Answer: (b)

\(\displaystyle {{S}_{1}}=4kmph\)

\(\displaystyle {{S}_{2}}=16.5kmph\)

\(\displaystyle {{T}_{1}}=2hrs.45\min =165\min \)

\(\displaystyle {{T}_{2}}=?\)

When distance is constant, then speed is inversely proportional

\(\displaystyle {{S}_{1}}:{{S}_{2}}={{T}_{2}}:{{T}_{1}}\)

\(\displaystyle 4:16.5={{T}_{2}}:165\)

or \(\displaystyle \frac{4}{{16.5}}=\frac{{{{T}_{2}}}}{{165}}\)

\(\displaystyle {{T}_{2}}=\frac{{165\times 4}}{{16.5}}=40\min \)

Alternate method

Distance = Speed \(\displaystyle \times \) time = \(\displaystyle 4\times \frac{{11}}{4}\)  = 11 km`

New Speed = 16.5 kmph

Therefore,

time = Distance/Speed = 11/16.5 =0.67 hours =  40 min

Answer: (b)

Let the distance between X and Y be x km. Then, the speed of A is \(\displaystyle \frac{x}{4}km/h\) and that of B is \(\displaystyle \frac{{2x}}{7}km/h\)

\(\displaystyle \frac{{2x}}{7}km/h\overset{{xkm}}{\longleftrightarrow}\frac{x}{4}km/h\)

Relative speeds of the trains = \(\displaystyle (\frac{x}{4}+\frac{{2x}}{7})=\frac{{15x}}{{28}}km/h\)

Therefore the distance between the trains at 7 a.m. = \(\displaystyle x-\frac{x}{2}=\frac{x}{2}km\)

Hence, time taken to cross each other = \(\displaystyle \frac{{\frac{x}{2}}}{{\frac{{15x}}{{28}}}}=\frac{x}{2}\times \frac{{28}}{{15x}}=\frac{{14}}{{15}}\times 60=56\min \)

Thus, both of them meet at 7 : 56 a.m

Answer: (c)

Let the man’s upstream speed be \(\displaystyle {{S}_{u}}kmph\)  and downstream speed be \(\displaystyle {{S}_{d}}kmph\)

Then, Distance covered upstream in 8 hrs 48 min.

d = Distance covered downstream in 4 hrs

\(\displaystyle ({{S}_{u}}\times 8\frac{4}{5})=({{S}_{d}}\times 4)\Rightarrow \frac{{44}}{5}{{S}_{u}}=4{{S}_{d}}=\frac{{11}}{5}{{S}_{u}}\)

Therefore, Required ratio is

\(\displaystyle (\frac{{{{S}_{d}}+{{S}_{u}}}}{2}):(\frac{{{{S}_{d}}-{{S}_{u}}}}{2})=(\frac{{16{{S}_{u}}}}{5}\times \frac{1}{2}):(\frac{{6{{S}_{u}}}}{5}\times \frac{1}{2})=\frac{8}{5}:\frac{3}{5}\)

= 8 : 3

Alternate method

Let the speed of the boat be x km/h

And the speed of the stream be y km/h

Distance travel in Upstream is

\(\displaystyle \Rightarrow (x-y)\times 8\frac{4}{5}=(x-y)\times \frac{{44}}{5}\)

Distance travel in Downstream is

\(\displaystyle \Rightarrow (x+y)\times 4\)

Since the distance is the same

\(\displaystyle \Rightarrow (x-y)\times 8\frac{4}{5}=(x-y)\times \frac{{44}}{5}\)

\(\displaystyle \Rightarrow 11x-11y=5x+5y\)

\(\displaystyle \Rightarrow 6x=16y\)

\(\displaystyle \Rightarrow x:y=8:3\)

The ratio of boat speed and speed of the stream is 8 : 3

Answer: (b)

Let the distance it travelled at the speed of 160 kmph be ‘x’ km.

\(\displaystyle \frac{x}{{160}}+\frac{{(560-x)}}{{40}}=9.5\)

\(\displaystyle \frac{{x+4(560-x)}}{{160}}=9.5\)

x + 2240 – 4x = 1520

 3x = 720

 x = 240 km

Answer: (c)

Speed of train = \(\displaystyle \frac{{lengthoftrain}}{{timetakenincros\sin g}}\)

\(\displaystyle \frac{{320}}{{16}}=20m/\sec =20\times \frac{{18}}{5}=72kmph\)

Total period of stoppage = \(\displaystyle 5\times 18\)

= 90 minutes = \(\displaystyle \frac{3}{2}hours\)

Total time taken in covering a distance of 576 km

\(\displaystyle (\frac{{576}}{{72}}+\frac{3}{2})hours=8+\frac{3}{2}=9\frac{1}{2}hours\)