Time : 60 minutes
Passing mark : 70 %
Answer : End of the problem
Problem 1
(3)
(4)
(5)
(6)
Problem 2
(1) Find the spreed ratio of Taro and Jiro.
Problem 3
Problem 4
Problem 5
Problem 6
Salt solutions 500g with 8% of concentration are contained in the vessel A and salt solutions 500g with unknown concentration are contained in the vessel B.
(1)
Find X.
3% of 0.4 km2 is X a (are).
(2)
Find Y.
(Y + 0.375) × 2/7 + 0.75 =1
(3)
Among fractions whose numerator is 8 and which is between 5/9 and 11/13, find the fractions which cannot be reduced.
(4)
The graph below shows the relation of the volume of the water that entered in the tank and time when the water was poured by A pipe first and by both A and B sometime later.
Water is poured by A pipe and B pipe with a fixed rate, respectively.
Find the volume(L) of the water that is poured by B pipe per minute.
(5)
I decided the list price of an article with expecting a profit of 20 % of the cost price 700 yen.
However it was not sold, I sold it at 12.5% discounted price of the list price.
Find the profit in this case.
(6)
The figure below is a diagram that connects the vertex and the middle point of each side of the square whose one side is 10cm.
Find the area of the shaded part.
Problem 2
While Taro takes five steps forward, Jiro takes six steps forward.
Jiro goes to the distance by five steps where Taro goes by three steps.
When Jiro started and walked 56 steps, Taro pursues Jiro.
(1) Find the spreed ratio of Taro and Jiro.
(2) How many steps does Taro take to catch up Jiro?
Problem 3
Right-angled triangle ABC as shown in a lower figure was rotated in the direction of an arrow by 60 degrees on a center of C.
Answer the following questions. ∠ADC and ∠ACB are 90 degrees.
(1) Find the length of CD.
(2) Find the area of the figure of the shadow area where the line segment AD moved.
Problem 4
The first grader of a certain junior high school will hold a tournament of soccer and basketball.
I took the questionnaire which do they want to play and examined the number of students.
The ratio of the number of students who want to play soccer and basketball 3 : 2.
However, because on the day of tournament three students who had been wanting to play basketball were absent, the ratio of the number of students who played soccer and basketball became 8 : 5.
Answer the following questions.
(1) Find the ratio of the number of students who wanted to play basketball and who actually played basketball.
(2) Find the number of the first graders of this junior high school.
Problem 5
A figure below is the solid R which cylinder Q with 4 cm in diameter EF of the bottom is hollowed out of the cylinder P with 6 cm diameter CD of the bottom and is cut aslant by the plane which passes along the points A and B.
The points C, E, and F and D are on the same straight line.
Answer the following questions.
(1) When the center of the bottom of the cylinders P and Q overlapped, find the volume of the solid R after hollowed out.
(2) When the points C and E overlap, find for the volume of the solid after hollowed out.
Problem 6
Salt solutions were put into the vessel C at a rate of 2 : 3 from A and B.
Salt solutions were put into the vessel D at a rate of 3 : 2 from A and B.
The ratio of the concentration of C and D became 13 : 12.
Answer the following questions.
(1) Find the concentration (%) of the salt solution of the vessel B.
(2) The water of the same weight as the salt solution taken out was put into the vessel B.
The concentration of the salt solution of the vessel B became 8.4%.
Find the weight (g) of the salt solution taken out of the vessel B.
<Answer>
Problem 1
(2)
(3)
(4)
(5)
(6)
Problem 2
(1) Find the spreed ratio of Taro and Jiro.
Problem 3
Problem 4
Problem 5
Problem 6
Answer
(2)
(1)
Find X.
3% of 0.4 km2 is X a (are).
Answer
120
(2)
Find Y.
(Y + 0.375) × 2/7 + 0.75 =1
Answer
1/2
(3)
Among fractions whose numerator is 8 and which is between 5/9 and 11/13, find the fractions which cannot be reduced.
Answer
8/13,
8/11
Solution
5/9 = 8/14.4----
11/13 = 8/9.4----
Between above two fractions, there are 8/14, 8/13, 8/12, 8/11, 8,10.
8/13 and 8/11 cannot be reduced.
(4)
The graph below shows the relation of the volume of the water that entered in the tank and time when the water was poured by A pipe first and by both A and B sometime later.
Water is poured by A pipe and B pipe with a fixed rate, respectively.
Find the volume(L) of the water that is poured by B pipe per minute.
Answer
5.7 L
Solution
16/5 = 3.2 L which is volume per minute from A.
(19.2 - 16) / 3.2 = 1 minute.
In 8 - 6 = 2 minutes 37 - 19.2 = 17.8 L of water from A and B
17.8 / 2 = 8.9 L per minute from A and B.
B is 8.9 - 3.2 = 5.7 L
(5)
I decided the list price of an article with expecting a profit of 20 % of the cost price 700 yen.
However it was not sold, I sold it at 12.5% discounted price of the list price.
Find the profit in this case.
Answer
35 yen
Solution
List price = 700 × (1 + 0.2) = 840 yen.
Discounted price = 840 × (1 - 0.125) = 735 yen.
Profit = 735 - 700 = 35 yen.
(6)
The figure below is a diagram that connects the vertex and the middle point of each side of the square whose one side is 10cm.
Find the area of the shaded part.
Answer
20 cm2
Solution
The figure is transformed by equivalence transfer as shown in figures below.
The area of shaded part is 10 × 10 × 1/5 = 20 cm2.
Problem 2
While Taro takes five steps forward, Jiro takes six steps forward.
Jiro goes to the distance by five steps where Taro goes by three steps.
When Jiro started and walked 56 steps, Taro pursues Jiro.
(1) Find the spreed ratio of Taro and Jiro.
(2) How many steps does Taro take to catch up Jiro?
Answer
(1) 25 : 18
(2) 120 steps
Solution
(1)
Ratio of steps of Taro : Jiro = 5 : 6.
Ratio of step length of Taro : Jiro = 5 : 3.
Speed ration of Taro : Jiro = 5 × 5 : 6 × 3 = 25 : 18.
(2)
Comparing the distance until Taro catch up with Jiro, Taro's distance = 25 and Jiro's = 56 steps + 18.
25 - 18 = 7 is equivatlent to Jiro's 56 steps.
1 = 56 / 7 = Jiro's 8 steps.
25 = 8 × 25 = Jiro's 200 steps.
Number of Taro's steps is 200 × 3/5 = 120 steps.
Problem 3
Right-angled triangle ABC as shown in a lower figure was rotated in the direction of an arrow by 60 degrees on a center of C.
Answer the following questions. ∠ADC and ∠ACB are 90 degrees.
(1) Find the length of CD.
(2) Find the area of the figure of the shadow area where the line segment AD moved.
Answer
(1) 4.8 cm
(2) 6.7824 cm2
Solution
(1)
CD = AC × 8/10 = 6 × 8/10 = 4.8 cm.
(2)
Shadow area = △ACD + Sector ACA´ - (△A´CD´ + Sector DCD´)
= Sector ACA´ - Sector DCD´
= 6 × 6 × 3.14 × 60/360 - 4.8 × 4.8 × 3.14 × 60/360
= (36 - 23.04) × 3.14 × 1/6
= 6.7824
Problem 4
The first grader of a certain junior high school will hold a tournament of soccer and basketball.
I took the questionnaire which do they want to play and examined the number of students.
The ratio of the number of students who want to play soccer and basketball 3 : 2.
However, because on the day of tournament three students who had been wanting to play basketball were absent, the ratio of the number of students who played soccer and basketball became 8 : 5.
Answer the following questions.
(1) Find the ratio of the number of students who wanted to play basketball and who actually played basketball.
(2) Find the number of the first graders of this junior high school.
Answer
(1) 16 : 15
(2) 120 persons
Solution
(1)
The number of students who wanted to play soccer is equal to the number who actually played soccer.
It is set to be 3 × 8 = 24.
The number of students who wanted to play basketball is 24 × 2/3 = 16.
The number of students who actually played basketball is 24 × 5/8 = 15.
(2)
16 - 15 = 1 is equivalent to 3 persons.
Thus the number of the first graders is 3 × (24 + 16) = 120 persons.
Problem 5
A figure below is the solid R which cylinder Q with 4 cm in diameter EF of the bottom is hollowed out of the cylinder P with 6 cm diameter CD of the bottom and is cut aslant by the plane which passes along the points A and B.
The points C, E, and F and D are on the same straight line.
Answer the following questions.
(1) When the center of the bottom of the cylinders P and Q overlapped, find the volume of the solid R after hollowed out.
(2) When the points C and E overlap, find for the volume of the solid after hollowed out.
Answer
(1) 94.2 cm2
(2) 81.64 cm2
Solution
(1)
When combining two solid R, it would be a cylinder with 3 + 9 = 12 cm in height.
Therefore the volume of the solid R = (3 × 3 × 3.14 - 2 × 2 × 3.14) × 12 × 1/2
= (9 - 4) × 3.14 × 6
= 94.2 cm2.
(2)
The volume of the solid before hollowing is 3 × 3 × 3.14 × (9 + 3) × 1/3 = 54 × 3.14 cm3.
In the Fig.2 below, AI = 9 - 3 = 6 cm = IB.
△GHB and △AIB are homothetic.
As GH : HB = AI : IB = 1 : 1, GH = 2 cm.
GF = 2 + 3 = 5 cm.
The volume of the solid hollowed out is 2 × 2 × 3.14 × (9 + 5) × 1/2 = 28 × 3.14 cm3.
Therefore the volume to be found is (54 - 28) × 3.14 = 81.64 cm3.
Problem 6
Salt solutions 500g with 8% of concentration are contained in the vessel A and salt solutions 500g with unknown concentration are contained in the vessel B.
Salt solutions were put into the vessel C at a rate of 2 : 3 from A and B.
Salt solutions were put into the vessel D at a rate of 3 : 2 from A and B.
The ratio of the concentration of C and D became 13 : 12.
Answer the following questions.
(1) Find the concentration (%) of the salt solution of the vessel B.
(2) The water of the same weight as the salt solution taken out was put into the vessel B.
The concentration of the salt solution of the vessel B became 8.4%.
Find the weight (g) of the salt solution taken out of the vessel B.
(1) 12 %
(2) 150 g
Solution
(1)
Fig. 1 shows the condition of C and Fig.2 shows the condition of D.
According to these figures, 13 - 12 = 1 = ③ - ② = ①.
As ③ = 1, 8% + ③ = 8% + 3 = 13.
13 - 3 = 10 = 8%
1 = 0.8%
X = 8% + 0.8% × 5 = 12%
The original amount of salt in B is 500 × 12 = 60 g.
The amount of salt remained in B is 500 × 8.4% = 42 g.
Therefore the amount of salt solution taken out of B is 500 × (60 - 42)/60 = 150 g.
