Aptitude - Time and Distance - Discussion
Discussion Forum : 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:
Answer: Option
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 |
 |
75 | - | 50 | = | 5 |
| x | x | 24 |
| x = |
|
25 x24 | |
= 120 kmph. |
| 5 |
Discussion:
342 comments Page 33 of 35.
Pradeep said:
1 decade ago
Can you exlpain how 150x/100 came ?
Deepak porwal said:
1 decade ago
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;
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;
Faizan Akhtar said:
1 decade ago
75/x - 75/3x/2 = 125/10*60
can you tell me how this equations is derived?
can you tell me how this equations is derived?
Sushil said:
1 decade ago
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
ie,the car speed is 100x kmph
Then train speed is 150kmph
equation is 75/1.5x+12.5/60=75/x
x=120kmph
Sonam jain said:
1 decade ago
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
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
Saravan said:
1 decade ago
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.
plz explain me why we subtract the train speed from car speed i didnt understood this step. can anyone explain me.
Singh said:
1 decade ago
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.
Nitesh Nandwana said:
1 decade ago
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
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
Santhiraju said:
1 decade ago
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
Train speed=(car speed + (50*carspeed)/100)
ie train speed =(X+(50X/100))
Take LCM and then(100X+50X)/100 .....ie 150X/100
Muthuraj said:
1 decade ago
125/10*60 here we are converting 12.5 min into hrs..12.5*10/10=125/10(min)--125/(10*60) hrs
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