Time : 60 minutes
Passing mark : 70%
Answer : End of the problem
Problem 1
Problem 2
Problem 5
Problem 6
Problem 11
Problem 12
Find X.
2005 × 1/17 = X × 1/119 + 48 × 17/7
Problem 2
A singer performed concert 3 times and sang ten songs in each concert.
Out of ten songs in each concert he sang five songs which he did not sing in other two concerts.
In case the same song sung even twice or more is counted as one song, how many songs the most did this singer sing in three concerts ?
Problem 3
Out of ten songs in each concert he sang five songs which he did not sing in other two concerts.
In case the same song sung even twice or more is counted as one song, how many songs the most did this singer sing in three concerts ?
Problem 3
There are four integers of two digit.
Problem 4
The sum and difference of two numbers of these four numbers were all investigated.
The largest number in sum is 187 and the smallest number is 137.
The largest number in difference is 40 and the smallest number is 10.
Find the 2nd smallest number among four numbers.
Problem 4
There are three cars A, B, and C.
These cars can run 10 km, 15 km, and 20 km with the gasoline of 1L, respectively.
A, B, and C ran total of 100 km consuming the gasoline of 8L in total.
C ran 3 times as much distance as A.
Find the distance where B ran.
Problem 5
There is a track and total weight of cargo must be set 2000 kg or less and total volume of cargo must be set 24 m3 or less.
The weight of one piece of the product A is 100 kg and the volume is 0.8 m3.
The weight of one piece of the product B is 50 kg and the volume is 3 m3.
When you load A and B in this track, find the maximum number of pieces of A and B in total to be loaded.
Problem 6
There are three different integers and the product of the three integers is larger than the sum of the three integers by four.
There are two sets of group of such three integers.
Largest integer is 4 in one group.
Find the each product of two groups.
Problem 7
There are two sets of group of such three integers.
Largest integer is 4 in one group.
Find the each product of two groups.
Problem 7
A certain product are sold at A store, B store, and C store.
Problem 8
Total of 6867 pieces sold in three stores this month and the number of product increased more than the previous month was same in each store.
It means that in this month total sales of this product was increased by 15% in A store, 10% in B store and 6% in C store.
Find the sum total number of the sales of three stores in the previous month.
Problem 8
There are eight color cards with number, three blue cards blue1, blue1, blue1 and red cards red1, red1, red2, red2, red3.
I arrange them in a line.
Answer the following questions about the way of arrangement that there are just four places where the card of different color adjoins each other like red2-blue1-blue2-red1-red1-red3-blue1-red2 for example.
(1) When distinguishing the color of a card and not distinguishing a number, how many ways of arrangement is there?
(2) When distinguishing both the color of a card and a number, how many ways of arrangement is there?
Problem 9
I arrange them in a line.
Answer the following questions about the way of arrangement that there are just four places where the card of different color adjoins each other like red2-blue1-blue2-red1-red1-red3-blue1-red2 for example.
(1) When distinguishing the color of a card and not distinguishing a number, how many ways of arrangement is there?
(2) When distinguishing both the color of a card and a number, how many ways of arrangement is there?
Problem 9
The quadrangle ABCD in a figure is a parallelogram. QS and BC are parallel and RT and CD are also parallel.
The point P which is an intersection of QS and RT is on the diagonal line BD.
(The length of BP) : (length of PD) = 2 : 1.
Find the area ratio between the sum of the area of the triangle AQT and CSR and the area of parallelogram ABCD.
Problem 10
Both triangle ABC and ADE in a figure are a right-angled isosceles triangle.
(length of BD) : (length of DC) = 1 : 3.
Find the area ratio of triangle DCE and triangle ABC.
The point P which is an intersection of QS and RT is on the diagonal line BD.
(The length of BP) : (length of PD) = 2 : 1.
Find the area ratio between the sum of the area of the triangle AQT and CSR and the area of parallelogram ABCD.
Problem 10
(length of BD) : (length of DC) = 1 : 3.
Find the area ratio of triangle DCE and triangle ABC.
Problem 11
In a figure below, the length of AC is 10 cm and that of AF is 6 cm.
The ratio of the length of AD to BD is 3 : 2.
The ratio of the length of BE to EC is 5 : 2.
Find the angle of X.
Problem 12
Make a large cube of 5cm one side by piling up 125 dices of 1cm one side without a gap.
This large cube is cut by the plane passing through the three vertex of the cube as shown in a figure.
Problem 13
This large cube is cut by the plane passing through the three vertex of the cube as shown in a figure.
As for the portion of the triangular pyramid below the cutting plane, find the following each number.
(1)The number of the dices which is not cut.
(2)The number of a solid with the volume larger than 0.5 cm3 in the solid made by cutting a dice.
(3)The number of a solid with the volume smaller than 0.5 cm3 in the solid made by cutting a dice.
Problem 13
Find the volume of the solid which is made by assembling the development view of the figure.
Each triangular face is an equilateral triangle.
Each face of a hexagon is a made figure where one side cuts out two right-angled isosceles triangles whose equal lengths of two sides are 1 cm from the square which is 2 cm one side.
Each triangular face is an equilateral triangle.
Each face of a hexagon is a made figure where one side cuts out two right-angled isosceles triangles whose equal lengths of two sides are 1 cm from the square which is 2 cm one side.
<Answer>
Problem 1
Problem 2
Problem 5
There is a track and total weight of cargo must be set 2000 kg or less and total volume of cargo must be set 24 m3 or less.
Therefore considering above combinations meeting two conditions, maximum number is 21 pieces.
Problem 6
Problem 11
In a figure below, the length of AC is 10 cm and that of AF is 6 cm.
The ratio of the length of AD to BD is 3 : 2.
The ratio of the length of BE to EC is 5 : 2.
Find the angle of X.
Answer
67 degrees
Solution
As for the figure below, AB and CG is parallel and G is a point on the extension of the line segment AE.
As △ABE and △CGE are homothetic and BE : CE = 5 : 2, AB : CG = 5 : 2.
Furthermore △ADF and △CGF are homothetic and AD : CG = 3 : 2, AF : GF = 3 : 2.
Thus GF = AF × 2/3 = 6 cm × 2/3 = 4 cm.
As AC = AG = 10 cm, ∠AGC = (180 - 46) / 2 = 67 degrees.
Therefore X = ∠AGC = 67 degrees.
Problem 12
Find X.
2005 × 1/17 = X × 1/119 + 48 × 17/7
Answer
163
Problem 2
A singer performed concert 3 times and sang ten songs in each concert. Out of ten songs in each concert he sang five songs which he did not sing in other two concerts.
In case the same song sung even twice or more is counted as one song, how many songs the most did this singer sing in three concerts ?
Answer
22 songs
Solution
The number of times of singing in three concerts is 10 times × 3= 30 times.
Since he sang five songs which are not sung in other two concerts, at least 5 times × 3= 15 times are different songs.
While of 30 - 15 times = 15 times, the same song is sung at least 2 times respectively in each concert.
In order to have the most number of songs, songs sung three times should be less as much as possible.
When the number of the song sung 3 times is 0 time, it is useless since 15 times cannot be divisible by 2.
When the number of song sung 3 times is 1 time, 15 - 3 × 1 = 12.
12 can be divisible by 2 and it is 12/2 = 6.
Therefore, most numbers of song are 15 + 6 + 1 = 22 songs.
Problem 3
In case the same song sung even twice or more is counted as one song, how many songs the most did this singer sing in three concerts ?
Answer
22 songs
Solution
The number of times of singing in three concerts is 10 times × 3= 30 times.
Since he sang five songs which are not sung in other two concerts, at least 5 times × 3= 15 times are different songs.
While of 30 - 15 times = 15 times, the same song is sung at least 2 times respectively in each concert.
In order to have the most number of songs, songs sung three times should be less as much as possible.
When the number of the song sung 3 times is 0 time, it is useless since 15 times cannot be divisible by 2.
When the number of song sung 3 times is 1 time, 15 - 3 × 1 = 12.
12 can be divisible by 2 and it is 12/2 = 6.
Therefore, most numbers of song are 15 + 6 + 1 = 22 songs.
Problem 3
There are four integers of two digit.
The sum and difference of two numbers of these four numbers were all investigated.
The largest number in sum is 187 and the smallest number is 137.
The largest number in difference is 40 and the smallest number is 10.
Find the 2nd smallest number among four numbers.
Answer
78
Solution
Four integers of two digit are set to A, B, C, and D at large order.
Since the largest sum is 187, combination of A and B are considered as (A, B) = (99, 88), (98, 89) (97, 90), -------.
Since the smallest difference is 10, A and B are determined as A= 99 and B= 88.
Since the largest difference is 59, D = 99 - 40 = 59.
Since the smallest sum is 137, C = 137 - 59 = 78.
Problem 4
There are three cars A, B, and C.
These cars can drive 10 km, 15 km, and 20 km with the gasoline of 1L, respectively.
A, B, and C drove total of 100 km consuming the gasoline of 8L in total.
C drove 3 times as much distance as A.
Find the distance where B drove.
Answer
36 km
Solution
Since the distance ratio of A : C = 1 : 3, volume ratio of gasoline used of A : C = 1/20 : 3/10 = 1 : 6.
The distance of A and C per 1L = (20 × 1 + 10 × 6) / (1 + 6) = 80/7 km.
If A and C use 8L gasoline, total distance is to be 80/7 × 8 = 640/7 km.
B can drive 15 - 80/7 = 25/7 km per 1L more than A and C.
Thus the volume of gasoline B used is (100 - 640/7) / 25/7 = 12/5L.
Therefore the distance B drove is 15 × 12/5 = 36km.
The largest number in sum is 187 and the smallest number is 137.
The largest number in difference is 40 and the smallest number is 10.
Find the 2nd smallest number among four numbers.
Answer
78
Solution
Four integers of two digit are set to A, B, C, and D at large order.
Since the largest sum is 187, combination of A and B are considered as (A, B) = (99, 88), (98, 89) (97, 90), -------.
Since the smallest difference is 10, A and B are determined as A= 99 and B= 88.
Since the largest difference is 59, D = 99 - 40 = 59.
Since the smallest sum is 137, C = 137 - 59 = 78.
Problem 4
These cars can drive 10 km, 15 km, and 20 km with the gasoline of 1L, respectively.
A, B, and C drove total of 100 km consuming the gasoline of 8L in total.
C drove 3 times as much distance as A.
Find the distance where B drove.
Answer
36 km
Solution
Since the distance ratio of A : C = 1 : 3, volume ratio of gasoline used of A : C = 1/20 : 3/10 = 1 : 6.
The distance of A and C per 1L = (20 × 1 + 10 × 6) / (1 + 6) = 80/7 km.
If A and C use 8L gasoline, total distance is to be 80/7 × 8 = 640/7 km.
B can drive 15 - 80/7 = 25/7 km per 1L more than A and C.
Thus the volume of gasoline B used is (100 - 640/7) / 25/7 = 12/5L.
Therefore the distance B drove is 15 × 12/5 = 36km.
Problem 5
The weight of one piece of the product A is 100 kg and the volume is 0.8 m3.
The weight of one piece of the product B is 50 kg and the volume is 3 m3.
When you load A and B in this track, find the maximum number of pieces of A and B in total to be loaded.
Answer
21 pieces
Solution
As for the weight, A can be loaded up to 2000 / 100 = 20 pieces.
B can be increased 100 / 50 = 2 pieces by decreasing 1 piece of A.
Thus combination of number of A and B is (20,9), (19,2), (18,4), (17,6), -----.
As for the volume, when A is loaded 20 pieces, B can be loaded 2 pieces according to the calculation of (24 - 0.8 × 20) /3 = 2 remainder 2 m3.
Calculating the combination of number of A and B, it is (20,2), (19,2), (18,3), (17,3), (16,3), (15,4),------.B can be increased 100 / 50 = 2 pieces by decreasing 1 piece of A.
Thus combination of number of A and B is (20,9), (19,2), (18,4), (17,6), -----.
As for the volume, when A is loaded 20 pieces, B can be loaded 2 pieces according to the calculation of (24 - 0.8 × 20) /3 = 2 remainder 2 m3.
Therefore considering above combinations meeting two conditions, maximum number is 21 pieces.
Problem 6
There are three different integers and the product of the three integers is larger than the sum of the three integers by four.
There are two sets of group of such three integers.
Largest integer is 4 in one group.
Find the each product of two groups.
Answer
12 and 14
Solution
Three integers are set to A, B, and C and assuming A = 4, then 4 × B × C = 4 + B + C + 4.
Since the product of 4 × B × C is an even number, the sum of 4 + B + C + 4 is also even number and the sum of B + C is also an even number.
Combination of B + C is an odd + odd.
Then searching for a combination of B and C, it is 1 and 3.
Therefore, the product of the three integers is 4 × 3 × 1 = 12.
Next, start considering the product 13 following 12.
There is a combination of the product of 13 = 1 × 1 × 13, but it is not suitable to the condition.
There are two combinations of the products of 14 = 1 × 1 × 14 and 1 × 2 × 7.
1 × 1 × 14 is not applied.
1 × 2 × 7 = 1 +2 +7 +4 = 14.
Then the product of the three integers is 14.
Problem 7
There are two sets of group of such three integers.
Largest integer is 4 in one group.
Find the each product of two groups.
Answer
12 and 14
Solution
Three integers are set to A, B, and C and assuming A = 4, then 4 × B × C = 4 + B + C + 4.
Since the product of 4 × B × C is an even number, the sum of 4 + B + C + 4 is also even number and the sum of B + C is also an even number.
Combination of B + C is an odd + odd.
Then searching for a combination of B and C, it is 1 and 3.
Therefore, the product of the three integers is 4 × 3 × 1 = 12.
Next, start considering the product 13 following 12.
There is a combination of the product of 13 = 1 × 1 × 13, but it is not suitable to the condition.
There are two combinations of the products of 14 = 1 × 1 × 14 and 1 × 2 × 7.
1 × 1 × 14 is not applied.
1 × 2 × 7 = 1 +2 +7 +4 = 14.
Then the product of the three integers is 14.
Problem 7
A certain product are sold at A store, B store, and C store.
Total of 6867 pieces sold in three stores this month and the number of product increased more than the previous month was same in each store.
It means that in this month total sales of this product was increased by 15% in A store, 10% in B store and 6% in C store.
Find the sum total number of the sales of three stores in the previous month.
Answer
6300 pieces
Solution
If the sales number of the previous month is set to A, B, and C, respectively, A × 15% = B × 10% = C × 6%.
Thus, A : B : C = 1/15 : 1/10 : 1/6 = 2 : 3 : 5.
It is reset to A = 20, B = 30, and C = 50 in order to make it easy to calculate.
As for sales this month, 20 × 1.15 = 23, 30 × 1.1 = 33, and 50 × 1.06 = 53.
23 + 33 + 53 = 109 is equivalent to 6867 pieces.
Since the sum total of sales of the previous month is set to 20 + 30 + 50 =100, it is 6867 × 100/109 = 6300 pieces.
Problem 8
6300 pieces
Solution
If the sales number of the previous month is set to A, B, and C, respectively, A × 15% = B × 10% = C × 6%.
Thus, A : B : C = 1/15 : 1/10 : 1/6 = 2 : 3 : 5.
It is reset to A = 20, B = 30, and C = 50 in order to make it easy to calculate.
As for sales this month, 20 × 1.15 = 23, 30 × 1.1 = 33, and 50 × 1.06 = 53.
23 + 33 + 53 = 109 is equivalent to 6867 pieces.
Since the sum total of sales of the previous month is set to 20 + 30 + 50 =100, it is 6867 × 100/109 = 6300 pieces.
Problem 8
There are eight color cards with number, three blue cards blue1, blue1, blue1 and red cards red1, red1, red2, red2, red3.
I arrange them in a line.
Answer the following questions about the way of arrangement that there are just four places where the card of different color adjoins each other like red2-blue1-blue2-red1-red1-red3-blue1-red2 for example.
(1) When distinguishing the color of a card and not distinguishing a number, how many ways of arrangement is there?
(2) When distinguishing both the color of a card and a number, how many ways of arrangement is there?
Answer
(1) 16 ways
(2) 480 ways
Solution
(1) It is considered that putting five red cards in order as shown in a figure and arranging blue cards in A~F.
Case 1 : A set of two blue and one blue are put in order like the example of problem sentence.
If a set of blue or one blue are put in A or F, there are only three places where the card of different color adjoins each other.
Thus, in this case, both ends of the arrangement should be reds.
Then ways of arranging the one blue and a set of two blue in the B ~ E are 4 × 3 = 12 ways.
Case 2 : Three blue cards are put in separately in A ~ F.
As for the way of arranging which suits the conditions in question, the both ends must be blue.
Since the remaining one blue card should be in either of B ~ E, it is four ways.
Therefore there are 12 + 4 = 16 ways in total.
(2) Since blue cards are all blue1, it examines how many ways of arrangement of red cards which are put in order by distinguishing a red number.
There are five ways for red3 to be put on one of five places.
And there are （4 × 3） / (2 × 1) = six ways for two cards of red1 to be put on two of four remaining places.
Thus, 5 × 6 = 30 ways.
Since there is 30 ways of arranging red cards in each 16 ways as determined in (1), total ways of arrangement to be obtained as 16 × 30 = 480 ways.
Problem 9
I arrange them in a line.
Answer the following questions about the way of arrangement that there are just four places where the card of different color adjoins each other like red2-blue1-blue2-red1-red1-red3-blue1-red2 for example.
(1) When distinguishing the color of a card and not distinguishing a number, how many ways of arrangement is there?
(2) When distinguishing both the color of a card and a number, how many ways of arrangement is there?
Answer
(1) 16 ways
(2) 480 ways
Solution
(1) It is considered that putting five red cards in order as shown in a figure and arranging blue cards in A~F.
Case 1 : A set of two blue and one blue are put in order like the example of problem sentence.
If a set of blue or one blue are put in A or F, there are only three places where the card of different color adjoins each other.
Thus, in this case, both ends of the arrangement should be reds.
Then ways of arranging the one blue and a set of two blue in the B ~ E are 4 × 3 = 12 ways.
Case 2 : Three blue cards are put in separately in A ~ F.
As for the way of arranging which suits the conditions in question, the both ends must be blue.
Since the remaining one blue card should be in either of B ~ E, it is four ways.
Therefore there are 12 + 4 = 16 ways in total.
(2) Since blue cards are all blue1, it examines how many ways of arrangement of red cards which are put in order by distinguishing a red number.
There are five ways for red3 to be put on one of five places.
And there are （4 × 3） / (2 × 1) = six ways for two cards of red1 to be put on two of four remaining places.
Thus, 5 × 6 = 30 ways.
Since there is 30 ways of arranging red cards in each 16 ways as determined in (1), total ways of arrangement to be obtained as 16 × 30 = 480 ways.
Problem 9
The quadrangle ABCD in a figure is a parallelogram. QS and BC are parallel and RT and CD are also parallel.
The point P which is an intersection of QS and RT is on the diagonal line BD.
(The length of BP) : (length of PD) = 2 : 1.
Find the area ratio between the sum of the area of the triangle AQT and CSR and the area of parallelogram ABCD.
The point P which is an intersection of QS and RT is on the diagonal line BD.
(The length of BP) : (length of PD) = 2 : 1.
Find the area ratio between the sum of the area of the triangle AQT and CSR and the area of parallelogram ABCD.
Answer 2 : 9
Solution
In a figure, since BP : PD = 2 : 1, AT : TD = BR : RC = BQ : QA = CS : SD = 2 : 1.
Each area of parallelograms AQPT and PRCS is set to be 2 × 1 = 2.
Then, the sum of the area of △AQT and △CSR can also be set to 2.
The area of parallelogram ABCD can be set to (2+1) × (2+1) = 9.
Therefore, the area ratio is 2 : 9.
Problem 10
Both triangle ABC and ADE in a figure are a right-angled isosceles triangle.
(length of BD) : (length of DC) = 1 : 3.
Find the area ratio of triangle DCE and triangle ABC.
Answer
3 : 8
Solution
As for △ABD and △ACE in a figure, AB = AC and AD = AE according to the conditions of the problem sentence.
Moreover, ∠BAD = ∠CAE = 90 degrees - ∠DAC.
Thus, △ABD and △ACE are congruent.
If it is set that CE = BD =1, since CE = BD, CE = 1.
Since ∠ACE = ∠ABD = 45 degrees and ∠ACB = 45 degree, it turns out to be ∠ECD = 90 degrees.
Therefore, as for △ABC, length of base = BD + DC = 1 + 3 = 4, height is half of BC, it is 4 × 1/2 = 2 and the area is 4 × 2 / 2 = 4.
As for △DCE, lenght of base = DC = 3, height is CE = 1 and the area is 3 × 1 / 2 = 1.5.
△DCE : △ABC = 1.5 : 4 = 3 : 8.
Solution
In a figure, since BP : PD = 2 : 1, AT : TD = BR : RC = BQ : QA = CS : SD = 2 : 1.
Each area of parallelograms AQPT and PRCS is set to be 2 × 1 = 2.
Then, the sum of the area of △AQT and △CSR can also be set to 2.
The area of parallelogram ABCD can be set to (2+1) × (2+1) = 9.
Therefore, the area ratio is 2 : 9.
Problem 10
(length of BD) : (length of DC) = 1 : 3.
Find the area ratio of triangle DCE and triangle ABC.
Answer
3 : 8
Solution
As for △ABD and △ACE in a figure, AB = AC and AD = AE according to the conditions of the problem sentence.
Moreover, ∠BAD = ∠CAE = 90 degrees - ∠DAC.
Thus, △ABD and △ACE are congruent.
If it is set that CE = BD =1, since CE = BD, CE = 1.
Since ∠ACE = ∠ABD = 45 degrees and ∠ACB = 45 degree, it turns out to be ∠ECD = 90 degrees.
Therefore, as for △ABC, length of base = BD + DC = 1 + 3 = 4, height is half of BC, it is 4 × 1/2 = 2 and the area is 4 × 2 / 2 = 4.
As for △DCE, lenght of base = DC = 3, height is CE = 1 and the area is 3 × 1 / 2 = 1.5.
△DCE : △ABC = 1.5 : 4 = 3 : 8.
Problem 11
The ratio of the length of AD to BD is 3 : 2.
The ratio of the length of BE to EC is 5 : 2.
Find the angle of X.
Answer
67 degrees
Solution
As for the figure below, AB and CG is parallel and G is a point on the extension of the line segment AE.
As △ABE and △CGE are homothetic and BE : CE = 5 : 2, AB : CG = 5 : 2.
Furthermore △ADF and △CGF are homothetic and AD : CG = 3 : 2, AF : GF = 3 : 2.
Thus GF = AF × 2/3 = 6 cm × 2/3 = 4 cm.
As AC = AG = 10 cm, ∠AGC = (180 - 46) / 2 = 67 degrees.
Therefore X = ∠AGC = 67 degrees.
Problem 12
Make a large cube of 5cm one side by piling up 125 dices of 1cm one side without a gap.
This large cube is cut by the plane passing through the three vertex of the cube as shown in a figure.
As for the portion of the triangular pyramid below the cutting plane, find the following each number.
The 2nd layer of cutting plane is indicated in Fig. 4 in the 2nd step of a right-angled isosceles triangle.
Moreover, the 5th layer of cutting plane is indicated in Fig. 5 in the 5th step of a right-angled isosceles triangle.
As mentioned above, the number of small cubes, the number of the shadow triangles (A), the number of yellow triangles (B), and the number of dices which are not cut will become as it is shown in a lower table.
Problem 13
This large cube is cut by the plane passing through the three vertex of the cube as shown in a figure.
As for the portion of the triangular pyramid below the cutting plane, find the following each number.
(1)The number of the dices which is not cut.
(2)The number of a solid with the volume larger than 0.5 cm3 in the solid made by cutting a dice.
(3)The number of a solid with the volume smaller than 0.5 cm3 in the solid made by cutting a dice.
Answer(1) 10 pieces
(2) 10 pieces
(3) 15 pieces
Solution
The dices being cut are two kinds of shapes of Fig. 2.
(2) 10 pieces
(3) 15 pieces
Solution
The dices being cut are two kinds of shapes of Fig. 2.
The upper part of (A) is separated and the lower part (A) with smaller volume one remains.
The volume of (A) is smaller than 0.5 cm3.
The upper part of (B) is also separated and the lower part (B) with larger volume one remains.
The volume of (B) is larger than 0.5 cm3.
In Fig.1, the 1st step (Fig. 3) of big shadow and yellow right-angled isosceles triangle becomes a cutting plane of the 1st layer from the top of the large cube with 5th layer.
The volume of (A) is smaller than 0.5 cm3.
The upper part of (B) is also separated and the lower part (B) with larger volume one remains.
The volume of (B) is larger than 0.5 cm3.
In Fig.1, the 1st step (Fig. 3) of big shadow and yellow right-angled isosceles triangle becomes a cutting plane of the 1st layer from the top of the large cube with 5th layer.
As mentioned above, the number of small cubes, the number of the shadow triangles (A), the number of yellow triangles (B), and the number of dices which are not cut will become as it is shown in a lower table.
1st layer
|
2nd
|
3rd
|
4th
|
5th
|
Sum
| |
Number of (A)
|
1
|
2
|
3
|
4
|
5
|
15
|
Number of (B)
|
0
|
1
|
2
|
3
|
4
|
10
|
Number of dices
|
1
|
3
|
6
|
10
|
15
|
35
|
Number of dices
with no cut
|
0
|
0
|
1
|
3
|
6
|
10
|
Problem 13
Find the volume of the solid which is made by assembling the development view of the figure.
Each triangular face is an equilateral triangle.
Each face of a hexagon is a made figure where one side cuts out two right-angled isosceles triangles whose equal lengths of two sides are 1 cm from the square which is 2 cm one side.
Each triangular face is an equilateral triangle.
Each face of a hexagon is a made figure where one side cuts out two right-angled isosceles triangles whose equal lengths of two sides are 1 cm from the square which is 2 cm one side.
Answer
22/3 cm3
Solution
It is required to surmise that this solid is made by cutting off four corners of the cube with 2cm one side based on the statement that there are six hexagon faces in the development as a hint to solve this problem.
As shown in a figure, the solid which is made by assembling this developed view becomes what cut out the triangular pyramid from four corners of the cube wit 2cm one side.
Therefore, the volume of this solid is 2 × 2 × 2 - 1 × 1 × 1/2 × 1 × 1/3 × 4 = 8 - 23 = 22/3cm3.
It is required to surmise that this solid is made by cutting off four corners of the cube with 2cm one side based on the statement that there are six hexagon faces in the development as a hint to solve this problem.
As shown in a figure, the solid which is made by assembling this developed view becomes what cut out the triangular pyramid from four corners of the cube wit 2cm one side.
Therefore, the volume of this solid is 2 × 2 × 2 - 1 × 1 × 1/2 × 1 × 1/3 × 4 = 8 - 23 = 22/3cm3.
