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\({\left( {\frac{1}{2}} \right)^{ - \;6}} = \frac{1}{{{{\left( {\frac{1}{2}} \right)}^6}}} = \frac{1}{{\frac{1}{{64}}}} = 64\)

\({\left( {\frac{1}{4}} \right)^{ - \;3}} = \frac{1}{{{{\left( {\frac{1}{4}} \right)}^3}}} = \frac{1}{{\frac{1}{{64}}}} = 64\)

\({\left( {\frac{1}{3}} \right)^{ - \;4}} = \frac{1}{{{{\left( {\frac{1}{3}} \right)}^4}}} = \frac{1}{{\frac{1}{{81}}}} = 81\)

\({\left( {\frac{1}{6}} \right)^{ - \;2}} = \frac{1}{{{{\left( {\frac{1}{6}} \right)}^2}}} = \frac{1}{{\frac{1}{{36}}}} = 36\)

∴ Answer is Option 3

Given:

The some fractions

Calculation:

⇒ 2/3 = 0.66, 2/5 = 0.40, 2/9 = .22, 4/5 = 0.80

⇒ 0.80 > 0.66 > 0.40 > .22

∴ The required result will be 4/5.

Given:

5^{3a – 1} × 125 = 25^{2a –} 1

Concept:

A^x × A^y = A^{x + y}

If B^p = B^q, then p = q

Calculation:

5^{3a – 1} × 125 = 25^{2a – 1}

⇒ 5^{3a – 1} × 5³ = 5^{2(2a – 1)}

⇒ 5^{3a – 1 + 3} = 5^{4a – 2}

⇒ 5^{3a + 2} = 5^{4a – 2}

Now, B^p = B^q

⇒ 3a + 2 = 4a - 2

⇒ 4a – 3a = 2 + 2

⇒ a = 4

∴ The value of a is 4.

Given:

43a + 2 ÷ 42a = 42a - 3

Concept:

A^x ÷ A^y = A^{x – y}

If B^p = B^q, then p = q

Calculation:

4^{3a + 2} ÷ 4^2a = 4^{2a – 3}

⇒ 4^{3a + 2 – 2a} = 4^{2a – 3}

⇒ 4^{a + 2} = 4^{2a - 3}

Now, Comparing exponents we'll get

a + 2 = 2a – 3

⇒ a = 3 + 2

⇒ a = 5

∴ The value of a is 5.

Given:

3^a × 81^2a = 27⁶

Concept:

A^x × A^y = A^{x + y}

If B^p = B^q, then p = q

Calculation:

3^a × 81^2a = 27⁶

⇒ 3^a × 3^4(2a) = 3^{3 × 6}

⇒ 3^a × 3^8a = 3¹⁸

⇒ 3^{a + 8a} = 3¹⁸

Now, B^p = B^q

⇒ 9a = 18

⇒ a = 2

∴ The value of a is 2.

Given:

2^{2a + 3} ÷ 64 = 2^{a + 1}

Concept:

A^x ÷ A^y = A^{x – y}

If B^p = B^q, then p = q

Calculation:

2^{2a + 3} ÷ 16 = 2^{a + 1}

⇒ 2^{2a + 3} ÷ 2⁴ = 2^{a + 1}

⇒ 2^{2a + 3 – 4} = 2^{a + 1}

⇒ 2^{2a – 1} = 2^{a + 1}

Now, B^p = B^q

⇒ 2a – 1 = a + 1

⇒ a = 2

∴ The value of a is 2.

Given:

3^{2a + 3} ÷ 27 = 3^{a + 1}

Concept:

A^x ÷ A^y = A^{x – y}

If B^p = B^q, then p = q

Calculation:

3^{2a + 3} ÷ 27 = 3^{a + 1}

⇒ 3^{2a + 3} ÷ 3³ = 3^{a + 1}

⇒ 3^{2a + 3 – 3} = 3^{a + 1}

⇒ 3^2a = 3^{a + 1}

Now, B^p = B^q              

⇒ 2a = a + 1

⇒ a = 1

Given:

3^{2a + 1} × 81^{a – 2} = 3^{4a + 1}

Concept:

A^x × A^y = A^{x + y}

If B^p = B^q, then p = q

Calculation:

3^{2a + 1} × 81^{a – 2} = 3^{4a + 1}

⇒ 3^{2a + 1} × 3^{4(a – 2)} = 3^{4a + 1}

⇒ 3^{(2a + 1) + (4a – 8)} = 3^{4a + 1}

⇒ 3^{6a – 7} = 3^{4a + 1}

Now, B^p = B^q

⇒ 6a – 7 = 4a + 1

⇒ 2a = 8

⇒ a = 4

∴ The value of a is 4.

Given:

The ratio between the expenditure of A and B = 5 : 7

Calculation:

Let be assume expenditure of A = 5p

Expenditure of B = 7p

⇒ 2000 = 7p – 5p

⇒ p = 1000

⇒ Expenditure of B  = 7 × 1000 = 7000

⇒ The income of A = (125/100) × 7000

⇒ 8750

∴ The required result is 8750.

Given:

(21 × 3)/7 – 4 × (15/5) – (30/2) × 15

Calculation:

⇒ 63/7 – 4 × 3 – 15 × 15

⇒ 9 – 12 – 225 

⇒ 9 – 237

⇒ – 228

∴ The required answer is – 228. 

Given: 

15.5 × 1.2 

Calculation: 

15.5 × 1.2 = 155/10 × 12/10 

⇒ 1860/100 

⇒ 18.6 

∴ The correct answer is 18.6

Given:

Expression: x³ + 3x² + 3x + 1

Concept used:

Quadratic surd: An expression that contains a square root under which is a rational number which is not a perfect square.

Calculation:

x³ + 3x² + 3x + 1 = (x + 1)³

The expression (x + 1)³ itself is not a perfect square.

Then, to represent it in quadratic surd form, we just need to write it in the form of a perfect square under the square root.

(x + 1)³ = √(x + 1)⁶

∴ The quadratic surd form of the above expression is √(x + 1)⁶.

Given

a/√2 = 12√2

Calculation

a/√2 = 12√2

⇒ a = 12√2 × √2

⇒ a = 24

Calculations :

Square of 40 = (40)² = 1600 

 A quarter of 40% of 1600 = 25% of 40% of 1600   

⇒ (1/4) of (2/5) of 1600 

⇒  (2/20) of 1600 

⇒ 160 

∴ The result will be 160 

Let the total no of workers be x 

No of women =x/3 

No of men =x-(x/3)=2x/3 

No of women having children =1/3 of 1/2 of x/3=x/18 

No of men having children=2/3 of 3/4 of 2x/3=x/3 

No of workers having children = x/8 + x/3 =7x/18 

Workers having no children=x-7x/18=11x/18=11/18 of all workers

Let the total length be xm 

Then black part =x/8cm 

The remaining part=(x-x/8)cm=7x/8cm 

White part=(1/2 *7x/8)=7x/16 cm 

Remaining part=(7x/8-7x/16)=7x/16cm 

7x/16=7/2 

x=8cm

Suppose x bottles can fill the tin completely 

Then 4/5x-3/4x=6-4 

X/20=2 

X=40 

Therefore required no of bottles =40