Aptitude - Time and Distance - Discussion

For people who still has confusion with the 150*x/100:

Why don't you remember the LCM formula?.. if speed of car is x then speed of train is more than speed of car by 50% of it. so after we calculate 50% of speed of car, we should not forget to add the result of it to the actual speed of car, i.e-
[50% of x]+x
= [(50/100)*x]+x
= [50x/100]+x... now, this is actually [50x/100]+[x/1] the two denominators '100' and '1' here are 100, and 1 and the LCM of the two is 100. This makes 100 the common denominator. now following the rule of LCM, divide the common denominator by the first actual denominator, i.e 100/100, the result is '1', and now multiply the first numerator by this result '1' that keeps 50x*1 that is the same, so basically the first numerator does not change. now, do the same with the second denominator, multiply this common denominator 100 by the second actual denominator, i.e 100/1, the result is '100', and now multiply the second numerator by this result '100' that makes it 100*x that is 100x.. so after the LCM now it looks like:
(50x+100x)/100.

Now take x as common and comprehend this numerator as-
x(50+100)/100
=x(150)/100
=150x/100
DID U GET THIS NOW? THE TRICK LIES IN THE LCM FORMULA !!

This can again be derived to 3x/2, applying the ratio formula.

------------------
Here is another simpler method.. when we calculate speed of train as 50% more than that of the car, in a lay-man's language it simply means speed of train is more than speed of car by half of car's speed, so instead of 50% of car's speed, we can also do half of car's speed, and if car's speed is 'x', then train's speed is

(x/2)+x, applying the same LCM formula, take 2 as the common denominator and the first numerator remains 'x' and the second numerator becomes '2*x'
= (x+2x)/2, now if u remember, 'x' is equivalent to '1*x', as per basic algebra logic- so,
= 3x/2.

GOT THE SAME ANSWER ??

Another shortcut yet simple method I would like to share wherein forget the hassle of x and y so the method is as follow:

Time taken by the car and train to reach the destination is the same and if you read the question carefully it is mentioned that the train has halted for 12.5 minutes at the stations that means just because of 12.5 minutes halt the train and the car covered 75 kilometers at the same time.

Just because of 12.5 minutes the train misses to cover 50% of the route that means the total time taken by the train and the car to reach the destination is 37.5 minutes ( working- 75km/2 =37.5, so in 12.5 minutes stop it missed 37.5km, hence the train took total 25 mins to reach the destination + 12.5 mins halt = 37.5 mins. )

Now comes the cross multiplication part. 37.5 mins = 75 kms
60 mins = ? (working- we already know how much kilometers it can cover in minutes but we want to find km/hour, so 1 hour = 60 mins.)
= (60*75)/37.5
=120.

To solve in this way you require a bit logic and good knowledge of cross multiplication. Still if you don't understand this method I suggest you to go for the Admins method.

Let car speed = x*100%.
= x * (100/100) = x kmph. // just for understanding

Train is 50% more than car
So train speed = x*(100+50)/100 = 150x/100 = 3x/2 kmph.

In the second step. Let us consider an example a train can reach the destination without stopping at the stations in 10 mins. So the actual time is 10 mins. If it takes extra 2 mins while stopping at the stations. So 10+2=12 mins.

A car if totally takes 12 mins to reach the destination. Train Time taken by train i.e. 12 mins = Time taken by car i.e. 12 mins //Both takes equal time.

10(actual time)+2(extra time) //train = 12 //car.

12-10 = 2 ---->1

NOW COMPARING 1 with our problem. The Extra time taken for the train is given that is 12.5 mins.
Hence the equation will be:
(time taken by car) - (actual time taken by train) = extra time taken by the train.

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

Since we don't know the time taken by car. Time = distance/speed. So, 75/x similarly for train as 75/(3x/2).

So 75/x -75/(3x/2) = 12.5/60. By solving get the x.

Let's start by assuming that the speed of the car is x km/h.

Since the train can travel 50% faster than the car, its speed would be 1.5x km/h.
Now, let's calculate the time taken by both the train and the car to travel from point A to point B.
Time taken by the car = Distance/Speed
= 75/x

Time taken by the train = Distance/Speed
= 75/(1.5x)

However, we know that the train lost 12.5 minutes while stopping at the stations. So, we need to add this time to the total time taken by the train.
Total time taken by the train = 75/(1.5x) + 12.5/60.

We know that both the train and the car reach point B at the same time. Therefore, we can equate the time taken by the car and the total time taken by the train (including the stoppage time).

75/x = 75/(1.5x) + 12.5/60.

Simplifying this equation, we get:
5/x = 4 / (1.5x) + 1/6.

Multiplying both sides by 6x, we get:
30 = 16 + x.

Therefore, the speed of the car (x) is:
x = 14 km/h.

So, the speed of the car is 14 km/h.

Hi, now see a simple and driveway.
Train speed is 50% more than car speed it"s given

Now,
Lets the speed of car be x and 50% of x be x*50/100=x/2.
The speed of the train is x+50% more than car (x/2) =3x/2

Now,
At the same time, both cover same distance ie. 75km so,

75/x=75/3x/2 .

But it is also given that train stop at any station nd waste time 12.5 minutes (and we have to change it in km/hr so it becomes 125/10 same as 12.5 divides it by 60 to make it in hour=125/10*60).

From question, we understand speed of the train is more than car so it reaches first but both reach at same time due to train stop at any station so,

To calculate speed of car,

75/x- (75/3x/2+125/10*60) =0 (according to question (75/3x/2+125/10*60) is the time taken by train to calculate same distance) now,

75/x-75/3x/2=125/600.
75/x-50/x=25/120.
25/x=25/120.
1/x=1/120.
X=120km/h is the speed of car now,
I think everyone understands and get the answer. thank you.

***Assume the speed of car be 'a'.

TIME OF THE CAR = distance of car/speed of car.
= 75/a.


Since train is 50% faster than car, therefore
50% of a=====50/100 X a===1/2 X a====a/2 (this is 50% of the car speed),
a + a/2====3a/2 (train is 50% faster than car).

TIME OF THE TRAIN = distance of train/speed of train.
= 75/(3a/2),
= 150/3a,
= 50/3a.

Time of the car - time of the train = 12.5 min ----------------------> MAIN FORMULA.

75/a - 150/3a = 12.5/60 (since we are calculating in km/hr, so divide it by 60),
75/a - 50/a = 12.5/60,
= 120.

The train can travel 50% faster than car. So we are assuming car speed as X kmph,then we want to know the train speed so we are adding car assumed speed + train speed (data given 50% faster so it was in percentage we are converting into normal form as 50/100)
Train speed= (X+ 50/100)
= (100X +50)/100
=150X/100
= 3/2 X
=1.5X



2.next step as we see
we want to find car speed only and if u see in the data he didn't ask exact train speed. See both car and train reach at point B 75kms away from point A.

75/x -------> car speed
75/(3/2)X ----->train speed
75/X-75/(3/2)X=12.5min (we have to add the train delayed time to get car speed)
75/X-75/(3/2)X=12.5/60 ------------>1min = 60 secs
Finally car speed X=120 kmph
Train speed = (3/2)X =(3/2)120
=180 kmph.

Let the speed of car be x.

Since the speed of the train is 50% more than the speed of the car, speed of train will be x+(50% of x).

i.e x+x/2 which is 3x/2.

now, let t be the time taken by car. therefore t=75/x --->eq(1)

(using time= distance/speed).
given that train loses 12.5 mins. therefore the total time travelled by train is t-12.5 min (note that both train and car took the same time to reach B. that's why we can take t as the time taken by both car and train)

using time=distance/speed formula, we will get;

t-12.5=75/(3/2x).
which can be rewritten as t-12.5=75/1.5x ----> eq 2
from eq1, we have x=75/t , sub in eq 2.

t-12.5=75/1.5x.
t-12.5=(75*t)/(1.5*75),
t-12.5=t/1.5,
solving which we get.
t=37.5 min=37.5/60 hrs.

as we have x=75/t (from eq1).
x=(75*60)/37.5,
x=120km/hr.

Let's start by assuming that the speed of the car is x km/h.

Since the train can travel 50% faster than the car, its speed would be 1.5x km/h.
Now, let's calculate the time taken by both the train and the car to travel from point A to point B.
Time is taken by the car = Distance/Speed
= 75/x

Time taken by the train = Distance/Speed.
= 75/(1.5x).

However, we know that the train lost 12.5 minutes while stopping at the stations. So, we need to add this time to the total time taken by the train.
Total time taken by the train = 75/(1.5x) + 12.5/60.

We know that both the train and the car reach point B at the same time. Therefore, we can equate the time taken by the car and the total time taken by the train (including the stoppage time).

75/x = 75/(1.5x) + 12.5/60.

The ratio of speeds of train and car is 1.5 : 1

Therefore the ratio of time taken by them to cover the same distance is 1 : 1.5

The train lost 12.5 mins while stopping at stations. This means if the train didn't stop at stations it would reach the destination early by 12.5 mins.

That means the difference between the time taken by train and car is 12.5 mins.

Let consider x as the time taken by train.
Then as per the time taken ratio 1 : 1.5, the difference between 1.5x and x is 12.5 mins,
Then x = 25 mins,
Then 1.5 x = 37.5 mins.

Now the car traveled 75 kms in 37.5 mins,
The speed of car is 75/37.5 kms per min.
Speed of car = 75000/2250 m/s
i.e., 100/3 m/s.
Converting m/s into km/hr-
i.e., (100/3) * (18/5) = 120 kmph.