Exponents and Powers Test

Result

Exponents and Powers Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

If the exponent of a negative integer is odd, then the result is a .......... integer.

A
positive

B
negative

C
zero

D
None of these

Solution

If the exponent of a negative integer is odd, then the result is a negative integer.
$$\Rightarrow$$ $$(-2)^{3}=(-2)\times (-2)\times (-2)=-8$$
$$\Rightarrow$$ Here, $$-2$$ is negative base and $$3$$ is odd power.
$$\Rightarrow$$ Then result $$-8$$ is a negative integer.
Question 2
1 / -0

Simplify: $$1 \div 7 \div 7$$

A
$$7^{-2}$$

B
$$7^{-1}$$

C
$$0.7$$

D
$$0.777$$

Solution

$$1 \div 7 \div 7 = \dfrac 17 \div 7 = \dfrac 17 \times \dfrac 17$$

$$=\dfrac {1}{7^2} = 7^{-2}$$

So, option $$A$$ is correct.
Question 3
1 / -0

$$-2^{-3}$$ can also be expressed as:

A
$$\dfrac{1}{8}$$

B
$$-\dfrac{1}{8}$$

C
$$8$$

D
$$-8$$

Solution

$$-2^{-3} = \dfrac {1}{-2 \times -2 \times -2} = -\dfrac 18$$

So, option B is correct.
Question 4
1 / -0

$$3^{-3}$$ can also be expressed as

A
$$\dfrac{1}{27}$$

B
$$\dfrac{-1}{27}$$

C
$$27$$

D
$$-27$$

E
None of these

Solution

$$a^{-b}$$ is expressed as: $$\dfrac1{a^b}$$
In the same manner,
$$3^{-3}$$ can be expressed as: $$\dfrac1{3^3} = \dfrac1{27}$$
Question 5
1 / -0

If the exponent of a negative integer is even then the result is a ............ integer.

A
Positive

B
Negative

C
0

D
None

Solution

If the exponent of a negative integer is even, then the result is a $$positive$$ integer.
$$\Rightarrow$$ $$(-2)^{4}=(-2)\times (-2)\times (-2)\times(-2)=16$$
$$\Rightarrow$$ Here, $$-2$$ is negative base and $$4$$ is even power.
$$\Rightarrow$$ Then result $$16$$ is a positive integer.
Question 6
1 / -0

Which of the following has an exponent with negative index?

A
$$-3^{4}$$

B
$$4^{3}$$

C
$$\dfrac {1}{3^{4}}$$

D
$$-3$$

Solution

$$\dfrac {1}{3^{4}} = 3^{-4}$$

$$\therefore \dfrac {1}{3^{4}}$$ has a negative index.
So, option $$C$$ is correct.
Question 7
1 / -0

Evaluate : $$(-4)^{-2}$$---

A
$$\displaystyle \frac{1}{-16}$$

B
$$\displaystyle \frac{1}{16}$$

C
$$-16$$

D
$$16$$

Solution

$$(-4)^{-2}=\dfrac{1}{(-4)^2}$$ $$\because a^{-m}=\dfrac{1}{a^m}$$

$$=\dfrac{1}{-4*-4}$$

$$=\dfrac1{16}$$

$$\therefore(-4)^{-2}=\dfrac1{16}$$
Question 8
1 / -0

$$\dfrac{1}{5}$$ can also be expressed as:

A
$$5^{-1}$$

B
$$\dfrac{1}{25}$$

C
$$\dfrac{-1}{25}$$

D
None of these

Solution

$$\dfrac 15 = \dfrac {1}{5^1}=5^{-1}$$
Question 9
1 / -0

$$5^{-2}$$ can also be expressed as

A
$$\dfrac{1}{25}$$

B
$$\dfrac{1}{5^{-2}}$$

C
$$\dfrac{1}{5}$$

D
None of these

Solution

$$5^{-2}=5^{-1*2}$$

$$=(5^2)^{-1}$$ $$\because (a)^{mn}=(a^m)^n$$

$$=\dfrac{1}{5^2}$$ $$\because a^{-n}=\dfrac{1}{a^n}$$

$$=\dfrac{1}{25}$$
Question 10
1 / -0

The value of $$\left (\dfrac {32}{243}\right )^{-3/5}$$ is _____.

A
$$\dfrac {27}{8}$$

B
$$\dfrac {8}{27}$$

C
$$\dfrac {16}{27}$$

D
$$\dfrac {27}{16}$$

Solution

We need to find value of $$\left (\dfrac {32}{243}\right )^{-3/5}$$
It can be written as,
$$\left(\dfrac {2^5}{3^5}\right)^{-3/5}$$
$$\Rightarrow \left (\dfrac {3}{2}\right)^{3} = \dfrac {27}{8}$$

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Re-Attempt Weekly Quiz Competition

Exponents and Powers Test - 15

Result

Exponents and Powers Test - 15

Score

-

Rank

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SHARING IS CARING

Question 1

positive

negative

zero

None of these

Solution

Question 2

$$7^{-2}$$

$$7^{-1}$$

$$0.7$$

$$0.777$$

Solution

Question 3

$$\dfrac{1}{8}$$

$$-\dfrac{1}{8}$$

$$8$$

$$-8$$

Solution

Question 4

$$\dfrac{1}{27}$$

$$\dfrac{-1}{27}$$

$$27$$

$$-27$$

None of these

Solution

Question 5

Positive

Negative

0

None

Solution

Question 6

$$-3^{4}$$

$$4^{3}$$

$$\dfrac {1}{3^{4}}$$

$$-3$$

Solution

Question 7

$$\displaystyle \frac{1}{-16}$$

$$\displaystyle \frac{1}{16}$$

$$-16$$

$$16$$

Solution

Question 8

$$5^{-1}$$

$$\dfrac{1}{25}$$

$$\dfrac{-1}{25}$$

None of these

Solution

Question 9

$$\dfrac{1}{25}$$

$$\dfrac{1}{5^{-2}}$$

$$\dfrac{1}{5}$$

None of these

Solution

Question 10

$$\dfrac {27}{8}$$

$$\dfrac {8}{27}$$

$$\dfrac {16}{27}$$

$$\dfrac {27}{16}$$

Solution

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