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:
100 kmph
110 kmph
120 kmph
130 kmph
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:
345 comments Page 19 of 35.

Sagar Bhamare said:   9 years ago
The total speed of a car 100% + 50% extra,
So the speed of the train is 150%.

Which can be written as (150/100)X.

Tabish said:   1 decade ago
Actually it is 50% more.

i.e (100% + 50 %) * x.

((100/100) + (50/100)) * x.

(1 + (50/100)) * x.

(150/100) * x.

Bhumi said:   10 years ago
Car speed = x.

Train speed is 50% more.

Train speed = x+ (50/100) x = 100x + 50x/100.

= 150x/100.
= 3/2x Kmph.

Rias said:   8 years ago
75km/s * 60 second = 4500sec,
50%-12.5 the train lost =37.5,

Speed = Distance/time.

then 4500sec/37.5 = 120km.
(1)

Yash said:   6 years ago
Let Car speed is x.
The train speed is 50% faster than x.
Then the train speed is x+50%of x.
x+50x/100 =150x/100.

Suja said:   1 decade ago
@ jamal:

We know that time=(distance/speed).

Here distance=75kmph and speed=3x/2.

Therefore time=75/(3x/2)

NITISH GULERIA said:   1 decade ago
Hello friends don't confuse 50% more means x+50% = x+50/100, = x+1/2.

= LCM are 2 and whole equation are 3x/2.

Srivishnu said:   8 years ago
It's 12.5 minutes.

To convert it to an hour it should be divided by 60.
i.e 12.5/60 hrs (or)125/(60*10) hrs.

Shashidhar said:   6 years ago
Given.

Time delay by 12.5 (minutes),
1hr = 60minutes,
1min = (1/60)hr, 125/10 = 12.5.
125/(10*60) = (5/24).

Sravani said:   5 years ago
Why to subtract the speed of the train from speed of car (75/x-75x/[3/2])?

Can anyone help me by explaining?
(1)


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