Time : 50 minutes
Passing mark : 70%
Answer : End of the problem
Problem 1
Problem 2
Problem 6
Problem 7
When a dray moves 900 m, Taro in A point start to pursue the dray with a ball.
If he catches up with the dray, he put the ball in the dray and he will return toward A point immediately.
He receives a ball at A point and start to pursue the dray again.
Taro repeats this motion until the dray arrives at B point.
The speed of Taro is 90 m/m.
Find the time until he finally puts a ball in the dray since he begins to move at first.
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Problem 10
Find X.
7/200 × (1/2 + 1/3 + 1/7 - 1/X) = (1/4 + 1/5 + 1/6) / (19 - 1/2)
Problem 2
Three clocks, A, B and C were set at noon of one day.
Problem 3
When it was 6:00 p.m. by A, it was 5:50 p.m. by B In the same day.
When it was 7:00 p.m. by B, it was 7:20 p.m. by C.
When it is 11:00 p.m. by C of the day, find the time of A and B.
Problem 3
Integers of which top is 2, end is 7 and all interval numbers are 0 such as 207, 2007, 20007, ‥ ‥ ‥, are divided by 27 and 81.
Problem 4
Find the smallest number among such numbers which is divisible by 27 and indivisible by 81.
Problem 4
Ten students roll three dice respectively and consider a number of sums which come out as the person's score.
The sum total of the ten persons' score was 100.
Moreover, after dividing total score of each student by 3 and omitting below the decimal point, ten students’ sum total was 30.
After dividing total score of each student by 3 and rounding off the number of 1st decimal digit, ten students’ sum total was 34.
In these ten students, find the number of students whose score is the number when it is divided by three, one remains.
Problem 5
40 g of salt solutions are in both of two beakers A and B each.
The ratio of the concentration of salt solution of A and B is 3 : 2.
I add 60 g of water to B and mixed it well and transferred some of salt solutions in B to A.
Furthermore, I add some water in both of A and B to become 100 g of salt solutions.
The ratio of the concentration of salt solution of A and B was to be 7 : 3.
Find the weight of the salt solution I transferred from B to A.
The sum total of the ten persons' score was 100.
Moreover, after dividing total score of each student by 3 and omitting below the decimal point, ten students’ sum total was 30.
After dividing total score of each student by 3 and rounding off the number of 1st decimal digit, ten students’ sum total was 34.
In these ten students, find the number of students whose score is the number when it is divided by three, one remains.
Problem 5
The ratio of the concentration of salt solution of A and B is 3 : 2.
I add 60 g of water to B and mixed it well and transferred some of salt solutions in B to A.
Furthermore, I add some water in both of A and B to become 100 g of salt solutions.
The ratio of the concentration of salt solution of A and B was to be 7 : 3.
Find the weight of the salt solution I transferred from B to A.
Problem 6
The figure below is a graphic made by arranging 36 equilateral triangles in order whose length of one side is 1 cm.
The point P moves for 4 seconds with the speed of 1 cm in one second along the side of the figure.
How many ways of movement returning to A first 4 seconds after leaving Point A?
P moves straight on between the two vertex.
P may pass along the same side 2 times or more.
Problem 7
As for the integer of 6 figures made using six numbers, 1, 2, 3, 4, 5, 6 and the integer is a multiple of 64, the smallest integer is 123456.
Then find the largest integer.
Problem 8
A dray moves at the speed of 30 m/m to B point 6000 m away from A point. When a dray moves 900 m, Taro in A point start to pursue the dray with a ball.
If he catches up with the dray, he put the ball in the dray and he will return toward A point immediately.
He receives a ball at A point and start to pursue the dray again.
Taro repeats this motion until the dray arrives at B point.
The speed of Taro is 90 m/m.
Find the time until he finally puts a ball in the dray since he begins to move at first.
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Problem 9
There are two points of P and Q.
Point P is a starting point and balls are put at point Q.
There is a game repeat carrying balls from Q to P after starting P and competing for the number of balls carried up to point P from point Q.
In this game, robots A and B competed each other.
A carries 3 balls at a time and it takes 15 seconds for going back and forth.
B carries 5 balls at a time and it takes 25 seconds for going back and forth.
Although B began to move immediately after the game started, it took 10 seconds for A to begin to move.
Find the time while A leads among 420 seconds from the start.
Moreover, find the time while B leads.
Problem 10
As shown in a figure, there are two squares and there is a circle in a small square.
Find the area of a shadow area.
Problem 11
Find the area of a shadow area.
Pi is assumed to be 3.14.
Problem 11
The quadrangle ABCD in a figure is a square with 5 cm one-side.
All the length of AE, BF, CG, and DH is 2 cm.
Find the area of a shadow area.
Problem 12
All the length of AE, BF, CG, and DH is 2 cm.
Find the area of a shadow area.
Problem 12
ABCD-EFGH in a figure is a box of rectangular prism 60cm in length, 80cm in width and 40cm in height and there is no lid.
There is a 40cm high pillar is straightly erected on the intersection M of a diagonal line of the bottom.
The vertex O of the pillar and chalk are tied up with string 50cm in length.
The length of MG is 50cm.
Pi is assumed to be 3.14.
(1) Find the area of the part painted with chalk in the outside of the box.
(2) Find the area of the part painted with chalk in the inside of the box.
Problem 13
There is a 40cm high pillar is straightly erected on the intersection M of a diagonal line of the bottom.
The vertex O of the pillar and chalk are tied up with string 50cm in length.
The length of MG is 50cm.
Pi is assumed to be 3.14.
(2) Find the area of the part painted with chalk in the inside of the box.
Problem 13
There is a square mat board one side 5cm as shown in the figure and it is partitioned into grid square 25.
I pile up cubes of one side 1cm on the square by the number written in the each square.
In faces of the cube, attach a red square seal of one side 1cm on each surface visible from the outside.
I do not attach the seal on other surfaces (the surface in contact with the mat board or cubes , etc.).
I pile up cubes of one side 1cm on the square by the number written in the each square.
In faces of the cube, attach a red square seal of one side 1cm on each surface visible from the outside.
I do not attach the seal on other surfaces (the surface in contact with the mat board or cubes , etc.).
(1) Find the number of seal attached.
(2) Find the number of cubes just three seals attached .
(2) Find the number of cubes just three seals attached .
Answer
Problem 1
Problem 2
Three clocks, A, B and C were set at noon of one day.
When it was 6:00 p.m. by A, it was 5:50 p.m. by B In the same day.
When it was 7:00 p.m. by B, it was 7:20 p.m. by C.
When it is 11:00 p.m. by C of the day, find the time of A and B.
Problem 6
The figure below is a graphic made by arranging 36 equilateral triangles in order whose length of one side is 1 cm.
The point P moves for 4 seconds with the speed of 1 cm in one second along the side of the figure.
How many ways of movement returning to A first 4 seconds after leaving Point A?
P moves straight on between the two vertex.
P may pass along the same side 2 times or more.
Answer
54 ways
Solution
A is located at the center of regular hexagon.
In the case that P moves to B after 1 second.
There are 5 ways for P to move from B which is to C, D, G, H, I after 2 seconds.
There are 9 ways from C, D, G, H, I for P to move after 3 seconds.
Those are C→B, C→E, D→B, D→F, G→B, G→C, H→B, I→B, I→D.
There are 9 ways as well when P moves to other 5 vertex of regular hexagon after 1 second.
Therefore there are 9 × 6 = 54 ways.
Problem 7
As for the integer of 6 figures made using six numbers, 1, 2, 3, 4, 5, 6 and the integer is a multiple of 64, the smallest integer is 123456.
Then find the largest integer.
Answer
645312
Solution
As 64 = 8 × 8, consider a number of a multiple of 8.
The number which lower 3 figures can be divided by 8 is a multiple of 8.
Upper 2 figures is assumed to be 65.
Considerable multiples of 8 of three figures among 1, 2, 3, 4 are 312 and 432.
However both 654312 and 651432 can not be divided by 64.
Thus both number is not suitable.
Next, Upper 2 figures is assumed to be 64.
Considerable multiples of 8 of three figures among 1, 2, 3, 5 are 312, 512, 152 and 352.
Among 645312, 643512, 643152 and 641352, only 645312 can be divided by 64.
Therefore the number to be found is 645312.
Problem 9
There are two points of P and Q.
Point P is a starting point and balls are put at point Q.
There is a game repeat carrying balls from Q to P after starting P and competing for the number of balls carried up to point P from point Q.
In this game, robots A and B competed each other.
A carries 3 balls at a time and it takes 15 seconds for going back and forth.
B carries 5 balls at a time and it takes 25 seconds for going back and forth.
Although B began to move immediately after the game started, it took 10 seconds for A to begin to move.
Find the time while A leads among 420 seconds from the start.
Moreover, find the time while B leads.
Answer
80 seconds,
240 seconds
Solution
25 seconds after, A and B bring balls back to point P at the same time.
As LCM of 15 and 25 is 75, check the number of balls until 25 + 75 = 100 seconds as the table below.
Find X.
7/200 × (1/2 + 1/3 + 1/7 - 1/X) = (1/4 + 1/5 + 1/6) / (19 - 1/2)
Answer
42
Three clocks, A, B and C were set at noon of one day.
When it was 6:00 p.m. by A, it was 5:50 p.m. by B In the same day.
When it was 7:00 p.m. by B, it was 7:20 p.m. by C.
When it is 11:00 p.m. by C of the day, find the time of A and B.
Answer
Solution
Because when A moves six hours, B moves five hours 50 minutes, the ratio of the speed of A and B is 6 hours : 5 5/6 hours = 36 : 35.
Because when B moves seven hours, C moves seven hours 20 minutes, the ratio of the speed of B and C is 7 hours : 7 1/3 hours = 21 : 22.
Thus, when C moves 11 hours, B moves 11 × 21/22 = 10.5 hours.
The time of B is 10:30.
When B moves 10.5 hours, A moves 10.5 × 36/35 = 10.8 hours.
The time of A is 10:48.
Problem 3
A : 10:48 p.m. B : 10:30 p.m.
Because when A moves six hours, B moves five hours 50 minutes, the ratio of the speed of A and B is 6 hours : 5 5/6 hours = 36 : 35.
Because when B moves seven hours, C moves seven hours 20 minutes, the ratio of the speed of B and C is 7 hours : 7 1/3 hours = 21 : 22.
Thus, when C moves 11 hours, B moves 11 × 21/22 = 10.5 hours.
The time of B is 10:30.
When B moves 10.5 hours, A moves 10.5 × 36/35 = 10.8 hours.
The time of A is 10:48.
Problem 3
Integers of which top is 2, end is 7 and all interval numbers are 0 such as 207, 2007, 20007, ‥ ‥ ‥, are divided by 27 and 81.
Find the smallest number among such numbers which is divisible by 27 and indivisible by 81.
Find the smallest number among such numbers which is divisible by 27 and indivisible by 81.
Answer
Solution
81 and 27 are factorized into prime factor respectively.
27 = 3 × 3 × 3
81 = 3 × 3 × 3 × 3.
207,2007 and 20007 are factorized into prime factor respectively.
207 = 3 × 3 × 23 · · · ①
2007 = 3 × 3 × 223 · · · ②
20007 = 3 × 3 × 2223 · · · ③
Thus, in order for 200 .. 07 is divisible by 27 and is indivisible by 81, it is understood that in 200 · · · 07 = 3 × 3 × A, A is divisible by 3 and indivisible by 9.
As can be seen in ① ~ ③ above, A is such a number as 22 ... 3, it is investigated in order from smaller one as follows.
20000007
81 and 27 are factorized into prime factor respectively.
27 = 3 × 3 × 3
81 = 3 × 3 × 3 × 3.
207,2007 and 20007 are factorized into prime factor respectively.
207 = 3 × 3 × 23 · · · ①
2007 = 3 × 3 × 223 · · · ②
20007 = 3 × 3 × 2223 · · · ③
Thus, in order for 200 .. 07 is divisible by 27 and is indivisible by 81, it is understood that in 200 · · · 07 = 3 × 3 × A, A is divisible by 3 and indivisible by 9.
As can be seen in ① ~ ③ above, A is such a number as 22 ... 3, it is investigated in order from smaller one as follows.
It is found that 2222223 is divisible by 3 and indivisible by 9.
Number to be obtained is 3 × 3 × 2222223.
If you look at the ① ~ ③, the number of 0 between 2 and 7 is same as the number of 2.
Therefore this number is 20000007.
Problem 4
Number to be obtained is 3 × 3 × 2222223.
If you look at the ① ~ ③, the number of 0 between 2 and 7 is same as the number of 2.
Therefore this number is 20000007.
Problem 4
Ten students roll three dice respectively and consider a number of sums which come out as the person's score.
The sum total of the ten persons' score was 100.
Moreover, after dividing total score of each student by 3 and omitting below the decimal point, ten students’ sum total was 30.
After dividing total score of each student by 3 and rounding off the number of 1st decimal digit, ten students’ sum total was 34.
In these ten students, find the number of students whose score is the number when it is divided by three, one remains.
Answer
Two persons
Solution
Range of scores are from 3 (1+1+1) to 18 (6+6+6) and these numbers are categorized by the remainder when the number is divided by 3 as following table.
As for numbers in 2 remainder group, in case they are divided by 3 and omitting below the decimal point or rounding off the number of 1st decimal digit, the number after rounding off is larger than the number after omitting by one.
The sum total at the time of omitting is 30 and the sum total in rounding off is 34.
It turns out that there was the four numbers of 2 remainder group.
Moreover, if 30 which is the sum total of the number after omitting is multiplied by 3, it will be set to 30 × 3 = 90.
There is a 100 - 90 = 10 point difference in 100 which is score's sum total of all the students.
This difference indicates that the sum of remainders when each number of 1 remainder group and 2 remainder group was divided by 3 was ten.
The sum of remainders when each number of 2 remainder group was divided by 3, is 2 × 4 = 8 points since there are four numbers of 2 remainder group.
10 - 8 =2 points.
This 2 is the sum of remainders of the number 1 remainder group being divided by 3.
Therefore the number of students to be found is 2 / 1 = 2 persons.
Problem 5
40 g of salt solutions are in both of two beakers A and B each.
The ratio of the concentration of salt solution of A and B is 3 : 2.
I add 60 g of water to B and mixed it well and transferred some of salt solutions in B to A.
Furthermore, I add some water in both of A and B to become 100 g of salt solutions.
The ratio of the concentration of salt solution of A and B was to be 7 : 3.
Find the weight of the salt solution I transferred from B to A.
Answer
25 g
Solution
All amount of the salt being set to 10 which is the sum of 7 and 3, as for the amount of the salt after a series of operation, the amount of salt in A is 7 and that in B is 3.
In addition, as the ratio of salt amount at the beginning was 3 : 2, the amount of salt in A was 6 and that in B was 4.
It is because the total amount of salt was never changed in the series of operation.
In other words I understand that the amount of salt transferred to A from B was 1.
Because the amount of the beginning of B was 4 and the amount after transfer was 3, 1/4 of the salt of B transferred.
1/4 of the salt solution should move for 1/4 of the salt to move.
Because the amount of the salt solution of B before the transfer was 40 + 60 = 100 g, the amount I transferred is 100 × 1/4 = 25g.
The sum total of the ten persons' score was 100.
Moreover, after dividing total score of each student by 3 and omitting below the decimal point, ten students’ sum total was 30.
After dividing total score of each student by 3 and rounding off the number of 1st decimal digit, ten students’ sum total was 34.
In these ten students, find the number of students whose score is the number when it is divided by three, one remains.
Answer
Two persons
Solution
Range of scores are from 3 (1+1+1) to 18 (6+6+6) and these numbers are categorized by the remainder when the number is divided by 3 as following table.
1st decimal digit
| |||||||
0 remainder group
|
0
|
3
|
6
|
9
|
12
|
15
|
18
|
1 remainder group
|
3
|
4
|
7
|
10
|
13
|
16
| |
2 remainder group
|
6
|
5
|
8
|
11
|
14
|
17
|
The sum total at the time of omitting is 30 and the sum total in rounding off is 34.
It turns out that there was the four numbers of 2 remainder group.
Moreover, if 30 which is the sum total of the number after omitting is multiplied by 3, it will be set to 30 × 3 = 90.
There is a 100 - 90 = 10 point difference in 100 which is score's sum total of all the students.
This difference indicates that the sum of remainders when each number of 1 remainder group and 2 remainder group was divided by 3 was ten.
The sum of remainders when each number of 2 remainder group was divided by 3, is 2 × 4 = 8 points since there are four numbers of 2 remainder group.
10 - 8 =2 points.
This 2 is the sum of remainders of the number 1 remainder group being divided by 3.
Therefore the number of students to be found is 2 / 1 = 2 persons.
Problem 5
The ratio of the concentration of salt solution of A and B is 3 : 2.
I add 60 g of water to B and mixed it well and transferred some of salt solutions in B to A.
Furthermore, I add some water in both of A and B to become 100 g of salt solutions.
The ratio of the concentration of salt solution of A and B was to be 7 : 3.
Find the weight of the salt solution I transferred from B to A.
Answer
25 g
Solution
All amount of the salt being set to 10 which is the sum of 7 and 3, as for the amount of the salt after a series of operation, the amount of salt in A is 7 and that in B is 3.
In addition, as the ratio of salt amount at the beginning was 3 : 2, the amount of salt in A was 6 and that in B was 4.
It is because the total amount of salt was never changed in the series of operation.
In other words I understand that the amount of salt transferred to A from B was 1.
Because the amount of the beginning of B was 4 and the amount after transfer was 3, 1/4 of the salt of B transferred.
1/4 of the salt solution should move for 1/4 of the salt to move.
Because the amount of the salt solution of B before the transfer was 40 + 60 = 100 g, the amount I transferred is 100 × 1/4 = 25g.
Problem 6
The point P moves for 4 seconds with the speed of 1 cm in one second along the side of the figure.
How many ways of movement returning to A first 4 seconds after leaving Point A?
P moves straight on between the two vertex.
P may pass along the same side 2 times or more.
54 ways
Solution
A is located at the center of regular hexagon.
In the case that P moves to B after 1 second.
There are 5 ways for P to move from B which is to C, D, G, H, I after 2 seconds.
There are 9 ways from C, D, G, H, I for P to move after 3 seconds.
Those are C→B, C→E, D→B, D→F, G→B, G→C, H→B, I→B, I→D.
There are 9 ways as well when P moves to other 5 vertex of regular hexagon after 1 second.
Therefore there are 9 × 6 = 54 ways.
Problem 7
Then find the largest integer.
Answer
645312
Solution
As 64 = 8 × 8, consider a number of a multiple of 8.
The number which lower 3 figures can be divided by 8 is a multiple of 8.
Upper 2 figures is assumed to be 65.
Considerable multiples of 8 of three figures among 1, 2, 3, 4 are 312 and 432.
However both 654312 and 651432 can not be divided by 64.
Thus both number is not suitable.
Next, Upper 2 figures is assumed to be 64.
Considerable multiples of 8 of three figures among 1, 2, 3, 5 are 312, 512, 152 and 352.
Among 645312, 643512, 643152 and 641352, only 645312 can be divided by 64.
Therefore the number to be found is 645312.
Problem 8
A dray moves at the speed of 30 m/m to B point 6000 m away from A point.
When a dray moves 900 m, Taro in A point start to pursue the dray with a ball.
If he catches up with the dray, he put the ball in the dray and he will return toward A point immediately.
He receives a ball at A point and start to pursue the dray again.
Taro repeats this motion until the dray arrives at B point.
The speed of Taro is 90 m/m.
Find the time until he finally puts a ball in the dray since he begins to move at first.
Answer
150 minutes
Solution
The time for Taro to catch up with the dray at the 1st time is 900 / (90 - 30) = 15 minutes.
The time of returning to A point is 15 minute × 2 = 30 minutes after leaving.
At this time, the dray is moving 900 × 2 = 1800 m from A point.
Since the distance to the dray is twice, the time to catch up with it 2nd time is 15 minutes × 2 = 30 minutes.
The time to return to A point is also 30 minutes.
It is considered as the 2nd time in the same way, the time to catch up 3rd time is 30 minute × 2 = 60 minutes.
When he catches up the 3rd time, the dray is 900 + 30 × (30 + 60 + 60) = 5400 m from A point.
From this, it turns out that the 4th time is impossible.
Therefore, the time until he finally puts a ball in the dray since he begins to move at first is 30 + 60 + 60 = 150 minutes.
When a dray moves 900 m, Taro in A point start to pursue the dray with a ball.
If he catches up with the dray, he put the ball in the dray and he will return toward A point immediately.
He receives a ball at A point and start to pursue the dray again.
Taro repeats this motion until the dray arrives at B point.
The speed of Taro is 90 m/m.
Find the time until he finally puts a ball in the dray since he begins to move at first.
Answer
150 minutes
Solution
The time for Taro to catch up with the dray at the 1st time is 900 / (90 - 30) = 15 minutes.
The time of returning to A point is 15 minute × 2 = 30 minutes after leaving.
At this time, the dray is moving 900 × 2 = 1800 m from A point.
Since the distance to the dray is twice, the time to catch up with it 2nd time is 15 minutes × 2 = 30 minutes.
The time to return to A point is also 30 minutes.
It is considered as the 2nd time in the same way, the time to catch up 3rd time is 30 minute × 2 = 60 minutes.
When he catches up the 3rd time, the dray is 900 + 30 × (30 + 60 + 60) = 5400 m from A point.
From this, it turns out that the 4th time is impossible.
Therefore, the time until he finally puts a ball in the dray since he begins to move at first is 30 + 60 + 60 = 150 minutes.
There are two points of P and Q.
Point P is a starting point and balls are put at point Q.
There is a game repeat carrying balls from Q to P after starting P and competing for the number of balls carried up to point P from point Q.
In this game, robots A and B competed each other.
A carries 3 balls at a time and it takes 15 seconds for going back and forth.
B carries 5 balls at a time and it takes 25 seconds for going back and forth.
Although B began to move immediately after the game started, it took 10 seconds for A to begin to move.
Find the time while A leads among 420 seconds from the start.
Moreover, find the time while B leads.
Answer
80 seconds,
240 seconds
Solution
25 seconds after, A and B bring balls back to point P at the same time.
As LCM of 15 and 25 is 75, check the number of balls until 25 + 75 = 100 seconds as the table below.
The time when A leads in the above table is 15 seconds of 40~50 and 70~75.
According to (420 - 25) / 75 = 5 remainder 20, it is 15 × 5 + 5 = 80 seconds.
The time when B leads is 45 seconds of 25~40, 50~70, 75~85.
It is 45 × 5 + 15 = 240 seconds.
Problem 10
According to (420 - 25) / 75 = 5 remainder 20, it is 15 × 5 + 5 = 80 seconds.
The time when B leads is 45 seconds of 25~40, 50~70, 75~85.
It is 45 × 5 + 15 = 240 seconds.
Problem 10
As shown in a figure, there are two squares and there is a circle in a small square.
Find the area of a shadow area.
Pi is assumed to be 3.14.
Find the area of a shadow area.
Pi is assumed to be 3.14.
Answer
When the radius of a circle is set to 1, (area of small square) : (area of circle) = (2 × 2) : (1 × 1 × 3.14) = 4 : 3.14.
Problem 11
7.31 cm2
Solution
The area of a shadow portion is determined by (area of a small square) - (area of a circle).
Since the area of a small square is (big square) - (right triangle × 4) = 8 × 8 - 3 × 5 / 2 × 4 = 34 cm2.
The area of a shadow portion is determined by (area of a small square) - (area of a circle).
Since the area of a small square is (big square) - (right triangle × 4) = 8 × 8 - 3 × 5 / 2 × 4 = 34 cm2.
Thus, the area of a circle is 34 × 3.14/4 = 26.69 cm2.
Therefore, the area to be found is 34 - 26.69 = 7.31 cm2.
Problem 11
The quadrangle ABCD in a figure is a square with 5 cm one-side.
All the length of AE, BF, CG, and DH is 2 cm.
Find the area of a shadow area.
All the length of AE, BF, CG, and DH is 2 cm.
Find the area of a shadow area.
Answer
Solution
In the figure, since AE and GC are parallel and AG and EC are parallel, it turns out that a quadrangle AECG is a parallelogram.
Thus, AP : PS = 3 : (5 - 3) = 3 : 2----<1>
Moreover,
since HP : DS = 3 : 5 and AP = DS, HP = SG, SG : AP = 3 : 5----<2>
According to <1> and <2>, if the continuous ratio is arranged as AP = 3 × 5 =15, it will be set that AP : PS : SG = 15 : 10 : 9.
Therefore, the area ratio of a shadow area and parallelogram AECG is (15 + 10 + 9) : 10 = 17 : 5.
Problem 12
(1) Find the area of the part painted with chalk in the outside of the box.
(2) Find the area of the part painted with chalk in the inside of the box.
Next, the painted range is checked in side face ABFE.
The most distant place from O in side face ABFE is the point Q in Fig. 5 which looked at the box from the direction of BCGF.
That is, on all the sides of the box, the area of the range painted is the area of every two semicircles with BC and AB a diameter.
Therefore, the area of the part painted is (area of circle centering on M with a radius of 30 cm) +(area of semicircle with BC a diameter) × 2 + (area of semicircle with AB a diameter) × 2.
Problem 13
50/17 cm2
In the figure, since AE and GC are parallel and AG and EC are parallel, it turns out that a quadrangle AECG is a parallelogram.
This area is 2 × 5 = 10cm2.
Since HB and DF are parallel, △APH and △ASD are homothetic and a homothetic ratio is AH : AD = 3 : 5. Thus, AP : PS = 3 : (5 - 3) = 3 : 2----<1>
Moreover,
since HP : DS = 3 : 5 and AP = DS, HP = SG, SG : AP = 3 : 5----<2>
According to <1> and <2>, if the continuous ratio is arranged as AP = 3 × 5 =15, it will be set that AP : PS : SG = 15 : 10 : 9.
Therefore, the area ratio of a shadow area and parallelogram AECG is (15 + 10 + 9) : 10 = 17 : 5.
The area of a shadow area is 10 cm2 × 5/17 = 50/17 cm2.
Problem 12
ABCD-EFGH in a figure is a box of rectangular prism 60cm in length, 80cm in width and 40cm in height and there is no lid.
There is a 40cm high pillar is straightly erected on the intersection M of a diagonal line of the bottom.
The vertex O of the pillar and chalk are tied up with string 50cm in length.
The length of MG is 50cm.
Pi is assumed to be 3.14.
There is a 40cm high pillar is straightly erected on the intersection M of a diagonal line of the bottom.
The vertex O of the pillar and chalk are tied up with string 50cm in length.
The length of MG is 50cm.
Pi is assumed to be 3.14.
(1) Find the area of the part painted with chalk in the outside of the box.
(2) Find the area of the part painted with chalk in the inside of the box.
Answer
Solution
(1) The development view of the box as shown in Fig. 1 is drawn.
Both the length of the string and the length of the diagonal line at the bottom is 50 cm.
The most length that chalk reaches on the outside of the box is 50 cm.
The part painted with chalk is an inner side of the circle centering on O in Fig. 1 and it is turns into the shadow area except the inside of rectangle ABCD.
Next, the range side face BCGF can be painted is checked.
The chalk passes along the middle point P of the side FG.
Furthermore, since OB = OC = 50cm, it passes along the points B and C.
Thus, the inner side of the semicircle BCP with BC a diameter can be painted as shown in Fig. 4.
(1) 3050 cm2
(2) 10676 cm2
(2) 10676 cm2
(1) The development view of the box as shown in Fig. 1 is drawn.
Both the length of the string and the length of the diagonal line at the bottom is 50 cm.
The most length that chalk reaches on the outside of the box is 50 cm.
The part painted with chalk is an inner side of the circle centering on O in Fig. 1 and it is turns into the shadow area except the inside of rectangle ABCD.
The area is 50 × 50 × 3.14 - 60 × 80 = 3050cm2.
(2) The part painted in the inner side of the box is considered by dividing a box into the bottom surface and the side surface.
The length of the pillar is 40cm and the length of the string is 50cm.
Thus, when you look at the box from the direction of ABFE, the most distant place that chalk reaches on the bottom of the box is the place of 30cm from O, as shown in Fig. 2.
As shown in Fig. 3, the part when the box is looked at from the top is the inside of a circle centering M with a radius of 30cm.
The chalk passes along the middle point P of the side FG.
Furthermore, since OB = OC = 50cm, it passes along the points B and C.
Thus, the inner side of the semicircle BCP with BC a diameter can be painted as shown in Fig. 4.
Next, the painted range is checked in side face ABFE. The most distant place from O in side face ABFE is the point Q in Fig. 5 which looked at the box from the direction of BCGF.
As shown in Fig. 6, the inner side of the semicircle ABQ with radius of AR = BR = RQ = 30cm can be painted.
Therefore, the area of the part painted is (area of circle centering on M with a radius of 30 cm) +(area of semicircle with BC a diameter) × 2 + (area of semicircle with AB a diameter) × 2.
It becomes 30 × 30 × 3.14 + 40 × 40 × 3.14 / 2 × 2 + 30 × 30 × 3.14 / 2 × 2 = 10676cm2.
Problem 13
There is a square mat board one side 5cm as shown in the figure and it is partitioned into grid square 25.
I pile up cubes of one side 1cm on the square by the number written in the each square.
In faces of the cube, attach a red square seal of one side 1cm on each surface visible from the outside.
I do not attach the seal on other surfaces (the surface in contact with the mat board or cubes , etc.).
(1) Find the number of seal attached.
(2) Find the number of cubes just three seals attached .
I pile up cubes of one side 1cm on the square by the number written in the each square.
In faces of the cube, attach a red square seal of one side 1cm on each surface visible from the outside.
I do not attach the seal on other surfaces (the surface in contact with the mat board or cubes , etc.).
(1) Find the number of seal attached.
(2) Find the number of cubes just three seals attached .
Answer
(1) 147 seals
(2) 15 pieces
(2) 15 pieces
Solution
(1) In order to enumerate the number of seals attached, the following procedure is taken.
First, the number of the cube faces of this solid made by piling up cubes is seen from up-down, front-rear and right-left sides and counted.
Next, the number of the faces which are not in sight even if it is seen from up-down, front-rear and right-left sides is counted.
Since a seal is attached on the upper face of every cube which is on the top of cubes piled up in each grid, the number of seal is 5 × 5 = 25 seals.
First, the number of the cube faces of this solid made by piling up cubes is seen from up-down, front-rear and right-left sides and counted.
Next, the number of the faces which are not in sight even if it is seen from up-down, front-rear and right-left sides is counted.
Since a seal is attached on the upper face of every cube which is on the top of cubes piled up in each grid, the number of seal is 5 × 5 = 25 seals.
Since there is no seal when it is seen from down side, it is 0.
Since the number of the faces of the cube which is visible from front and rear side in the case of this problem is the same, only one side is counted.
Moreover, since the number of the faces of the cube which is visible from right and left side is the same, only one side is counted.
The red number in Fig. 1 is the number of the faces of the cubes which is visible from front and right sides, the sum total is (5 + 5 + 4 + 5 + 5) × 2 + (5 + 4 + 5 + 5 + 5) × 2 = 96.
The red number in Fig. 1 is the number of the faces of the cubes which is visible from front and right sides, the sum total is (5 + 5 + 4 + 5 + 5) × 2 + (5 + 4 + 5 + 5 + 5) × 2 = 96.
Adding the number of the faces which are in sight from up side is added, it is 96 + 25 = 121.
Next, the blue number in Fig. 1 shows the number of the faces which are not in sight from up-down, front-rear and right-left sides and the sum is 26.
Therefore, the grand total of the number of seals is 121 + 26 = 147 seals.
(2) In order to count the number of the cube in which three seals are attached exactly, divides this solid into five layers and count the number from the top layer or the 5th layer in order.
There is no cube which three seals attached in the 5th layer. As shown in Fig. 2, there are three pieces in the 4th layer.
As shown in Fig. 3, there are five pieces in the 3rd layer.
As shown in Fig. 4, there are five pieces in the 2nd layer.
As shown in Fig. 5, there are two pieces in the 1st layer.
There are in total 3 + 5 + 5 + 2 = 15 pieces.
