Algebraic Expressions and Identities Test

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Algebraic Expressions and Identities Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

The product of $$(3x^2\, -\, 5x\, +\, 6)$$ and $$-8x^3$$ when $$x=0$$ is

A
$$\displaystyle\frac{1}{2}$$

B
$$2$$

C
$$1$$

D
$$0$$

Solution

$$(3x^2\,-\,5x\,+\,6)\, \times\, -8x^3$$
= $$-24x^5\,+\, 40x^4\,-\, 48x^3$$
Put $$x=0,$$ we get value as $$0$$
Question 2
1 / -0

$$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$$=______

A
$$\displaystyle \left ( 2a+2b \right )$$

B
$$\displaystyle \left ( 2a-2b \right )$$

C
$$\displaystyle 4ab$$

D
$$\displaystyle -4ab$$

Solution

Given, $$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$$.

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

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

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

The value of $$(501)^2\, -\, (500)^2$$ is :

A
$$1$$

B
$$101$$

C
$$1,001$$

D
None of these

Solution

Given, $$(501)^{2}- (500)^{2}$$.

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

Then,
$$(501)^{2}- (500)^{2}$$
$$=(501+500)(501-500)$$
$$=(1001)(1) $$
$$= 1001$$.

Therefore, option $$C$$ is correct.
Question 4
1 / -0

If$$(3x\, -\,4)\,(5x\, +\, 7)\, =\, 15x^2\, -\, ax\, -\, 28$$ then a = .........

A
$$1$$

B
$$-1$$

C
$$-2$$

D
none of these

Solution

(3x - 4) (5x + 7) = $$15x^2 \, -\, ax\, - 28$$
$$15x^2\, + \, x\, -\, 28\,=\, 15x^2\,-\, ax\,-\, 28$$
Comparing we get a = -1
Question 5
1 / -0

$$\displaystyle \left ( \frac{2}{5}ab+c \right )\left ( \frac{2}{5}ab-c \right )$$ is equal to

A
$$\displaystyle \frac{4}{25}a^{2}b^{2}-\frac{4}{5}abc+c^{2}$$

B
$$\displaystyle \frac{4}{25}a^{2}b^{2}+\frac{4}{5}abc+c^{2}$$

C
$$\displaystyle \frac{4}{25}a^{2}b^{2}-c^{2}$$

D
$$\displaystyle \frac{4}{25}a^{2}b^{2}+c^{2}$$

Solution

$$\displaystyle \left ( \frac{2}{5}ab+c \right )\left ( \frac{2}{5}ab-c \right )$$

= $$\dfrac 25 ab \left( \dfrac 25ab -c\right) + c \left(\dfrac 25 ab - c\right)$$

= $$\dfrac{4}{25}a^2b^2 - \dfrac 25 abc + \dfrac 25 abc - c^2$$

= $$\dfrac{4}{25}a^2b^2-c^{2}$$
Question 6
1 / -0

$$\displaystyle \left ( -a-b \right )\left ( b-a \right )$$ is equal to

A
$$\displaystyle a^{2}+b^{2}$$

B
$$\displaystyle b^{2}-a^{2}$$

C
$$\displaystyle a^{2}-b^{2}$$

D
$$\displaystyle -\left ( b^{2}+a^{2} \right )$$

Solution

Given, $$(-a-b)(b-a)$$
Taking the minus sign outside and rearranging
$$=-(b+a)(b-a)$$
Using the identity, $$x^2 - y^2 = (x+y)(x-y)$$
$$-(b+a)(b-a)$$
$$= - (b^2 - a^2)$$
$$= a^2 - b^2$$
Question 7
1 / -0

$$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$$ is equal to:

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

B
$$\displaystyle x^{4}-64$$

C
$$\displaystyle x^{4}-256$$

D
$$\displaystyle x^{2}-256$$

Solution

Given, $$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$$.

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

Then,
$$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$$
= $$\displaystyle \left ( x^{2}-4^{2} \right )\left ( x^{2}+16 \right )$$
= $$\displaystyle \left ( x^{2}-16 \right )\left ( x^{2}+16 \right )=\left ( x^{2} \right )^{2}-\left ( 16 \right )^{2}$$
= $$\displaystyle x^{4}-256 $$.

Therefore, option $$C$$ is correct.
Question 8
1 / -0

$$a^{2} + 4a + 4$$ =

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

B
$$(a + 1)^{2}$$

C
$$(a - 2)^{2}$$

D
$$(a - 1)^{2}$$

Solution

$$a^{2} + 4a + 4 = a^{2} + 2(a) (2)+ (2)^{2}$$
$$=(a + 2)^{2}$$
Question 9
1 / -0

Find the missing term in the following problem:
$$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, =\, \displaystyle \frac{9x^2}{16}\, +\, ..........\, +\, \displaystyle \frac{16y^2}{9}$$.

A
$$2xy$$

B
$$- 2xy$$

C
$$12xy$$

D
$$- 12xy$$

Solution

Given, $$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, $$.
We know, $$(a-b)^2=a^2-2ab+b^2$$.
Then,
$$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, $$
$$=\left( \displaystyle \frac{3x}{4} \right )^2\, -\, 2\, \left(\displaystyle \frac{3x}{4} \right )\left(\displaystyle \frac{4y}{3} \right )\, +\, \left(\displaystyle \frac{4y}{3}\right)^2$$
$$=\, \displaystyle \frac{9x^2}{16}\, -\, 2xy\, +\, \displaystyle \frac{16y^2}{9}$$
$$=\, \displaystyle \frac{9x^2}{16}\, +\, (-\, 2xy)\, +\, \displaystyle \frac{16y^2}{9}$$.
Hence, the missing term is $$-2xy$$.
Therefore, option $$B$$ is correct.
Question 10
1 / -0

Find the missing term in the following problem.
$$\displaystyle {\left (\frac{3x}{4}\, -\, \frac{4y}{3} \right )^2\, =\, \frac{9x^2}{16}\, +\, .........\, +\, \frac{16y^2}{9}}$$

A
$$2xy$$

B
$$-2xy$$

C
$$12xy$$

D
$$-12xy$$

Solution

The L.H.S. of the algebraic expression is of the form $${ \left( a-b \right) }^{ 2 }$$
where, $$a=\dfrac { 3x }{ 4 } \quad \& \quad b=\dfrac { 4y }{ 3 } $$.
$${ \left( a-b \right) }^{ 2 }={ a }^{ 2 }-2ab+{ b }^{ 2 }\\ \implies \left(\dfrac{3x}{4}-\dfrac{4y}{3}\right)={ \left( \dfrac { 3x }{ 4 } \right) }^{ 2 }-2\times \dfrac { 3x }{ 4 } \times \dfrac { 4y }{ 3 } +{ \left( \dfrac { 4y }{ 3 } \right) }^{ 2 }\\ =\dfrac { 9{ x }^{ 2 } }{ 16 } -2xy+\dfrac { 16{ y }^{ 2 } }{ 9 } $$.
So the middle term $$-2xy$$ is missing.

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Re-Attempt Weekly Quiz Competition

Algebraic Expressions and Identities Test - 28

Result

Algebraic Expressions and Identities Test - 28

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SHARING IS CARING

Question 1

$$\displaystyle\frac{1}{2}$$

$$2$$

$$1$$

$$0$$

Solution

Question 2

$$\displaystyle \left ( 2a+2b \right )$$

$$\displaystyle \left ( 2a-2b \right )$$

$$\displaystyle 4ab$$

$$\displaystyle -4ab$$

Solution

Question 3

$$1$$

$$101$$

$$1,001$$

None of these

Solution

Question 4

$$1$$

$$-1$$

$$-2$$

none of these

Solution

Question 5

$$\displaystyle \frac{4}{25}a^{2}b^{2}-\frac{4}{5}abc+c^{2}$$

$$\displaystyle \frac{4}{25}a^{2}b^{2}+\frac{4}{5}abc+c^{2}$$

$$\displaystyle \frac{4}{25}a^{2}b^{2}-c^{2}$$

$$\displaystyle \frac{4}{25}a^{2}b^{2}+c^{2}$$

Solution

Question 6

$$\displaystyle a^{2}+b^{2}$$

$$\displaystyle b^{2}-a^{2}$$

$$\displaystyle a^{2}-b^{2}$$

$$\displaystyle -\left ( b^{2}+a^{2} \right )$$

Solution

Question 7

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

$$\displaystyle x^{4}-64$$

$$\displaystyle x^{4}-256$$

$$\displaystyle x^{2}-256$$

Solution

Question 8

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

$$(a + 1)^{2}$$

$$(a - 2)^{2}$$

$$(a - 1)^{2}$$

Solution

Question 9

$$2xy$$

$$- 2xy$$

$$12xy$$

$$- 12xy$$

Solution

Question 10

$$2xy$$

$$-2xy$$

$$12xy$$

$$-12xy$$

Solution

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