Algebraic Expressions and Identities Test

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Algebraic Expressions and Identities Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$(7.2)^2$$ is (use an identity to expand):

A
$$49.9$$

B
$$14.4$$

C
$$51.84$$

D
$$49.04$$

Solution

We will use, $$(a+b)^2$$= $$a^2+2ab+b^2$$ to find the required result.

Substitute $$7$$ for $$a$$ and $$0.2$$ for $$B$$ in the above identity,
$$(7+0.2)^2=(7)^2+2(7)(0.2)+(0.2)^2$$
$$=49+2.8+0.04$$
$$=51.84$$

Hence, option $$C$$ is correct.
Question 2
1 / -0

The expansion of $$(2x - 3y)^2$$ is:

A
$$2x^2+3y^2+6xy$$

B
$$4x^2+9y^2-12xy$$

C
$$2x^2+3y^2-6xy$$

D
$$4x^2+9y^2+12xy$$

Solution

Given, $$(2x-3y)^2$$.

We know, $$(a-b)^2$$ $$ =$$ $$a^2-2ab+b^2$$.

Then,
$$(2x-3y)^2$$ $$ =$$ $$(2x)^2-2 (2x)(3y)+(3y)^2$$
$$=$$ $$4x^2-12xy+9y^2$$.

Therefore, option $$B$$ is correct.
Question 3
1 / -0

Which of the following relation is correct.

A
$$3(x - 9) = 3x - 27$$

B
$$3(x - 9) = 3x - 24$$

C
$$3(x - 8) = 3x - 27$$

D
None

Solution

Let us multiply the given monomial $$3$$ by a binomial $$x-9$$ as shown below:

$$3(x-9)=(3\times x)-(3\times 9)=3x-27$$

Hence, $$3(x-9)=3x-27$$.
Question 4
1 / -0

Which of the following relation is correct.

A
$$(3x + 2) = 3x^2 + 2x$$

B
$$x(3x + 2) = 3x^2 - 2x$$

C
$$x(3x + 2) = 3x^2 + 2x$$

D
None

Solution

Let us multiply the given monomial $$x$$ by a binomial $$3x+2$$ as shown below:

$$x(3x+2)=(x\times 3x)+(x\times 2)=3x^{ 2 }+2x$$

Hence, $$x(3x+2)=3x^{ 2 }+2x$$.
Question 5
1 / -0

Find the product of the following pairs:
$$-3l,-2m$$

A
$$6lm$$

B
$$-6lm$$

C
$$4lm$$

D
None of these

Solution

We multiply the given monomials $$-3l$$ and $$-2m$$ as shown below:

$$(-3l)\times (-2m)\\ =(-3\times -2)\times (l\times m)\\ =6\times lm\\ =6lm$$

Hence, the product of $$-3l$$ and $$-2m$$ is $$6lm$$.
Question 6
1 / -0

Find the product of the following pairs: $$6,7k$$

A
$$42k$$

B
$$21k$$

C
$$23k$$

D
None of these

Solution

We multiply the given monomials $$6$$ and $$7k$$ as shown below:

$$\Rightarrow 6\times 7k\\ =(6\times 7)\times k\\ =42\times k\\ =42k$$

Hence, the product of $$6$$ and $$7k$$ is $$42k$$.
Question 7
1 / -0

$$(a + b)^{2} - (a - b)^{2}=$$ _____.

A
$$4ab$$

B
$$2ab$$

C
$$a^{2} + 2ab + b^{2}$$

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

Solution

Given, $$(a+b)^{2}-(a-b)^{2}$$.

We know, $$(a+b)^{2}$$ $$=\left ( a^{2}+2ab+b^{2} \right )$$
and $$(a-b)^{2}$$ $$=(a^{2}-2ab+b^{2})$$.

Then,
$$(a+b)^{2}-(a-b)^{2}$$
$$=\left ( a^{2}+2ab+b^{2} \right )-(a^{2}-2ab+b^{2})$$
$$=a^{2}+2ab+b^{2}-a^{2}+2ab-b^{2}\\=4ab$$.

Therefore, the required answer is $$4ab$$.

Hence, option $$A$$ is correct.
Question 8
1 / -0

The product of $$(x - 1)(2x - 3)$$ is _____

A
$$2x^{2} - 5x - 3$$

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

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

D
$$2x^{2} + 5x + 3$$

Solution

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

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

After combining the like terms we get:

$$=2x^{2}-5x+3$$
Question 9
1 / -0

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:

A
$$2$$

B
$$3$$

C
$$5$$

D
$$1$$

Solution

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.
Question 10
1 / -0

Find the product of $$\left(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\right)$$ and $$\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\right)$$.

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

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

C
$$4x^2-9y^4$$

D
$$\dfrac{1}{4}x^4+\dfrac{1}{9}y^4$$

Solution

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.

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Algebraic Expressions and Identities Test - 38

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Algebraic Expressions and Identities Test - 38

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SHARING IS CARING

Question 1

$$49.9$$

$$14.4$$

$$51.84$$

$$49.04$$

Solution

Question 2

$$2x^2+3y^2+6xy$$

$$4x^2+9y^2-12xy$$

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

$$4x^2+9y^2+12xy$$

Solution

Question 3

$$3(x - 9) = 3x - 27$$

$$3(x - 9) = 3x - 24$$

$$3(x - 8) = 3x - 27$$

None

Solution

Question 4

$$(3x + 2) = 3x^2 + 2x$$

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

$$x(3x + 2) = 3x^2 + 2x$$

None

Solution

Question 5

$$6lm$$

$$-6lm$$

$$4lm$$

None of these

Solution

Question 6

$$42k$$

$$21k$$

$$23k$$

None of these

Solution

Question 7

$$4ab$$

$$2ab$$

$$a^{2} + 2ab + b^{2}$$

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

Solution

Question 8

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

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

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

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

Solution

Question 9

$$2$$

$$3$$

$$5$$

$$1$$

Solution

Question 10

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

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

$$4x^2-9y^4$$

$$\dfrac{1}{4}x^4+\dfrac{1}{9}y^4$$

Solution

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