial?
(a) 7 × a + a
(b) 6a2 + 7b + 2c
(c) 4a × 3b × 2c
(d) 6 (a2 + b)
Ans: d
Solution: that contain exactly two terms are called binomials.
= 6 (a2 + b)
= 6a2 + b
Q.8. Sum of a – b + ab, b + c – bc and c – a – ac is
(a) 2c + ab – ac – bc
(b) 2c – ab – ac – bc
(c) 2c + ab + ac + bc
(d) 2c – ab + ac + bc
Ans: a
Solution: We have,
= (a – b + ab) + (b + c – bc) + (c – a – ac)
= a – b + ab + b + c – bc + c – a – ac
Now, grouping like terms
= (a – a) + (-b + b) + (c + c) + ab – bc – ac
= 2c + ab – bc – ac
Q.9. Product of the following monomials 4p, – 7q3, –7pq is
(a) 196 p2q4
(b) 196 pq4
(c) – 196 p2q4
(d) 196 p2q3
Ans: a
Solution: = 4p × (– 7q3) × (–7pq)
= (4 × (-7) × (-7)) × p × q3 × pq
= 196p2q4
Q.10. Area of a rectangle with length 4ab and breadth 6b2 is
(a) 24a2b2
(b) 24ab3
(c) 24ab2
(d) 24ab
Ans: b
Solution: We know that, area of rectangle = length × breadth
Given, length = 4ab, breadth = 6b2
= 4ab × 6b2
= 24ab3
Q.11. Volume of a rectangular box (cuboid) with length = 2ab, breadth = 3ac and height = 2ac is
(a) 12a3bc2
(b) 12a3bc
(c) 12a2bc
(d) 2ab +3ac + 2ac
Ans: a
Solution: We know that, volume of cuboid = length × breadth × height
Given, length = 2ab, breadth = 3ac, height = 2ac
= 2ab × 3ac × 2ac
= (2 × 3 × 2) × ab × ac × ac
= 12a3bc2
Q.12. Product of 6a2 – 7b + 5ab and 2ab is
(a) 12a3b – 14ab2 + 10ab
(b) 12a3b – 14ab2 + 10a2b2
(c) 6a2 – 7b + 7ab
(d) 12a2b – 7ab2 + 10ab
Ans: b
Solution: Now we have find product of trinomial and monomial,
= (6a2 – 7b + 5ab) × 2ab
= (2ab × 6a2) – (2ab × 7b) + (2ab × 5ab)
= 12a3b – 14ab2 + 10a2b2
Q.13. Square of 3x – 4y is
(a) 9x2 – 16y2
(b) 6x2 – 8y2
(c) 9x2 + 16y2 + 24xy
(d) 9x2 + 16y2 – 24xy
Ans: d
Solution: As per the condition in the question, (3x – 4y)2
The standard identity = (a – b)2 = a2 – 2ab + b2
Where, a = 3x, b = 4y
Then,
(3x – 4y)2 = (3x)2 – (2 × 3x × 4y) + (4y)2
= 9x2 – 24xy + 16y2
Q.14. Which of the following are like terms?
(a) 5xyz2, – 3xy2z
(b) – 5xyz2, 7xyz2
(c) 5xyz2, 5x2yz
(d) 5xyz2, x2y2z2
Ans: b
Solution: Like terms are formed from the same variables and the powers of these variables are also the same. But coefficients of like terms need not be the same.
Q.15. Coefficient of y in the term –y/3 is
(a) – 1
(b) – 3
(c) -1/3
(d) 1/3
Ans: c
Solution: -y/3 can also be written as y × (-1/3)
So, Coefficient of y is -1/3
Q.16. a2 – b2 is equal to
(a) (a – b)2
(b) (a – b) (a – b)
(c) (a + b) (a – b)
(d) (a + b) (a + b)
Ans: c
Solution: (a2 – b2) = (a + b) (a – b) is one of the standard identity.
Q.17. Common factor of 17abc, 34ab2, 51a2b is
(a) 17abc
(b) 17ab
(c) 17ac
(d) 17a2b2c
Ans: b
Solution: The given factors can be written in expanded form as,
17abc = 17 × a × b × c
34ab2 = 2 × 17 × a × b × b
51a2b = 3 × 17 × a × a × b
So, common factors in the above is 17 × a × b
= 17ab
Q.18. Square of 9x – 7xy is
(a) 81x2 + 49x2y2
(b) 81x2 – 49x2y2
(c) 81x2 + 49x2y2 –126x2y
(d) 81x2 + 49x2y2 – 63x2y
Ans: c
Solution: As per the condition in the question, (9x – 7xy)2
The standard identity = (a – b)2 = a2 – 2ab + b2
Where, a = 9x, b = 7xy
Then,
(9x – 7xy)2 = (9x)2 – (2 × 9x × 7xy) + (7xy)2
= 81x2 – 126x2y + 49x2y2
Q.19. Factorised form of 23xy – 46x + 54y – 108 is
(a) (23x + 54) (y – 2)
(b) (23x + 54y) (y – 2)
(c) (23xy + 54y) (– 46x – 108)
(d) (23x + 54) (y + 2)
Ans: a
Solution: Factorised form of 23xy – 46x + 54y – 108 is = 23xy – (2 × 23x) + 54y – (2 × 54)
Take out the common factors,
= 23x (y – 2) + 54 (y – 2)
Again take out the common factor,
= (y – 2) (23x + 54)
Q.20. Factorised form of r2 – 10r + 21 is
(a) (r – 1) (r – 4)
(b) (r – 7) (r – 3)
(c) (r – 7) (r + 3)
(d) (r + 7) (r + 3)
Ans: b
Solution: Factorised form of r2 – 10r + 21 is = r2 – 7r – 3r + 21
Take out the common factors,
= r (r – 7) – 3 (r – 7)
Again take out the common factor,
= (r – 7) (r – 3)
Q.21. Factorised form of p2 – 17p – 38 is
(a) (p – 19) (p + 2)
(b) (p – 19) (p – 2)
(c) (p + 19) (p + 2)
(d) (p + 19) (p – 2)
Ans: a
Solution: Factorised form of p2 – 17p – 38 is = p2 – 19p + 2p – 38
Take out the common factors,
= p (p – 19) + 2 (p – 19)
Again take out the common factor,
= (p – 19) (p + 2)
Q.22. On dividing 57p2qr by 114pq, we get
(a) ¼pr
(b) ¾pr
(c) ½pr
(d) 2pr
Ans: c
Solution: On dividing 57p2qr by 114pq,
It can be expanded as = (57 × p × p × q × r)/(114 × p × q)
= 57pr/114 … [divide both numerator and denominator by 57]
= ½pr
Q.23. On dividing p (4p2 – 16) by 4p (p – 2), we get
(a) 2p + 4
(b) 2p – 4
(c) p + 2
(d) p – 2
Ans: c
Solution: On dividing p (4p2 – 16) by 4p (p – 2)
= (p((2p)2 – (4)2))/ (4p(p – 2))
= ((2p – 4) × (2p + 4))/(4(p – 2))
Take out the common factors
= ((2(p – 2)) × (2 (p + 4)))/(4(p -2))
= (4(p – 2)(p + 2))/ (4(p – 2))
= p + 2
Q.24. The common factor of 3ab and 2cd is
(a) 1
(b) – 1
(c) a
(d) c
Ans: a
Solution: Considering the two monomials 3ab and 2cd there is no common factor except 1.
Q.25. An irreducible factor of 24x2y2 is
(a) x2
(b) y2
(c) x
(d) 24x
Ans: c
Solution: An irreducible factor is a factor which cannot be expressed further as a product of factors. Such a factorisation is called an irreducible factorisation.
24x2y2 = 2 × 2 × 2 × 3 × x × x × y × y
Therefore an irreducible factor is x.
Q.26. Number of factors of (a + b)2 is
(a) 4
(b) 3
(c) 2
(d) 1
Ans: c
Solution: Number of factors of (a + b)2 is = (a + b) (a + b) no further factorisation is possible.
Q.27. The factorised form of 3x – 24 is
(a) 3x × 24
(b) 3 (x – 8)
(c) 24 (x – 3)
(d) 3(x – 12)
Ans: b
Solution: The factorised form of 3x – 24 is,
Take out 3 as common,
= 3 (x – 8)
Q.28. The factors of x2 – 4 are
(a) (x – 2), (x – 2)
(b) (x + 2), (x – 2)
(c) (x + 2), (x + 2)
(d) (x – 4), (x – 4)
Ans: b
Solution: The factors of x2 – 4 are,
X2 – 4 = x2 – 22
= (x + 2) (x – 2)
Q.29. The value of (– 27x2y) ÷ (– 9xy) is
(a) 3xy
(b) – 3xy
(c) – 3x
(d) 3x
Ans: d
Solution: The value of (– 27x2y) ÷ (– 9xy) = (-27 × x × x × y)/(- 9 × x × y)
= (27/9)x … [divide both numerator, denominator by 3]
= 3x
Q.30. The value of (2x2 + 4) ÷ 2 is
(a) 2x2 + 2
(b) x2 + 2
(c) x2 + 4
(d) 2x2 + 4
Ans: b
Solution: The value of (2x2 + 4) ÷ 2 = (2x2 + 4)/2
= (2(x2 + 2))/2
= x2 + 2
Q.31. The value of (3x3 +9x2 + 27x) ÷ 3x is
(a) x2 +9 + 27x
(b) 3x3 +3x2 + 27x
(c) 3x3 +9x2 + 9
(d) x2 +3x + 9
Ans: d
Solution: The value of (3x3 +9x2 + 27x) ÷ 3x = (3x3 + 9x2 + 27x)/3x
Takeout 3x as common,
= 3x (x2 + 3x + 9)/3x
= x2 + 3x + 9
Q.32. The value of (a + b)2 + (a – b)2 is
(a) 2a + 2b
(b) 2a – 2b
(c) 2a2 + 2b2
(d) 2a2 – 2b2
Ans: c
Solution: (a + b)2 + (a – b)2 = (a2 + b2 + 2ab) + (a2 + b2 – 2ab)
= (a2 + a2) + (b2 + b2) + (2ab – 2ab)
= 2a2 + 2b2
Q.33. The value of (a + b)2 – (a – b)2 is
(a) 4ab
(b) – 4ab
(c) 2a2 + 2b2
(d) 2a2 – 2b2
Ans: a
Solution: The value of (a + b)2 – (a – b)2 = (a2 + b2 + 2ab) – (a2 + b2 – 2ab)
= a2 – a2 + b2 – b2 + 2ab + 2ab
= 4ab
In questions 34 to 58, fill in the blanks to make the statements true:
Q.34. The product of two terms with like signs is a term.
Ans: The product of two terms with like signs is a positive term.
Let us assume two like terms are, 3p and 2q
= 3p × 2q
= 6pq
Q.35. The product of two terms with unlike signs is a term.
Ans: The product of two terms with unlike signs is a negative term.
Let us assume two unlike terms are, – 3p and 2q
= -3p × 2q
= – 6pq
Q.36. a (b + c) = a × ______ + a × ______.
Ans: a (b + c) = a × b + a × c. … [by using left distributive law]
= ab + ac
Q.37. (a – b) ______ = a2 – 2ab + b2
Ans: (a – b) (a – b) = (a – b)2= a2 – 2ab + b2
(a – b) (a – b)= a × (a – b) – b × (a – b)
= a2 – ab – ba + b2
= a2 – 2ab + b2
Q.38. a2 – b2 = (a + b ) ______.
Ans: a2 – b2 = (a + b) (a – b) … [from the standard identities]
Q.39. (a – b)2 + ______ = a2 – b2
Ans: (a – b)2 + (2ab – 2b2) = a2 – b2
= (a – b)2 + (2ab – 2b2)
= a2 + b2 – 2ab + 2ab – 2b2
= a2 – b2
Q.40. (a + b)2 – 2ab = ______ + ______.
Ans: (a + b)2 – 2ab = a2 + b2
= (a + b)2 – 2ab
= a2 + 2ab + b2 – 2ab
= a2 + b2
Q.41. (x + a) (x + b) = x2 + (a + b) x + ______.
Ans: (x + a) (x + b) = x2 + (a + b) x + ab
= (x + a) (x + b)
= x × (x + b) + a × (x + b)
= x2 + xb + xa + ab
= x2 + x (b + a) + ab
Q.42. The product of two polynomials is a ______.
Ans: The product of two polynomials is a polynomials.
Q.43. Common factor of ax2 + bx is ______.
Ans: Common factor of ax2 + bx is x (ax + b)
Q.44. Factorised form of 18mn + 10mnp is ______.
Ans: Factorised form of 18mn + 10mnp is 2mn (9 + 5p)
= (2 × 9 × m × n) + (2 × 5 × m × n × p)
= 2mn (9 + 5p)
