Time : 50 minutes
Passing mark : 70%
Answer : The end of the problem
Problem 1
Problem 2
Problem 3
Problem 6
Problem 7
Problem 8
Find X.
1/17 - X/2006 = (2/17 + 17/59) × 1/22
Problem 2
Three different numbers are selected and put in order from four different numbers 1, 3, X, and 9 so that the integer of triple figures is made.
The number of triple figures is 24 pieces and the average is 555.
Find X.
Problem 3
The distance between A point and B point is 10 km.
Taro walks from A point at 4 km/h to B point.
Problem 4
Taro walks from A point at 4 km/h to B point.
He walks for 30 minutes and take a rest for 5 minutes and he repeat this time cycle.
Jiro goes to A point from B point at 12 km/h by a bicycle without taking a rest and return to B point.
Taro and Jiro left at the same time.
Find the time (time after they left) for Jiro to pass Taro.
Moreover, find the distance from A point to the point Jiro passed.
Taro and Jiro left at the same time.
Find the time (time after they left) for Jiro to pass Taro.
Moreover, find the distance from A point to the point Jiro passed.
Problem 4
The questionnaire survey was conducted on 40 sixth graders in an elementary school.
The questionnaire was as for three subjects, language, mathematics (Sansue) and science, if they like it, mark ○, if they do not like it, mark ×.
The grand total of ○ was 100 and that of × was 20.
There was no pupil who marked × to all three subjects.
There are 35 pupils who marked ○ to mathematics.
Among these, there are two pupils who marked ○ only to mathematics and four students who marked × to mathematics and ○ to science.
Answer the following questions.
(1) Find the number of the pupil who marked ○ only to language.
(2) Find the number of pupil at most who marked ○ to all three subjects.
Problem 5
The questionnaire was as for three subjects, language, mathematics (Sansue) and science, if they like it, mark ○, if they do not like it, mark ×.
The grand total of ○ was 100 and that of × was 20.
There was no pupil who marked × to all three subjects.
There are 35 pupils who marked ○ to mathematics.
Among these, there are two pupils who marked ○ only to mathematics and four students who marked × to mathematics and ○ to science.
Answer the following questions.
(1) Find the number of the pupil who marked ○ only to language.
(2) Find the number of pupil at most who marked ○ to all three subjects.
Problem 5
There is an integer of 5 figures which is a multiple of 36.
In addition, the integer contains 2, 3 and 5 in either digit.
For example, it is 53928 etc.
Find the smallest number in such integers.
Problem 6
When 100 is divided by 35, a quotient is 2 and remainder is 30.
When 100 is divided by 40, a quotient is 2 and remainder is 20.
Find the number of three digits integer including 100 which a quotient becomes the same, when it is divided by 35 and 40.
And also find the largest integer among them.
Problem 7
As for the admission ticket of a certain theater, the advance tickets and the day tickets are available at a rate of 7 : 3.
The advance tickets are on sale three weeks before to the previous day and the day tickets are on sale 2 hours before opening of the performance.
The day ticket audience of one day began to stand in a line 3 hours before opening and joined the sequence four persons per one minute.
The day tickets were sold at a sales window of the theater to six persons per one minute.
Answer the following questions.
(1) How many day tickets audience had stood in a line 10 minutes before opening of the performance?
(2) How many minutes did the person take to enter the theater who is the day tickets visitor and was in the 120th from the beginning of the line?
(Find the time this person was in the line.)
(Find the time this person was in the line.)
(3) Before opening of the performance, ten percent of scheduled number of the advance tickets remained unsold and 80 percent of schedule number of of the day tickets were sold.
Find the total number of audience this day.
Find the total number of audience this day.
Problem 8
In trapezoid ABCD as shown in a figure, the point O is an intersection of a diagonal line.
Problem 9
Problem 10
The area of △ AOB is 10 cm2 and △ BOC is 25 cm2.
Find the area of trapezoid ABCD.
Problem 9
Fig. 1 and Fig. 2 are development views of the cube in which the numbers from 1 to 6 were written to each face.
It is considered that the sum of numbers written to three faces gathering in the one vertex of each cube.
As for Fig.1, the greatest sum of numbers is 15.
In the case of Fig. 2, find the greatest sum of numbers.
It is considered that the sum of numbers written to three faces gathering in the one vertex of each cube.
As for Fig.1, the greatest sum of numbers is 15.
In the case of Fig. 2, find the greatest sum of numbers.
As shown in a figure, there are the right hexagon ABCDEF and a right heptagon ABGHIJK.
Find the angle of X and Y.
Problem 11
Problem 11
As shown in the figure, five pieces equilateral triangle of the same size are put in order without a gap.
Among quarters point of BC, point D is the point closest to the B.
The length of AE is 9 cm.
Find the length of AD.
Among quarters point of BC, point D is the point closest to the B.
The length of AE is 9 cm.
Find the length of AD.
Problem 12
Triangle ABC is a right triangle of AB = 18 cm, AC = 24 cm and BC = 30 cm.
The points P and Q are points on the bisectors of angle B and angle C, respectively.
PQ is parallel to BC. PH and AB, QK and AC are vertical respectively.
The area of pentagon AHPQK is half of the area of triangle ABC.
Find the length of PH and the length of PQ.
The points P and Q are points on the bisectors of angle B and angle C, respectively.
PQ is parallel to BC. PH and AB, QK and AC are vertical respectively.
The area of pentagon AHPQK is half of the area of triangle ABC.
Find the length of PH and the length of PQ.
Problem 13
The shadow portion in the figure is made of cutting off four isosceles triangles 2cm in height from the square which is 8 cm one side.
Find the volume of the quadrangular pyramid made by assembling this development.
<Answer>
Problem 1
Problem 2
Problem 3
Problem 6
Problem 7
Problem 8
Find X.
1/17 - X/2006 = (2/17 + 17/59) × 1/22
Answer
81
Problem 2
Three different numbers are selected and put in order from four different numbers 1, 3, X, and 9 so that the integer of triple figures is made.
The number of triple figures is 24 pieces and the average is 555.
Find X.
Answer
7
Solution
In the each digit of the integer of triple figures four numbers of 1, 3, X, 9 are used 24 / 4 = 6 times respectively.
The sum of the number of each digit is calculated.
The sum of the number of hundreds digit is (1 + 3 + X + 9) × 6 × 100.
The sum of the number of tens digit is (1 + 3 + X + 9) × 6 × 10.
The sum of the number of ones digit is (1 + 3 + X + 9) × 6 × 1.
All these three are added to be (1 + 3 + X + 9) × 6 × 111.
Since this is equal to 555 × 24, it is (1 + 3 + X + 9) × 6 × 111 = 555 × 24.
13 + X = 20 and X = 7.
Problem 3
The distance between A point and B point is 10 km.
Taro walks from A point at 4 km/h to B point.
Taro walks from A point at 4 km/h to B point.
He walks for 30 minutes and take a rest for 5 minutes and he repeat this time cycle.
Jiro goes to A point from B point at 12 km/h by a bicycle without taking a rest and return to B point. Taro and Jiro left at the same time.
Find the time (time after they left) for Jiro to pass Taro.
Moreover, find the distance from A point to the point Jiro passed.
Find the time (time after they left) for Jiro to pass Taro.
Moreover, find the distance from A point to the point Jiro passed.
Answer
Problem 4
70 minutes
4 km
4 km
Solution
The time concerning Jiro reaching A point is 60 × 10/12 = 50 minutes.
He will turn up at A point and pursues Taro.
Taro walks 4 × 30/60 = 2 km in 30 minutes and he walks another 4 × 30/60 = 2 km until 65 minutes after 35 minutes.
He takes the 2nd rest at the point 4 km from A and he walks again after 70 minutes.
By the way, as for Jiro, he is in the point of 4 km from A in 70 minutes (12 × (70 - 50)/60 = 4) that means he catches up with Taro exactly 70 minutes after leaving.
Problem 4
The questionnaire survey was conducted on 40 sixth graders in an elementary school.
The questionnaire was as for three subjects, language, mathematics (Sansue) and science, if they like it, mark ○, if they do not like it, mark ×.
The grand total of ○ was 100 and that of × was 20.
There was no pupil who marked × to all three subjects.
There are 35 pupils who marked ○ to mathematics.
Among these, there are two pupils who marked ○ only to mathematics and four students who marked × to mathematics and ○ to science.
Answer the following questions.
(1) Find the number of the pupil who marked ○ only to language.
(2) Find the number of pupil at most who marked ○ to all three subjects.
The questionnaire was as for three subjects, language, mathematics (Sansue) and science, if they like it, mark ○, if they do not like it, mark ×.
The grand total of ○ was 100 and that of × was 20.
There was no pupil who marked × to all three subjects.
There are 35 pupils who marked ○ to mathematics.
Among these, there are two pupils who marked ○ only to mathematics and four students who marked × to mathematics and ○ to science.
Answer the following questions.
(1) Find the number of the pupil who marked ○ only to language.
(2) Find the number of pupil at most who marked ○ to all three subjects.
Answer
(2) The table as shown below can be created from problem sentence and the result of (1).
In order to increase the number of the pupil who marked ○ to all three subjects as many as possible, it is to lessen the number of × as many as possible in 3~35 pupils in a table.
Since there are 20 × in all, if language of the pupil of 36~39 is also set to ×, the number of × will be decreased in the frame of 3~35.
Problem 5
(1) one
(2) 27
(2) 27
Solution
(1) There are 35 pupils who marked ○ to mathematics and four pupils who marked × to mathematics and marked ○ to science in total 40 pupils.
This shows that there are 35 + 4 = 39 pupils who marked ○ to mathematics or science.
There is no person who marked × to all three subjects.
This shows that there are 35 + 4 = 39 pupils who marked ○ to mathematics or science.
There is no person who marked × to all three subjects.
Therefore, the number of the pupil who marked ○ only to language is 40 - 39 = 1 person.
(2) The table as shown below can be created from problem sentence and the result of (1).
In order to increase the number of the pupil who marked ○ to all three subjects as many as possible, it is to lessen the number of × as many as possible in 3~35 pupils in a table.
Since there are 20 × in all, if language of the pupil of 36~39 is also set to ×, the number of × will be decreased in the frame of 3~35.
Thus, × will be 14 pieces in all into 1, 2, and 36~40, and the remainder is 20 - 14 = 6 pieces.
Since there are two persons who marked ○ only to mathematics, × is marked only one piece to any of 33 persons among 3~35.
Six remaining × will be only one piece to one person.
Therefore, there are 33 - 6 = 27 persons at most who marked ○ to all three subjects.
Problem 5
There is an integer of 5 figures which is a multiple of 36.
In addition, the integer contains 2, 3 and 5 in either digit.
For example, it is 53928 etc.
Find the smallest number in such integers.
Answer
13572
Solution
36= 4 × 9.
As for the multiple of 9, the sum of the number of each digit is divided by 9.
As 18 - (2 + 3 + 5) = 8, the sum of two numbers is 8.
As for ten thousands digit, 1 is good for the number of 5 figures becoming the minimum.
Thus, the two remaining numbers are found 1 and 7.
Next, since this number is a multiple of 4, the lower 2 figure can be considered 12, 32, 52, and 72.
In order for this 5 digit number to be minimum, the lower 2 figure should be maximum.
That is, lower 2 figure is set to 72.
Remaining 1, 3, and 5 are put in order from the smaller one and the minimum number is found 13572.
Problem 6
When 100 is divided by 35, a quotient is 2 and remainder is 30.
When 100 is divided by 40, a quotient is 2 and remainder is 20.
Find the number of three digits integer including 100 which a quotient becomes the same, when it is divided by 35 and 40.
And also find the largest integer among them.
Answer
55 pieces
244
244
Solution
According to that 35 × 2 = 70 and 35 × 3 = 105, the range of the integer from which a quotient is 2 if it is divided by 35 is 70 ≦ X ≦ 104.
According to that 40 × 2 = 80 and 40 × 3 = 120, the range of the integer from which a quotient is 2 if it is divided by 40 is 80 ≦ X ≦ 119.
According to that 40 × 2 = 80 and 40 × 3 = 120, the range of the integer from which a quotient is 2 if it is divided by 40 is 80 ≦ X ≦ 119.
Thus, the three digit integers from which a quotient is 2 are five pieces in 100 or more and 104 or less.
Arranges the range of the three digit integers of which quotient becomes 3, 4, 5, ....., it will become as it is shown in the table below.
Arranges the range of the three digit integers of which quotient becomes 3, 4, 5, ....., it will become as it is shown in the table below.
It is 20 pieces among 120 or more and 139 or less that a quotient is 3.
It is 15 pieces among 160 or more and 174 or less that a quotient is 4
It is 10 pieces among 200 or more and 209 or less that a quotient is 5
It is 5 pieces among 240 or more and 244 or less that a quotient is 6.
There is no integer overlapped that a quotient is 7.
Therefore total number of integer is 5 + 20 + 15 + 10 + 5 = 55 pieces.
And it comes out among these and the biggest integer is 244.
It is 15 pieces among 160 or more and 174 or less that a quotient is 4
It is 10 pieces among 200 or more and 209 or less that a quotient is 5
It is 5 pieces among 240 or more and 244 or less that a quotient is 6.
There is no integer overlapped that a quotient is 7.
Therefore total number of integer is 5 + 20 + 15 + 10 + 5 = 55 pieces.
And it comes out among these and the biggest integer is 244.
Problem 7
As for the admission ticket of a certain theater, the advance tickets and the day tickets are available at a rate of 7 : 3.
The advance tickets are on sale three weeks before to the previous day and the day tickets are on sale 2 hours before opening of the performance.
The day ticket audience of one day began to stand in a line 3 hours before opening and joined the sequence four persons per one minute.
The day tickets were sold at a sales window of the theater to six persons per one minute.
Answer the following questions.
(1) How many day tickets audience had stood in a line 10 minutes before opening of the performance?
(2) How many minutes did the person take to enter the theater who is the day tickets visitor and was in the 120th from the beginning of the line?
(Find the time this person was in the line.)
(Find the time this person was in the line.)
(3) Before opening of the performance, ten percent of scheduled number of the advance tickets remained unsold and 80 percent of schedule number of of the day tickets were sold.
Find the total number of audience this day.
Find the total number of audience this day.
Answer
(1) 20 persons
(2) 50 minutes
(3) 2610 persons
(2) 50 minutes
(3) 2610 persons
Solution
(1) Since line up at the rate of four persons per one minute from three hours before the curtain, the number of people lined up in 2 hours 50 minutes = 170 minutes which is the time of 10 minutes before the opening, 4 persons × 170 minutes = 680 persons..
Because the day tickets are on sale 2 hours before opening, during 1 hour and 50 minutes which is the time of 10 minutes before opening the number of persons whom tickets are sold is 6 persons × 110 minutes = 660 persons.
Because the day tickets are on sale 2 hours before opening, during 1 hour and 50 minutes which is the time of 10 minutes before opening the number of persons whom tickets are sold is 6 persons × 110 minutes = 660 persons.
Therefore there are 20 persons in the line 10 minutes before opening as 680 - 660 = 20.
(2) There are one hour = 60 minutes until the time of the day tickets on sale.
According to the calculation of 120 persons / four persons = 30 minutes, the 120th person was stood in the line before the time of the day tickets sale.
Since six persons enter per one from a sale start, this person needs to take 120 persons / six person = 20 minutes to enter.
Above all, the time taken the 120th person to enter is 30 minutes + 20 minutes = 50 minutes.
（3） Since the selling hours of the day tickets are 120 minutes and tickets were sold at the rate of six persons per one minute, the number of tickets sold is six sheets × 120 minutes = 720 sheets.
Since 720 sheets is equivalent to 80 percent of schedule number of sheets, schedule number of sheets is 720 / 0.8 = 900 sheets.
Since the ratio of the sales schedule number of the advance tickets and the day tickets was 7 : 3 and there was 900 schedule number of the day tickets, the schedule number of the advance tickets is 900 / 3 × 7 = 2100 sheets.
Since 720 sheets is equivalent to 80 percent of schedule number of sheets, schedule number of sheets is 720 / 0.8 = 900 sheets.
Since the ratio of the sales schedule number of the advance tickets and the day tickets was 7 : 3 and there was 900 schedule number of the day tickets, the schedule number of the advance tickets is 900 / 3 × 7 = 2100 sheets.
Since the sales number of the advance tickets was 90 percent of schedule number, it is 2100 sheets × 0.9= 1890 sheets.
Above all, the total number of audience is 720 + 1890 = 2610 persons.
Problem 8
In trapezoid ABCD as shown in a figure, the point O is an intersection of a diagonal line.
Problem 10
The area of △ AOB is 10 cm2 and △ BOC is 25 cm2.
Find the area of trapezoid ABCD.
Answer
Problem 9
49 cm2
Solution
In △AOB (P) and △BOC(Q), the ratio of triangular base is equal to the ratio of area of triangle, it is AO : CO = 10 : 25 = 2 : 5.
Moreover, since the area of △ABD and △ACD is equal, it is P + S = R + S.
Thus, it is found that R = P = 10 cm2.
Furthermore, since the area ratio of S and R turns into a ratio of a base, it is 2 : 5.
Thus, the area of S is 10 cm2 × 2/5 = 4 cm2.
Therefore, the area of trapezoid ABCD is 10 + 25 + 10 + 4 = 49 cm2.
Problem 9
Fig. 1 and Fig. 2 are development views of the cube in which the numbers from 1 to 6 were written to each face.
It is considered that the sum of numbers written to three faces gathering in the one vertex of each cube.
As for Fig.1, the greatest sum of numbers is 15.
In the case of Fig. 2, find the greatest sum of numbers.
It is considered that the sum of numbers written to three faces gathering in the one vertex of each cube.
As for Fig.1, the greatest sum of numbers is 15.
In the case of Fig. 2, find the greatest sum of numbers.
Answer
Vertex of one face of the development view in the problem is given for ABCD, as shown in Fig. 4.
The face including the side BC is only a face of BCGF in addition to ABCD.
The next of B is F, since the next of C is G, it turns out that P is G and Q is F.
Since a face including the side BF is only a face of ABFE in addition to BCGF, it is found that R is A and S is E.
As a result of this work, as shown in Fig. 5, alphabet is assigned to all vertex.
Based on the alphabet number is put into the face, it will become as it is shown in Fig. 6.
In Fig. 6, the number with O is a number of the face which is not seen.
The greatest sum among a number of sums written in three faces gathering in one vertex according to this figure is 6 + 5 + 2 = 13.
13
Solution
As shown in Fig. 3, a sketch is drawn as a first step and alphabet A~H is assigned to each vertex. Vertex of one face of the development view in the problem is given for ABCD, as shown in Fig. 4.
The face including the side BC is only a face of BCGF in addition to ABCD.
The next of B is F, since the next of C is G, it turns out that P is G and Q is F.
Since a face including the side BF is only a face of ABFE in addition to BCGF, it is found that R is A and S is E.
As a result of this work, as shown in Fig. 5, alphabet is assigned to all vertex.
Based on the alphabet number is put into the face, it will become as it is shown in Fig. 6.
In Fig. 6, the number with O is a number of the face which is not seen.
The greatest sum among a number of sums written in three faces gathering in one vertex according to this figure is 6 + 5 + 2 = 13.
As shown in a figure, there are the right hexagon ABCDEF and a right heptagon ABGHIJK.
Find the angle of X and Y.
Answer
Problem 11
x = 60/7 degrees
y = 240/7 degrees
y = 240/7 degrees
Solution
One of Interior angle of the right hexagon is 120 degrees and that of right heptagon is 180 - (360/7) = 900/7 degrees.
X = 900/7 - 120 = 60/7 degrees.
Figure is an enlarged figure of the upper part of the figure in question.
The point L is an intersection of extended upwards the side EF and DC of the right hexagon.
The point O is the intersection of the side ED and the line which is drawn by being extended the line being connected L and vertex I of right heptagon.
Since ∠MEO=∠NDO = 60 degrees, △ LED is an equilateral triangle.
Thus, as for △LMI, ∠MLI = 60 / 2 = 30 degree and ∠MIL = (360 - One of interior angle of the right heptagon) / 2 = (360 - 900/7) / 2 = 810/7 degrees.
Therefore, Y = ∠LMI = 180 - 30 - 810/7 = 240/7 degrees.
Problem 11
As shown in the figure, five pieces equilateral triangle of the same size are put in order without a gap.
Among quarters point of BC, point D is the point closest to the B.
The length of AE is 9 cm.
Find the length of AD.
Among quarters point of BC, point D is the point closest to the B.
The length of AE is 9 cm.
Find the length of AD.
Answer
9/4 cm
Solution
In Fig. 1, △ACF and △BEF are homothetic and a homothetic ratio is AC : BE = 1 : 3.
CF : FB = 1 : 3 as well.
The length of AF is 9cm × 1 / (1+3) = 9/4 cm.
D is a point closest to B among quarters point of BC.
Furthermore, since it is BG = CG, according to Fig.2, it is found that BD = DG = GF = FC.
In Fig. 1, since AH and BC are the diagonal lines of rhombus ABHC, ∠AGD = ∠AGF = 90°.
Since as for △ADG and △AFG, these are right triangles with DG = GF and AG = AG, △ADG and △AFG are congruent.
Thus, AD = AF.
Therefore, AD = AF = 9/4 cm.
Problem 12
Triangle ABC is a right triangle of AB = 18 cm, AC = 24 cm and BC = 30 cm.
The points P and Q are points on the bisectors of angle B and angle C, respectively.
PQ is parallel to BC. PH and AB, QK and AC are vertical respectively.
The area of pentagon AHPQK is half of the area of triangle ABC.
Find the length of PH and the length of PQ.
The points P and Q are points on the bisectors of angle B and angle C, respectively.
PQ is parallel to BC. PH and AB, QK and AC are vertical respectively.
The area of pentagon AHPQK is half of the area of triangle ABC.
Find the length of PH and the length of PQ.
Answer
△BPH and △BPD, △CQK and △CQE become congruence, respectively.
Thus, it also turns out that PH = PD = QE = QK.
When equivalence movement of △BPH and △CQK is carried out as shown in Fig. 2, a quadrangle HBCK will turn into a rectangle.
PH = 3.6 cm
PQ = 12 cm
PQ = 12 cm
Solution
Since the area of pentagon AHPQK is a half of the area of triangle ABC, it is 18 × 24 / 2 = 108 cm2.
The area of the remaining portion is also 108 cm2 .
As shown in Fig. 1, two lines vertical to BC is drawn from P and Q respectively and each intersection is set to D and E, respectively.
Thus, it also turns out that PH = PD = QE = QK.
When equivalence movement of △BPH and △CQK is carried out as shown in Fig. 2, a quadrangle HBCK will turn into a rectangle.
Since the area of rectangle HBCK is also 108 cm2, HP in Fig.1 = HB in Fig.2 = 108 / 30 = 3.6 cm.
As shown in Fig. 3, the end of the line segment PQ is made to extend and an intersection with AB and AC is set to F and G. PQ = FG - (FP+QG).
As shown in Fig. 3, the end of the line segment PQ is made to extend and an intersection with AB and AC is set to F and G. PQ = FG - (FP+QG).
If BC is set as a base of △ABC, the height is 216 × 2 / 30 =14.4 cm.
Since HP in Fig.1 = HB in Fig.2 = 3.6 cm according to (1), the height of △AFG is 14.4 - 3.6 = 10.8 cm.
The homothetic ratio of △ABC and △AFG is equal to the ratio of height.
As 14.4 : 10.8 = 4 : 3, FG = 30 × 3/4 = 22.5 cm.
Since △HFP and △ABC are homothetic, HP : FP = 24 : 30 = 4 : 5.
FP = 3.6 × 5/4 = 4.5 cm.
Moreover, △KQG and △ABC are homothetic, KQ : QG = 18 : 30 = 3 : 5.
QG = 3.6 × 5/3 = 6 cm.
Therefore, PQ = FG - (FP + QG) = 22.5 - (4.5 + 6) = 12 cm.
Problem 13
The shadow portion in the figure is made of cutting off four isosceles triangles 2cm in height from the square which is 8 cm one side.
Find the volume of the quadrangular pyramid made by assembling this development.
Answer
32/3 cm3
Solution
Fig. 1 is a sketch of the solid which assembled the portion of the shadow.
When the solid of Fig. 1 is equally divided into four, one solid will become triangular pyramid A-BCD of Fig. 3.
The development view of triangular pyramid A-BCD becomes the square which divided the development view in the problem into four equally as shown in Fig. 2.
The height of the solid of Fig. 1 is equal to the length of one side of the small square of Fig. 2.
It is 8cm / 2 = 4cm.
The area of a square at the bottom is 4cm × 4cm / 2 = 8cm2.
The volume of the solid is 8cm2 × 4cm × 1/3 = 32/3cm3.
