Taro and Jiro are in A and B, respectively in the figure at first.
They begin to walk at the same time with the same speed and they play rock-paper-scissors at the place where they met.
The person who won at rock-paper-scissors walks as it is and the person who lost runs with speed 3 times the speed of walking and returns to a starting point and walks with the original speed to partners.
And they play rock-paper-scissors at the place where they meet again.
After repeating as above several times and the points they met are set to C, D, E, and F at order.
Taro won at C and E and Jiro won at D.
Answer the following questions.
(1) Answer the ratio of the distance between EB and between AB by the ratio of the least integer.
(2) The distance of between AF was 10 m.
Find the distance of between AB.
Round off the 2nd decimal place and find to the 1st decimal place.
Answer
(1) 13 : 18
(2) 13.2 m
Solution
(1) Since the person who lost runs with speed 3 times the speed of walking and returns to a starting point, Fig. 1 shows that Jiro having lost and having returned with the speed of 3 to B.
Taro's speed is 1 and since the ratio of the speed in the same time is equal to the ratio of distance, if distance between CB is set to 3, Taro moves distance of 1 as it is 1/3 between CB by the time Jiro returns to B.
Fig. 2 shows that after Jiro returning to B, he walks to meet Taro at D and the distance which moved by the time of the meeting is 1, respectively.
Thus, if between CB is set to 3, CD is 2.
Next, Fig.3 shows that Taro lost at D and he returned to A and he walks from A to meet Jiro at E.
Since the distance between AC is equal to that of between BC, the distance of between AC is 3 when between BC is set to 3.
Since between AD is set to 3 + 2 = 5, between AE is 5 × 1/3 = 5/3.
Moreover, between EB is 6 - 5/3 = 13/3.
Therefore, the ratio of the distance of between EB and between AB is 13/3 : 6 = 13 : 18.
(2) Fig.4 shows that Jiro who lost at E returned to B and meets Taro again at F.
Since between EB was 13/3, between EF is 13/3 × 2/3 = 26/9.
Between AF is AE + EF, it is 5/3 + 26/ 9 = 41/9.
Since the distance of between AF is 10 m, the distance of between AB which are 6 is 10m / 41/9 × 6 =10 × 9/41 × 6 = 13.17----.
The 2nd decimal place is rounded off and it is 13.2 m.
They begin to walk at the same time with the same speed and they play rock-paper-scissors at the place where they met.
The person who won at rock-paper-scissors walks as it is and the person who lost runs with speed 3 times the speed of walking and returns to a starting point and walks with the original speed to partners.
And they play rock-paper-scissors at the place where they meet again.
After repeating as above several times and the points they met are set to C, D, E, and F at order.
Taro won at C and E and Jiro won at D.
Answer the following questions.
(1) Answer the ratio of the distance between EB and between AB by the ratio of the least integer.
(2) The distance of between AF was 10 m.
Find the distance of between AB.
Round off the 2nd decimal place and find to the 1st decimal place.
Answer
(1) 13 : 18
(2) 13.2 m
Solution
(1) Since the person who lost runs with speed 3 times the speed of walking and returns to a starting point, Fig. 1 shows that Jiro having lost and having returned with the speed of 3 to B.
Taro's speed is 1 and since the ratio of the speed in the same time is equal to the ratio of distance, if distance between CB is set to 3, Taro moves distance of 1 as it is 1/3 between CB by the time Jiro returns to B.
Fig. 2 shows that after Jiro returning to B, he walks to meet Taro at D and the distance which moved by the time of the meeting is 1, respectively.
Thus, if between CB is set to 3, CD is 2.
Next, Fig.3 shows that Taro lost at D and he returned to A and he walks from A to meet Jiro at E.
Since the distance between AC is equal to that of between BC, the distance of between AC is 3 when between BC is set to 3.
Since between AD is set to 3 + 2 = 5, between AE is 5 × 1/3 = 5/3.
Moreover, between EB is 6 - 5/3 = 13/3.
Therefore, the ratio of the distance of between EB and between AB is 13/3 : 6 = 13 : 18.
(2) Fig.4 shows that Jiro who lost at E returned to B and meets Taro again at F.
Since between EB was 13/3, between EF is 13/3 × 2/3 = 26/9.
Between AF is AE + EF, it is 5/3 + 26/ 9 = 41/9.
Since the distance of between AF is 10 m, the distance of between AB which are 6 is 10m / 41/9 × 6 =10 × 9/41 × 6 = 13.17----.
The 2nd decimal place is rounded off and it is 13.2 m.
