Time : 60 minutes
Passing mark : 70 %
Answer : End of the problem
Problem 1
As shown in a figure, there are three water tank of different sizes. Every tank is a rectangular prism.
Problem 2
Problem 5
When a certain fraction was denoted by the decimal and the 3rd decimal place was rounded off, it was to be 0.32.
Find the smallest denominator among such fractions.
Problem 6
In addition, A and B, B and C are connected by a thin pipe with a cock X, Y respectively.
When the cock is opened, water in connected two water tanks moves through the pipe until height of the water surface in each tank is same.
Answer the following questions.
Pay attention that the amount of the water in the pipe shall not be considered.
(1) At first, water is contained up to a height of 100cm from bottom in A and no water in B and C.
When the cock X is opened while cock Y being closed, the height of the water surface of A and B comes to be 40 cm.
Indicate the ratio of the amount of water in the tank of A and B as the integral ratio at this moment.
(2) Next, when the cock X is closed and the cock Y is opened, the height of the water surface of B and C comes to be 25cm.
Indicate the ratio of the amount of water in the tank of B and C as the integral ratio at this moment.
(3) Next, with cock Y being opened, cock X is opened again, the height of the water surface of the water in the three tanks comes to be same.
Find the height of the water surface at this moment.
Pay attention that the amount of the water in the pipe shall not be considered.
(1) At first, water is contained up to a height of 100cm from bottom in A and no water in B and C.
When the cock X is opened while cock Y being closed, the height of the water surface of A and B comes to be 40 cm.
Indicate the ratio of the amount of water in the tank of A and B as the integral ratio at this moment.
(2) Next, when the cock X is closed and the cock Y is opened, the height of the water surface of B and C comes to be 25cm.
Indicate the ratio of the amount of water in the tank of B and C as the integral ratio at this moment.
(3) Next, with cock Y being opened, cock X is opened again, the height of the water surface of the water in the three tanks comes to be same.
Find the height of the water surface at this moment.
Problem 2
As shown in a figure, there is a point A on the straight line ① and the straight line ② is drawn from point A.
Problem 3
Line ② is drawn as the angle with line ① to be 10 degrees.
Next, the point B is set on the straight line ② and the straight line ③ is drawn from point B.
Line ③ is drawn as the angle with line ② to be 20 degrees.
In the same way, the straight line ④ is drawn increasing 10 degrees of angles at a time.
Moreover, the point on a straight line and its straight line is set like the point D on line ④.
When this operation was continued several times, the straight line drawn newly overlapped with the straight line ①.
There was no line that is parallel to the straight line ① in the continuous operation.
Answer the following questions.
(1) Find the number of the straight line drawn.
The last straight line which has overlapped with the straight line ① is not counted.
(2) There are some straight lines which are in a vertical position each other.
Answer all numbers of pairs of lines.
(3) There are some straight lines which are in a parallel position each other.
Answer all numbers of pairs of lines.
Problem 3
According to the order as shown in a figure, a figure is made by adding a square whose one side is 1cm one by one.
(1) Find the number of all the squares in the figure which added the 9th square.
For example, in the case of the figure which added the 6th square, a square will be eight pieces in all.
(2) Find the number of all the squares in the figure which added the 75th square.
(3) What number of square is added when there are 76 pieces of square in all ?
Problem 4
For example, the figure which added the 23rd square becomes as follows.
Answer the following questions.
For example, in the case of the figure which added the 6th square, a square will be eight pieces in all.
(3) What number of square is added when there are 76 pieces of square in all ?
Problem 4
Tournament game was held in four persons, A, B, C, and D.
This game is held in every two persons' groups.
There are three matches of AB group versus CD group, AC group versus BD group, and AD group versus BC group.
Two winning points in case of win, one point of draw and zero point of lose is goven to every person of the group by one match.
When three matches are completed, answer the following questions.
(1) Find the sum total of four persons' winning point.
(2) When the number of a certain person's winning points is six, what are four persons' winning point? Answer all the cases.
Answer should be in the large order of a winning point, for example as [6, 4, 2, 2].
(3) When the number of a certain person's winning points is three, what are four persons' winning point?
Answer all the cases according to (2).
(4) What kinds of cases are there in four persons' winning point?
Answer all the cases other than (2) and (3) according to (2).
Problem 5
Find the smallest denominator among such fractions.
Problem 6
There is a circular clock.
There is a scale showing the “minutes” which divided the circumference into 60 equally in the dial plate.
(2) At a certain time in a.m., the 3rd hand pointed a certain scale.
(1) Choose all the cases where both the hour hand and the long hand have indicated the scale among the following time.
①0:15 a.m. ②2:36 a.m. ③4:44 a.m. ④8:03 a.m. ⑤10:12 a.m.
①0:15 a.m. ②2:36 a.m. ③4:44 a.m. ④8:03 a.m. ⑤10:12 a.m.
Furthermore, the 3rd hand (it is not the second hand) that rotates with fixed speed is supposed.
The center of rotation and the direction of this hand are the same as the hand of the clock.
Answer the following questions.
(2) At a certain time in a.m., the 3rd hand pointed a certain scale.
At this time, the hour hand pointed the 12th scale counterclockwise from the 3rd hand.
Moreover, the long hand pointed the 12th scale clockwise from the 3rd needle.
Find this time.
(3) Afterward at a certain time within 12 hours of the time of (2), the 3rd hand pointed a certain scale.
At this time the hour hand pointed the 12th scale clockwise from the 3rd hand.
Moreover, the long hand pointed the 12th scale counterclockwise from the 3rd hand.
Find this time.
(4) Between the time of (2) and (3), the 3rd hand passed the hour hand 10 times.
Find the time for one time rotation of the 3rd hand.
<Answer>
Problem 1
As shown in a figure, there are three water tank of different sizes. Every tank is a rectangular prism.
Problem 2
(3) When winning point of a certain person is three by three matches is in case of one win, one loss and one draw or in case of three draws.
It is assumed that A was in one win, one loss and one draw, and B won at AB group.
As B is in one win and one draw at playing with another person against A, a winning point is to be 2 + 2 + 1 = 5 points.
Moreover, as the person who was with A when A lost becomes one loss and one draw at playing against A, a winning points is one.
Since the person who was with A when A draw becomes be one win and one loss at playing against A, a winning point is 1 + 2 = 3 points. Therefore, four persons' winning points in this case are [5, 3, 3, 1] and they are [3, 3, 3, 3] in case of three draws.
Problem 5
Problem 6
In addition, A and B, B and C are connected by a thin pipe with a cock X, Y respectively.
When the cock is opened, water in connected two water tanks moves through the pipe until height of the water surface in each tank is same.
Answer the following questions.
Pay attention that the amount of the water in the pipe shall not be considered.(1) At first, water is contained up to a height of 100cm from bottom in A and no water in B and C.
When the cock X is opened while cock Y being closed, the height of the water surface of A and B comes to be 40 cm.
Indicate the ratio of the amount of water in the tank of A and B as the integral ratio at this moment.
(2) Next, when the cock X is closed and the cock Y is opened, the height of the water surface of B and C comes to be 25cm.
Indicate the ratio of the amount of water in the tank of B and C as the integral ratio at this moment.
(3) Next, with cock Y being opened, cock X is opened again, the height of the water surface of the water in the three tanks comes to be same.
Find the height of the water surface at this moment.
Pay attention that the amount of the water in the pipe shall not be considered.(1) At first, water is contained up to a height of 100cm from bottom in A and no water in B and C.
When the cock X is opened while cock Y being closed, the height of the water surface of A and B comes to be 40 cm.
Indicate the ratio of the amount of water in the tank of A and B as the integral ratio at this moment.
(2) Next, when the cock X is closed and the cock Y is opened, the height of the water surface of B and C comes to be 25cm.
Indicate the ratio of the amount of water in the tank of B and C as the integral ratio at this moment.
(3) Next, with cock Y being opened, cock X is opened again, the height of the water surface of the water in the three tanks comes to be same.
Find the height of the water surface at this moment.
Answer
(1) 3:2
(2) 5:3
(3) 500/17 cm
(2) 5:3
(3) 500/17 cm
Solution
(1) The height of the water surface of A is to 40cm down by 60cm from 100cm, the height of the water surface of B becomes 40cm from 0cm.
The height of the surface of the water of A having fallen 60cm, 60/100 = 3/5 of the amount of the water in A moved to B, and 1 - 3/5 = 2/5 was left in A.
Therefore, the ratio of the quantity of the water of A and B of this time is 3/5 : 2/5 = 3 : 2.
(2) When the cock Y is opened, the height of the surface of the water of B is down by 15cm from 40cm and becomes to 25cm.
15/40 = 3/8 of the amount of the water in B moved to C and 1 - 3/8 = 5/8 was left in B.
Therefore, the ratio of the amount of the water of B and C is 5/8 : 3/8 = 5 : 3.
(3) When cock X and Y are opened, water in A goes into B and C and some remain in A.
Then the height of the water surface of the water in the three tanks comes to be same.
The height of the water surface of this case is same as the height of the water surface when water is transferred to the water tank with the combined bottom area of A, B and C.
That is , to find the height of the water level at this time, the amount of water in A first may be divided by the total of the bottom area of A, B, and C.
The ratio of the base of A and B is same as the ratio of the volume of A and B found in (1), it is 2 : 3.
The height of the surface of the water of A having fallen 60cm, 60/100 = 3/5 of the amount of the water in A moved to B, and 1 - 3/5 = 2/5 was left in A.
Therefore, the ratio of the quantity of the water of A and B of this time is 3/5 : 2/5 = 3 : 2.
(2) When the cock Y is opened, the height of the surface of the water of B is down by 15cm from 40cm and becomes to 25cm.
15/40 = 3/8 of the amount of the water in B moved to C and 1 - 3/8 = 5/8 was left in B.
Therefore, the ratio of the amount of the water of B and C is 5/8 : 3/8 = 5 : 3.
(3) When cock X and Y are opened, water in A goes into B and C and some remain in A.
Then the height of the water surface of the water in the three tanks comes to be same.
The height of the water surface of this case is same as the height of the water surface when water is transferred to the water tank with the combined bottom area of A, B and C.
That is , to find the height of the water level at this time, the amount of water in A first may be divided by the total of the bottom area of A, B, and C.
The ratio of the base of A and B is same as the ratio of the volume of A and B found in (1), it is 2 : 3.
The ratio of the base of B and C is 5 : 3 as (2).
2 sets of these ratios are made into a continuous ratio, it is A : B : C = 10 : 15 : 9.
Because the height of the surface of the water of A of the beginning was 100cm, the volume of the water becomes 10 × 100 = 1000.
Total of base area of the three is 10 + 15 + 9 = 34.
2 sets of these ratios are made into a continuous ratio, it is A : B : C = 10 : 15 : 9.
Because the height of the surface of the water of A of the beginning was 100cm, the volume of the water becomes 10 × 100 = 1000.
Total of base area of the three is 10 + 15 + 9 = 34.
Therefore the height is 1000/34 = 500/17.
Problem 2
As shown in a figure, there is a point A on the straight line ① and the straight line ② is drawn from point A.
Problem 3
Line ② is drawn as the angle with line ① to be 10 degrees.
Next, the point B is set on the straight line ② and the straight line ③ is drawn from point B.
Line ③ is drawn as the angle with line ② to be 20 degrees.
In the same way, the straight line ④ is drawn increasing 10 degrees of angles at a time.
Moreover, the point on a straight line and its straight line is set like the point D on line ④.
When this operation was continued several times, the straight line drawn newly overlapped with the straight line ①.
There was no line that is parallel to the straight line ① in the continuous operation.
Answer the following questions.(1) Find the number of the straight line drawn.
The last straight line which has overlapped with the straight line ① is not counted.
(2) There are some straight lines which are in a vertical position each other.
Answer all numbers of pairs of lines.
(3) There are some straight lines which are in a parallel position each other.
Answer all numbers of pairs of lines.
The last straight line which has overlapped with the straight line ① is not counted.
(2) There are some straight lines which are in a vertical position each other.
Answer all numbers of pairs of lines.
(3) There are some straight lines which are in a parallel position each other.
Answer all numbers of pairs of lines.
Answer
(1) Eight lines
(2) ② and ⑤、② and ⑧、④ and ⑥
(3) ③ and ⑦、⑤ and ⑧
(2) ② and ⑤、② and ⑧、④ and ⑥
(3) ③ and ⑦、⑤ and ⑧
Solution
(1)
The sum of the polygonal exterior angle is 360 degrees.
As 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 = 360, it turns out that the last angle is 80 degrees.
Since the straight line of which angle becomes 80 degrees with ① is equivalent to the 8th of above formula, the number of straight line is eight in all.
(2) Make the following table (it is not so tough work).
Since straight lines that are vertical mutually are that the angle in the table should be 90 degrees or 270 degrees, those are lines correspond to the red circle.
(3) Make the following table.
Since parallel straight lines are straight lines with 180 degrees in angle, those are lines correspond to the red rectangle.
Problem 3
According to the order as shown in a figure, a figure is made by adding a square whose one side is 1cm one by one.
(1) Find the number of all the squares in the figure which added the 9th square.
For example, in the case of the figure which added the 6th square, a square will be eight pieces in all.
(2) Find the number of all the squares in the figure which added the 75th square.
(3) What number of square is added when there are 76 pieces of square in all ?
For example, the figure which added the 23rd square becomes as follows.
Answer the following questions.
For example, in the case of the figure which added the 6th square, a square will be eight pieces in all.
(3) What number of square is added when there are 76 pieces of square in all ?
Answer
Problem 4
(1) 14 pieces
(2) 246 pieces
(3) the 33rd
(2) 246 pieces
(3) the 33rd
Solution
(1) The figure which added the 9th square becomes as it is shown in a figure.
The square whose one side is 1 cm in this figure is nine pieces.
The square whose one side is 2 cm is 2 x 2 = 4 pieces.
The square whose one side is 3 cm is one piece. In total there are 9 + 4 + 1 = 14 pieces.
(2) Looking at the leftmost column, it turns out that 1, 4, 9, 16, 25, 36, 49, 64, 81 -----.
The figure with 75th square is what the 65th ~ the 75th square was added to the big square made with the 8 × 8 = 64 pieces square as shown in a figure.
The figure with 75th square is what the 65th ~ the 75th square was added to the big square made with the 8 × 8 = 64 pieces square as shown in a figure.
There are eight squares 1cm one side in length and nine in width and are three in a line upper right.
The square whose one side is 1 cm is 75 pieces.
The square whose one side is 2 cm is 7 × 8 = 56 pieces in an 8 cm long and 9 cm wide rectangle.
The square whose one side is 1 cm is 75 pieces.
The square whose one side is 2 cm is 7 × 8 = 56 pieces in an 8 cm long and 9 cm wide rectangle.
Furthermore, as shown in a figure, two pieces (red and blue) are made using the grid of 73 ~ 75. Sum total 56 + 2 = 58 pieces.
The square whose one side is 3 cm is 6 × 7 = 42 pieces in an 8 cm long and 9 cm wide rectangle.
The square whose one side is 3 cm is 6 × 7 = 42 pieces in an 8 cm long and 9 cm wide rectangle.
Furthermore, since one piece of square is made using the grid of 73 ~ 75 as shown in a figure, it is sum total 42 + 1 = 43 pieces.
The square whose one side is 4 cm is 5 × 6 = 30 pieces.
The square whose one side is 5 cm is 4 × 5 = 20 pieces.
The square whose one side is 6 cm is 3 × 4 = 12 pieces.
The square whose one side is 7 cm is 2 × 3 = 6 pieces.
The square whose one side is 8 cm is 1 × 2 = 2 pieces.
The above is totaled, it is 75 + 58 + 43 + 30 + 20 + 12 + 6 + 2 = 246 pieces.
The square whose one side is 4 cm is 5 × 6 = 30 pieces.
The square whose one side is 5 cm is 4 × 5 = 20 pieces.
The square whose one side is 6 cm is 3 × 4 = 12 pieces.
The square whose one side is 7 cm is 2 × 3 = 6 pieces.
The square whose one side is 8 cm is 1 × 2 = 2 pieces.
The above is totaled, it is 75 + 58 + 43 + 30 + 20 + 12 + 6 + 2 = 246 pieces.
(3) When the square 1cm one side is put as the square 5cm one side, the number of all squares is counted.
The number of the squares whose one side is 1 cm, 2 cm, 3 cm, 4 cm, and 5 cm will be sum total 25 + 16 + 9 + 4 + 1 = 55 pieces calculated by 5 × 5 = 25 pieces, 4 × 4 = 16 pieces, 3 × 3 = 9 pieces, 2 × 2 = 4 pieces, and 1 × 1 = 1 piece, respectively.
Next, while one piece is added from the 26th piece, it would be counted the sum total will be 76 pieces by how many 1cm squares are added. When the 26th piece is added, one 1 cm square will increase.
When the 27th piece is added, two 1cm square and one 2 cm square will increase.
When the 28th piece is added, three 1 cm, two 2cm and one 3cm square will increase.
When the 29th piece is added, four 1 cm, three 2cm, two 3cm and one 4cm square will increase.
When the 30th piece is added, five 1cm, four 2cm, three 3cm, two 4cm and one 5cm square will increase.
When the 30th piece is added, it will increase by 5 + 4 + 3 + 2 + 1 = 15 pieces, and will become sum total 55 + 15 = 70 pieces.
When the 27th piece is added, two 1cm square and one 2 cm square will increase.
When the 28th piece is added, three 1 cm, two 2cm and one 3cm square will increase.
When the 29th piece is added, four 1 cm, three 2cm, two 3cm and one 4cm square will increase.
When the 30th piece is added, five 1cm, four 2cm, three 3cm, two 4cm and one 5cm square will increase.
When the 30th piece is added, it will increase by 5 + 4 + 3 + 2 + 1 = 15 pieces, and will become sum total 55 + 15 = 70 pieces.
Furthermore, when the 31st piece is added, the square of 1 cm increases by one piece, it is the same as above, and it turns out by the 32nd piece 1 + 2 = 3 pieces increase and by the 33rd 1 + 2 + 3 = 6 pieces increase.
Therefore, the number of square when it becomes 76 pieces in all is the 33rd.
Therefore, the number of square when it becomes 76 pieces in all is the 33rd.
Problem 4
Tournament game was held in four persons, A, B, C, and D.
This game is held in every two persons' groups.
There are three matches of AB group versus CD group, AC group versus BD group, and AD group versus BC group.
Two winning points in case of win, one point of draw and zero point of lose is goven to every person of the group by one match.
When three matches are completed, answer the following questions.
(1) Find the sum total of four persons' winning point.
(2) When the number of a certain person's winning points is six, what are four persons' winning point? Answer all the cases.
Answer should be in the large order of a winning point, for example as [6, 4, 2, 2].
(3) When the number of a certain person's winning points is three, what are four persons' winning point?
Answer all the cases according to (2).
(4) What kinds of cases are there in four persons' winning point?
Answer all the cases other than (2) and (3) according to (2).
Answer
(1) 12 points
(2) [6, 2, 2, 2]
(3) [5, 3, 3, 1], [3, 3, 3, 3]
(4) [4, 4, 2, 2], [4, 4, 4, 0]
Solution
(1) Calculate the sum total of four persons' winning point by one match. In case of win and loss settled, it is 2 × 2 = 4 points and in case of draw, it is also 1 × 4 = 4 points.
Therefore, the sum total of four persons' winning point when three matches are completed is 4 × 3 = 12 point.
(2) In order for a winning point to be six points by three matches, you have to win 6 / 2= 3 times.
When winning points of A are six, three remaining persons can win only with A.
Therefore, three persons' winning points are two points as shown in the table below.
It is assumed that A was in one win, one loss and one draw, and B won at AB group.
As B is in one win and one draw at playing with another person against A, a winning point is to be 2 + 2 + 1 = 5 points.
Moreover, as the person who was with A when A lost becomes one loss and one draw at playing against A, a winning points is one.
Since the person who was with A when A draw becomes be one win and one loss at playing against A, a winning point is 1 + 2 = 3 points. Therefore, four persons' winning points in this case are [5, 3, 3, 1] and they are [3, 3, 3, 3] in case of three draws.
(4) Other than (2) and (3), case of winning points are four can be considered.
That is in case of two wins and one loss as shown in the left table below or in case of one win and two draws as shown in the right table.
Therefore four person's winning points are [4, 4, 4, 0] and [4, 4, 2, 2].
Problem 5
When a certain fraction was denoted by the decimal and the 3rd decimal place was rounded off, it was to be 0.32.
Find the smallest denominator among such fractions.
Find the smallest denominator among such fractions.
Answer
6/19
Solution
The fraction to be found is that it is 0.315 or more and smaller than 0.325 when it is changed to a decimal.
In order to find the fraction of which denominator is smallest, the numerator should also be the smallest number as less as possible.
When the numerator is 1, the fraction close to 0.315 is 1/3.
When the numerator is 1, the fraction close to 0.315 is 1/3.
Since 1/3 = 0.3333---, it does not suit conditions.
When a numerator is 2, the fraction close to 0.315 is 2/6 and as checked above, it does not suit conditions.
When a numerator is 3, the fraction close to 0.315 is 3/10, it is 3/10=0.3 and does not suit conditions.
when a numerator is 4, the denominator which is likely to be applied to conditions is 13. 4/13 = 0.30----.
When a numerator is 2, the fraction close to 0.315 is 2/6 and as checked above, it does not suit conditions.
When a numerator is 3, the fraction close to 0.315 is 3/10, it is 3/10=0.3 and does not suit conditions.
when a numerator is 4, the denominator which is likely to be applied to conditions is 13. 4/13 = 0.30----.
It does not suit conditions.
When a numerator is 5, the denominator which is likely to be applied to conditions is 16. 5/16 = 0.312---.
When a numerator is 5, the denominator which is likely to be applied to conditions is 16. 5/16 = 0.312---.
It does not suit conditions.
When a numerator is 6, the denominator which is likely to be applied to conditions is 19.
6/19 = 0.315----.
When a numerator is 6, the denominator which is likely to be applied to conditions is 19.
6/19 = 0.315----.
This will be set to 0.32 if a 3rd decimal place is rounded off.
Therefore the fraction to be found is 6/19.
Problem 6
There is a circular clock.
There is a scale showing the “minutes” which divided the circumference into 60 equally in the dial plate.
(2) At a certain time in a.m., the 3rd hand pointed a certain scale.
(1) Choose all the cases where both the hour hand and the long hand have indicated the scale among the following time.
①0:15 a.m. ②2:36 a.m. ③4:44 a.m. ④8:03 a.m. ⑤10:12 a.m.
①0:15 a.m. ②2:36 a.m. ③4:44 a.m. ④8:03 a.m. ⑤10:12 a.m.
Furthermore, the 3rd hand (it is not the second hand) that rotates with fixed speed is supposed.
The center of rotation and the direction of this hand are the same as the hand of the clock.
Answer the following questions.
(2) At a certain time in a.m., the 3rd hand pointed a certain scale.
At this time, the hour hand pointed the 12th scale counterclockwise from the 3rd hand.
Moreover, the long hand pointed the 12th scale clockwise from the 3rd needle.
Find this time.
(3) Afterward at a certain time within 12 hours of the time of (2), the 3rd hand pointed a certain scale.
At this time the hour hand pointed the 12th scale clockwise from the 3rd hand.
Moreover, the long hand pointed the 12th scale counterclockwise from the 3rd hand.
Find this time.
(4) Between the time of (2) and (3), the 3rd hand passed the hour hand 10 times.
Find the time for one time rotation of the 3rd hand.
Answer
(1) ②，⑤
(2) 4:48
(3) 7:12
(4) 13 minutes and 20 seconds (40/3 minutes)
(2) 4:48
(3) 7:12
(4) 13 minutes and 20 seconds (40/3 minutes)
Solution
(1) Since the scale expresses the minute, at the time of ①~⑤, the long hand always points the scale.
At this time, it is checked whether the hour hand points the scale.
One scale is 360 degrees / 60 = 6 degrees.
The hour hand moves 360 degrees / 12 = 30 degrees in 1 hour.
In 1 minute, it moves 30 degrees / 60 minute = 0.5 degree.
Check the degree that the hour hand moves at the time of ①~⑤ respectively.
① 0.5 degree × 15 minute = 7.5 degrees which is not a multiple of 6, the scale is not pointed.
① 0.5 degree × 15 minute = 7.5 degrees which is not a multiple of 6, the scale is not pointed.
② 0.5 degree × 36 minute =18 degrees which is a multiple of 6, the scale is pointed.
③ 0.5 degree × 44 minute =22 degrees which is not a multiple of 6, the scale is not pointed.
④ 0.5 degree × 3 minute =1.5 degrees which is not a multiple of 6, the scale is not pointed.
⑤ 0.5 degree × 12 minute =6 degrees which is a multiple of 6, the scale is pointed.
Therefore, the time when both the hour hand and the long hand pointing the scale is ② and ⑤.
③ 0.5 degree × 44 minute =22 degrees which is not a multiple of 6, the scale is not pointed.
④ 0.5 degree × 3 minute =1.5 degrees which is not a multiple of 6, the scale is not pointed.
⑤ 0.5 degree × 12 minute =6 degrees which is a multiple of 6, the scale is pointed.
Therefore, the time when both the hour hand and the long hand pointing the scale is ② and ⑤.
(2) A problem sentence shows that the hour hand pointed the scale.
According to (1), since the time when the hour hand points are 12, 24, 36, 48, 60 minutes, the time to be found is either of X o’clock and 0 minute, 12 minutes, 24 minutes, 36 minutes, and 48 minutes.
The hour hand points the 12th scale counterclockwise and the long hand points the 12th scale clockwise from the 3rd hand.
The hour hand points the 12th scale counterclockwise and the long hand points the 12th scale clockwise from the 3rd hand.
Thus, It turns out that the hour hand is 24 scale behind the long hand.
Based on above, the position of the long hand and the hour hand are investigated for 0 minute ~ 48 minutes as it is shown in a lower figure.
The time of possible position of hands in a figure is checked.
① In case the long hand points 0 minute, the time when the hour hand point 26 minutes is not possible.
② In case the long hand points 12 minutes, the time when the hour hand point 48 minutes is not possible.
③ In case the long hand points 24 minutes, the time when the hour hand point 0 minute is not possible.
④ In case the long hand points 36 minutes, the time when the hour hand point 12 minutes is not possible.
⑤ In case the long hand points 48 minutes, the time when the hour hand point 24 minutes is possible.
Therefore, the time to be found is at 4:48.
② In case the long hand points 12 minutes, the time when the hour hand point 48 minutes is not possible.
③ In case the long hand points 24 minutes, the time when the hour hand point 0 minute is not possible.
④ In case the long hand points 36 minutes, the time when the hour hand point 12 minutes is not possible.
⑤ In case the long hand points 48 minutes, the time when the hour hand point 24 minutes is possible.
Therefore, the time to be found is at 4:48.
(3) Same as (2), the position of the long hand and the hour hand should be shown in the figure to confirm.
According to the problem sentences, it turns out that the hour hand is 24 scale ahead the long hand.
The position of the long hand and the hour hand are investigated for 0 minute ~ 48 minutes as it is shown in a lower figure.
The time of possible position of hands in a figure is checked.
① In case the long hand points 0 minute, the time when the hour hand point 24 minutes is not possible.
② In case the long hand points 12 minutes, the time when the hour hand point 36 minutes is possible.
③ In case the long hand points 24 minutes, the time when the hour hand point 48 minute is not possible.
④ In case the long hand points 36 minutes, the time when the hour hand point 0 minute is not possible.
⑤ In case the long hand points 48 minutes, the time when the hour hand point 12 minutes is not possible.
Therefore, the time to be found is at 7:12.
① In case the long hand points 0 minute, the time when the hour hand point 24 minutes is not possible.
② In case the long hand points 12 minutes, the time when the hour hand point 36 minutes is possible.
③ In case the long hand points 24 minutes, the time when the hour hand point 48 minute is not possible.
④ In case the long hand points 36 minutes, the time when the hour hand point 0 minute is not possible.
⑤ In case the long hand points 48 minutes, the time when the hour hand point 12 minutes is not possible.
Therefore, the time to be found is at 7:12.
(4) The time from (2) to (3) is 7:12 - 4:48= 144 minutes.
In 144 minutes, the 3rd hand passes the short hand 10 times.
It means that the 3rd hand rotated 10 round from the position at 4:48 and also it moved to the position at 7:12.
The 3rd hand is a position for 36 minutes at 4:48 and is a position for 24 minutes at 7:12.
The angle is 6 degrees × (60 - 36 + 24) = 6 degrees × 48 minute = 288 degrees.
Thus, the angle by which the 3rd hand moved is to be 360 degrees × 10 + 288 degrees = 3888 degrees.
It took 144 minutes to move 3888 degrees.
The time to move 360 degrees is 144 minutes × 360/3888 = 40/3 minutes = 13 minutes and 20 seconds.
