Discussion :: 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:
| [A]. | 100 kmph | [B]. | 110 kmph | [C]. | 120 kmph | [D]. | 130 kmph | Answer: Option C 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 |
| Mythili said: (Jun 30, 2010) | | Can you exlpain how 150x/100 came? | Chandra Mouli said: (Jul 8, 2010) | | Let car speed be xkmph
then train speed is 50% more than car speed
ie 50/100*x | Komal said: (Jul 9, 2010) | | I think there should be 50x/100 & not 150x/100 is it? | Xyz said: (Jul 16, 2010) | | When we take 50x/100 we get x=[-75* 24]/5 = - 360kmph
its not in the option | Mohit said: (Jul 18, 2010) | | Hey look it say train travels 50% faster then car. Suppose car travels at x speed than train travels at 50x. So its 150/100 thats as a % age. | Shalini said: (Jul 19, 2010) | | Hae even though its % it must me 50x/100 only na? | Paresh said: (Aug 3, 2010) | | Can any one please tell me how that 125/(10x60) came? | Vidu said: (Aug 4, 2010) | | Please some one let me know how 150/100 came. | Rakesk said: (Aug 13, 2010) | | 50 % more means not 50/100. car has already 100 %.50 % more means 150/100 | Raj said: (Aug 23, 2010) | | Hi, actually we get speed as 25 only since Speed = Distance/(Time x Extra time)
Speed = 25/(125/600) => (25 x 24)/5 = 120. | Bhavana said: (Aug 29, 2010) | | Car seed is 100. So train speed in 50% more.
So it is 150/100. | Hari said: (Sep 11, 2010) | | Car speed be x then train speed is 50% more so. It must be 100+50 if we take car speed in percentile.
So train speed is 150/100 x. | Muthuraj said: (Sep 21, 2010) | | 125/10*60 here we are converting 12.5 min into hrs..12.5*10/10=125/10(min)--125/(10*60) hrs | Santhiraju said: (Oct 4, 2010) | | Car speed is X KMPH then Train speed=(car speed + (50*carspeed)/100) ie train speed =(X+(50X/100))
Take LCM and then(100X+50X)/100 .....ie 150X/100 | Nitesh Nandwana said: (Oct 9, 2010) | | Hi guys think simple not complicated. Let speed of the Car be x kmph. so, 50% of x = x/2 Train speed = Car speed + 50% of Car speed
Train speed = x + x/2 = 3/2x | Singh said: (Oct 29, 2010) | | Here percentage is given means we have to take any unknown quantity as 100 thats why x is taken as 100. here train speed is 50% more than car means it travels with 100+100/2 speed.total speed is 150.now covert 150 into percentages as 150/100.now 150/100=3/2.if car speed is let xkm/h.then train speed is 3/2x km/h. | Saravan said: (Nov 11, 2010) | | 75/x - 75/(3/2)x = 125 /10*60
plz explain me why we subtract the train speed from car speed i didnt understood this step. can anyone explain me. | Sonam Jain said: (Nov 19, 2010) | | In 125/10*60 Actually what happen in the question they said train lost 12.5 min. so first of all we change it in hour thats y divided by 60. now bcz it lost thats y it takes the difference b/w boths time is equal to the lost time. 12.5 is equal to the 125/10
So final eq is diff b/w time of train and car is = to lost time i.e. 125/10*60 | Sushil said: (Dec 10, 2010) | | Let the car speed is x kmph
ie,the car speed is 100x kmph Then train speed is 150kmph
equation is 75/1.5x+12.5/60=75/x
x=120kmph | Faizan Akhtar said: (Dec 30, 2010) | | 75/x - 75/3x/2 = 125/10*60
can you tell me how this equations is derived? | Deepak Porwal said: (Jan 6, 2011) | | As the train speed is 50% more than car So train speed is (car speed + 50% of car speed)
i.e. (x+(50/100)*x)= 3x/2;
now time diffrence between them is 12.5 minute converting it in to hour 12.5/60;
now (distance/speed of train) - (distance /speed of car)= 12.5/60;
as time diffrence between them is 12.5 min; | Pradeep said: (Jan 7, 2011) | | Can you exlpain how 150x/100 came ? | Rohita said: (Jan 27, 2011) | | Explain the simple way, I can't understand train speed. | Senhil said: (Feb 1, 2011) | | Given train is 50% greater than car. Let car speed will be 50km/hr then train speed ll be 100km/hr. So 100+50=150 km/hr. | Prasad said: (Feb 22, 2011) | | Train speed is 50% more than car, so it is 1/2 % more speed than car .
So it is 1 and 1/2 == 3/2 ( as car 100% it will be 1). | Jamal Akhter said: (Feb 22, 2011) | | Could you explain how time for train you had taken 75/(3/2.x) ? | Suja said: (Feb 23, 2011) | | @ jamal: We know that time=(distance/speed). Here distance=75kmph and speed=3x/2.
Therefore time=75/(3x/2) | Shunmuga Priya said: (Mar 3, 2011) | | Sonam Jain said that " (125/10*60)it takes the difference b/w boths time is equal to the lost time " But in question it is clearly given as "from point A at the same time and reach point B 75 kms away from A at the same time". The car and train starting and ending time is same so no need of subtraction but they subtracted here.
I cant get this. please explain | Sushil Pal said: (Mar 5, 2011) | | Let Car's speed be x Kmph
Train speed = (Carspeed + (50*Carspeed)/100) " = (x+(50x)/100) " = 150x/100 " = 3/2xKmph Now the train lost about 12.5min, so time subtraction betn Car and train will give us distance covered by car in that much period of time (lost period)and to know the speed we have to equate it with lost period
Distance covered = lostPeriod
75/x-75/(3/2)x = 125/600
75/x-50/x = 5/24
x = (25*24/5)=120 Kmph. | Jinto said: (Apr 14, 2011) | | 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 reaches 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 ther if both the vehicles reaches at the same time pls help | Reddy said: (Apr 29, 2011) | | Hi friend that is not delay thats the time wasted by the train due to its stoppings if the train don't stop any where it will come first than car. | Divs said: (May 9, 2011) | | Raj would you please tell me how come the speed be 25? | Hemant said: (May 9, 2011) | | Explain the simple way, (75/x)-(75/(3/2)x)=12.5/60 | Atul said: (Jun 8, 2011) | | (75/x)denotes time taken by car.
this means time taken by the car to cover the distance of 75 km with speed x.
(75/(3/2)x)denotes time taken by train to cover the 75 km distance with speed 3/2x.
in the problem it is said that train rest for 12.5 min at station and both reach 75 km at same time.
There fore difference between the time is (75/x)-(75/(3/2)x)=12.5/60 in hrs. | Rameez said: (Jun 13, 2011) | | Let t' is the exact time of the train without stopping...
Let t is the exact time of the car
12.5 minutes is the wastage of time because of train stopping Since in given problem it is said that both car & train arrives at same time.... Therefore t'+(125/600)=t
75/(3x/2)+125/600=75/x
75/x-75/(3x/2)=125/600
By solving it we get x=120Kmph | Hemant Sharma said: (Jun 25, 2011) | | Hi, its right, I'm little confuse that why calculate difference b/w both but car lost time 12.5.so the difference b/w boths time is equal to the lost time | i.e 12.5 is equal to the 125/10 and change into hrs so 125/(60*10) . | Prawinn said: (Jul 5, 2011) | | Simple train and car reach the place in a same time if car takes 8 hours to reach the place train also take 8 hours that 8 hour be t+somewaste of time the diff b/w both train and car is 0. So, t1-(t+x) = 0
=> t1-t = x . | Kiru said: (Jul 6, 2011) | | Thanks senhil I understand easily. | Sreekanth said: (Jul 8, 2011) | | Think... Y 3/2x came with an example...
if u think car has x speed, then train ll have 2x.
then substitute x with simple value.. x:2x
if x=10 then
10:20
it mean train has double value of car speed. but x:3/2x
then 10:15x
In which train has 50% faster than car.
50% of car is 5. so 10+5=15. | Arun said: (Jul 13, 2011) | | Rameez your explanation is great. | Mysterious said: (Jul 25, 2011) | | Friends...... let speed of car be x
thus, speed of train (150/100)x (stated that train is 50% faster than car) now since the train lost 12.5 mins. i.e. the difference between time taken by car and time taken by train is 12.5 mins, and since TIME=DISTANCE/SPEED time taken by car= 75/x
time taken by train= 75/(150/100)x or 75/(3/2)x and the difference between time taken by car and time taken by train is 75/x - 75/(3/2)x = 125/10*60.
x=120 kmph | Srihari said: (Jul 27, 2011) | | Tell me about 12.5/100*60 ? | Hab said: (Jul 27, 2011) | | Since the time taken is in kilometers per hour then we need to convert 12.5 minutes into hours.
So just to 12.5/60 = 5/24 . | Swetha said: (Aug 10, 2011) | | @mysterious
well explanation mysterious..thank you!!! | Priyanga said: (Aug 10, 2011) | | Please explain me in a easy manner I could n't understand it. | Yesoda said: (Aug 11, 2011) | | How do you get 150/100? | R.Sreekanth said: (Aug 14, 2011) | | @Yesoda They given speed is 50% more than car so, We don't know car speed,let consider car speed as X. So, the train speed is 50% more than car(i.e, when ever we talking about anything in the form of % we must add or subtract from 100. Then convert it into [total/100], total= % + 100 ). Therefore: the train speed is [(50+100)/100] more speed of car.
So, the train speed is = (150/100)*x. | Kasi Srinivas said: (Sep 1, 2011) | | @ mysterious.
Thank you dude :) I understood. | Raul said: (Sep 2, 2011) | | 100*(t)=150*(t+12.5)
t=37.5min
t
=37.5/60hr
speed=(75*60)/37.5 =120 | Bhupat said: (Sep 6, 2011) | | All righ but my question is. No body has taken 12.5 minus loss in train speed calculation. Why? | Khanchana said: (Sep 10, 2011) | | Suppose speed of the car = xkm/hr(1)
speed of the train = 50% more than that of train = x + 50/100x = (100x + 50x)/100 = 150x/100 = 3/2x ------------>(2)
delay of train = 12.5 mins= 12.5 * 10/60*10 hrs | Sravya said: (Sep 12, 2011) | | Thank you so much @kanchana. | Sravya said: (Sep 12, 2011) | | @kanchana
How did x+50/100 became 100x+50x as it will become (100x+50)/100x ? | Amit said: (Sep 19, 2011) | | Sarvya Kanchna is right:-) dont consider x with 100 consider with 50 like this.
=x+50x/100
=(100x+50x)/100
=150x/100 If you consider x with 100 then when we take LCM then eq.
=(100x^2+50)/100x
:-) | Dhruva & Hitesh said: (Oct 6, 2011) | | First we assume that the speed of car is 100% so as per Q.speed of train is 150%. | Ajit said: (Nov 12, 2011) | | In given problem ,u ve to equate total or exact time of both... Let t be time tkn by car to cover 75kms... n t' be time takn by train...
i.e. train's total time=stoppage time+time in which it covers 75kms...nw t=t'.... | Santhosh said: (Nov 17, 2011) | | Here,in this problem if we take 50/100 then it becomes 1/2 So we have to take 50% more..
So,assuming car speed is 100 %,then increase it to 50%..means 100+50/100 = 150/100 | SONAI SHEIKH said: (Dec 16, 2011) | | Why 10 is multiplied with 12.5? Can anybody explain? | Solanki said: (Dec 16, 2011) | | A is faster than B by 50 percent means, take a is 100 percent and more 20 percent faster than 100+20=120. And take B is 100 percent. Then 150/100. |
Nagesh said: (Dec 27, 2011) | | How we get 125/10*60?
Can any one explain easy way? | Monideepa Ganguly said: (Jan 6, 2012) | | 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 ?? | Ranjeet said: (Jan 8, 2012) | | Let suppose car speed = x km/h
As per problem train speed 50% more than car so = x+50/100x = 3/2x. | Firoz Rahman said: (Jan 8, 2012) | | The Train lost 12.5 min to reach the destination.By this we understood that train took more time than car to reach the destination. So the car time should be subtracted from train's time and the result we get is -120. As speed will not be negative, the answer is 120. Is this right one? | Shro said: (Feb 17, 2012) | | Thanks Santhiraju :). | Shro said: (Feb 17, 2012) | | I am not understanding the main calculation part i.e. steps after finding speed pf the train.
Somebody please explain me the steps of main calculation | Laxmi Priya said: (Mar 6, 2012) | | Let the speed of the car be 100% and the train is 50% more
100%+50%=150%
that is 150/100 | Dhinesh said: (Jul 5, 2012) | | Let Speed of the car=x. Train Speed=Speed of car+50% (Speed of car). => x + 50/100 (x). => x + x/2.
=> 3x/2. | Rajan said: (Aug 12, 2012) | | Say, the speed of the car be 1 km/hr (0. 5 +0. 5). And if there is 50% increase in speed means (1 +0. 5) = 1. 5 km /hr.
That's how 3/2 came. | Rahul said: (Aug 20, 2012) | | It's a very easy question
suppose train running 12.5 min. & lost 12.5 mint in station
s=d/t====>
s=(75/25*60)km/s
s=1/20km/s change in m/sec
s=1000/20 m/sec =50 m/sec
change in km/hr
50*18/5=180km/h train speed.....in option of this question
only 120 car speed 50% is satisfy train speed 120+120/2=180 that's it...... | Maheshwaran said: (Aug 30, 2012) | | HOW 150 comes in the initial step ? | Saranya said: (Sep 2, 2012) | | @maheshwaran take x as a spd of car
T=75/x -->*
spd of train is x+x/2(train 50% faster than car)
and time reduce by 12.5 min( by qus)
12.5min=12.5/60hr= 2.5/12hr T-2.5/12=75/(x+x/2)
T=(75/x+0.5x)+ 2.5/12
-->#
by solving * and# we will get ans as 120 | Meetali said: (Oct 4, 2012) | | If speed is x, how do you'll apply the formula in 75/x i.e distance upon speed gives time isn't it? | Lalit, said: (Oct 5, 2012) | | Yes, Meetali T=d/s. The difference between the the time of car and train is 12. 5 min. i.e. we are calculating 75/x-75/3/2= 125/10*60. | Pavan said: (Oct 22, 2012) | | 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. | Neo said: (Dec 14, 2012) | | 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. | Chinnu said: (Jan 5, 2013) | | train lost 12.5 min , convert into hrs = 12/60 = 1/5. Remaining time = 4/5. 4/5*75 = 60.
As it is 50% then total speed of car is 120. | Preethi said: (Jan 23, 2013) | | 75/x - 75/(3/2)x is in KMPH and the term 12.5 mins is converted into secs. how could the answer can be given in KMPH by equating these two terms? please anyone who knows the explanation, explain me? | Ashish Katoch said: (Feb 24, 2013) | | 150/100 come using this strategy. <-------------simple logic:-------------> Suppose car speed =x km/hr. Then train speed= x + 0.5x. 1.5x or 150/100 or 3/2.
{0.5x comes by breaking the 1x into half (50%) i.e 1/2 i.e 0.5}. | Smith Matsiko said: (Mar 31, 2013) | | Since the train is 50% faster than than the car and also delayed for 12.5 minutes, It implies it would have used the time delayed to cover half of the distance=75/2 km. Therefore its speed =180 km/hr (75/2*60/12.5). Since trains' speed assuming x to be the speed of the car, Then x + 0.5x = 180 km,
Thus x = 120 km/hr. | Tabish said: (Apr 6, 2013) | | Actually it is 50% more. i.e (100% + 50 %) * x. ((100/100) + (50/100)) * x. (1 + (50/100)) * x.
(150/100) * x. | Sankalp said: (May 1, 2013) | | Can anyone please explain me this solution easily? | Sowmi said: (May 7, 2013) | | 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. | Anisha Patel said: (Jul 20, 2013) | | Hi, Actually we get speed as 25 only since. Speed = Distance/(Time x Extra time).
Speed = 25/(125/600) => (25 x 24)/5 = 120. | Gannu said: (Sep 1, 2013) | | Hi guys let me explain in this way. Given that both reached the distance 75 km at the same time. That indicates the time taken by car is equal to the time taken by train. However the time taken by train also includes the delay times due to stations. Thus, Train time = distance/speed + 12.5 min (total time). Cars time = distance/speed (total time). As distance = 75 km it is clear but. The speed of train is 50% faster, means 50% more. If car speed = S then 50% more is. S+(50/100)S.
Substitute the equation and you can see the answer your self. | Pravin said: (Sep 3, 2013) | | Time taken for car - time taken for train = 5/24 hr.
This equation is derived because both start at same time and end at same time. | Manmohan Pal said: (Sep 5, 2013) | | Let speed of car = c kmph. So speed of train be c + (50/100)c = (3/2)c. Time taken by car to cover 75 km distance t1 = 75/c. Time taken by train to cover 75 km distance t2 = (75*2/3c). But it is given train lost 12.5 m while stopping at the stations. So, Actual time taken by train when stopping at stations to cover 75 km = (75*2/3c) + (12.5/60). It is given that. Both start from point A at the same time and reach point B 75 kms away from A at the same time. 75/c = (75*2/3c) + (12.5/60).
c = 120 kmph. | Rajesh said: (Oct 15, 2013) | | Assume the car speed as 120 km/hr.
If cars speed is 120 km/hr then it reaches 75 km in 37.5 min(explained below). Since 120 km/hr is 120 km/60 min, which is 2 km/1 min.
Therefore for 75 km it takes 37.5 min. Train speed is 50% higher so it will be 180km/hr.
180 km/hr is 180 km/60 min, which is 3km/min. Therefore for 75 km it takes 25 min.
It lost 12.5 min in stopping so total time taken by train is (25+12.5) = 37.5 min.
Therefore our assumption is correct. Answer is 120 km/hr. | Somya said: (Nov 17, 2013) | | @Neo, your method is good. I just want to know how you got that 32.5? please elaborate? | Sandeep said: (Nov 29, 2013) | | Let as take car speed as =100 than train speed is 50% faster than car so 100+50=150. | K. Srinivasan said: (Dec 9, 2013) | | Let the speed of the car be 10km/h and so as the speed of the train is 50% more ie 50% of 10 is 5 and so the speed of the train can be assumed to be 15km/h. Based on this assumption also this problem can be solved. | Sandy said: (Jan 29, 2014) | | Please any one can explain the equation [75/x]-[75/1.5x] = [12.5/60]. | Gauravnagayach said: (Feb 22, 2014) | | please explain: (x+(50/100)*x)= 3x/2
Where 3x/2 & how 150/100? | Neo said: (Feb 28, 2014) | | Here 150/100 how it come means. Actually they said train 50%more than speed of car.
So train actually speed was 100% means 100/100 and he said 50 more so 150/100 that's it. | Shivu said: (Mar 5, 2014) | | Can you please tell me how that 125/(10x60) came? | Sri said: (Mar 7, 2014) | | How 125/ (10*60) came?. Anyone please explain this point. | Vidhya said: (Apr 6, 2014) | | We need to convert 12.5 min to hours because the options given are in kmph (kilometre per hour). To convert 12.5 min to hours you divide it by 60. To make calculation easier 12.5 is converted into 125 by multiplying it by 10. If you multiply numerator with a number, then you also have to multiply denominator by that number. Hence in this case it will be 12.5*10/60*10 = 125/600 which is actually the same as 12.5/60. 125/600 = 5/24.
12.5/60 = 5/24. | Mancy said: (Apr 8, 2014) | | Still I have a one confusion in my mind that why we are taking speed of a train as,
(50% of car+x) instead of 50% more i.e (50%+x). | Justin said: (Apr 10, 2014) | | Hi @Mancy, OK here the answer for your question, Ex:1% of 2 = 1/100*2 = 1/50. 10% of 100 = 10/100*100 = 10. So now come to our problem ok. Take speed of the car is X ok. (Consider 100). Train is 50% faster ok so 100+50 = 150;. So we need find that 150%X which gives 150/100*X.
All the best. | Vibha said: (Apr 15, 2014) | | Since the Car covered the half distance while train was halting, the speed of car becomes 75/2 * 60/12.5 = 180KM/hr
Now Train is 50% faster than car so train's speed should be 180 + 90 = 270 KM/hr | Pachimbare said: (May 4, 2014) | | 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. | Sachin said: (May 30, 2014) | | And what about that 12.5 min? | Sharad Bhosale said: (Jun 11, 2014) | | Suppose car speed is 'x' km/hr and speed of train is 50% higher than car speed, i.e. [ x + 50/100x]. Hence [ x + 0.5x] = 1.5x.
i.e. 150/100x this is speed of car. | Sadhana said: (Jul 4, 2014) | | Hi,
Why 150/100 why not 100/150? | Mudassir said: (Jul 11, 2014) | | 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. | Anonymous said: (Jul 29, 2014) | | Let the speed of the car be x. Then 50 percent of x = x/2.
Speed of the train = x + x/2 = 3x/2. | Ramesh said: (Aug 1, 2014) | | Why the LHS is equated with RHS? | Jothi said: (Aug 18, 2014) | | What you mean by percentage ? hey its not the additional value to add with existent one number. How its possible to 3/2 for 50% ? 1/2 is only possible for 50%. | Sriram Kalyan said: (Aug 23, 2014) | | Can you please explain the first step. How did you take 150x/100? | Manu said: (Aug 28, 2014) | | 50% more than the car speed means lets car speed is 100 usual assumption 50% more means 150%. | Arun Sanap said: (Aug 30, 2014) | | Hey. I can't get how they take 125/10*60. In that I understand 125/60 but. How they take 10? Please help me. | Prasad Chowdary said: (Sep 5, 2014) | | Hi, @Arun to remove the decimal form 12.5. We multiply and divide 10.
(12.5*10)/10 = 125/10. | GIS said: (Sep 13, 2014) | | Let Speed of Car = x. Train is 50% faster than car = x + x/2 = 3x/2. Time difference = 12. 5 mins = 12. 5/60 hours = 125/600 = 5/24. Distance = 75 km. (Distance/Car speed) - (Distance/Train speed) = 12. 5 min = 5/24 hr. (75/x) - (75/ (3/2x) ) = 5/24. (75/x) - (150/3x) = 5/24. 25/x = 5/24.
X = (25*24) /5 = 120kmph. | Anonymous said: (Oct 6, 2014) | | Can someone explain this because I don't understand where all the fractions just popped out. | Karthik said: (Oct 12, 2014) | | It says 50% more than Car's speed. So if car's speed is X the train's speed is = (X+(50/100)X).
Hence (150/100)X. | Vipul said: (Nov 4, 2014) | | Can you explain how 125/10*6? | Jeeva said: (Jan 31, 2015) | | I cannot understand this problem. Briefly explain. | Kulwant said: (Feb 22, 2015) | | 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. | Kamesh said: (Mar 2, 2015) | | I don't understand this problem please solve easily. | Diwakar.I said: (Mar 4, 2015) | | Can anyone explain how(150/100) came? | ANAND said: (Apr 9, 2015) | | 50% car faster means 100%+50% = 150. | Pramod said: (Apr 21, 2015) | | 50% faster means 1+1/2 = 3/2. | Ashish said: (May 5, 2015) | | 50% more than x means 50/100 + x = 150/100x. | Tarun said: (May 8, 2015) | | Its easy guys OK I explain. Let speed of car is X. Then speed of train is 150/100 multiply by X. Then after solve this 3/2X kmph. Taking difference between car and train speed minutes.
= 75/x-75/3/2x = 12.5/60. Then speed of car means X is equal to 120 kmph.
| Tarun said: (May 8, 2015) | | Lost time is equal to difference between the time taken by car in X kmph and time taken by train in 150/100X kmph distance is same for both 75 kms.
= 75/X-75/(150/100)X = 12.5/60.X = 120 kmph. | Gourav said: (May 9, 2015) | | How to solve this question please tell me right formulas? | Sarath said: (May 14, 2015) | | In one way it can be understood as difference of times is 12.5 min. i.e 75/x-50/x = 12.5 min. But, if the time taken by train is excess 12.5 mins then car reaches before 12.5 mins.
We write this as car time + 12.5 mins excess = Train time i.e 75/x+ 12.5 min = 50/x. | Bhu said: (May 20, 2015) | | Time of train+12.5 = Time of car. | Satyajit said: (Jun 10, 2015) | | Train is faster so it should have been 75/(3/2*x)-75/x = 125/10*60. | Sowmiya.k said: (Jun 12, 2015) | | Please anybody explain it briefly I can't understand. | Deepak said: (Jun 24, 2015) | | Please explain in easy way can't understand please. | Rohit said: (Jun 29, 2015) | | Let speed of car be = x km/hr. Then, speed of train will be = 50% more than car. 50% means half of speed of car. Half of car speed can be written as "x/2. Then, speed of train will be = x+x/2 = 3x/2. Total distance = 75km. Then, according to question Time = Distance/Speed, 75/3x/2-75/x = 12.5.
300/3x-75/x = 125/10/60.
300-225/3x = 125/600. 75/3x = 5/24.
3x*5 = 75*24.
3x = 75*24/5. Therefore, x = 75*24/15.
x = 5*24 = 120. | Fazil said: (Jul 23, 2015) | | Can't we say speed of train is 50% faster than car.
If speed of car is x, then train must move with 2x speed. Is it? | Mukund said: (Aug 22, 2015) | | I think the question is technically wrong. It is given that both the car & train reach point B at the same time. Then what is the significance of lagging of train by 12.5 min. And if it is so (if they reach in same time) then the equation this should be valid. 75/x = 75/x+ (x/2). Where x = Speed of car.
x+(x/2) = Speed of train. | Aliya said: (Aug 31, 2015) | | Its not in same time it is at same time if we consider car starts at 10 to reach at 12 then train also have at same time but it is only possible with different speed if have time lag. | NITISH GULERIA said: (Sep 1, 2015) | | 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. | Shyam said: (Sep 10, 2015) | | Can anyone explain in detail? | Kishan B said: (Oct 13, 2015) | | Here is simple way of solving this problem. We have Velocity = Distance/Time. For car let x be the speed, then speed of the train is 3/2x as per the problem. Now x = 75/t for car. 1.5x = 75/(t-(5/24)) as train stops for a duration of 5/24 hours. 5/24 obtained by converting 12.5 minute to hour. Now divide two equations resulting in: 1/1.5 = (24t-5)/24t which upon solving gives t = 0.625. Put t value in v = d/t where d = 75 km and t = 0.625. Thus you get v = 120 kmph.
Hope this was useful. | Abhishek Banik said: (Dec 19, 2015) | | A to B=75 km. Distance T = Time. S = Speed, D = Distance. Let S of car = x kmph. S of train = 50% of car speed = x+{(50/100)*x}. = x+{(1/2)*x}.
= x+x/2 = 3x/2. T = D/S. T of car - T of train = 12.5 mins as car speed is low that of train speed. = 75/x-75/(3x/2) = 12.5/60 (converting 12.5 mins to 12.5/60 hr).
= 75/x-75*2/3x = 12.5/60. = 75/x-50/x = 12.5/60.
= 25/x = 12.5/60. x = 25*60/12.5 = 25*60*10/125 = 120.
x = 120 kmph. | Ravi said: (Feb 11, 2016) | | Can you explain please? | Pallavi said: (Feb 13, 2016) | | Can you explain from second step? | SHOBA said: (Feb 18, 2016) | | Can you explain about 12.5*10/60*10? | Eleanor said: (Feb 27, 2016) | | @Abhishek Banik. Can you please explain from the T of car - T of train?
You have lost me there. | Rakesh said: (Mar 1, 2016) | | If you take speed of train 50/ more than car, let us assume car speed 100 the train speed is 150. Hello Shobha, minutes convert in seconds multiply with 60 SND. | Yogeshwar Yaduwanshi said: (Mar 18, 2016) | | Suppose car speed : x. Than 50% more speed train : 3x/2.
Than formula car is taken time : 75/x = T....(1). And train is taken time : 75*2/3x = T-12.5/60....(2).
Both equation solve: x = 120 km/hr. | Bhumi said: (Mar 29, 2016) | | Car speed = x. Train speed is 50% more. Train speed = x+ (50/100) x = 100x + 50x/100. = 150x/100.
= 3/2x Kmph. | Snehesh said: (Apr 3, 2016) | | Car speed = x. Train speed is 50% more.
Train speed = x + (50/100)x = x + 1/2x = 3/2x. | Snehesh said: (Apr 3, 2016) | | An alternative Method:
Speed of Train = S Train; Speed of Car = S Car; Time of Train = T Train; Time of Car = T Car;
S Train = 3/2 S Car.
Therefore, T Train = 2/3 T Car (Speed inversely Proportional to Time since Distance is constant). T Car - T Train = 12.5 mins. Substituting, T Car - 2/3.
T Car = 12.5.
T Car = 37.5 minutes. S Car = Distance/ T Car.
S Car = 75/(37.5/60). Converting minutes to hours.
S Car = 120 kmph. | Nisha said: (Apr 4, 2016) | | Can anybody tell me why we multiply the speed with 25? | Koushik Roy said: (Apr 5, 2016) | | Can anybody explain me how to came 3/2 ? | Rahul said: (Apr 27, 2016) | | Since the ratio of speed => train : car.
150 : 100 -----> A. Then the ratio of time will be 100 : 150 ----> B.
(speed = distance/time). Now given 12.5 min is the difference of time as the train stopped for the same then, from equation B. => 50 = 12.5 (ratio difference).
then 1=12.5/5.
So time taken by train = (12.5/50)x100.
=25 min (25/60 hr) ----> C. Now the speed of train = distance covered by train/time taken.
= (75km) x 25/60 hr (from C).
= 180km/hr.
Hence, the answer will be 180km/hr. | Bdboy Hasan said: (May 2, 2016) | | Let, Time needed for car = 3x minute.
Time needed for train = 2x minute.
3x - 2x = 12.5 minute.
So, x = 12.5 minute. Time needed for car = 3*12.5.
= 37.5 minute. So, speed of car = 75km/37.5 minute.
= 2 km per minute.
= 60 * 2kmph.
= 120 kmph. | Gayuverma said: (Jul 4, 2016) | | Can this question be solved using relative speed concept? If no, why? If yes, please show it. | Yeshi said: (Jul 20, 2016) | | Can anyone suggest me the logic? Please. | Siva said: (Aug 12, 2016) | | 12.5/10 * 60, = time/speed.
= 12.5/10 in minutes.
= 12.5/10 * 60 in seconds. | Rajeev said: (Aug 26, 2016) | | Actually, here x = (25 x 15)/5? | Ashok said: (Aug 28, 2016) | | 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. | Shaheer Sheik said: (Aug 30, 2016) | | 50% more than x means 50/100 + x = 150/100x. | Tanshi said: (Sep 6, 2016) | | The formula to convert from minutes to hours is:
hours = minutes ÷ 60. | Praveenkumar said: (Sep 14, 2016) | | Hi guys,
Can anyone explain why we are subtracting car speed 75/x from train speed i.e. 75/(3/2)x? | Kranthi said: (Sep 19, 2016) | | If there is an another method please explain me. | Amit Chauhan LPU said: (Sep 30, 2016) | | 150/100 X it came because. 50% more than X ie : => (50/100) * x + x.
==> 150/100 * x. | Gowtham Sunny said: (Oct 6, 2016) | | Here, Train and car reach at the same time. So, T1(Car) = T2(train).
Then, T1(Car) = T2(train) + 12.5 So, T1 - T2 = 12.5.
Given Distance = 75, Let speed be x.
Speed = x -----> Car speed.
Speed = x + 50/100 of x = 3x/2 -----> Train speed. Therefore, T1 - T2 = 12.5. 75/x - 75/(3x/2) = 12.5 min.
(75 - 150)/x = 12.5/60 hr (Since we are cal in kmph we should convert min to hrs).
x = 120 kmph. | Anonymous said: (Oct 25, 2016) | | Thank u@Prawinn. | Dip said: (Dec 12, 2016) | | Please anyone can explain me how 125/600 came? | Palani said: (Dec 17, 2016) | | But in question (50% more) is not given, only 50% faster is given so explain me the correct solution. | Ankit said: (Dec 23, 2016) | | Let me explain : assume car travel at speed x so speed of train 50% more than car 50/100 * x or 0.5 x.
So total speed of train faster than car is x + 0.5x which is 1.5x. Meaning of 150/100 * x = 1.5x. | Hong Kong said: (Dec 27, 2016) | | But can anyone explain to me why can't I think like this: Let x be the time for Bus to arrive point B, then x = 12.5x2 (since train 50% faster) = 25.
So we know the train bus spend 25 mins to arrive B, and 75km/25mins = 3km/mins = 180kmph? | Vishnu B Nair said: (Jan 17, 2017) | | The speed of both of them is 100 per cent and the train is 50% more faster than car it means the train have the 100 + 50 = 150%. | Rakshak said: (Jan 24, 2017) | | Can you explain, how 125/(10 * 60)? | Tesit said: (Feb 7, 2017) | | Can you explain, how 125/ (10 * 60)? | Rajesh.r.kzm said: (Feb 10, 2017) | | 12.5 can be taken as 125/10. So it will be multiplied by no of minutes in an hour. So, it is 125/(10*60). | Priya said: (Feb 25, 2017) | | Can you exlpain how 150x/100 came? | Sagar Bhamare said: (Mar 22, 2017) | | 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. | Ravi Kataria said: (Mar 29, 2017) | | As a train is much faster, then time difference must be [distance /train speed - distance /car speed =12.5min].
Then, how it does inversely? | Nayem said: (Mar 31, 2017) | | Let speed of car = x (km/h).
So, speed of train = x + 50% of x = x + x/2 = 3x/2 (km/h),
To travel 75 km, car needs time of 75/x hours,
To travel 75 km, train needs time of 75/(3x/2) hours + 12.5/60 hours,
As the two reach destination on same time on the clock, so these times should be equal. That is:
75/x = 75/(3x/2) + 12.5/60,
=> 75/x = (75*2)/3x + 12.5/60,
=> 75/x = 50/x + 12.5/60,
=> 75/x - 50/x = 12.5/60,
=> 25/x = 12.5/60,
=> 12.5x = 25 * 60,
=> x = 1500/12.5.
So, x = 120 (km/h). | Mithlesh said: (Apr 6, 2017) | | Nice explanation @Sushil. | Rias said: (Apr 8, 2017) | | 75km/s * 60 second = 4500sec,
50%-12.5 the train lost =37.5, Speed = Distance/time.
then 4500sec/37.5 = 120km. | Harika said: (Apr 18, 2017) | | I understood. Good explanation @Dhinesh. | JIGNESH said: (Apr 22, 2017) | | How can 125/ (10 x 60) came? Please Explain me. | Rahul said: (May 9, 2017) | | Let Car speed = x kmph. Train's speed is 50% more than car.
So, train's speed = x + 50x/100. = 150x / 100. = 3x/2. | Rahul said: (May 9, 2017) | | @Jignesh. Time is given in minutes. So to change it in hours, it is divided by 60.
And 10 has been placed for the decimal point. | Raihan Bangladesh said: (May 11, 2017) | | Let, car speed = X km/min. Train speed = X + x* (50/100) Km/min. =X+X/2 km/min. =3X/2 km/min. Time of car t =75/X. Time of Train t =75/ (3X/2). According to the question. 75/X=75/ (3X/2) +12. 5. >> 75/X-150/3X=12. 5. >> (225-150) /3X=12. 5. >> 75 = 37. 5X. >> X = 2 Km/min. So car speed 1 minute to 2 km.
Car Speed 60 minutes to (60*2) = 120 km/h. | Madhu said: (May 11, 2017) | | @Bdboy Hasan.
How do you say car's speed is 3x? | Nng said: (May 30, 2017) | | Hi, can you please explain why we are subtracting the time of car and train to get 12.5 minutes? | Sanju said: (Jun 9, 2017) | | Good explanation, thanks @Raihan. | Chirag Painter said: (Jun 9, 2017) | | Let speed ratio=car:train=100:150=2:3. Here distance is same so as per the rule "if distance is same then speed ratio is inversely proportional to time" Speed ratio = 2:3
Time ratio = 3:2 Therefore car usual time = 3 * 12.5 = 37.5 min.
And train usual time = 2 * 12.5 = 25 min. Now, speed of car = Distance/time,
= 75kmph / 37.5 min,
= 2.
Now convert into hour=2 * 60 = 120kmphr. | Suvasish Singha said: (Jun 21, 2017) | | Let, Speed of the train be X kmph. Than, Speed of the train= x + (x * 50/100). = x + x/2, = (2x+x) / 2, = 3x/2. | Ravikumar J said: (Jun 27, 2017) | | Good explanation, thank you @Rihan. | Shakeeb said: (Jul 3, 2017) | | ***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.
| Gurbir said: (Jul 29, 2017) | | Train and car covers the same distance in same time. But train 50%faster than the car. The Train stops at stations 12.5. It means they cover 75km in 25minutes. So convert minutes into hr (75÷60=12/5).
So that the speed of the train = 75*12÷5=180, 50% of 120 is 60,
180 - 60 = 120.
So Speed of car is 120km/hr.
Because train is faster than car 50% is given above. | Soham said: (Aug 7, 2017) | | Many have asked how 3/2x came? According to me "train travels 50% faster than car"
Let the speed of car be x.
Then the speed of train is, The speed of car+ 50%speed of car (as per question)
ie,
x+ 1/2x.
=3/2x. | Ankur Bravo said: (Aug 13, 2017) | | 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. | SSS said: (Aug 31, 2017) | | 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. | Vickivignesh said: (Sep 8, 2017) | | speed : car=x;train =y;
x=100% of speed;y=150% of speed; from this wkt, train covers a 150 km per hour but as we know from the detail.
It loses the 12.5 min due to stoppages.so it covers the 150km in 47.5min. Speed=distance/time.(150*60/47.5)=180kmph.so train speed is 180kmph.
Hence car speed is 120kmph. | Shivani said: (Oct 4, 2017) | | Could we take 75/ (3/2) x before 75/x ? | Nilakshi said: (Oct 11, 2017) | | Time taken by train minus time taken by the car is how come that? | Noshiya said: (Oct 24, 2017) | | Can you please explain 150/100x=3/2x? | MOHAMMAD MOHSIN said: (Oct 28, 2017) | | 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. | Frustrated Engineer said: (Nov 2, 2017) | | A train can travel 50% faster than a car. That means IF CAR SPEED IS X, THEN TRAIN SPEED SHOULD BE 2X. Double speed of car. So how can you guys say 50% more speed?
How this 150x/100 comes? | MikesMatE said: (Dec 1, 2017) | | @ALL. 150x/100 is to say that if the car travelled at x speed which is 100%x=x, then the train was 50% faster than that, 2x is an addition of 100% to x not 50%.
Therefore to add 50% to x would be to add 1/2x which makes it 3/2x. | Loganayaki said: (Dec 5, 2017) | | Car speed is X , and train speed is 50percent faster. Therefore train speed is,
=>car's speed + 50percent of car's speed.
=> x + (50*x)/100,
=>(100x+50x)/100,
=> 150x/100,
=>3x/2 is the speed of train. | Arvindra Shukla said: (Dec 21, 2017) | | How come 125 and why divided by 60? | Srivishnu said: (Jan 4, 2018) | | 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. | Dan said: (Feb 10, 2018) | | Train spent extra 12.5 minutes then a car. So we have to subtract bigger time from lesser time to get 12.5. It means that time spent by car must be subtracted from the time taken by car to get the excess time of 12.5 minutes. Thus, it should be like this:
75/(1.5*x)-75/x=0.21 (in hours). | DevB007 said: (Apr 13, 2018) | | Train speed = 50% faster than car speed. Train speed = car speed + extra 50 % of car speed. (car speed = X), (50% extra of X = X/2),
Train speed = X + X/2 = (X/1 + X/2) = (2X+X)/2;
Train speed = 3X/2. | Chinna said: (May 7, 2018) | | Here why we used 75/x-75/(3/2)x=125/10*60.
Please anyone can explain me?. | Saggi said: (Jun 5, 2018) | | I think there should be 50/100 not 15/100 right? | Harika Ravuri said: (Jun 19, 2018) | | Actually he gave 50% more not 50% times. so, the speed of train = speed of car + 50% speed of car. = x+(50/100)*x = 3/2x. Given that, time of car - time of train = 12.5 min, (75/x) - (75/(3/2*x)) =125/10 min, 75/x(1 - (2/3)) =125/(10*60) hrs, 1 = 125x/600, x = 120 kmph. | Viji said: (Jun 22, 2018) | | The speed of the train is 50% more than a car.
If the car travels at x kmph then 50%of x is x/2kmph.
Note train travels 50% more which means actual x and 50% of x is x/2 together (x+x/2) =3x/2. | Hemanth said: (Jul 1, 2018) | | 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. | Asawari said: (Jul 6, 2018) | | Anyone explain me from 2nd step. | Vivek said: (Jul 23, 2018) | | The speed of car : speed of train :100:150=2:3.
The time needed for the car: time needed for train =3:2.
i.e., the train only takes 23 of the time taken by car. Since both the car and train start and reach at the same time,
13 of the time needed by car is 12.5 minutes. Time needed by the car =3*12.5 min.
Therefore, speed of the car =75(3 * 12.560)=120 km/hr. | Radha said: (Aug 3, 2018) | | Can anybody tell me why we need to subtract 75/x with 75÷3/2x instead we can add it also right? | Jyothi said: (Aug 11, 2018) | | Here we take 150/100 because it is given 50% more and always we have to take 100 as a base value here it is 50%more so 100+50/100 = 150/100 = 3/2. (If it is given less than 50% then 100-50/100 = 50/100 = 1/2.
I hope you all understand. | Vijay Kumar said: (Aug 12, 2018) | | Both reach same time but suppose TI but train losses 12.5 so train time is TI-12.5 and car time is TI. Now WKT time=distance/speed, x be the speed of car and 3x/2 speed of the train, both distance is same. Substitue in formula for train time=TI-12.5 dis=75 speed 3x/2 we get (TI-12.5)=75/3x/2
for car time=TI speed=x dis=75 we get TI=75/x now substitue TI In above trian equation we get;
75/x-12.5 = 75 * 2/3x convert time into 12.5/60 and the result:120km/hr | Ajit said: (Aug 16, 2018) | | The train lost about 12.5 minutes while stopping at the stations. How the time loss is calculated? Please anyone tell me. | Dr John said: (Aug 18, 2018) | | @All. The time lost by the train (12.5min) was already given in the question. We can't calculate the lost time without knowing other quantities( speed of car, etc). | Astik Kumar said: (Aug 25, 2018) | | Let the speed of the car be 100 then speed of the train be 150.
The ratio of the speed of car and train be = 2/3.
When the distance is constant the time is inversly proprotional,
So ratio of time = 3:2,
Diffrence in time =12.5min=5/24hours,
Total time is taken by car=3*5/24=15/24,
Speed = d/t => dis=75km,
Speed of car=75/(15/24)=120km/h. | Trishita said: (Sep 4, 2018) | | Dist=speed*time.
Let, speed of car=100x.
Speed of train=150x (as 50% faster).
So..Time taken by car= dist/speed=75/100x.
And by train=75/150x.
And train lost 12.5 min=12.5/60 hr =125/600 hr.
This time is equal to the time taken b/w car and train i.e;
75/100x - 75/150x =125/600.
Solving this eq we get x=6/5.
So, the speed of car is=100*(6/5)= 120km/hr. | Ershaad Shah said: (Sep 12, 2018) | | @Trishita.
Can you please explain, how do you put 75/100x-75/150x = 125/600? | Ratul said: (Sep 18, 2018) | | Actually, it is time of car to cover the required distance and train take same time but including 12.5 minutes. So both take same time you can rearrange the formula to understand easily.
(75/x) = (12.5/60)+-(75/(3/2)x). | Ashish Patil said: (Sep 25, 2018) | | Dist = speed * time. Let, the speed of car=100x.
Speed of train=150x (as 50% faster).
So, Time taken by car= dist/speed=75/100x.
And by train=75/150x.
And train lost 12.5 min=12.5/60 hr =125/600 hr.
This time is equal to the time taken b/w car and train i.e;
75/100x - 75/150x =125/600.
Solving this eq we get x=6/5.
So, the speed of car is=100*(6/5)= 120km/hr. | Bhavi said: (Oct 8, 2018) | | Speed of train =( 1 + 50% )of speed of car.
75/t-12.5 = (1+1/2)75/t,
t=37.5 minutes,
Car speed = (2/3) * train speed,
Car speed = (2/3)*75/(37.5-12.5),
Car speed = 2 kmpm,
2 * 60 = 120 kmph. | Sameer Azmi said: (Oct 15, 2018) | | The train speed of the train is 50% more the speed of the car = x + 50%x/100.
Then, (100x+50x) /100. | Gete Basu Chakma said: (Oct 15, 2018) | | The speed of car = x.
speed of train = 1.5x.
The lost time=12.5/60 = 5/24 hr. The time taken by car = lost time +running time of train.
75/x = (5/24) +75/1.5x.
Then, x =120km/hr. | Zama said: (Oct 28, 2018) | | @All. Explanation of this part, 12.5/10 * 60.
12.5 is 125/10.
=> 125/10*60,
=> 5*25/10*60, #since 125 can be written as 5*25.
=> 25/10*12, #after cancelling denominator 60 by numerator 5 as 5*12=60.
=>25/120, #here cancellation goes as 5*5=25 & 5*24=120.
=>5/24.
Hence, we got 5/24. | RAMJAN RAEEN said: (Nov 23, 2018) | | Can anyone explain to me how 75/X - 75/ (3x/2) = 12.5?
How the time is the same in both case car and train if a halt is not considering? | Ansh Gupta said: (Dec 29, 2018) | | Can you explain how 150/100x came? | Ivdurp said: (Jan 3, 2019) | | We always consider percentage to 100 right. They have given that train speed 50 % more than a car. here they take car speed as x.
We already we knew the percentage is (100+50)*x/100. | Ameer said: (Mar 6, 2019) | | The speed of the car - speed of the train = loss time. | Abraham said: (Mar 7, 2019) | | Speed of car be x.
Speed of train= x+x/2 = 3x/2 ie; 1.5x. Train didnt run for 12.5min ie; 12.5/60 hrs. Difference between time ran by both is 12.5/60hr.
ie: (75/x) - (75/1.5x) = 12.5/60.
Upon solving x= 120kmph. | Neranjan said: (May 6, 2019) | | A to B have 75km. We think, Train not stopped, It can go 112.5km. So, train speed =112.5km-75km/12.5m. 12.5m covert to hour 12.5/60. = 37.5km * 60/12.5, = 180kmph. So car speed =180kmph * 100/150. =120kmph. | Sandip said: (Jun 10, 2019) | | Speed of train:speed of car = 3/2:1 = 3:2.
Then time should be in 2:3,
The difference is 12.5min. That means 1 part=12.5.
3pt=37.5min=37.5/60,
Speed=75/37.5/60= 120. | Varsha said: (Jun 14, 2019) | | How can we assume that car speed is 100 not other than that speed? | Bittu said: (Jun 26, 2019) | | Let car speed is x. Train speed is 50% faster than car it means, Car speed(x) +50%of car speed=x+50%x,
50% means 0.5, i.e, x+0.5x = 1.5x or 3/2x. | Helip said: (Jul 4, 2019) | | If we have to convert the minute in the second we will multiply with 60 then what does it mean in solution 125/10*60? Please tell me. | VIJAY said: (Jul 15, 2019) | | Here, we can use Car travel time = train travel time + stoppage time. | Shashidhar said: (Jul 28, 2019) | | Given. Time delay by 12.5 (minutes), 1hr = 60minutes,
1min = (1/60)hr, 125/10 = 12.5.
125/(10*60) = (5/24). | Aishwarya said: (Aug 1, 2019) | | Let the Speed of car be=x km/hr.
Speed of train = 1.5x km/hr. Time taken by car to travel 75km = 75/x h.
Time taken by train to travel 75 km = 75/1.5 x-12.5/60.
Both are equal. 75/x = 75/1.5x-12.5/60.
1/x(75"75/1.5) = 12.5/60. 25/x = 12.5/60
1500 = 12.5x.
So, x = 1500/12.5 = 120 km/hr is the Answer. | Sharada said: (Aug 20, 2019) | | How can you get that 150/100? please explain that. | Anomie said: (Aug 29, 2019) | | Let the car speed be x. 50% of car speed =50/100*x = x/2.
The train speed = x+x/2 = 3x/2. | Anandhu said: (Sep 6, 2019) | | Can anybody tell me that, In the question, it states that the car and the train start at the same time from A and reach at the same time a B. Which means they took the same time to reach point B from A, so how can we take the difference of time taken by them? | Jasmine said: (Nov 21, 2019) | | Distance between A and B = 75kms.
Let 'x' be the speed of the car. Given that the train can travel 50% faster than the car i.e., Car speed = x.
Train speed = x+x*50/100 = 3/2*x.
Distance = speed*time. Tc(Time of the car) = 75/x.
Tt(Time of the train) = 75/(3/2*x)+12.5 min(while stopping at the station). =>75/(3/2*x)+12.5/60(into hours). =>75/x=75/(3/2*x)+12.5/60.
By solving x = 120kmph. | Sandhya said: (Nov 22, 2019) | | Can you explain 125/10*60 how it got or is there any simple method for answering this question. | Anas said: (Nov 29, 2019) | | Both trains start at the same time and reach the destination at the same time. So the time taken by both trains should be the same right? So the difference between the times should be zero. How come it is 125/10*60? | Sreenadh Reddy said: (Dec 11, 2019) | | The train speed is 50% faster than the car. If the car is moving in 100 then the train moves 50% fast i.e. (100car speed+ 50faster speed) so they taken 150/100. | Hannu said: (Jan 8, 2020) | | If the speed of car is 100 then 50% of 100 is 50. And the speed of the train is 50% more than of car speed.
so 150*x/100. | KArthik said: (Jan 18, 2020) | | How did 75/x -75/(3/2x) = 125/(10*60) came? Please explain this. | Sandhu said: (Feb 13, 2020) | | x+50% of x's speed.
x+50x/100 = 150x/100. |
Sameer said: (Feb 17, 2020) | | 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. | Jarvis said: (Feb 26, 2020) | | If it's 50x.
It says that train travels in 50% less speed than the car. | Purvi said: (Mar 5, 2020) | | How can we assume that car speed is 100 not other than that speed? | Yash said: (Mar 12, 2020) | | 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. | Sulbha said: (Mar 24, 2020) | | How can we decide that train speed is more than 50% means? Car speed is 100? how come.
We can assume anything why 100? | Ramesh said: (Apr 14, 2020) | | Thanks @Bhavana. | Rohith said: (Apr 30, 2020) | | Can someone give a clear explanation for this problem? | Surya said: (May 4, 2020) | | 75/x - 75/(3/2) =125/10*60.
Anyone can explain this?
How this 125/10*60 came? | Rezviii said: (May 9, 2020) | | Why dividing 125 by 10* 60?
Please explain it. | Sravani said: (Jun 1, 2020) | | Why to subtract the speed of the train from speed of car (75/x-75x/[3/2])?
Can anyone help me by explaining? | Kalyan said: (Jun 15, 2020) | | Can you explain how 150/100 came? | Adla said: (Jun 17, 2020) | | Car speed is x. Speed of train = 150/100x= 3/2xkmph(explain 25*3/25*2).
75/x-75/3/2x = 12.5/60.
75/x-75/3/2x = 125/600.
75/x-75*2/3x = 5/24.
75/x-150/3x = 5/24.
75/x-50/x = 5/24.
25/x = 5/24.
x= 25*24/5.
x= 5*24.
x= 120. | Jen said: (Jun 18, 2020) | | Thanks all. | Anjali said: (Jun 19, 2020) | | How 125/10 * 60 come at the initial step? Please explain. | Rishika said: (Jun 27, 2020) | | Why are we subtracting 75/x - 75/3/2x? Please explain. | Naga Sai said: (Jul 15, 2020) | | @Rishika.
We are subtracting it because the time difference between them is 12.5 min. | Anandhi said: (Jul 17, 2020) | | @Anjali. 12.5 minutes lost in stopping so(time difference). 12.5/60 (dividing by 60 for converting minutes to hours). To remove decimal dividing by 10.
125/10 * 60. | Piyush said: (Aug 3, 2020) | | Can anyone please explain whole solution again? | Dimple said: (Aug 13, 2020) | | How come 3/2? | Aditya Mohanty said: (Aug 19, 2020) | | How come is the speed 125/10 * 60? | Nahid said: (Aug 21, 2020) | | Let, Speed of the car= x kph.
So, speed of the train = x + 50% of x = (3x/2) kph. | Priyam said: (Aug 26, 2020) | | @Aditya Mohanty.
For converting the minute into the hour we take 12.5/60 and when we remove the decimal point it became 125/60*10. | Jay said: (Aug 31, 2020) | | How 3/2 come? Please explain. | Shavra Yaqub Shah said: (Sep 1, 2020) | | T1 - T2 = 12.5 minutes
speed1/distance1 - speed2/distance2 = 12.5 minutes . @All.
This is how you form the equation and solve the problem related to the time difference (early/late than usual),or speed difference or distance difference. | Hani said: (Sep 3, 2020) | | @Shavra Yaqub Shah.
I'm not getting your answer method, please explain to me in detail. | Navas said: (Sep 9, 2020) | | I'm not getting your answer method, please explain to me in detail. | Aldrin said: (Sep 14, 2020) | | Also, they assumed that the car didn't stop anywhere and was continuously travelling. | XiYoZ said: (Sep 21, 2020) | | How (24 *25)/5 came? | Vicky045 said: (Sep 30, 2020) | | By Change Minute to hours 12.5*60.
Am I right, then how 12.5/60? Please explain me. | Anonymous said: (Oct 3, 2020) | | @VIcky045. Actually,
To take delay time of train we need to have express time in hr(given is in min).
To convert Minute to Hour (x minutes/60).
To convert Hour to Minute (x hours*60).
So,
Converting 12.5 mins to 12.5/60 hour.
125/10*60 also correct since it only removes decimals. | Sahil Tiwari said: (Dec 14, 2020) | | Speed = D/T.
i.e- 75/12.5 = 60km.
The speed is 50% more;
60 is 50% of 120 = 120 kmph. | Gabbar said: (Dec 29, 2020) | | Car speed is X KMPH then,
Train speed=(car speed + (50*carspeed)/100), ie train speed =(X+(50X/100)),
Take LCM and then(100X+50X)/100 . ie 150X/100, | Aditya said: (Jan 2, 2021) | | Those who are not understanding 75/ 3/2.
For them (75*2) /3 = 50. | Rakkesh said: (Jan 19, 2021) | | The car and train reached 75km at the same time.the train was late for 12.5 mins.
The train could reach 12.5 mins faster than the car.
* 75/x is the timing of car to reach 75kms.
*75/(3x/2) +12.5/60(time for stops) is the time taken by train to reach 75kms.
*Both the timings are equal since the time taken by both car and train are equal so then you can get the equation.
75/x = 75/(3x/2) + 12.5/60.
75/x - 75/(3x/2) = 12.5/60. | Aiswarya C V said: (Feb 3, 2021) | | Why was it taken 125/10 * 60?
Please explain. | Pintu said: (Feb 6, 2021) | | I think the answer is 110. | Arvind Gupta said: (Feb 7, 2021) | | Late Train speed = X km/hrs.
Than Car speed will be = X+X/2 =3X/2 km/hrs,
Time_train - Time_car = Time - train stop.
75/X - 75/(3/2)X = 12.5/60, 75/X - 50/X = 25/120,
25/X = 25/120.
X = 120 km/hrs Ans. | Shubham Boni said: (Feb 12, 2021) | | The Answer method, I correct with 1 simple mistake in calculations. 75/x - 75/(3x/2) = 12.5 minutes.
Now, the equation is correct but Now since time is in Minutes we need to convert it into seconds.
We Know, 1min = 60seconds this means, 12.5min = (12.5X 60)seconds. | PALAK said: (Jun 10, 2021) | | Let the car speed be x. Speed of train = 150/100x = 3/2xkmph(explain 25*3/25*2).
75/x-75/3/2x = 12.5/60.
75/x-75/3/2x = 125/600.
75/x-75*2/3x = 5/24.
75/x-150/3x = 5/24.
75/x-50/x = 5/24.
25/x = 5/24.
x = 25*24/5.
x = 5*24.
x = 120. | Yogi said: (Jul 2, 2021) | | @Yogi, Train increase speed 1/2, means time decrease 1/3,
This 1/3 =12.5 minute. = 12.5/60 = 5/24 hour,
Speed = distance/time.
Speed(train)= 75km/5/24 = 360km/h,
Car speed = train speed/2,
Car speed =120kmph. | Yogi said: (Jul 3, 2021) | | The Speed of train increase by 1/2, means time reduce by 1/3, this 1/3 time reduction is 12.5 minute. So, 1/3 = 12.5/37.5. Means the car will take three-time of 12.5 minutes which 37.5 minutes.
75 x 60/37.5 = 120kmph. | Rajuuu said: (Jul 5, 2021) | | 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. | Insane said: (Jul 11, 2021) | | If you guys wonder how did 12.5 converts in 125/60*10 then; As 12.5 min should be converted into hours. So, 12.5/60 (to remove decimal from 12.5 bcos we cant divide),
We add * 10,
125/ 60*10 or 125/600. | Arvind said: (Jul 18, 2021) | | T = D/S
T = 75/X-75/1.5X = 12.5/60. | Bishal Subba said: (Aug 26, 2021) | | @All. It is just 12.5/60 in the RHS eliminate the 10.
You will get the answer as 120kmph. | Shubham said: (Sep 2, 2021) | | It's given that both Car and the train covers 75kms at the same time (t1=t2)
t=d/s Let's assume car speed as 'x' km/hr.
Train Speed = x + x/2. =3/2x. t1=75/x.
t2=75/(3/2)*x. t1 = t2+12.5mins extra time.
Then solve by converting 12.5 mins to hr by dividing by 60. | Navin said: (Sep 24, 2021) | | @Sahil Tiwari.
Thanks for explaining the answer in the simplest way. | Shruti said: (Oct 28, 2021) | | Let speed of car be x km/hr,
100% speed of car is x. 150% speed of train is x/100*150,
= 150x/100. | Golu said: (Nov 8, 2021) | | Thanks everyone for explaining the answer in a better way. | Vivekanand Vyas said: (Dec 4, 2021) | | 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. | Vidhyasaagar said: (Dec 21, 2021) | | @All. The soultion is; Speed of car = x =>1x=>100/100x.
(All are literally same, in terms of percentage it's 100/100x).
Speed of train = 50%+ x=> (50/100)+(100/100x)=150/100x=3/2x. | Shivam said: (Jan 8, 2022) | | Thanks for explaining the answer. | Piyush said: (Jan 23, 2022) | | @All. So many of us are confused in where the hell this (150x/100) came from right. So, my simple explanation is that see the train is 50% faster than car. Now, consider car speed is X so the train speed will be (X+ 50% of X) Now calculate the train speed. And you will get (X + 50X/100) = 150X/100.
Hope this helps. | Ayush said: (Mar 3, 2022) | | Please anyone can explain since the speed of the train is 1.5 * speed of the car, so how in the solution we are taking that difference of time is 12.5 min. How we come to know that if 12.5min added to the time taken by car will equal to the time taken by train?
Since the speed of both is different so how can they have time taken by car = T and by train =T+12.5 * 60. | Tharun Kumar said: (Mar 19, 2022) | | Let the speed of the car be x. As per the data the train is 50% faster than a car (not 50 times) which means [ (50/100) x +x].
So, (150/100) x. | Naz said: (Apr 20, 2022) | | 75/x * 1.5 = 75/x-12.5 (speed of car * 1.5 = speed pf train).
X= 37.5 time taken (min).
Car speed 75/(37.5/60) as time was in mins. | Rakesh Selvam A said: (Jun 28, 2022) | | By ratio of Speed. T :C = 150: 100, = 3:2.
Since, Distance is constant.
Time ratio = 2:3.
Diff between time ratios 2 and 3 is 1, which means 1 = 12.5 mins. 1 -----> 12.5 min,
3------> 37.5min,
2 ------> 25 min,
,
Speed of Car = Distance/Time. = (75 /37.5) * 60( For min to hr), = 120km/hr. | Bhushan said: (Sep 12, 2022) | | Let the speed of car be x.
now, the train is 50% faster than a car,
i.e. speed of the train = x + (50/100)x. = 150x/100. | Charles said: (Sep 26, 2022) | | I did not understand. Please help me to get it. | Rohan said: (Oct 12, 2022) | | Someone help me in this by explaining this. | Rushi Gadi said: (Oct 13, 2022) | | n any of the time distance sum out of time, distance, speed one of the quantities common Here the time is common; So we; d = s*t.
Car Time = train Time. Dist of car/Speed of car = Dist of train/Speed of train.
But here is train takes extra time for pause which must be equal.
Time train = Time of train running + Time train pass.
Time train running + Time train pass = Dist of car/Speed of the car.
Time train running - Dist car/Speed car = Time train pass.
75/1.5(x) -75/x = 12.5/60. FYI (12.5/60 = 125/10*60),
x = 120Km/hr. | |