Objective Type Questions - Multiple Choice Questions The literal coefficient of − 9 x y z 2 -9xyz^2 − 9 x y z 2 is
-9
x y xy x y
x y z 2 xyz^2 x y z 2
− 9 x y -9xy − 9 x y
Answer
Any factor of a term is called the coefficient of the remaining factor of the term. The constant part is the numerical coefficient and the remaining part is the literal coefficient.
In − 9 x y z 2 -9xyz^2 − 9 x y z 2 , the numerical coefficient is -9 and the remaining part is x y z 2 xyz^2 x y z 2 .
Hence, option 3 is the correct option.
Which of the following algebraic expression is not a polynomial?
3 x 2 − 2 x + 7 3x^2 - 2x + 7 3 x 2 − 2 x + 7
5 x 3 2 x + 3 x 2 + 8 \dfrac{5x^3}{2x} + 3x^2 + 8 2 x 5 x 3 + 3 x 2 + 8
3 x + 2 x + 7 3x + \dfrac{2}{x} + 7 3 x + x 2 + 7
2 x 2 + 3 x + 6 \sqrt{2}x^2 + \sqrt{3}x + \sqrt{6} 2 x 2 + 3 x + 6
Answer
An algebraic expression is a polynomial if the powers of the variables in each term are non-negative integers.
Option 1: In 3 x 2 − 2 x + 7 3x^2 - 2x + 7 3 x 2 − 2 x + 7 the powers of x x x are 2, 1 and 0. So, it is a polynomial.
Option 2: 5 x 3 2 x + 3 x 2 + 8 = 5 2 x 2 + 3 x 2 + 8 \dfrac{5x^3}{2x} + 3x^2 + 8 = \dfrac{5}{2}x^2 + 3x^2 + 8 2 x 5 x 3 + 3 x 2 + 8 = 2 5 x 2 + 3 x 2 + 8 , in which the powers of x x x are 2 and 0. So, it is a polynomial.
Option 3: In 3 x + 2 x + 7 3x + \dfrac{2}{x} + 7 3 x + x 2 + 7 , the term 2 x = 2 x − 1 \dfrac{2}{x} = 2x^{-1} x 2 = 2 x − 1 , in which the power of x x x is -1, a negative integer. So, it is not a polynomial.
Option 4: In 2 x 2 + 3 x + 6 \sqrt{2}x^2 + \sqrt{3}x + \sqrt{6} 2 x 2 + 3 x + 6 the powers of x x x are 2, 1 and 0. So, it is a polynomial.
Hence, option 3 is the correct option.
Which of the following algebraic expression is not a monomial?
3 x × y × z 3x \times y \times z 3 x × y × z
− 5 p q -5pq − 5 pq
8 m 2 × n ÷ 3 l 8m^2 \times n \div 3l 8 m 2 × n ÷ 3 l
7 x + y − z 7x + y - z 7 x + y − z
Answer
An algebraic expression having only one term is called a monomial. Multiplication and division do not separate the terms of an algebraic expression.
Option 1: 3 x × y × z 3x \times y \times z 3 x × y × z has only one term, so it is a monomial.
Option 2: − 5 p q -5pq − 5 pq has only one term, so it is a monomial.
Option 3: 8 m 2 × n ÷ 3 l 8m^2 \times n \div 3l 8 m 2 × n ÷ 3 l has only one term, so it is a monomial.
Option 4: 7 x + y − z 7x + y - z 7 x + y − z has three terms, so it is not a monomial. It is a trinomial.
Hence, option 4 is the correct option.
Degree of the polynomial 7 x 2 y z 2 + 6 x 3 y 2 z 2 − 5 x + 8 y 7x^2yz^2 + 6x^3y^2z^2 - 5x + 8y 7 x 2 y z 2 + 6 x 3 y 2 z 2 − 5 x + 8 y is
4
5
6
7
Answer
In a polynomial in two or more variables, the degree is the greatest sum of the powers of the variables in a term.
The given polynomial is 7 x 2 y z 2 + 6 x 3 y 2 z 2 − 5 x + 8 y 7x^2yz^2 + 6x^3y^2z^2 - 5x + 8y 7 x 2 y z 2 + 6 x 3 y 2 z 2 − 5 x + 8 y
Sum of powers of 7 x 2 y z 2 = 2 + 1 + 2 = 5 7x^2yz^2 = 2 + 1 + 2 = 5 7 x 2 y z 2 = 2 + 1 + 2 = 5
Sum of powers of 6 x 3 y 2 z 2 = 3 + 2 + 2 = 7 6x^3y^2z^2 = 3 + 2 + 2 = 7 6 x 3 y 2 z 2 = 3 + 2 + 2 = 7
Sum of powers of − 5 x = 1 -5x = 1 − 5 x = 1
Sum of powers of 8 y = 1 8y = 1 8 y = 1
The greatest sum is 7.
Hence, option 4 is the correct option.
a ( b − c ) + b ( c − a ) + c ( a − b ) a(b - c) + b(c - a) + c(a - b) a ( b − c ) + b ( c − a ) + c ( a − b ) is equal to
a b + b c + c a ab + bc + ca ab + b c + c a
0
2 ( a b + b c + c a ) 2(ab + bc + ca) 2 ( ab + b c + c a )
none of these
Answer
Solving,
⇒ a ( b − c ) + b ( c − a ) + c ( a − b ) ⇒ a b − a c + b c − a b + a c − b c ⇒ ( a b − a b ) + ( − a c + a c ) + ( b c − b c ) ⇒ 0 \Rightarrow a(b - c) + b(c - a) + c(a - b)\\[1em] \Rightarrow ab - ac + bc - ab + ac - bc\\[1em] \Rightarrow (ab - ab) + (-ac + ac) + (bc - bc)\\[1em] \Rightarrow 0 ⇒ a ( b − c ) + b ( c − a ) + c ( a − b ) ⇒ ab − a c + b c − ab + a c − b c ⇒ ( ab − ab ) + ( − a c + a c ) + ( b c − b c ) ⇒ 0
Hence, option 2 is the correct option.
7 5 x y − 2 3 x y + 8 9 x y \dfrac{7}{5}xy - \dfrac{2}{3}xy + \dfrac{8}{9}xy 5 7 x y − 3 2 x y + 9 8 x y is equal to
73 45 x y \dfrac{73}{45}xy 45 73 x y
− 73 45 x y -\dfrac{73}{45}xy − 45 73 x y
x y xy x y
none of these
Answer
Solving,
⇒ 7 5 x y − 2 3 x y + 8 9 x y ⇒ ( 7 5 − 2 3 + 8 9 ) x y ⇒ 63 − 30 + 40 45 x y ⇒ 73 45 x y \Rightarrow \dfrac{7}{5}xy - \dfrac{2}{3}xy + \dfrac{8}{9}xy\\[1em] \Rightarrow \Big(\dfrac{7}{5} - \dfrac{2}{3} + \dfrac{8}{9}\Big)xy\\[1em] \Rightarrow \dfrac{63 - 30 + 40}{45}xy\\[1em] \Rightarrow \dfrac{73}{45}xy ⇒ 5 7 x y − 3 2 x y + 9 8 x y ⇒ ( 5 7 − 3 2 + 9 8 ) x y ⇒ 45 63 − 30 + 40 x y ⇒ 45 73 x y
Hence, option 1 is the correct option.
( 3 p 2 q r 3 ) × ( − 4 p 3 q 2 r 2 ) × ( 7 p q 2 r ) (3p^2qr^3) \times (-4p^3q^2r^2) \times (7pq^2r) ( 3 p 2 q r 3 ) × ( − 4 p 3 q 2 r 2 ) × ( 7 p q 2 r ) is equal to
84 p 6 q 6 r 6 84p^6q^6r^6 84 p 6 q 6 r 6
− 84 p 6 q 6 r 6 -84p^6q^6r^6 − 84 p 6 q 6 r 6
84 p 6 q 5 r 6 84p^6q^5r^6 84 p 6 q 5 r 6
− 84 p 6 q 5 r 6 -84p^6q^5r^6 − 84 p 6 q 5 r 6
Answer
Solving,
⇒ ( 3 p 2 q r 3 ) × ( − 4 p 3 q 2 r 2 ) × ( 7 p q 2 r ) ⇒ ( 3 × ( − 4 ) × 7 ) × ( p 2 × p 3 × p ) × ( q × q 2 × q 2 ) × ( r 3 × r 2 × r ) ⇒ − 84 p ( 2 + 3 + 1 ) q ( 1 + 2 + 2 ) r ( 3 + 2 + 1 ) ⇒ − 84 p 6 q 5 r 6 \Rightarrow (3p^2qr^3) \times (-4p^3q^2r^2) \times (7pq^2r)\\[1em] \Rightarrow (3 \times (-4) \times 7) \times (p^2 \times p^3 \times p) \times (q \times q^2 \times q^2) \times (r^3 \times r^2 \times r)\\[1em] \Rightarrow -84p^{(2+3+1)}q^{(1+2+2)}r^{(3+2+1)}\\[1em] \Rightarrow -84p^6q^5r^6 ⇒ ( 3 p 2 q r 3 ) × ( − 4 p 3 q 2 r 2 ) × ( 7 p q 2 r ) ⇒ ( 3 × ( − 4 ) × 7 ) × ( p 2 × p 3 × p ) × ( q × q 2 × q 2 ) × ( r 3 × r 2 × r ) ⇒ − 84 p ( 2 + 3 + 1 ) q ( 1 + 2 + 2 ) r ( 3 + 2 + 1 ) ⇒ − 84 p 6 q 5 r 6
Hence, option 4 is the correct option.
3 m × ( 2 m 2 − 5 m n + 4 n 2 ) 3m \times (2m^2 - 5mn + 4n^2) 3 m × ( 2 m 2 − 5 mn + 4 n 2 ) is equal to
6 m 3 + 15 m 2 n − 12 m n 2 6m^3 + 15m^2n - 12mn^2 6 m 3 + 15 m 2 n − 12 m n 2
6 m 3 − 15 m 2 n + 12 m n 2 6m^3 - 15m^2n + 12mn^2 6 m 3 − 15 m 2 n + 12 m n 2
6 m 3 − 15 m 2 n − 12 m n 2 6m^3 - 15m^2n - 12mn^2 6 m 3 − 15 m 2 n − 12 m n 2
6 m 3 + 15 m 2 n + 12 m n 2 6m^3 + 15m^2n + 12mn^2 6 m 3 + 15 m 2 n + 12 m n 2
Answer
Solving,
⇒ 3 m × ( 2 m 2 − 5 m n + 4 n 2 ) ⇒ 3 m × 2 m 2 − 3 m × 5 m n + 3 m × 4 n 2 ⇒ 6 m 3 − 15 m 2 n + 12 m n 2 \Rightarrow 3m \times (2m^2 - 5mn + 4n^2)\\[1em] \Rightarrow 3m \times 2m^2 - 3m \times 5mn + 3m \times 4n^2\\[1em] \Rightarrow 6m^3 - 15m^2n + 12mn^2 ⇒ 3 m × ( 2 m 2 − 5 mn + 4 n 2 ) ⇒ 3 m × 2 m 2 − 3 m × 5 mn + 3 m × 4 n 2 ⇒ 6 m 3 − 15 m 2 n + 12 m n 2
Hence, option 2 is the correct option.
Volume of a rectangular box whose adjacent edges are 3 x 2 y , 4 y 2 z 3x^2y, 4y^2z 3 x 2 y , 4 y 2 z and 5 z 2 x 5z^2x 5 z 2 x respectively is
60 x y z 60xyz 60 x yz
60 x 2 y 2 z 2 60x^2y^2z^2 60 x 2 y 2 z 2
60 x 3 y 3 z 3 60x^3y^3z^3 60 x 3 y 3 z 3
none of these
Answer
Volume of a rectangular box = length × breadth × height
⇒ 3 x 2 y × 4 y 2 z × 5 z 2 x ⇒ ( 3 × 4 × 5 ) × ( x 2 × x ) × ( y × y 2 ) × ( z × z 2 ) ⇒ 60 x 3 y 3 z 3 \Rightarrow 3x^2y \times 4y^2z \times 5z^2x\\[1em] \Rightarrow (3 \times 4 \times 5) \times (x^2 \times x) \times (y \times y^2) \times (z \times z^2)\\[1em] \Rightarrow 60x^3y^3z^3 ⇒ 3 x 2 y × 4 y 2 z × 5 z 2 x ⇒ ( 3 × 4 × 5 ) × ( x 2 × x ) × ( y × y 2 ) × ( z × z 2 ) ⇒ 60 x 3 y 3 z 3
Hence, option 3 is the correct option.
( x + 1 ) ( x + 2 ) (x + 1)(x + 2) ( x + 1 ) ( x + 2 ) is equal to
2 x + 3 2x + 3 2 x + 3
x 2 + 2 x + 2 x^2 + 2x + 2 x 2 + 2 x + 2
x 2 + 3 x + 2 x^2 + 3x + 2 x 2 + 3 x + 2
x 2 + 2 x + 3 x^2 + 2x + 3 x 2 + 2 x + 3
Answer
( x + 1 ) ( x + 2 ) (x + 1)(x + 2) ( x + 1 ) ( x + 2 )
Using the identity ( x + a ) ( x + b ) = x 2 + ( a + b ) x + a b (x + a)(x + b) = x^2 + (a + b)x + ab ( x + a ) ( x + b ) = x 2 + ( a + b ) x + ab ,
⇒ x 2 + ( 1 + 2 ) x + 1 × 2 ⇒ x 2 + 3 x + 2 \Rightarrow x^2 + (1 + 2)x + 1 \times 2\\[1em] \Rightarrow x^2 + 3x + 2 ⇒ x 2 + ( 1 + 2 ) x + 1 × 2 ⇒ x 2 + 3 x + 2
Hence, option 3 is the correct option.
If x + 1 x = 2 x + \dfrac{1}{x} = 2 x + x 1 = 2 , then x 2 + 1 x 2 x^2 + \dfrac{1}{x^2} x 2 + x 2 1 is equal to
4
2
0
none of these
Answer
Given, x + 1 x = 2 x + \dfrac{1}{x} = 2 x + x 1 = 2
Squaring both sides,
⇒ ( x + 1 x ) 2 = 2 2 ⇒ x 2 + 2 × x × 1 x + 1 x 2 = 4 ⇒ x 2 + 2 + 1 x 2 = 4 ⇒ x 2 + 1 x 2 = 4 − 2 ⇒ x 2 + 1 x 2 = 2 \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 2^2\\[1em] \Rightarrow x^2 + 2 \times x \times \dfrac{1}{x} + \dfrac{1}{x^2} = 4\\[1em] \Rightarrow x^2 + 2 + \dfrac{1}{x^2} = 4\\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 4 - 2\\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 2 ⇒ ( x + x 1 ) 2 = 2 2 ⇒ x 2 + 2 × x × x 1 + x 2 1 = 4 ⇒ x 2 + 2 + x 2 1 = 4 ⇒ x 2 + x 2 1 = 4 − 2 ⇒ x 2 + x 2 1 = 2
Hence, option 2 is the correct option.
If x 2 + y 2 = 9 x^2 + y^2 = 9 x 2 + y 2 = 9 and x y = 8 xy = 8 x y = 8 , then x + y x + y x + y is equal to
25
5
-5
± 5 \pm5 ± 5
Answer
Given,
x 2 + y 2 = 9 x^2 + y^2 = 9 x 2 + y 2 = 9 and x y = 8 xy = 8 x y = 8
We know that,
( x + y ) 2 = x 2 + y 2 + 2 x y (x + y)^2 = x^2 + y^2 + 2xy ( x + y ) 2 = x 2 + y 2 + 2 x y
Substituting the values,
⇒ ( x + y ) 2 = 9 + 2 × 8 ⇒ ( x + y ) 2 = 9 + 16 ⇒ ( x + y ) 2 = 25 ⇒ x + y = ± 25 ⇒ x + y = ± 5 \Rightarrow (x + y)^2 = 9 + 2 \times 8\\[1em] \Rightarrow (x + y)^2 = 9 + 16\\[1em] \Rightarrow (x + y)^2 = 25\\[1em] \Rightarrow x + y = \pm\sqrt{25}\\[1em] \Rightarrow x + y = \pm 5 ⇒ ( x + y ) 2 = 9 + 2 × 8 ⇒ ( x + y ) 2 = 9 + 16 ⇒ ( x + y ) 2 = 25 ⇒ x + y = ± 25 ⇒ x + y = ± 5
Hence, option 4 is the correct option.
(102)2 - (98)2 is equal to
200
400
600
800
Answer
Using the identity,
a 2 − b 2 = ( a + b ) ( a − b ) a^2 - b^2 = (a + b)(a - b) a 2 − b 2 = ( a + b ) ( a − b )
⇒ (102)2 - (98)2
⇒ (102 + 98)(102 - 98)
⇒ 200 × 4
⇒ 800
Hence, option 4 is the correct option.
− 50 x 3 y 2 z 2 -50x^3y^2z^2 − 50 x 3 y 2 z 2 divided by − 5 x y z -5xyz − 5 x yz is equal to
10 x y z 10xyz 10 x yz
10 x 2 y z 10x^2yz 10 x 2 yz
− 10 x y z -10xyz − 10 x yz
− 10 x 2 y z -10x^2yz − 10 x 2 yz
Answer
Solving,
⇒ − 50 x 3 y 2 z 2 ÷ ( − 5 x y z ) ⇒ − 50 x 3 y 2 z 2 − 5 x y z ⇒ 50 5 × x 3 − 1 y 2 − 1 z 2 − 1 ⇒ 10 x 2 y z \Rightarrow -50x^3y^2z^2 \div (-5xyz)\\[1em] \Rightarrow \dfrac{-50x^3y^2z^2}{-5xyz}\\[1em] \Rightarrow \dfrac{50}{5} \times x^{3-1}y^{2-1}z^{2-1}\\[1em] \Rightarrow 10x^2yz ⇒ − 50 x 3 y 2 z 2 ÷ ( − 5 x yz ) ⇒ − 5 x yz − 50 x 3 y 2 z 2 ⇒ 5 50 × x 3 − 1 y 2 − 1 z 2 − 1 ⇒ 10 x 2 yz
Hence, option 2 is the correct option.
96 × 104 96 \times 104 96 × 104 is equal to
9984
9974
9964
none of these
Answer
96 × 104 = ( 100 − 4 ) ( 100 + 4 ) 96 \times 104 = (100 - 4)(100 + 4) 96 × 104 = ( 100 − 4 ) ( 100 + 4 )
Using the identity ( a − b ) ( a + b ) = a 2 − b 2 (a - b)(a + b) = a^2 - b^2 ( a − b ) ( a + b ) = a 2 − b 2 ,
⇒ ( 100 ) 2 − ( 4 ) 2 ⇒ 10000 − 16 ⇒ 9984 \Rightarrow (100)^2 - (4)^2\\[1em] \Rightarrow 10000 - 16\\[1em] \Rightarrow 9984 ⇒ ( 100 ) 2 − ( 4 ) 2 ⇒ 10000 − 16 ⇒ 9984
Hence, option 1 is the correct option.
If the area of a rectangle is 24 ( x 2 y z + x y 2 z + x y z 2 ) 24(x^2yz + xy^2z + xyz^2) 24 ( x 2 yz + x y 2 z + x y z 2 ) and its length is 8 x y z 8xyz 8 x yz , then its breadth is
3 ( x + y + z ) 3(x + y + z) 3 ( x + y + z )
3 x y z 3xyz 3 x yz
3 ( x + y − z ) 3(x + y - z) 3 ( x + y − z )
none of these
Answer
Area of rectangle = length × breadth
⇒ Breadth = Area of rectangle ÷ length
⇒ 24 ( x 2 y z + x y 2 z + x y z 2 ) 8 x y z ⇒ 24 x y z ( x + y + z ) 8 x y z ⇒ 3 ( x + y + z ) \Rightarrow \dfrac{24(x^2yz + xy^2z + xyz^2)}{8xyz}\\[1em] \Rightarrow \dfrac{24xyz(x + y + z)}{8xyz}\\[1em] \Rightarrow 3(x + y + z) ⇒ 8 x yz 24 ( x 2 yz + x y 2 z + x y z 2 ) ⇒ 8 x yz 24 x yz ( x + y + z ) ⇒ 3 ( x + y + z )
Hence, option 1 is the correct option.