Time : 60 minutes
Problem 1
As for two integers, calculation by < , > is defined as the following [Example].
Passing mark : 70%
Answer : End of the problem
Problem 1
As for two integers, calculation by < , > is defined as the following [Example].
[Example]
< 3, 4 > = 1 / 3×(3+4) =1 / 3×7 = 1/21
< 5, 3 > =1 / 5×(5+3) =1 / 5×8= 1/40
(1) Execute the following calculation in accordance with the rule of a [Example].
① < 8,3 > + < 3,8 >
② < 3,6 > + < 7,2 > + < 6,3 > + < 5,2 >
(2) Find all Integer A which is applicable to the formula,
< A,B > = 1 / 72.
Answer the following questions.
< 3, 4 > = 1 / 3×(3+4) =1 / 3×7 = 1/21
< 5, 3 > =1 / 5×(5+3) =1 / 5×8= 1/40
(1) Execute the following calculation in accordance with the rule of a [Example].
① < 8,3 > + < 3,8 >
② < 3,6 > + < 7,2 > + < 6,3 > + < 5,2 >
(2) Find all Integer A which is applicable to the formula,
< A,B > = 1 / 72.
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Problem 2Answer the following questions.
(1) Taro continues walking first 3 km at 4 km/h and afterwards at 3 km/h.
Find the distance he walks in 2 hours.
(2) 2 hours after Taro leaves, Jiro pursues Taro from the same point.
Jiro moves first 4 km at 15 km/h and afterwards at 12 km/h by bicycle.
Find the time when Jiro catches up with Taro after Jiro’s leaving.
Problem 3
There are 48 balls.
These balls are put into five boxes so that it may be applied to the following conditions 1 and 2.
Find the distance he walks in 2 hours.
(2) 2 hours after Taro leaves, Jiro pursues Taro from the same point.
Jiro moves first 4 km at 15 km/h and afterwards at 12 km/h by bicycle.
Find the time when Jiro catches up with Taro after Jiro’s leaving.
Problem 3
These balls are put into five boxes so that it may be applied to the following conditions 1 and 2.
<Condition 1>
Five or more balls are put into every box.
Five or more balls are put into every box.
<Condition 2>
As for every two boxes, the common divisor of the number of ball in each box is 1 only.
Find all groups of the each number of balls in five boxes.
Problem 4
As for every two boxes, the common divisor of the number of ball in each box is 1 only.
Find all groups of the each number of balls in five boxes.
Problem 4
Taro walks around a certain pond with fixed speed and Jiro runs with fixed speed to opposite direction.
Taro passed by Jiro again 10 minutes after passing by Jiro.
Just after first passing, Taro also started running increasing the speed by 120 m/m more than the speed at walking.
Then he passed by Jiro 6 minutes afterward again.
Answer the following questions.
(1) Find the surrounding length of the pond.
(2) After passing first, in order to pass again in 4 minutes and 48 seconds, find the speed per minute which Taro increases from the speed at walking first.
(3) If Taro runs with speed twice the speed of walking first after passing first, he will pass again in 7 minutes and 12 seconds.
Find the speed per minute of Jiro.
Problem 5
Taro passed by Jiro again 10 minutes after passing by Jiro.
Just after first passing, Taro also started running increasing the speed by 120 m/m more than the speed at walking.
Then he passed by Jiro 6 minutes afterward again.
Answer the following questions.
(1) Find the surrounding length of the pond.
(2) After passing first, in order to pass again in 4 minutes and 48 seconds, find the speed per minute which Taro increases from the speed at walking first.
(3) If Taro runs with speed twice the speed of walking first after passing first, he will pass again in 7 minutes and 12 seconds.
Find the speed per minute of Jiro.
Problem 5
The side AB of the right hexagon ABCDEF is equally divided into two and the side CD is equally divided into four.
Find the ratio of the area of the quadrangle BCNM and the hexagon AMNDEF in the figure.
The answer should be by the least integer.
Problem 6
Answer the following questions.
Express an answer by the ratio of the least integer.
(1) There is a waterway through which it flows downward from upside on as shown in the figure below.
The water through the entrance A is divided into the same volume at the fork.
There are five exits located in four stories of waterways but the volume of the water which comes out of exits is not the same.
Find the ratio of the volume of the water of the least and the most.
Find the ratio of the area of the quadrangle BCNM and the hexagon AMNDEF in the figure.
The answer should be by the least integer.
Problem 6
Express an answer by the ratio of the least integer.
(1) There is a waterway through which it flows downward from upside on as shown in the figure below.
The water through the entrance A is divided into the same volume at the fork.
There are five exits located in four stories of waterways but the volume of the water which comes out of exits is not the same.
Find the ratio of the volume of the water of the least and the most.
(2) Next, as shown in Fig. P, the water through A is divided into three and there is a waterway of the solid shape which the same volume of water comes out of B, C, and D.
Waterway as shown in Fig.Q made by combining four of waterways is called waterway of two stories.
Like this, one waterway after another is connected.
The exits of one story, two stories, three stories and four stories of waterways becomes as it is shown in the Fig.1 ~ Fig.4, respectively.
① As for the volume of the water which comes out of exits of the waterway of three stories, find the ratio of the volume of the water of the least and the most.
② As for the volume of the water which comes out of exits of the waterway of four stories, find the ratio of the volume of the water of the least and the most.
Answer
Problem 1
As for two integers, calculation by < , > is defined as the following [Example].
As for two integers, calculation by < , > is defined as the following [Example].
[Example]
< 3, 4 > = 1 / 3×(3+4) =1 / 3×7 = 1/21
< 5, 3 > =1 / 5×(5+3) =1 / 5×8= 1/40
(1) Execute the following calculation in accordance with the rule of a [Example].
① < 8,3 > + < 3,8 >
② < 3,6 > + < 7,2 > + < 6,3 > + < 5,2 >
(2) Find all Integer A which is applicable to the formula,
< A,B > = 1 / 72.
< 3, 4 > = 1 / 3×(3+4) =1 / 3×7 = 1/21
< 5, 3 > =1 / 5×(5+3) =1 / 5×8= 1/40
(1) Execute the following calculation in accordance with the rule of a [Example].
① < 8,3 > + < 3,8 >
② < 3,6 > + < 7,2 > + < 6,3 > + < 5,2 >
(2) Find all Integer A which is applicable to the formula,
< A,B > = 1 / 72.
Answer
Problem 2
Problem 3
Problem 4
(1) ① 1/24
② 1 /10
(2) 1,2,3,4,6,8
② 1 /10
(2) 1,2,3,4,6,8
Solution
(1) ① It calculates correctly as a rule.
< 8,3 > + < 3,8 > = 1 / 8×(8+3) + 1 / 3×(3+8) = 11 / 8×3×11 = 1 / 8×3 = 1/24.
According to the result of ①, it turns out that
< 8,3 > + < 3,8 > = 1 / 8×(8+3) + 1 / 3×(3+8) = 11 / 8×3×11 = 1 / 8×3 = 1/24.
According to the result of ①, it turns out that
< 3,6 > + < 6,3 > =1 / 3×6 = 1/18.
Therefore,
Therefore,
< 3,6 > + < 7,2 > + < 6,3 > + < 5,2 >
=1 / 18 + 1 / 7×9 + 1 / 5×7
= 1 / 2×9 + 1 / 7×9 + 1 / 5×7
= 35+10+18 / 2×9×5×7
= 63 / 2×9×5×7= 1 / 2×5 = 1/10.
(2) < A,B > = 1 / A × (A+B) = 1 / 72.
Integer A which is applicable to the formula A × (A+B) =72 is a divisor of 72.
Divisors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
Then A should be applicable to the formula A × (A+B), A is 1, 2, 3, 4, 6, 8.
Integer A which is applicable to the formula A × (A+B) =72 is a divisor of 72.
Divisors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
Then A should be applicable to the formula A × (A+B), A is 1, 2, 3, 4, 6, 8.
Problem 2
Answer the following questions.
(1) Taro continues walking first 3 km at 4 km/h and afterwards at 3 km/h.
Find the distance he walks in 2 hours.
(2) 2 hours after Taro leaves, Jiro pursues Taro from the same point.
Jiro moves first 4 km at 15 km/h and afterwards at 12 km/h by bicycle.
Find the time when Jiro catches up with Taro after Jiro’s leaving.
Find the distance he walks in 2 hours.
(2) 2 hours after Taro leaves, Jiro pursues Taro from the same point.
Jiro moves first 4 km at 15 km/h and afterwards at 12 km/h by bicycle.
Find the time when Jiro catches up with Taro after Jiro’s leaving.
Answer
(1) 6.75 km
(2) 119/3 minutes
(2) 119/3 minutes
Solution
(1) When Taro moves 3 km at 4 km/h, it takes 3 km/4 km/hr = 3/4 hours.
As he moves at 3km/h in the remaining time, total distance in 2 hours is 3km + 3km/h × (2 - 3/4) = 3 + 3.75 = 6.75km.
(2) The time Jiro moves 4 km at 15 km/h is 4 km / 15 km/h = 4/15 hours.
The distance between Taro and Jiro when Jiro moved 4km is 6.75km + 3 km/h × 4/15 hour - 4km = 3.55km.
Jiro pursues Taro for this distance at 12 km/h.
The time for Jiro to catch up with Taro is 3.55 km / (12 - 3) km/h = 71/180 hours.
Therefore the time when Jiro catches up with Taro after Jiro’s leaving is (4/15 + 71/180) × 60 =119/3 minutes.
Problem 3
There are 48 balls.
These balls are put into five boxes so that it may be applied to the following conditions 1 and 2.
These balls are put into five boxes so that it may be applied to the following conditions 1 and 2.
<Condition 1>
Five or more balls are put into every box.
Five or more balls are put into every box.
<Condition 2>
As for every two boxes, the common divisor of the number of ball in each box is 1 only.
Find all groups of the each number of balls in five boxes.
As for every two boxes, the common divisor of the number of ball in each box is 1 only.
Find all groups of the each number of balls in five boxes.
Answer
Solution
According to the condition 2, the number of the balls which put into five boxes is three kinds of numbers which are multiple of 2, multiple of 3, and a prime number.
① In case first two numbers are 5 and 6, the next candidate number is 7.
① In case first two numbers are 5 and 6, the next candidate number is 7.
Since 6 is the multiple of 2 and 3, it is only a prime number that put into the last two boxes.
48 - (5 + 6 + 7) = 30.
The group of two prime numbers with 30 of the sum is 11 + 19 and 13 + 17.
② In case first two numbers are 5 and 6, the next candidate numbers are 8, 9, 10.
However these numbers are multiple of 2 and/or 3, these are unsuitable.
② In case first two numbers are 5 and 6, the next candidate numbers are 8, 9, 10.
However these numbers are multiple of 2 and/or 3, these are unsuitable.
The next candidate is 11 and the following prime number is 13 according to 48 - (5 + 6 + 11) = 25.
There is no applied number based on 25 - 13 = 12.
Therefore, there is no combination of numbers in case first two numbers are 5 and 6.
③ In case first two numbers are 5 and 7, the next candidate number is 8.
③ In case first two numbers are 5 and 7, the next candidate number is 8.
48 - (5 + 7 + 8) = 28.
In the group of two numbers with the sum of 28, the group of (multiple of 3) + (prime number) or (prime number) + (prime number) should be found.
There are 2 groups, 9+19 and 11+17.
④ In case first two numbers are 5 and 7, the next candidate number is 9.
48 - (5 + 7 + 9) = 27.
In the group of two numbers with the sum of 27, the group of (multiple of 2 with no multiple of 5) + (prime number) or (prime number) + (prime number) should be found.
There is one group, 11+16.
⑤ In case first two numbers are 5 and 7, the next candidate number is 11.
48 - (5 + 7 + 11) = 25.
In the group of two numbers with the sum of 25, the group of (multiple of 2 with no multiple of 5) + (prime number) or (prime number) + (prime number) should be found.
There is one group, 12+13.
⑥ In case first two numbers are 5 and 7, the next candidate number is 13.
48 - (5 + 7 + 13) = 23.
Since the 4th number should be bigger than 13, there is no suitable group of number.
There is no combination of numbers in case first two numbers are 5 and 7.
⑦ In case first two numbers are 5 and 8, the next candidate number is 9.
There is no combination of numbers in case first two numbers are 5 and 7.
⑦ In case first two numbers are 5 and 8, the next candidate number is 9.
48 - (5 + 8 + 9) = 26.
There is no group of two prime numbers of 11 or more with the sum of 26.
The next candidate number is 11.
There is no group of two prime numbers of 11 or more with the sum of 26.
The next candidate number is 11.
48 - (5 + 8 + 11) = 24.
Since the 4th number should be 13 or more, the sum total cannot be set to 24 by two.
There is no suitable group of number.
There is no combination of numbers in case first two numbers are 5 and 8.
There is no combination of numbers in case first two numbers are 5 and 8.
⑧ In case first two numbers are 5 and 9, there is no combination of numbers according to the same reason as ⑦.
⑨ In case first two numbers are 6 and 7, the next candidate number is 11.
48 - (6 + 7 + 11) = 24.
This is same as ⑦ and there is no combination of numbers in case first two numbers are 6 and 7.
⑩ In case first two numbers are 7 and 8, the next candidate number is 9.
48 - (7 + 8 + 9) = 24.
In the group of two numbers with the sum of 24, the group of (prime number) + (prime number) should be found.
There is one group, 11+13.
⑪ As for the group of 7, 8 and 11 or more, there is no combination of numbers can not be found.
Therefore there are seven groups in total to be found.
Problem 4
Taro walks around a certain pond with fixed speed and Jiro runs with fixed speed to opposite direction.
Taro passed by Jiro again 10 minutes after passing by Jiro.
Just after first passing, Taro also started running increasing the speed by 120 m/m more than the speed at walking.
Then he passed by Jiro 6 minutes afterward again.
Answer the following questions.
(1) Find the surrounding length of the pond.
(2) After passing first, in order to pass again in 4 minutes and 48 seconds, find the speed per minute which Taro increases from the speed at walking first.
(3) If Taro runs with speed twice the speed of walking first after passing first, he will pass again in 7 minutes and 12 seconds.
Find the speed per minute of Jiro.
Taro passed by Jiro again 10 minutes after passing by Jiro.
Just after first passing, Taro also started running increasing the speed by 120 m/m more than the speed at walking.
Then he passed by Jiro 6 minutes afterward again.
Answer the following questions.
(1) Find the surrounding length of the pond.
(2) After passing first, in order to pass again in 4 minutes and 48 seconds, find the speed per minute which Taro increases from the speed at walking first.
(3) If Taro runs with speed twice the speed of walking first after passing first, he will pass again in 7 minutes and 12 seconds.
Find the speed per minute of Jiro.
Answer
Problem 5
(1) 1800 m
(2) 195 m/m
(3) 110 m/m
(2) 195 m/m
(3) 110 m/m
Solution
(1) Set the surrounding length of a pond to 30 of the least common multiple of 6 and 10.
The sum total of two persons' speed when it passes first is 30 / 10 minutes = 3.
Moreover, the sum total of speed when Taro increases speed 120 m/m is 30 / 6 minutes = 5.
Only Taro increased speed.
The sum total of two persons' speed increased from 3 to 5 and the difference 2 is equivalent to 120 m/m.
The speed of 3 which is the sum of two persons' first speed is 120 / 2 × 3 = 180m/m.
Therefore, since it took 10 minutes to pass first, the surrounding length of the pond is 180 m/m × 10 minutes = 1800m.
(2) They pass 1800 m in 4 minute 48 second = 24/5 minutes.
The sum total of two persons' speed is 1800 m / (24/5) minutes = 375 m/m.
Since the sum total of two persons' first speed is 180 m/m, the speed which Taro increases is 375 - 180 = 195 m/m.
(3) They pass 1800 m in 7 minute 12 second = 36/5 minutes.
The sum of two persons' speed becomes 1800 m / (36/5)5 minutes = 250 m/m.
Since the sum total of two persons' first speed is 180 m/m, the speed which Taro increases is 250 - 180 = 70 m/m.
Since Taro increased speed twice, the first speed of Taro is also 70 m/m.
Since the sum total of two persons' first speed was 180 m/m, the speed of Jiro is 180 - 70 = 110 m/m.
Problem 5
The side AB of the right hexagon ABCDEF is equally divided into two and the side CD is equally divided into four.
Find the ratio of the area of the quadrangle BCNM and the hexagon AMNDEF in the figure.
The answer should be by the least integer.
Problem 6
Find the ratio of the area of the quadrangle BCNM and the hexagon AMNDEF in the figure.
The answer should be by the least integer.
Answer
The area of the right hexagon ABCDEF = 6.
△MNO = 1 × 3/2 × 5/4 = 15/8.
The area of quadrangle BCNM = 15/8 - 1 = 7/8.
The area of hexagon AMNDEF = 6 - 7/8 = 41/8.
Thus 7/8 : 41/8 = 7 : 41.
7 : 41
Solution
The area of △BCO is set to 1.The area of the right hexagon ABCDEF = 6.
△MNO = 1 × 3/2 × 5/4 = 15/8.
The area of quadrangle BCNM = 15/8 - 1 = 7/8.
The area of hexagon AMNDEF = 6 - 7/8 = 41/8.
Thus 7/8 : 41/8 = 7 : 41.
Problem 6
Answer the following questions.
Express an answer by the ratio of the least integer.
(1) There is a waterway through which it flows downward from upside on as shown in the figure below.
The water through the entrance A is divided into the same volume at the fork.
There are five exits located in four stories of waterways but the volume of the water which comes out of exits is not the same.
Find the ratio of the volume of the water of the least and the most.
(2) Next, as shown in Fig. P, the water through A is divided into three and there is a waterway of the solid shape which the same volume of water comes out of B, C, and D.
Waterway as shown in Fig.Q made by combining four of waterways is called waterway of two stories.
Like this, one waterway after another is connected.
The exits of one story, two stories, three stories and four stories of waterways becomes as it is shown in the Fig.1 ~ Fig.4, respectively.
① As for the volume of the water which comes out of exits of the waterway of three stories, find the ratio of the volume of the water of the least and the most.
② As for the volume of the water which comes out of exits of the waterway of four stories, find the ratio of the volume of the water of the least and the most.
Answer
(1) 1 : 6
(2) ① 1 : 6
② 1 : 12
Solution
(1)
The volume of the water into A is set to be 1.
The volume of the water passing the fork is shown as Fig.5 below.
Therefore the ratio of the volume of the water of the least and the most = 1/16 : 6/16 = 1 : 6.
(2)
① As in case of waterway of (1) the least volume can be calculated 1/2 1/2 1/2 1/2 = 1/16, in case of the waterway of 3 stories, the least volume can be calculated as 1/3 × 1/3 × 1/3 = 1/27.
Then the volume of the water into A is set to be 27 and water would be divided as shown in Fig. 6 below.
The least volume is 1 and the most is 6.
② In case of 4th stories, the volume of the water unto A is set to be 3 × 3 × 3 × 3 = 81.
The volume of the exit of the 3rd stories is calculated as shown in the left figure of Fig.7 based on ①.
Then the most volume can be calculated as 18 × 1/3 + 9 × 1/3 + 9 × 1/3 = 6 + 3 + 3 = 12.
