Algebraic Expressions and Identities Test

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Algebraic Expressions and Identities Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

$$(a+1)(a-2) = a^2-a-2$$, this equation is true for

A
$$a = 1 $$only

B
$$a = 0$$ only

C
$$a > 0$$ only

D
every value of $$a$$

Solution

$$(a+1)(a-2) = a^2-2a+a-2 = a^2-a-2 = RHS$$
This is correct for every value of $$a$$
Question 2
1 / -0

Find the value of $$99\times 101$$ using standard identity.

A
$$9999$$

B
$$9989$$

C
$$9979$$

D
$$1009$$

Solution

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.
Question 3
1 / -0

Using standard identity, find the value of $$102^2$$.

A
$$10402$$

B
$$10408$$

C
$$10204$$

D
$$10404$$

Solution

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.
Question 4
1 / -0

Using standard identity, find $$(2x-y)^2$$.

A
$$4x^2+y^2-4xy$$

B
$$4x^2+y^2+4xy$$

C
$$2x^2+y^2-4xy$$

D
$$4x^2+y^2-2xy$$

Solution

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.
Question 5
1 / -0

Using standard identity, find the value of $$(a+2b)^2$$.

A
$$a^2+2b^2+4ab$$

B
$$a^2+4b^2+2ab$$

C
$$a^2+4b^2+4ab$$

D
$$2a^2+4b^2+4ab$$

Solution

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.
Question 6
1 / -0

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

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

B
$$2x^5y^3z+2x^3y^2$$

C
$$2x^5y^3z+2x^4y^2$$

D
$$x^5y^3z+2x^4y^2z$$

Solution

$$2x^2y\times(x^2y^2z+xyz)$$
$$=2x^4y^3z+2x^3y^2z$$
Question 7
1 / -0

Using standard identity, find the value of $$99^2$$.

A
9901

B
9801

C
1001

D
9701

Solution

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.
Question 8
1 / -0

Find the value of $$995^2-5^2$$.

A
$$99000$$

B
$$9900$$

C
$$99005$$

D
$$990000$$

Solution

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.
Question 9
1 / -0

Find the value of $$49^2$$ using standard identity.

A
$$2391$$

B
$$2401$$

C
$$2411$$

D
$$2421$$

Solution

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.
Question 10
1 / -0

Square of $$3a-4b$$ is:

A
$$9a^2+16b^2-24ab$$

B
$$6a^2-8b^2$$

C
$$9a^2-16b^2$$

D
$$9a^2+16b^2+24ab$$

Solution

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.

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Re-Attempt Weekly Quiz Competition

Algebraic Expressions and Identities Test - 33

Result

Algebraic Expressions and Identities Test - 33

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SHARING IS CARING

Question 1

$$a = 1 $$only

$$a = 0$$ only

$$a > 0$$ only

every value of $$a$$

Solution

Question 2

$$9999$$

$$9989$$

$$9979$$

$$1009$$

Solution

Question 3

$$10402$$

$$10408$$

$$10204$$

$$10404$$

Solution

Question 4

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

$$4x^2+y^2+4xy$$

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

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

Solution

Question 5

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

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

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

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

Solution

Question 6

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

$$2x^5y^3z+2x^3y^2$$

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

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

Solution

Question 7

9901

9801

1001

9701

Solution

Question 8

$$99000$$

$$9900$$

$$99005$$

$$990000$$

Solution

Question 9

$$2391$$

$$2401$$

$$2411$$

$$2421$$

Solution

Question 10

$$9a^2+16b^2-24ab$$

$$6a^2-8b^2$$

$$9a^2-16b^2$$

$$9a^2+16b^2+24ab$$

Solution

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