11. A 10-km long straight road connects two towns A and
B. Two cyclists simultaneously start one from town A
and the other from town B. On reaching the opposite
town, a cyclist immediately returns to his starting town
whereas the other cyclist takes somerest and then returns
to his starting town. Both of them can ride at speed
20 km hl in absence of wind but during their whole
journey uniform wind from town A to B increases the
speed of the cyclist going into the wind by the same
amount as it decreases the speed of the cyclist going
against the wind. Both the cyclists meet twice, first at
2 km and then 6 km away from one of the towns. For
what period does a cyclist rest?
answer aata hoga to hi dena bakwaas answer deke apni auqaat mt dikha dena
Answers
➡➡➡The wind favours the cyclist going from A to B, so on the outward journey they meet 2 km from B - hence 8 km from A.➡➡➡
The wind favours the cyclist going from A to B, so on the outward journey they meet 2 km from B - hence 8 km from A.
If the wind speed is v kph then the speed of the cyclist going from A to B is 20+v and the speed of the cyclist going from B to A is 20-v kph.
When they first meet they have travelled distances of 8 and 2 km respectively in the same time t.
Therefore :
t = 8/(20+v) = 2/(20-v)
80-4v = 20+v
60 = 5v
v = 12 kph.
So the cyclist going from A to B travels at 20+12=32 kph and takes a time of 10km/32kph = 5/16 h = 5*60/16 min = 75/4 min = 18.75 min.
The cyclist going from B to A travels at 20-12=8 kph and takes a time of 10km/8kph = 5/4 h = 5*60/4 min = 75 min.
On the return journey they meet 6km from B hence 4km from A.
The cyclist from B travels against the wind at 8 kph and gets to the meeting point P after 6km/8kph = 3/4 h = 45 min.
The cyclist from A travels with the wind at 32 kph and gets to the meeting point after 4km/32kph = 1/8 h = 7.5 min.
The total cycling times of the cyclists from the start up to the 2nd meeting point P are 18.75+45 = 63.75 min for the journey ABP and 75+7.5 = 82.5 min for the journey BAP.
The cyclists started at the same time and arrived at the meeting point at the same time, so the difference of 82.5-63.75 = 18.75 min is the time spent resting at B to make the journey ABP as long as that of BAP.
ANSWER : 18.75 min rest at B.
Answer:
The wind favours the cyclist going from A to B, so on the outward journey they meet 2 km from B - hence 8 km from A.➡➡➡
The wind favours the cyclist going from A to B, so on the outward journey they meet 2 km from B - hence 8 km from A.
If the wind speed is v kph then the speed of the cyclist going from A to B is 20+v and the speed of the cyclist going from B to A is 20-v kph.
When they first meet they have travelled distances of 8 and 2 km respectively in the same time t.
Therefore :
t = 8/(20+v) = 2/(20-v)
80-4v = 20+v
60 = 5v
v = 12 kph.
So the cyclist going from A to B travels at 20+12=32 kph and takes a time of 10km/32kph = 5/16 h = 5*60/16 min = 75/4 min = 18.75 min.
The cyclist going from B to A travels at 20-12=8 kph and takes a time of 10km/8kph = 5/4 h = 5*60/4 min = 75 min.
On the return journey they meet 6km from B hence 4km from A.
The cyclist from B travels against the wind at 8 kph and gets to the meeting point P after 6km/8kph = 3/4 h = 45 min.
The cyclist from A travels with the wind at 32 kph and gets to the meeting point after 4km/32kph = 1/8 h = 7.5 min.
The total cycling times of the cyclists from the start up to the 2nd meeting point P are 18.75+45 = 63.75 min for the journey ABP and 75+7.5 = 82.5 min for the journey BAP.
The cyclists started at the same time and arrived at the meeting point at the same time, so the difference of 82.5-63.75 = 18.75 min is the time spent resting at B to make the journey ABP as long as that of BAP.
ANSWER : 18.75 min rest at B.