# Trains, Time and Distance – Speed Maths Tricks

- 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.

- 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.

- 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.

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

- Distance = 2[(S1*S2)/(S1+S2)]* Total Time.

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

- 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,

d | 1.5d |

t | 1.5t |

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 **

s | 4/3 s |

t | ¾ 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**

s | 150/100 S i.e. 3/2 s |

d | 150/100d i.e. 3/2 d |

** **

**Diff of distance = 15 miles.**

**3/2d-d =15****D = 30miles.**

** **

**Problem 4: David travels at 30 kmph for 1 ^{st} 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**

s | 3/4 s |

t | 4/3 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 s | New s | |

s | 10* 14/10 | 4/3 t |

d | d * 14/10 | 7d/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**

s | 5/7 s |

t | 7/5 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 ] / (d_{1}y +d_{2}x)

= [(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**

S1 | S2 | |

s | 5 | 3 |

t | d/5 | d/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**

S1 | S2 | |

s | 3 | 2 |

t | d/3 | d/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 V _{A }**

_{ }**and V**

_{b }m/s in the same direction.They will meet once in a time interval of [d / (V_{A} – V_{B})] sec.

They will meet at the starting point in a time interval of LCM [d/VA, d/V_{B}]

**If two persons A and B are running around a circular track of the length d mts with speed V _{A }**

_{ }**and V**

_{b }m/s in the Opposite direction.**They will meet once in a time interval of **[d / (V_{A} + V_{B})] sec.

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

**If three persons A, B, and C are running around a circular track of the length d mts with speed V _{A, }V_{B }**

_{ }**and V**

_{C }m/s in the same direction.Then, All three meet once in a time interval of LCM**{ (**d / (V_{A }– V_{B}) , d/ (V_{A}– V_{C})} .

Meeting at starting point => LCM[d/V_{A} , d/V_{B , } d/ V_{C}].

**If three persons A, B, and C are running around a circular track of the length d mts with speed V _{A, }V_{B }**

_{ }**and V**

_{C }m/s in the Opposite direction.Then, All three meet once in a time interval of LCM

**{ (**d / (V

_{A }+ V

_{B}) , d/ (V

_{A }+ V

_{C})} .

Meeting at starting point => LCM[d/V

_{A}, d/V

_{B , }d/ V

_{C}].