Time : 45 minutes
Passing mark : 70%
Answer : End of the problem
Problem 1
Problem 2
(2)
Problem 5
(2) Find N in the tournament in which the 6th round game is the final game.
(3) N was 4083 in the tournament of a certain number of teams.
In the tournament of twice as many number of teams as this, N is to be 8178.
Find the number of teams when N is 4083.
Problem 6
Find X, Y and Z.
(1)
16/3 × 19/4 - 21/8 / 7/12 = X
(2)
5 / (Y × 1.75 - 0.625) = 2.5
(3)
2.3 minutes + 132 seconds - 1 minute and 18 seconds = Z minutes
Problem 2
(1)
There are train A of 400m in length and train B of 240m in length.
It takes eight seconds while both trains start passing each other and leaving completely.
It takes 16 seconds while train A catches up train B and passing by completely.
Find the speed per hour of train A.
(2)
Taro bought in 150 goods whose one cost price is 3000 yen.
(3)
He started to sell goods at the fixed price with the expected profit of 60 percent of the cost price, but only 50 pieces were sold.
Then, he decided to make the price of remaining goods discounted by 20 percent of the fixed price and sell them.
Find the number of the goods at least which must be sold at the discounted price in order that Taro may not lose.
Find the number of the goods at least which must be sold at the discounted price in order that Taro may not lose.
(3)
A quadrangle ABCD is a parallelogram in the figure.
Find the area of the shadow portion.
Problem 3
Problem 4
Find the area of the shadow portion.
Problem 3
As shown in Fig. 1, there is cube ABCD-EFGH whose length of one side is 6 cm.
The points I and J are middle points of the sides AE and BF, respectively.
The points L and K are on the sides DH and CG, respectively and DL : LH = CK : KG.
This cube is cut by the plane which passes along four point I, J, K, and L, and it is divided into two solids.
Answer the following questions.
(1) When DL : LH = 1 : 2, find the volume of the solid including the vertex A.
(2) Draw the sides of the figure of the cut surface of (1) in the development view of Fig. 2.
(3) The ratio of the volume of a solid including the vertex A and a solid including the vertex E was 7 : 9.
Find the length of DL.
The points I and J are middle points of the sides AE and BF, respectively.
The points L and K are on the sides DH and CG, respectively and DL : LH = CK : KG.
This cube is cut by the plane which passes along four point I, J, K, and L, and it is divided into two solids.
Answer the following questions.
(1) When DL : LH = 1 : 2, find the volume of the solid including the vertex A.
(2) Draw the sides of the figure of the cut surface of (1) in the development view of Fig. 2.
(3) The ratio of the volume of a solid including the vertex A and a solid including the vertex E was 7 : 9.
Find the length of DL.
Problem 4
There are vessels of pillar form A of 20cm in height and B of 6cm in height.
The ratio of radius of the bottom in vessel A and B is 3 : 2.
There is water in A up to 20㎝ in height and is no water in B.
B is kept upward and sunk in water of A with the bottom of B being parallel to the bottom of A.
The thickness of the vessel is not considered.
Answer the following questions.
(1) Find the height of the surface of the water in A when the water just begins to enter in B.
(2) After the water began to enter in B, the height of the surface of the water in A became 12cm.
At this time, find the height of the surface of the water in B measured from the bottom of B.
Problem 5
The tournament game in which number of teams participated is multiplied some 2 like 2, 4, 8, 16, -----, is considered.
It is expressed as N which is the number added all of several numbers that shows which round game is each game held on one tournament.
For example, when the number of teams is 8, the tournament becomes as shown in the figure.
As for the number of games, there are four games in the 1st round, two games in the 2nd round and one game in the 3rd round.
Therefore, when the number in ○ is added, N = 1 + 1 + 1 + 1 + 2 + 2 + 3 = 11.
It is expressed as N which is the number added all of several numbers that shows which round game is each game held on one tournament.
For example, when the number of teams is 8, the tournament becomes as shown in the figure.
As for the number of games, there are four games in the 1st round, two games in the 2nd round and one game in the 3rd round.
Therefore, when the number in ○ is added, N = 1 + 1 + 1 + 1 + 2 + 2 + 3 = 11.
(1) Find N in the tournament whose number of teams is 16.
(2) Find N in the tournament in which the 6th round game is the final game.
(3) N was 4083 in the tournament of a certain number of teams.
In the tournament of twice as many number of teams as this, N is to be 8178.
Find the number of teams when N is 4083.
Problem 6
There is a ship which moves at 45 m/m with a load and moves at 60 m/m without a load under no flow of the water.
This ship goes back and forth the river whose speed of a flow is 15 m/m.
(1) One day this ship carried the load from A point to downstream B point.
When the load was taken down at B point and it moved to downstream C point further, it took a total of 2 hours.
It is 7.8 km from A point to C point.
Find the distance from A point to B point.
The time concerning taking down a load is not considered.
(2) On the other day, although the load has been dropped into the river as this ship is carrying the load toward downstream E point from D point, it moved as it is without noticing.
However, it noticed that it dropped before arriving at E point, it changed the direction and it went to the upper stream.
The load which has flowed from the dropped point were picked up, the ship changed the direction again and went to E point.
As a result, it took time 12 minutes more to carry the load than the schedule.
Find the time until the ship noticed dropping after dropping the load.
Time to change direction of the ship and time to pick up loads are not considered.
When the load was taken down at B point and it moved to downstream C point further, it took a total of 2 hours.
It is 7.8 km from A point to C point.
Find the distance from A point to B point.
The time concerning taking down a load is not considered.
(2) On the other day, although the load has been dropped into the river as this ship is carrying the load toward downstream E point from D point, it moved as it is without noticing.
However, it noticed that it dropped before arriving at E point, it changed the direction and it went to the upper stream.
The load which has flowed from the dropped point were picked up, the ship changed the direction again and went to E point.
As a result, it took time 12 minutes more to carry the load than the schedule.
Find the time until the ship noticed dropping after dropping the load.
Time to change direction of the ship and time to pick up loads are not considered.
<Answer>
Problem 1
Problem 2
(2)
Problem 5
(2) Find N in the tournament in which the 6th round game is the final game.
(3) N was 4083 in the tournament of a certain number of teams.
In the tournament of twice as many number of teams as this, N is to be 8178.
Find the number of teams when N is 4083.
Find X, Y and Z.
(1)
16/3 × 19/4 - 21/8 / 7/12 = X
(2)
5 / (Y × 1.75 - 0.625) = 2.5
(3)
2.3 minutes + 132 seconds - 1 minute and 18 seconds = Z minutes
Answer
(1) 125/6
(2) 3/2
(3) 3.2
Problem 2
(1)
There are train A of 400m in length and train B of 240m in length.
It takes eight seconds while both trains start passing each other and leaving completely.
It takes 16 seconds while train A catches up train B and passing by completely.
Find the speed per hour of train A.
Answer
216 km/h
Solution
Sum of speed of both trains is (400 + 240) / 8 = 80 m/s.
Difference of speed of both trains is (400 + 240) / 16 = 40 m/s.
The speed per second of A is (80 + 40) / 2 = 60 m/s.
The speed per hour is 60 × 3.6 = 216 km/h.
(2)
Taro bought in 150 goods whose one cost price is 3000 yen.
He started to sell goods at the fixed price with the expected profit of 60 percent of the cost price, but only 50 pieces were sold.
Then, he decided to make the price of remaining goods discounted by 20 percent of the fixed price and sell them.
Find the number of the goods at least which must be sold at the discounted price in order that Taro may not lose.
Answer
(3)
55 pieces
Solution
Total cost price is 3000 × 150 = 450000 yen.
List price is 3000 × 1.6 = 4800 yen and total sales of the list price is 4800 × 50 = 240000 yen.
450000 - 240000 = 210000 yen.
Discounted price is 4800 0.8 = 3840 yen.
210000 / 3840 = 54.6----.
Thus Taro must sell 55 pieces.
(3)
A quadrangle ABCD is a parallelogram in the figure.
Find the area of the shadow portion.
Find the area of the shadow portion.
Answer
Problem 3
2 cm2
Solution
The height of the shadow triangle is 4 cm × 3/(3+6) = 4/3 cm.
Thus the area = 3 × 4/3 × 1/2 = 2 cm2.
Problem 3
As shown in Fig. 1, there is cube ABCD-EFGH whose length of one side is 6 cm.
The points I and J are middle points of the sides AE and BF, respectively.
The points L and K are on the sides DH and CG, respectively and DL : LH = CK : KG.
This cube is cut by the plane which passes along four point I, J, K, and L, and it is divided into two solids.
Answer the following questions.
(1) When DL : LH = 1 : 2, find the volume of the solid including the vertex A.
(2) Draw the sides of the figure of the cut surface of (1) in the development view of Fig. 2.
(3) The ratio of the volume of a solid including the vertex A and a solid including the vertex E was 7 : 9.
Find the length of DL.
The points I and J are middle points of the sides AE and BF, respectively.
The points L and K are on the sides DH and CG, respectively and DL : LH = CK : KG.
This cube is cut by the plane which passes along four point I, J, K, and L, and it is divided into two solids.
Answer the following questions.
(1) When DL : LH = 1 : 2, find the volume of the solid including the vertex A.
(2) Draw the sides of the figure of the cut surface of (1) in the development view of Fig. 2.
(3) The ratio of the volume of a solid including the vertex A and a solid including the vertex E was 7 : 9.
Find the length of DL.
Answer
This statement means that the area ratio of trapezoid DAIL : HEIL = 7 : 9.
This also means that (DL+AI) : (HL+EI) = 7 : 9.
DH + AE = 6 + 6 = 12 cm.
DL + AI = 12 × 7/(7+9) = 21/4 cm.
As AI = 3cm, DL = 21/4 - 3 = 9/4 cm.
Problem 4
There are vessels of pillar form A of 20cm in height and B of 6cm in height.
The ratio of radius of the bottom in vessel A and B is 3 : 2.
There is water in A up to 20㎝ in height and is no water in B.
B is kept upward and sunk in water of A with the bottom of B being parallel to the bottom of A.
The thickness of the vessel is not considered.
Answer the following questions.
(1) 90 cm3
(2)
(2)
(3) 9/4 cm
Solution
(1)
DL = 6 cm × 1/(1+2) = 2 cm.
The volume = (2 + 3) × 6 / 2 × 6 = 90 cm3.
(2)
Write vertex as shown below to find the cut lines.
(3)
The ratio of the volume of a solid including the vertex A and a solid including the vertex E was 7 : 9.This statement means that the area ratio of trapezoid DAIL : HEIL = 7 : 9.
This also means that (DL+AI) : (HL+EI) = 7 : 9.
DH + AE = 6 + 6 = 12 cm.
DL + AI = 12 × 7/(7+9) = 21/4 cm.
As AI = 3cm, DL = 21/4 - 3 = 9/4 cm.
Problem 4
The ratio of radius of the bottom in vessel A and B is 3 : 2.
There is water in A up to 20㎝ in height and is no water in B.
B is kept upward and sunk in water of A with the bottom of B being parallel to the bottom of A.
The thickness of the vessel is not considered.
Answer the following questions.
(1) Find the height of the surface of the water in A when the water just begins to enter in B.
(2) After the water began to enter in B, the height of the surface of the water in A became 12cm.
At this time, find the height of the surface of the water in B measured from the bottom of B.
Answer
(1) 38/3 cm
(2) 1.5 cm
Solution
(1)
The bottom area of vessel A and B is set to be 3 × 3 = 9 and 2 × 2 = 4, respectively.
The height of the water surface of Fig.2 below = (9 × 10cm + 4 × 6cm) / 9 = 114/9 = 38/9 cm.
(2)
In the FIg.3 below, 9 × 12cm - X = 9 × 10cm = 90.
X = 108 - 90 = 18.
18 / 4 = 4.5 cm.
Therefore the height of water surface in B is 6 - 4.5 = 1.5 cm.
Problem 5
The tournament game in which number of teams participated is multiplied some 2 like 2, 4, 8, 16, -----, is considered.
It is expressed as N which is the number added all of several numbers that shows which round game is each game held on one tournament.
For example, when the number of teams is 8, the tournament becomes as shown in the figure.
As for the number of games, there are four games in the 1st round, two games in the 2nd round and one game in the 3rd round.
Therefore, when the number in ○ is added, N = 1 + 1 + 1 + 1 + 2 + 2 + 3 = 11.
It is expressed as N which is the number added all of several numbers that shows which round game is each game held on one tournament.
For example, when the number of teams is 8, the tournament becomes as shown in the figure.
As for the number of games, there are four games in the 1st round, two games in the 2nd round and one game in the 3rd round.
Therefore, when the number in ○ is added, N = 1 + 1 + 1 + 1 + 2 + 2 + 3 = 11.
(1) Find N in the tournament whose number of teams is 16.
(2) Find N in the tournament in which the 6th round game is the final game.
(3) N was 4083 in the tournament of a certain number of teams.
In the tournament of twice as many number of teams as this, N is to be 8178.
Find the number of teams when N is 4083.
Answer
Problem 6
(1) 26
(2) 120
(3) 2048
(2) 120
(3) 2048
Solution
(1)
The number of games in 1st round is 16 / 2 = 8.
2nd round is 8 / 2 = 4.
3rd round is 4 / 2 = 2.
4th round is 2 / 2 = 1.
Therefore N is calculated as 1 × 8 + 2 × 4 + 3 × 2 + 4 × 1 = 26.
(2)
Organize the conditions of the problem in the table below.
| Number of team | 2 | 4 | 8 | 16 | 32 | 64 |
| Final round | 1st | 2nd | 3rd | 4th | 5th | 6th |
| N | 1 | 4 | 11 | 26 | 57 |
The number of N is increased by (Number of team) - 1 as shown above.
Therefore N in blank is 57 + (64 - 1) =120.
(3)
When N = 4083, number of team is set to be X.
The number of team when N = 8178 is X × 2.
According to (2), 8178 - 4083 = 4095 = X × 2 - 1.
X = (4095 + 1) / 2 = 2048.
Problem 6
There is a ship which moves at 45 m/m with a load and moves at 60 m/m without a load under no flow of the water.
This ship goes back and forth the river whose speed of a flow is 15 m/m.
(1) One day this ship carried the load from A point to downstream B point.
When the load was taken down at B point and it moved to downstream C point further, it took a total of 2 hours.
It is 7.8 km from A point to C point.
Find the distance from A point to B point.
The time concerning taking down a load is not considered.
(2) On the other day, although the load has been dropped into the river as this ship is carrying the load toward downstream E point from D point, it moved as it is without noticing.
However, it noticed that it dropped before arriving at E point, it changed the direction and it went to the upper stream.
The load which has flowed from the dropped point were picked up, the ship changed the direction again and went to E point.
As a result, it took time 12 minutes more to carry the load than the schedule.
Find the time until the ship noticed dropping after dropping the load.
Time to change direction of the ship and time to pick up loads are not considered.
This ship goes back and forth the river whose speed of a flow is 15 m/m.
(1) One day this ship carried the load from A point to downstream B point.
When the load was taken down at B point and it moved to downstream C point further, it took a total of 2 hours.
It is 7.8 km from A point to C point.
Find the distance from A point to B point.
The time concerning taking down a load is not considered.
(2) On the other day, although the load has been dropped into the river as this ship is carrying the load toward downstream E point from D point, it moved as it is without noticing.
However, it noticed that it dropped before arriving at E point, it changed the direction and it went to the upper stream.
The load which has flowed from the dropped point were picked up, the ship changed the direction again and went to E point.
As a result, it took time 12 minutes more to carry the load than the schedule.
Find the time until the ship noticed dropping after dropping the load.
Time to change direction of the ship and time to pick up loads are not considered.
Answer
(1) 4.8 km
(2) 8 minutes
(2) 8 minutes
Solution
(1)
The speed from A to B is 45 + 15 = 60 m/m.
The speed from B to C is 60 + 15 = 75 m/m.
The time of moving at the speed of 60 m/m is (75 × 120 - 7800) / (75 - 60) = 80 minutes.
Therefore the distance from A to B is 60 × 80 = 4800 m = 4.8 km.
(2)
In the figure below,
F is a point the ship dropped the load,
G is a point the ship returned,
H is a point the load was flowing when the ship returned,
I is a point the ship picked up the load.
FG : FH = 75 : 15 = 5 : 1.
GI : HI = 45 : 15 = 3 : 1.
As HG = 5 - 1 = 3 + 1 = 4, FH : HI : GI = 1 : 1 : 3.
The time the ship moved F-G-I is 5 / 75 + 3 / 45 = 2/15.
The time the ship would move F-I with the load is (1 + 1) / 60 = 1/30.
2/15 - 1/30 = 1/10 which is equivalent to 12 minutes.
1 = 12 / 1/10 = 120 minutes.
As the time from F to G is 5 / 75 = 1/15, 120 × 1/15 = 8 minutes.
