Algebraic Expressions and Identities Test

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Algebraic Expressions and Identities Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Multiply the binomials $(2x + 5)$ and $(4x - 3)$

A
$8x^2+ 14x -15$

B
$7x^2-14x -15$

C
$x^2+ 14x -18$

D
None of these

Solution

$(2x + 5)\times (4x - 3)= 2x(4x - 3) + 5(4x - 3)$
$= 8x^2- 6x + 20x-15$
$= 8x^2+ 14x -15$
Question 2
1 / -0

Using identities, evaluate $8.9^2$ .

A
$74.53$

B
$73.42$

C
$79.21$

D
$73.21$

Solution

Given, $8.9^2$
$= (9 -0.1)^2$ .

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

Then,
$8.9^2$
$= (9 -0.1)^2$
$= (9)^2 -2(9) (0.1) + (0.1)^2$
$= 81 -1.8 + 0.01 = 79.20+0.01=79.21$ .

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

Using identities, evaluate $12.1^2 -7.9^2$ .

A
$84$

B
$72$

C
$69$

D
$87$

Solution

Given, $(12.1)^2-(7.9)^2$ .

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

Then,
$(12.1)^2-(7.9)^2$

$=(12.1+7.9)(12.1-7.9)$

$=20\times 4.2$

$=84$ .

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

Using $(x + a) (x + b) = x^2+(a+b)x+ab$ , find $103 \times 104$ .

A
$14982$

B
$11992$

C
$10712$

D
$11482$

Solution

Given, $103 \times 104$
$=(100+3)(100+4)$ .

Using, $(x+a)(x+b)=x^2+(a+b)x+ab$ .

Then,
$103 \times 104$
$=(100+3)(100+4)$
$=(100)^2+(3+4)(100) +(3)(4)$
$=(100)^2+(7)(100) +12$
$= 10000 +700 + 12$
$= 10712$ .

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

Simplify: $(7m -8n)^2 + (7m + 8n)^2$ .

A
$49m^2+164n^2$

B
$98m^2+128n^2$

C
$49m^2+64n^2$

D
$128m^2+64n^2$

Solution

Given, $(7m -8n)^2 + (7m + 8n)^2$ .

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

Then,
$(7m -8n)^2 + (7m + 8n)^2$
$= ((7m)^2-2(7m)(8n) + (8n)^2)+((7m)^2 + 2(7m)(8n) + (8n)^2)$
$= (49m^2-112mn + 64n^2)+(49m^2 + 112mn + 64n^2)$
$= 49m^2-112mn + 64n^2+49m^2 + 112m + 64n^2$
$= 98m^2 + 128n^2$ .

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

Using identities, evaluate $5.2^2$ .

A
$26.04$

B
$2704$

C
$27.04$

D
$2604$

Solution

Given, $(5.2)^2$

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

Then,
$(5.2)^2$
$=(5+0.2)^2$
$=5^2+(0.2)^2+2(5)(0.2)$
$=25+0.04+2(1)$
$=25+0.04+2$
$=27.04$ .

Therefore, option $C$ is correct.
Question 7
1 / -0

Simplify $(x^2-5)(x+5)+ 25$

A
$x^2(x+5) -5(x+5)+15$

B
$x^2(x+5) -5(x-5)+25$

C
$x^2(x+5) -5(x+5)+25$

D
$x^2(x+5) -(x+5)+25$

Solution

Given, $(x^2-5)(x + 5) + 25$

While multiplying a binomial with another binomial, each term of the first binomial must be multiplied by each term of the second binomial.

$= x^2(x+5) -5(x+5)+25$
Question 8
1 / -0

Simplify: $(a^2 + 5)(b^3 + 3) + 5$

A
$a^3(b^3+3)+ 5(b^3 + 3)+ 5$

B
$a^2(b^3+3)+ 5(b^3 + 3)+ 5$

C
$a^3(b^3+3)+ 5(b^3 + 3)- 5$

D
$a^2(b^3+3)+ 5(b^3 - 3)+ 5$

Solution

$(a^2+ 5)(b^3+ 3) + 5$
$= a^2(b^3+3)+ 5(b^3 + 3)+ 5$
$= a^2b^3+ 3a^2+ 5b^3+ 15 + 5$
$= a^2b^3 + 3a^2+ 5b^3+ 20$
Question 9
1 / -0

$3x \times (4x + 2)$

A
$12x^{2} + 6x$

B
$6 (2x^{2} + x)$

C
$x (12x + 6)$

D
All of the above

Solution

$3x \times (4x + 2)$
$= 12x^{2} + 6x$
Hence option A is correct.

$3x \times (4x + 2)$
$=3x \times 2 (2x+1)$
$=6(2x^2+x)$
Hence option B is correct.

$3x \times (4x + 2)$
$= 12x^{2} + 6x$
$=x(12x+6)$
Hence option C is correct.
Question 10
1 / -0

$(4x + 1)(4x - 1) =$

A
$16x^{2} + 8x + 2$

B
$16x^{2} + 8x + 1$

C
$16x^{2} - 1$

D
$16x^{2} + 1$

Solution

Given that: $(4x+1)\times(4x-1)$
We can see that: $(4x+1)\times(4x-1)=$ $4x\times(4x-1)+1\times (4x-1)$
$\Rightarrow (4x\times 4x)-(4x\times 1)+(1\times 4x)-(1\times 1)$
$\Rightarrow 16x^2-4x+4x-1$
$\Rightarrow 16x^2-1$

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Algebraic Expressions and Identities Test - 30

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Algebraic Expressions and Identities Test - 30

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Question 1

8x2+14x−15 8x^2+ 14x -15 8x2+14x−15

7x2−14x−15 7x^2-14x -15 7x2−14x−15

x2+14x−18 x^2+ 14x -18x2+14x−18

None of these

Solution

Question 2

74.5374.5374.53

73.4273.4273.42

79.2179.2179.21

73.2173.2173.21

Solution

Question 3

848484

727272

696969

878787

Solution

Question 4

149821498214982

119921199211992

107121071210712

114821148211482

Solution

Question 5

49m2+164n249m^2+164n^249m2+164n2

98m2+128n298m^2+128n^298m2+128n2

49m2+64n249m^2+64n^249m2+64n2

128m2+64n2128m^2+64n^2128m2+64n2

Solution

Question 6

26.0426.0426.04

270427042704

27.0427.0427.04

260426042604

Solution

Question 7

x2(x+5)−5(x+5)+15 x^2(x+5) -5(x+5)+15x2(x+5)−5(x+5)+15

x2(x+5)−5(x−5)+25 x^2(x+5) -5(x-5)+25x2(x+5)−5(x−5)+25

x2(x+5)−5(x+5)+25 x^2(x+5) -5(x+5)+25x2(x+5)−5(x+5)+25

x2(x+5)−(x+5)+25 x^2(x+5) -(x+5)+25x2(x+5)−(x+5)+25

Solution

Question 8

a3(b3+3)+5(b3+3)+5 a^3(b^3+3)+ 5(b^3 + 3)+ 5a3(b3+3)+5(b3+3)+5

a2(b3+3)+5(b3+3)+5 a^2(b^3+3)+ 5(b^3 + 3)+ 5a2(b3+3)+5(b3+3)+5

a3(b3+3)+5(b3+3)−5 a^3(b^3+3)+ 5(b^3 + 3)- 5a3(b3+3)+5(b3+3)−5

a2(b3+3)+5(b3−3)+5 a^2(b^3+3)+ 5(b^3 - 3)+ 5a2(b3+3)+5(b3−3)+5

Solution

Question 9

12x2+6x12x^{2} + 6x12x2+6x

6(2x2+x)6 (2x^{2} + x)6(2x2+x)

x(12x+6)x (12x + 6)x(12x+6)

All of the above

Solution

Question 10

16x2+8x+216x^{2} + 8x + 216x2+8x+2

16x2+8x+116x^{2} + 8x + 116x2+8x+1

16x2−116x^{2} - 116x2−1

16x2+116x^{2} + 116x2+1

Solution

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$8x^2+ 14x -15$

$7x^2-14x -15$

$x^2+ 14x -18$

$74.53$

$73.42$

$79.21$

$73.21$

$84$

$72$

$69$

$87$

$14982$

$11992$

$10712$

$11482$

$49m^2+164n^2$

$98m^2+128n^2$

$49m^2+64n^2$

$128m^2+64n^2$

$26.04$

$2704$

$27.04$

$2604$

$x^2(x+5) -5(x+5)+15$

$x^2(x+5) -5(x-5)+25$

$x^2(x+5) -5(x+5)+25$

$x^2(x+5) -(x+5)+25$

$a^3(b^3+3)+ 5(b^3 + 3)+ 5$

$a^2(b^3+3)+ 5(b^3 + 3)+ 5$

$a^3(b^3+3)+ 5(b^3 + 3)- 5$

$a^2(b^3+3)+ 5(b^3 - 3)+ 5$

$12x^{2} + 6x$

$6 (2x^{2} + x)$

$x (12x + 6)$

$16x^{2} + 8x + 2$

$16x^{2} + 8x + 1$

$16x^{2} - 1$

$16x^{2} + 1$