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:
Let car speed be xkmph
then train speed is 50% more than car speed
ie 50/100*x
Komal said:
(Fri, Jul 9, 2010 11:34:38 AM)
I think there should be 50x/100 & not 150x/100 is it?
Xyz said:
(Fri, Jul 16, 2010 02:20:58 PM)
When we take 50x/100
we get x=[-75* 24]/5 = - 360kmph
its not in the option
Mohit said:
(Sun, Jul 18, 2010 09:37:25 AM)
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:
(Mon, Jul 19, 2010 01:13:31 PM)
Hae even though its % it must me 50x/100 only na?
Paresh said:
(Tue, Aug 3, 2010 11:54:01 AM)
Can any one please tell me how that 125/(10x60) came?
Vidu said:
(Wed, Aug 4, 2010 06:58:32 AM)
Please some one let me know how 150/100 came.
Rakesk said:
(Fri, Aug 13, 2010 12:10:16 AM)
50 % more means not 50/100. car has already 100 %.50 % more means 150/100
Raj said:
(Mon, Aug 23, 2010 10:45:00 PM)
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:
(Sun, Aug 29, 2010 10:21:09 AM)
Car seed is 100.
So train speed in 50% more.
So it is 150/100.
Hari said:
(Sat, Sep 11, 2010 12:10:35 AM)
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:
(Tue, Sep 21, 2010 01:08:45 PM)
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:
(Mon, Oct 4, 2010 10:30:19 AM)
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:
(Sat, Oct 9, 2010 06:38:12 AM)
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:
(Fri, Oct 29, 2010 11:46:10 PM)
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:
(Thu, Nov 11, 2010 12:17:52 AM)
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:
(Fri, Nov 19, 2010 01:35:51 AM)
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:
(Fri, Dec 10, 2010 01:11:08 PM)
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:
(Thu, Dec 30, 2010 01:31:26 AM)
75/x - 75/3x/2 = 125/10*60
can you tell me how this equations is derived?
Deepak Porwal said:
(Thu, Jan 6, 2011 02:55:07 AM)
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:
(Fri, Jan 7, 2011 09:55:16 AM)
Can you exlpain how 150x/100 came ?
Rohita said:
(Thu, Jan 27, 2011 06:10:59 AM)
Explain the simple way, I can't understand train speed.
Senhil said:
(Tue, Feb 1, 2011 11:01:29 AM)
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:
(Tue, Feb 22, 2011 06:05:46 AM)
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:
(Tue, Feb 22, 2011 02:37:41 PM)
Could you explain how time for train you had taken 75/(3/2.x) ?
Suja said:
(Wed, Feb 23, 2011 03:30:41 AM)
@ jamal:
We know that time=(distance/speed).
Here distance=75kmph and speed=3x/2.
Therefore time=75/(3x/2)
Shunmuga Priya said:
(Thu, Mar 3, 2011 06:35:08 AM)
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:
(Sat, Mar 5, 2011 05:20:37 AM)
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:
(Thu, Apr 14, 2011 06:52:00 PM)
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:
(Fri, Apr 29, 2011 03:57:23 AM)
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:
(Mon, May 9, 2011 02:44:44 AM)
Raj would you please tell me how come the speed be 25?
Hemant said:
(Mon, May 9, 2011 09:15:16 AM)
Explain the simple way, (75/x)-(75/(3/2)x)=12.5/60
Atul said:
(Wed, Jun 8, 2011 08:50:45 AM)
(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:
(Mon, Jun 13, 2011 08:17:49 AM)
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....
Hemant Sharma said:
(Sat, Jun 25, 2011 09:31:17 AM)
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:
(Tue, Jul 5, 2011 11:20:29 PM)
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:
(Wed, Jul 6, 2011 02:42:36 PM)
Thanks senhil I understand easily.
Sreekanth said:
(Fri, Jul 8, 2011 01:45:46 AM)
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:
(Wed, Jul 13, 2011 10:53:20 AM)
Rameez your explanation is great.
Mysterious said:
(Mon, Jul 25, 2011 05:05:21 PM)
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:
(Wed, Jul 27, 2011 09:41:31 AM)
Tell me about 12.5/100*60 ?
Hab said:
(Wed, Jul 27, 2011 09:20:20 PM)
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:
(Wed, Aug 10, 2011 11:49:33 AM)
@mysterious
well explanation mysterious..thank you!!!
Priyanga said:
(Wed, Aug 10, 2011 09:42:56 PM)
Please explain me in a easy manner I could n't understand it.
Yesoda said:
(Thu, Aug 11, 2011 10:55:34 AM)
How do you get 150/100?
R.Sreekanth said:
(Sun, Aug 14, 2011 08:53:15 PM)
@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.
100*(t)=150*(t+12.5)
t=37.5min
t
=37.5/60hr
speed=(75*60)/37.5
=120
Bhupat said:
(Tue, Sep 6, 2011 03:19:47 PM)
All righ but my question is. No body has taken 12.5 minus loss in train speed calculation. Why?
Khanchana said:
(Sat, Sep 10, 2011 01:51:55 PM)
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:
(Mon, Sep 12, 2011 11:07:57 AM)
Thank you so much @kanchana.
Sravya said:
(Mon, Sep 12, 2011 11:11:08 AM)
@kanchana
How did x+50/100 became 100x+50x as it will become (100x+50)/100x ?
Amit said:
(Mon, Sep 19, 2011 01:39:28 PM)
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 &Amp; Hitesh said:
(Thu, Oct 6, 2011 01:53:06 AM)
First we assume that the speed of car is 100% so as per Q.speed of train is 150%.
Ajit said:
(Sat, Nov 12, 2011 07:36:01 AM)
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:
(Thu, Nov 17, 2011 06:48:59 PM)
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:
(Fri, Dec 16, 2011 08:04:08 AM)
Why 10 is multiplied with 12.5? Can anybody explain?
Solanki said:
(Fri, Dec 16, 2011 11:20:06 AM)
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:
(Tue, Dec 27, 2011 05:02:22 PM)
How we get 125/10*60?
Can any one explain easy way?
Monideepa Ganguly said:
(Fri, Jan 6, 2012 09:25:01 AM)
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:
(Sun, Jan 8, 2012 05:28:42 PM)
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:
(Sun, Jan 8, 2012 10:00:04 PM)
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:
(Fri, Feb 17, 2012 11:18:19 PM)
Thanks Santhiraju :).
Shro said:
(Fri, Feb 17, 2012 11:28:47 PM)
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:
(Tue, Mar 6, 2012 04:55:47 PM)
Let the speed of the car be 100% and the train is 50% more
100%+50%=150%
that is 150/100
Dhinesh said:
(Thu, Jul 5, 2012 03:28:43 PM)
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:
(Sun, Aug 12, 2012 04:10:46 PM)
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:
(Mon, Aug 20, 2012 01:13:06 AM)
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:
(Thu, Aug 30, 2012 12:00:05 PM)
HOW 150 comes in the initial step ?
Saranya said:
(Sun, Sep 2, 2012 05:12:10 PM)
@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:
(Thu, Oct 4, 2012 07:29:39 PM)
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:
(Fri, Oct 5, 2012 07:00:06 PM)
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:
(Mon, Oct 22, 2012 12:01:47 AM)
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:
(Fri, Dec 14, 2012 09:47:25 AM)
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:
(Sat, Jan 5, 2013 01:00:40 AM)
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:
(Wed, Jan 23, 2013 09:28:29 AM)
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:
(Sun, Feb 24, 2013 12:30:22 AM)
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:
(Sun, Mar 31, 2013 08:39:53 PM)
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:
(Sat, Apr 6, 2013 09:22:41 AM)
Actually it is 50% more.
i.e (100% + 50 %) * x.
((100/100) + (50/100)) * x.
(1 + (50/100)) * x.
(150/100) * x.
Sankalp said:
(Wed, May 1, 2013 02:57:00 PM)
Can anyone please explain me this solution easily?
Sowmi said:
(Tue, May 7, 2013 02:04:12 PM)
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.
kmph.
