# Aptitude - Time and Distance - Discussion

Discussion Forum : Time and Distance - General Questions (Q.No. 4)

4.

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

Answer: Option

Explanation:

Let speed of the car be *x* kmph.

Then, speed of the train = | 150 | x |
= | 3 | x |
kmph. | |

100 | 2 |

75 | - | 75 | = | 125 | |

x |
(3/2)x |
10 x 60 |

75 | - | 50 | = | 5 | |

x |
x |
24 |

x = |
25 x24 | = 120 kmph. | ||

5 |

Discussion:

325 comments Page 1 of 33.
Ravi said:
2 months ago

Anyone, please explain how can be the term (150/100) mean?

(18)

Pavan said:
3 months ago

Let time is t.

Suppose Car is going at speed x,

Train speed will be x+(x/2) = 3x/2 -------> 1

Speed of Train(dist/time) = 75/(t-12.5) ------> 2

1 =2.

3x/2 = 75/t ----->3

Speed of car = 75/t.

x = 75/t ----> 4.

substitute 4 in 3.

By solving you get;

time of car(t) = 37.5(in min)

By converting into hr.

t = 0.625 hr.

Speed of car = 75/0.625.

Ans : 120kmph

Suppose Car is going at speed x,

Train speed will be x+(x/2) = 3x/2 -------> 1

Speed of Train(dist/time) = 75/(t-12.5) ------> 2

1 =2.

3x/2 = 75/t ----->3

Speed of car = 75/t.

x = 75/t ----> 4.

substitute 4 in 3.

By solving you get;

time of car(t) = 37.5(in min)

By converting into hr.

t = 0.625 hr.

Speed of car = 75/0.625.

Ans : 120kmph

(17)

Navik said:
3 months ago

T:C ( speed) = 3:2.

T:C ( time) = 2:3.

Time taken by car = 12.5× 3.

Speed of the car = 120 km/h.

T:C ( time) = 2:3.

Time taken by car = 12.5× 3.

Speed of the car = 120 km/h.

(29)

Bindu sree said:
4 months ago

The speed of train and car = 150 : 100 = 3 : 2.

Speed = 1/time,

Time = 2 : 3.

Delay = 12.5 for train,

So, car time = train delay × 3 = 3 × 12.5 = 37.5min = 37.5/60 = 5/8h.

Finally speed = d/t =75/5/8 =120km/h.

Speed = 1/time,

Time = 2 : 3.

Delay = 12.5 for train,

So, car time = train delay × 3 = 3 × 12.5 = 37.5min = 37.5/60 = 5/8h.

Finally speed = d/t =75/5/8 =120km/h.

(48)

Rutuja P said:
5 months ago

Speed of car=Sc ; speed of train =St

Sc:St= 100:150 ( A train can travel 50% faster than a car)

= 2:3 -----> (1)

and therefore Tc:Tt= 3:2 -----> (2)

Delay given= 12.5 min = 12.5/60hrs.

Therefore time for car, Tc= 3*12.5/60 = 12.5/20 (refer (2)Tc:Tt= 3:2 )

Distance given= 75km.

We know, Speed= Distance/Time.

Therefore, the speed of the car,

Sc= 75/(12.5/20),

= 120 km/hr.

Sc:St= 100:150 ( A train can travel 50% faster than a car)

= 2:3 -----> (1)

and therefore Tc:Tt= 3:2 -----> (2)

Delay given= 12.5 min = 12.5/60hrs.

Therefore time for car, Tc= 3*12.5/60 = 12.5/20 (refer (2)Tc:Tt= 3:2 )

Distance given= 75km.

We know, Speed= Distance/Time.

Therefore, the speed of the car,

Sc= 75/(12.5/20),

= 120 km/hr.

(71)

Animesh Biswas said:
5 months ago

T:C ( speed) = 3:2.

T:C ( time) = 2:3.

Time is taken by car = 12.5× 3.

Speed of the car = 120 km/h.

T:C ( time) = 2:3.

Time is taken by car = 12.5× 3.

Speed of the car = 120 km/h.

(35)

Saloni Adhikari said:
5 months ago

Here, How can we say given time in the question is equal to the value obtained after subtraction? Please explain me.

(10)

AP bhai said:
5 months ago

Case1: For car;

D = TV.

75=TV.

Case2:

75={T+(12.5/60)}*1.5V.

75={(60T+12.5)*V}/40.

Now, compare case 1&2.

You get, T = 12.5/20;

As. D = TV.

75 = 12.5/20*V,

V = 120.

D = TV.

75=TV.

Case2:

75={T+(12.5/60)}*1.5V.

75={(60T+12.5)*V}/40.

Now, compare case 1&2.

You get, T = 12.5/20;

As. D = TV.

75 = 12.5/20*V,

V = 120.

(2)

Sai Durga said:
6 months ago

Let the speed of the car be x.

Now, the train is 50% faster than a car,

i.e. speed of the train = x + (50/100)x.

= 150x/100.

Now, the train is 50% faster than a car,

i.e. speed of the train = x + (50/100)x.

= 150x/100.

(43)

Sai Durga said:
6 months ago

Let the speed of the train be x.

So, train speed 50% faster than car means x + x/2 = 3x/2, speed = 75km/hr,

time delayed = 12.5 mins==12.5/60 sec.

Now;

D/s = T; (here speed = speed of the car -speed of the train),

So,

75/x -75/3x/2 = 12.5/60.

75/x - 50/x = 5/24.

x = 25 x 24/5

x = 120 km/hr.

So, train speed 50% faster than car means x + x/2 = 3x/2, speed = 75km/hr,

time delayed = 12.5 mins==12.5/60 sec.

Now;

D/s = T; (here speed = speed of the car -speed of the train),

So,

75/x -75/3x/2 = 12.5/60.

75/x - 50/x = 5/24.

x = 25 x 24/5

x = 120 km/hr.

(22)

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