Algebraic Expressions and Identities Test

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Algebraic Expressions and Identities Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Find the square of:
$$607 $$

A
368549

B
368449

C
368349

D
368249

Solution

$$607$$ can be written as $$600+7$$
$$\therefore {607}^{2} = {(600 + 7)}^{2} $$
It is the form of $$ {(a+b)}^{2} $$, where $$ a = 600, b = 7 $$
Applying the formula $$ { (a+b) }^{ 2 } = {a}^{2} + { b }^{ 2 } + 2ab $$
$$ {607}^{2} = {(600 + 7)}^{2} = {600}^{2} + { 7 }^{ 2 } +2\times 600 \times 7 = 360000 + 49 + 8400 = 368449 $$
Question 2
1 / -0

What is the missing term in the following product
$$\displaystyle \left ( 2a^{3}-3 \right )\left ( 5a^{3}-2 \right )=10a^{6}+$$_____+6

A
$$\displaystyle 19a^{3}$$

B
$$\displaystyle -19a^{3}$$

C
$$\displaystyle 16a^{3}$$

D
$$\displaystyle -16a^{3}$$

Solution

$$\displaystyle \left ( 2a^{3}-3 \right )\left ( 5a^{3}-2 \right )$$
$$\displaystyle =2a^{3}\times \left ( 5a^{3}-2 \right )+\left ( -3 \right )\times \left ( 5a^{3}-2 \right )$$
$$\displaystyle 10a^{6}-4a^{3}-15a^{3}+6$$
$$\displaystyle =10a^{6}-19a^{3}+6$$
$$\displaystyle =-19a^{3}$$
Question 3
1 / -0

Multiply $$2x$$, $$\dfrac{5y}{2}$$ and $$z^2$$

A
$$5x y z^2$$

B
$$2x y z^2$$

C
$$5x y^2 z$$

D
none of these

Solution

$$2x \times \dfrac{5y}{2} \times z^2 = 5xyz^2$$
Question 4
1 / -0

Find the product: $$\displaystyle \left( -3axy \right) \times \left( -15{ a }^{ 2 }{ xy }^{ 2 }z \right) $$

A
$$\displaystyle -45{ a }^{ 3 }{ x }^{ 2 }{ y }^{ 3 }z$$

B
$$\displaystyle -45{ a }^{ 2 }{ x }^{ 2 }{ y }^{ 2 }$$

C
$$\displaystyle 45{ a }^{ 3 }{ x }^{ 2 }{ y }^{ 3 }z$$

D
$$\displaystyle -45{ a }{ x }^{ 2 }{ y }^{ 2 }z$$

Solution

Required product $$=\displaystyle \left( -3axy \right) \times \left( -15{ a }^{ 2 }{ xy }^{ 2 }z \right) $$
=$$\displaystyle \left( -3\times -15 \right) \times \left( a\times { a }^{ 2 } \right) \times \left( x\times x \right) \times \left( y\times { y }^{ 2 } \right) \times z$$
$$\displaystyle =45{ a }^{ 3 }{ x }^{ 2 }{ y }^{ 3 }z$$.
Question 5
1 / -0

$$\displaystyle \left( x+8 \right) \left( x-8 \right) $$ is equal to

A
$$\displaystyle { x }^{ 2 }-16x+64$$

B
$$\displaystyle { x }^{ 2 }-64$$

C
$$\displaystyle { x }^{ 2 }+64$$

D
$$\displaystyle { x }^{ 2 }+16x-64$$

Solution

Required multiplication is
$$=\displaystyle \left( x+8 \right) \left( x-8 \right) $$
$$\displaystyle =x\left( x-8 \right) +8\left( x-8 \right) $$
$$\displaystyle ={ x }^{ 2 }-8x+8x-64$$
$$\displaystyle ={ x }^{ 2 }-64$$
Question 6
1 / -0

$$\displaystyle \left( \frac { 1 }{ 5 } x-\frac { 1 }{ 6 } y \right) \left( 5x+6y \right) $$ is equal to

A
$$\displaystyle { x }^{ 2 }-\frac { 11xy }{ 30 } -{ y }^{ 2 }$$

B
$$\displaystyle { x }^{ 2 }+\frac { 11xy }{ 30 } -{ y }^{ 2 }$$

C
$$\displaystyle { x }^{ 2 }+\frac { 11xy }{ 30 } +{ y }^{ 2 }$$

D
$$\displaystyle { x }^{ 2 }-\frac { 11xy }{ 30 } +{ y }^{ 2 }$$

Solution

Required multiplication
$$=\displaystyle \left( \frac { 1 }{ 5 } x-\frac { 1 }{ 6 } y \right) \left( 5x+6y \right) $$
$$\displaystyle =\frac { 1 }{ 5 } x\left( 5x+6y \right) -\frac { 1 }{ 6 } y\left( 5x+6y \right) $$
$$\displaystyle =\left( \frac { 1 }{ 5 } x \right) \left( 5x \right) +\left( \frac { 1 }{ 5 } x \right) \left( 6y \right) +\left( -\frac { 1 }{ 6 } y \right) \left( 5x \right) +\left( -\frac { 1 }{ 6 } y \right) \left( 6y \right) $$
$$\displaystyle ={ x }^{ 2 }+\frac { 6 }{ 5 } xy-\frac { 5 }{ 6 } xy-{ y }^{ 2 }$$

$$\displaystyle ={ x }^{ 2 }+\frac { 36xy-25xy }{ 30 } -{ y }^{ 2 }$$

$$={ x }^{ 2 }+\dfrac { 11xy }{ 30 } -{ y }^{ 2 }$$
Question 7
1 / -0

Evaluate: $$\displaystyle \left(\frac{2x}{7}\, -\, \frac{7y}{4}\right)^{2}$$.

A
$$\displaystyle \left(\frac{4x^{2}}{32}\,+\, \frac{49y^{2}}{16}\, +\, xy\right)$$

B
$$\displaystyle \left(\frac{4x^{2}}{15}\,+\, \frac{49y^{2}}{16}\, +\, xy\right)$$

C
$$\displaystyle \left(\frac{4x^{2}}{49}\,+\, \frac{49y^{2}}{16}\, -\, xy\right)$$

D
$$\displaystyle \left(\frac{4x^{2}}{19}\,+\, \frac{49y^{2}}{16}\, -\, xy\right)$$

Solution

Given, $$\left(\dfrac{2x}{7}-\dfrac{7y}{4}\right)^{2}$$.
We know, $$(a-b)^2=a^2+2ab+b^2$$.
Then,
$$\left(\dfrac{2x}{7}-\dfrac{7y}{4}\right)^{2}$$

$$=(\dfrac{2x}{7})^2-2(\dfrac{2x}{7})(\dfrac{7y}{4})+(\dfrac{7y}{4})^2$$

$$=\dfrac{4x^{2}}{49}- xy+ \dfrac{49y^{2}}{16} $$.
Therefore, ption $$C$$ is correct.
Question 8
1 / -0

Evaluate: $$\displaystyle\left(\dfrac{7}{8}{x}\, +\, \dfrac{4}{5}{y}\right )^{2}$$.

A
$$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{6}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$$

B
$$\displaystyle \frac{18}{77}{x^{2}}\, +\, \frac{16}{5}{y^{2}}\, +\, \frac{1}{5}{xy}$$

C
$$\displaystyle \frac{13}{22}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{1}{5}{xy}$$

D
$$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$$

Solution

Given, $$\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$$.
We know, $$(x+y)^2=x^2+2xy+y^2$$.
Then,
$$\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$$

$$=(\dfrac{7}{8}{x})^2+2(\dfrac{7}{8}{x})(\dfrac{4}{5}{y})+(\dfrac{4}{5}{y})^2$$

$$=\dfrac{49}{64}{x^{2}} +\dfrac{7}{5}{xy}+\dfrac{16}{25}{y^{2}} $$.

Therefore, option $$D$$ is correct.
Question 9
1 / -0

Evaluate: $$(4a\, +\, 3b)^{2}\, -\, (4a\, -\, 3b)^{2}\, +\, 48\, ab$$.

A
$$76 ab$$

B
$$96 ab$$

C
$$46 ab$$

D
$$106 ab$$

Solution

Given, $$(4a+3b)^{2} -(4a-3b)^{2}+48 ab$$.
We know, $$(a+b)^2=a^2+2ab+b^2$$
and $$(a-b)^2=a^2-2ab+b^2$$.
Then,
$$(4a+3b)^{2} -(4a-3b)^{2}+48 ab$$
$$=((4a)^2+2(4a)(3b)+(3b)^2)-((4a)^2-2(4a)(3b)+(3b)^2)+48ab$$
$$=(4a)^2+2(4a)(3b)+(3b)^2-(4a)^2+2(4a)(3b)-(3b)^2+48ab$$
$$=24ab+24ab+48ab$$
$$=48ab+48ab$$
$$=96ab$$.

Therefore, option $$B$$ is correct.
Question 10
1 / -0

Find the square of the following number: $$998$$

A
$$9,96,004$$

B
$$9,15,004$$

C
$$9,77,004$$

D
$$9,83,004$$

Solution

$$(998)^2=(1000-2)^2$$

Using, $$(a-b)^2=a^2-2ab+b^2$$, we get
$$(998)^2 = (1000)^2-2(1000)2+2^2$$
$$=1000000-4000+4$$
$$=996004$$

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Re-Attempt Weekly Quiz Competition

Algebraic Expressions and Identities Test - 26

Result

Algebraic Expressions and Identities Test - 26

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SHARING IS CARING

Question 1

368549

368449

368349

368249

Solution

Question 2

$$\displaystyle 19a^{3}$$

$$\displaystyle -19a^{3}$$

$$\displaystyle 16a^{3}$$

$$\displaystyle -16a^{3}$$

Solution

Question 3

$$5x y z^2$$

$$2x y z^2$$

$$5x y^2 z$$

none of these

Solution

Question 4

$$\displaystyle -45{ a }^{ 3 }{ x }^{ 2 }{ y }^{ 3 }z$$

$$\displaystyle -45{ a }^{ 2 }{ x }^{ 2 }{ y }^{ 2 }$$

$$\displaystyle 45{ a }^{ 3 }{ x }^{ 2 }{ y }^{ 3 }z$$

$$\displaystyle -45{ a }{ x }^{ 2 }{ y }^{ 2 }z$$

Solution

Question 5

$$\displaystyle { x }^{ 2 }-16x+64$$

$$\displaystyle { x }^{ 2 }-64$$

$$\displaystyle { x }^{ 2 }+64$$

$$\displaystyle { x }^{ 2 }+16x-64$$

Solution

Question 6

$$\displaystyle { x }^{ 2 }-\frac { 11xy }{ 30 } -{ y }^{ 2 }$$

$$\displaystyle { x }^{ 2 }+\frac { 11xy }{ 30 } -{ y }^{ 2 }$$

$$\displaystyle { x }^{ 2 }+\frac { 11xy }{ 30 } +{ y }^{ 2 }$$

$$\displaystyle { x }^{ 2 }-\frac { 11xy }{ 30 } +{ y }^{ 2 }$$

Solution

Question 7

$$\displaystyle \left(\frac{4x^{2}}{32}\,+\, \frac{49y^{2}}{16}\, +\, xy\right)$$

$$\displaystyle \left(\frac{4x^{2}}{15}\,+\, \frac{49y^{2}}{16}\, +\, xy\right)$$

$$\displaystyle \left(\frac{4x^{2}}{49}\,+\, \frac{49y^{2}}{16}\, -\, xy\right)$$

$$\displaystyle \left(\frac{4x^{2}}{19}\,+\, \frac{49y^{2}}{16}\, -\, xy\right)$$

Solution

Question 8

$$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{6}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$$

$$\displaystyle \frac{18}{77}{x^{2}}\, +\, \frac{16}{5}{y^{2}}\, +\, \frac{1}{5}{xy}$$

$$\displaystyle \frac{13}{22}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{1}{5}{xy}$$

$$\displaystyle \frac{49}{64}{x^{2}}\, +\, \frac{16}{25}{y^{2}}\, +\, \frac{7}{5}{xy}$$

Solution

Question 9

$$76 ab$$

$$96 ab$$

$$46 ab$$

$$106 ab$$

Solution

Question 10

$$9,96,004$$

$$9,15,004$$

$$9,77,004$$

$$9,83,004$$

Solution

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