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Let’s assume that the original speed is x km/hr.

∴ Original time taken = 3000/x hrs

When the speed reduces by 200 km/hr, its speed = (x – 200) km/hr

This time, the flight takes 30 more minutes, i.e. ½ hour more.

\(\begin{array}{l} \frac{{3000}}{{x - 200}} - \frac{{3000}}{x} = \frac{1}{2}\ \Rightarrow \frac{{3000\left[ {x - \left( {x - 200} \right)} \right]}}{{x\left( {x - 200} \right)}} = \frac{1}{2}\ \Rightarrow \frac{{3000 \times 200}}{{{x^2} - 200x}} = \frac{1}{2} \end{array}\)

⇒ x² – 200x – 1200000 = 0

⇒ (x – 1200)(x + 1000) = 0

⇒ x = 1200 or x = -1000

Taking only positive value, x = 1200 km/hr

Original time taken = 3000/1200 = 2.5 hours

We know that, (a^b)^c = a^bc

∴ 3⁶ⁿ – 6³ⁿ = (3⁶)ⁿ – (6³)ⁿ = 729ⁿ – 216ⁿ

We also know that, from binomial expansion: (aⁿ – bⁿ) is divisible by (a – b).

∴ Here, 729ⁿ – 216ⁿ is divisible by (729 – 216) = 513

Also, 3 is a factor of 513, hence the given number is also divisible by 3.

Let the rational number be p/q.

As per the given condition: if 2 is added to the denominator, it becomes 1/3.

\(\therefore \frac{p}{{q + 2}} = \frac{1}{3}\)

⇒ 3p – q = 2        ----(i)

Also, if 4 is added to the numerator it becomes ½.

\(\therefore \frac{{p + 4}}{q} = \frac{1}{2}\)

⇒ 2p – q = - 8        ----(ii)

Solving (i) and (ii), we get:

p = 10 and q = 28

sum of the numerator and the denominator of the rational number = p + q = 10 + 28 = 38

As we can see, the given triangle is an isosceles triangle. The altitude in this triangle divides the triangle in two equal parts, as show in figure.

In right angled triangle ADC, AD² + DC² = AC²

AD² = 34² – 16² = 900

AD = 30 cm

∴ Area of triangle ABC = ½ × BC × AD = ½ × 32 × 30 = 480 cm²

The population of bacteria increases by 9% per annum and 1% due to additional factors, so every year, there is a net increase of 10% in yeast population.

Let’s assume that the population today is x.

∴ Population after 2 years \(= x \times \frac{{110}}{{100}} \times \frac{{110}}{{100}} = \frac{{121x}}{{100}} = 1.21x\)

% increase in population = (1.21x – x)/x × 100 = 21

The given expression 2y – 3x will be minimum when y is minimum and x is maximum.

Minimum value of y = -1

Maximum value of x = 1

Minimum value of given expression = 2 × (-1) – 3 × (1) = -2 – 3 = -5

The curved surface area of hemisphere with radius r is given as:

CSA = 2πr²

∴ 8π = 2πr²

⇒ r = 2 cm

Since their HCF is 25, let the two numbers be 25x and 25y, where x and y are coprime numbers.

We know that, product of numbers = LCM × HCF

∴ 25x × 25y = 300 × 25

⇒ xy = 12 ⇒ y = 12/x

Also, 25x + 25y = 175 ⇒ x + y = 7

⇒ x + 12/x = 7

⇒ x² – 7x + 12 = 0

⇒ (x – 3)(x – 4) = 0

⇒ x = 3 and y = 4 OR x = 4 and y = 3

∴ the two numbers are: 25 × 3 = 75 and 25 × 4 = 100

∴ difference of two numbers = 100 – 75 = 25

Let’s assume that pipe B alone can the tank in x minutes, so pipe B fills 1/x^th part of tank in one minute.

Pipe A can fill the tank 2 times faster than pipe B, so pipe A fills 2/x^th part of tank in one minute.

Together, the two pipes fill the tank in 24 minutes, so together they fill 1/24^th part in one minute.

\(\therefore \frac{1}{x} + \frac{2}{x} = \frac{1}{{24}}\) 

⇒ 3/x = 24

⇒ x = 72 minutes

∴ Pipe B alone fills the tank in 72 minutes.

We know that, for a quadratic equation ax² + bx + c,

Sum of its roots = -b/a and product of roots = c/a

For x² + 2x – 15 = 0:

x₁ + x₂ = -2 and x₁x₂ = -15

For the equation whose roots are 1/x₁ and 1/x₂,

Sum of roots \(= \frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} = \frac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} = \frac{{ - 2}}{{ - 15}}; = \frac{2}{{15}}\)

Product of roots \( = \frac{1}{{{x_1}{x_2}}} = - \frac{1}{{15}}\)

∴ the quadratic equation becomes: \({x^2} - \frac{2}{{15}}x - \frac{1}{{15}} = 15{x^2} - 2x - 1\)