Time : 40 minutes
Passing mark : 70 %
Answer : End of the problem
Problem 1
Problem 2
Problem 5
Problem 6
① Find X
5/6 - {3.5 / (4.875 + X)- 1/9} = 0.5
② Find Y
2 + 4 + 6 + 8 + ----- + 48 + 50 = Y × Y + Y
Problem 2
Many people gathered in the party.
Problem 3
I prepared three kinds of fruit to be distributed to each person.
Rate of allocation was that one melon for seven people, one apple for three people and one pear for four people.
There was no rest and all fruit could be distributed. The total number of the fruit distributed was 183.
How many people were there who gathered in this party?
Problem 3
There are 2400 m between A point and B point.
Problem 4
Taro left at 60 m/m and Jiro left at 50 m/m A point at the same time and they went to B point.
When Taro arrived at intermediate C point, it was 320 m between Taro and Jiro.
(1) Taro went to B point as he is.
When Jiro arrives at C point, find the distance with Taro.
(2) Taro noticed the thing left behind at C point and he returned to A point with the same speed.
Without resting he left A point at 270 m/m by bicycle and went to B point.
Jiro returned from B point without resting and went to A point with the same speed.
Find the distance from A point of the position where Taro and Jiro meet.
(1) Taro went to B point as he is.
When Jiro arrives at C point, find the distance with Taro.
(2) Taro noticed the thing left behind at C point and he returned to A point with the same speed.
Without resting he left A point at 270 m/m by bicycle and went to B point.
Jiro returned from B point without resting and went to A point with the same speed.
Find the distance from A point of the position where Taro and Jiro meet.
Problem 4
In the water tank of the rectangular prism of Fig. 1, four solids as shown in Fig. 2 are placed for the shadow face to be as the bottom.
Then, water is poured into this water tank every 10 cm3/s.
The volume of the solid of Fig. 2 can be calculated as (base area) × (height) × 1/3.
(1) When the depth of water is to be 2 cm, find the area of the water surface.
(2) After (1), find the time when it will take until the water fills up the tank. The volume of the solid of Fig. 2 can be calculated as (base area) × (height) × 1/3.
Problem 5
There is a ground of a rectangle as shown in a figure.
Taro and Jiro left A point at the same time and Taro walked to B point and Jiro walked to E point, respectively.
When they walked with the same speed, they arrived at B point and E point at the same time.
Jiro who arrived at E point turned to the right so that Taro might be seen to the front.
Find the angle at which Jiro turned to the right.
Taro and Jiro left A point at the same time and Taro walked to B point and Jiro walked to E point, respectively.
When they walked with the same speed, they arrived at B point and E point at the same time.
Jiro who arrived at E point turned to the right so that Taro might be seen to the front.
Find the angle at which Jiro turned to the right.
Problem 6
A certain solid was made as shown Fig.1 of eight equilateral triangles.
Also as shown in Fig. 1, the point in the middle of each side of this solid was connected by the straight line and this solid was developed as shown in Fig. 2.
Draw the remaining straight lines drawn on the solid on the development view of Fig. 2.
AB of Fig. 2 is the same as AB of Fig. 1.
Problem 7
There were one yen and five yen coins in total 55 pieces.
They were exchanged into ten yen coins so that the number of coins might decrease as many as possible.
The number of coins became 15 pieces and the number of one yen coins was 2.
The one yen coin which suited first should ask for number of sheets.
Find the number of pieces of one yen coin at first.
Problem 8
Draw the remaining straight lines drawn on the solid on the development view of Fig. 2.
AB of Fig. 2 is the same as AB of Fig. 1.
Problem 7
They were exchanged into ten yen coins so that the number of coins might decrease as many as possible.
The number of coins became 15 pieces and the number of one yen coins was 2.
The one yen coin which suited first should ask for number of sheets.
Find the number of pieces of one yen coin at first.
Problem 8
Six persons, A, B, C, D, E, and F, sat down groups of three face to face as shown in the figure.
These six persons explained how to sit down as follows.
A : I sat on the edge.
B : A was on the right hand side of the front looked at from me.
C : I sat on the front of E.
D : F was sitting on the left edge of the opposite side looked at from me.
E : C was sitting on the same side as F.
F : I and B were not facing each other.
Write the seat of A to F in a figure.
Problem 9
These six persons explained how to sit down as follows.
A : I sat on the edge.
B : A was on the right hand side of the front looked at from me.
C : I sat on the front of E.
D : F was sitting on the left edge of the opposite side looked at from me.
E : C was sitting on the same side as F.
F : I and B were not facing each other.
Write the seat of A to F in a figure.
Problem 9
There is a circle with radius 6 cm as shown in the figure and there are 12 points are on the circumference arranged with the same interval.
① Find X
5/6 - {3.5 / (4.875 + X)- 1/9} = 0.5
② Find Y
2 + 4 + 6 + 8 + ----- + 48 + 50 = Y × Y + Y
② 25
(2+50) × 25 / 2 = 52 × 25 / 2 = 26 × 25 = (25+1) × 25 = 25 × 25 + 25
Problem 2
Many people gathered in the party.
I prepared three kinds of fruit to be distributed to each person.
Rate of allocation was that one melon for seven people, one apple for three people and one pear for four people.
There was no rest and all fruit could be distributed. The total number of the fruit distributed was 183.
How many people were there who gathered in this party?
Answer
252 people
Solution
Because I distribute at the rate that one melon to seven people, one apple to three people and one pear to four people, it is found that the ratio of the number of melon, apple and pear is 1/3 : 1/4 : 1/7 = 28 : 21 : 12.
Then the number of apple is identified as 183 × 28 /(28+21+12) = 84 pieces.
Therefore, the number of people of the party was 84 × 3 = 252 people.
Problem 3
There are 2400 m between A point and B point.
Taro left at 60 m/m and Jiro left at 50 m/m A point at the same time and they went to B point.
When Taro arrived at intermediate C point, it was 320 m between Taro and Jiro.
(1) Taro went to B point as he is.
When Jiro arrives at C point, find the distance with Taro.
(2) Taro noticed the thing left behind at C point and he returned to A point with the same speed.
Without resting he left A point at 270 m/m by bicycle and went to B point.
Jiro returned from B point without resting and went to A point with the same speed.
Find the distance from A point of the position where Taro and Jiro meet.
Answer
(1) 64 m
(2) 1350 m
Solution
(1)
Speed ratio of Taro and Jiro = 60 : 50 = 6 : 5.
While Jiro walks 320m, Taro walks 320 × 6/5 = 384m.
384 - 320 = 64 m
(2)
The time when Taro noticed the thing behind is 320 / (60 - 50) = 32 minutes after.
The distance between Taro and Jiro when Taro returned back to A is 2400 × 2 - 50 × (32 × 2) = 4800 - 3200 = 1600m.
Speed ratio of Taro : Jiro = 270 : 50 = 27 : 5.
They meet at 1600 × 27/(27+50) = 1600 × 27/32 = 1350m.
Problem 4
In the water tank of the rectangular prism of Fig. 1, four solids as shown in Fig. 2 are placed for the shadow face to be as the bottom.
Then, water is poured into this water tank every 10 cm3/s.
(1) When the depth of water is to be 2 cm, find the area of the water surface.
(2) After (1), find the time when it will take until the water fills up the tank.
The volume of the solid of Fig. 2 can be calculated as (base area) × (height) × 1/3.
Problem 5
There is a ground of a rectangle as shown in a figure.
Taro and Jiro left A point at the same time and Taro walked to B point and Jiro walked to E point, respectively.
When they walked with the same speed, they arrived at B point and E point at the same time.
Jiro who arrived at E point turned to the right so that Taro might be seen to the front.
Find the angle at which Jiro turned to the right.
Answer
105 degrees
Solution
AE = AB = 100m
AD : AE = 50 : 100 = 1 : 2
As △ADE is half of equilateral triangle, ∠DAE = 60 degrees.
∠BAE = 90 - 60 =30 degrees.
∠AEB = (180 - 30) / 2 = 75 degrees
180 - 75 = 105 degrees
Problem 6
A certain solid was made as shown Fig.1 of eight equilateral triangles.
Also as shown in Fig. 1, the point in the middle of each side of this solid was connected by the straight line and this solid was developed as shown in Fig. 2.
Draw the remaining straight lines drawn on the solid on the development view of Fig. 2.
AB of Fig. 2 is the same as AB of Fig. 1.
The height of the equilateral triangle whose length of one side is 6 cm is assumed to be 5.2 cm and Pi is assumed to be 3.14.
(1) Find the length of AB.
(2) Find the area of the shadow portion.
<Answer>
Problem 1
① Find X
5/6 - {3.5 / (4.875 + X)- 1/9} = 0.5
② Find Y
2 + 4 + 6 + 8 + ----- + 48 + 50 = Y × Y + Y
Answer
① 3② 25
(2+50) × 25 / 2 = 52 × 25 / 2 = 26 × 25 = (25+1) × 25 = 25 × 25 + 25
Many people gathered in the party.
I prepared three kinds of fruit to be distributed to each person.
Rate of allocation was that one melon for seven people, one apple for three people and one pear for four people.
There was no rest and all fruit could be distributed. The total number of the fruit distributed was 183.
How many people were there who gathered in this party?
Answer
252 people
Solution
Because I distribute at the rate that one melon to seven people, one apple to three people and one pear to four people, it is found that the ratio of the number of melon, apple and pear is 1/3 : 1/4 : 1/7 = 28 : 21 : 12.
Then the number of apple is identified as 183 × 28 /(28+21+12) = 84 pieces.
Therefore, the number of people of the party was 84 × 3 = 252 people.
There are 2400 m between A point and B point.
Taro left at 60 m/m and Jiro left at 50 m/m A point at the same time and they went to B point.
When Taro arrived at intermediate C point, it was 320 m between Taro and Jiro.
(1) Taro went to B point as he is.
When Jiro arrives at C point, find the distance with Taro.
(2) Taro noticed the thing left behind at C point and he returned to A point with the same speed.
Without resting he left A point at 270 m/m by bicycle and went to B point.
Jiro returned from B point without resting and went to A point with the same speed.
Find the distance from A point of the position where Taro and Jiro meet.
Answer
(1) 64 m
(2) 1350 m
Solution
(1)
Speed ratio of Taro and Jiro = 60 : 50 = 6 : 5.
While Jiro walks 320m, Taro walks 320 × 6/5 = 384m.
384 - 320 = 64 m
(2)
The time when Taro noticed the thing behind is 320 / (60 - 50) = 32 minutes after.
The distance between Taro and Jiro when Taro returned back to A is 2400 × 2 - 50 × (32 × 2) = 4800 - 3200 = 1600m.
Speed ratio of Taro : Jiro = 270 : 50 = 27 : 5.
They meet at 1600 × 27/(27+50) = 1600 × 27/32 = 1350m.
In the water tank of the rectangular prism of Fig. 1, four solids as shown in Fig. 2 are placed for the shadow face to be as the bottom.
Then, water is poured into this water tank every 10 cm3/s.
(1) When the depth of water is to be 2 cm, find the area of the water surface.
(2) After (1), find the time when it will take until the water fills up the tank.
The volume of the solid of Fig. 2 can be calculated as (base area) × (height) × 1/3.
Answer
DB : GE = AD : AG = 8 : 6 = 4 : 3
GE = 6cm × 3/4 = 9/2 cm
(1) 103.5 cm2
(2) 78.3 seconds
Solution
(1)DB : GE = AD : AG = 8 : 6 = 4 : 3
GE = 6cm × 3/4 = 9/2 cm
The area of △GEF = 9/2 × 9/2 / 2 = 81/8 cm2
The area of bottom of water tank is 12 × 12 = 144 cm2
144 - 81/8 × 4 = 103.5 cm2
(2)
The volume of triangular pyramid APQRis 81/8 × (8-2) × 1/3 = 81/4 cm3
12 × 12 × (8 - 2) - 81/4 × 4 = 864 - 81 = 783 cm3
The time is 783 / 10 = 78.3 seconds.There is a ground of a rectangle as shown in a figure.
Taro and Jiro left A point at the same time and Taro walked to B point and Jiro walked to E point, respectively.
When they walked with the same speed, they arrived at B point and E point at the same time.
Jiro who arrived at E point turned to the right so that Taro might be seen to the front.
Find the angle at which Jiro turned to the right.
105 degrees
Solution
AE = AB = 100m
AD : AE = 50 : 100 = 1 : 2
As △ADE is half of equilateral triangle, ∠DAE = 60 degrees.
∠BAE = 90 - 60 =30 degrees.
∠AEB = (180 - 30) / 2 = 75 degrees
180 - 75 = 105 degrees
A certain solid was made as shown Fig.1 of eight equilateral triangles.
Also as shown in Fig. 1, the point in the middle of each side of this solid was connected by the straight line and this solid was developed as shown in Fig. 2.
Draw the remaining straight lines drawn on the solid on the development view of Fig. 2.
AB of Fig. 2 is the same as AB of Fig. 1.
Answer
There were one yen and five yen coins in total 55 pieces.
They were exchanged into ten yen coins so that the number of coins might decrease as many as possible.
The number of coins became 15 pieces and the number of one yen coins was 2.
The one yen coin which suited first should ask for number of sheets.
Find the number of pieces of one yen coin at first.
Problem 8
Solution
Attach alphabet on each vertex and corner as shown in Fig.3 and attach in the development view as well as shown in Fig.4
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Problem 7There were one yen and five yen coins in total 55 pieces.
They were exchanged into ten yen coins so that the number of coins might decrease as many as possible.
The number of coins became 15 pieces and the number of one yen coins was 2.
The one yen coin which suited first should ask for number of sheets.
Find the number of pieces of one yen coin at first.
Answer
37 pieces
Solution
Sum total number of one yen and five yen coins is 55 means total amount of money is to be odd number since number of one kind of coin is odd number and another is even number.
Since the number of coins after exchange is 15 pieces and that of one yen coin is 2 pieces, total number of five and ten yen coins are 15 -2 = 13 pieces.
Number of five yen coin is 0 or 1 pieces.
Sum total is 1 × 2 + 10 × 13 = 132 yen or 1 × 2 + 5 × 1 + 10 × 12 = 127 yen
As 132 is even number, the amount is 127 yen.
The number of one yen coin at first is (5 × 55 - 127) / (5 - 1) = 37 pieces.
Problem 8
Six persons, A, B, C, D, E, and F, sat down groups of three face to face as shown in the figure.
These six persons explained how to sit down as follows.
A : I sat on the edge.
B : A was on the right hand side of the front looked at from me.
C : I sat on the front of E.
D : F was sitting on the left edge of the opposite side looked at from me.
E : C was sitting on the same side as F.
F : I and B were not facing each other.
Write the seat of A to F in a figure.
These six persons explained how to sit down as follows.
A : I sat on the edge.
B : A was on the right hand side of the front looked at from me.
C : I sat on the front of E.
D : F was sitting on the left edge of the opposite side looked at from me.
E : C was sitting on the same side as F.
F : I and B were not facing each other.
Write the seat of A to F in a figure.
Answer
Problem 9
Solution
Two patterns
① ②
Based on two patterns of A and B, considering the explanation of each member, it is found two kinds of seat as in answer.
Problem 9
There is a circle with radius 6 cm as shown in the figure and there are 12 points are on the circumference arranged with the same interval.
The height of the equilateral triangle whose length of one side is 6 cm is assumed to be 5.2 cm and Pi is assumed to be 3.14.
(1) Find the length of AB.
(2) Find the area of the shadow portion.
Answer
(1) 2.2 cm
(2) 64.08 cm2
(2) 64.08 cm2
Solution
(1)
△OAC = △OEF = equilateral triangle central white quadrangle is square.
AD = 5.2cm and BD = half of EF = 6 / 2 = 3cm
AB = 5.2 - 3 = 2.2cm
(2)
(Shadow area) = (Circle area) - (Sum of white area)
= 6 × 6 × 3.14 - {6 × 6 + 6 × 6 × 3.14 × 60/360 - (6 × 5.2 / 2) × 4}
= 113.4 - (36 + 12.96)
= 64.08 cm2
