Speed, Distance, and Time | problems tricks | fast track arithmetic formulae for problem solving.

Speed, Distance, and Time | problems tricks | fast track arithmetic formulae for problem solving.

SPREAD TO HELP OTHERS
  • 2
  •  
  •  
  •  
  • 1
  •  
  •  
  •  
    3
    Shares

SPEED, DISTANCE, AND TIME PROBLEMS TRICKS IN HINDI

Speed , Distance and Time problems tricks in Hindi | fast track arithmetic formulae for problem solving.

1). distance = speed × time

2). time = distance/speed

3). speed = distance / time

4). Kilometers are multiplied to make miles = \frac{5}{8}

5). Miles is multiplied to make kilometers = \frac{8}{5}

6). Feet – seconds are multiplied to make mile – hour = \frac{15}{22}

7). Mile – hour is multiplied to make foot – second = \frac{22}{15}

8). meter – second is multiplied to make km – hour = \frac{18}{5}

9). km – hour is multiplied to make meter – second =\frac{5}{18}

10). If a man covers the distance between two fixed places with a speed of a km/h, he reaches t1 hour late, and when he travels at a speed of b km/h, he reaches t2 hours early, then the distance between the two places =

    \[\frac{ab(t_1+t_2)}{(b-a)}\; km\]

11). If a man walks at a km/h, he reaches his destination t1 hour late, next time he increases his speed by b km/h, he reaches t2 hour late, then the distance covered by him =

    \[\frac{a (a+b)(t_1-t_2)}{b}\]

12). Two persons X and Y start to cover a distance D km from point A to B with the speed of x km/h and y km/h respectively, if Y reaches point B and returns soon, and meets X at point C, then-
* distance traveled by X = 

    \[\frac{2Dx}{(x+y)}\; km\]

* distance traveled by Y =

    \[\frac{2Dy}{(x+y)} , (x<y)\]

13). A policeman sees a thief at a distance of d km, when he starts chasing the thief, the thief also starts running, if the speed of the thief and the policeman is a km/h and b km/h respectively, (b> a), then the distance traveled by the thief before being caught=

    \[\frac{d\cdot a}{(b-a)}\; km\]

14). If the ratio of the speeds of two objects is x:y, then the ratio of time taken by them is y:x

15). If an object covers two equal distances with speeds of x km/h and y km/h respectively, then the average speed for the whole journey is =

    \[\frac{2xy}{(x+y)}\]

16). If a person walks at the speed x/y of his original speed, and reaches the destination t time before/after, then the time taken to reach the destination with the new speed = \frac{xt}{(xy)} If (x>y), and \frac{xt }{(yx)}, if (y>x)

17). If two objects move in the same direction with a speed of a km/h and b km/h (a>b), then their relative speed = (a-b) km/h

18). If two objects move in opposite directions with a speed of a km/h and b km/h, (a>b), then their relative speed = (a+b) km/h

19). If an object P departs from A to B and another object Q leaves from B to A at the same time and takes t1 and t2 respectively to reach B and A respectively from the time after meeting, Then P’s speed / Q’s speed =

    \[\sqrt{\left(\frac{t_2}{t_1}\right)}\]

20). When a train crosses a stationary object, tree, pole, or person, then = The train has to travel a distance equal to its own length.

21). When a train crosses a long object such as a platform, bridge, or any other train, then =  the train has to cover a distance equal to the sum of the length of the object and itself.

22). A train crosses a person in t1 time and a platform of length x in time t2, then the length of the train =

    \[\frac{xt_1}{(t_2-t_1)}\]

23). Time is taken by the train to cross a pole or tree = \frac{l}{s}

24). time is taken by the vehicle to cross a bridge or platform =

    \[\frac{(l+L)}{s}\]

Where l is the length of the train/train, and L is the length of the object/platform.
25). If the car is going in the same direction then –

*.Time is taken to catch the slow-moving train =
(Distance covered by it in extra time)/(difference of their speed)

*. time taken to cross each other =

    \[\frac{(l_1 - l_2)}{(s_1- s_2)}\]

26). If a train covers a distance D1 with a speed of x1 km/h, a distance D2 with a speed of x2 km/h and a distance D3 at a speed of x3 km/h, then their average speed is =

    \[\left(\frac{D_1+D_2+D_3+...+D_n)}{(\frac{D_1}{x_1} +\frac{D_2}{x_2} +\frac{D_3}{x_3} +...+\frac{D_n}{x_n}}\right)\]

27). If the speed is taken to be constant, then the distance covered by a person is proportional to the time, then

    \[\frac{D_1}{t_1} = \frac{D_2}{t_2}\]

28). If time is taken as a constant, then the distance covered by a person is proportional to the speed, then

    \[\frac{D_1}{s_1} = \frac{D_2}{s_2}\]

29). If distance is kept constant, then speed is inversely proportional to time, then

    \[s_1t_1 = s_2t_2\]

Was this helpful


SPREAD TO HELP OTHERS
  • 2
  •  
  •  
  •  
  • 1
  •  
  •  
  •  
    3
    Shares

Leave a Reply