If a car travels from city A to B with 1 kmph speed and another travels from city B to A with 5 kmph speed and they start their journey simultaneously. Then, after how much time will they meet for the first and second time? The distance between the cities is 100 km.

Open in App

Solution

They will first meet at

$5 T + T = 100$
$T = \frac{100}{6}$

As shown in the Picture They will again meet at
$T = \frac{300}{6}$

Suggest Corrections

Similar questions

Q. A car travels 40 km in an hour. How much time will it take to reach city B from city A if the distance between the 2 cities is 800 km ?

Q. Two persons P and Q start at the same time from city A for city B, 60 km away. P travels 4 kmph slower than Q. Q reaches city B and at once turns back meeting P, 12 km from city B. What is speed of P?

Jahangir travelled from his town to city. Jahangir went to city by bicycle at the speed of 20 kmph and came back at the speed of 5 kmph. If Jahangir took 5 hours and 45 min to complete his journey, what is the distance between town and city?

Q. The distance between two cities

A

and

B

330

km. A train starts from

A

8

a . m .

and travels towards

B

60

km/hr. Another train starts from

B

9

a . m .

and travels towards

A

75

km/hr. At what time do they meet?

Q. There are two cities P and Q. A car A starts from city P at 6:00 am and reaches city Q at noon. Another car B starts from city Q at 7:00 am and reaches city P at 3:00 pm.
Another car C starts from city P, an hour earlier than car A. If car C meets car A and car B simultaneously, then how much time would the car C require traveling from city P to city Q?

Join BYJU'S Learning Program

If a car travels from city A to B with 1 kmph speed and another travels from city B to A with 5 kmph speed and they start their journey simultaneously. Then, after how much time will they meet for the first and second time? The distance between the cities is 100 km.

They will first meet at5T+T=100T=1006As shown in the Picture They will again meet at T=3006

They will first meet at

$5 T + T = 100$
$T = \frac{100}{6}$

As shown in the Picture They will again meet at
$T = \frac{300}{6}$