Time and Distance – Solved Aptitude Questions and Formulas

1 Kmph = $5/18$ m/s

1 m/s = $18/5$ Kmph

Speed(S) = $\text"Distance(d)"/\text"Time(t)"$

Average Speed = $\text"Total distance"/\text"Total Time"$ = ${d_1+d_2}/{t_1+t_2}$

When $d_1 = d_2$, Average Speed = ${2S_1S_2}/{S_1+S_2}$, where $S_1$ and $S_2$ are the speeds for covering $d_1$ and $d_2$ respectively.

When $t_1 = t_2$, Average Speed = ${S_1+S_2}/2$, where $S_1$ and $S_2$ are the speeds during $t_1$ and $t_2$ respectively.

 Relative speed when moving in opposite direction is $S_1+S_2$

 Relative speed when moving in same direction is $S_1-S_2$

A person goes certain distance (A to B) at a speed of $S_1$ kmph and returns back (B to A) at a speed of $S_2$ kmph. If he takes T hours in all, the distance between A and B is ${T ×S_1S_2}/{S_1+S_2}$

When two trains of lengths $L_1$ and $L_2$ respectively travelling at the speeds of $S_1$ & $S_2$ respectively cross each other in time t, then the equation is given as $S_1+S_2$ = ${L_1+L_2}/t$

When a train of length $L_1$ travelling at a speed of $S_1$ overtakes another train of length $L_2$ travelling at speed $S_2$ in time t, then the equation is given as ${S_1-S_2} = {L_1+L_2}/t$

When a train of length $L_1$ travelling at a speed of $S_1$ crosses a platform/bridge/tunnel of length $L_2$ in time t, then the equation is given as $S_1$ = ${L_1+L_2}/t$

When a train of lengths L travelling at a speed s crosses a pole/pillar/flag post in time t, then the equation is given as s = $L/t$

If two persons A and B start at the same time from two points P and Q towards each other and after crossing they take $T_1$ and $T_2$ hours in reaching Q and P respectively, then (A’s speed) / (B’s speed) = ${√T_2} / {√T_1}$