MCQ Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 7

Find the product

$(a+b)(a-2b)$ , when

$a=1$ ,

$b=2$ .

0%
$8$
0%
$10$
0%
$9$
0%
$-9$

Explanation

$=(a+b)(a-2b) =a\times a+b\times (-2b)+b\times a+a\times (-2b)$

$=a^2-2b^2-ab$

$a=1, b=2$

$=1-2\times 2^2-1\times 2$

$=-9$

$(4x + 1)\times (4x + 1) =$

0%
$16x^{2} + 8x + 2$
0%
$16x^{2} + 8x + 1$
0%
$16x^{2} + 8x + 3$
0%
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

$(x+4)(x^2+2x+1)$

0%
$x^3+4x^2+9x+4$
0%
$x^3+6x^2+9x+4$
0%
$x^3+4x^2+8x+4$
0%
$x^3+6x^2+8x+4$

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.

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

$102^2$ .

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

$(2x-y)^2$ .

0%
$4x^2+y^2-4xy$
0%
$4x^2+y^2+4xy$
0%
$2x^2+y^2-4xy$
0%
$4x^2+y^2-2xy$

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

$\dfrac{x}{2}\times (2x+4xy+6z)$ .

0%
$2x^2+4x^2y+3xz$
0%
$624x^2y+3xz$
0%
$2x^2+2x^2y+6xz$
0%
$x^2+2x^2y+3xz$

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

$12m\times (n+2p+3m)$ .

0%
$36m^2+12mn+12mp$
0%
$36m^2+12mn+24mp$
0%
$36m^2+mn+mp$
0%
$m^2+12mn+24mp$

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

$(a+2b)^2$ .

0%
$a^2+2b^2+4ab$
0%
$a^2+4b^2+2ab$
0%
$a^2+4b^2+4ab$
0%
$2a^2+4b^2+4ab$

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.

$2x^2y\times (x^2y^2z+xyz)$

0%
$2x^4y^3z+2x^3y^2z$
0%
$2x^5y^3z+2x^3y^2$
0%
$2x^5y^3z+2x^4y^2$
0%
$x^5y^3z+2x^4y^2z$

Explanation

$2x^2y\times(x^2y^2z+xyz)$

$=2x^4y^3z+2x^3y^2z$

Multiply

$\displaystyle \left((\frac{2}{5}x+y)(x+\frac{3}{5}y)\right)$

0%
$\displaystyle \frac{2}{5}x^2+\frac{3}{5}y^2$
0%
$\displaystyle \frac{2}{5}x^2+\frac{3}{5}y^2+\frac{3}{5}xy$
0%
$\displaystyle \frac{2}{5}x^2+\frac{3}{5}y^2+\frac{31}{25}xy$
0%
$\displaystyle \frac{2}{5}x^2+\frac{3}{5}y^2+\frac{xy}{5}$

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

$2p(p+q)-2p(p-q)$

0%
$4p^2$
0%
$4q^2$
0%
$2pq$
0%
$4pq$

Explanation

$2p(p+q) - 2p(p-q)$

$=2p^2+2pq-2p^2+2pq=4pq$

Using standard identity, find the value of

$99^2$ .

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?

0%
$(x-y)^2=x^2+2xy-y^2$
0%
$(x-y)^2=x^2-2xy+y^2$
0%
$(x-y)^2=x^2-y^2$
0%
$(x+y)^2=x^2+2xy-y^2$

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.

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:

0%
$9a^2+16b^2-24ab$
0%
$6a^2-8b^2$
0%
$9a^2-16b^2$
0%
$9a^2+16b^2+24ab$

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

$47\times 53$ .

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

$87^2-13^2$ .

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

0%
$120x^2y^2$
0%
$120xy^4$
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

0%
$12a^3b-14ab^2+10ab$
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.

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

$mn\times np\times mp \times mnp$

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

$995^2-5^2$ .

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

$p=0, q=1$

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:

0%
$4ab-b^2$
0%
$2ab-b^2$
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.

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

$199\times 201$ .

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

$(-2x^2)\times (-3y^2)\times (-6z^2)$ .

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

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

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

$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)$ .

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.

CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 7 - 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 7 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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