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On this section of Time Speed and Distance questions have been categorized in three difficulty levels. Detailed Solution to all the problems have been given. Use timer to keep check on speed performance. This section is useful for those who are preparing for Exams like: CAT, MAT, XAT, SNAP, TISS, GRE, Bank PO, GATE aptitude.

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Important Formulae:

i) $\text" Speed" = \text"Distance"/\text"Time"$

ii) $\text" Time" = \text"Distance"/\text"speed"$

iii) $\text" Distance = speed*time"$

iv) $1{km}/{hr} = 5/18 m/s$

v) $1 m/s = 18/5 {Km}/{hr}$

vi) If the ratio of the speed of A and B is $a:b$,then the ratio of the time taken by them to cover the same distance is $1/a : 1/b$ or $b:a$

vii)  Suppose  a  man  covers  a  distance  at  x  kmph  and  an  equal  distance  at  y kmph, then the AVERAGE SPEED during the whole journey is $({2xy}/{x+y})$ kmph
Out of time, speed and distance we can compute any one of the quantities when we happen to know the other two. For example, suppose we drive for 2 hours at 30 miles per hour, for a total of 60 miles.

If we know the time and the speed, we can find the $\text"distance:"$
$\text" 2 hour * 30 miles"/\text"hour" = \text"60 miles"$
If we know the time and the distance, we can find the $\text"speed:"$
$\text" 60 miles"/\text"2 hours" = 30 \text"miles"/\text"hour"$


Distance is directly proportional to Velocity when time is constant:

Example: A car travels at 30km/hr for the first 2 hrs & then 40km/hr for the next 2hrs. Find the ratio of distance travelled

Example: Two cars leave simultaneously from points A & B (100km apart) & they meet at a point 40 km from A. What is $V_a/V_b$?
Time is constant so $V_1/V_2=S_1/S_2=40/60=4/6$

Example: A train meets with an accident and moves at $(3/4)^{th}$ its original speed. Due to this, it is 20 min late. Find the original time for the journey beyond the point of accident?

Method1: Think about 2 diff. situations, 1st with accident and another w/o accident. As distance in both the cases is constant
So $V_1/V_2=T_2/T_1$                       
=>$V_1/[{3/4}*V_1]= {T_1+20}/ T_1$
=> $4/3={T_1+20}/ T_1$ =>$T_1=60$

Method 2: Velocity decreases by 25% ($3/4$ of original speed => decrement by $1/4$) so time will increase by 33.3% ($4/3$ of original time => increment by $1/3$)
now, $33.3%=20$ min =>$100%=60$ min


Relative Speed:

Case1: Two bodies are moving in opposite directions at speed $V_1$ & $V_2$ respectively. The relative speed is defined as $V_r = V_1 + V_2$

Case2: Two bodies are moving in same directions at speed $V_1$ & $V_2$ respectively. The relative speed is defined as $V_r = |V_1 – V_2|$


Train Problems:

The basic equation in train problem is the same $\text" Speed" = \text"Distance"/\text"Time"$
The following things need to be kept in mind while solving the train related problems.

  • When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object.
  • The distance to be covered when crossing an object, whenever trains crosses an object will be equal to: Length of the train + Length of the object

Boats & Streams:

Let U= Velocity of the boat in still water
V=Velocity of the stream.
While moving in upstream, $\text"distance covered, " S= (U-V) T$
In case of downstream, $\text"distance covered ,"S= (U+V) T$



For clock problems consider the clock as a circular track of 60km.
Min. hand moves at the speed of 60km/hr (think min. hand as a point on the track) and hour hand moves at 5km/hr and second hand at the speed of 3600 km/hr.
Relative speed between HOUR hand and MINUTE hand = 55