MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 2 - MCQExams.com
CBSE
Class 8 Maths
Algebraic Expressions And Identities
Quiz 2
State whether True or False.
Multiply: $$ -\dfrac{2}{3}a^7b^2 $$ and $$-\dfrac{9}{4}ab^5 $$.
The answer is $$\dfrac{3}{2}a^8b^7$$.
Report Question
0%
True
0%
False
Explanation
$$-\frac{2}{3}a^7b^2 \times -\frac{9}{4}ab^5$$
$$=(-\frac{2}{3})(-\frac{9}{4})a^{7+1}b^{2+5}$$
$$=\frac{3}{2}a^8b^7$$
Multiply:
$$ -5cd^2 $$ by $$-5cd^2 $$
Report Question
0%
$$25c^2d^5 $$
0%
$$25c^3d^4 $$
0%
$$25c^2d^3 $$
0%
$$25c^2d^4 $$
Explanation
$$-5cd^2 \times -5cd^2$$
$$=(-5)(-5)c^{1+1}d^{2+2}$$
$$=25c^2d^4$$
Multiply:
$$ \frac{2}{3}ab $$ by $$ -\frac{1}{4}a^2b $$
Report Question
0%
$$-\frac{1}{6}a^3b^3 $$
0%
$$-\frac{1}{6}a^3b^2 $$
0%
$$-\frac{1}{6}a^4b^2 $$
0%
$$-\frac{1}{6}a^3b^4 $$
Explanation
$$\frac{2}{3}ab \times -\frac{1}{4}a^2b$$
$$=\frac{2}{3}(-\frac{1}{4})a^{1+2}b^{1+1}$$
$$=-\frac{2}{12}a^3b^2$$
State whether True or False.
Multiply: $$ 4a$$ and $$ 6a+7 $$.
The answer is $$24a^2+28a $$.
Report Question
0%
True
0%
False
Explanation
$$\\ 4a\times (6a+7)\\ =24{ a }^{ 2 }+28a\\ \\ $$
State whether True or False.
Multiply: $$ 2a^2-5a-4$$ and $$ -3a $$.
The anwser is $$-6a^3+15a^2+12a $$.
Report Question
0%
True
0%
False
Explanation
$$(2a^2-5a-4)\times (-3a)$$
$$=2a^2(-3a)-5a(-3a)-4(-3a)$$
$$=2(-3)a^{2+1}-(5)(-3)a^{1+1}-4(-3)a$$
$$=-6a^3+15a^2+12a$$
Use identities to solve: $$(97)^{2}$$
Report Question
0%
$$9,659$$
0%
$$9,409$$
0%
$$9,009$$
0%
$$9,209$$
Explanation
$$(97)^2$$
$$=(100-3)^2$$
using, $$(a-b)^2=a^2-2ab+b^2$$
$$=(100)^2-2(100)3+(3)^2$$
$$=10000-600+9$$
$$=9409$$
Simplify: $$\displaystyle { x }^{ 2 }\left( 3-{ 5y }^{ 2 } \right) +x\left( { xy }^{ 2 }-3x \right) -2y\left( y-{ 2x }^{ 2 }y \right) $$
Report Question
0%
$$\displaystyle { 6x }^{ 2 }-{ 2y }^{ 2 }-{ 3x }^{ 2 }y$$
0%
$$\displaystyle { 6x }^{ 2 }$$
0%
$$\displaystyle { 3x }^{ 2 }y$$
0%
$$\displaystyle { -2y }^{ 2 }$$
Explanation
We have to simplify $$\displaystyle { x }^{ 2 }\left( 3-{ 5y }^{ 2 } \right) +x\left( { xy }^{ 2 }-3x \right) -2y\left( y-{ 2x }^{ 2 }y \right) $$
$$\displaystyle ={ 3x }^{ 2 }-{ 5x }^{ 2 }{ y }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 }-{ 3x }^{ 2 }-{ 2y }^{ 2 }+{ 4x }^{ 2 }{ y }^{ 2 }$$
$$\displaystyle =\left( { 3x }^{ 2 }-{ 3x }^{ 2 } \right) +\left( -{ 5x }^{ 2 }{ y }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 }+{ 4x }^{ 2 }{ y }^{ 2 } \right) -{ 2y }^{ 2 }$$
$$=\displaystyle 0-{ 2y }^{ 2 }=-{ 2y }^{ 2 }$$.
Hence given expression simplified to $$-2y^2$$
Use the identity $$ (a+b)(a-b) = a^2-b^2$$ to evaluate:
$$33\times 27 $$.
Report Question
0%
$$891$$
0%
$$881$$
0%
$$981$$
0%
$$841$$
Explanation
We know, $$ 33 \times 27 = (30 + 3) \times (30 - 3) $$ .
Applying the formula $$ (a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 } $$,
where $$ a = 30 , b = 3 $$,
we get,
$$ 33 \times 27 = (30 + 3) \times (30 - 3) = { 30 }^{ 2 }-{ 3 }^{ 2 } = 900 - 9 = 891 $$ .
Therefore, option $$A$$ is correct.
Use the identity $$ (a+b)(a-b) = a^2-b^2$$ to evaluate:
$$21\times 19 $$.
Report Question
0%
$$389$$
0%
$$399$$
0%
$$289$$
0%
$$429$$
Explanation
We know, $$ 21 \times 19 = (20+ 1) \times (20 - 1) $$ .
Applying the formula $$ (a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 } $$,
where $$ a = 20 , b = 1 $$,
we get,
$$ 21 \times 19 = (20+ 1) \times (20 - 1) = { 20 }^{ 2 }-{ 1 }^{ 2 } = 400 - 1 = 399 $$ .
Therefore, option $$B$$ is correct.
Solve for $$x$$:
$$(502)^{2}$$
Report Question
0%
$$2,52,004$$
0%
$$2,62,104$$
0%
$$2,22,004$$
0%
$$2,52,864$$
Explanation
$$(502)^2$$
$$=(500+2)^2$$
using, $$(a+b)^2=a^2+2ab+b^2$$
$$=(500)^2+2(500)(2)+(2)^2$$
$$=250000+2000+4$$
$$=252004$$
Use identities to evaluate:
$$(101)^{2}$$
Report Question
0%
$$11,601$$
0%
$$12,761$$
0%
$$10,111$$
0%
$$10,201$$
Explanation
$$(101)^2$$
$$=(100+1)^2$$
using, $$(a+b)^2=a^2+2ab+b^2$$
$$=(100)^2+2(100)(1)+1^2$$
$$=10000+200+1$$
$$=10201$$
If $$49x^2-b=\left (7x+\dfrac {1}{2}\right )\left (7x-\dfrac {1}{2}\right )$$, then the value of $$b$$ is :
Report Question
0%
$$0$$
0%
$$\dfrac {1}{\sqrt 2}$$
0%
$$\dfrac {1}{4}$$
0%
$$\dfrac {1}{2}$$
Explanation
$$(7x+\frac12)(7x-\frac {1}{2})=49x^2-b$$
$$\Rightarrow 49x^2-\frac14=49x^2-b$$ ($$\because (a-b)(a+b)=a^2-b^2$$)
$$\Rightarrow b=\frac14$$
Option C is correct.
Evaluate using the expansion of $$(a+b)^2$$ or $$(a-b)^2$$ :
$$(92)^2$$
Report Question
0%
8444
0%
8464
0%
8474
0%
8414
Explanation
Given: $$ {92}^{2} = {(90 + 2)}^{2} $$
It is in the form of $$ {(a+b)}^{2} $$, where $$ a = 90, b = 2 $$.
Applying the formula $$ { (a+b) }^{ 2 } = {a}^{2} + { b }^{ 2 } + 2ab $$ we get,
$$ {92}^{2} = {(90 + 2)}^{2} $$
$$= {90}^{2} + { 2 }^{ 2 } + 2\times 90 \times 2 $$
$$= 8100 + 4 + 360 = 8464 $$
Evaluate using expansion of $$(a+b)^2$$ or $$(a-b)^2$$ :
$$(188)^2$$
Report Question
0%
35444
0%
35484
0%
35344
0%
35384
Explanation
$$ {188}^{2} = {(200 - 12)}^{2} $$
It is the form of $$ {(a - b)}^{2} $$, where $$ a = 200, b = 12 $$
Applying the formula $$ { (a - b) }^{ 2 } = {a}^{2} + { b }^{ 2 } - 2ab $$
$$ {188}^{2} = {(200 - 12)}^{2} = {200}^{2} + { 12 }^{ 2 } - 2\times 200 \times 12 = 40000 + 144 - 4800 = 35344 $$
Obtain the product of
$$2, 4y, 8y^2, 16y^3$$
Report Question
0%
$$1045y^5$$
0%
$$1024y^6$$
0%
$$1634y^3$$
0%
$$102y^5$$
Explanation
Product of $$2,4y,8y^2,16y^3$$
$$=(2)\times (4y)\times (8y^2)\times (16y^3)$$
$$=2\times 4 \times 8\times 16 \times y^{(1+2+3)}$$ $$[\because a^m \times a^n =a^{m+n} ]$$
$$=1024y^6$$
The product of $$\displaystyle\frac{2}{3}xy$$ and $$\dfrac{3}{2}xz$$ is equal to
Report Question
0%
$$\displaystyle\frac{1}{6}$$ xyz
0%
$$x^2yz$$
0%
$$6x^2yz$$
0%
none of these
Explanation
$$\dfrac{2}{3} xy \times \dfrac{3}{2}xz$$ = $$x^2yz$$
Obtain the product of
$$ a,-a^2,a^3$$
Report Question
0%
$$a^5$$
0%
$$-a^6$$
0%
$$a^3$$
0%
$$-a^5$$
Explanation
$$a,a^2,a^3$$
$$=(a)\times (-a^2)\times (a^3)$$
$$=-a^6$$
Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
$$xy, 2x^2y, 2xy^2$$
Report Question
0%
$$16x^3y^4$$
0%
$$4x^2y^3$$
0%
$$4x^4y^4$$
0%
$$4x^2y^2$$
Explanation
Given: $$xy, 2x^2y, 2xy^2$$
Volume$$=l\times b\times h$$
$$=(xy\times 2x^2y\times 2xy^2)$$
$$=4x^4y^4$$
Find the product of
$$4p, 0$$
Report Question
0%
4
4
0%
$$1$$
0%
$$4p$$
0%
$$0$$
Explanation
$$4p\times(0)$$
$$=0$$
Obtain the volume of rectangular boxes with the given length, breadth and height respectively.
$$2p, 4q, 8r$$
Report Question
0%
$$64pqr$$
0%
$$16pqr$$
0%
$$32pqr$$
0%
$$8pr + r$$
Explanation
Given : $$l,b,h$$ of the rectangular box respectively are: $$2p, 4q, 8r$$
Volume$$=l\times b\times h$$
$$=(2\times 4\times 8)pqr$$
$$=64pqr$$
Obtain the product of
$$a, 2b, 3c, 6abc$$
Report Question
0%
$$36a^2b^2c^3$$
0%
$$36a^3b^2c^2$$
0%
$$36a^2b^2c^2$$
0%
$$36a^2b^3c^2$$
Explanation
$$a, 2b, 3c, 6abc$$
$$=(a)\times (2b)\times (3c)\times (6abc)$$
$$=36a^2b^2c^2$$
Carry out the multiplication of the expressions of a given pair.
$$ab,a -b$$
Report Question
0%
$$ab-b$$
0%
$$a^2b+ab$$
0%
$$ab-ab^2$$
0%
$$a^2b-ab^2$$
Explanation
Given: $$ab,a -b$$
$$=ab\times (a-b)$$
$$=ab\times a-ab\times b$$
$$=a^2b-ab^2$$
Simplify:
$$x(3x+2)$$
Report Question
0%
$$3x^2+2$$
0%
$$3x+2$$
0%
$$3x^2+2x$$
0%
$$3x^2+x$$
Explanation
$$x(3x+2)=3x^2+2$$
$$L.H.S=x(3x+2)$$
$$=3x^2+2x \neq R.H.S$$
The correct statement is $$x(3x+2)=3x^2+2x$$
Carry out the multiplication of the given expressions.
$$a + b, 7a^2b^2$$
Report Question
0%
$$a^3b^2 - 7a^2b^3$$
0%
$$a^3b^2 + 7a^2b^3$$
0%
$$7a^3b^3 + 7a^2b^2$$
0%
$$7a^3b^2 + 7a^2b^3$$
Explanation
Given: $$a + b, 7a^2b^2$$
Product$$=(a+b)\times 7a^2b^2$$
$$=(a\times 7a^2b^2) + (b\times 7a^2b^2)$$
$$=7a^3b^2+7a^2b^3$$
The value of $$(x^2y)(2x)(3y^3)$$ is
Report Question
0%
$$12x^2y^4$$
0%
$$6x^3y^4$$
0%
$$3x^3y^4$$
0%
$$x^3y^4$$
Explanation
$$(x^2y)(2x)(3y^3)$$
First: $$(2)(3) = 6$$
Then: $$(x^3)(y^4)$$
So, $$(x^2y)(2x)(3y^3)= 6x^3y^4$$
$$(a + b)^2-(a -b)^2$$ will be equal to $$2ab$$:
Report Question
0%
True
0%
False
0%
Ambiguous
0%
Data insufficient
Explanation
$$(a+b)^2 -(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)$$
$$=a^2+2ab+b^2-a^2+2ab-b^2$$
$$=4ab$$.
Hence, answer is $$4ab$$ but not $$2ab$$.
Option $$B$$ is correct.
Find the solution of $$(2x+6)^2$$.
Report Question
0%
$$x^2+24x+36$$
0%
$$4x^2+24x+6$$
0%
$$4x^2+24x+36$$
0%
$$4x^2-24x+36$$
Explanation
$$(2x+6)^2 = (2x+6)(2x+6)$$
= $$2x.2x+2x.6+6.2x+6.6$$
= $$4x^2+12x+12x+36$$
= $$4x^2+24x+36$$
What is the coefficient of $$x$$ when $$-x + 6$$ is multiplied by $$2x - 3$$?
Report Question
0%
$$-15$$
0%
$$15$$
0%
$$-9$$
0%
$$9$$
Explanation
We have, $$(-x+6)(2x-3) $$
$$=-x(2x-3)+6(2x-3)$$
$$=-2x^2+3x+12x-18 $$
$$= -2x^2+15x-18$$
Hence, coefficient of $$x$$ is $$15$$.
The coefficient of the product of two monomials is not equal to the product of their coefficient.
Report Question
0%
True
0%
False
0%
Sometimes
0%
Data insufficient
Explanation
The given statement is not true. We take an example.
$$2x, 3y$$ are two monomials with coefficients $$2$$ and $$3$$ respectively.
Now, $$(2x)(3y)=6xy$$, it is also monomial with coefficient $$6,$$ which is a product of $$2$$ and $$3.$$
Hence, option $$B$$ is correct.
$$107\times 107+93\times93=$$?
Report Question
0%
$$19578$$
0%
$$19418$$
0%
$$20098$$
0%
$$21908$$
0%
None of these
Explanation
$$107\times 107+93\times 93={107}^{2}+{93}^{2}$$
$$={(100+7)}^{2}+{(100-7)}^{2}$$
$$=2\times [{(100)}^{2}+{7}^{2}]$$ ...................... [Ref: $${(a+b)}^{2}+{(a-b)}^{2}=2({a}^{2}+{b}^{2})$$]
$$=20098$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 8 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
