\begin{align} & {\text{Speed of first Jogger}} \cr & = \left[ {\frac{{\left( {x - 1} \right)}}{{42}}} \right] \times 60\, {\text{kmph}} \cr & {\text{Speedof}}\, {2^{nd}}\, {\text{jogger}} \cr & = \left[ {\frac{{\left( {x - 2} \right)}}{{52}}} \right] \times 60\, {\text{kmph}} \cr & {\text{Then}}, \cr & \left( {\frac{{x - 2}}{{{s...b}}}} \right) - \left( {\frac{{x - 1}}{{{s...b}}}} \right) \cr & \cr\end{align} Now, we use option checking method which gives us that option (b) is correct.

Let the speeds of boat and stream was S and V km/hr respectively.

Then, Actual Speed Downstream = (S + V) km/hr.

Actual Speed upstream = (S - V) km/hr.

Given,

\begin{align} & \left\{ {\frac{d}{{\left( {S + V} \right)}}} \right\} + \left\{ {\frac{d}{{\left( {S - V} \right)}}} \right\} \cr & = 5\, {\text{hrs}}\, 15\, {\text{minutes}} \cr & = \frac{{21}}{4}\, {\text{hours}}\, .....\left( 1 \right) \cr & {\text{and}}\left\{ {\frac{{2d}}{{\left( {S + V} \right)}}} \right\} = 7\, .....(2) \cr & {\text{Now}}, \cr & \frac{d}{{\left( {S + V} \right)}} = \frac{7}{4} \cr & Or, \, \frac{{2d}}{{\left( {S + V} \right)}} = \frac{7}{2} \cr\end{align} Hence, he takes ⁷/₂ hours to row 2d km distance downstream.

\begin{align} & {\text{Let}}\, {\text{the}}\, {\text{speed}}\, {\text{of}}\, {\text{two}}\, {\text{trains}}\, {\text{be}}\, 7x\, {\text{and}}\, 8x\, {\text{km/hr}} \cr & {\text{Then}}, \, 8x = \left( {\frac{{400}}{4}} \right) = 100 \cr & \Rightarrow x = \left( {\frac{{100}}{8}} \right) = 12.5 \cr & \therefore {\text{Speed}}\, {\text{of}}\, {\text{first}}\, {\text{train}} \cr & = \left( {7 \times 12.5} \right){\text{km/hr}} \cr & = 87.5\, {\text{km/hr}} \cr\end{align}

In a 6 *6 grid of a chessboard, each row and each column contains 3 white and 3 black squares placed alternatively.

There are a total of 18 black and 18 white squares. For every black square chosen to put one coin, we cannot choose any white square present in its row or column. There are 3 white squares in its row and 3 white square in its column for every black square.

Hence for every black square chosen, we can choose (18 -6)=12 white squares.

Total number of possibilities where a black square and a white square can be chosen so that they do not fall in the same row or in the same column,

=18 *12 =216.

So, there are 216 ways of placing the coins that are identical.

(6 - x) = (8 - 1.5x)

Or, x = 4cm.

So, It will take 4 hours to burn it in such a way that they will remain equal in lengths.