Solution:
A _______ B ________ C Let both the trains start from point A. At point B third train overtook the first train. To overtake the first train by third train, Third train needs to cover, Distance covered by first train in $$1\frac{1}{2}$$ h = distance covered by third train in 1 h. [as third train has started 30 minutes later] In this situation distance is constant, then from, s × t = d; we get, S α $$\frac{1}{{\text{t}}}$$ Now,
$$\eqalign{ \frac{{{\text{t}} + \frac{1}{2}}}{{\text{t}}} = \frac{{\text{s}}}{{40}} \cr \frac{{2{\text{t}} + 1}}{{\text{t}}} = \frac{{\text{s}}}{{40}}\,.\,.\,.\,.\,.\,.\,.\,.\left( 1 \right) \cr} $$ From equation (1), It is clear that time is in the ratio 3 : 2 then speed will be in 2 : 3 ratios. Hence, Speed of the Third train will be 60 km/h.

Solution:
$$\eqalign{ {\text{Speed of train :}} \cr = \frac{{900 + 300}}{{40}} \cr = 30{\text{ m/s}} \cr {\text{Speed (in km/hr):}} \cr = 30 \times \frac{{18}}{5} \cr = 108{\text{ km/hr}} \cr} $$

Solution:
Let distance = x km and usual rate = y kmph. Then,
$$\eqalign{ \frac{x}{y} - \frac{x}{{y + 10}} = 1 \cr {\text{or,}}\,y\left( {y + 10} \right) = 10x\,.\,.\,.\,.\,.\,.\,.\,.\,\,\left( 1 \right) \cr} $$
Now, in the 2^nd scenario with a further increase in speed the driver could have saved another 45 min = $$\frac{3}{4}$$ hrs.
Therefore, total time saved
$$ = 1 + \frac{3}{4} = \frac{7}{4}\,{\text{hrs}}{\text{.}}$$
Putting it in equation, we get
$$\eqalign{ \frac{x}{y} - \frac{x}{{y + 20}} = \frac{7}{4} \cr {\text{or,}}\,y\left( {y + 20} \right) = \frac{{80x}}{7}\,.\,.\,.\,.\,.\,.\,.\,.\,.\left( 2 \right) \cr} $$
On dividing (1) by (2), we get y = 60
Substituting y = 60 in (1), we get :
x = 420 km.

Solution:
A ____100 km____ P₁ ________300 km_____ P₂ ____X______ B If speed becomes $$\frac{5}{6}$$ then time taken will be $$\frac{6}{5}$$ of original time. Thus, extra time = 2 hour $$\frac{1}{5}$$ = 2 hour Thus unusual time which is require to complete the journey = 5 × 2 = 10 hours. It means It covers 300 km distance in 5 hours. then speed = $$\frac{{300}}{5}$$ = 60 km/h

Solution:
let required speed be S km/h 50 × 30 = S × 20 S = 75 km/h

Solution:
Let the speed of the man be x m/sec
Then, distance travelled by the man in 5 min 52 sec
= Distance travelled by sound in 8 sec
$$\eqalign{ \Leftrightarrow x \times 352 = 330 \times 8 \cr \Leftrightarrow x = \left( {\frac{{330 \times 8}}{{352}}} \right){\text{m/sec}} \cr \Leftrightarrow x = \left( {\frac{{330 \times 8}}{{352}} \times \frac{{18}}{5}} \right){\text{km/hr}} \cr \Leftrightarrow x = 27{\text{ km/hr}} \cr} $$

Solution:
$$\eqalign{ {\text{Average speed}} = \frac{{2 \times 90 \times 126}}{{216}} \cr = \frac{{2 \times 5 \times 126}}{{12}} \cr = 105{\text{ km/h}} \cr {\text{Time}} = \frac{{525}}{{105}} = 5{\text{ hours}} \cr} $$