Aptitude - Time and Distance - Discussion

Let me clear this.

We can assume the speed of car as X
Now the speed of train (as per question) = X+X/2 = 3X/2
Distance (as per question) = 75 km
So time = distance/speed
For car its= 75/x
For train = 75/3x/2 now multiply 2 to 75 = 150/3x = 50/x
Now we have to add waist time to the train speed = 50/x+ 12.5/60
12.5/60 because time is given in minutes and we need it in hour so multiply 12.5 with 1/60 So it's = 12.5/60 or we can write it like 125/600= 5/24

Now the new speed of the train is 50/x+ 5/24.

put car speed = trains speed

75/x = 50/x+5/24.
or
75/x -50/x =5/24.
or
75 - 50
--------- = 5/24.
x
or
25/x = 5/24.
cross multiply.

24*25 = 5x.
600 = 5x.
x= 120 kmph.

Just read the question carefully.

Both car and train starting and finishing at a same time that means they are taking the same time to reach A to B.

now we know that,
time=distance/speed

distance= 75
speed of car = x
speed of train = speed of car + 50% of speed of car = x+[50/100]x or 3x/2

now
travelling time of car = time taken in travelling +no waiting
75/x + no waiting
time of train = time taken in the travelling + waiting
= 75/[3x/2] + 12.5/60 hr

and both times are equal as in question it is given they are starting and reaching at the same time let we compare them.

75/x =75/[3x/2] + 12.5/60
which will give x= 50.

Let the speed of the car be x km/hr.

The train travels 50% faster than the car, so its speed is;
1.5x km/hr.

Travel time of car = 75/x hours.

Travel time of train = 75/1.5 x hours = 50/x hours (time=distance/speed).

But the total travel time taken by train is;

The train lost 12.5 minutes (which is Equal to 12.5/60 hour=5/24 hour) due to stops.
Hence, the total time for the train is the Total time taken by the train = 50/x + 5/24 hour.
In a question given that;
Both the car and the train reach point B at the same time.

Therefore:
75/x = 50/x+50/24.
75/x-50/x = 50/24.
25/x = 50/24.
Solve for x by cross-multiplying.
X = (25*24)/5.
X = 120km/hr.

Let car complete distance of 75 km in time t with speed v kmhr

Distance = time x velocity

i.e. 75 km = t * v ------ equation 1

Train waste 12.5 min hence time required for train is t-12.5min

Convert 12.5 min in to hour = 12.5 ÷ 60 = 5 / 24.

i.e. time required for train is t- 5/24.

Speed of train is 50% more than car i.e. 3/2 v.

Distance = time x velocity

Now equation for train is 75 = (t - 5/24) (3 / 2 v) ---------- equation 2.

Now equate equations 1 and 2,

t x v = t - ( 5/24) (3/2 v).

After solving this equation we get,

t = 5 /8.

NOW put t= 5/8 value in equation 1 we get V = 120 i. e. Car speed.

Simple:

Total Distance is 75kms.

Both train and car reached at same time in-spite of train having speed advantage of 50% over car i.e. 3/2.

Train lost this advantage due to 12.5 minutes stoppage.

That means this 12.5 minutes lost time equals to the speed advantage of train, had it not lost this time train would have reached earlier.

Therefore 12.5 *3 = 37.5 is the time taken by car to cover 75 kms.

Train took similar time to cover 75 distance stopping for 12.5 minutes. Had it not stopped it would have covered in 25 minutes (37.5-12.5 = 25).

S=d/t : Speed of car = 75/37.5 = 2 km per hour 2*60 = 120 km per hour.

Can anybody explain if both the vehicle reaches at the same time then where is the delay arises I am talking about if two vehicles reach at the same time then how this equation came?

(75/x) - (75/ (3/2) x) =12.5/60, 12.5 minutes delay will not be there if both the vehicles reach at the same time please help,

Let me explain, the time taken by car to reach B point is 40 min (just assume). Then the time taken by train is 27.5 min without stopping in stations. Hence 12.5 equal to the time taken by car (-) time taken by train without a stop in stations.

Hope you understand, thank you.

Logic is simple guys

Let car speed is x kmph = 100%.
Train speed is 50% more than car speed.

i.e x + 50% of x, 50% of 100% is 50%.

We assume that x = 100% ,so x + 50% will be 150%.
So, speed of train T = 150% of x, i.e (150/100)x, i.e (3/2)x or 3x/2.

We have t = d/s, since they are moving in same direction there difference of d/s is taken and should equal to given time.

ie, (75/x) - (75/(3x/2)) = (12.5/60)hrs (12.5 min to hours).

Solving above equation;

(75/x) - (75 * 2/3x) = 12.5/60.
(3 * 75 - 2 * 75)/3x = 12.5/60.
75/x = 12.5/20.
x = 75 * 20/12.5.
x = 6 * 20 = 120kmph.

Let the speed of the car=x kmph.

Then, speed of the train is 50% more than car = 50x/100 + x = 3x/2 kmph. (50x/100+x/1=50x+100x/100 =150x/100= 3x/2).

Time is taken by the car to travel from A to B= 75x hours.

Time is taken by the train to travel from A to B=75/ (3x/2) +12.5/60 hours. 1 min = 1/60 hrs. So 12.5/60 hrs.

Since both start from A at the same time and reach point B at the same time,

75/x =75/ (3x/2) +12.5/60.
75/x = 150/3x + 12.5/60.
75/x= 50/x + 12.5/60.
75/x-50/x = 12.5/60.
25/x =12.5/60.
x=25*60/12.5...... 1/x=12. 5/60*25 so, x = 60*25/12.5.
X=2*60.
x = 120.

Hi,

S=D/T
D=75, T=?, S=?

We need to find out the Time T, to get the speed S.

Time difference given as 12.5 mins, within 12.5 mins car is able to cover the same distance as train.

If the speed of the car be 100, then the train speed be 150.
It means 12.5 minus = 1/3 km.
12.5=75/3=25.

Therefore in 12.5 mins distance covered is 25km.
Here distance is given in km.

Therefore convert 12.5 minutes to hour i.e., 12.5/60=0.20833.
0.20833 hr = 25 km.
? = 75 km.

By cross multiplication the time taken to reach 75 km is 0.6249.
.
. . s = d/t.
s = 75/0.6249 = 120km.

I have an easier and simple solution.

The train is 50% faster than a car so for a 75km distance when the train reaches (without stopping) at B car will be some distance behind the train.

Well, train is 50% faster so the car should be at 50km when the train reaches point B (75km) (because 50 + 50% of 50 = 75).

And it's given that the train is 12.5 faster than a car (if it stops for 12.5 min and then after they reach at point B together).

Means rest of the distance (25km) car will cover in 12.5 min.

So, the speed of car = 50km/12. 5/60 hr = 120km/hr.