Algebraic Expressions and Identities Test

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Algebraic Expressions and Identities Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of $$47\times 53$$.

A
$$2451$$

B
$$2471$$

C
$$2491$$

D
$$2501$$

Solution

Given, $$47\times 53 $$
$$= (50-3)(50+3)$$.

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

Then,
$$47\times 53 = (50-3)(50+3)$$
$$=(50)^2-3^2=2500-9=2491$$.

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

Find the value of $$87^2-13^2$$.

A
$$7300$$

B
$$7350$$

C
$$7400$$

D
$$7450$$

Solution

Given, $$87^2-13^2$$.

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

Then,
$$=87^2-13^2$$
$$=(87+13)(87-13)$$
$$=100\times 74=7400$$.

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

Product of the following monomials $$4x, -5y^3, 6xy$$ is

A
$$120x^2y^2$$

B
$$120xy^4$$

C
$$-120x^2y^4$$

D
$$-120x^2y^3$$

Solution

$$4x\times (-5y^3)\times (6xy) = (-20xy^3)\times (6xy)$$
$$=-120x^2y^4$$
Question 4
1 / -0

Product of $$6a^2-7b+5ab$$ and $$2ab$$ is

A
$$12a^3b-14ab^2+10ab$$

B
$$12a^3b-14ab^2+10a^2b^2$$

C
$$6a^2-7b+7ab$$

D
$$12a^2b-7ab+10ab$$

Solution

$$2ab(6a^2-7b+5ab)=2ab\times 6a^2-2ab\times 7b+2ab\times 5ab$$
$$= 12a^3b-14ab^2+10a^2b^2$$
Question 5
1 / -0

Find the value of $$52^2$$ using standard identity.

A
$$2604$$

B
$$2704$$

C
$$2804$$

D
$$2904$$

Solution

Given, $$52^2$$
$$=(50+2)^2$$.

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

Then,
$$52^2=(50+2)^2$$
$$=50^2+2^2+2\times 50\times 2$$
$$=2500+4+200$$
$$=2704$$.

Therefore, option $$B$$ is correct.
Question 6
1 / -0

Find the value of $$2p(p+q)-2p(p-q)$$

A
$$4p^2$$

B
$$4q^2$$

C
$$2pq$$

D
$$4pq$$

Solution

$$2p(p+q) - 2p(p-q)$$
$$=2p^2+2pq-2p^2+2pq=4pq$$
Question 7
1 / -0

Find the value of $$mn\times np\times mp \times mnp$$

A
$$m^2n^2p^2$$

B
$$mnp$$

C
$$m^3n^3p^3$$

D
1

Solution

We know $$a^n \times a^m = a^{(m+n)} ]$$

Hence $$mn\times np\times pm\times mnp $$ $$=m^{(1+1+1)} \times n^{(1+1+1)} \times p^{(1+1+1)}$$

$$=m^3n^3p^3$$
Question 8
1 / -0

Find the product of $$(x+4)(x+10)$$

A
$$x^2+40$$

B
$$x^2+4x+10$$

C
$$x^2+6x+40$$

D
$$x^2+14x+40$$

Solution

$$(x+4)(x+10) = x(x+10)+4(x+10)$$
$$ = x^2+10x+4x+40$$
$$ =x^2+14x+40$$
Question 9
1 / -0

Simplify: $$\cfrac { { m }^{ 8 }{ p }^{ 7 }{ r }^{ 12 } }{ { m }^{ 3 }{ r }^{ 9 }{ p }^{ } } \times { p }^{ 2 }{ r }^{ 3 }{ m }^{ 4 }$$

A
$${m}^{5}{p}^{8}{r}^{3}$$

B
$${m}^{9}{p}^{8}{r}^{7}$$

C
$${m}^{9}{p}^{7}{r}^{4}$$

D
$${m}^{9}{p}^{8}{r}^{6}$$

Solution

$$\cfrac { { m }^{ 8 }{ p }^{ 7 }{ r }^{ 12 } }{ { m }^{ 3 }{ r }^{ 9 }{ p }^{ } } \times { p }^{ 2 }{ r }^{ 3 }{ m }^{ 4 }$$
$$\Rightarrow$$
$$\cfrac { { m }^{ 12 }{ p }^{ 9 }{ r }^{ 1 5} }{ { m }^{ 3 }{ r }^{ 9 }{ p }^{1 } } $$
$$\Rightarrow$$
$$ m ^9 p ^8 r ^{ 6 } $$ (option D)
Question 10
1 / -0

Find the value of $$3p(p^2+q^2)$$ at $$ p=0, q=1$$

A
$$0$$

B
$$1$$

C
$$3$$

D
$$5$$

Solution

We need to find value of $$3p(p^2+q^2)$$ at $$p=0, q=1$$

Therefore, $$3p(p^2+q^2)=3p^3+3pq^2$$

Now, on substituting values of $$p$$ and $$q$$, we get

$$3p^2+3pq^2=3(0)^2+3(0)(1)^2$$

$$=0+0=0$$

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Re-Attempt Weekly Quiz Competition

Algebraic Expressions and Identities Test - 34

Result

Algebraic Expressions and Identities Test - 34

Score

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Rank

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SHARING IS CARING

Question 1

$$2451$$

$$2471$$

$$2491$$

$$2501$$

Solution

Question 2

$$7300$$

$$7350$$

$$7400$$

$$7450$$

Solution

Question 3

$$120x^2y^2$$

$$120xy^4$$

$$-120x^2y^4$$

$$-120x^2y^3$$

Solution

Question 4

$$12a^3b-14ab^2+10ab$$

$$12a^3b-14ab^2+10a^2b^2$$

$$6a^2-7b+7ab$$

$$12a^2b-7ab+10ab$$

Solution

Question 5

$$2604$$

$$2704$$

$$2804$$

$$2904$$

Solution

Question 6

$$4p^2$$

$$4q^2$$

$$2pq$$

$$4pq$$

Solution

Question 7

$$m^2n^2p^2$$

$$mnp$$

$$m^3n^3p^3$$

1

Solution

Question 8

$$x^2+40$$

$$x^2+4x+10$$

$$x^2+6x+40$$

$$x^2+14x+40$$

Solution

Question 9

$${m}^{5}{p}^{8}{r}^{3}$$

$${m}^{9}{p}^{8}{r}^{7}$$

$${m}^{9}{p}^{7}{r}^{4}$$

$${m}^{9}{p}^{8}{r}^{6}$$

Solution

Question 10

$$0$$

$$1$$

$$3$$

$$5$$

Solution

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