Time : 50 minutes
Passing mark : 70%
Answer : End of the problem
Problem 1
Problem 2
Problem 3
Problem 5
Problem 7
Problem 12
Problem 14
Find X
5/7 × (X × 119/15 + 1/4) / 11/8 = 16/11
Problem 2
There are 3 kind coins of A, B, and C.
Each weight of combination of A 6 sheets, B 1 sheet, C 1 sheet and A 1 sheet, B 4 sheets, C 1 sheet and A 1 sheet, B 1 sheet, C 3 sheets is 61g.
Find the weight of 1 sheet of C.
Problem 3
The memorial day of the 100th anniversary of the foundation of A junior high school is October 24, 2027.
Problem 4
When the years are a multiple of 4 when the years of A.D. cannot be divided by 100, it is a leap year.
January 31, 2004 is Saturday.
When February 1, 2004 is set to the 1st day, what number is October 24, 2027? and what day of the week is it?
Problem 4
When salt solution Xg with 8% concentration, salt solution Yg with 12% concentration and the salt 10g are mixed and stirred well, salt solution 800g with 10% concentration will be made.
Find X and Y, respectively.
Problem 5
When the integer A was divided by the integer B, the quotient was 32 and remainder was 10.
Furthermore, when division process was continued to the 3rd decimal digit, it was 32.322 and it was not divisible yet.
Problem 6
Furthermore, when division process was continued to the 3rd decimal digit, it was 32.322 and it was not divisible yet.
Find the integers A and B.
Problem 6
Taro was born in May, Jiro was born in July and Hanako was born in September, respectively.
In March, 1992, Taro was 14 years old and Hanako was 23 years old.
At a certain date in several years after March, 1992, Taro was 23 years old and the sum total of the age of Taro, Jiro and Hanako was 1.4 times of the sum total in the time of March, 1992.
Find the age of Jiro at a certain date in several years after March, 1992.
Problem 7
Taro reads a book of 151 pages, Jiro reads a book of 216 pages and Hanako reads a book of 294 pages.
Problem 8
For example, if everyone reads 4 pages a day, as for the number of pages read on the day which each finishes reading, it is 3 pages for Taro, 4 pages for Jiro and 2 pages for Hanako.
How many pages should be read a day in order to become the same number of pages which three persons read on the day which each finishes reading?
Moreover, find the number of pages read on the day which each finishes reading.
Noted that the number of pages read on the last day of reading is less than the number of pages read every day.
Problem 8
There was water full in the tank. I scheduled to drain away the water in the tank at the rate of 6m3 per minute.
After started draining, the volume of water in the tank became 3/4 of the full and it was rescheduled.
After the change, I drained at the ratio of 7m3 per minute until the volume of the water in the tank becomes half of the drainage started.
I stopped draining afterwards for seven minutes.
Next, I drained at the ratio of 14m3 per minute afterwards. After rescheduled, the time of the tank emptied was earlier by two minutes than first plan.
Find the capacity of this tank.
Problem 9
After started draining, the volume of water in the tank became 3/4 of the full and it was rescheduled.
After the change, I drained at the ratio of 7m3 per minute until the volume of the water in the tank becomes half of the drainage started.
I stopped draining afterwards for seven minutes.
Next, I drained at the ratio of 14m3 per minute afterwards. After rescheduled, the time of the tank emptied was earlier by two minutes than first plan.
Find the capacity of this tank.
Problem 9
The paper of the square whose length of a diagonal line is 20 cm is placed on the desk.
This paper is rotated on the desk by 45 degrees centering on the one vertex.
Find the area of the portion which this paper passes.
Pi is assumed to be 3.14.
Problem 10
A figure is a rectangular prism and AB = 5cm, AD = 6cm, and AE = 8cm.
Problem 11
Moreover, BF = 4 cm, AG = 2cm.
A rectangular prism is divided into two solids by the plane which passes along three point C, F, and G.
Find the ratio of the volume of a large solid to the volume of a small solid.
Problem 11
Fig.1 and Fig.2 show development views of solids.
The development view of Fig.1 consists of one rectangle, two equilateral triangles, and two trapezoids.
The developed view of Fig.2 consists of four equilateral triangles.
Find the volume ratio of the solid made by assembling Fig.1 and the solid made by assembling Fig.2.
Problem 12
3 sets of sides of the hexagon in the figure faced each other are parallel.
Problem 13
As for each of 3 sets, the ratio of the length of a short side and a long side is 1 : 3.
Find the area ratio of the area of a shadow area, and the area of a hexagon.
In a figure, AB and CD are vertical.
AE = 24 cm, BE = 6 cm, CE = 18 cm, DE = 8 cm.
The area of a circle is 785 cm2.
Find the sum of the area of a shadow area.
Problem 14
In a figure, AG is vertical to BC.
Problem 15
The points D and E are the middle points of AG and AC, respectively.
The point F is an intersection of BE and CD.
The length of DE is 2 cm.
The area of triangle ABC is 36 cm2.
The area of the triangle CEF is 2 cm2.
Find the length of AG.
Problem 15
There is a right-angled isosceles triangle ABC.
There is another right triangle DBE width is shorter than ABC by 3 cm and length is longer than ABC by 6 cm.
The figure shows two triangles overlap.
The point F is an intersection of AC and DE.
The ratio of the area of triangle CEF to ADF is 3 : 8.
Find the area of triangle ABC.
<Answer>
Problem 1
Problem 2
Problem 3
Problem 5
Problem 7
Problem 12
Problem 14
Find X
5/7 × (X × 119/15 + 1/4) / 11/8 = 16/11
Answer
9/28
Problem 2
There are 3 kind coins of A, B, and C.
Each weight of combination of A 6 sheets, B 1 sheet, C 1 sheet and A 1 sheet, B 4 sheets, C 1 sheet and A 1 sheet, B 1 sheet, C 3 sheets is 61g.
Find the weight of 1 sheet of C.
Answer
15 g
Solution
A B C
① 6 1 1 61g
② 1 4 1 61g
③ 1 1 3 61g
①=②，5A = 3B. Then A : B = 3 : 5.
②=③, 3B = 2C. Then B : C = 2 : 3.
Above all, A : B : C = 6 : 10 : 15.
61g is equivalent to 6 × 6 + 10 × 1 + 15 × 1 = 61.
Therefore C is 61g / 61 × 15 = 15g.
Problem 3
The memorial day of the 100th anniversary of the foundation of A junior high school is October 24, 2027.
When the years are a multiple of 4 when the years of A.D. cannot be divided by 100, it is a leap year.
January 31, 2004 is Saturday.
When February 1, 2004 is set to the 1st day, what number is October 24, 2027? and what day of the week is it?
Answer
Problem 4
8667th day
Sunday
Sunday
Solution
There is no year which the year can be divided by 100 from 2004 to 2027.
2027 - 2003 = 24 and if 24 is divided by 4, the number of times of a leap year can be found.
24 / 4 = 6 and there are six leap years.
Thus, number of days from February 1, 2004 to January 31, 2027 are (2027 - 2004) × 365 + 6 = 8401 days.
Number of days from February 1, 2027 to October 24, is 28(February) + 31(March) + 30(April) + 31(May) + 30(June) + 31(July) + 31(August) + 30(September) + 24(October) = 266 days.
It is in total, there are 8401 + 266 = 8667 days until October 24, 2027.
February 1, 2004 is Sunday.
Calculation is made assuming that Sunday is the 1st day and Saturday is the last day in the week, it is 8667 / 7 = 1238 remainder 1.
Remainder 1 means that October 24, 2027 is the 1st day of one week of beginning on Sunday.
Remainder 1 means that October 24, 2027 is the 1st day of one week of beginning on Sunday.
Therefore, it is Sunday.
Problem 4
When salt solution Xg with 8% concentration, salt solution Yg with 12% concentration and the salt 10g are mixed and stirred well, salt solution 800g with 10% concentration will be made.
Find X and Y, respectively.
Answer
X=620 g,
Y=170 g
Solution
800 × 0.1 = 80 g
80 - 10 = 70 g both in X and Y.
X + Y = 800 - 10 = 790g
If all of 790g is 8%, weight of salt is 790 × 0.08 = 63.2g.
Salt solution 1 g with 12 % concentration contain salt of 1g × (0.12 - 0.08) = 0.04g more than 8% salt solution.
(70 - 63.2) / 0.04 = 170 g which is weight of Y.
X = 790 - 170 = 620 g
Problem 5
When the integer A was divided by the integer B, the quotient was 32 and remainder was 10.
Furthermore, when division process was continued to the 3rd decimal digit, it was 32.322 and it was not divisible yet.
Furthermore, when division process was continued to the 3rd decimal digit, it was 32.322 and it was not divisible yet.
Find the integers A and B.
Answer
Problem 6
A = 1002
B = 31
B = 31
Solution
According to the problem sentences, when 10 is divided by B, the quotient becomes 0.322---.
10 / 0.322 = 31. --- and 10 / 0.323 = 30.----
Based on this calculation, it is found B = 31.
A = 32 × B + 10 = 32 × 31 + 10 = 1002.
10 / 0.322 = 31. --- and 10 / 0.323 = 30.----
Based on this calculation, it is found B = 31.
A = 32 × B + 10 = 32 × 31 + 10 = 1002.
Problem 6
Taro was born in May, Jiro was born in July and Hanako was born in September, respectively.
In March, 1992, Taro was 14 years old and Hanako was 23 years old.
At a certain date in several years after March, 1992, Taro was 23 years old and the sum total of the age of Taro, Jiro and Hanako was 1.4 times of the sum total in the time of March, 1992.
Find the age of Jiro at a certain date in several years after March, 1992.
Answer
37 years old
Solution
The age of Jiro in March, 1992 is set to X years old.
Sum total of three persons is 14 + X + 23 = X + 37.
As after several years Taro took 23 - 14 = 9 years, Jiro and Hanko might take 8 or 9 years.
Sum total of three persons increased by 25, 26 or 27 years.
The ratio of three persons is 1 : 1.4 = 5 : 7.
The increased years are equivalent to 7 - 5 = 2 which is even number.
The increased years are to be found 26 years which is even number as well.
Thus Jiro took 9 years and Hanako took 8 years.
Sum total of three persons in March, 1992 is 26 / 2 × 5 = 65 years old.
The above X was 65 - 37 = 28 and he became 28 + 9 = 37 years old.
Problem 7
Taro reads a book of 151 pages, Jiro reads a book of 216 pages and Hanako reads a book of 294 pages.
For example, if everyone reads 4 pages a day, as for the number of pages read on the day which each finishes reading, it is 3 pages for Taro, 4 pages for Jiro and 2 pages for Hanako.
How many pages should be read a day in order to become the same number of pages which three persons read on the day which each finishes reading?
Moreover, find the number of pages read on the day which each finishes reading.
Noted that the number of pages read on the last day of reading is less than the number of pages read every day.
Answer
Problem 8
13 pages
8 pages
8 pages
Solution
The story in this problem can be denoted as following statement.
151, 216 and 294 is divided by the same divisor and the remainder of each division should be the same number.
151, 216 and 294 is divided by the same divisor and the remainder of each division should be the same number.
According to this statement we find that the difference between two numbers out of three should be divided by the same divisor.
The difference between 151 and 216 is 65.
The difference between 151 and 294 is 143.
The difference between 216 and 294 is 78.
65, 143 and 78 are divided by the same divisor or these three numbers have common divisor.
It is found 1 and 13 but 1 is not suitable for this case.
Therefore the answer is 13.
151 / 13 = 11 remainder 8.
151 / 13 = 11 remainder 8.
Then the number of pages read last day is 8.
Problem 8
There was water full in the tank. I scheduled to drain away the water in the tank at the rate of 6m3 per minute.
After started draining, the volume of water in the tank became 3/4 of the full and it was rescheduled.
After the change, I drained at the ratio of 7m3 per minute until the volume of the water in the tank becomes half of the drainage started.
I stopped draining afterwards for seven minutes.
Next, I drained at the ratio of 14m3 per minute afterwards. After rescheduled, the time of the tank emptied was earlier by two minutes than first plan.
Find the capacity of this tank.
After started draining, the volume of water in the tank became 3/4 of the full and it was rescheduled.
After the change, I drained at the ratio of 7m3 per minute until the volume of the water in the tank becomes half of the drainage started.
I stopped draining afterwards for seven minutes.
Next, I drained at the ratio of 14m3 per minute afterwards. After rescheduled, the time of the tank emptied was earlier by two minutes than first plan.
Find the capacity of this tank.
Answer
Problem 9
168 m3
Solution
Capacity of the tank is set to 1.
Time planned for drainage is 1 / 6 = ⅙ minutes.
The time after rescheduled is 1/4 / 6 + 1/4 / 7 + 7 + 1/4 × 2 / 14 = 1/24 + 1/28 +7 + 1/28.
Since this time was 2 minutes earlier than first schedule which was 1/6 minutes, 1/24 + 1/28 +7 + 1/28 + 2 = 1/6.
The time after rescheduled is 1/4 / 6 + 1/4 / 7 + 7 + 1/4 × 2 / 14 = 1/24 + 1/28 +7 + 1/28.
Since this time was 2 minutes earlier than first schedule which was 1/6 minutes, 1/24 + 1/28 +7 + 1/28 + 2 = 1/6.
Calculating this formula, 9/168 = 9 minutes. 1 = 9 / (9/168) = 168.
Therefore capacity of the tank is 168 m3.
Problem 9
The paper of the square whose length of a diagonal line is 20 cm is placed on the desk.
This paper is rotated on the desk by 45 degrees centering on the one vertex.
Find the area of the portion which this paper passes.
Pi is assumed to be 3.14.
Answer
357 cm2
Solution
When rotating 45 degrees of squares ABCD centering on the vertex C, as shown in a figure, a square ABCD moves to A'B'CD'.
The portion which a square passes is the sum total of red and blue portion.
The area of red portion is the same as the area of a square ABCD, it is (20cm × 20cm) / 2 = 200 cm2.
The area of red portion is the same as the area of a square ABCD, it is (20cm × 20cm) / 2 = 200 cm2.
The area of blue portion is a sector with 20 cm in radius and 45 degrees of central angles, it is 20cm × 20cm × 3.14 × 45/360 = 157 cm2.
The area to be found is 200 + 157 = 357 cm2.
Problem 10
A figure is a rectangular prism and AB = 5cm, AD = 6cm, and AE = 8cm.
Moreover, BF = 4 cm, AG = 2cm.
A rectangular prism is divided into two solids by the plane which passes along three point C, F, and G.
Find the ratio of the volume of a large solid to the volume of a small solid.
Answer
Problem 11
41 : 7
Solution
The cut surface becomes Fig.1 which is cut by the plane which passes along three point C, F, and G.
The volume of the rectangular prism is 5 × 6 × 8 = 240cm3.
(Volume of large solid) = (Volume of rectangular prism) - (Volume of a small solid).
As shown in Fig.2, AB, GF and HC are made to extend.
The volume of the small solid can be found by subtracting the volume of triangular pyramid I-AGH from the volume of triangular pyramid I-BFC. BF = 4 cm, BC = 6 cm.
△BFC and △AGH are homothetic and since AG = 2cm, AH = 6cm × 2/4 = 3cm.
Moreover, △IBF and △IAG are homothetic and a homothetic ratio is BF : AG = 4cm : 2cm = 2 : 1.
As IA = AB = 5cm, it turns out IB = 10cm.
Thus, the volume of triangular pyramid I-BFC is 4 × 6 / 2 × 10 × 1/3 = 40cm3.
The volume of triangular pyramid I-AGH is 2 × 3 / 2 × 5 × 1/3 = 5cm3.
Thus, the volume of a small solid is 40 - 5 = 35cm3.
The volume of a large solid is 240 - 35 = 205cm3.
Therefore, a volume ratio is 205 : 35 = 41 : 7.
Problem 11
Fig.1 and Fig.2 show development views of solids.
The development view of Fig.1 consists of one rectangle, two equilateral triangles, and two trapezoids.
The development view of Fig.2 consists of four equilateral triangles.
Find the volume ratio of the solid made by assembling Fig.1 and the solid made by assembling Fig.2.
Answer
13 : 1
Solution
Fig.3 is the solid made by assembling Fig.1.
Fig.4 is the solid made by assembling Fig.2.
The area of the shadow portion in the figure is set to A cm2.
The volume of Fig.3 is A × (40 + 40 + 50) / 3 = A × 130/3.
The volume of Fig.4 is A × 5 × 1/3 × 2 = A × 10/3.
Therefore the volume ratio of Fig.3 and Fig.4 is 130/3 : 10/3 =13 : 1.
Problem 12
3 sets of sides of the hexagon in the figure faced each other are parallel.
Problem 13
As for each of 3 sets, the ratio of the length of a short side and a long side is 1 : 3.
Find the area ratio of the area of a shadow area, and the area of a hexagon.
Answer
13 : 22
Solution
As shown in Fig. 1, three triangles are made on the outside of a hexagon by drawing an auxiliary line and three triangles are congruent.
△ADE and △ABC are homothetic and a homothetic ratio is DE : BC = 1 : (3 + 1 + 1) = 1 : 5.
An area ratio is 1 × 1 : 5 × 5 = 1 : 25.
If the area of △ADE is set to 1, △ABC is 25.
Comparing the area of △ADE and △DEH, since FG = AE, AE : EH = 1 : 3 and an area ratio is 1 : 3.
The area of △DEH is 3.
In the same way, the area of △FGD and △HGI is also 3.
Therefore, the area of a shadow area is 25 - (1 + 3) × 3 = 13.
The area of a hexagon is 25 - 1 × 3 = 22.
In a figure, AB and CD are vertical.
AE = 24 cm, BE = 6 cm, CE = 18 cm, DE = 8 cm.
The area of a circle is 785 cm2.
Find the sum of the area of a shadow area.
Answer
302.5 cm2
Solution
As shown in Fig. 1, the auxiliary line of HI and GF is drawn to be that AB = HI and CD = GF.
EJ =24 - 6 =18cm. EL =18 - 8 = 10cm.
The area of rectangle EJKL is 18 × 10 =180cm2.
If equivalence transfer of the part of a shadow area is carried out as shown in Fig. 1, the area of a shadow area and the area of a red portion will become equal.
Therefore, since the area of a shadow area is 1/2 of the remaining area after the area of the rectangle EJKL being subtracted from the area of a circle, the area of a shadow area is (785 - 180) × 12 = 302.5cm2.
Problem 14
In a figure, AG is vertical to BC.
The points D and E are the middle points of AG and AC, respectively.
The point F is an intersection of BE and CD.
The length of DE is 2 cm.
The area of triangle ABC is 36 cm2.
The area of the triangle CEF is 2 cm2.
Find the length of AG.
Answer
Problem 15
4.5 cm
Solution
Since the area of △ABC is known, in order to find the length of AG which is the height of △ABC, the length of BC which is a base should just be known.
Since point E is the middle point of AC, the area of △BCE is a half of the area of △ABC, and it is 36 cm2 / 2 = 18 cm2.
Since the area of △CEF is 2 cm2, the area of △CBF is 18 - 2 = 16 cm2.
Since the area ratio of △CEF and △CBF is 2 : 16 = 1 : 8 and the height of both triangle is same, EF : BF which is a base of △CEF and △CBF is also 1 : 8.
△DEF and △BCF are homothetic and a homothetic ratio is EF : BF = 1 : 8.
DE : BC is also 1 : 8.
As DE = 2cm, BC = 2 cm × 8/1 = 16 cm.
Therefore, AG = 36 cm2 × 2 / 16 cm = 4.5 cm.
Therefore, AG = 36 cm2 × 2 / 16 cm = 4.5 cm.
Problem 15
There is a right-angled isosceles triangle ABC.
There is another right triangle DBE width is shorter than ABC by 3 cm and length is longer than ABC by 6 cm.
The figure shows two triangles overlap.
The point F is an intersection of AC and DE.
The ratio of the area of triangle CEF to ADF is 3 : 8.
Find the area of triangle ABC.
Answer
220.5 cm2
Solution
Line EG is drawn in parallel with AB from E and △GEC is created.
△GEC is also a right-angled isosceles triangle and GE = CE = 3cm.
Moreover, since GE and DB are parallel, △ADF and △GEF are homothetic and a homothetic ratio is DA : GE = 6cm : 3cm = 2 : 1.
An area ratio is 2 × 2 : 1 × 1 = 4 : 1.
The area ratio of △ADF and △CEF is 8 : 3 and the area ratio of △ADF and △GEF is 4 : 1 = 8 : 2.
When the area of △GEF is set to 2, the area of △GCE is 3 - 2 = 1.
Thus, the area ratio of △GEF and △GCE is 2 : 1.
FG : GC is also 2 : 1.
Next, as shown in Fig. 2, FH is drawn in parallel with GE from F.
Since FG : GC = 2 : 1, HE : EC is also 2 : 1, then HE = 3cm × 2 = 6cm.
Moreover, as for the homothetic ratio of △ADF and △GEF is 2 : 1, BH : HE which is a ratio of the height of two triangles is also 2 : 1.
Thus, it is understood that BH = HE × 2 = 6cm × 2 =12cm.
Therefore, as BC = AB = 12 + 6 + 3 = 21cm, the area of△ABC is 21 × 21 / 2 = 220.5cm2.
