Time : 50 minutes
Answer is the End of the problem
Problem 1
Problem 2
Problem 3
Problem 4
(2) Find all integers x of two digits that become "[[x]+21] =36.
Problem 5
Problem 6
Answer is the End of the problem
Problem 1
(1) Calculation
1 - {11/12 - (3/8 / 1.5 + 0.6)} × 5/3 - 5/6 =
(2) Find X
1.625 × 4/3 / (1/2 - X) = 39/7
(3) Find Y
{54/13 + 2/3 / ( Y - 8/27)} × 3.25 = 18
(4) When 5/7 is represented in decimal, find the number of the 2014th decimal place.
Problem 2
(1)
In a certain shop, it stocked 90 pieces of cup at 7200 yen.
They sold at the list price of 150 yen per one piece, but some remained unsold.
The profit was 3000 yen.
Find the number of the cut remained unsold.
(2)
The number of students of girls in a certain junior high school is 95% of the number of students of boys.
The number of absentees on a certain day were four and all of them were girls.
The ratio of the number of girls' attendant in this day became 11/12 of the boy.
Find the number of all students of this junior high school.
(3)
10 pieces of dice are piled up on the desk so that the direction of each face might become same as shown in the figure below.
Find the total of the number of dice spots not to be seen from anywhere.
(4)
As shown in the figure below, eight circles with radius 4cm each are lined in touch with the next circle.
When the center of each circle is connected by the line, find the area of the shaded area.
Pi is assumed to be 3.14.
(5)
In the figure below, the area of triangle ADF, triangle DEF, triangle CEF, and triangle BCE is equal.
Find the length of segment AD.
(6)
A salt solution of 400g is in each container A, B, and C and density in A, B , and C is 16%, 8%, and 4% respectively.
When total of 100g salt solution was taken out of container A and B and it added to container C, the density of container C became 5%.
Find the weight of the salt solution taken out of container A.
Problem 3
(1)
There is a container whose all corners are the right angle as shown in the Fig.1 below.
Water is put in 2/3 of the container.
And the container is put on with the shaded face being placed in the bottom as shown in Fig.2.
In this case find the height of the water.
(2)
As shown in the figure below, there is a trapezoid ABCD, and AD and BC are parallel, and AD = 2cm, BC = 3cm.
P and Q are points on the side AB, DC and AP : PB = DQ : QC = 2 : 1.
R is an intersection of PQ and AC.
S is an intersection of CD and the extended line of BR.
T is an intersection of the extended lines of BR and AD.
In this case, find DS : SQ : QC by the ratio of the simplest integer.
Problem 4
Sign [x] is assumed as a sign to calculate the product of each digit of a certain number x like Example.
Example: [46] = 24, [88] = 64
Example: [46] = 24, [88] = 64
Answer the following questions.
(1) Find all integers x of two digits that become "[54]+[35]=[x]".
(2) Find all integers x of two digits that become "[[x]+21] =36.
Problem 5
Cows are pastured at some grassland.
When 6 cows are pastured, grasses are eaten up in 30 days.
When 10 cows are pastured, grasses are eaten up in 12 days.
At this time, answer the next questions.
Noted that grasses grow at a fixed rate every day in this grassland.
In addition the amount of the grass each cow eats per day is same.
(1) Find the number of cow to be pastured without the grass of the grassland being eaten up.
(2) How many cows at least should be pastured in order to eat all the grass of the grassland within six days?
Problem 6
As shown in the figure below, there is a road along the surroundings of a pond.
Taro and Hanako leave A point and B point respectively at the same time.
They go to the park through the post in the middle without returning.
They met 14 minutes after they left.
Taro passes through the point B in 10 minutes after they met and after 24 minutes further they arrived at the park at the same time.
Answer the following questions.
Noted we assume that Taro and Hanako walk at a fixed speed respectively.
(1) How many minutes did Hanako take from A point to the park?
(2) Both of them leave the park at the same time and come back to the starting point without passing the course they came.
Taro walks with the same speed as he came to the park.
For two people back to the original location at the same time, by how many times of first speed should Hanako walk?
Answer
Problem 1
Problem 2
Solution
Problem 3
Problem 4
(2) Find all integers x of two digits that become "[[x]+21] =36.
Problem 5
Problem 6
(1) Calculation
1 - {11/12 - (3/8 / 1.5 + 0.6)} × 5/3 - 5/6 =
Answer
1/18
(2) Find X
1.625 × 4/3 / (1/2 - X) = 39/7
Answer
1/9
(3) Find Y
{54/13 + 2/3 / ( Y - 8/27)} × 3.25 = 18
Answer
7/9
(4) When 5/7 is represented in decimal, find the number of the 2014th decimal place.
Answer
2
Solution
5/7 = 07142857142-----
One cycle is six figures of 714285.
2014 / 6 = 335 and remainder is 4.
The number of the 4th decimal place is 2.
Problem 2
(1)
In a certain shop, it stocked 90 pieces of cup at 7200 yen.
They sold at the list price of 150 yen per one piece, but some remained unsold.
The profit was 3000 yen.
Find the number of the cut remained unsold.
(2)
Answer
22 pieces
Solution
7200 + 3000 = 10200
The number of the cup sold is 10200 / 150 = 68 pieces.
Thus the number of unsold is 90 - 68 = 22 pieces.
The number of students of girls in a certain junior high school is 95% of the number of students of boys.
The number of absentees on a certain day were four and all of them were girls.
The ratio of the number of girls' attendant in this day became 11/12 of the boy.
Find the number of all students of this junior high school.
Answer
234 persons
Solution
The number of boys is set to be X.
X : X × 0.95 - 4 = 12 : 11
X × 12 × 0.95 - 48 = X × 11
Thus X × 0.4 = 48
Then X = 120.
120 + 120 × 0.95 = 120 + 114 = 234
(3)
10 pieces of dice are piled up on the desk so that the direction of each face might become same as shown in the figure below.
Find the total of the number of dice spots not to be seen from anywhere.
Answer
120
Solution
The number of face of each die is six.
Total number of faces of 10 dice is 6 × 10 = 60.
The number of faces can be seen is (1 + 2 + 3) × 5 = 30.
The number of faces being attached on the desk is six.
The number of points where the face of dice are attached each other is (30 - 6) / 2 = 12.
Therefore the total of the number of dice spots not to be seen from anywhere is 7 × 12 + 6 × 6 = 120.
(4)
As shown in the figure below, eight circles with radius 4cm each are lined in touch with the next circle.
When the center of each circle is connected by the line, find the area of the shaded area.
Pi is assumed to be 3.14.
Answer
150.72 cm2
The total of the interior angles of the octagon is 180 × (8 - 2) = 1080 = 360 × 3.
Thus the area to be found is 4 × 4 × 3.14 × 3 = 150.72 cm2.
(5)
In the figure below, the area of triangle ADF, triangle DEF, triangle CEF, and triangle BCE is equal.
Find the length of segment AD.
Answer
3 cm
Solution
AD : DE = 1 : 1 and AE : EB = 3 : 1
Thus AD : DE : EB = 1.5 : 1.5 : 1 = 3 : 3 : 2
AD = 8 cm × 3 / (3+3+2) = 3 cm
(6)
A salt solution of 400g is in each container A, B, and C and density in A, B , and C is 16%, 8%, and 4% respectively.
When total of 100g salt solution was taken out of container A and B and it added to container C, the density of container C became 5%.
Find the weight of the salt solution taken out of container A.
Answer
12.5 g
Solution
400g × 4% = 16g
(400g + 100g) × 5% = 25 g
25g - 16g = 9g
9 / 100 = 9%
(16 - 9) : ( 9 - 8) = 7 : 1
Thus the weight of salt solution from A is 100 × 1/(7+1) = 12.5g
Problem 3
(1)
There is a container whose all corners are the right angle as shown in the Fig.1 below.
Water is put in 2/3 of the container.
And the container is put on with the shaded face being placed in the bottom as shown in Fig.2.
In this case find the height of the water.
Answer
17/3 cm
Solution
The volume of the water is (6 × 8 - 3 × 2) × 3 × 2/3 = 84 cm3.
The volume of A in Fig.3 is 3 × 6 × 3 = 54 cm3.
The volume of B is 2 × 3 × 3 = 18 cm3.
84 - (54 + 18) = 12 cm3
12 / (3 × 6) = 2/3 cm
Therefore the height of the water is 3 + 2 + 2/3 = 17/3 cm
(2)
As shown in the figure below, there is a trapezoid ABCD, and AD and BC are parallel, and AD = 2cm, BC = 3cm.
P and Q are points on the side AB, DC and AP : PB = DQ : QC = 2 : 1.
R is an intersection of PQ and AC.
S is an intersection of CD and the extended line of BR.
T is an intersection of the extended lines of BR and AD.
In this case, find DS : SQ : QC by the ratio of the simplest integer.
Answer
12 : 2 : 7
Solution
AR : RC = AP : PB = 2 : 1
△ART and △CRB are homothetic and homothetic ratio is 2 : 1.
AT = 3cm × 2 = 6cm
OT = 6 - 2 = 4cm
DS : SC = DT : CB = 4 : 3
DC is set to to be (4 + 3) × (2 + 1) = 21, DS : SQ : QC = 12 : 2 : 7 as shown below.
Problem 4
Sign [x] is assumed as a sign to calculate the product of each digit of a certain number x like Example.
Example: [46] = 24, [88] = 64
Example: [46] = 24, [88] = 64
Answer the following questions.
(1) Find all integers x of two digits that become "[54]+[35]=[x]".
(2) Find all integers x of two digits that become "[[x]+21] =36.
Answer
(1) 57,75
(2) 47, 74, 59, 95
Solution
(1) [54] + [35] = 20 + 15 = 35 = [57], [75]
(2) 36 = 4 × 9 =6 × 6 = 9 × 4
In case [[x] + 21} = [49], [x] = 49 - 21 = 28 = [47], [74]
In case [[x] + 21} = [66], [x] = 66 - 21 = 45 = [59], [95]
In case [[x] + 21} = [94], [x] = 94 - 21 = 73 which is not available.
Problem 5
Cows are pastured at some grassland.
When 6 cows are pastured, grasses are eaten up in 30 days.
When 10 cows are pastured, grasses are eaten up in 12 days.
At this time, answer the next questions.
Noted that grasses grow at a fixed rate every day in this grassland.
In addition the amount of the grass each cow eats per day is same.
(1) Find the number of cow to be pastured without the grass of the grassland being eaten up.
(2) How many cows at least should be pastured in order to eat all the grass of the grassland within six days?
Answer
(1) Three
(2) 17
Solution
(1) The amount of the grass that one cow eats in a day is assumed to be one.
The amount of the grass growing in this grassland per day is （180 - 120）/（30 - 12） = 10/3.
According to 1 × 3 < 10/3 and 1 × 4 > 10/3, the number of cow to be found is three.
(2) The total amount of the grass grown in this grassland, 180 - 10/3 × 3 = 80.
80 / 6 = 40/3
(40/3) + (10/3) = 50/3 = 16.66---
Therefore the number of cow at least is 17.
Problem 6
As shown in the figure below, there is a road along the surroundings of a pond.
Taro and Hanako leave A point and B point respectively at the same time.
They go to the park through the post in the middle without returning.
They met 14 minutes after they left.
Taro passes through the point B in 10 minutes after they met and after 24 minutes further they arrived at the park at the same time.
Answer the following questions.
Noted we assume that Taro and Hanako walk at a fixed speed respectively.
(1) How many minutes did Hanako take from A point to the park?
(2) Both of them leave the park at the same time and come back to the starting point without passing the course they came.
Taro walks with the same speed as he came to the park.
For two people back to the original location at the same time, by how many times of first speed should Hanako walk?
Answer
(1) 72/5 minutes
(2) 49/15 times
Solution
(1) The speed ratio of walking of Taro and Hanako is 14 : 10 = 7 : 5.
Hanako walked in 14 × 7/5 = 98/5 minutes from the place they met to A point.
Hanako took 10 + 24 = 34 minutes to the park after the meeting.
Therefore, Hanako took 34 - 98/5 = 72 / 5 minutes from A point to the park.
(2) Taro took 72/5 × 5/7= 72 / 7 minutes from the park to A point.
When the speed of Taro is assumed to be 7, the distance from the park to B point is 7 × 24 = 168.
The speed for Hanako to walk this distance in 72/7 minutes is 168 ÷ 72/7 = 49/3 m/minute.
Because the first speed of Hanako was 5, (49/3) / 5 = 49/15 times.
