Suppose the distance is \(a\), then

In the same time A covers \(a\), B covers \(a - 45\), hence ratio of speeds of A and B is \(\)\(\frac{{{S_A}}}{{{S_B}}} = \frac{a}{{a - 45}}\)

In the same time B covers \(a\), C covers \(a - 50\), hence ratio of speeds of A and B is \(\)\(\frac{{{S_B}}}{{{S_C}}} = \frac{a}{{a - 50}}\)

In the same time A covers \(a\), C covers \(a - 90\), hence ratio of speeds of A and B is \(\)\(\frac{{{S_A}}}{{{S_C}}} = \frac{a}{{a - 90}}\)

As \(\frac{{{S_A}}}{{{S_C}}} = \frac{{{S_A}}}{{{S_B}}} \times \frac{{{S_B}}}{{{S_C}}}\) 

or \(\frac{a}{{a - 90}} = \left( {\frac{a}{{a - 45}}} \right) \times \left( {\frac{a}{{a - 50}}} \right)\)

\( \Rightarrow {a^2} - 95a + 2250 = {a^2} - 90a\) 

or \(a = 450\)

Suppose the students are \({x_1},\,{x_2},\,{x_3},......{x_9},\,{x_{10}}\) and their weights are in ascending order, then
\({x_1} + {x_2} + {x_3} + ..... + {x_9} = 42 \times 9 = 378\) and
\({x_2} + {x_3} + ..... + {x_9} + {x_{10}} = 47 \times 9 = 423\)
Adding both the equations we have
\(({x_1} + {x_2} + ...... + {x_{10}}) + ({x_2} + {x_3} + {x_4} + .... + {x_9}) = 801\)...(1)
If the mean of all the numbers is \(m\), then \(({x_1} + {x_2} + ...... + {x_{10}}) = 10m\)
Also note that minimum value of \(({x_2} + {x_3} + {x_4} + .... + {x_9})\) is 42×8 and maximum value is 47×8

Therefore from the equation (1), there are two extreme values of \(m\):
\(10{m_1} + 42 \times 8 = 801\) and
\(10{m_2} + 47 \times 8 = 801\)
\( \Rightarrow {m_1} - {m_2} = 4\).

Suppose t years after Veeru's investment, their balances will be same, then Veeru invested for t years and Joy invested for (t – 2) years.

10000 + 10000×5×t/100 = 8000 + 8000×10×(t – 2)/100 

=> 10000 + 500t = 8000 + 800(t – 2)

or t = 12 years

When train travelled at one third speed, then the time taken would have been three times of the usual time. Suppose the usual time is T, then

3T – T = 30

Let us assume total distance is 15x and each minute train covers x units at usual speed. In the return journey, in the first 9 minutes, the distance covered is 5x. So in the remaining 6 minutes, the train has to cover the distance of 10x.  So the speed required = 10x/6 = 5x/3  and percentage increase in the speed is \(\frac{{\frac{{5x}}{3} - x}}{x} \times 100 = 66.67\% \)

We know that for a positive number a,
a + (1/a) ≥ 2 

And the value is 2 at a = 1.

Here (2^x + 1/2^x) ≥ 2 and the value is 2 at x = 0.
But the value of 2 – (x – 2)² is always less than or equal to 2.
Therefore the values of 2^x + 2^-x and  2 – (x – 2)² can never be same and so there is no solution to the equation. 

x = v₁×T and y = v₂×T
=> (x/y) = v₁/v₂
After meeting at C, the cars take 45 and 20 minutes to reach the points B and A. 
y = v₁×45 and x = v₂×20
= x/y = (20v₂)/(45v₁)

Equating x/y, we get
v₁/v₂ = (20v₂)/(45v₁)
=> (v₁/v₂)² = 20/45 or v₁/v₂ = 2/3

Given that v₁ = 60, hence v₂= 90

From the first condition,
A + (B + C)/2 = 5 or 2A + B + C = 10

From the second condition,
B + (A + C)/2 = 7 or 2B + A + C = 14

Subtracting the first equation from the second, we have B – A = 4

The possible values of (B, A) are (5, 1), (6, 2), (7, 3) etc. Note that if we take B = 6 and A = 2, then the value of C becomes zero or negative, hence (B, A) can only be (5, 1).
Thus A + B = 6.

We know that \(4096 = {64^2}\)

\(x = {(4096)^{7 + 4\sqrt 3 }} = {(64)^{2(7 + 4\sqrt 3 )}}\)

\( \Rightarrow 64 = {x^{\frac{1}{{2(7 + 4\sqrt 3 )}}}} = {x^{\frac{{(7 - 4\sqrt 3 )}}{{2(7 + 4\sqrt 3 )(7 - 4\sqrt 3 )}}}}\)

\( \Rightarrow 64 = {x^{\frac{7}{2} - 2\sqrt 3 }} = \frac{{{x^{7/2}}}}{{{x^{2\sqrt 3 }}}}\)

\(\frac{{0.20x - 0.10y}}{{x + y}} \times 100 = 2\)

\( \Rightarrow 20x - 10y = 2x + 2y\)

Since x₀ = max(x₁, x₂, ..., x₁₂) and we want smallest possible value of x0, so we have to keep all x₁, x₂, ..., x₁₂ as low as possible.

Also x₁ +  x₂  +  ...+ x₁₂ = 100

If we keep all x₁, x₂, ..., x₁₂ ≤ 8, then their sum will not reach to 100 (as 12×8 = 96) and hence it is not possible. In other words some of x₁, x₂, ..., x₁₂ will be more than 8. 

We can manage the sum as 100 by taking some of x₁, x₂, ..., x₁₂ as 9.

For example 8,8,8,8,8,8,8,8,9,9,9,9.

So minimum value of x₀ = 9

Given that \({({x^2} - 7x + 11)^{{x^2} - 13x + 42}} = 1\)

We know that  \(a^0 = 1\), when \(a\) is a non zero number.

case 1: \(x^2 - 13x + 42 = 0\)

or \((x – 6)(x – 7) = 0\) or \(x = 6, 7\)

case 2: We know that 1^{any number} = 1, hence

\(x^2 - 7x + 11 = 1\) or \(x^2 - 7x + 10 = 0\)

or \(x = 2, 5\)

case 3: We know that (– 1)^{even  number} = 1, hence

\(x^2- 7x + 11 = - 1\) or \(x^2-7x + 12 = 0\)

or \(x = 3, 4\)

Note that at \(x =3, 4\) the value of  \(x^2-13x + 42\) is even. Hence there are 6 solutions.

There are 3 graphs y ≥ |x| – 1, y ≥ 0 and y ≤ 1 

When y = 1, |x| – 1 = 1 or x = 2, – 2

Area of the trapezium ABDE = Area of the ▲ ABC – Area of the ▲ CDE = (4×2)/2 – (2×1)/2 = 3

Given that ab = 432 and bc = 96, thus a : c = 432 : 96 = 9 : 2

Let us assume a = 9k, c = 2k and b = 48/k

=> a + b + c = 11k + 48/k

The value of 11k + 48/k will be minimum when 11k ≈ 48/k or k = 2

=> a + b + c = 11×2 + 48/2 = 46

Alternate Solution: Given that c < 9, hence k ≤ 4.

k = 1, a + b + c = 11 + 48 = 59

k = 2, a + b + c = 11×2 + 48/2 = 46

k = 3, a + b + c = 11×3 + 48/3 = 49

k = 4, a + b + c = 11×4 + 48/4 = 56

Smallest value = 46 

When his speed becomes 8 to 15, the time taken in the journey will become 8/15. If the time taken when the speed is 8 = T, then time taken now will be 8T/15

Saving in time = T – 8T/15 = 7T/15

Here the time decreases by 35 minutes, thus

7T/15 = 35 or T = 75 minutes.

Now we need that the time taken in the journey should be 10: 00 – 9:10 = 50 minutes, therefore the speed required = (75/50)×8 = 12 kmph

Taking log both sides

\({y^2} {\log_3}5\log 2 = {\log _2}3\log 5\)

\( \Rightarrow \frac{{{y^2}\log 5}}{{\log 3}} \times \log 2 = \frac{{\log 3}}{{\log 2}} \times \log 5\)

\( \Rightarrow {y^2} = {\left( {\frac{{\log 3}}{{\log 2}}} \right)^2}\)

\( \Rightarrow y =  - \frac{{\log 3}}{{\log 2}} =  - {\log _2}3 = {\log _2}\left( {\frac{1}{3}} \right)\)

The ratio of the weights of A, B and C in the mixture is 3×5 : 4×2 : 7×6 = 15 : 8 : 42

Weight of the mixture = 130 kg.
Hence weight of C \(=\frac{42}{15+8+42}\times 130=84\)

\(3\sqrt \pi  \left( {\frac{{12}}{\pi } + 5} \right)\)