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Mathematical operation problem
(1) When B finally catches up with A, B must make one more turn than A. Then the distance for B to pursue A should be100+120× =120m. Regardless of the same turning point between Party B and Party A, Party B needs to pass120 ÷ (150-120) = 4 minutes, so the distance that Party B needs to run is 150× 4 = 600 meters. B needs to go through five inflection points to run 600 (note that it is five, not six, because there is no need to turn an inflection point when catching up at last). So the time for B is 4 minutes running time plus 5 inflection points, so the standard answer is 4 minutes and 50 seconds.

Method 2: Process: B needs to catch up with A after starting on the straight road100/(150-120) =10/3 (minutes).

A is gone150 * (10/3) = 500m; B gone120 * (10/3) = 400 m.

According to the topic: A will be delayed for 30 seconds. B it takes 40 seconds.

Seconds usage time: (10/3)*60+40=240 (seconds), that is, 4 minutes.

(2) 150-90 = 60m 100 divided by 60 = 5/3min =100s 150x5/3 = 250m.

250 divided by 100=2. . . . 50

10x2=20 seconds 100+20= 120 seconds a: If the speed of A is changed to 90 meters per minute, B will catch up with A at the midpoint of BC.