MCQ Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 9

The product of

$(x - 1)(2x - 3)$ is _____

0%
$2x^{2} - 5x - 3$
0%
$2x^{2} - 5x + 3$
0%
$2x^{2} + 5x - 3$
0%
$2x^{2} + 5x + 3$

Explanation

Expanding:

$(x-1)\times(2x-3)$

$=2x^{2}-2x-3x+3$

After combining the like terms we get:

$=2x^{2}-5x+3$

Find the product of

$\left(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\right)$ and

$\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\right)$ .

0%
$\dfrac{1}{9}y^4-\dfrac{1}{4}x^4$
0%
$\dfrac{1}{4}x^4-\dfrac{1}{9}y^4$
0%
$4x^2-9y^4$
0%
$\dfrac{1}{4}x^4+\dfrac{1}{9}y^4$

Explanation

Given, product of

$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)$ and

$\Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$

i.e.

$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)\times \Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$ .

We know, the identity

$(a+b)(a-b)=a^2-b^2$ .

Then,

$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)\times \Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$

$=\Bigl(\dfrac{1}{2}x^2\Bigr)^2-\Bigr(\dfrac{1}{3}y^2\Bigl)^2$

$=\dfrac{1}{4}x^4-\dfrac{1}{9}y^4$ .

Therefore, option

$B$ is correct.

The value of

$(-5{x}^{2}y) \times (-\dfrac{2}{3} x{y}^{2}z) \times (\dfrac{8}{15} xy{z}^{2}) \times (-\dfrac{1}{4} z)$ is ______.

0%
$-\dfrac{4}{9} {x}^{4}{y}^{4}{z}^{4}$
0%
$\dfrac{4}{9} {x}^{4}{y}^{4}{z}^{4}$
0%
$-\dfrac{4}{9} {x}^{3}{y}^{3}{z}^{3}$
0%
$\dfrac{4}{9} {x}^{3}{y}^{3}{z}^{3}$

Explanation

We have,

$\quad \quad \left( -5{ x }^{ 2 }y \right) \times \left( -\dfrac { 2 }{ 3 } x{ y }^{ 2 }z \right) \times \left( \dfrac { 8 }{ 15 } xy{ z }^{ 2 } \right) \times \left( -\dfrac { 1 }{ 4 } z \right) \\ =\left( -5\times \left( -\dfrac { 2 }{ 3 } \right) \times \dfrac { 8 }{ 15 } \times \left( -\dfrac { 1 }{ 4 } \right) \right) \times \left( { x }^{ 2 }y\quad \times \quad x{ y }^{ 2 }z\quad \times \quad xy{ z }^{ 2 }\quad \times \quad z \right) \quad \\ =\left( -\dfrac { 2\times 2 }{ 3\times 3 } \right) \times \left( \left( { x }^{ 2 }\times x\times x \right) \times \left( y\times { y }^{ 2 }\times y \right) \times \left( z\times { z }^{ 2 }\times z \right) \right) \\ =\left( -\dfrac { 4 }{ 9 } \right) \times \left( { x }^{ 4 } \right) \times \left( { y }^{ 4 } \right) \times \left( { z }^{ 4 } \right) \\ =\left( -\dfrac { 4 }{ 9 } \right) \left( { x }^{ 4 } \right) \left( { y }^{ 4 } \right) \left( { z }^{ 4 } \right)$

Answer is A

The value of

$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$ is:

0%
$2$
0%
$3$
0%
$5$
0%
$1$

Explanation

Given,

$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$ .

We know,

$(a+b)(a-b)=a^2-b^2$ .

Then,

$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$

$=\displaystyle 5^{\tfrac {1}{2}\times2} - 3^{\tfrac {1}{2}\times2}$

$=\displaystyle 5^{1} - 3^{1}$

$=\displaystyle 5- 3$

$=\displaystyle 2$ .

Hence, option

$A$ is correct.

Using identities, evaluate:

$71^2$ .

0%
$5041$
0%
$3041$
0%
$6041$
0%
$2041$

Explanation

Given,

$71^2$ ,

i.e.

$71^2$

$=(70+1)^2$ .

We know,

$(x+y)^{ 2 }=x^{ 2 }+y^{ 2 }+2xy$ .

Then,

$71^{ 2 }=(70+1)^{ 2 }\\ =70^{ 2 }+(2\times 70\times 1) + 1^2\\ =4900+140+1\\ =5040+1 \\=5041.$

Hence,

$71^2=5041$ .

Therefore, option

$A$ is correct.

Find the product

$24x^2(1-2x)$ and evaluate if for

$x=2$

0%
$288$
0%
$384$
0%
$-288$
0%
$-384$

Explanation

Let us multiply the given monomial and binomial as shown below:

$24{ x }^{ 2 }(1-2x)\\ =\left( 24{ x }^{ 2 }\times 1 \right) -\left( 24{ x }^{ 2 }\times 2x \right) \\ =24{ x }^{ 2 }-48{ x }^{ 3 }$

Now, let us substitute

$x=2$ in

$24{ x }^{ 2 }-48{ x }^{ 3 }$ as follows:

$24{ (2) }^{ 2 }-48{ (2) }^{ 3 }=(24\times 4)-(48\times 8)=96-384=-288$

Hence,

$24{ x }^{ 2 }(1-2x)=-288$ for

$x=2$ .

The value of

${ \left( 1.02 \right) }^{ 2 }+{ \left( 0.98 \right) }^{ 2 }$ , corrected to three decimal places is:

0%
$2.001$
0%
$2.004$
0%
$2.006$
0%
$1.995$

Explanation

Given,

$(1.02)^2+(0.98)^2$ .

We know,

$(a+b)^2= a^2+2ab+b^2$ .

Then,

$(1.02)^2+(0.98)^2$

$=(1.02)^2+(0.98)^2+(2\times 1.02\times 0.98)-(2\times 1.02\times 0.98$ )

$=(1.02+0.98)^2-1.9992$

$=(2)^2-1.9992$

$=4-1.9992$

$=2.0008$ .

When corrected to three decimal digit,

$2.0008$

$=2.001$ .

Therefore, option

$A$ is correct.

Factorise :

${ (ax+by) }^{ 2 }+{ (2bx-2ay) }^{ 2 }-6abxy$

0%
${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+2{ b }^{ 2 }{ x }^{ 2 }+2{ a }^{ 2 }{ y }^{ 2 }+6abxy$
0%
${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }-12abxy$
0%
${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }+4abxy$
0%
${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }+6abxy$

Explanation

$\left( ax+by\right)^2+\left( 2bx-2ay\right)^2-babxy$

Here we will use the following identity,

$\Rightarrow \left( m+n\right)^2=m^2+n^2+2mn$

and

$\left(m-n\right)^2=m^2+n^2-2mn$

Using this we have :

$= \left(ax\right)^2+\left(by\right)^2+2axby+4\left(bx\right)^2+4\left(ay\right)^2-8axby-6abxy$

$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-14axby+2axby$

$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby$

Hence, the answer is

$a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby.$

$xy=20$ and

$(x+y)^{2}=70$ , then

$x^{2}+y^{2}$ is equal to

0%
$25$
0%
$35$
0%
$30$
0%
$50$

Explanation

$(x+y)^2=x^2+y^2+2xy$

It is given that

$(x+y)^2=70$ and

$xy=20$

$\Rightarrow 70=x^2+y^2+(2\times 20)$

$x^2+y^2=70-40$

$=30$

The product of

$\left( 7a-8b \right)$ and

$\left( 7a-8b \right)$ is _______________.

0%
${ 14a }^{ 2 }-112ab+{ 16b }^{ 2 }$
0%
${ 49a }^{ 2 }+112ab+{ 64b }^{ 2 }$
0%
${ 49a }^{ 2 }-112ab+{ 64b }^{ 2 }$
0%
${ 49a }^{ 2 }-112ab-{ 64b }^{ 2 }$

Explanation

We know that

$(x-y)^2 = x^2 -2xy + y^2$

So,

$(7a-8b)(7a-8b)$

$= (7a)^2-2(7a)(8b)+(8b)^2$

$=49a^2-112ab+64b^2$

Thus, C is the correct answer.

Product of the following monomials

$4p$ ,

$- 7q^{3}$ ,

$-7pq$ is

0%
$196 p^{2}q^{4}$
0%
$196 pq^{4}$
0%
$-196p^{2}q^{4}$
0%
$196 p^{2}q^{3}$

Explanation

Given monomials are :

$4p$ ,

$- 7q^{3}$ ,

$-7pq$

$\therefore$ Product

$=(4p)\times(- 7q^{3})\times(-7pq)=4\times (-7)\times (-7)\times p\times q^{3} \times pq =196p^{2}q^{4}$

Hence, option (A) is correct.

The value of

$(a +b)^{2} + (a-b)^{2}$ is

0%
$2a + 2b$
0%
$2a - 2b$
0%
$2a^{2} + 2b^{2}$
0%
$2a^{2} - 2b^{2}$

Explanation

$\textbf{Step-1: Apply the relevant formula & simplify.}$

$\text{We have,}$

$(a + b)^{2} + (a - b)^{2}$

$\text{Now,}$

$(a+b)^2 = a^2+2ab+b^2$

$......(i)$

$(a-b)^2= a^2-2ab+b^2$

$......(ii)$

$\textbf{Step-2: Add the above expressions to get the result}$

$(a + b)^{2} + (a - b)^{2}$

$a^2+2ab+b^2 + a^2-2ab+b^2$

$=2a^2 +2b^2$

$\textbf{Hence, option - C is the answer}$

State whether the statement is true (T) or false (F).
The product of one negative and one positive term is a negative term.

0%
True
0%
False

Explanation

The given statement is true.

The product of one negative and a positive term is always a negative term.

For example,

$(-2)\times3=-6$

or,

$-a\times b = -ab$

What is the product of

$(x+2)(x-3) ?$

0%
$2 x-6$
0%
$3 x-2$
0%
$x^{2}-x-6$
0%
$x^{2}-6 x$

Explanation

$(x+2)(x-3)$

$x(x-3)+2(x-3)$

$x^2-3x+2x-6$

$x^{2}-x-6$

Simplify the following expression:

$(3x^2 * 7x^7)+ (2y^3 * 9y^{12})$

0%
$21x^{14} + 18y{26}$
0%
$10x^{9} + 11y{15}$
0%
$21x^{14} + 18y{15}$
0%
$21x^{9} + 18y{15}$

$x + y = 2013$ and

$\frac{1}{x}+\frac{1}{y}=2013$ , what is the value of xy?

0%
$\frac{1}{2013}$
0%
4026
0%
0
0%
1

Find the product of

$(p+2q)(p+3q)$ when

$p=1$ ,

$q=0$ .

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

$p\times p+2q\times p+p\times 3q+2q\times 3q$

$=p^2+6q^2+5pq$
at

$p=1, q=0$
product

$= 1+0+0=1$

Simplify the following expression:

$(2x^4y^7m^2z) ^{TM} (5x^2y^3m^8)$

0%
$10x^6y^9m^{10}z$
0%
$7x^6y^{10}m^{10}z$
0%
$10x^5y^{10}m^{10}z$
0%
$10x^6y^{10}m^{10}z$

$Let\,x = \sqrt {3 - \sqrt 5 } \,and\,y = \sqrt {3 + \sqrt 5 } .$ If the value of expression

$x - y + 2{x^2}y + 2x{y^2} - {x^4}y + x{y^4}$ can be expressed in the form

$\sqrt p + \sqrt q$

$where\,p,\,q \in N,$ then

$(p+q)$ has the value equal

0%
$448$
0%
$610$
0%
$510$
0%
$540$

Multiply:

$\frac{2}{3},xy,6{x^2}{y^2}\,\,and\,\,\frac{1}{4}{y^3}\,\,is\underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,}$

0%
${x^3}{y^3}$
0%
$\frac{2}{3}{x^3}{y^6}$
0%
${x^3}{y^6}$
0%
$\frac{2}{3}$

Multiply:

$4x, - 3y\,\,and\,\, - 2xy$

0%
$24xy$
0%
$24{x^2}{y^2}$
0%
$- 24{x^2}{y^2}$
0%
$- 24xy$

Multiply:

$3{x^2},2xy,4x{y^2}$

0%
$24{x^4}{y^3}$
0%
$12x{y^2}$
0%
$24{x^2}{y^2}$
0%
$24xy$

Find the value of

$\dfrac { 10.5 \times 10.5 - 15 \times 10.5 + 7.5 \times 7.5 } { 10.5 - 7.5 }.$

0%
$17$
0%
$3$
0%
$6$
0%
$22$

The product of a monomial and a binomial is a

0%
Monomial
0%
binomial
0%
Trinomial
0%
None of these

$A = \left[ \overset { -4 }{ 3 } \quad \overset { -1 }{ 1 } \right]$ , then the determinant of the matrix

$(A^{2016} - 2A^{2015} - A^{2014})$ is :

0%
$2014$
0%
$- 175$
0%
$2016$
0%
$-25$

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 9 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 9 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

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