Solution:

Distance same
Speed = 1 : 2
Time = 2 : 1
1 unit → 2.5
2 unit → 2 × 2.5 = 5 hr
A = 5 hr
B = 1 hr
Speed of B = $$ = \frac{{80}}{1}$$ = 80 km/hr
Time $$ = \frac{{90}}{{80}} = 1\frac{1}{8}{\text{hr}}$$

Solution:
Suppose the trains meet after x hrs
Then,
Distance covered by 1^st train in x hrs + Distance covered by 2^nd train in $$\left( {x - \frac{3}{4}} \right)$$  hrs = 232 km
$$\eqalign{ \Rightarrow 48x + 50\left( {x - \frac{3}{4}} \right) = 232 \cr \Rightarrow 98x = 232 + \frac{{75}}{2} \cr \Rightarrow 98x = \frac{{539}}{2} \cr \Rightarrow x = \frac{{539}}{{196}}{\text{ hrs}} \cr} $$
Required distance = Distance travelled by 1^st train in $$\left( {\frac{{539}}{{196}}} \right)$$  hrs
$$\eqalign{ = \left( {48 \times \frac{{539}}{{196}}} \right){\text{km}} \cr = 132{\text{ km}} \cr} $$

Solution:
Time taken by man if he did not stop :
$$\eqalign{ = \frac{{5{\text{ km}}}}{{10{\text{ kmph}}}} \cr = \frac{1}{2}{\text{hr}} \cr = 30{\text{ min}} \cr} $$
$$\because $$ Man takes rest for 5 minutes on each km
Total rest time = 5 × 4 = 20 min
Total travelling time :
= 30 min + 20 min
= 50 min

Solution:
Int e question it is given that, a man who sit in the slower train cross the faster train it means faster train cross the man 18 sec.
⇒ Relative speed of faster train and man in the same direction
= (58 - 30)
= 28 kmph
So, the distance covered by faster train in 18 sec :
= 28 kmph × 18 sec
= 28 × $$\frac{5}{18}$$ × 18
= 140 metres

Solution:
Total distance covered by man in (1.30 pm - 10.00 am)
= $$3\frac{1}{2}$$ hours at a speed of 12 km/hr
= 12 × $$3\frac{1}{2}$$
= 42 km (Total distance)
Time taken by his elder brother to catch him :
= $$3\frac{1}{2}$$ - 1 hour 15 min
∴ Brother's time
= 3 hr 30 min - 1 hr 15 min
= 2 hr 15 min
= $$2\frac{15}{60}$$
= $$2\frac{1}{4}$$
= $$\frac{9}{4}$$ hours
⇒ Brother's speed :
= $$\frac{{42}}{{\frac{9}{4}}}\,\,\left( {{\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right)$$
= $$18\frac{2}{3}{\text{ km/hr}}$$

Solution:
Relative speed = (200 + 200) km/hr = 400 km/hr
Distance covered in 1 hr 30 min
$$\eqalign{ = \left( {400 \times \frac{1}{2} + 200 \times \frac{1}{2} + 100 \times \frac{1}{2}} \right){\text{km}} \cr = \left( {200 + 100 + 50} \right){\text{km}} \cr = 350{\text{ km}} \cr} $$
[$$\because $$ Speed reduces by half every half an hour]
Hence, distance between the trains :
= (425 - 350) km
= 75 km

Solution:
$$\eqalign{ {\text{Average speed :}} \cr = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr = \frac{50 + 40 + 90}{2 + 2 + 6} \cr = \frac{180}{10} \cr = 18{\text{ km/hr}} }$$