Taro and Jiro left A town at the same time and went to B town.
Taro went by bicycle and Jiro went on foot.
The ratio of the speed of Taro and Jiro is 5 : 1.
As Taro noticed that he left something at point P on the way, he returned toward A town and passed each other with Jiro 4 minutes afterward.
Taro returned to A town and went to B town again.
Taro caught up with Jiro at Q point and he arrived at B town 24 minutes earlier than Jiro.
Two persons shall move with fixed speed, respectively.
Answer the following questions.
(1) Find the time after leaving when Taro returned at P point.
(2) If the distance from A town to Q point sets to 800 m, find the distance from A town to B town.
Answer
(1) 6 minutes (after leaving)
(2) 2.4 km
Solution
(1) Fig. 1 is a figure which shows that Taro moved at the speed of 5 and returned at P then 4 minutes after met Jiro at Q point who moves with the speed of 1.
By the time of meeting, Jiro followed the blue line at the speed of 1 and Taro followed the red line at the speed of 1.
The sum total of the distance which they moved by the time of meeting can be expressed as 5 + 1 = 6 by using the ratio of speed.
According to this line segment figure, the distance of 6 is twice the distance of AP.
The distance of AP is 6/2=3 and the distance of RP is set to 3 - 1 = 2.
That is, since Taro moved 2, the distance of RP in 4 minutes, the time to move 3, the distance of AP with the same speed is 4 minutes / 2 × 3 = 6 minutes.
Therefore, it is 6 minutes Taro returned after leaving.
(2) Fig. 3 is a figure which Jiro arrived at B 24 minutes later than Taro.
Taro and Jiro are in the point of ①, ② and ③ at the same time, respectively.
According to that Taro caught up with Jiro at Q point, Taro moved to B from Q and Jiro moved to ③ from Q in the same time.
Since the ratio of speed is 5 : 1, the ratio of a distance advanced in the same time is also 5 : 1.
When QB is set to 5, the distance from Q to ③ is 1.
Thus, it turned out that Jiro moved the distance from ③ to B which is 5 - 1 = 4 in 24 minutes.
Since he moved the distance of 4 in 24 minute, he moved between QB, the distance of 5 in 24 minutes / 4 × 5 = 30 minutes.
As a next step, according to Fig. 3, when Taro returns to A, Jiro moved to the point of ①.
The time Jiro reached ③ is the same time as the time Taro went and come back between AP which is 6 minute × 2 = 12 minutes.
The ratio of the speed is 5 : 1 and Taro caught up with Jiro at Q.
If the distance of AQ is set to 5, the distance of Jiro's from ① to ② will be 1, and also the distance from A to ① will be 5 - 1 = 4.
That is, since Jiro moved the distance of 4 in 12 minutes, it turns out that he moved between AQ, 5 in 12 minute / 4 × 5 = 15 minutes.
Since Jiro moved between AQ in 15 minutes and between QB in 30 minutes, the distance between AB is 800m × 45 minutes /15 minutes = 2400m = 2.4km.
Taro went by bicycle and Jiro went on foot.
The ratio of the speed of Taro and Jiro is 5 : 1.
As Taro noticed that he left something at point P on the way, he returned toward A town and passed each other with Jiro 4 minutes afterward.
Taro returned to A town and went to B town again.
Taro caught up with Jiro at Q point and he arrived at B town 24 minutes earlier than Jiro.
Two persons shall move with fixed speed, respectively.
Answer the following questions.
(1) Find the time after leaving when Taro returned at P point.
(2) If the distance from A town to Q point sets to 800 m, find the distance from A town to B town.
Answer
(1) 6 minutes (after leaving)
(2) 2.4 km
Solution
(1) Fig. 1 is a figure which shows that Taro moved at the speed of 5 and returned at P then 4 minutes after met Jiro at Q point who moves with the speed of 1.
By the time of meeting, Jiro followed the blue line at the speed of 1 and Taro followed the red line at the speed of 1.
The sum total of the distance which they moved by the time of meeting can be expressed as 5 + 1 = 6 by using the ratio of speed.
According to this line segment figure, the distance of 6 is twice the distance of AP.
The distance of AP is 6/2=3 and the distance of RP is set to 3 - 1 = 2.
That is, since Taro moved 2, the distance of RP in 4 minutes, the time to move 3, the distance of AP with the same speed is 4 minutes / 2 × 3 = 6 minutes.
Therefore, it is 6 minutes Taro returned after leaving.
(2) Fig. 3 is a figure which Jiro arrived at B 24 minutes later than Taro.
Taro and Jiro are in the point of ①, ② and ③ at the same time, respectively.
According to that Taro caught up with Jiro at Q point, Taro moved to B from Q and Jiro moved to ③ from Q in the same time.
Since the ratio of speed is 5 : 1, the ratio of a distance advanced in the same time is also 5 : 1.
When QB is set to 5, the distance from Q to ③ is 1.
Thus, it turned out that Jiro moved the distance from ③ to B which is 5 - 1 = 4 in 24 minutes.
Since he moved the distance of 4 in 24 minute, he moved between QB, the distance of 5 in 24 minutes / 4 × 5 = 30 minutes.
As a next step, according to Fig. 3, when Taro returns to A, Jiro moved to the point of ①.
The time Jiro reached ③ is the same time as the time Taro went and come back between AP which is 6 minute × 2 = 12 minutes.
The ratio of the speed is 5 : 1 and Taro caught up with Jiro at Q.
If the distance of AQ is set to 5, the distance of Jiro's from ① to ② will be 1, and also the distance from A to ① will be 5 - 1 = 4.
That is, since Jiro moved the distance of 4 in 12 minutes, it turns out that he moved between AQ, 5 in 12 minute / 4 × 5 = 15 minutes.
Since Jiro moved between AQ in 15 minutes and between QB in 30 minutes, the distance between AB is 800m × 45 minutes /15 minutes = 2400m = 2.4km.
