Polynomials Test

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Polynomials Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is correct?

A
$$(x-y)^2=x^2+2xy-y^2$$

B
$$(x-y)^2=x^2-2xy+y^2$$

C
$$(x-y)^2=x^2-y^2$$

D
$$(x+y)^2=x^2+2xy-y^2$$

Solution

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

Then,
$$(x-y)^2$$
$$=x^2-2(x)(y)+y^2$$ $$= x^2-2xy+y^2$$.

To verify this, let us consider:
$$(x-y)^2$$ $$=(x-y)$$$$(x-y)$$
$$=x(x-y)-y(x-y)$$
$$=x^2-xy-yx+y^2$$
$$=x^2-xy-xy+y^2$$
$$=x^2-2xy+y^2$$.

Hence, $$(x-y)^2$$ $$= x^2-2xy+y^2$$

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

Find the value of $$49^2$$ using standard identity.

A
$$2391$$

B
$$2401$$

C
$$2411$$

D
$$2421$$

Solution

As we know that,
$$(a-b)^2=a^2-2ab+b^2$$

Substitute $$50$$ for $$a$$ and $$1$$ for $$b$$ in above formula,
$$\begin{aligned}{}{(50 - 1)^2}& = {50^2} - 2 \times 50 \times 1 + {1^2}\\{49^2}& = 2500 - 100 + 1\\ &= 2401\end{aligned}$$

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

Square of $$3a-4b$$ is:

A
$$9a^2+16b^2-24ab$$

B
$$6a^2-8b^2$$

C
$$9a^2-16b^2$$

D
$$9a^2+16b^2+24ab$$

Solution

Given, square of $$3a-4b$$
i.e. $$(3a-4b)^2$$.

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

Then,$$(3a-4b)^2$$
$$= (3a)^2-2(3a)(4b)+(4 b)^2$$
$$=9a^2-24ab+16b^2$$.

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

Find the value of $$1.05\times 0.95$$ using standard identity.

A
$$0.9985$$

B
$$0.9975$$

C
$$0.9875$$

D
$$0.9995$$

Solution

Given, $$1.05\times 0.95$$
$$=(1+0.05)(1-0.05)$$.

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

Then,
$$1.05\times 0.95$$
$$=(1+0.05)(1-0.05)$$
$$=1^2-(0.05)^2$$
$$=1-0.0025$$
$$=0.9975$$.

Therefore, option $$B$$ is correct.
Question 5
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 6
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 7
1 / -0

Find the value of $$199\times 201$$.

A
$$39989$$

B
$$40001$$

C
$$39999$$

D
$$400011$$

Solution

Given, $$199\times 201$$
$$ =(200-1)(200+1) $$.

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

Then,
$$199\times 201$$
$$ =(200-1)(200+1) $$
$$= 200^2-1^2$$
$$= 40000-1$$
$$=39999$$.

Therefore, option $$C$$ is correct.
Question 8
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 9
1 / -0

Find the value of $$1.03^2$$ using identity.

A
$$1.0409$$

B
$$1.0609$$

C
$$1.0009$$

D
$$1.0309$$

Solution

Given, $$1.03^2 = (1+0.03)^2$$.

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

Then,
$$1.03^2 = (1+0.03)^2$$
$$=1^2+(0.03)^2+2\times 1\times 0.03$$
$$=1+0.0009+0.06$$
$$=1.0609$$.

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

Find the value of $$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)$$.

A
$$abc$$

B
$$a^2+b^2+c^2$$

C
$$0$$

D
$$1$$

Solution

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

$$\therefore (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)$$
$$=(a^2-b^2)+(b^2-c^2)+(c^2-a^2)$$
$$=a^2-b^2+b^2-c^2+c^2-a^2$$
$$=0$$.

Therefore, option $$C$$ is correct.

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Polynomials Test - 34

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Polynomials Test - 34

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Question 1

$$(x-y)^2=x^2+2xy-y^2$$

$$(x-y)^2=x^2-2xy+y^2$$

$$(x-y)^2=x^2-y^2$$

$$(x+y)^2=x^2+2xy-y^2$$

Solution

Question 2

$$2391$$

$$2401$$

$$2411$$

$$2421$$

Solution

Question 3

$$9a^2+16b^2-24ab$$

$$6a^2-8b^2$$

$$9a^2-16b^2$$

$$9a^2+16b^2+24ab$$

Solution

Question 4

$$0.9985$$

$$0.9975$$

$$0.9875$$

$$0.9995$$

Solution

Question 5

$$2451$$

$$2471$$

$$2491$$

$$2501$$

Solution

Question 6

$$7300$$

$$7350$$

$$7400$$

$$7450$$

Solution

Question 7

$$39989$$

$$40001$$

$$39999$$

$$400011$$

Solution

Question 8

$$2604$$

$$2704$$

$$2804$$

$$2904$$

Solution

Question 9

$$1.0409$$

$$1.0609$$

$$1.0009$$

$$1.0309$$

Solution

Question 10

$$abc$$

$$a^2+b^2+c^2$$

$$0$$

$$1$$

Solution

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