Exponents and Powers Test

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Exponents and Powers Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Compare the folllowing:
$$0.000000038$$ _______ $$\displaystyle 3\cdot 8\times 10^{-8}$$

A
$$<$$

B
$$>$$

C
$$=$$

D
None of these

Solution

$$\displaystyle 0\cdot 000000038= 3\cdot 8\times 10^{-8}$$
Question 2
1 / -0

The value of $${ \left[ { \left( { 3 }^{ 2 } \right) }^{ 2 } \right] }^{ -1 }$$ is-

A
$$81$$

B
$$-81$$

C
$$-0.0123$$

D
$$0.0123$$

Solution

$$\cfrac { 1 }{ { \left( { 3 }^{ 2 } \right) }^{ 2 } } $$ $$\because (a^m)^n=a^{mn}$$
$$=\cfrac { 1 }{ { 3 }^{ 4 } }$$
$$ =\cfrac { 1 }{ 81 }$$
$$ =0.0123$$
Question 3
1 / -0

The thickness of paper is $$0.004\ m$$ and that of another paper is $$0.008\ m$$. Compare their sizes.

A
Paper-1 is double thicker than paper-2.

B
Thickness of paper-1 is half the thickness of paper-2.

C
Thickness of paper-1 is one-fourth of the thickness of paper-2.

D
Paper-1 and paper-2 are both equally thick.
Question 4
1 / -0

$$For \ a \ natural \ number \ n , \ 2n(n-1)!\leqslant n^n, if$$:

A
$$n<2$$

B
$$n>2$$

C
$$n\ge2$$

D
Never

Solution
Question 5
1 / -0

The value of $$(-2)^{-5}$$

A
$$-32$$

B
$$\dfrac{-1}{32}$$

C
$$32$$

D
$$\dfrac{1}{32}$$

Solution

$${\textbf{Step - 1: Use negative exponent rule}}$$

$${\text{If the number written in the power of a number is negative like }}{{\text{x}}^{ - n}}$$$${\text{ then we write it as }}\dfrac{1}{{{x^n}}}.$$
$$\therefore{( - 2)^{ - 5}} = \dfrac{1}{{{{( - 2)}^{^5}}}} $$

$${\textbf{Step - 2: Find the value}}$$

$$\dfrac{1}{{{{( - 2)}^{^5}}}} =\dfrac{1}{{{{- 2}^{^5}}}}= - \dfrac{1}{{32}}$$

$${\textbf{Hence, the correct option is B.}}$$
Question 6
1 / -0

The value of $$(-3)^{-4}$$ is

A
$$12$$

B
$$81$$

C
$$\dfrac{-1}{12}$$

D
$$\dfrac{1}{81}$$

Solution

Given $$(-3)^{-4}$$
We know according to the law of exponents $$a^{-m}={ \left( \dfrac { 1 }{ { a }^{ m } } \right) }$$
We can write given equation as $$(-3)^{-4} ={ \left( \dfrac { 1 }{ { -3 }^{ 4 } } \right) }$$
But we also know that $$3^4=81$$
substituting in above we get $$(-3)^{-4}=\dfrac{1}{81}$$
[$$\therefore $$ Since the multiplication of negative sign is even so it becomes positive]
Question 7
1 / -0

The value of $$\left(\dfrac{2}{5}\right)^{-3}$$ is

A
$$\dfrac{-8}{125}$$

B
$$\dfrac{25}{4}$$

C
$$\dfrac{125}{8}$$

D
$$\dfrac{-2}{5}$$

Solution

Given $$\left(\dfrac{2}{5}\right)^{-3}$$

We know according to the law of exponents $${ \left( \dfrac { a }{ b } \right) }^{ -m } = { \left( \dfrac { b }{ a } \right) }^{ m }$$

We can write given equation as $$\left(\dfrac{2}{5}\right)^{-3} = \left(\dfrac{5}{2}\right)^{3}$$

And also we know that $$2^3$$ is $$8$$ and $$5^3$$ is $$125$$

By substituting in above we get $$\left(\dfrac{2}{5}\right)^{-3} = \left(\dfrac{5}{2}\right)^{3} = \dfrac{125}{8}$$
Question 8
1 / -0

Mark tick against the correct answer in each of the following:
$$(1/2)^{-2} +(1/3)^{-2} + (1/4)^{-2}=$$?

A
$$(61/144)$$

B
$$29$$

C
$$(144/61)$$

D
none of these

Solution

We know that,
$$\Rightarrow (1/2)^{-2}=(2/1)^{2}\quad [\because (a/b)^{-n}=(b/a)^n]$$
also, $$\Rightarrow (1/3)^{-2}=(3/1)^{2}\quad [\because (a/b)^{-n}=(b/a)^n]$$
and $$\Rightarrow (1/4)^{-2}=(4/1)^{2}\quad [\because (a/b)^{-n}=(b/a)^n]$$
Now add,
$$=(2)^2 +(3)^2+(4)^2$$
$$=4+9+16$$
$$=29$$
Question 9
1 / -0

$$\left( \dfrac { 1 }{ 2 } \right) ^{ -2 }+\quad \left( \dfrac { 1 }{ 3 } \right) ^{ -2 }+\left( \dfrac { 1 }{ 4 } \right) ^{ -2 }$$

A
$$\dfrac{61}{144}$$

B
$$\dfrac{144}{61}$$

C
$$29$$

D
$$\dfrac{1}{29}$$

Solution

Given $$\left( \dfrac { 1 }{ 2 } \right) ^{ -2 }+\quad \left( \dfrac { 1 }{ 3 } \right) ^{ -2 }+\left( \dfrac { 1 }{ 4 } \right) ^{ -2 }$$

We know according to the law of exponents $$a^{-m}={ \left( \dfrac { 1 }{ a^m } \right) }$$

$$\left( \dfrac { 1 }{ 2 } \right) ^{ -2 }+\quad \left( \dfrac { 1 }{ 3 } \right) ^{ -2 }+\left( \dfrac { 1 }{ 4 } \right) ^{ -2 }=\left( \dfrac { 2 }{ 1 } \right) ^{ 2 }+\quad \left( \dfrac { 3 }{ 1 } \right) ^{ 2 }+\left( \dfrac { 4 }{ 1 } \right) ^{ -2 } =2^2+3^2+4^2=4+9+16=29$$
Question 10
1 / -0

Mark tick against the correct answer in each of the following:
$$\left\{6^{-1} +(3/2)^{-1}\right\}^{-1}$$

A
$$(2/3)$$

B
$$(5/6)$$

C
$$(6/5)$$

D
None of these

Solution

We know that,
$$=(6)^{-1}=(1/6)\quad [\because (a/b)^{-n}=(b/a)^n]$$
$$=(3/2)^{-1}=(2/3)\quad [\because (a/b)^{-n}=(b/a)^n]$$
Now add,
$$=\left\{(1/6)+(2/3)\right\}^{-1}$$
$$=\left\{(1+4)/6\right\}^{-1}$$
$$=\left\{5/6\right\}^{-1}$$
$$=\left\{6/5\right\}$$

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Re-Attempt Weekly Quiz Competition

Exponents and Powers Test - 17

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Exponents and Powers Test - 17

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SHARING IS CARING

Question 1

$$<$$

$$>$$

$$=$$

None of these

Solution

Question 2

$$81$$

$$-81$$

$$-0.0123$$

$$0.0123$$

Solution

Question 3

Paper-1 is double thicker than paper-2.

Thickness of paper-1 is half the thickness of paper-2.

Thickness of paper-1 is one-fourth of the thickness of paper-2.

Paper-1 and paper-2 are both equally thick.

Question 4

$$n<2$$

$$n>2$$

$$n\ge2$$

Never

Solution

Question 5

$$-32$$

$$\dfrac{-1}{32}$$

$$32$$

$$\dfrac{1}{32}$$

Solution

Question 6

$$12$$

$$81$$

$$\dfrac{-1}{12}$$

$$\dfrac{1}{81}$$

Solution

Question 7

$$\dfrac{-8}{125}$$

$$\dfrac{25}{4}$$

$$\dfrac{125}{8}$$

$$\dfrac{-2}{5}$$

Solution

Question 8

$$(61/144)$$

$$29$$

$$(144/61)$$

none of these

Solution

Question 9

$$\dfrac{61}{144}$$

$$\dfrac{144}{61}$$

$$29$$

$$\dfrac{1}{29}$$

Solution

Question 10

$$(2/3)$$

$$(5/6)$$

$$(6/5)$$

None of these

Solution

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