Time : 60 minutes
Passing marks : 70%
Answer : End of the problem
Problem 1
Problem 2
Problem 3
Problem 4
Problem 2
Problem 3
Water is contained in the three tanks A, B, and C.
First, after moving 1/3 of the water of A to B, 1/3 of the water remaining in A was moved to C.
Next, after moving 1/3 of the water of B to C, 1/3 of the water remaining in B was moved to A.
As a result, the volume ratio of water remaining in A, B, and C was 2 : 3 : 4.
Find the volume ratio of the water which was contained in the tank of A, B, and C at first by the least integer ratio.
Problem 2
The gas rate of every month of a certain gas company is the sum total of the fixed amount of basic charge and the charge corresponding to the amount of the gas used.
(1) Find the basic charge of gas and the unit price of x yen of a secondary stage.
The charge corresponding to the amount used is calculated from the unit price per m3 provided in stages corresponding to the amount used.
The unit price is as follows.
The amount used Up to 10 m3 160 yen per 1 m3
Over 10 m3 up to 30 m3 x yen per 1m3
Over 30 m3 240 yen per 1 m3
In addition, the gas rate is calculated by 1 m3 increments.
The amount used Up to 10 m3 160 yen per 1 m3
Over 10 m3 up to 30 m3 x yen per 1m3
Over 30 m3 240 yen per 1 m3
In addition, the gas rate is calculated by 1 m3 increments.
For example, when it is used 40 m3, the charge is calculated as (basic charge) + 10 × 160 yen + 20 × x yen + 10 × 240 yen.
At Taro’s house, 28 m3 gas was used in October and the charge was 6,310 yen.
Moreover, 40 m3 gas was used in December and the charge was 9,150 yen.
Answer the following questions.
(1) Find the basic charge of gas and the unit price of x yen of a secondary stage.
(2) As for Taro’s house, the least amount of the gas used in a year is in August and the most is in February.
In February the amount of gas used was exactly 3 times in August.
Moreover, the gas rate in February also became exactly 3 times in August.
Find the amount of gas used and the gas rate in February.
Problem 3
In a volleyball tournament the league match (round-robin matches) of 5 teams, A, B, C, D, and E is held.
There is one court and the schedule is for two days.
The game of this tournament is organized in accordance with the following administration rules.
<Rule 1>
In addition to two teams under game, there are one team that referee a game and two teams that are waiting to play a next game.
<Rule 2>
There are five games held a day.
<Rule 3>
One team does not hold two games continuously among one day.
There is one court and the schedule is for two days.
The game of this tournament is organized in accordance with the following administration rules.
<Rule 1>
In addition to two teams under game, there are one team that referee a game and two teams that are waiting to play a next game.
<Rule 2>
There are five games held a day.
<Rule 3>
One team does not hold two games continuously among one day.
One team may hold the 5th game on Day1 and the first game on Day2.
<Rule 4>
Each team will referee a game one time with the 1st day on the 2nd, respectively.
A part of schedule organized according to this rule is presented as it is shown in the next table.
Fill in all blanks in the table for answer.
<Rule 4>
Each team will referee a game one time with the 1st day on the 2nd, respectively.
A part of schedule organized according to this rule is presented as it is shown in the next table.
Fill in all blanks in the table for answer.
Problem 4
Quadrangle PQRSA was made by extending each side of quadrangle ABCD as follows.
① The length of the side AP is 3 times of the length of the side AB.
② The length of the side BQ is twice of the length of the side BC.
③ The length of the side CR is 3 times of the length of side CD.
④ The length of the side DS is twice of the length of the side DA.
Answer the following questions.
(1) Find the area ratio of the area of triangle APS and the area of the triangle ABD.
(2) Find the area ratio of the area of quadrangle PQRS and the area of quadrangle ABCD.
(3) When the multiplying factor is 50 times of ① and ③, and 100 times of ② and ④, find the area ratio of the area of quadrangle PQRS and the area of quadrangle ABCD.
Problem 5
Problem 6
The rows of letters can be put in order using a Lottery “Amidakuji”.
① The length of the side AP is 3 times of the length of the side AB.
② The length of the side BQ is twice of the length of the side BC.
③ The length of the side CR is 3 times of the length of side CD.
④ The length of the side DS is twice of the length of the side DA.
Answer the following questions.
(1) Find the area ratio of the area of triangle APS and the area of the triangle ABD.
(2) Find the area ratio of the area of quadrangle PQRS and the area of quadrangle ABCD.
(3) When the multiplying factor is 50 times of ① and ③, and 100 times of ② and ④, find the area ratio of the area of quadrangle PQRS and the area of quadrangle ABCD.
Problem 5
By putting the equilateral triangles with 1cm one side in order without any gap nor overlap, a big equilateral triangle is made.
(1) How many equilateral triangles with 1cm one side are used for the equilateral triangle with 4cm one side?
(2) The equilateral triangle of 1cm, 2cm, 3cm, 4cm ---- one side is named as the 1st, the 2nd, the 3rd, the 4th, ----- equilateral triangle, respectively.
Answer the following questions.
① How many equilateral triangles with 1cm one side are used for the 26th equilateral triangle?
② I made two equilateral triangles, one is a certain numerative number and another is the following.
There are in total 1013 pieces of equilateral triangles with 1cm one side are used for the two equilateral triangles.
Find the numerative number of each equilateral triangle.
(1) How many equilateral triangles with 1cm one side are used for the equilateral triangle with 4cm one side?
(2) The equilateral triangle of 1cm, 2cm, 3cm, 4cm ---- one side is named as the 1st, the 2nd, the 3rd, the 4th, ----- equilateral triangle, respectively.
Answer the following questions.
① How many equilateral triangles with 1cm one side are used for the 26th equilateral triangle?
② I made two equilateral triangles, one is a certain numerative number and another is the following.
There are in total 1013 pieces of equilateral triangles with 1cm one side are used for the two equilateral triangles.
Find the numerative number of each equilateral triangle.
Problem 6
For example, the row ABCDE of letters is replaced along with the row of letters ECDBA with the Lottery of Fig. 1.
The thick line in the figure is a route showing the way of A.
Answer the following questions.
(1) Connect two Lotteries of Fig. 1 lengthwise and make a Lottery as shown in Fig. 2.
How is the row of letters ABCDE located in a line is replaced with this Lottery?
Answer the new row of letters.
(2) With the Lottery which connected some Lotteries of Fig. 1 lengthwise, when the row of letters ABCDE are rearranged, it became the same row of letters as original ABCDE.
How many Lotteries of Fig. 1 are connected lengthwise in this case?
The thick line in the figure is a route showing the way of A.
Answer the following questions.
(1) Connect two Lotteries of Fig. 1 lengthwise and make a Lottery as shown in Fig. 2.
How is the row of letters ABCDE located in a line is replaced with this Lottery?
Answer the new row of letters.
(2) With the Lottery which connected some Lotteries of Fig. 1 lengthwise, when the row of letters ABCDE are rearranged, it became the same row of letters as original ABCDE.
How many Lotteries of Fig. 1 are connected lengthwise in this case?
Answer the least number.
(3) There is a Lottery which replaces the row of letters ABCDEFGHIJ along with the row of letters BCAJFGHIED.
With the Lottery which connected this Lottery lengthwise, when the row of letters ABCDEFGHIJ are rearranged, it became the same row of letters as original ABCDEFGHIJ.
How many Lotteries of this Lottery are connected lengthwise in this case?
Answer the least number.
(3) There is a Lottery which replaces the row of letters ABCDEFGHIJ along with the row of letters BCAJFGHIED.
With the Lottery which connected this Lottery lengthwise, when the row of letters ABCDEFGHIJ are rearranged, it became the same row of letters as original ABCDEFGHIJ.
How many Lotteries of this Lottery are connected lengthwise in this case?
Answer the least number.
<Answer>
Problem 1
Water is contained in the three tanks A, B, and C.
First, after moving 1/3 of the water of A to B, 1/3 of the water remaining in A was moved to C.
Next, after moving 1/3 of the water of B to C, 1/3 of the water remaining in B was moved to A.
As a result, the volume ratio of water remaining in A, B, and C was 2 : 3 : 4.
Find the volume ratio of the water which was contained in the tank of A, B, and C at first by the least integer ratio.
Answer
3 : 17 : 4
Solution
Assuming the volume of the water remaining in A, B, and C is 2, 3, and 4 respectively as a result of moving operation, it is traced back in order.
Since the volume of water in B became 3 after moving 1/3 of the water remaining in B to A, the volume in B before moving is 3 / (1 - 1/3) = 9/2, and A is 2 - (9/2 × 1/3) = 1/2.
At the previous operation, 1/3 of the water in B was moved to C, the volume in B before moving is 9/2 / (1 - 1/3) = 27/4 and C is 4 - (27/4 × 1/3) = 7/4.
At the previous operation, 1/3 of the water in A was moved to C, the volume in A before moving is 1/2 / (1 - 1/3) = 3/4 and C is 7/4 - (3/4 × 1/3) = 3/2.
At the previous operation, 1/3 of the water in A was moved to B, the volume in A before moving is 3/4 / (1 - 1/3) = 9/8 and B is 27/4 - (9/8 × 1/3) = 51/8.
Therefore, the volume ratio of the water at first is 9/8 : 51/8 : 3/2 = 9 : 51 : 12 = 3 : 17 : 4.
At the previous operation, 1/3 of the water in A was moved to B, the volume in A before moving is 3/4 / (1 - 1/3) = 9/8 and B is 27/4 - (9/8 × 1/3) = 51/8.
Therefore, the volume ratio of the water at first is 9/8 : 51/8 : 3/2 = 9 : 51 : 12 = 3 : 17 : 4.
A
|
2
|
1/2
|
1/2
|
3/4
|
9/8
|
B
|
3
|
9/2
|
27/4
|
27/4
|
51/8
|
C
|
4
|
4
|
7/4
|
3/2
|
3/2
|
Problem 2
The gas rate of every month of a certain gas company is the sum total of the fixed amount of basic charge and the charge corresponding to the amount of the gas used.
(1) Find the basic charge of gas and the unit price of x yen of a secondary stage.
The charge corresponding to the amount used is calculated from the unit price per m3 provided in stages corresponding to the amount used.
The unit price is as follows.
The amount used Up to 10 m3 160 yen per 1 m3
Over 10 m3 up to 30 m3 x yen per 1m3
Over 30 m3 240 yen per 1 m3
In addition, the gas rate is calculated by 1 m3 increments.
The amount used Up to 10 m3 160 yen per 1 m3
Over 10 m3 up to 30 m3 x yen per 1m3
Over 30 m3 240 yen per 1 m3
In addition, the gas rate is calculated by 1 m3 increments.
For example, when it is used 40 m3, the charge is calculated as (basic charge) + 10 × 160 yen + 20 × x yen + 10 × 240 yen.
At Taro’s house, 28 m3 gas was used in October and the charge was 6,310 yen.
Moreover, 40 m3 gas was used in December and the charge was 9,150 yen.
Answer the following questions.
(1) Find the basic charge of gas and the unit price of x yen of a secondary stage.
(2) As for Taro’s house, the least amount of the gas used in a year is in August and the most is in February.
In February the amount of gas used was exactly 3 times in August.
Moreover, the gas rate in February also became exactly 3 times in August.
Find the amount of gas used and the gas rate in February.
Answer
(1) Basic charge = 750 yen,
Unit price = 220 yen
(2) The amount of gas used = 45 m3,
(2) The amount of gas used = 45 m3,
The gas rate = 10350 yen
Problem 3
In a volleyball tournament the league match (round-robin matches) of 5 teams, A, B, C, D, and E is held.
There is one court and the schedule is for two days.
The game of this tournament is organized in accordance with the following administration rules.
<Rule 1>
In addition to two teams under game, there are one team that referee a game and two teams that are waiting to play a next game.
<Rule 2>
There are five games held a day.
<Rule 3>
One team does not hold two games continuously among one day.
There is one court and the schedule is for two days.
The game of this tournament is organized in accordance with the following administration rules.
<Rule 1>
In addition to two teams under game, there are one team that referee a game and two teams that are waiting to play a next game.
<Rule 2>
There are five games held a day.
<Rule 3>
One team does not hold two games continuously among one day.
One team may hold the 5th game on Day1 and the first game on Day2.
<Rule 4>
Each team will referee a game one time with the 1st day on the 2nd, respectively.
A part of schedule organized according to this rule is presented as it is shown in the next table.
Fill in all blanks in the table for answer.
<Rule 4>
Each team will referee a game one time with the 1st day on the 2nd, respectively.
A part of schedule organized according to this rule is presented as it is shown in the next table.
Fill in all blanks in the table for answer.
Answer
Since the waiting teams of the first game are B and E, the first game is C-D.
Since the 3rd game referee is a team other than A, C, D, and E, it is B.
The 4th game referee is D according to Rule 4.
Since the waiting teams of the 3rd game are C and E, the 4th game is C-E.
Since the waiting team of the 4th game is A and B, the 5th game is A-B.
The 2nd game on Day2 is A-E or D-E.
If it would be A-E, since the waiting teams of the first game are also A and E, the first game will be C-D and will become the same as the first game on Day1.
Therefore, the 2nd game on Day2 should be D-E and a referee is A. Since the waiting teams of the first game are D and E, the first game is A-C.
Since the 3rd game referee is D, the 4th game is A-E and a referee is C. Finally, the 5th game turns out to be B-D.
Problem 4
Solution
The 3rd game on Day1 is A-D and since the waiting team of the 2nd game is also A and D, the 2nd game is B-E.Since the waiting teams of the first game are B and E, the first game is C-D.
Since the 3rd game referee is a team other than A, C, D, and E, it is B.
The 4th game referee is D according to Rule 4.
Since the waiting teams of the 3rd game are C and E, the 4th game is C-E.
Since the waiting team of the 4th game is A and B, the 5th game is A-B.
The 2nd game on Day2 is A-E or D-E.
If it would be A-E, since the waiting teams of the first game are also A and E, the first game will be C-D and will become the same as the first game on Day1.
Therefore, the 2nd game on Day2 should be D-E and a referee is A. Since the waiting teams of the first game are D and E, the first game is A-C.
Since the 3rd game referee is D, the 4th game is A-E and a referee is C. Finally, the 5th game turns out to be B-D.
Problem 4
Quadrangle PQRSA was made by extending each side of quadrangle ABCD as follows.
① The length of the side AP is 3 times of the length of the side AB.
② The length of the side BQ is twice of the length of the side BC.
③ The length of the side CR is 3 times of the length of side CD.
④ The length of the side DS is twice of the length of the side DA.
Answer the following questions.
(1) Find the area ratio of the area of triangle APS and the area of the triangle ABD.
(2) Find the area ratio of the area of quadrangle PQRS and the area of quadrangle ABCD.
(3) When the multiplying factor is 50 times of ① and ③, and 100 times of ② and ④, find the area ratio of the area of quadrangle PQRS and the area of quadrangle ABCD.
Problem 6
The rows of letters can be put in order using a Lottery “Amidakuji”.
① The length of the side AP is 3 times of the length of the side AB.
② The length of the side BQ is twice of the length of the side BC.
③ The length of the side CR is 3 times of the length of side CD.
④ The length of the side DS is twice of the length of the side DA.
Answer the following questions.
(1) Find the area ratio of the area of triangle APS and the area of the triangle ABD.
(2) Find the area ratio of the area of quadrangle PQRS and the area of quadrangle ABCD.
(3) When the multiplying factor is 50 times of ① and ③, and 100 times of ② and ④, find the area ratio of the area of quadrangle PQRS and the area of quadrangle ABCD.
Answer
Problem 5
(1) 3 : 1
(2) 8 : 1
(3) 9851 : 1
(2) 8 : 1
(3) 9851 : 1
Reference
Problem 5
By putting the equilateral triangles with 1cm one side in order without any gap nor overlap, a big equilateral triangle is made.
(1) How many equilateral triangles with 1cm one side are used for the equilateral triangle with 4cm one side?
(2) The equilateral triangle of 1cm, 2cm, 3cm, 4cm ---- one side is named as the 1st, the 2nd, the 3rd, the 4th, ----- equilateral triangle, respectively.
Answer the following questions.
① How many equilateral triangles with 1cm one side are used for the 26th equilateral triangle?
② I made two equilateral triangles, one is a certain numerative number and another is the following.
There are in total 1013 pieces of equilateral triangles with 1cm one side are used for the two equilateral triangles.
Find the numerative number of each equilateral triangle.
(1) How many equilateral triangles with 1cm one side are used for the equilateral triangle with 4cm one side?
(2) The equilateral triangle of 1cm, 2cm, 3cm, 4cm ---- one side is named as the 1st, the 2nd, the 3rd, the 4th, ----- equilateral triangle, respectively.
Answer the following questions.
① How many equilateral triangles with 1cm one side are used for the 26th equilateral triangle?
② I made two equilateral triangles, one is a certain numerative number and another is the following.
There are in total 1013 pieces of equilateral triangles with 1cm one side are used for the two equilateral triangles.
Find the numerative number of each equilateral triangle.
Answer
(1) The equilateral triangle with 4cm one side and the equilateral triangle with 1cm one side are homothetic and a homothetic ratio is 4 : 1.
(1) 16 pieces
(2) ① 676 pieces
② 22nd and 23rd
Solution(2) ① 676 pieces
② 22nd and 23rd
(1) The equilateral triangle with 4cm one side and the equilateral triangle with 1cm one side are homothetic and a homothetic ratio is 4 : 1.
An area ratio is 4 × 4 : 1 × 1 = 16 : 1.
The number of the equilateral triangles of the area 1 required to make the equilateral triangle of the area 16 is 16 / 1 = 16 piece.
(2) ① Since the length of one side of the 26th equilateral triangle is 26cm, the homothetic ratio with the equilateral triangle 1cm one side is 16 : 1.
Considering the same way as (1), the number of the equilateral triangles with 1cm one side is 26 × 26 = 676 pieces.
② Since 1013 / 2 = 506.5, it is found that the number which makes a certain numerative number of equilateral triangle is 506 or less, and the following numerative number is 506 or more.
Since the number which makes an equilateral triangle is a square number, examine the square number near 506.
According to the calculation as 21 × 21 = 441, 22 × 22 = 484, 23 × 23 = 529, and 24 × 24 = 576, it turns out that 484 + 529 =1013 is applied to conditions.
Therefore, these two triangles are the 22nd and the 23rd.
(2) ① Since the length of one side of the 26th equilateral triangle is 26cm, the homothetic ratio with the equilateral triangle 1cm one side is 16 : 1.
Considering the same way as (1), the number of the equilateral triangles with 1cm one side is 26 × 26 = 676 pieces.
② Since 1013 / 2 = 506.5, it is found that the number which makes a certain numerative number of equilateral triangle is 506 or less, and the following numerative number is 506 or more.
Since the number which makes an equilateral triangle is a square number, examine the square number near 506.
According to the calculation as 21 × 21 = 441, 22 × 22 = 484, 23 × 23 = 529, and 24 × 24 = 576, it turns out that 484 + 529 =1013 is applied to conditions.
Therefore, these two triangles are the 22nd and the 23rd.
Problem 6
For example, the row ABCDE of letters is replaced along with the row of letters ECDBA with the Lottery of Fig. 1.
The thick line in the figure is a route showing the way of A.
Answer the following questions.
(1) Connect two Lotteries of Fig. 1 lengthwise and make a Lottery as shown in Fig. 2.
How is the row of letters ABCDE located in a line is replaced with this Lottery?
Answer the new row of letters.
(2) With the Lottery which connected some Lotteries of Fig. 1 lengthwise, when the row of letters ABCDE are rearranged, it became the same row of letters as original ABCDE.
How many Lotteries of Fig. 1 are connected lengthwise in this case?
The thick line in the figure is a route showing the way of A.
Answer the following questions.
(1) Connect two Lotteries of Fig. 1 lengthwise and make a Lottery as shown in Fig. 2.
How is the row of letters ABCDE located in a line is replaced with this Lottery?
Answer the new row of letters.
(2) With the Lottery which connected some Lotteries of Fig. 1 lengthwise, when the row of letters ABCDE are rearranged, it became the same row of letters as original ABCDE.
How many Lotteries of Fig. 1 are connected lengthwise in this case?
Answer the least number.
(3) There is a Lottery which replaces the row of letters ABCDEFGHIJ along with the row of letters BCAJFGHIED.
With the Lottery which connected this Lottery lengthwise, when the row of letters ABCDEFGHIJ are rearranged, it became the same row of letters as original ABCDEFGHIJ.
How many Lotteries of this Lottery are connected lengthwise in this case?
Answer the least number.
Answer
Solution
(1) A table is used to arrange the problem in order.
The row of letters ABCDE is replaced along with the row of letters ECDBA with the Lottery of Fig. 1.
As shown in a table, it is that each letter in the row of letters ABCDE is replaced with each of ECDBA, respectively by one operation. connecting two Lotteries of Fig. 1 means that the same operation is repeated twice.
(3) There is a Lottery which replaces the row of letters ABCDEFGHIJ along with the row of letters BCAJFGHIED.
With the Lottery which connected this Lottery lengthwise, when the row of letters ABCDEFGHIJ are rearranged, it became the same row of letters as original ABCDEFGHIJ.
How many Lotteries of this Lottery are connected lengthwise in this case?
Answer the least number.
Answer
(1) ADBCE
(2) Six pieces
(3) 30 pieces
(2) Six pieces
(3) 30 pieces
(1) A table is used to arrange the problem in order.
The row of letters ABCDE is replaced along with the row of letters ECDBA with the Lottery of Fig. 1.
As shown in a table, it is that each letter in the row of letters ABCDE is replaced with each of ECDBA, respectively by one operation. connecting two Lotteries of Fig. 1 means that the same operation is repeated twice.
As shown in a table, E is changed to A, C to D, D to B, B to C, and A to E, respectively.
Therefore, ABCDE is replaced with ADBCE.
(2) As shown in a table, operation is repeated following (1).
When a letter returns to the original position, the operation of the letter is stopped.
For example, since A and E have returned to the original position by the 2nd time, operations are stopped.
As you can see the table the result of operation is filled out, there are two kinds of letter, one is the letter which returns to the original position by 2 times, and another is the letter which returns by 3 times.
Therefore, ABCDE is replaced with ADBCE.
(2) As shown in a table, operation is repeated following (1).
When a letter returns to the original position, the operation of the letter is stopped.
For example, since A and E have returned to the original position by the 2nd time, operations are stopped.
As you can see the table the result of operation is filled out, there are two kinds of letter, one is the letter which returns to the original position by 2 times, and another is the letter which returns by 3 times.
Therefore, it is a time of operating by the number of times of the common multiple of 2 and 3 that all letters return to the original position at the same time.
Since the least number of times is the least common multiple which is six, there are six pieces of lotteries to be connected.
(3) Operation is being continued like (2), the result is shown as in a table.
The number of times that each character returns to the original position will be three kinds, 2 times, 3 times, and 5 times.
Therefore, since the minimum number of times which all letters return to the original position simultaneously turns into the least common multiple of 2, 3, and 5, which is 30, the number of lotteries connected is 30 pieces.
Since the least number of times is the least common multiple which is six, there are six pieces of lotteries to be connected.
(3) Operation is being continued like (2), the result is shown as in a table.
The number of times that each character returns to the original position will be three kinds, 2 times, 3 times, and 5 times.
Therefore, since the minimum number of times which all letters return to the original position simultaneously turns into the least common multiple of 2, 3, and 5, which is 30, the number of lotteries connected is 30 pieces.
