Time : 60 minutes
Passing mark : 70 %
Answer : End of the problem
Problem 1
Answer the following questions about a right dodecagon.
(1) Find the number of a diagonal line.
Problem 2
Problem 3
There is a paper tape 1 m in length.
After marking the points where this tape is divided into three equal parts , five equal parts and seven equal parts, respectively, this tape is cut off at the point marked.
Find all the kinds of length of the tape after cutting.
Problem 4
(1) Find the number of a diagonal line.
(2) Write in all the right dodecagons that have every side on the dotted line in the figure.
Problem 2
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.
(2) If the distance from A town to Q point sets to 800 m, find the distance from A town to B town.
Problem 3
After marking the points where this tape is divided into three equal parts , five equal parts and seven equal parts, respectively, this tape is cut off at the point marked.
Find all the kinds of length of the tape after cutting.
Problem 4
Black and white squares whose length of one side is 1 cm are spread as shown in a figure.
In each figure of (1) to (3) shown by the bold line, comparing the area of black(shadow) and white, which is larger by how much or same?
In each figure of (1) to (3) shown by the bold line, comparing the area of black(shadow) and white, which is larger by how much or same?
Answer
Problem 1
Answer the following questions about a right dodecagon.
(1) Find the number of a diagonal line.
(1) Find the number of a diagonal line.
(2) Write in all the right dodecagons that have every side on the dotted line in the figure.
Answer
(1) 54
(2)
(12 - 3) × 12 / 2 =54
(2)
As sum of interior angles of a right dodecagon is 180 × (12 - 2) = 1800 degrees, one interior angle is 1800 / 12 = 150 degrees.
Then, in the figure angle of 150 degrees should be found as Fig.1 ~ Fig. 6 below.
Problem 2
Problem 3
Problem 4
(1) 54
(2)
Solution
(1)(12 - 3) × 12 / 2 =54
(2)
As sum of interior angles of a right dodecagon is 180 × (12 - 2) = 1800 degrees, one interior angle is 1800 / 12 = 150 degrees.
Then, in the figure angle of 150 degrees should be found as Fig.1 ~ Fig. 6 below.
Problem 2
Taro and Jiro left A town at the same time and went to B town.
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.
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
(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.
Problem 3
There is a paper tape 1 m in length.
After marking the points where this tape is divided into three equal parts , five equal parts and seven equal parts, respectively, this tape is cut off at the point marked.
Find all the kinds of length of the tape after cutting.
Answer
1/7 m, 2/35 m, 3/35 m, 1/21 m, 1/15 m, 1/35 m
Solution
It is hard to compare the size of fractions in Fig. 1.
Then, the length of a paper tape is set to 105 which is the least common multiple of 3, 5 and 7 and cutting part is expressed as shown in Fig.2.
If this is made into the number straight line, it will become as it is shown in Fig. 3.
The length of a between is investigated, it is 15 - 0 = 15, 21 - 15 = 6, 30 - 21 = 9, 35 - 30 = 5, 42 - 35 = 7, 45 - 42 = 3.
It turns out that there are six kinds of length.
Therefore, actual length is 15/105= 1/7 m, 6/105 = 2/35 m, 9/105 = 3/35 m, 5/105 = 1/21 m, 7/105 = 1/15 m, and 3/105 = 1/35 m.
Problem 4
Black and white squares whose length of one side is 1 cm are spread as shown in a figure.
In each figure of (1) to (3) shown by the bold line, comparing the area of black(shadow) and white, which is larger by how much or same?
In each figure of (1) to (3) shown by the bold line, comparing the area of black(shadow) and white, which is larger by how much or same?
Answer
(1) Black is larger by ½ cm2
(2) Same
(3) Black is larger by ½ cm2
(2) Same
(3) Black is larger by ½ cm2
Solution
(1)
The area of △ABC = 1 × 4 / 2 = 2 cm2.
Black = 2 cm2 (3+7) / (1+3+5+7) = 5/4 cm2
White = 2 cm2 (1+5) / (1+3+5+7) = 3/4 cm2
Black is larger than white by 5/4 - 1/2 = 1/2 cm2.
(2)
Number of full square of black is same as white as shown below.
Area transfer as shown below.
The result is that the area of black = white.
(3)
① Red portion
Black = white
② Blue portion
White area is bigger than black by 1/2 cm2.
③ Yellow portion
Black = white
The area of GHIJ = △DEF of (2) + Red - (Blue + Yellow).
This means that Black is larger than white by 1/2 cm2 in GHIJ.
