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CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 7 - MCQExams.com
CBSE
Class 8 Maths
Algebraic Expressions And Identities
Quiz 7
Find the product $$(a+b)(a-2b)$$, when $$a=1$$, $$ b=2$$.
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0%
$$8$$
0%
$$10$$
0%
$$9$$
0%
$$-9$$
Explanation
$$=(a+b)(a-2b) =a\times a+b\times (-2b)+b\times a+a\times (-2b)$$
$$=a^2-2b^2-ab$$
$$a=1, b=2$$
$$=1-2\times 2^2-1\times 2$$
$$=-9$$
$$(4x + 1)\times (4x + 1) =$$
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$$16x^{2} + 8x + 2$$
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$$16x^{2} + 8x + 1$$
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$$16x^{2} + 8x + 3$$
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None of the above
Explanation
Given that: $$ (4x+1)\times(4x+1)$$
No we can see that:
$$ (4x+1)\times(4x+1) =$$
$$ 4x\times(4x+1)+1\times (4x+1)$$
$$\Rightarrow (4x\times 4x)+(4x\times 1)+(1\times 4x)+(1\times 1)$$
$$\Rightarrow 16x^2+4x+4x+1$$
$$\Rightarrow 16x^2+8x+1$$
Find the product $$(x+4)(x^2+2x+1)$$
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$$x^3+4x^2+9x+4$$
0%
$$x^3+6x^2+9x+4$$
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$$x^3+4x^2+8x+4$$
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$$x^3+6x^2+8x+4$$
Explanation
$$(x+4)(x^2+2x+1) = x\times(x^2+2x+1)+4(x^2+2x+1)$$
$$=x^3+2x^2+x+4x^2+8x+4$$
$$=x^3+6x^2+9x+4$$
Find the value of $$99\times 101$$ using standard identity.
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0%
$$9999$$
0%
$$9989$$
0%
$$9979$$
0%
$$1009$$
Explanation
Given, $$99\times 101=(100-1)(100+1)$$.
We know, $$(a+b)(a-b) = a^2-b^2$$.
Then,
$$99\times 101=(100-1)(100+1)$$
$$=(100)^2-1^2$$
$$=10000-1$$
$$=9999$$.
Therefore, option $$A$$ is correct.
Using standard identity, find the value of $$102^2$$.
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0%
$$10402$$
0%
$$10408$$
0%
$$10204$$
0%
$$10404$$
Explanation
Given, $$102^2$$
$$=(100+2)^2$$.
We know, $$(a+b)^2=a^2+2ab+b^2$$.
Then,
$$102^2$$
$$=(100+2)^2$$
$$=100^2+2^2+2(2)(100) $$
$$=10000+4+400 $$
$$= 10404$$.
Therefore, option $$D$$ is correct.
Using standard identity, find $$(2x-y)^2$$.
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$$4x^2+y^2-4xy$$
0%
$$4x^2+y^2+4xy$$
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$$2x^2+y^2-4xy$$
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$$4x^2+y^2-2xy$$
Explanation
Given,
$$(2x-y)^2$$
We know, $$(a-b)^2=a^2-2ab+b^2$$.
Then,
$$(2x-y)^2$$
$$=(2x)^2+y^2-2\times 2x\times y$$
$$= 4x^2+y^2-4xy$$.
Therefore, option $$A$$ is correct.
Find the product $$\dfrac{x}{2}\times (2x+4xy+6z)$$.
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$$2x^2+4x^2y+3xz$$
0%
$$624x^2y+3xz$$
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$$2x^2+2x^2y+6xz$$
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$$x^2+2x^2y+3xz$$
Explanation
$$\dfrac{x}{2} \times(2x+4xy+6x)$$
$$=\dfrac{x}{2}\times 2x+\dfrac{x}{2}\times 4xy+\dfrac{x}{2}\times 6z$$
$$=x^2+2x^2y+3xz$$
Find the product $$12m\times (n+2p+3m)$$.
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$$36m^2+12mn+12mp$$
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$$36m^2+12mn+24mp$$
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$$36m^2+mn+mp$$
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$$m^2+12mn+24mp$$
Explanation
$$12m(n+2p+3m)$$
$$=12m\times n +12 m\times 2p+12m\times 3m$$
$$= 12mn+24mp+36m^2$$
Using standard identity, find the value of $$(a+2b)^2$$.
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0%
$$a^2+2b^2+4ab$$
0%
$$a^2+4b^2+2ab$$
0%
$$a^2+4b^2+4ab$$
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$$2a^2+4b^2+4ab$$
Explanation
Given,
$$(a+2b)^2$$.
We know, $$(x+y)^2=x^2+2xy+y^2$$.
Then,
$$(a+2b)^2$$
$$=a^2+(2b)^2 +2\times a\times 2b$$
$$=a^2+4b^2+4ab$$.
Therefore, option $$C$$ is correct.
$$2x^2y\times (x^2y^2z+xyz)$$
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$$2x^4y^3z+2x^3y^2z$$
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$$2x^5y^3z+2x^3y^2$$
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$$2x^5y^3z+2x^4y^2$$
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$$x^5y^3z+2x^4y^2z$$
Explanation
$$2x^2y\times(x^2y^2z+xyz)$$
$$=2x^4y^3z+2x^3y^2z$$
Multiply $$\displaystyle \left((\frac{2}{5}x+y)(x+\frac{3}{5}y)\right)$$
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$$\displaystyle \frac{2}{5}x^2+\frac{3}{5}y^2$$
0%
$$\displaystyle \frac{2}{5}x^2+\frac{3}{5}y^2+\frac{3}{5}xy$$
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$$\displaystyle \frac{2}{5}x^2+\frac{3}{5}y^2+\frac{31}{25}xy$$
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$$\displaystyle \frac{2}{5}x^2+\frac{3}{5}y^2+\frac{xy}{5}$$
Explanation
$$((\dfrac{2}{5}x+y)(x+\dfrac{3}{5}y)$$
=$$\dfrac{2}{5}x^2+\dfrac{6}{25}xy+xy+\dfrac{3}{5}x^2$$
=$$\dfrac{2}{5}x^2+\dfrac{31}{25}xy+\dfrac{3}{5}y^2$$
Find the value of $$2p(p+q)-2p(p-q)$$
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0%
$$4p^2$$
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$$4q^2$$
0%
$$2pq$$
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$$4pq$$
Explanation
$$2p(p+q) - 2p(p-q)$$
$$=2p^2+2pq-2p^2+2pq=4pq$$
Using standard identity, find the value of $$99^2$$.
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0%
9901
0%
9801
0%
1001
0%
9701
Explanation
Given,
$$=99^2$$
$$=(100-1)^2$$
We know, $$(a-b)^2=a^2-2ab+b^2$$
Then,
$$99^2$$
$$=(100-1)^2$$
$$=100^2+1^2 - 2\times 100\times 1$$
$$= 10000+1- 200$$
$$=10001-200$$
$$=9801$$
Therefore, option $$B$$ is correct.
Which of the following is correct?
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$$(x-y)^2=x^2+2xy-y^2$$
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$$(x-y)^2=x^2-2xy+y^2$$
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$$(x-y)^2=x^2-y^2$$
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$$(x+y)^2=x^2+2xy-y^2$$
Explanation
We know, $$(a-b)^2=a^2-2ab+b^2$$.
Then,
$$(x-y)^2$$
$$=x^2-2(x)(y)+y^2$$ $$= x^2-2xy+y^2$$.
To verify this, let us consider:
$$(x-y)^2$$
$$=(x-y)$$
$$(x-y)$$
$$=x(x-y)-y(x-y)$$
$$=x^2-xy-yx+y^2$$
$$=x^2-xy-xy+y^2$$
$$=x^2-2xy+y^2$$.
Hence,
$$(x-y)^2$$
$$= x^2-2xy+y^2$$
Therefore, option $$B$$ is correct.
Find the value of $$49^2$$ using standard identity.
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0%
$$2391$$
0%
$$2401$$
0%
$$2411$$
0%
$$2421$$
Explanation
As we know that,
$$(a-b)^2=a^2-2ab+b^2$$
Substitute $$50$$ for $$a$$ and $$1$$ for $$b$$ in above formula,
$$\begin{aligned}{}{(50 - 1)^2}& = {50^2} - 2 \times 50 \times 1 + {1^2}\\{49^2}& = 2500 - 100 + 1\\ &= 2401\end{aligned}$$
Therefore, $$B$$ is correct.
Square of $$3a-4b$$ is:
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0%
$$9a^2+16b^2-24ab$$
0%
$$6a^2-8b^2$$
0%
$$9a^2-16b^2$$
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$$9a^2+16b^2+24ab$$
Explanation
Given, square of $$3a-4b$$
i.e.
$$(3a-4b)^2$$.
We know, $$(a-b)^2=a^2-2ab+b^2$$.
Then,$$(3a-4b)^2$$
$$= (3a)^2-2(3a)(4b)+(4 b)^2$$
$$=9a^2-24ab+16b^2$$.
Therefore, option $$A$$ is correct.
Find the value of $$47\times 53$$.
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0%
$$2451$$
0%
$$2471$$
0%
$$2491$$
0%
$$2501$$
Explanation
Given,
$$47\times 53 $$
$$= (50-3)(50+3)$$.
We know, $$(a+b)(a-b)=a^2-b^2$$.
Then,
$$47\times 53 = (50-3)(50+3)$$
$$=(50)^2-3^2=2500-9=2491$$.
Therefore, option $$C$$ is correct.
Find the value of $$87^2-13^2$$.
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0%
$$7300$$
0%
$$7350$$
0%
$$7400$$
0%
$$7450$$
Explanation
Given,
$$87^2-13^2$$.
We know, $$a^2-b^2=(a+b)(a-b)$$.
Then,
$$=87^2-13^2$$
$$=(87+13)(87-13)$$
$$=100\times 74=7400$$.
Therefore, option $$C$$ is correct.
Product of the following monomials $$4x, -5y^3, 6xy$$ is
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0%
$$120x^2y^2$$
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$$120xy^4$$
0%
$$-120x^2y^4$$
0%
$$-120x^2y^3$$
Explanation
$$4x\times (-5y^3)\times (6xy) = (-20xy^3)\times (6xy)$$
$$=-120x^2y^4$$
Product of $$6a^2-7b+5ab$$ and $$2ab$$ is
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$$12a^3b-14ab^2+10ab$$
0%
$$12a^3b-14ab^2+10a^2b^2$$
0%
$$6a^2-7b+7ab$$
0%
$$12a^2b-7ab+10ab$$
Explanation
$$2ab(6a^2-7b+5ab)=2ab\times 6a^2-2ab\times 7b+2ab\times 5ab$$
$$= 12a^3b-14ab^2+10a^2b^2$$
Find the value of $$52^2$$ using standard identity.
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0%
$$2604$$
0%
$$2704$$
0%
$$2804$$
0%
$$2904$$
Explanation
Given,
$$52^2$$
$$=(50+2)^2$$.
We know, $$(a+b)^2=a^2+b^2+2ab$$.
Then,
$$52^2=(50+2)^2$$
$$=50^2+2^2+2\times 50\times 2$$
$$=2500+4+200$$
$$=2704$$.
Therefore, option $$B$$ is correct.
Find the value of $$mn\times np\times mp \times mnp$$
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0%
$$m^2n^2p^2$$
0%
$$mnp$$
0%
$$m^3n^3p^3$$
0%
1
Explanation
We know $$a^n \times a^m = a^{(m+n)} ]$$
Hence
$$mn\times np\times pm\times mnp $$
$$=m^{(1+1+1)} \times n^{(1+1+1)} \times p^{(1+1+1)}$$
$$=m^3n^3p^3$$
Find the value of $$995^2-5^2$$.
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0%
$$99000$$
0%
$$9900$$
0%
$$99005$$
0%
$$990000$$
Explanation
Given,
$$995^2-5^2$$.
We know, $$a^2-b^2=(a+b)(a-b)$$.
Then,
$$995^2-5^2=(995+5)(995-5)$$
$$=1000\times 990$$
$$=990000$$.
Therefore, option $$D$$ is correct.
Find the value of $$3p(p^2+q^2)$$ at $$ p=0, q=1$$
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0%
$$0$$
0%
$$1$$
0%
$$3$$
0%
$$5$$
Explanation
We need to find value of $$3p(p^2+q^2)$$ at $$p=0, q=1$$
Therefore, $$3p(p^2+q^2)=3p^3+3pq^2$$
Now, on substituting values of $$p$$ and $$q$$, we get
$$3p^2+3pq^2=3(0)^2+3(0)(1)^2$$
$$=0+0=0$$
The value of $$(a+b)^2-2(a-b)^2+(a-b)(a+b)$$ is:
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0%
$$4ab-b^2$$
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$$2ab-b^2$$
0%
$$3ab-b^2$$
0%
$$6ab-2b^2$$
Explanation
Given, $$(a+b)^2-2(a-b)^2+(a-b)(a+b)$$.
We know,
$$(a+b)^{2}=a^2+2ab+b^2$$,
$$(a-b)^{2}=a^2-2ab+b^2$$
and $$(a+b)(a-b)=a^2-b^2$$.
Then,
$$(a+b)^2-2(a-b)^2+(a-b)(a+b)$$
$$=(a^2+b^2+2ab)-2(a^2+b^2-2ab)+(a^2-b^2)$$
$$=a^2+b^2+2ab-2a^2-2b^2+ 4ab+a^2-b^2$$
$$=2a^2+b^2+6ab-2a^2-3b^2$$
$$=6ab-2b^2$$.
Therefore, option $$D$$ is correct.
Find the value of $$1.05\times 0.95$$ using standard identity.
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0%
$$0.9985$$
0%
$$0.9975$$
0%
$$0.9875$$
0%
$$0.9995$$
Explanation
Given, $$1.05\times 0.95$$
$$=(1+0.05)(1-0.05)$$.
We know,
$$(a+b)(a-b)=a^2-b^2$$.
Then,
$$1.05\times 0.95$$
$$=(1+0.05)(1-0.05)$$
$$=1^2-(0.05)^2$$
$$=1-0.0025$$
$$=0.9975$$.
Therefore, option $$B$$ is correct.
Find the value of $$199\times 201$$.
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0%
$$39989$$
0%
$$40001$$
0%
$$39999$$
0%
$$400011$$
Explanation
Given,
$$199\times 201$$
$$ =(200-1)(200+1) $$.
We know, $$a^2-b^2=(a-b)(a+b)$$.
Then,
$$199\times 201$$
$$ =(200-1)(200+1) $$
$$= 200^2-1^2$$
$$= 40000-1$$
$$=39999$$.
Therefore, option $$C$$ is correct.
Find the value of $$(-2x^2)\times (-3y^2)\times (-6z^2)$$.
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0%
$$-36xyz$$
0%
$$36xyz$$
0%
$$-36x^2y^2z$$
0%
$$-36x^2y^2z^2$$
Explanation
$$-2x^2\times -3y^2\times -6z^2$$
$$=-2\times -3\times -6\times x^2\times y^2\times z^2$$
$$=-36x^2y^2z^2$$
Find the value of $$(a+b)^2-(a-b)^2$$.
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0%
$$ab$$
0%
$$2ab$$
0%
$$3ab$$
0%
$$4ab$$
Explanation
Given, $$(a+b)^2-(a-b)^2$$.
We know,
$$(a+b)^2$$
$$=a^2+b^2+2ab$$
and
$$(a-b)^2$$
$$a^2+b^2-2ab$$.
Then,
$$(a+b)^2-(a-b)^2$$
$$=a^2+b^2+2ab-(a^2+b^2-2ab)$$
$$=a^2+b^2+2ab-a^2-b^2+2ab$$
$$= 4ab$$.
Therefore, option $$D$$ is correct.
Find the value of $$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)$$.
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0%
$$abc$$
0%
$$a^2+b^2+c^2$$
0%
$$0$$
0%
$$1$$
Explanation
We know, $$(a-b)(a+b)=a^2-b^2$$.
$$\therefore (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)$$
$$=(a^2-b^2)+(b^2-c^2)+(c^2-a^2)$$
$$=a^2-b^2+b^2-c^2+c^2-a^2$$
$$=0$$.
Therefore, option $$C$$ is correct.
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