Exponents and Powers Test

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Exponents and Powers Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

$${ \left( \dfrac { -1 }{ 2 } \right) }^{ -3 }=?$$

A
$$\dfrac{-1}{6}$$

B
$$\dfrac{1}{6}$$

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

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

Solution

Given $${ \left( \dfrac { -1 }{ 2 } \right) }^{ -3 }$$
We know according to the law of exponents $$a^{-m}={ \left( \dfrac { 1 }{ a^m } \right) }$$

By applying the above law we get $${ \left( \dfrac { -1 }{ 2 } \right) }^{ -3 }=\dfrac{-2}{1}^3$$

But we know that $$(-2)^3=-8$$
By substituting this we get $${ \left( \dfrac { -1 }{ 2 } \right) }^{ -3 }=-\dfrac{1}{8}$$
Question 2
1 / -0

$${ \left( \dfrac { -5 }{ 3 } \right) }^{ -1 }=?$$

A
$$\dfrac{5}{3}$$

B
$$\dfrac{3}{5}$$

C
$$\dfrac{-3}{5}$$

D
none of these

Solution

Given $${ \left( \dfrac { -5 }{ 3 } \right) }^{ -1 }$$
We know according to the law of exponents $$a^{-m}={ \left( \dfrac { 1 }{ a^m } \right) }$$
By applying the above law we get $${ \left( \dfrac { -5 }{ 3 } \right) }^{ -1 }=\dfrac{1}{{ \dfrac { -5 }{ 3 } }^{ -1 }}$$
On rearranging the above we get $${ \left( \dfrac { -5 }{ 3 } \right) }^{ -1 }=\dfrac{-3}{5}$$
Question 3
1 / -0

$$(-3/7)^{-1} = ?$$

A
$$7/3$$

B
$$-7/3$$

C
$$3/7$$

D
None of these

Solution

$$(-3/7)^{-1}$$
We know, $$x^{-1}=\dfrac{1}{x}$$

$$\Rightarrow$$ $$(-3/7)^{-1}$$ $$=\dfrac{1}{\dfrac{-3}{7}}$$

$$=1\times\dfrac{7}{-3}$$

$$=\dfrac{-7}{3}$$

$$\therefore$$ $$(-3/7)^{-1}$$ $$=\dfrac{-7}{3}$$
Question 4
1 / -0

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

A
$$\dfrac{19}{64}$$

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

C
$$\dfrac{64}{19}$$

D
$$\dfrac{16}{25}$$

Solution

Given $$\left\{ \left( \dfrac { 1 }{ 3 } \right) ^{ -3 }-\quad \left( \dfrac { 1 }{ 2 } \right) ^{ -3 } \right\} \div \quad \left( \dfrac { 1 }{ 4 } \right) ^{ -3 }$$
We know according to the law of exponents $$a^{-m}={ \left( \dfrac { 1 }{ a^m } \right) }$$
$$\left\{ \left( \dfrac { 1 }{ 3 } \right) ^{ -3 }-\quad \left( \dfrac { 1 }{ 2 } \right) ^{ -3 } \right\} \div \quad \left( \dfrac { 1 }{ 4 } \right) ^{ -3 }=\left\{ \left( \dfrac { 3 }{ 1 } \right) ^{ 3 }-\quad \left( \dfrac { 2 }{ 1 } \right) ^{ 3 } \right\} \div \quad \left( \dfrac { 4 }{ 1 } \right) ^{ 3 }=(3^3-2^3) \div 4^3$$
But we know that $$3^3$$ is $$27,2^3$$ is $$8$$ and $$4^3$$ is $$64$$, on substituting
$$\Rightarrow (3^3-2^3) \div 4^3=(27-8) \div \ 64$$
$$\Rightarrow 19\ \div 64=\dfrac{19}{64}$$
Question 5
1 / -0

Mark tick against the correct answer in each of the following:
$$(6^{-1}-8^{-1})^{-1}=$$?

A
$$-\dfrac{1}{24}$$

B
$$-24$$

C
$$\dfrac{1}{24}$$

D
$$24$$

Solution

We know that,
$$=(6)^{-1}=\left(\dfrac{1}{6}\right)^{1}\quad [\because \left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^n]$$
$$=(8)^{-1}=\left(\dfrac{1}{8}\right)^{1}\quad $$
Now $$(6^{-1}-8^{-1})^{-1}=\left(\dfrac{1}{6}-\dfrac{1}{8}\right)^{-1}$$
$$=\left(\dfrac{4-3}{24}\right)^{-1}$$
$$=\left(\dfrac{1}{24}\right)^{-1}$$
$$=24$$
Question 6
1 / -0

Mark tick against the correct answer in each of the following:
$$(-1/2)^{-6}=$$?

A
$$-64$$

B
$$64$$

C
$$(1/64)$$

D
$$(-1/64)$$

Solution

We know that,
$$=(-1/2)^{-6}=(-2/1)^6\quad [\because (a/b)^{-n}=(b/a)^n]$$
$$=(-2)^6$$
$$=64$$
Question 7
1 / -0

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

A
$$(3/8)$$

B
$$(3/16)$$

C
$$(8/3)$$

D
$$(-8/3)$$

Solution

We know that,
$$=(3/4)^{-1}=(4/3)^1\quad [\because (a/b)^{-n}=(b/a)^n]$$
$$=(1/4)^{-1}=(4/1)^1 \quad [\because (a/b)^{-n}=(b/a)^n]$$
Now subtract,
$$=\left\{(4/3)+(4/1)\right\}^{-1}$$
$$=\left\{(4+12)/3\right\}^{-1}$$
$$=\left\{16/3\right\}^{-1}$$
$$=\left\{3/16\right\}$$
Question 8
1 / -0

The value of $$\dfrac {1}{4^{-2}}$$ is

A
$$16$$

B
$$8$$

C
$$\dfrac {1}{16}$$

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

Solution

We will use, law of exponent $$a^{-m}=\dfrac {1}{a^m}$$,
Therefore, $$\dfrac {1}{4^{-2}}=\dfrac {\dfrac {1}{1}}{4^{2}}=\dfrac {\dfrac {1}{1}}{16} =1\times 16=16$$
Question 9
1 / -0

For a fixed base(10) if the exponent decreases by $$1$$, the number becomes

A
One-tenth of the previous number.

B
Ten times of the previous number

C
Hundred of the previous number

D
Hundred times of the previous number

Solution

If the exponent is decreased by $$1$$, then for the fixed base, the number becomes one-tenth of the previous number.
For $$10^5$$, exponent decreases by $$1$$
$$\Rightarrow \ 10^{5-1}=10^4$$
$$\Rightarrow \ \dfrac {10^4}{10^5}=\dfrac {1}{10}$$
Question 10
1 / -0

Mark tick against the correct answer of the following:
$$\left(\dfrac23\right)^{-5}=$$?

A
$$\dfrac{32}{243}$$

B
$$\dfrac{243}{32}$$

C
$$-\dfrac{32}{243}$$

D
$$-\dfrac{243}{32}$$

Solution

We know that,
$$ \left(\dfrac ab\right)^{-n}=\left(\dfrac ba\right)^{n}$$

$$\therefore\left(\dfrac23\right)^{-5}\\=\left(\dfrac32\right)^5$$
$$=\dfrac{3^5 }{2^5}$$ $$\left[\because \left(\dfrac ab\right)^{n}=\left(\dfrac{a^n}{b^n}\right)\right]$$
$$=\dfrac{243}{32}$$

Hence the correct answer is $$Opt:[B]$$

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Re-Attempt Weekly Quiz Competition

Exponents and Powers Test - 18

Result

Exponents and Powers Test - 18

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SHARING IS CARING

Question 1

$$\dfrac{-1}{6}$$

$$\dfrac{1}{6}$$

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

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

Solution

Question 2

$$\dfrac{5}{3}$$

$$\dfrac{3}{5}$$

$$\dfrac{-3}{5}$$

none of these

Solution

Question 3

$$7/3$$

$$-7/3$$

$$3/7$$

None of these

Solution

Question 4

$$\dfrac{19}{64}$$

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

$$\dfrac{64}{19}$$

$$\dfrac{16}{25}$$

Solution

Question 5

$$-\dfrac{1}{24}$$

$$-24$$

$$\dfrac{1}{24}$$

$$24$$

Solution

Question 6

$$-64$$

$$64$$

$$(1/64)$$

$$(-1/64)$$

Solution

Question 7

$$(3/8)$$

$$(3/16)$$

$$(8/3)$$

$$(-8/3)$$

Solution

Question 8

$$16$$

$$8$$

$$\dfrac {1}{16}$$

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

Solution

Question 9

One-tenth of the previous number.

Ten times of the previous number

Hundred of the previous number

Hundred times of the previous number

Solution

Question 10

$$\dfrac{32}{243}$$

$$\dfrac{243}{32}$$

$$-\dfrac{32}{243}$$

$$-\dfrac{243}{32}$$

Solution

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