Xiaohong and Xiao Qiang set out from home at the same time and walked in the opposite direction. Xiaohong walks 52 meters per minute and Xiao Qiang walks 70 meters per minute. They meet on the way. If Xiaohong leaves four minutes early and the speed remains the same, and Xiao Qiang walks 90 meters every minute, then the two will still meet at point A. How many meters away is Xiaohong from Xiao Qiang's home?
Answer: Because Xiaohong's speed and meeting place remain the same, the time from departure to meeting is the same for Xiaohong twice. In other words, Xiao Qiang walked four points less than the first time. From (70× 4) ÷ (90-70) = 14 (point), we can know that Xiao Qiang left at 14 for the second time, and it is inferred that he left at 18 for the first time, and the distance between their homes is (52+70) ×/kl.