Aptitude - Problems on Trains - Discussion

7/2= 3*2+1/2;

But i have one doubt

Why don't we convert miles into meters can anyone explain this?

In this problem we need time only that's why we are changing time only into minutes.

Because train to pass through the tunnel from the moment the front enters to the moment the rear emerges. So we should add double the length of train then answer should be 3.2 min.

Yes aditya, you are absolutely rite, the answer should be 3. 2 min.

Answer is 3 min because.

Given is speed = 75 miles per hour
Total distance is 7/2 + 1/4 = 15/4

Distance is 15/4 because in question it is mentioned that calculate time from entry of tunnel to the point when train's rear left the tunnel.

                        7/2 miles              1/4 miles
           <---------------------------------><------->
            _________________________________   train
 entry of  [               tunnel            ]__________
  tunnel   [_________________________________]__________]--> -->

Hope so this will help.

mph--->meters/hour
Mph---->miles/our

We have given 3(1/2) mile length of the tunnel and train length as 1/4 mile.
So we relate tunnel length and train length

We have to see in 3 1/2 how many 1/4's are there or how many 1/4's make a 3 1/2
3 1/2 is a mixed fraction to make into a proper fration we do as follows
first multiply 2*3=6
next add the numerator 1 to previous results
6+1=7

We have numerator as '7'and keep the denominator '2' as it is
now the proper fraction for 3 1/2 is 7/2

So here total length 7/2+1/4=15/4miles
time taken=distance/speed
time taken=(15/4miles)/75miles/hr
15/4*75=15/300=>1/20hrs
to convert into min v multiply by 60 .
because 1 hr = 60 min
1/20 *60=3mins
here ends the problem

No need of convert miles to kms because it makes the problem complex and even though all units are in miles they get cancelled.

Please can any one clearly explain the formation of expression.

(7/2) + (1/4).

Can't we just do it this way

3 1/2 =3.50 and 1/4 = .25

So 3.50+0.25 = 3.75

(3.75/75 as t = dist/speed) t = 3.75/75 = 0.05

Now 0.05 is in hr so 0.05*60 = 3

This method is better then above.

According to me answer will be c)3.2min

Because the lenth of train has to be taken twice bec. intially train of length 1/4 will enter and then it wil leave the tunnel

Therefore distance travelled=7/2+1/4+1/4=4

Please check it out