What are the options?

Here's how I solved:

Let t be the time taken by Ken to reach the summit.

500t = 350t + 50

Solving for t we get t as 5 hours. Therefore, Ken reached at 1 pm.

Here's my response from the other thread:

From the times given, we can figure out that Sari is climbing at 350 meters per hour; from 6:00 to 8:00, Sari climbs 700 meters. 700 meters/2 hours = 350 meters per hour.

If Ken is climbing at 500 meters per hour, we know that Ken is making up 150 meters per hour on Sari.

At 8:00, Sari is 700 meters ahead.

At 9:00, Sari is 550 meters ahead.

At 10:00, Sari is 400 meters ahead.

At 11:00, Sari is 250 meters ahead.

At 12:00, Sari is 100 meters ahead.

At 1:00, Sari is 50 meters behind. This must be the time at which Ken reached the summit.

bobdylan wrote:Sari and Ken climb up a mountain. At night, they camp together. On the day they are supposed to reach the summit, Sari wakes up at 06:00 and starts climbing at a constant pace. Ken starts climbing only at 08:00, when Sari is already 700 meters ahead of him. Nevertheless, Ken climbs at a constant pace of 500 meters per hour, and reaches the summit before Sari. If Sari is 50 meters behind Ken when he reaches the summit, at what time did Ken reach the summit?

From 6 to 8pm, Sari climbs 700 meters.
In order for Ken to finish 50 meters ahead of Sari, he must catch up by 700 meters and then climb 50 more meters than Sari, implying a total distance of 700+50 = 750 meters.

Sari's rate from 6 to 8pm = d/t = 700/2 = 350 meters per hour.
Ken's rate = 500 meters per hour.

When elements compete, SUBTRACT THEIR RATES.
Ken's rate - Sari's rate = 500-350 = 150 meters per hour.
Thus, every hour, Ken climbs 150 more meters than Sari.

Time for Ken to get 50 meters ahead of Sari = d/r = 750/150 = 5 hours.
Thus, the hour at which Ken reaches the summit = 8am + 5 hours = 1pm.