Time : 60 minutes
(2)
(4)
(5)
Problem 3
Problem 5
Problem 6
Passing mark : 70 %
Answer : End of the problem
Problem 1
Problem 2
(1)
Calculation
5/3 / 25/16 - {1 - 1/3 × (2/3 - 1/6)} =
(2)
Find X.
12 × (X - 0.3) - 0.15 = 1/4
(3)
Find Y.
6 × 6 × 3.14 - 32 × 1.57 + Y × 62.8 = 314
Problem 2
(1)
Find all the fractions in the fractions cannot reduced to the lowest terms of which a denominator is 60 and which is larger than 3/5 and smaller than 13/18. (2)
The integer which cannot be divided by 3 or 5 is lined up in ascending order as 1, 2, 4, 7, 8....
Find the 240th number in this sequence.
(3)
Find the 240th number in this sequence.
(3)
The total cost price of articles was 132000 yen and one third of the whole articles was sold at list price.
Furthermore, two fifths of the whole articles was sold at 15% discounted price.
All the remaining articles were sold at 25% discounted price.
Total net profit became less than the profit selling all articles at list price by 30400 yen.
Find the amount of actual profit.
(4)
As shown in a figure, triangle BDE and the triangle CFD are made in equilateral triangle ABC.
BE = CD and BD = CF.
The angle of X is 87 degrees.
Find the angle of Y.
(5)
In the figure, triangle ABC is an equilateral triangle and all three circles are 3 cm in radius.
Find the circumferential length and area of a shadow area, respectively.
Pi is assumed to be 3.14.
Find the circumferential length and area of a shadow area, respectively.
Pi is assumed to be 3.14.
Problem 3
Taro began to walk toward the place 3200 m far from the starting point.
When he reduced the walking speed by 1/5 at the place where he walked 1/4 of the distance, he reached 10 minutes later than the schedule.
Answer the following questions.
(1) After reducing speed, how many times is the time to walk the same distance comparing the time before reducing speed?
(2) How many minutes did it take after reducing speed for him to reach?
(3) Find the first speed.
Problem 4
When he reduced the walking speed by 1/5 at the place where he walked 1/4 of the distance, he reached 10 minutes later than the schedule.
Answer the following questions.
(1) After reducing speed, how many times is the time to walk the same distance comparing the time before reducing speed?
(2) How many minutes did it take after reducing speed for him to reach?
(3) Find the first speed.
Problem 4
As shown in a figure, there are three circles centering on the point O and radii are 1m, 2m, and 3m, respectively.
Point A on the most inside circumference, point B on the next circumference, point C on the most outside circumference are moving clockwise, respectively.
The ratio of the speed of A : B : C is 1 : 3 : 2.
Three points start from the position where O, A, B, and C are on a straight line at this order and point A goes one round in 12 seconds.
Answer the following questions.(1) What round is each point moved round by the time when three points A, B, and C return to the position of the start simultaneously for the first time?
(2) During the time of (1), how many times does the angle made by OA and OB become right-angled?
(3) Consider in 10 minutes after starting.
(3)-1 How many times does the angle made by OA and OB become right-angled?
(3)-2 How many times does the angle made by OA and OC become right-angled?
(2) During the time of (1), how many times does the angle made by OA and OB become right-angled?
(3) Consider in 10 minutes after starting.
(3)-1 How many times does the angle made by OA and OB become right-angled?
(3)-2 How many times does the angle made by OA and OC become right-angled?
Problem 5
In a figure, AE : EC = 2 : 3, AF : FB = 3 : 1 and the area ratio of the triangle ABC and the triangle DEF is 40 : 9.
Answer the following questions.
(1) Find the area ratio of the triangle BEF and the triangle DEF by the ratio of the least integer.
(2) Find the area ratio of the triangle DEF and the triangle CEF by the ratio of the least integer.
(3) Find BD : DC by the ratio of the least integer.
Answer the following questions.
(1) Find the area ratio of the triangle BEF and the triangle DEF by the ratio of the least integer.
(2) Find the area ratio of the triangle DEF and the triangle CEF by the ratio of the least integer.
(3) Find BD : DC by the ratio of the least integer.
Problem 6
As shown in Fig. 1, make a rectangular prism by arranging cubes P, Q, and R whose one side is 6 cm and a rectangular prism with 6 cm in length, 18 cm in width, and 6 cm in height.
This rectangular prism is cut with the plane which passes along three point of the figure A, B, and C.
Answer the following questions.
(1) Show the cut surface in Fig. 2 with slash shadow made in the cube Q.
(2) The cube Q is divided into two solids.
Find the volume ratio of the two solids.
The volume of a pyramid is calculated as (base area) × height × 1/3.
<Answer>
Problem 1
Problem 2
(2)
(4)
(5)
Problem 5
Problem 6
(2)
(1)
Calculation
5/3 / 25/16 - {1 - 1/3 × (2/3 - 1/6)} =
Answer
7/30
(2)
Find X.
12 × (X - 0.3) - 0.15 = 1/4
Answer
1/3
(3)
Find Y.
6 × 6 × 3.14 - 32 × 1.57 + Y × 62.8 = 314
Answer
4
Problem 2
(1)
Find all the fractions in the fractions cannot reduced to the lowest terms of which a denominator is 60 and which is larger than 3/5 and smaller than 13/18.
Answer
37/60, 41/60, 43/60
Solution
3/5 = 36/60
13/18 = 43.3---/60
Numerators of fractions between 36/60 and 43.3/60 are 37, 38, 39, 40, 41, 42, 43.
Among these fractions, fractions cannot be reduced are 37/60, 41/60, 43/60.
(2)
The integer which cannot be divided by 3 or 5 is lined up in ascending order as 1, 2, 4, 7, 8....
Find the 240th number in this sequence.
Find the 240th number in this sequence.
Answer
(3)
449
Solution
LCM of 3 and 5 is 15.
From 1 to 15 number of integers which can not be divided by 3 or 5 are eight of 1, 2, 4, 7, 8, 11, 13, 14.
One cycle in the sequence is 8 pieces of integers and there are 240 / 8 = 30 cycles.
The number of the end of the last cycle is calculated as 15 × 30 - 1 = 450 - 1 = 449.
(3)
The total cost price of articles was 132000 yen and one third of the whole articles was sold at list price.
Furthermore, two fifths of the whole articles was sold at 15% discounted price.
All the remaining articles were sold at 25% discounted price.
Total net profit became less than the profit selling all articles at list price by 30400 yen.
Find the amount of actual profit.
Answer
77600 yen
Solution
The list price is set to be 1 and the number of all articles is also set to be 1.
The number sold at 25 % discounted price is 1 - 1/3 - 2/5 = 4/15.
Total sales is calculated as 1 × 1/3 + (1 - 0.15) × 2/5 + (1 - 0.25) × 4/15 = 131/150.
1 × 1 - 131/150 = 19/150 which is equivalent to 30400 yen.
1 = 30400 / 19/150 = 2400000 yen which is total sales sold at list price.
Profit is 240000 - 132000 = 108000 yen.
Therefore actual profit = 108000 - 30400 = 77600 yen.
(4)
As shown in a figure, triangle BDE and the triangle CFD are made in equilateral triangle ABC.
BE = CD and BD = CF.
The angle of X is 87 degrees.
Find the angle of Y.
Answer
153 degrees
Solution
According to BE = CD, BD = CF and ∠B = ∠C = 60 degrees, △BDE and △CFD are congruent.
Thus ∠BDE = ∠CFD.
As the angle of X = 87 degrees, ∠CFD = ∠BDE = 87 - 60 = 27 degrees.
Therefore the angle of Y = 180 - 27 = 153 degrees.
(5)
In the figure, triangle ABC is an equilateral triangle and all three circles are 3 cm in radius.
Find the circumferential length and area of a shadow area, respectively.
Pi is assumed to be 3.14.
Find the circumferential length and area of a shadow area, respectively.
Pi is assumed to be 3.14.
Answer
3 × 2 × 3.14 × 120/360 × 4 = 25.12 cm.
According to the equivalence transfer as shown in the right figure below, the area of a shadow area = total area of sector OAQB and OBPC.
Problem 3
25.12 cm
18.84 cm2
18.84 cm2
Solution
The circumferential length of a shadow area = total length of arc AQB, AOB, BOC, BPC in the left figure below.3 × 2 × 3.14 × 120/360 × 4 = 25.12 cm.
According to the equivalence transfer as shown in the right figure below, the area of a shadow area = total area of sector OAQB and OBPC.
3 × 3 × 3.14 × 120/360 × 2 = 18.84 cm2.
Problem 3
Taro began to walk toward the place 3200 m far from the starting point.
When he reduced the walking speed by 1/5 at the place where he walked 1/4 of the distance, he reached 10 minutes later than the schedule.
Answer the following questions.
(1) After reducing speed, how many times is the time to walk the same distance comparing the time before reducing speed?
(2) How many minutes did it take after reducing speed for him to reach?
(3) Find the first speed.
When he reduced the walking speed by 1/5 at the place where he walked 1/4 of the distance, he reached 10 minutes later than the schedule.
Answer the following questions.
(1) After reducing speed, how many times is the time to walk the same distance comparing the time before reducing speed?
(2) How many minutes did it take after reducing speed for him to reach?
(3) Find the first speed.
Answer
Problem 4
(1) 5/4 times
(2) 50 minutes
(3) 60 m/m
(2) 50 minutes
(3) 60 m/m
Solution
(1)
When the speed is reduced by 1/5, the reduced speed is 4/5 times of the first speed.
The time to take for moving same distance is to be 1 / 4/5 = 5/4 times.
(2)
The time ratio of walking at 1 and at 4/5 is 4 : 5.
The difference of 5 - 4 = 1 is equivalent to 10 minutes.
5 is equivalent to 10 × 5 = 50 minutes.
(3)
If Taro would keep walking at first speed, it would take 10 × 4 = 40 minutes to walk 3200 × (1 - 1/4) = 2400 m.
Therefore the first speed is 2400 / 40 = 60 m/m.
Problem 4
As shown in a figure, there are three circles centering on the point O and radii are 1m, 2m, and 3m, respectively.
Point A on the most inside circumference, point B on the next circumference, point C on the most outside circumference are moving clockwise, respectively.
The ratio of the speed of A : B : C is 1 : 3 : 2.
Three points start from the position where O, A, B, and C are on a straight line at this order and point A goes one round in 12 seconds.
Answer the following questions.
(1) What round is each point moved round by the time when three points A, B, and C return to the position of the start simultaneously for the first time?
(2) During the time of (1), how many times does the angle made by OA and OB become right-angled?
(3) Consider in 10 minutes after starting.
(3)-1 How many times does the angle made by OA and OB become right-angled?
(3)-2 How many times does the angle made by OA and OC become right-angled?
Point A on the most inside circumference, point B on the next circumference, point C on the most outside circumference are moving clockwise, respectively.
The ratio of the speed of A : B : C is 1 : 3 : 2.
Three points start from the position where O, A, B, and C are on a straight line at this order and point A goes one round in 12 seconds.
Answer the following questions.
(1) What round is each point moved round by the time when three points A, B, and C return to the position of the start simultaneously for the first time?
(2) During the time of (1), how many times does the angle made by OA and OB become right-angled?
(3) Consider in 10 minutes after starting.
(3)-1 How many times does the angle made by OA and OB become right-angled?
(3)-2 How many times does the angle made by OA and OC become right-angled?
Answer
(1) A : 6 round, B : 9 round, C : 4 round
(2) Six times
(3)-1 50 times
(3)-2 33 times
(2) Six times
(3)-1 50 times
(3)-2 33 times
Solution
(1)
The speed ratio of A : B : C = 1 : 3 : 2.
The length ration of three circles is 1 : 2 : 3.
The time ratio of one round of A : B : C = 1/1 : 2/3 : 3/2 = 6 : 4 : 9.
As A takes 12 seconds one round, B takes 12 sec. × 4/6 = 8 seconds and C takes 12 sec. × 9/6 = 18 seconds.
The time when three points return to the starting point at the same time is LCM of 12, 8 and 18 which is 72 seconds.
72 / 12 = 6 round for A, 72 / 8 = 9 round for B and 72 / 18 = 4 round for C.
(2)
The speed of A is 360 / 12 = 30 degrees/s.
The speed of B is 360 / 8 = 45 degrees/s.
The speed of C is 360 / 18 = 20 degrees/s.
If A is fixed at the starting point, the speed of B is to be 45 - 30 = 15 degrees/s.
1st time and 2nd time for the angle to be 90 degrees is as shown in the figures below.
1st time is 90 / 15 = 6 seconds and 2nd time is 180 / 15 = 12 seconds after the 1st time.
Then after the 2nd time, the angle becomes 90 degrees in every 12 second.
As 6 + 12 × 5 = 66, it is 5 + 1 = 6 times in 72 seconds.
(3)
(3)-1
10 minutes = 600 seconds.
According to the same calculation of (2), (600 - 6) / 12 = 49.5, then 6 + 12 × 49 = 594.
Thus 49 + 1 = 50 times.
(3)-2
If C is fixed at the starting point, the speed of A is to be 30 - 20 = 10 degrees/s.
90 / 10 = 9 seconds and 180 / 10 = 18 seconds.
(600 - 9) / 18 = 32.8---
Then 9 + 18 × 32 = 585.
Therefore 32 + 1 = 33 times.
Problem 5
In a figure, AE : EC = 2 : 3, AF : FB = 3 : 1 and the area ratio of the triangle ABC and the triangle DEF is 40 : 9.
Answer the following questions.
(1) Find the area ratio of the triangle BEF and the triangle DEF by the ratio of the least integer.
(2) Find the area ratio of the triangle DEF and the triangle CEF by the ratio of the least integer.
(3) Find BD : DC by the ratio of the least integer.
Answer the following questions.
(1) Find the area ratio of the triangle BEF and the triangle DEF by the ratio of the least integer.
(2) Find the area ratio of the triangle DEF and the triangle CEF by the ratio of the least integer.
(3) Find BD : DC by the ratio of the least integer.
Answer
(1) 4 : 9
(2) 1 : 2
(3) 5 : 9
(2) 1 : 2
(3) 5 : 9
Solution
(1)
△AEF = △ABC × 3/(1+3) × 2/(2+3) = 40 × 3/4 × 2/5 = 12。
△BEF = △AEF × 1/3 = 4
Thus △BEF : △DEF = 4 : 9.
(2)
△CEF = △AEF × 3/2 = 12 × 3/2 = 18.
Thus △DEF : △CEF = 9 : 18 = 1 : 2.
(3)
According to (1) and (2), △BEF : △DEF △CEF = 4 : 9 : 18.
Base of three triangles are same, it is EF.
As shown in the Fig.2, BG, DH and CI are vertical lines to JI.
Then BG : DH : CI = 4 : 9 : 18.
In this case, JB : JD : JC = 4 : 9 : 18.
Thus BD : DC = (9 - 4) : (18 - 9) = 5 : 9.
Problem 6
As shown in Fig. 1, make a rectangular prism by arranging cubes P, Q, and R whose one side is 6 cm and a rectangular prism with 6 cm in length, 18 cm in width, and 6 cm in height.
This rectangular prism is cut with the plane which passes along three point of the figure A, B, and C.
Answer the following questions.
(1) Show the cut surface in Fig. 2 with slash shadow made in the cube Q.
(2) The cube Q is divided into two solids.
Find the volume ratio of the two solids.
The volume of a pyramid is calculated as (base area) × height × 1/3.
Answer
(1)
161 : 55
Solution
(1)
The cut surface is shown in Fig.3 below.
Thus the cut surface in Q is shown in Fig.4.
(2)
In the Fig.5 below, three triangular pyramids, D-EFC, X, Y are homothetic.
The homothetic ratio is the same as the height of each triangular pyramid, it is 4 : 2 : 1.
Then the volume ratio is (4 × 4 × 4) : (2 × 2 × 2) : (1 × 1 × 1) = 64 : 8 : 1.
According to Fig.3, EF = 6 × 2 × 2/3 = 8 cm and CE = 12 cm.
The volume of D-EFC = 8 × 12 × 1/2 × 4 × 1/3 = 64 cm3.
Thus The volume of X = 64 × 8/64 = 8 cm3 and the volume of Y = 64 × 1/64 = 1 cm3.
Therefore the volume of the solid under the cut surface is 64 - 8 - 1 = 55 cm3.
The volume of another solid = 6 × 6 × 6 - 55 = 161 cm3.
