## How to solve Time, Speed and Distance aptitude questions

The terms Time and Distance are related to the speed of a moving object.

Speed We define the speed of an object as the distance coverd in a unit time intervel. It is obtained by the distance covered by the object, by the time it take to
cover that distance,

Thus, Speed=Distance traveled/Time taken

### Some tricks for solving Time, Speed and Distance Aptitude Questions

1. If the time taken is constant, the distance traveled is proportional to the speed, that is more the speed more distance is traveled in the same time

2. If the speed is constant, the distance traveled is proportional to the time taken, that is more the distance traveled more the time taken at the same speed.

3. If the distance traveled is constant, the speed is inversely propotional to the time taken, that is more the speed, less the time taken for the same distance
traveled.

### Basic Formulae for Time and Distance Aptitude Questions

1. Speed = Distance/Time.

2. Distance = Speed X TIme.

3. Time = Distance / Speed.

### Units Of Measurement

Generally, if the distance is measured in kilometer, the we measure time in hour and speed in kilometer per hour and is written in km/hr. If the distance is measured
in meter then we denote time in second and speed in meter and it is written in m/sec.

 One km/hr = [\frac{1000m}{60 \times 60 sec}=\frac{5}{18}m/sec\] There for one m/sec= 18/5 (km/hr) Thus, $xkm/hr=\left(x \times \frac{5}{18}\right) m/sec$ and, $xm/sec=\left(x \times \frac{18}{5}\right) km/hr$

### Short Cut Methods for Time and Distance Aptitude Questions

1.(a) If B covers a distance d1 km at s1 km/hr and then d2 at s2 km/hr, then the average speed during the whole journey is given by

Average speed = $\frac{s1 s2 \left( d1 + d2 \right)}{s1 d2+ s2 d1} km/hr$

(b) If B goes from x to y at s1 km/hr and comes back from y to x at s2 km/hr. Then the average speed during the whole journey is given by

Average speed = $\frac{2s1s2}{s1+s2}$

2. A person goes certain distance at a speed of s1 and returns back to the same place at a speed of s2 km/hr. If he take T hours in all, then the distance between
first place to the second place is

$T \left(\frac{s1s2}{s1+s2}\right)$

3. If two persons A and B start at the same time from two points X and Y towords each other and after crossing they take T1 and T2 hours in reaching Y and X
respectively, then

$\frac{Aspeed}{Bspeed}=\frac{\sqrt{T_{2}}}{\sqrt{T_{1}}}$

4. If a person travels d1, d2, d3, ...., dn meters with different speeds s1, s2, s3, ......... sn m/sec in time t1, t2, t3, ....,tn seconds respectively. Then the
average speed of the person through out the journey is given by

Average Speed = Total distance traveled / Total time taken.

5. If the new speed is x/y of the original speed, then the change in time to cover the same distannce is given by

Change in time =$\left(\frac{y}{x}-1\right)$ X Original Time

6. An object covers a distance d in time T1 with speed s1, but when it travels with the speed of s2 covers the same distance in time T2,

Product of speed / d = s1/T2 = s2/T1 = Difference of speed / Difference of time .

Equating any two of the above we can find the unknowns.

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