Time : 50 minutes, Passing mark : 70%
Answer : Please click "Answer >>" at the end of the problem.
Problem 1
Find X.
(2003 + X ) × 1/4 × 1/5 × 1/6 × 1/8 + 7/10 = 17/6
Problem 2
There are some one-yen coins.
When I change these into five-yen coins as many as possible, the total number of coins is decreased by 60 pieces.
Furthermore, when I changes those coins into ten-coins as many as possible, the total number of coins becomes ten pieces.
Find the number of one-yen coins first.
Problem 3
Find X.
(2003 + X ) × 1/4 × 1/5 × 1/6 × 1/8 + 7/10 = 17/6
Problem 2
There are some one-yen coins.
When I change these into five-yen coins as many as possible, the total number of coins is decreased by 60 pieces.
Furthermore, when I changes those coins into ten-coins as many as possible, the total number of coins becomes ten pieces.
Find the number of one-yen coins first.
Problem 3
The ratio of the amount of the money that Taro and Jiro had was 7 : 4.
When Taro gave 150 yen to Jiro, the ratio of the money that Taro and Jiro had became 8 : 5.
Find the amount of the money Taro had first.
Problem 4
As for a price of an apple, a persimmon, and an orange, the price of apple is highest and that of orange is cheapest.
When Taro gave 150 yen to Jiro, the ratio of the money that Taro and Jiro had became 8 : 5.
Find the amount of the money Taro had first.
Problem 4
As for a price of an apple, a persimmon, and an orange, the price of apple is highest and that of orange is cheapest.
When I purchase five pieces in all including at least one of each fruit, the total amount differs in six kinds of 580 yen, 600 yen, 620 yen, 640 yen, 660 yen, and 700 yen depending on the combination of five pieces.
Find the sum total when I purchase one apple, one persimmon, and one orange.
Moreover, find the price of one persimmon.
Problem 5
I plan to distribute notes in a certain class.
Problem 5
I plan to distribute notes in a certain class.
Supposing I distribute four notes to to every boy and six to every girl, seven notes will remain.
Supposing I distribute six to every boy and four to every girl, nine notes fall short.
Supposing I distribute five to every boy and seven to every girl, 33 notes fall short.
Find the number of boys in this class.
Moreover, find the number of notes.
Problem 6
Yesterday I could exchange some amount of Japanese yen I had into 100 US dollars.
Today the Japanese yen of the same amount as yesterday can be exchanged for 1.02 times as many dollars, but I exchanged 12750 yen into 100 US dollars.
Find the amount of money of the Japanese yen which it had yesterday.
Problem 6
Yesterday I could exchange some amount of Japanese yen I had into 100 US dollars.
Today the Japanese yen of the same amount as yesterday can be exchanged for 1.02 times as many dollars, but I exchanged 12750 yen into 100 US dollars.
Find the amount of money of the Japanese yen which it had yesterday.
Problem 7
When integer of 6 digits 5ABC15 becomes a multiple of 999, find the integer ABC of 3 digits.
Problem 8
When integer of 6 digits 5ABC15 becomes a multiple of 999, find the integer ABC of 3 digits.
Problem 8
When doing some shopping, the consumption tax is 5% of sales price and less than 1 yen is omitted.
For example, if sales price is 219 yen, 10 yen consumption tax will be cost and 229 yen will be paid.
When sales price is 220 yen, 231 yen will be paid.
Therefore, 230 yen does not emerge with the amount of money including the consumption tax.
Find an amount of money nearest to 1000 yen among those which do not emerge with the amount of money of including consumption tax.
Problem 9
I arrange four number of 1,2,3,4 to one line.
For example, I arrange it as "2,4,1,3".
Considering a set of two numbers, there are three cases that big number is on the left side than small number which is "2 and 1", "4 and 1", and "4 and 3".
This case is called that “the number of inverse order of “2,4,1,3” is 3.”
For example, if sales price is 219 yen, 10 yen consumption tax will be cost and 229 yen will be paid.
When sales price is 220 yen, 231 yen will be paid.
Therefore, 230 yen does not emerge with the amount of money including the consumption tax.
Find an amount of money nearest to 1000 yen among those which do not emerge with the amount of money of including consumption tax.
Problem 9
I arrange four number of 1,2,3,4 to one line.
For example, I arrange it as "2,4,1,3".
Considering a set of two numbers, there are three cases that big number is on the left side than small number which is "2 and 1", "4 and 1", and "4 and 3".
This case is called that “the number of inverse order of “2,4,1,3” is 3.”
The table shows the number of inverse order and arrangement ways of each number of inverse order about all 24 ways of arrangement of 1,2,3,4.
When I arrange five numbers of 1,2,3,4,5 to one line, I think about the number of inverse order in the same way as a case of the number of four.
Answer the following questions.
(1) What is the number of inverse order when I arrange it as "5,2,4,1,3"?
(2) In reference to an upper table, considering 120 arrangement ways of 1,2,3,4,5, how many ways are there the number of inverse order is 5?
Problem 10
In a figure, a quadrangle ABCD is a parallelogram.
(length of AE) : (length of ED) = 2 : 3.
The area of the triangle ABE is 80 cm2.
Find the area of triangle CDF.
Problem 11
I make a large equilateral triangle by arranging small equilateral triangle 1cm of the length of one side side by side with no gap nor overlapping.
How many small equilateral triangles do I need to make a large equilateral triangle of 20cm of one side?
Problem 12
There is a paper of rectangle as shown in a figure.
The length of vertical side, horizontal side and diagonal line are 15 cm, 20 cm and 25 cm respectively.
This paper is folded so that A and C may overlap.
Find the sum total of a triangular area with which paper has not overlapped in this case.
Problem 13
The shadow area of a figure is a figure which is a combination of three equilateral triangles with 6 cm one side cm and three sectors 6 cm in radius.
When a circle 1 cm in radius takes one round touching the circumference of this figure, find the area of the portion along which this circle passes.
Pi is assumed to be 3.14.
Problem 14
There is a big wall which stands vertically to the ground.
The bottom of the square pillar of Fig.2 is a trapezoid as shown in Fig.1.
When I arrange five numbers of 1,2,3,4,5 to one line, I think about the number of inverse order in the same way as a case of the number of four.
Answer the following questions.
(1) What is the number of inverse order when I arrange it as "5,2,4,1,3"?
(2) In reference to an upper table, considering 120 arrangement ways of 1,2,3,4,5, how many ways are there the number of inverse order is 5?
Problem 10
In a figure, a quadrangle ABCD is a parallelogram.
(length of AE) : (length of ED) = 2 : 3.
The area of the triangle ABE is 80 cm2.
Find the area of triangle CDF.
Problem 11
I make a large equilateral triangle by arranging small equilateral triangle 1cm of the length of one side side by side with no gap nor overlapping.
How many small equilateral triangles do I need to make a large equilateral triangle of 20cm of one side?
Problem 12
There is a paper of rectangle as shown in a figure.
The length of vertical side, horizontal side and diagonal line are 15 cm, 20 cm and 25 cm respectively.
This paper is folded so that A and C may overlap.
Find the sum total of a triangular area with which paper has not overlapped in this case.
Problem 13
The shadow area of a figure is a figure which is a combination of three equilateral triangles with 6 cm one side cm and three sectors 6 cm in radius.
When a circle 1 cm in radius takes one round touching the circumference of this figure, find the area of the portion along which this circle passes.
Pi is assumed to be 3.14.
Problem 14
There is a big wall which stands vertically to the ground.
The bottom of the square pillar of Fig.2 is a trapezoid as shown in Fig.1.
The height is 30 cm.
It is placed as the side CD is parallel to the wall and the distance to the wall is 10 cm.
The angle to look up at the sun from the ground is 45 degrees.
The shadow area of Fig.2 expresses the shadow made on the ground of this quadratic prism.
Find the area of the whole shadow made to the wall by this quadratic prism.
Problem 15
Solid O-ABCD in a figure is a right quadrangular pyramid.
The bottom is a square and all the length of OA, OB, OC and OD is equal.
Moreover, E and F divide OB and OD into the ratio of 3 : 1, respectively.
The intersection of the plane which passes along three point A, E, F and the side OC is set to G.
Find OG : GC.
Moreover, in two portions of the right quadrangular pyramid divided by this plane, find the volume ratio of the volume of a portion containing O and the volume of the right quadrangular pyramid.
It is placed as the side CD is parallel to the wall and the distance to the wall is 10 cm.
The angle to look up at the sun from the ground is 45 degrees.
The shadow area of Fig.2 expresses the shadow made on the ground of this quadratic prism.
Find the area of the whole shadow made to the wall by this quadratic prism.
Problem 15
Solid O-ABCD in a figure is a right quadrangular pyramid.
The bottom is a square and all the length of OA, OB, OC and OD is equal.
Moreover, E and F divide OB and OD into the ratio of 3 : 1, respectively.
The intersection of the plane which passes along three point A, E, F and the side OC is set to G.
Find OG : GC.
Moreover, in two portions of the right quadrangular pyramid divided by this plane, find the volume ratio of the volume of a portion containing O and the volume of the right quadrangular pyramid.
<Problem & Answer>
Problem 1
Find X.
(2003 + X ) × 1/4 × 1/5 × 1/6 × 1/8 + 7/10 = 17/6
Problem 2
There are some one-yen coins.
When I change these into five-yen coins as many as possible, the total number of coins is decreased by 60 pieces.
Furthermore, when I changes those coins into ten-coins as many as possible, the total number of coins becomes ten pieces.
Find the number of one-yen coins first.
Answer
77 pieces
Solution
Whenever changing five pieces of one-yen coins into one piece of five-yen coin, the number of coins decreases by 5 - 1 = four pieces.
Since the total number of coins are decreased by 60 pieces, the number of five-yen coins changed in total is 60 / 4 = 15 pieces.
Two pieces of five-yen coins become one piece of ten-yen coin.
In case 15 pieces of five-yen coins are changed into ten-yen coins, there are seven pieces of ten-yen coins and one piece of five-yen coin based on the calculation that 15 / 2 = 7 remainder 1.
Since total number of coins are ten pieces, it is found that there are 10 - 7 - 1 = 2 pieces of one-yen coins remained.
Therefore the number of one-yen coins is 10 × 7 + 5 × 1 + 2 = 77 pieces.
Problem 3
The ratio of the amount of the money that Taro and Jiro had was 7 : 4.
When Taro gave 150 yen to Jiro, the ratio of the money that Taro and Jiro had became 8 : 5.
Find the amount of the money Taro had first.
Problem 4
As for a price of an apple, a persimmon, and an orange, the price of apple is highest and that of orange is cheapest.
When I purchase five pieces in all including at least one of each fruit, the total amount differs in six kinds of 580 yen, 600 yen, 620 yen, 640 yen, 660 yen, and 700 yen depending on the combination of five pieces.
Find the sum total when I purchase one apple, one persimmon, and one orange.
Problem 5
I plan to distribute notes in a certain class.
Problem 6
Yesterday I could exchange some amount of Japanese yen I had into 100 US dollars.
Today the Japanese yen of the same amount as yesterday can be exchanged for 1.02 times as many dollars, but I exchanged 12750 yen into 100 US dollars.
Find the amount of money of the Japanese yen which it had yesterday.
Problem 14
There is a big wall which stands vertically to the ground.
The bottom of the square pillar of Fig.2 is a trapezoid as shown in Fig.1.
The height is 30 cm.
It is placed as the side CD is parallel to the wall and the distance to the wall is 10 cm.
The angle to look up at the sun from the ground is 45 degrees.
The shadow area of Fig.2 expresses the shadow made on the ground of this quadratic prism.
Find the area of the whole shadow made to the wall by this quadratic prism.
Problem 15
Solid O-ABCD in a figure is a right quadrangular pyramid.
The bottom is a square and all the length of OA, OB, OC and OD is equal.
Moreover, E and F divide OB and OD into the ratio of 3 : 1, respectively.
The intersection of the plane which passes along three point A, E, F and the side OC is set to G.
Find OG : GC.
Moreover, in two portions of the right quadrangular pyramid divided by this plane, find the volume ratio of the volume of a portion containing O and the volume of the right quadrangular pyramid.
(2003 + X ) × 1/4 × 1/5 × 1/6 × 1/8 + 7/10 = 17/6
Answer
45
Problem 2
There are some one-yen coins.
When I change these into five-yen coins as many as possible, the total number of coins is decreased by 60 pieces.
Furthermore, when I changes those coins into ten-coins as many as possible, the total number of coins becomes ten pieces.
Find the number of one-yen coins first.
77 pieces
Solution
Whenever changing five pieces of one-yen coins into one piece of five-yen coin, the number of coins decreases by 5 - 1 = four pieces.
Since the total number of coins are decreased by 60 pieces, the number of five-yen coins changed in total is 60 / 4 = 15 pieces.
Two pieces of five-yen coins become one piece of ten-yen coin.
In case 15 pieces of five-yen coins are changed into ten-yen coins, there are seven pieces of ten-yen coins and one piece of five-yen coin based on the calculation that 15 / 2 = 7 remainder 1.
Since total number of coins are ten pieces, it is found that there are 10 - 7 - 1 = 2 pieces of one-yen coins remained.
Therefore the number of one-yen coins is 10 × 7 + 5 × 1 + 2 = 77 pieces.
Problem 3
The ratio of the amount of the money that Taro and Jiro had was 7 : 4.
When Taro gave 150 yen to Jiro, the ratio of the money that Taro and Jiro had became 8 : 5.
Find the amount of the money Taro had first.
Answer
4550 yen
Solution
Since the sum total of two persons' money in hand does not change, the sum total is set to (7 + 4) × (8 + 5) = 143.
The money in hand of Taro is 143 × 7/11 = 91.
After giving 150 yen to Jiro, it is 143 × 8/13 =88.
4550 yen
Solution
Since the sum total of two persons' money in hand does not change, the sum total is set to (7 + 4) × (8 + 5) = 143.
The money in hand of Taro is 143 × 7/11 = 91.
After giving 150 yen to Jiro, it is 143 × 8/13 =88.
91 - 88 = 3 is equivalent to 150 yen.
The amount of the money Taro had first was 150 yen / 3 × 91 = 4550 yen.
The amount of the money Taro had first was 150 yen / 3 × 91 = 4550 yen.
As for a price of an apple, a persimmon, and an orange, the price of apple is highest and that of orange is cheapest.
When I purchase five pieces in all including at least one of each fruit, the total amount differs in six kinds of 580 yen, 600 yen, 620 yen, 640 yen, 660 yen, and 700 yen depending on the combination of five pieces.
Find the sum total when I purchase one apple, one persimmon, and one orange.
Moreover, find the price of one persimmon.
Answer
380 yen,
Answer
380 yen,
120 yen
Problem 5
I plan to distribute notes in a certain class.
Supposing I distribute four notes to to every boy and six to every girl, seven notes will remain.
Supposing I distribute six to every boy and four to every girl, nine notes fall short.
Supposing I distribute five to every boy and seven to every girl, 33 notes fall short.
Find the number of boys in this class. Moreover, find the number of notes.
Answer
24 persons,
199
Problem 6
Yesterday I could exchange some amount of Japanese yen I had into 100 US dollars.
Today the Japanese yen of the same amount as yesterday can be exchanged for 1.02 times as many dollars, but I exchanged 12750 yen into 100 US dollars.
Find the amount of money of the Japanese yen which it had yesterday.
Answer
Problem 7
When integer of 6 digits 5ABC15 becomes a multiple of 999, find the integer ABC of 3 digits.
13005 yen
Solution
As for the amount of Japanese yen which can be exchanged into 100 US dollar, it is 12750 yen / 100 = 127.5 yen.
Necessary amount of Japanese yen for exchanging 102 US dollar is 127.5 × 102 = 13005 yen.
And it is the amount I exchanged into 100 US dollar yesterday.
Problem 7
When integer of 6 digits 5ABC15 becomes a multiple of 999, find the integer ABC of 3 digits.
Answer
Problem 8
When doing some shopping, the consumption tax is 5% of sales price and less than 1 yen is omitted.
For example, if sales price is 219 yen, 10 yen consumption tax will be cost and 229 yen will be paid.
When sales price is 220 yen, 231 yen will be paid.
Therefore, 230 yen does not emerge with the amount of money including the consumption tax.
Find an amount of money nearest to 1000 yen among those which do not emerge with the amount of money of including consumption tax.
Answer
Problem 9
I arrange four number of 1,2,3,4 to one line.
For example, I arrange it as "2,4,1,3".
Considering a set of two numbers, there are three cases that big number is on the left side than small number which is "2 and 1", "4 and 1", and "4 and 3".
This case is called that “the number of inverse order of “2,4,1,3” is 3.”
844
Solution
When 5ABC15 is divided by 999, it is considered what digit a quotient will be.
Since it is 5ABC15/1000=5AB.C15, the quotient of 5ABC15 / 999 is also understood that it becomes three digits.
Assuming the three digits quotient is DEF and 999 = 1000 - 1, it is calculated as follows.
(1000 - 1) × DEF = DEF000 - DEF = 5ABC15
DEF000
- DEF
5ABC15
Based on this formula, it is found that D = 5, E = 8, F = 5.
Therefore, 585000 - 585 = 584415, then ABC = 844.
(1000 - 1) × DEF = DEF000 - DEF = 5ABC15
DEF000
- DEF
5ABC15
Based on this formula, it is found that D = 5, E = 8, F = 5.
Therefore, 585000 - 585 = 584415, then ABC = 844.
Problem 8
When doing some shopping, the consumption tax is 5% of sales price and less than 1 yen is omitted.
For example, if sales price is 219 yen, 10 yen consumption tax will be cost and 229 yen will be paid.
When sales price is 220 yen, 231 yen will be paid.
Therefore, 230 yen does not emerge with the amount of money including the consumption tax.
Find an amount of money nearest to 1000 yen among those which do not emerge with the amount of money of including consumption tax.
Answer
1007 yen
Solution
In the case of 19 yen of the sales price, the amount of money including the consumption tax is 19 yen as 19 × 1.05 = 19.95.
In the case of 20 yen of the sales price, the amount of money including the consumption tax is 21 yen as 20 × 1.05 = 21.
From this, the smallest amount of money not to emerge with the amount of money including the consumption tax is 20 yen.
Next, the second smallest amount of money not to emerge with the amount of money including the consumption is 41 yen according to the calculation of 39 × 1.05 = 40.95 and 40 × 1.05 = 42.
The third is 62 yen according to the calculation of 59 × 1.05 = 61.95 and 60 × 1.05 = 63.
In this way, the amount of money that does not emerge as the amount of money including the consumption tax is to be found every 21 yen from 20 yen to 41 yen, 62 yen, ... .
Thus, the amount nearest to 1000 yen can be found in the number of 20 yen +21 yen × □.
(1000 - 20) / 21 = 46.6.
In the case of 20 yen of the sales price, the amount of money including the consumption tax is 21 yen as 20 × 1.05 = 21.
From this, the smallest amount of money not to emerge with the amount of money including the consumption tax is 20 yen.
Next, the second smallest amount of money not to emerge with the amount of money including the consumption is 41 yen according to the calculation of 39 × 1.05 = 40.95 and 40 × 1.05 = 42.
The third is 62 yen according to the calculation of 59 × 1.05 = 61.95 and 60 × 1.05 = 63.
In this way, the amount of money that does not emerge as the amount of money including the consumption tax is to be found every 21 yen from 20 yen to 41 yen, 62 yen, ... .
Thus, the amount nearest to 1000 yen can be found in the number of 20 yen +21 yen × □.
(1000 - 20) / 21 = 46.6.
20 + 21 × 46 = 986, 20 + 21 × 47 = 1007.
It is 1,007 yen to be the nearest to 1,000 yen.
It is 1,007 yen to be the nearest to 1,000 yen.
Problem 9
I arrange four number of 1,2,3,4 to one line.
For example, I arrange it as "2,4,1,3".
Considering a set of two numbers, there are three cases that big number is on the left side than small number which is "2 and 1", "4 and 1", and "4 and 3".
This case is called that “the number of inverse order of “2,4,1,3” is 3.”
The table shows the number of inverse order and arrangement ways of each number of inverse order about all 24 ways of arrangement of 1,2,3,4.
When I arrange five numbers of 1,2,3,4,5 to one line, I think about the number of inverse order in the same way as a case of the number of four.
Answer the following questions.
(1) What is the number of inverse order when I arrange it as "5,2,4,1,3"?
(2) In reference to an upper table, considering 120 arrangement ways of 1,2,3,4,5, how many ways are there the number of inverse order is 5?
Answer
Problem 10
In a figure, a quadrangle ABCD is a parallelogram.
(length of AE) : (length of ED) = 2 : 3.
The area of the triangle ABE is 80 cm2.
Find the area of triangle CDF.
When I arrange five numbers of 1,2,3,4,5 to one line, I think about the number of inverse order in the same way as a case of the number of four.
Answer the following questions.
(1) What is the number of inverse order when I arrange it as "5,2,4,1,3"?
(2) In reference to an upper table, considering 120 arrangement ways of 1,2,3,4,5, how many ways are there the number of inverse order is 5?
Answer
(1) 7
(2) 22 ways
(2) 22 ways
Solution
(1) The set of two numbers that big number is on the left than small number is "5 and 2", "5 and 4", "5 and 1", "5 and 3", "2 and 1", "4 and 1",and "4 and 3".
The number of inverse order is 7.
The number of inverse order is 7.
(2) Consider by case analysis by the position 5.
That is, in the case of ① 5XXXX, ② X5XXX, ③ XX5XX, ④ XXX5X and ⑤ XXXX5.
Since 5 is the biggest number in five numbers, all numbers which are on the right side of 5 are counted as the number of inverse order.
That is, in the case of ① 5XXXX, ② X5XXX, ③ XX5XX, ④ XXX5X and ⑤ XXXX5.
Since 5 is the biggest number in five numbers, all numbers which are on the right side of 5 are counted as the number of inverse order.
① According to (1), the number of reverse order is four in case of 5 being in the leftmost.
Thus, the number of reverse order to become five in total, the number of inverse order in the arrangement of numbers of four other than 5, it is 5-4 = 1.
Thus, the number of reverse order to become five in total, the number of inverse order in the arrangement of numbers of four other than 5, it is 5-4 = 1.
How arranged in such a case, there are three ways from the table above.
② As there are three numbers on the right side of 5, the number of reverse order of 5 is found 3.
Then the number of reverse order of how to arrange in the numbers of other four numbers of 1 ~ 4 should be 5 - 3 = 2.
According to the table, there are five ways in case the number of reverse order is 2.
③ As there are two numbers on the right side of 5, the number of reverse order of 5 is found 2.
Then the number of reverse order of how to arrange in the numbers of other four numbers of 1 ~ 4 should be 5 - 2 = 3.
According to the table, there are six ways in case the number of reverse order is 3.
④ As there are one number on the right side of 5, the number of reverse order of 5 is found 1.
Then the number of reverse order of how to arrange in the numbers of other four numbers of 1 ~ 4 should be 5 - 1 = 4.
According to the table, there are five ways in case the number of reverse order is 4.
⑤ As there are no number on the right side of 5 which is onthe rightmost, the number of reverse order of 5 is found 0.
Then the number of reverse order of how to arrange in the numbers of other four numbers of 1 ~ 4 should be 5 - 0 = 5.
According to the table, there are three ways in case the number of reverse order is 5.
In total there are 3 + 5 + 6 + 5 + 3 = 22 ways.
Problem 10
In a figure, a quadrangle ABCD is a parallelogram.
(length of AE) : (length of ED) = 2 : 3.
The area of the triangle ABE is 80 cm2.
Find the area of triangle CDF.
Answer
75 cm2
Solution
Since the length ratio of a base of △ABE and △CED is AE : ED = 2 : 3, and height is equal, an area ratio is also set to 2 : 3.
Thus, the area of △CED =△ABE × 3/2 = 80 × 3/2 = 120cm2.
Next, △EFD and △CFB are homothetic and a homothetic ratio is ED : BC = 3 : (2 + 3) = 3 : 5.
Next, △EFD and △CFB are homothetic and a homothetic ratio is ED : BC = 3 : (2 + 3) = 3 : 5.
EF : CF is also set to 3 : 5.
Therefore, the area of △CDF is △CED × 5/(3 + 5) = 120 × 5/8 = 75cm2.
Problem 11
I make a large equilateral triangle by arranging small equilateral triangle 1cm of the length of one side side by side with no gap nor overlapping.
How many small equilateral triangles do I need to make a large equilateral triangle of 20cm of one side?
Answer
Problem 12
There is a paper of rectangle as shown in a figure.
The length of vertical side, horizontal side and diagonal line are 15 cm, 20 cm and 25 cm respectively.
This paper is folded so that A and C may overlap.
Find the sum total of a triangular area with which paper has not overlapped in this case.
I make a large equilateral triangle by arranging small equilateral triangle 1cm of the length of one side side by side with no gap nor overlapping.
How many small equilateral triangles do I need to make a large equilateral triangle of 20cm of one side?
Answer
400 pieces
Solution
In order to make a large equilateral triangle, small equilateral triangles are set out 1 piece, 3 pieces, 5 pieces, 7 pieces, -----.
In order to make the equilateral triangle of 2 cm one side, I need 1 + 3 = 4 pieces and that of 3 cm one side, I need 1 + 3 + 5 = 9 pieces.
That is N x N pieces are needed in order to make the equilateral triangle of N cm one side.
In case one side is 20 cm, I need 20 × 20 = 400 pieces.
In order to make the equilateral triangle of 2 cm one side, I need 1 + 3 = 4 pieces and that of 3 cm one side, I need 1 + 3 + 5 = 9 pieces.
That is N x N pieces are needed in order to make the equilateral triangle of N cm one side.
In case one side is 20 cm, I need 20 × 20 = 400 pieces.
Problem 12
There is a paper of rectangle as shown in a figure.
The length of vertical side, horizontal side and diagonal line are 15 cm, 20 cm and 25 cm respectively.
This paper is folded so that A and C may overlap.
Find the sum total of a triangular area with which paper has not overlapped in this case.
Answer
65.625 cm2
Solution
A figure is a figure of the rectangle is being folded so that the vertex A and C may overlap.
A fold is set to EF and the intersection of EF and AC is set to G.
The triangles with which paper has not overlapped are △ABE and △ADF.
Since △C(D) F was turned up to be △ADF, find the area of △ABE and △C(D) F.
The area is calculated as (rectangular area) - (area of a quadrangle AECF).
As for a quadrangle AECF, AF and EC, AE and FC are parallel and since the diagonal line crosses vertically, it is a rhombus.
The area is determined by EC × AB.
△GEC and △ABC are homothetic and BC : AC = GC : EC = 20 : 25 = 4 : 5.
EC = GC × 5/4 = 25/2 × 5/4 = 125/8 =15.625cm.
Therefore, as for the area of rhombus AECF, it is 15.625 × 15 = 234.375cm2.
The area of rectangles is 15 × 20 = 300cm2.
The area to find is 300 - 234.375 = 65.625cm2.
Problem 13
The shadow area of a figure is a figure which is a combination of three equilateral triangles with 6 cm one side cm and three sectors 6 cm in radius.
When a circle 1 cm in radius takes one round touching the circumference of this figure, find the area of the portion along which this circle passes.
Pi is assumed to be 3.14.
The shadow area of a figure is a figure which is a combination of three equilateral triangles with 6 cm one side cm and three sectors 6 cm in radius.
When a circle 1 cm in radius takes one round touching the circumference of this figure, find the area of the portion along which this circle passes.
Pi is assumed to be 3.14.
Answer
86.24 cm2
Solution
When a circle takes one round the shadow area, as shown in Fig. 1, there are three kinds of motions, A, B, and C.
Motion of the B portion becomes a sector which makes the diameter of a circle a radius, as shown in Fig. 2.
This sector is made in six all. The area along which the circle passed for three kinds of every motions is determined.
① In the case of A
It is a form of the rectangle shown in Fig. 1, and there are three, the area is 6 × 2 × 3 = 36cm2.
① In the case of A
It is a form of the rectangle shown in Fig. 1, and there are three, the area is 6 × 2 × 3 = 36cm2.
② In the case of C
It becomes a form of C shown in Fig.1. The sum total of the central angle of three sectors, 360° - (60° × 3) = 180°.
It becomes a form of C shown in Fig.1. The sum total of the central angle of three sectors, 360° - (60° × 3) = 180°.
The area is (8 × 8 - 6 × 6) × 3.14 × 180/360 = 14 × 3.14cm2.
③ In the case of B
The central angle of one sector is 180° - 60° - 90° = 30° as shown in Fig. 2.
The central angle of one sector is 180° - 60° - 90° = 30° as shown in Fig. 2.
Since there are six sectors, total area is 2 × 2 × 3.14 × 30/360 × 6 = 2 × 3.14cm2.
Total area overall is 36 + 14 × 3.14 + 2 × 3.14 = 86.24 cm2.
Problem 14
There is a big wall which stands vertically to the ground.
The bottom of the square pillar of Fig.2 is a trapezoid as shown in Fig.1.
The height is 30 cm.
It is placed as the side CD is parallel to the wall and the distance to the wall is 10 cm.
The angle to look up at the sun from the ground is 45 degrees.
The shadow area of Fig.2 expresses the shadow made on the ground of this quadratic prism.
Find the area of the whole shadow made to the wall by this quadratic prism.
Answer
175 cm2
Solution
It turns out that the shadow made to the wall as shown in Fig.3 becomes a pentagon.
You can recognize easily if you draw the figure seen from just beside as Fig.4.
The angle which looks up at the sun from the ground is 45 degrees.
The straight line FP and GQ of Fig.3 express the line of light and the angle to the ground is 45 degrees.
Since △FBP is a right-angled isosceles triangle and is FB=30cm, BP = FB = 30cm.
Since BC = CO =10cm, OP = 10cm and X is 10 cm as well.
Thinking the same way, since △GCQ is also a right-angled isosceles triangle, Y is 20 cm.
Therefore, the shadow made to a wall becomes like Fig.5 and the area is 10 x 20 - 5 × 10 / 2 = 175cm2.
Problem 15
Solid O-ABCD in a figure is a right quadrangular pyramid.
The bottom is a square and all the length of OA, OB, OC and OD is equal.
Moreover, E and F divide OB and OD into the ratio of 3 : 1, respectively.
The intersection of the plane which passes along three point A, E, F and the side OC is set to G.
Find OG : GC.
Moreover, in two portions of the right quadrangular pyramid divided by this plane, find the volume ratio of the volume of a portion containing O and the volume of the right quadrangular pyramid.
Answer
3 : 2
9 : 20
9 : 20
Solution
Fig. 1 is a cross section which cut right quadrangular pyramid by the plane which passes along OAC.
Fig. 2 is a cross section which cut by the plane which passes along OBD.
Since bottom ABCD is a square, the line OH which connected the intersection H of AC and BD and the vertex O becomes vertical to both AC and BD.
The intersection of OH and AG is set to I in Fig. 1.
I is a point on EF in Fig. 2, and OF : FD = 3 : 1.
Thus, OI : IH = 3 : 1.
In Fig. 1, in order to calculate OG : GC, △GCJ is made by drawing an auxiliary line.
As for △ACJ, since AH : CH = 1 : 1, IH : JC = 1 : 2.
Considering that OI : IH = 3 : 1, it turns out OI : JC = 3 : 2.
Thus, △GOI and △GCJ are homothetic and a homothetic ratio is 3 : 2.
Therefore, OG : GC = 3 : 2.
In order to compare the volume of the solid containing O with the volume of the right quadrangular pyramid, right quadrangular pyramid are divided into two at the plane OAC.
The volume of the triangular pyramid E-OAG and the triangular pyramid B-AOC is measured.
The bottom of two triangular pyramids is set to △OAG and △OAC, respectively.
Since OG : GC = 3 : 2, the ratio of the area of △OAG to △OAC is 3 : (3+2) = 3 : 5.
The ratio of the height of two triangular pyramids turns into a ratio of EO to BO and is set to 3 : (3+1) = 3 : 4.
The triangular pyramid E-OAG will be the 3/5 × 3/4 = 9/20 times of the volume of the triangular pyramid B-AOC.
Completely think in the same way and the triangular pyramid F-OAG will also be 9/20 times of the volume of the triangular pyramid D-AOC.
Therefore, the volume ratio to be found is 9 : 20.
