Algebraic Expressions and Identities Test

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Algebraic Expressions and Identities Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$(-5{x}^{2}y) \times (-\dfrac{2}{3} x{y}^{2}z) \times (\dfrac{8}{15} xy{z}^{2}) \times (-\dfrac{1}{4} z)$$ is ______.

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

B
$$\dfrac{4}{9} {x}^{4}{y}^{4}{z}^{4}$$

C
$$-\dfrac{4}{9} {x}^{3}{y}^{3}{z}^{3}$$

D
$$\dfrac{4}{9} {x}^{3}{y}^{3}{z}^{3}$$

Solution

We have,
$$\quad \quad \left( -5{ x }^{ 2 }y \right) \times \left( -\dfrac { 2 }{ 3 } x{ y }^{ 2 }z \right) \times \left( \dfrac { 8 }{ 15 } xy{ z }^{ 2 } \right) \times \left( -\dfrac { 1 }{ 4 } z \right) \\ =\left( -5\times \left( -\dfrac { 2 }{ 3 } \right) \times \dfrac { 8 }{ 15 } \times \left( -\dfrac { 1 }{ 4 } \right) \right) \times \left( { x }^{ 2 }y\quad \times \quad x{ y }^{ 2 }z\quad \times \quad xy{ z }^{ 2 }\quad \times \quad z \right) \quad \\ =\left( -\dfrac { 2\times 2 }{ 3\times 3 } \right) \times \left( \left( { x }^{ 2 }\times x\times x \right) \times \left( y\times { y }^{ 2 }\times y \right) \times \left( z\times { z }^{ 2 }\times z \right) \right) \\ =\left( -\dfrac { 4 }{ 9 } \right) \times \left( { x }^{ 4 } \right) \times \left( { y }^{ 4 } \right) \times \left( { z }^{ 4 } \right) \\ =\left( -\dfrac { 4 }{ 9 } \right) \left( { x }^{ 4 } \right) \left( { y }^{ 4 } \right) \left( { z }^{ 4 } \right)$$
Answer is A
Question 2
1 / -0

Using identities, evaluate:
$$71^2$$.

A
$$5041$$

B
$$3041$$

C
$$6041$$

D
$$2041$$

Solution

Given, $$71^2$$,
i.e. $$71^2$$ $$=(70+1)^2$$.

We know,
$$(x+y)^{ 2 }=x^{ 2 }+y^{ 2 }+2xy$$.

Then,
$$71^{ 2 }=(70+1)^{ 2 }\\ =70^{ 2 }+(2\times 70\times 1) + 1^2\\ =4900+140+1\\ =5040+1 \\=5041.$$

Hence, $$71^2=5041$$.

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

Find the product $$24x^2(1-2x)$$ and evaluate if for $$x=2$$

A
$$288$$

B
$$384$$

C
$$-288$$

D
$$-384$$

Solution

Let us multiply the given monomial and binomial as shown below:

$$24{ x }^{ 2 }(1-2x)\\ =\left( 24{ x }^{ 2 }\times 1 \right) -\left( 24{ x }^{ 2 }\times 2x \right) \\ =24{ x }^{ 2 }-48{ x }^{ 3 }$$

Now, let us substitute $$x=2$$ in $$24{ x }^{ 2 }-48{ x }^{ 3 }$$ as follows:

$$24{ (2) }^{ 2 }-48{ (2) }^{ 3 }=(24\times 4)-(48\times 8)=96-384=-288$$

Hence, $$24{ x }^{ 2 }(1-2x)=-288$$ for $$x=2$$.
Question 4
1 / -0

The value of $${ \left( 1.02 \right) }^{ 2 }+{ \left( 0.98 \right) }^{ 2 }$$, corrected to three decimal places is:

A
$$2.001$$

B
$$2.004$$

C
$$2.006$$

D
$$1.995$$

Solution

Given, $$(1.02)^2+(0.98)^2$$.

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

Then,
$$(1.02)^2+(0.98)^2$$
$$=(1.02)^2+(0.98)^2+(2\times 1.02\times 0.98)-(2\times 1.02\times 0.98$$)
$$=(1.02+0.98)^2-1.9992$$
$$=(2)^2-1.9992$$
$$=4-1.9992$$
$$=2.0008$$.

When corrected to three decimal digit,
$$2.0008$$ $$=2.001$$.

Therefore, option $$A$$ is correct.
Question 5
1 / -0

Factorise : $${ (ax+by) }^{ 2 }+{ (2bx-2ay) }^{ 2 }-6abxy$$

A
$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+2{ b }^{ 2 }{ x }^{ 2 }+2{ a }^{ 2 }{ y }^{ 2 }+6abxy$$

B
$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }-12abxy$$

C
$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }+4abxy$$

D
$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }+6abxy$$

Solution

$$\left( ax+by\right)^2+\left( 2bx-2ay\right)^2-babxy$$
Here we will use the following identity,
$$\Rightarrow \left( m+n\right)^2=m^2+n^2+2mn$$
and $$\left(m-n\right)^2=m^2+n^2-2mn$$

Using this we have :
$$= \left(ax\right)^2+\left(by\right)^2+2axby+4\left(bx\right)^2+4\left(ay\right)^2-8axby-6abxy$$
$$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-14axby+2axby$$
$$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby$$
Hence, the answer is $$a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby.$$
Question 6
1 / -0

If $$xy=20$$ and $$(x+y)^{2}=70$$, then $$x^{2}+y^{2}$$ is equal to

A
$$25$$

B
$$35$$

C
$$30$$

D
$$50$$

Solution

$$(x+y)^2=x^2+y^2+2xy$$

It is given that $$(x+y)^2=70$$ and $$xy=20$$

$$\Rightarrow 70=x^2+y^2+(2\times 20)$$

$$x^2+y^2=70-40$$

$$=30$$
Question 7
1 / -0

Product of the following monomials $$4p$$, $$- 7q^{3}$$, $$-7pq$$ is

A
$$ 196 p^{2}q^{4}$$

B
$$ 196 pq^{4}$$

C
$$-196p^{2}q^{4}$$

D
$$ 196 p^{2}q^{3}$$

Solution

Given monomials are : $$4p$$, $$- 7q^{3}$$, $$-7pq$$

$$\therefore$$ Product $$=(4p)\times(- 7q^{3})\times(-7pq)=4\times (-7)\times (-7)\times p\times q^{3} \times pq =196p^{2}q^{4}$$

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

If $$x + y = 2013$$ and $$\frac{1}{x}+\frac{1}{y}=2013$$, what is the value of xy?

A
$$\frac{1}{2013}$$

B
4026

C
0

D
1
Question 9
1 / -0

Simplify the following expression:
$$(3x^2 * 7x^7)+ (2y^3 * 9y^{12})$$

A
$$21x^{14} + 18y{26}$$

B
$$10x^{9} + 11y{15}$$

C
$$21x^{14} + 18y{15}$$

D
$$21x^{9} + 18y{15}$$
Question 10
1 / -0

Simplify the following expression:
$$(2x^4y^7m^2z) ^{TM} (5x^2y^3m^8)$$

A
$$10x^6y^9m^{10}z $$

B
$$7x^6y^{10}m^{10}z$$

C
$$10x^5y^{10}m^{10}z$$

D
$$10x^6y^{10}m^{10}z$$

Solution

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Re-Attempt Weekly Quiz Competition

Algebraic Expressions and Identities Test - 39

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Algebraic Expressions and Identities Test - 39

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SHARING IS CARING

Question 1

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

$$\dfrac{4}{9} {x}^{4}{y}^{4}{z}^{4}$$

$$-\dfrac{4}{9} {x}^{3}{y}^{3}{z}^{3}$$

$$\dfrac{4}{9} {x}^{3}{y}^{3}{z}^{3}$$

Solution

Question 2

$$5041$$

$$3041$$

$$6041$$

$$2041$$

Solution

Question 3

$$288$$

$$384$$

$$-288$$

$$-384$$

Solution

Question 4

$$2.001$$

$$2.004$$

$$2.006$$

$$1.995$$

Solution

Question 5

$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+2{ b }^{ 2 }{ x }^{ 2 }+2{ a }^{ 2 }{ y }^{ 2 }+6abxy$$

$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }-12abxy$$

$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }+4abxy$$

$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }+6abxy$$

Solution

Question 6

$$25$$

$$35$$

$$30$$

$$50$$

Solution

Question 7

$$ 196 p^{2}q^{4}$$

$$ 196 pq^{4}$$

$$-196p^{2}q^{4}$$

$$ 196 p^{2}q^{3}$$

Solution

Question 8

$$\frac{1}{2013}$$

4026

0

1

Question 9

$$21x^{14} + 18y{26}$$

$$10x^{9} + 11y{15}$$

$$21x^{14} + 18y{15}$$

$$21x^{9} + 18y{15}$$

Question 10

$$10x^6y^9m^{10}z $$

$$7x^6y^{10}m^{10}z$$

$$10x^5y^{10}m^{10}z$$

$$10x^6y^{10}m^{10}z$$

Solution

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