Time : 45 minutes
Passing mark : 80%
Answer : End of the problem
Problem 1
(1) 14/3 × 3.125 / 7/27 + 21/4 / 0.22 × 55/14 =
(2) 3707 + 3711 + 3715 + 3719 + ---- + 3743 =
(3) When 1570 was divided by the integer of 3 digit, remainder was 23.
Find the least integer among such integers.
(4) January 1, 2008 of the leap year was Tuesday.
What day is it the last Tuesday of January, 2010?
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Problem 2
(1) Taro runs 60 m in 8.4 seconds and Jiro runs 60 m in 9.6 seconds.
They started at the same time at 100 m race.
When Taro makes a goal, find the distance to the goal for Jiro.
(2) Salt solution 100g is taken out of 1000g of salt solution with 10% concentration and 100g of water is added to the remaining salt solution. After stirring well, the salt solution 200g is taken out and 200g of water is added.
Find the concentration of the salt solution at this time.
(3) The unit of British money is pound and the unit of South Korean money is won.
When 1 pound = 180 yen and 1 won = 0.11 yen, how much is 77 pounds in won?
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Problem 3
(1) As shown in the figure, five numbers are put in nine grids.
Other numbers are put into the remaining grids and it is arranged that products of three numbers in any vertical or horizontal row become same.
Find the number in A.
(2) The price for the ticket from A station to B station is 2000 yen.
By the coupon ticket of a six-sheet set, it becomes 18% discount and by the coupon ticket of a ten-sheet set, it becomes 20% discount.
You buy tickets for 48 persons combining these tickets or coupon tickets.
When you buy them by the cheapest combination, find the price per person.
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Problem 4
(1) There is a rectangle as shown in the figure.
AB = 3.5 cm, AD = 8.4 cm, BD = 9.1 cm. AE is vertical to BD.
Find the length of AE at this time.
(2) The figure shows two circles 8 cm in radius and rectangle ABCD whose two sides are diameters of the circles.
As for the shadow portion, when the sum total of the area of X and Z is equal to the area of Y, find the length of the side AD.
Pi is assumed to be 3.14.
Passing mark : 80%
Answer : End of the problem
Problem 1
(1) 14/3 × 3.125 / 7/27 + 21/4 / 0.22 × 55/14 =
(2) 3707 + 3711 + 3715 + 3719 + ---- + 3743 =
(3) When 1570 was divided by the integer of 3 digit, remainder was 23.
Find the least integer among such integers.
(4) January 1, 2008 of the leap year was Tuesday.
What day is it the last Tuesday of January, 2010?
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Problem 2
(1) Taro runs 60 m in 8.4 seconds and Jiro runs 60 m in 9.6 seconds.
They started at the same time at 100 m race.
When Taro makes a goal, find the distance to the goal for Jiro.
(2) Salt solution 100g is taken out of 1000g of salt solution with 10% concentration and 100g of water is added to the remaining salt solution. After stirring well, the salt solution 200g is taken out and 200g of water is added.
Find the concentration of the salt solution at this time.
(3) The unit of British money is pound and the unit of South Korean money is won.
When 1 pound = 180 yen and 1 won = 0.11 yen, how much is 77 pounds in won?
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Problem 3
(1) As shown in the figure, five numbers are put in nine grids.
Other numbers are put into the remaining grids and it is arranged that products of three numbers in any vertical or horizontal row become same.
Find the number in A.
(2) The price for the ticket from A station to B station is 2000 yen.
By the coupon ticket of a six-sheet set, it becomes 18% discount and by the coupon ticket of a ten-sheet set, it becomes 20% discount.
You buy tickets for 48 persons combining these tickets or coupon tickets.
When you buy them by the cheapest combination, find the price per person.
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Problem 4
(1) There is a rectangle as shown in the figure.
AB = 3.5 cm, AD = 8.4 cm, BD = 9.1 cm. AE is vertical to BD.
Find the length of AE at this time.
(2) The figure shows two circles 8 cm in radius and rectangle ABCD whose two sides are diameters of the circles.
As for the shadow portion, when the sum total of the area of X and Z is equal to the area of Y, find the length of the side AD.
Pi is assumed to be 3.14.
(3) As shown in the figure, there is a trapezoid ABCD whose sides AB = 5 cm, BC = 4 cm and CD = 8 cm.
Find the volume of the solid which is made by rotating the trapezoid one time around the side AB as a rotation axis.
Pi is assumed to be 3.14.
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Problem 5
As for the integer N, it is considered as A(N) = N, B(N) = N × N and C(N) = N × N × N.
For example, A(2) = 2, B(3) = 3 × 3 = 9 and C(4) = 4 × 4 × 4 = 64.
Find X when B(X) = C(B(2)) + B(2) × C(3) / A(3).
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Problem 6
Only one heavy ball is included in the balls whose form and size are same.
I think that I will find a heavy ball using an even balance by which heavier ball is found.
For example, when there is only one heavy ball in three balls, it is just once to find certainly a heavy ball by using even balance.
The number of times of using even balance should be lessened as much as possible.
Answer the following questions.
Noted that there is no limit of number of balls to be put on the even balance.
(1) In order to certainly find a heavy ball in 21 balls, find the minimum number of times using the even balance.
(2) Suppose that the even balance can be used only 4 times.
Find the most number of balls in which a heavy ball is certainly found.
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Problem 7
There are the two points A and B.
Taro leaves A point, Jiro leaves B point at the same time and they go back and forth between A and B repeatedly with fixed speed, respectively.
Taro walks faster than Jiro.
The graph expresses the relation of the distance between Taro and Jiro, and time.
Answer the following questions.
(1) Find the time after leaving when Taro and Jiro passed each other for the first time.
(2) Find the time after leaving when Taro passed Jiro for the first time.
Problem 1
(1) 14/3 × 3.125 / 7/27 + 21/4 / 0.22 × 55/14 =
(2) 3707 + 3711 + 3715 + 3719 + ---- + 3743 =
(3) When 1570 was divided by the integer of 3 digit, remainder was 23.
Find the least integer among such integers.
(4) January 1, 2008 of the leap year was Tuesday.
What day is it the last Tuesday of January, 2010?
Answer
(1) 150
(2) 37250
Find the volume of the solid which is made by rotating the trapezoid one time around the side AB as a rotation axis.
Pi is assumed to be 3.14.
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Problem 5
As for the integer N, it is considered as A(N) = N, B(N) = N × N and C(N) = N × N × N.
For example, A(2) = 2, B(3) = 3 × 3 = 9 and C(4) = 4 × 4 × 4 = 64.
Find X when B(X) = C(B(2)) + B(2) × C(3) / A(3).
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Problem 6
Only one heavy ball is included in the balls whose form and size are same.
I think that I will find a heavy ball using an even balance by which heavier ball is found.
For example, when there is only one heavy ball in three balls, it is just once to find certainly a heavy ball by using even balance.
The number of times of using even balance should be lessened as much as possible.
Answer the following questions.
Noted that there is no limit of number of balls to be put on the even balance.
(1) In order to certainly find a heavy ball in 21 balls, find the minimum number of times using the even balance.
(2) Suppose that the even balance can be used only 4 times.
Find the most number of balls in which a heavy ball is certainly found.
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Problem 7
There are the two points A and B.
Taro leaves A point, Jiro leaves B point at the same time and they go back and forth between A and B repeatedly with fixed speed, respectively.
Taro walks faster than Jiro.
The graph expresses the relation of the distance between Taro and Jiro, and time.
Answer the following questions.
(1) Find the time after leaving when Taro and Jiro passed each other for the first time.
(2) Find the time after leaving when Taro passed Jiro for the first time.
<Answer>
Problem 1
(1) 14/3 × 3.125 / 7/27 + 21/4 / 0.22 × 55/14 =
(2) 3707 + 3711 + 3715 + 3719 + ---- + 3743 =
(3) When 1570 was divided by the integer of 3 digit, remainder was 23.
Find the least integer among such integers.
(4) January 1, 2008 of the leap year was Tuesday.
What day is it the last Tuesday of January, 2010?
Answer
(1) 150
(2) 37250
Solution
3707 + 3711 + 3715 + 3719 + ---- + 3743
＝3700 × 10 + (7 + 11 + 15 + 19 + ---- + 43)
＝37000 + ( 7 + 43 ) × 10 / 2
= 37000 + 250
= 37250
3707 + 3711 + 3715 + 3719 + ---- + 3743
＝3700 × 10 + (7 + 11 + 15 + 19 + ---- + 43)
＝37000 + ( 7 + 43 ) × 10 / 2
= 37000 + 250
= 37250
(3) 119
Solution
1570 - 23 = 1547
1547 = 7 × 13 × 17
7 × 17 = 119
(4) 26th
Solution
Jan. 1, 2009 is Thursday.
Jan. 1, 2010 is Friday.
Jan. 31, 2010 is Sunday as 31 / 7 = 4 remainder 3.
31 - 5 = 26th.
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Problem 2
(1) Taro runs 60 m in 8.4 seconds and Jiro runs 60 m in 9.6 seconds.
They started at the same time at 100 m race.
When Taro makes a goal, find the distance to the goal for Jiro.
Answer
12.5 m
Solution
Time for Taro to goal 100m is 8.4 / 60 × 100 = 14 seconds.
At the time Jiro is 60 m × 14/9.6 = 87.5 m.
100 - 87.5 = 12.5 m
(2) Salt solution 100g is taken out of 1000g of salt solution with 10% concentration and 100g of water is added to the remaining salt solution. After stirring well, the salt solution 200g is taken out and 200g of water is added.
Find the concentration of the salt solution at this time.
Answer
7.2 %
Solution
1000 × 0.1 = 100 g
100 × (1 - 1/10) = 90 g
900 + 100 = 1000 g
1000 - 200 = 800 g
90 × (1 - 2/10) = 72 g
800 + 200 = 1000 g
72 / 1000 × 100 = 7.2%
(3) The unit of British money is pound and the unit of South Korean money is won.
When 1 pound = 180 yen and 1 won = 0.11 yen, how much is 77 pounds in won?
Answer
126000 won
Solution
Since 1 pound = 180 yen and 1 won = 0.11 yen, 1 pound = 180 / 0.11 = 18000/11 won.
Thus 77 pound = 18000/11 × 77 = 126000 won.
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Problem 3
(1) As shown in the figure, five numbers are put in nine grids.
Other numbers are put into the remaining grids and it is arranged that products of three numbers in any vertical or horizontal row become same.
Find the number in A.
Answer
22
Solution
12 × 77 × 15 = 33 × B × 12
B = (12 × 77 × 15) / (33 × 12) = 35
A = (33 × 35 × 12) / (35 × 18) = 22
(2) The price for the ticket from A station to B station is 2000 yen.
By the coupon ticket of a six-sheet set, it becomes 18% discount and by the coupon ticket of a ten-sheet set, it becomes 20% discount.
You buy tickets for 48 persons combining these tickets or coupon tickets.
When you buy them by the cheapest combination, find the price per person.
Answer
1615 yen
Solution
When purchasing coupon ticket of ten-sheet as many as possible and total number of tickets should not be more than 48 persons.
2000 yen × 10 × (1 - 0.2) × 4 = 64000 yen
2000 × 6 × (1 - 0.18) × 1 = 9840 yen
2000 × 2 = 4000 yen
64000 + 9840 + 4000 = 77840 yen
When purchasing 2000 yen of ticket as less as possible.
According to 48 = 10 × 3 + 6 × 3, 16000 × 3 = 48000 and 9840 × 3 = 29520.
48000 + 29520 = 77520 yen which is cheaper than above.
Thus 77520 / 48 = 1615 yen.
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Problem 4
(1) There is a rectangle as shown in the figure.
AB = 3.5 cm, AD = 8.4 cm, BD = 9.1 cm. AE is vertical to BD.
Find the length of AE at this time.
Answer
42/13 cm
Solution
The area of △ABD = AB × AD / 2 = 3.5 × 8.4 / 2 = BD × AE / 2
BD × AE = 3.5 × 8.4
As BD = 9.1, AE = (3.5 × 8.4) / 9.1 = 42/13.
(2) The figure shows two circles 8 cm in radius and rectangle ABCD whose two sides are diameters of the circles.
As for the shadow portion, when the sum total of the area of X and Z is equal to the area of Y, find the length of the side AD.
Pi is assumed to be 3.14.
Answer
12.56 cm
Solution
White portion in rectangle ABCD is set to P as below.
Solution
1570 - 23 = 1547
1547 = 7 × 13 × 17
7 × 17 = 119
(4) 26th
Solution
Jan. 1, 2009 is Thursday.
Jan. 1, 2010 is Friday.
Jan. 31, 2010 is Sunday as 31 / 7 = 4 remainder 3.
31 - 5 = 26th.
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Problem 2
(1) Taro runs 60 m in 8.4 seconds and Jiro runs 60 m in 9.6 seconds.
They started at the same time at 100 m race.
When Taro makes a goal, find the distance to the goal for Jiro.
Answer
12.5 m
Solution
Time for Taro to goal 100m is 8.4 / 60 × 100 = 14 seconds.
At the time Jiro is 60 m × 14/9.6 = 87.5 m.
100 - 87.5 = 12.5 m
(2) Salt solution 100g is taken out of 1000g of salt solution with 10% concentration and 100g of water is added to the remaining salt solution. After stirring well, the salt solution 200g is taken out and 200g of water is added.
Find the concentration of the salt solution at this time.
Answer
7.2 %
Solution
1000 × 0.1 = 100 g
100 × (1 - 1/10) = 90 g
900 + 100 = 1000 g
1000 - 200 = 800 g
90 × (1 - 2/10) = 72 g
800 + 200 = 1000 g
72 / 1000 × 100 = 7.2%
(3) The unit of British money is pound and the unit of South Korean money is won.
When 1 pound = 180 yen and 1 won = 0.11 yen, how much is 77 pounds in won?
Answer
126000 won
Solution
Since 1 pound = 180 yen and 1 won = 0.11 yen, 1 pound = 180 / 0.11 = 18000/11 won.
Thus 77 pound = 18000/11 × 77 = 126000 won.
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Problem 3
(1) As shown in the figure, five numbers are put in nine grids.
Other numbers are put into the remaining grids and it is arranged that products of three numbers in any vertical or horizontal row become same.
Find the number in A.
Answer
22
Solution
12 × 77 × 15 = 33 × B × 12
B = (12 × 77 × 15) / (33 × 12) = 35
A = (33 × 35 × 12) / (35 × 18) = 22
(2) The price for the ticket from A station to B station is 2000 yen.
By the coupon ticket of a six-sheet set, it becomes 18% discount and by the coupon ticket of a ten-sheet set, it becomes 20% discount.
You buy tickets for 48 persons combining these tickets or coupon tickets.
When you buy them by the cheapest combination, find the price per person.
Answer
1615 yen
Solution
When purchasing coupon ticket of ten-sheet as many as possible and total number of tickets should not be more than 48 persons.
2000 yen × 10 × (1 - 0.2) × 4 = 64000 yen
2000 × 6 × (1 - 0.18) × 1 = 9840 yen
2000 × 2 = 4000 yen
64000 + 9840 + 4000 = 77840 yen
When purchasing 2000 yen of ticket as less as possible.
According to 48 = 10 × 3 + 6 × 3, 16000 × 3 = 48000 and 9840 × 3 = 29520.
48000 + 29520 = 77520 yen which is cheaper than above.
Thus 77520 / 48 = 1615 yen.
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Problem 4
(1) There is a rectangle as shown in the figure.
AB = 3.5 cm, AD = 8.4 cm, BD = 9.1 cm. AE is vertical to BD.
Find the length of AE at this time.
Answer
42/13 cm
Solution
The area of △ABD = AB × AD / 2 = 3.5 × 8.4 / 2 = BD × AE / 2
BD × AE = 3.5 × 8.4
As BD = 9.1, AE = (3.5 × 8.4) / 9.1 = 42/13.
(2) The figure shows two circles 8 cm in radius and rectangle ABCD whose two sides are diameters of the circles.
As for the shadow portion, when the sum total of the area of X and Z is equal to the area of Y, find the length of the side AD.
Pi is assumed to be 3.14.
Answer
12.56 cm
Solution
White portion in rectangle ABCD is set to P as below.
When the sum total of the area of X and Z is equal to the area of Y, X + Z + P = Y + P.
The area of rectangle is twice of the area of semicircle and it is 8 × 8 × 3.14 × 1/2 × 2 = 64 × 3.14.
As AB = 16 cm, AD = 64 × 3.14 / 16 = 12,56 cm.
(3) As shown in the figure, there is a trapezoid ABCD whose sides AB = 5 cm, BC = 4 cm and CD = 8 cm.
Find the volume of the solid which is made by rotating the trapezoid one time around the side AB as a rotation axis.
Pi is assumed to be 3.14.
Answer
351.68 cm3
Solution
4 × 4 × 3.14 × 8 - 4 × 4 × 3.14 × 3 × 1/3
= 4 × 4 × 3.14 × 8 - 4 × 4 × 3.14
= 351.68 cm3
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Problem 5
As for the integer N, it is considered as A(N) = N, B(N) = N × N and C(N) = N × N × N.
For example, A(2) = 2, B(3) = 3 × 3 = 9 and C(4) = 4 × 4 × 4 = 64.
Find X when B(X) = C(B(2)) + B(2) × C(3) / A(3).
Answer
10
Solution
B(2) = 2 × 2 = 4
C(B(2)) = C(4) = 4 × 4 × 4 =64
C(3) = 3 × 3 × 3 = 27
A(3) = 3
Therefore, C(B(2)) + B(2) × C(3) / A(3) = 64 + 4 × 27 / 3 = 100
100 = 10 × 10
X = 10
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Problem 6
Only one heavy ball is included in the balls whose form and size are same.
I think that I will find a heavy ball using an even balance by which heavier ball is found.
For example, when there is only one heavy ball in three balls, it is just once to find certainly a heavy ball by using even balance.
The number of times of using even balance should be lessened as much as possible.
Answer the following questions.
Noted that there is no limit of number of balls to be put on the even balance.
(1) In order to certainly find a heavy ball in 21 balls, find the minimum number of times using the even balance.
(2) Suppose that the even balance can be used only 4 times.
Find the most number of balls in which a heavy ball is certainly found.
Answer
(1) 3 times
(2) 81 balls
Solution
(1)
When there are three balls, the heavy ball can be found by using the even balance one time.
Take two balls out of three to put on the even balance.
If it is not balanced, one of them are heavy ball and if it is balanced, remaining one is heavy ball.
This method is applicable to the case of 21 balls.
21 / 3 = 7 balls
21 balls can be reduced to 7 balls.
7 = 3 + 3 + 1
Then minimum number of times to find heavy ball certainly is three.
(2)
3 × 3 × 3 × 3 = 81 balls
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Problem 7
There are the two points A and B.
Taro leaves A point, Jiro leaves B point at the same time and they go back and forth between A and B repeatedly with fixed speed, respectively.
Taro walks faster than Jiro.
The graph expresses the relation of the distance between Taro and Jiro, and time.
Answer the following questions.
(1) Find the time after leaving when Taro and Jiro passed each other for the first time.
(2) Find the time after leaving when Taro passed Jiro for the first time.
Answer
(1) 10 seconds
(2) 40 seconds
Solution
(1)
According to the graph, the time for Taro and Jiro to pass each other second time is 30 minutes after.
Until this second meeting they walk in total three times of the distance between AB.
The time of first meeting is 1/3 of time of walking the above distance and it is 30 seconds × 1/3 = 10 seconds.
(2)
According to the graph, Taro arrive at B in 16 minutes.
After first meeting with Jiro, Taro takes 16 - 10 = 6 seconds.
Jiro takes 10 seconds to walk same distance.
Time ratio between two persons is 6 : 10 = 3 : 5, then speed ratio is 5 : 3.
The distance between AB is set to (5 + 3) × 10 = 80.
It takes 16 × 2 = 32 seconds for Taro to walk one round.
When Taro walk one round trip, the difference between two is 3 × 32 - 80 = 16.
Thus the time when Taro pass Jiro after round trip is 16 / (5 - 3) = 8 seconds after.
Therefore it is 32 + 8 = 40 seconds after starting.
As AB = 16 cm, AD = 64 × 3.14 / 16 = 12,56 cm.
(3) As shown in the figure, there is a trapezoid ABCD whose sides AB = 5 cm, BC = 4 cm and CD = 8 cm.
Find the volume of the solid which is made by rotating the trapezoid one time around the side AB as a rotation axis.
Pi is assumed to be 3.14.
Answer
351.68 cm3
Solution
4 × 4 × 3.14 × 8 - 4 × 4 × 3.14 × 3 × 1/3
= 4 × 4 × 3.14 × 8 - 4 × 4 × 3.14
= 351.68 cm3
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Problem 5
As for the integer N, it is considered as A(N) = N, B(N) = N × N and C(N) = N × N × N.
For example, A(2) = 2, B(3) = 3 × 3 = 9 and C(4) = 4 × 4 × 4 = 64.
Find X when B(X) = C(B(2)) + B(2) × C(3) / A(3).
Answer
10
Solution
B(2) = 2 × 2 = 4
C(B(2)) = C(4) = 4 × 4 × 4 =64
C(3) = 3 × 3 × 3 = 27
A(3) = 3
Therefore, C(B(2)) + B(2) × C(3) / A(3) = 64 + 4 × 27 / 3 = 100
100 = 10 × 10
X = 10
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Problem 6
Only one heavy ball is included in the balls whose form and size are same.
I think that I will find a heavy ball using an even balance by which heavier ball is found.
For example, when there is only one heavy ball in three balls, it is just once to find certainly a heavy ball by using even balance.
The number of times of using even balance should be lessened as much as possible.
Answer the following questions.
Noted that there is no limit of number of balls to be put on the even balance.
(1) In order to certainly find a heavy ball in 21 balls, find the minimum number of times using the even balance.
(2) Suppose that the even balance can be used only 4 times.
Find the most number of balls in which a heavy ball is certainly found.
Answer
(1) 3 times
(2) 81 balls
Solution
(1)
When there are three balls, the heavy ball can be found by using the even balance one time.
Take two balls out of three to put on the even balance.
If it is not balanced, one of them are heavy ball and if it is balanced, remaining one is heavy ball.
This method is applicable to the case of 21 balls.
21 / 3 = 7 balls
21 balls can be reduced to 7 balls.
7 = 3 + 3 + 1
Then minimum number of times to find heavy ball certainly is three.
(2)
3 × 3 × 3 × 3 = 81 balls
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Problem 7
There are the two points A and B.
Taro leaves A point, Jiro leaves B point at the same time and they go back and forth between A and B repeatedly with fixed speed, respectively.
Taro walks faster than Jiro.
The graph expresses the relation of the distance between Taro and Jiro, and time.
Answer the following questions.
(1) Find the time after leaving when Taro and Jiro passed each other for the first time.
(2) Find the time after leaving when Taro passed Jiro for the first time.
Answer
(1) 10 seconds
(2) 40 seconds
Solution
(1)
According to the graph, the time for Taro and Jiro to pass each other second time is 30 minutes after.
Until this second meeting they walk in total three times of the distance between AB.
The time of first meeting is 1/3 of time of walking the above distance and it is 30 seconds × 1/3 = 10 seconds.
(2)
According to the graph, Taro arrive at B in 16 minutes.
After first meeting with Jiro, Taro takes 16 - 10 = 6 seconds.
Jiro takes 10 seconds to walk same distance.
Time ratio between two persons is 6 : 10 = 3 : 5, then speed ratio is 5 : 3.
The distance between AB is set to (5 + 3) × 10 = 80.
It takes 16 × 2 = 32 seconds for Taro to walk one round.
When Taro walk one round trip, the difference between two is 3 × 32 - 80 = 16.
Thus the time when Taro pass Jiro after round trip is 16 / (5 - 3) = 8 seconds after.
Therefore it is 32 + 8 = 40 seconds after starting.
