Algebraic Expressions and Identities Test

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Algebraic Expressions and Identities Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Multiply:
$$ \frac{2}{3}ab $$ by $$ -\frac{1}{4}a^2b $$

A
$$-\frac{1}{6}a^3b^3 $$

B
$$-\frac{1}{6}a^3b^2 $$

C
$$-\frac{1}{6}a^4b^2 $$

D
$$-\frac{1}{6}a^3b^4 $$

Solution

$$\frac{2}{3}ab \times -\frac{1}{4}a^2b$$
$$=\frac{2}{3}(-\frac{1}{4})a^{1+2}b^{1+1}$$
$$=-\frac{2}{12}a^3b^2$$
Question 2
1 / -0

Multiply:
$$ -5cd^2 $$ by $$-5cd^2 $$

A
$$25c^2d^5 $$

B
$$25c^3d^4 $$

C
$$25c^2d^3 $$

D
$$25c^2d^4 $$

Solution

$$-5cd^2 \times -5cd^2$$
$$=(-5)(-5)c^{1+1}d^{2+2}$$
$$=25c^2d^4$$
Question 3
1 / -0

Use the identity $$ (a+b)(a-b) = a^2-b^2$$ to evaluate:
$$33\times 27 $$.

A
$$891$$

B
$$881$$

C
$$981$$

D
$$841$$

Solution

We know, $$ 33 \times 27 = (30 + 3) \times (30 - 3) $$ .

Applying the formula $$ (a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 } $$, where $$ a = 30 , b = 3 $$,
we get,
$$ 33 \times 27 = (30 + 3) \times (30 - 3) = { 30 }^{ 2 }-{ 3 }^{ 2 } = 900 - 9 = 891 $$ .

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

Use the identity $$ (a+b)(a-b) = a^2-b^2$$ to evaluate:
$$21\times 19 $$.

A
$$389$$

B
$$399$$

C
$$289$$

D
$$429$$

Solution

We know, $$ 21 \times 19 = (20+ 1) \times (20 - 1) $$ .

Applying the formula $$ (a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 } $$, where $$ a = 20 , b = 1 $$,
we get,
$$ 21 \times 19 = (20+ 1) \times (20 - 1) = { 20 }^{ 2 }-{ 1 }^{ 2 } = 400 - 1 = 399 $$ .
Therefore, option $$B$$ is correct.
Question 5
1 / -0

Solve for $$x$$: $$(502)^{2}$$

A
$$2,52,004$$

B
$$2,62,104$$

C
$$2,22,004$$

D
$$2,52,864$$

Solution

$$(502)^2$$
$$=(500+2)^2$$
using, $$(a+b)^2=a^2+2ab+b^2$$
$$=(500)^2+2(500)(2)+(2)^2$$
$$=250000+2000+4$$
$$=252004$$
Question 6
1 / -0

Use identities to evaluate: $$(101)^{2}$$

A
$$11,601$$

B
$$12,761$$

C
$$10,111$$

D
$$10,201$$

Solution

$$(101)^2$$
$$=(100+1)^2$$
using, $$(a+b)^2=a^2+2ab+b^2$$
$$=(100)^2+2(100)(1)+1^2$$
$$=10000+200+1$$
$$=10201$$
Question 7
1 / -0

Use identities to solve: $$(97)^{2}$$

A
$$9,659$$

B
$$9,409$$

C
$$9,009$$

D
$$9,209$$

Solution

$$(97)^2$$
$$=(100-3)^2$$
using, $$(a-b)^2=a^2-2ab+b^2$$
$$=(100)^2-2(100)3+(3)^2$$
$$=10000-600+9$$
$$=9409$$
Question 8
1 / -0

The product of $$\displaystyle\frac{2}{3}xy$$ and $$\dfrac{3}{2}xz$$ is equal to

A
$$\displaystyle\frac{1}{6}$$ xyz

B
$$x^2yz$$

C
$$6x^2yz$$

D
none of these

Solution

$$\dfrac{2}{3} xy \times \dfrac{3}{2}xz$$ = $$x^2yz$$
Question 9
1 / -0

Simplify: $$\displaystyle { x }^{ 2 }\left( 3-{ 5y }^{ 2 } \right) +x\left( { xy }^{ 2 }-3x \right) -2y\left( y-{ 2x }^{ 2 }y \right) $$

A
$$\displaystyle { 6x }^{ 2 }-{ 2y }^{ 2 }-{ 3x }^{ 2 }y$$

B
$$\displaystyle { 6x }^{ 2 }$$

C
$$\displaystyle { 3x }^{ 2 }y$$

D
$$\displaystyle { -2y }^{ 2 }$$

Solution

We have to simplify $$\displaystyle { x }^{ 2 }\left( 3-{ 5y }^{ 2 } \right) +x\left( { xy }^{ 2 }-3x \right) -2y\left( y-{ 2x }^{ 2 }y \right) $$
$$\displaystyle ={ 3x }^{ 2 }-{ 5x }^{ 2 }{ y }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 }-{ 3x }^{ 2 }-{ 2y }^{ 2 }+{ 4x }^{ 2 }{ y }^{ 2 }$$
$$\displaystyle =\left( { 3x }^{ 2 }-{ 3x }^{ 2 } \right) +\left( -{ 5x }^{ 2 }{ y }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 }+{ 4x }^{ 2 }{ y }^{ 2 } \right) -{ 2y }^{ 2 }$$
$$=\displaystyle 0-{ 2y }^{ 2 }=-{ 2y }^{ 2 }$$.
Hence given expression simplified to $$-2y^2$$
Question 10
1 / -0

Find the product of $$4p, 0$$

A
4

B
$$1$$

C
$$4p$$

D
$$0$$

Solution

$$4p\times(0)$$
$$=0$$

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Re-Attempt Weekly Quiz Competition

Algebraic Expressions and Identities Test - 19

Result

Algebraic Expressions and Identities Test - 19

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SHARING IS CARING

Question 1

$$-\frac{1}{6}a^3b^3 $$

$$-\frac{1}{6}a^3b^2 $$

$$-\frac{1}{6}a^4b^2 $$

$$-\frac{1}{6}a^3b^4 $$

Solution

Question 2

$$25c^2d^5 $$

$$25c^3d^4 $$

$$25c^2d^3 $$

$$25c^2d^4 $$

Solution

Question 3

$$891$$

$$881$$

$$981$$

$$841$$

Solution

Question 4

$$389$$

$$399$$

$$289$$

$$429$$

Solution

Question 5

$$2,52,004$$

$$2,62,104$$

$$2,22,004$$

$$2,52,864$$

Solution

Question 6

$$11,601$$

$$12,761$$

$$10,111$$

$$10,201$$

Solution

Question 7

$$9,659$$

$$9,409$$

$$9,009$$

$$9,209$$

Solution

Question 8

$$\displaystyle\frac{1}{6}$$ xyz

$$x^2yz$$

$$6x^2yz$$

none of these

Solution

Question 9

$$\displaystyle { 6x }^{ 2 }-{ 2y }^{ 2 }-{ 3x }^{ 2 }y$$

$$\displaystyle { 6x }^{ 2 }$$

$$\displaystyle { 3x }^{ 2 }y$$

$$\displaystyle { -2y }^{ 2 }$$

Solution

Question 10

44

$$1$$

$$4p$$

$$0$$

Solution

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