The distance between two stations is 100 km. 2 cars start from A and B at the same time and return t

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03 Nov 2011 station

The distance between two stations is 100 km. 2 cars start from A and B at the same time and return to the original station with speed 48 km/hr and 64 km/hr. The time gap between their meeting twice will be

OPtion

1) 10/7 hr
2) 20/17 hr
3) 25/14 hr
4) 30/7 hr
5) 27/14 hr
6) 3 hr
7) 20/7 hr
8) 4 hr
9) 5 hr
10) none of these

Solution

After first meeting total distance covered by the two cars upto second meeting = 200 km.
and relative velocity = 64+48 = 112
so time taken = 200/112 = 50/28 = 25/14

Correct Option: 3 Best Solution (16)

Dr.Dipin Singh   13 years AGO

Relative speed = 64+48=112 kmph

Two cars will meet first when they together travel 100 kms after time 100/(48+64) = 100/112 hrs

Two cars will meet second time when they together travel 300 kms after time 300/(48+64) = 300/112 hrs

Difference in two timings = 300/112 - 100/112= 200/112 = 25/14 hrs ... option 4

Like? Yes (5) | No (1)  

Atul   13 years AGO

Two cars will meet two times after travelling together a distance of 100 kms and 300kms.

Difference of time between 2 meetings will be time taken to travel together a distance of 200 kms = 200/112= 25/14 hrs

Like? Yes (4) | No   

Khanna   13 years AGO

Total distance covered between their meetings is 200 KM travelled together by 2 cars.
total combined speed = 64+48 = 112
Time take to cover 200KM with 112 KM/H speed is
200/112
=25/ 14 Hour

Like? Yes (2) | No   

Saranjit Singh   13 years AGO

The time gap between their meeting twice will be
time taken by two cars to travel distance of 200 kms jointly.
i.e 200/(64+48) = 200/112= 25/14 hrs

Like? Yes (2) | No   

vivek shekhar singh   13 years AGO

relative speed of one train with respect to the other is 64-(-48)=112..........nd the total distance covered by ny one of them is 200....therefore the time gap...is 200/112=25/14........

Like? Yes (1) | No   

Darshan   13 years AGO

Distance=200km

645m/hr+48km/hr=112.

200/112=25/14.

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Shyoram   13 years AGO

thier first meeting after 100/112 hr.
secnd meeting after 225/82 hr

time gap 25/14

Like? Yes (1) | No   

Priyanjali yadav   13 years AGO

300/(64+48)-100/(64+48)=25/14

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Honey Goel   13 years AGO

relative speed of car from A = 64+48=112kmph
first meet at station B i.e. t1=100/112hr
second meet at station B i.e. t2=300/112hr
time difference=200/112=25/14

Like? Yes (1) | No   

saurabh middha   13 years AGO

first meeting time is suppose t1.
so (48+64)*t1=100
it will give u t1=25/28.
similar t2=75/28.
so the time difference is 25/14.

Like? Yes (1) | No   

Karan Kosta   13 years AGO

they meet first time at 400/7 km from B .Now second time they meet at 500/7 from B.
x/64=(100-x)/48
x=400/7
now second time
(100+x)/64=(200-x)/48
x=500/7
time=75/28-25/28
time=25/14

Like? Yes (1) | No   

Nishit   13 years AGO

Distance between their meeting twice = 200
Sum of speeds = 48 + 64 = 112
So time = 200/112 = 25/14

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praveen dhanuka   13 years AGO

first time they will meet at=25/28 hr
Now,
distance between A and B when A reaches another end is = 200/3 km

which , they will cover relatively in = 25/42 hr
difference=100/48+25/42-25/28=25/14 hr

Like? Yes (1) | No   

Mohammed Akram   13 years AGO

Soln:
x1(48) + x2(64) = 100
Solving this we get 1.7857 which is option 3: 25/14 hour

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madmadhav   13 years AGO

200/(48+64)=25/14

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Karan   13 years AGO

A............>.100 km..B
A..100 km.

Like? Yes (1) | No (1)