Time : 40 minutes
Problem 2
Problem 3
Problem 4
Above all the time to be found is 8.4 - 12 seconds, 32.4 - 36 seconds.
Passing mark : 70 %
Answer : End of the problem
Problem 1
Problem 2
Problem 3
Problem 4
There is paper of the rectangle whose vertical length is 40 cm and width is 128 cm.
I cut this paper in parallel with the side of length or width.
I throw away paper with a small area and leaves paper with a large area.
When the area of two sheets of papers is the same, I leave one of papers.
This operation is counted as 1 time and I repeat same operation to the paper of the rectangle remaining.
The remaining paper is to become a square after some operations.
Answer the following questions.
(1) Where should be cut at the 1st operation in order for the remaining paper to become a square by two operations?
Answer by drawing the portion with slashes which may be cut in a rectangle below.
Moreover, also write in length so that the position of the portion is expressed.
(2) It was made for the length of one side of the square to become the shortest by three operations.
Find the length of one side of this square.
(3) I made 13 operations so that the length of one side of the square might become the shortest.
In this case, find the sum total of the area of the thrown-away paper.
Problem 2
There are some two-faced coins and piles up every coin of which heads face upward initially.
(1) In case there are three coins, as the 1st operation, flip one piece of the top of coins.
As the 2nd operation, flip piled two pieces from the top of coins.
As the 3rd operation, flip piled three pieces all.
Answer the following questions.
As the 2nd operation, flip piled two pieces from the top of coins.
As the 3rd operation, flip piled three pieces all.
As the 4th operation, flip one piece of the top again.
After this, continue the same cycle of operation of flipping two pieces from the top, three pieces all, and one piece of the top-------.
(1)-1 Find the number of times of operation in case heads of all coin face upward for the first time.
(1)-2 When you perform operation 100 times, find the number of pieces of the coin of which heads face upward.
(2) In case there are two coins, as the 1st operation, flip one upper piece.
As the 2nd operation, flip plied two all.
As the 3rd operation, flip one upper piece again.
After this, continue the same cycle of operation of flipping two all, one upper piece, two all,----------.
When you perform operation 100 times, which face of heads and/or tails of two coins face upwards respectively?
After this, continue the same cycle of operation of flipping two pieces from the top, three pieces all, and one piece of the top-------.
(1)-1 Find the number of times of operation in case heads of all coin face upward for the first time.
(1)-2 When you perform operation 100 times, find the number of pieces of the coin of which heads face upward.
(2) In case there are two coins, as the 1st operation, flip one upper piece.
As the 2nd operation, flip plied two all.
As the 3rd operation, flip one upper piece again.
After this, continue the same cycle of operation of flipping two all, one upper piece, two all,----------.
When you perform operation 100 times, which face of heads and/or tails of two coins face upwards respectively?
Problem 3
There is a crossing between A station and B station.
Trains bound for A station in every 5 minutes and trains bound for B station in every 6 minutes passes along this crossing.
The crossing is closed anytime a train passes.
While being closed, the display [A] and [B] is turned on in fixed time when trains for A station and B station pass respectively.
In 30 minutes from 8:00 to 8:30, the time when display either [A] or [B] is turned on was 10 minutes and 37 seconds in all.
Moreover, there were 1 minute and 34 seconds while both of displays were turned on simultaneously in all.
Answer the following questions.
(1) In case the time when [A] or [B] is displayed by one train is the same, find the time.
(2) In case the time when [B] is displayed by one train for B station is 22 seconds longer than the time when [A] is displayed by one train for A station,
(2)-1 Find the time when [A] is displayed by one train for A station.
(2)-2 B began to be displayed at 8:00 sharp and display [A] was turned off at 8:02:02 (2 seconds).
Find the time while the display of both [A] and [B] was turned on simultaneously in 100 minutes from 8:30 to 10:10.
Trains bound for A station in every 5 minutes and trains bound for B station in every 6 minutes passes along this crossing.
The crossing is closed anytime a train passes.
While being closed, the display [A] and [B] is turned on in fixed time when trains for A station and B station pass respectively.
In 30 minutes from 8:00 to 8:30, the time when display either [A] or [B] is turned on was 10 minutes and 37 seconds in all.
Moreover, there were 1 minute and 34 seconds while both of displays were turned on simultaneously in all.
Answer the following questions.
(1) In case the time when [A] or [B] is displayed by one train is the same, find the time.
(2) In case the time when [B] is displayed by one train for B station is 22 seconds longer than the time when [A] is displayed by one train for A station,
(2)-1 Find the time when [A] is displayed by one train for A station.
(2)-2 B began to be displayed at 8:00 sharp and display [A] was turned off at 8:02:02 (2 seconds).
Find the time while the display of both [A] and [B] was turned on simultaneously in 100 minutes from 8:30 to 10:10.
Problem 4
As shown in Fig. 1, there is a rectangle ABCD whose length of side AB is 18 cm and length of side AD is 42 cm.
The point P leaves A and moves at the speed of 3 cm/s clockwise rotation on the side of rectangle ABCD as A -> D -> C -> B -> A.
It will stop when it arrives at A again.
The point Q leaves B at the same time as P leaves A.
It moves counterclockwise until P stops at the speed of 2 cm/s on the side of rectangle.
Answer the following questions.
(1) In how many seconds after started is it that P and Q meet?
(2) The point R leaves A at the same time with P and moves counterclockwise on the side of the rectangle at the speed of 2 cm/s.
It moves until P stops.
The figure which connected three point P, Q, and R is considered.
The point P leaves A and moves at the speed of 3 cm/s clockwise rotation on the side of rectangle ABCD as A -> D -> C -> B -> A.
It will stop when it arrives at A again.
The point Q leaves B at the same time as P leaves A.
It moves counterclockwise until P stops at the speed of 2 cm/s on the side of rectangle.
Answer the following questions.
(1) In how many seconds after started is it that P and Q meet?
(2) The point R leaves A at the same time with P and moves counterclockwise on the side of the rectangle at the speed of 2 cm/s.
It moves until P stops.
The figure which connected three point P, Q, and R is considered.
(2)-1 A triangle is not made several times while three points are moving.
Find the time altogether when the triangle is not be made.
The time should be found the time since they begin to move.
For example, When the triangle is not be made all the time from x-second to y-second, you may answer as x~y seconds.
(2)-2 As shown in Fig. 2, the intersection of the line AC and the line BD is set to E.
Find the time altogether when E is on the inside and circumference of the triangle PQR.
In accordance with (2)-1, answer the time altogether.
Find the time altogether when the triangle is not be made.
The time should be found the time since they begin to move.
For example, When the triangle is not be made all the time from x-second to y-second, you may answer as x~y seconds.
(2)-2 As shown in Fig. 2, the intersection of the line AC and the line BD is set to E.
Find the time altogether when E is on the inside and circumference of the triangle PQR.
In accordance with (2)-1, answer the time altogether.
Answer
Problem 1
There is paper of the rectangle whose vertical length is 40 cm and width is 128 cm.
I cut this paper in parallel with the side of length or width.
I throw away paper with a small area and leaves paper with a large area.
When the area of two sheets of papers is the same, I leave one of papers.
This operation is counted as 1 time and I repeat same operation to the paper of the rectangle remaining.
The remaining paper is to become a square after some operations.
Answer the following questions.
(1) Where should be cut at the 1st operation in order for the remaining paper to become a square by two operations?
Answer by drawing the portion with slashes which may be cut in a rectangle below.
Moreover, also write in length so that the position of the portion is expressed.
(2) It was made for the length of one side of the square to become the shortest by three operations.
Find the length of one side of this square.
(3) I made 13 operations so that the length of one side of the square might become the shortest.
In this case, find the sum total of the area of the thrown-away paper.
Answer
(1)
(2) 32 cm
(3) 5119 cm2
(3) 5119 cm2
Solution
(1) When the paper remaining after two operations becomes a square, the length of one side is 40 cm.
After the 1st operation, the width must be below 40 × 2 = 80 cm.
Moreover, since the larger paper only remain, the width is more than 128/2 = 64 cm.
Therefore, the portion which may be cut at the 1st operation becomes as it is shown in the answer.
(2) In order for the length of one side of the square to be the shortest, the paper to be thrown away should be as much as possible.
That is, the paper is equally divided into two.
128/2 = 64, 64/2 = 32.
After the 2nd operation, the rectangle with 40 cm in length and 32 cm in width will remain.
Accordingly the length of one side of the square made after the 3rd operation is to be 32 cm.
(3) When cutting the longer side of the rectangle into two equally, the length of the side of length and width will change as shown in the table below.
The length of the length of the rectangle after 12th operation is 1.25 cm and the width is 1 cm.
Thus, the length of one side of the square after the 13th operation is to be 1 cm.
Therefore, the area of the square remained in this case is 1 × 1 = 1 cm2.
The sum total of the area of the thrown-away paper is 40 × 128 - 1 = 5120 - 1 = 5119 cm2.
Problem 2
There are some two-faced coins and piles up every coin of which heads face upward initially.
(1) In case there are three coins, as the 1st operation, flip one piece of the top of coins.
As the 2nd operation, flip piled two pieces from the top of coins.
As the 3rd operation, flip piled three pieces all.
Answer the following questions.
As the 2nd operation, flip piled two pieces from the top of coins.
As the 3rd operation, flip piled three pieces all.
As the 4th operation, flip one piece of the top again.
After this, continue the same cycle of operation of flipping two pieces from the top, three pieces all, and one piece of the top-------.
(1)-1 Find the number of times of operation in case heads of all coin face upward for the first time.
(1)-2 When you perform operation 100 times, find the number of pieces of the coin of which heads face upward.
(2) In case there are two coins, as the 1st operation, flip one upper piece.
As the 2nd operation, flip plied two all.
As the 3rd operation, flip one upper piece again.
After this, continue the same cycle of operation of flipping two all, one upper piece, two all,----------.
When you perform operation 100 times, which face of heads and/or tails of two coins face upwards respectively?
Answer
(1)-1 9 times
(1)-2 2 pieces
(2) Both tails of two coins face upward.
After this, continue the same cycle of operation of flipping two pieces from the top, three pieces all, and one piece of the top-------.
(1)-1 Find the number of times of operation in case heads of all coin face upward for the first time.
(1)-2 When you perform operation 100 times, find the number of pieces of the coin of which heads face upward.
(2) In case there are two coins, as the 1st operation, flip one upper piece.
As the 2nd operation, flip plied two all.
As the 3rd operation, flip one upper piece again.
After this, continue the same cycle of operation of flipping two all, one upper piece, two all,----------.
When you perform operation 100 times, which face of heads and/or tails of two coins face upwards respectively?
Answer
(1)-1 9 times
(1)-2 2 pieces
(2) Both tails of two coins face upward.
Solution
(1)-1
The coin whose head faces up is set to A and whose tail faces up is set to B.
Condition of three coins at the time of continuing operations becomes as it is shown in the table below.
According to this table, it is the 9th operation when all heads of coins face up.
(1)-2
Whenever being operated 9 times, it returns to the state of the beginning. 100 / 9 = 11 with remainder 1. When operation is turned 100 times, the number of sheets of the coin whose head faces up is the same as the case when it is operated once and there are two sheets.
(2)
Consider in the same way as (1).
Condition of the head and tail of coins becomes as it is shown in the table below.
Whenever being operated 8 times, it returns to the state of the beginning.
100 / 8 = 12 with remainder 4.
When it is operated 100 times, the condition is the same as the case when operation is turned four times.
As for both of two coins, the tails face upward.
Problem 3
There is a crossing between A station and B station.
Trains bound for A station in every 5 minutes and trains bound for B station in every 6 minutes passes along this crossing.
The crossing is closed anytime a train passes.
While being closed, the display [A] and [B] is turned on in fixed time when trains for A station and B station pass respectively.
In 30 minutes from 8:00 to 8:30, the time when display either [A] or [B] is turned on was 10 minutes and 37 seconds in all.
Moreover, there were 1 minute and 34 seconds while both of displays were turned on simultaneously in all.
Answer the following questions.
(1) In case the time when [A] or [B] is displayed by one train is the same, find the time.
(2) In case the time when [B] is displayed by one train for B station is 22 seconds longer than the time when [A] is displayed by one train for A station,
(2)-1 Find the time when [A] is displayed by one train for A station.
(2)-2 B began to be displayed at 8:00 sharp and display [A] was turned off at 8:02:02 (2 seconds).
Find the time while the display of both [A] and [B] was turned on simultaneously in 100 minutes from 8:30 to 10:10.
Answer
(1) 1 minute and 15 seconds
(2)-1 1 minute and 5 seconds
(2)-2 6 minutes and 14 seconds
(2)-2
Trains bound for A station in every 5 minutes and trains bound for B station in every 6 minutes passes along this crossing.
The crossing is closed anytime a train passes.
While being closed, the display [A] and [B] is turned on in fixed time when trains for A station and B station pass respectively.
In 30 minutes from 8:00 to 8:30, the time when display either [A] or [B] is turned on was 10 minutes and 37 seconds in all.
Moreover, there were 1 minute and 34 seconds while both of displays were turned on simultaneously in all.
Answer the following questions.
(1) In case the time when [A] or [B] is displayed by one train is the same, find the time.
(2) In case the time when [B] is displayed by one train for B station is 22 seconds longer than the time when [A] is displayed by one train for A station,
(2)-1 Find the time when [A] is displayed by one train for A station.
(2)-2 B began to be displayed at 8:00 sharp and display [A] was turned off at 8:02:02 (2 seconds).
Find the time while the display of both [A] and [B] was turned on simultaneously in 100 minutes from 8:30 to 10:10.
Answer
(1) 1 minute and 15 seconds
(2)-1 1 minute and 5 seconds
(2)-2 6 minutes and 14 seconds
Solution
(1)
In 30 minutes display A is turned on 30/ 5 = 6 times and display B is tunred on 30 / 6 = 5 times.
A total time when displays are turned on is 10 minutes and 37 seconds + (1 minute and 34 seconds) × 2 = 825 seconds.
Therefore the time of every 0ne time display is 825 / (6 + 5) = 75 seconds = 1 minute and 15 seconds.
(2)-1
When display time of A is set to ① second, display time of B is to be ①+ 22 seconds.
The number of display of A is 6 times and the number of display of B is 5 times.
The total time is expressed as ① × 6 + (① + 22) × 5 = ⑥ + ⑤ + 110 = ⑪ + 110 = 825 seconds.
Since ⑪ = 715 and ① = 715 / 11 = 65 seconds, the time when A is displayed is 1 minute and 5 seconds.
Since the display time of B is 1 minute and 5 seconds + 22 seconds = 1 minute and 27 seconds, the time while B was displayed was 8:00:00 - 8:01:27.
The display of A was off at 8:02:02.
The time when it started to display was 8:02:02 - 1 minute and 15 seconds = 8:00:57.
The time while A was displayed was 8:00:57 - 8:02:02.
Therefore the time cycle of display after 8:30:00 is stared from 8:30:57 - 8:32:02 for A and 8:30:00 - 8:31:27 for B.
Display time is arranged in the table as shown below.
The time cycle is 8:30 - 9:00 is to be one cycle.
There are three cycles of 8:30 - 9:00, 9:00 - 9:30, and 9:30 - 10:00 until 10:00.
In 10:00-10:10, the 1st display time and the 2nd display time overlap and it is 30 + 62 = 92 seconds.
Therefore the total time is (30 + 62 + 2) × 3 + 92 = 374 seconds = 6 minutes and 14 seconds.
Problem 4
As shown in Fig. 1, there is a rectangle ABCD whose length of side AB is 18 cm and length of side AD is 42 cm.
The point P leaves A and moves at the speed of 3 cm/s clockwise rotation on the side of rectangle ABCD as A -> D -> C -> B -> A.
It will stop when it arrives at A again.
The point Q leaves B at the same time as P leaves A.
It moves counterclockwise until P stops at the speed of 2 cm/s on the side of rectangle.
Answer the following questions.
(1) In how many seconds after started is it that P and Q meet?
(2) The point R leaves A at the same time with P and moves counterclockwise on the side of the rectangle at the speed of 2 cm/s.
It moves until P stops.
The figure which connected three point P, Q, and R is considered.
The point P leaves A and moves at the speed of 3 cm/s clockwise rotation on the side of rectangle ABCD as A -> D -> C -> B -> A.
It will stop when it arrives at A again.
The point Q leaves B at the same time as P leaves A.
It moves counterclockwise until P stops at the speed of 2 cm/s on the side of rectangle.
Answer the following questions.
(1) In how many seconds after started is it that P and Q meet?
(2) The point R leaves A at the same time with P and moves counterclockwise on the side of the rectangle at the speed of 2 cm/s.
It moves until P stops.
The figure which connected three point P, Q, and R is considered.
(2)-1 A triangle is not made several times while three points are moving.
Find the time altogether when the triangle is not be made.
The time should be found the time since they begin to move.
For example, When the triangle is not be made all the time from x-second to y-second, you may answer as x~y seconds.
(2)-2 As shown in Fig. 2, the intersection of the line AC and the line BD is set to E.
Find the time altogether when E is on the inside and circumference of the triangle PQR.
In accordance with (2)-1, answer the time altogether.
Find the time altogether when the triangle is not be made.
The time should be found the time since they begin to move.
For example, When the triangle is not be made all the time from x-second to y-second, you may answer as x~y seconds.
(2)-2 As shown in Fig. 2, the intersection of the line AC and the line BD is set to E.
Find the time altogether when E is on the inside and circumference of the triangle PQR.
In accordance with (2)-1, answer the time altogether.
Answer
(1) 20.4 seconds
(2)-1 20 ~ 21 seconds, 24 seconds
(2)-2 8.4 ~ 12 seconds, 32.4 ~ 36 seconds
(2)-1 20 ~ 21 seconds, 24 seconds
(2)-2 8.4 ~ 12 seconds, 32.4 ~ 36 seconds
Solution
(1) When P and Q meet is the case that the distance where P and Q moved becomes 42 × 2 + 18 = 102 cm as shown in the Fig.3.
The time is calculated as 102 / (3 + 2)= 20.4 seconds.
(2)-1 There are two cases that triangle can not made.
One is when two points are overlapped as (1) and another is when three points are on the straight line.
It's (42 + 18) / 20 = 20 seconds after leaving when P arrives at C.
In the meantime Q and R move 2 × 20 = 40 cm, respectively and that means Q and R are on the side of BC.
Thus triangle can not be made when P arrives at C.
Three points are on the same straight line until Q arrives at C.
Because time when Q arrives at C is 42 / 2 = 21 seconds, a triangle can not made from 20 seconds to 21 seconds.
After this time, three points can't be on the same line at the same time before P arrives at A.
Another case that triangle is not made is when P and R meet.
The time to take when P and R meet is the time when two points move the distance of 42 cm × 2 + 18 cm × 2 = 120 cm and meet.
The time to take when P and R meet is the time when two points move the distance of 42 cm × 2 + 18 cm × 2 = 120 cm and meet.
It is 120 cm / (3+2) = 24 seconds.
Above all, the time to be found is 20 seconds - 21 seconds and 24 seconds.
(2)-2 When E comes to the circumference of △PQR for the first time is the case that E comes on the line connecting P and Q as shown in the Fig.4.
The line connecting P and Q is the line dividing the area of the rectangle into two equally.
That will be when AP + BQ = 42cm.
The time to take is 42 cm/ (3 + 2) = 8.4 seconds.
That will be when AP + BQ = 42cm.
The time to take is 42 cm/ (3 + 2) = 8.4 seconds.
After that when E comes to the circumference of △PQR is the case that E comes on the line connecting P and R as shown in the Fig.5.
It is set to ① seconds when AP + BR is to be 42 cm.
Then, AP = 3 × ① and BR = 2 × ① - 18.
AP + BR = 3 × ① +2 × ① - 18 = 42cm.
AP + BR = 3 × ① +2 × ① - 18 = 42cm.
(3 + 2) × ① = 42 + 18 = 60 cm.
① = 60 / 5 = 12.
Therefore it is 12 seconds later.
Next, when E comes to the circumference of △PQR is the case that P is on the side BC and Q is on the side AD as shown in the Fig.6.
It is set to ① seconds when PC + QD is to be 42 cm.
Then, PC = 3 × ① - (42 + 18) and QD = 2 × ① - (42 + 18).
PC + QD = 3 × ① - 60 + 2 × ① - 60 = 42cm.
PC + QD = 3 × ① - 60 + 2 × ① - 60 = 42cm.
(3 + 2) × ① = 42 + 60 + 60 = 162 cm.
① = 162 / 5 = 32.4.
Therefore it is 32.4 seconds later.
Then, next, when E comes to the circumference of △PQR is the case that E comes on the line connecting P and R as shown in the Fig.7.
It is set to ① seconds when PB + CR is to be 18 cm.
Then, PB = 3 × ① - (42 + 18 + 42) and CR = 2 × ① - (18 + 42).
PB + CR = 3 × ① - 102 + 2 × ① - 60 = 18cm.
PB + CR = 3 × ① - 102 + 2 × ① - 60 = 18cm.
(3 + 2) × ① = 18 + 102 + 60 = 180 cm.
① = 180 / 5 = 36.
Therefore it is 36 seconds later.
