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Trains, Time and Distance – Speed Maths Tricks

Trains, Time and Distance – Speed Maths Tricks




  1. If a certain distance is covered at x km/hr and same distance is covered at y km/hr. Then, Average speed = 2xy/(x+y) km/hr.
  1. If the ratio of speeds 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.
  1. Required Distance = [(S1*S2)/(S1-S2)]*Difference in arrival times.

Eg: With S1 speed reaches 10 mins late and with s2 speed reaches 5 mins earlier.

  1. Distance = [(S1 * S2)/ (S1+S2)] * Total time A……………….>S1……………S2<……….B
  1. Distance = 2[(S1*S2)/(S1+S2)]* Total Time.

A…….>S1…….>S2 B (half distance with s1 and remaining half distance with s2 sped)

  1. Meeting Points distance from starting point = [(S1*S2)/ (S2-S1)]* Diff. in time.

A…………………………B
The 1st train starts 9am with S1 speed.
The 2nd train starts 2pm with S2 speed.
Speed = Distance/Time   i.e. S=D/T
If T= Constant or same , D α S, D is directly proportional to S ( that means if distance increases speed also increases)
If S= Cont. or same , D α T, D is directly proportional to Time ( that means If distance increases  speed also increases)
If D = Const. or same, S α 1/T. S is inversely proportional to Time( That means if speed increases time decreases as vice versa.) 

Practice Questions With Answers on Trains, Time & Distance:




Problem 1: If I travel 50% more then I take 1 hour more Calculate the actual time was taken to travel the distance. 

Sol: Here, S= Const. ,

D α Time,

d1.5d
t1.5t

Given, Diff of time = 1 hour.

=> 1.5t –t =1hr

=> 1/2t = 1hr

=> t = 2hrs.

Problem 2: If a speed of a car is multiplied by 4/3 then 10mins are saved. Calculate usual time is taken to cover the distance.

 Sol:D=const.,

  S α 1/t

s4/3 s
t¾ t

Diff of time = 10 mins.

  • T – ( ¾)T = 10 mins.
  • T/4 = 10.
  • T = 40 mins.

Problem 3: If I travel for 50% more speed then I travel 15miles more. Calculate     Actual distance.

Sol:   T=const., 

D α S

s150/100 S   i.e. 3/2 s
d150/100d    i.e.  3/2 d

 

Diff of distance = 15 miles.

  • 3/2d-d =15
  • D = 30miles.

 

Problem 4: David travels at 30 kmph for 1st 2 hours and then 40kmph for next 2 hrs. Find the ratio of distance traveled.

Sol: Here T= const.,  

 

S α D.

S1/S2   α   d1/d2

  • 30/40
  • ¾

Problem 5: Maggie and John leave simultaneously from A& B towards each other and they meet at a point 40km from A. If the distance between two points is 100 km. What is the ratio of Maggie and john speed?

Sol: Here t=const., D α S

d1/d2  α s1/s2

i.e. 40/60 = 2/3.

Problem 6: Nancy meets an accident and moves ¾ the of original speed and reaches 20 mins late. Find original time for the journey beyond the point of accident.

Sol:   D=const. , Sα1/t

s3/4  s
t4/3  t

Diff of time = 2o mins.

  • 4/3t –t =20 mins.
  • Actual t = 60 mins.

 

Actual time + 20 mins = 80 mins.

 

Problem 7: Sultan walks at 14 kmph instead of 10 kmph he would have walked 20 km more. find the actual distance traveled by him.

Sol:  t=const.,     S α d

 Old sNew  s
s10* 14/104/3  t
dd * 14/107d/5

Diff of distance = 20km

  • d -7/5d = 20
  • 2d =100
  • D =50km.

 

Problem 8: Ratio between speeds of A & B is 7: 8 and If A takes 20 mins more than  B .  Calculate time is taken by B to cover the distance.

Sol:   d= const., s α 1/t

S1/ s2  α t1/t2

i.e. 7/8 α 8/7

8x -7x =20mins

X= 20mins.

Actual time of B = 7x =7*20 =140mins.

Problem 9: Matt travels with 5/7 of his actual speed and covers 35km in 1 hr 20mins.         Calculate the actual speed of matt.

Sol: d= const.,   s α 1/t

s5/7  s
t7/5  t

 

7/5 t  = 80mins

  • T =80*5/7
  • T=400/7mins= (400/7 )/60 hr = 20/21 hrs.

S= d/t =35kmph/ (20/21) kmph.

 

Problem 10: A train covers a distance of 50km with  a speed of 40km/hr and next 60km with the speed of 30kmph in traveling A to B. Find Average speed.

Sol: Average Speed =[(d1+d2) xy ] /  (d1y +d2x)
= [(50+60) 40*30] / (1500 +2400)
= (110*40*30)/3900
=440/13
Problem 11: A train crosses 450km in 7hrs and 740km in 10hrs. find Avg speed?

Sol: Avg speed = Total Distance / Total time.
Problem 12: If a person travels a certain distance at X mph and returns at lymph. If the time was taken to cover the whole journey is t hours. Then one-way distance?
Sol: One way distance = [xy / (x+y)] *t
Problem 13: If A and B start the journey at the same time from two points P & Q towards each other they travel a and b hours in reaching Q and P.
Then A’s speed/B’s speed =.
Problem 14: If same distance is covered at two different speeds S1 and S2 and time took to cover the distance are T1 and T2 Then the distance is given by

Distance = [(S1*S2) / (S1-S2)]*(T1-T2)
Problem 15: A travels from home to office @5kmph and reaches 15mins earlier. After that comes back @3kmph and reaches 9 minutes late. Find the distance between home to school?

Sol:  d=const,

   s α 1/t

 S1S2
s53
td/5d/3

 

Diff of time= 24mins

  • d/3-d/5 =24mins
  • d= 24*(15/2) = 180/60 =3km

      

Problem 16: Pink goes to college @3kmph and returns @2kmph. He spends 5hours in the total journey. What is the distance?

Sol:  d=const,

   s α 1/t

 S1S2
s32
td/3d/2

 

Total Time =  d/3 +d/2 =5hrs.

=>        5d =5*6

=> d=6km.

Problem 17: Sonu walks @6kmph reaches 5min late. If she walks @5kmph also late by 30mins. find distance.

Sol:

Diff of time= 25mins

@6kmph ……….>  5 mins late

@5kmph ………..> 30 mins late

Total work = Lcm (6 , 5) = 30 units.  6-5= 1hr

i.e. 30km ß 1hr

30km <- 60mins

? <- 25mins

  • 25 * 30/60.
  • 5 km.

Problem 18: The length of the bridge, which a train with  130mts long is traveling 45kmph speed to cross the bridge in 30secs. Then what is bridge length?

Sol:  We know, Total Time= Total distance/ Total Speed.

i.e.,            30secs = (Bridge Length + Train Length) / [45*5/18]m/secs
=> Blength+130 = 30 * 25/2
=> 375 – 130 = 245 mts.
 
Circular Motions

If two persons A and B are running around a circular track of the length d mts with speed  VA  and Vm/s in the same direction.

They will meet once in a time interval of      [d / (VA – VB)] sec.

They will meet at the starting point in a time interval of     LCM [d/VA, d/VB]



If two persons A and B are running around a circular track of the length d mts with speed  VA  and Vm/s in the Opposite direction.

They will meet once in a time interval of [d / (VA + VB)] sec.

They will meet at the starting point in a time interval of     LCM [d/VA, d/VB].

If three  persons A, B, and C are running around a circular track of the length d mts with speed  VA, VB  and Vm/s in the same direction.

Then, All three meet once in a time interval of  LCM{ (d / (VA – VB) , d/ (VA– VC)} .

Meeting at starting point  => LCM[d/VA , d/VB ,  d/ VC].

If three  persons A, B, and C are running around a circular track of the length d mts with speed  VA, VB  and Vm/s in the Opposite direction.
Then, All three meet once in a time interval of  LCM{ (d / (VA + VB) , d/ (VA + VC)} .
Meeting at starting point  => LCM[d/VA , d/VB ,  d/ VC].



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