Time : 40 minutes
(2)
(3)
Problem 3
Three persons, Taro, Jiro, and Hanako carry out bagging work of a certain product.
In order to all products are bagged, it will take 150 minutes by two persons, Taro and Jiro, and it will take 125 minutes by two persons, Taro and Hanako and it will take 600 minutes by Jiro.
Find the time of bagging all products done by Taro.
When Taro and Jiro start bagging first and Hanako will join afterwards, find the minimum time when Hanako must work in order to finish bagging all within 120 minutes.
Answer should be rounded out the decimal point and calculated to the unit of minute.
Problem 4
Taro and Jiro participated in the triathlon carrying out 1.5 km of swimming, 40 km of bicycle, and 10 km of running in order.
(1) The time of swimming of Taro was 20 minutes and 50 seconds.
Find the speed per minute of swimming of Taro.
The speed ratio of swimming and pedaling a bicycle of Taro was 3 : 20.
Find the time of the bicycle of Taro.
(2) It was 5 minutes and 20 seconds later than Taro for Jiro to start running.
Since the speed of running of Jiro was quicker than Taro by 32 m/m, Jiro caught up with Taro 34 minutes and 40 seconds after Jiro started running.
Find the ratio of the speed per minute of running of Taro and Jiro.
Find the speed per minute of running of Jiro.
Find the distance from the point where Jiro caught up with the Taro to the goal.
(3) Find the sum total time of three events of Jiro.
Problem 5
Problem 6
Problem 7
Passing marks : 70 %
Answer : End of the problem
Problem 1
Problem 2
I can buy 90 juice exactly with the amount of the budget.
If it is sandwiches, I can buy 36 pieces exactly and if it is cakes, I can buy 40 pieces exactly.
When one juice, one sandwiches, and one cake are made into 1 set for one person, I can buy them for all the members within the budget and 360 yen will remain.
The budget is insufficient if there are more members.
(1)-1 Find the number of participants of the party.
(1)-2 Find the budget.
Find X
0.384 / 0.24 + 25/27 × 9/10 - (5 - 5/4 + 27/8) / X = 8/15
Problem 2
(1)
I became a manager of the party. I can buy 90 juice exactly with the amount of the budget.
If it is sandwiches, I can buy 36 pieces exactly and if it is cakes, I can buy 40 pieces exactly.
When one juice, one sandwiches, and one cake are made into 1 set for one person, I can buy them for all the members within the budget and 360 yen will remain.
The budget is insufficient if there are more members.
(1)-1 Find the number of participants of the party.
(1)-2 Find the budget.
(2)
Find the degree of angle X, Y, and Z in a figure.
In Fig. 1, quadrangle ABCD is a square and curve is a part of circle with a vertex of the square as a center.
In Fig. 2, triangle ABC is an equilateral triangle and pentagon ADEFG is an equilateral pentagon.
In Fig. 1, quadrangle ABCD is a square and curve is a part of circle with a vertex of the square as a center.
In Fig. 2, triangle ABC is an equilateral triangle and pentagon ADEFG is an equilateral pentagon.
(3)
In the figure, triangle ABC is an isosceles triangle and a quadrangle ACDE is a trapezoid.
Find the area of X and Y.
Problem 3
In order to all products are bagged, it will take 150 minutes by two persons, Taro and Jiro, and it will take 125 minutes by two persons, Taro and Hanako and it will take 600 minutes by Jiro.
Find the time of bagging all products done by Taro.
When Taro and Jiro start bagging first and Hanako will join afterwards, find the minimum time when Hanako must work in order to finish bagging all within 120 minutes.
Answer should be rounded out the decimal point and calculated to the unit of minute.
Problem 4
(1) The time of swimming of Taro was 20 minutes and 50 seconds.
Find the speed per minute of swimming of Taro.
The speed ratio of swimming and pedaling a bicycle of Taro was 3 : 20.
Find the time of the bicycle of Taro.
(2) It was 5 minutes and 20 seconds later than Taro for Jiro to start running.
Since the speed of running of Jiro was quicker than Taro by 32 m/m, Jiro caught up with Taro 34 minutes and 40 seconds after Jiro started running.
Find the ratio of the speed per minute of running of Taro and Jiro.
Find the speed per minute of running of Jiro.
Find the distance from the point where Jiro caught up with the Taro to the goal.
(3) Find the sum total time of three events of Jiro.
Problem 5
There are four kinds of iron balls A, B, C and D and each volume is different.
When balls were completely sunk into the vessels X and Y which are same form and with same amount of water poured in, and the height of the water surface was compared with them, it became as shown in (1), (2), and (3).
Arrange the ball of A~D in order with small volume.
When balls were completely sunk into the vessels X and Y which are same form and with same amount of water poured in, and the height of the water surface was compared with them, it became as shown in (1), (2), and (3).
Arrange the ball of A~D in order with small volume.
Problem 6
There are three kinds of tanks with equal capacity which are the rectangular prism as shown in the figure.
The side of the tank on which there is the same mark expresses the same length.
The graph expresses a relation with the depth of the water and time when the same amount of water per minute is put in three tanks.
The height of the tank represented by the graph of B is 40 cm.
(1) Find the height of the tank represented by the graph of A.
(2) Find the amount of the water put in per minute.
The side of the tank on which there is the same mark expresses the same length.
The graph expresses a relation with the depth of the water and time when the same amount of water per minute is put in three tanks.
The height of the tank represented by the graph of B is 40 cm.
(1) Find the height of the tank represented by the graph of A.
(2) Find the amount of the water put in per minute.
Problem 7
Find the area of the shaded portion in the rectangle of the figure.
A and B are the centers of each circle.
Pi is assumed to be 3.14 and the answer should round off the 2nd decimal place.
A and B are the centers of each circle.
Pi is assumed to be 3.14 and the answer should round off the 2nd decimal place.
<Answer>
Problem 1
Problem 2
I can buy 90 juice exactly with the amount of the budget.
If it is sandwiches, I can buy 36 pieces exactly and if it is cakes, I can buy 40 pieces exactly.
When one juice, one sandwiches, and one cake are made into 1 set for one person, I can buy them for all the members within the budget and 360 yen will remain.
The budget is insufficient if there are more members.
(1)-1 Find the number of participants of the party.
(1)-2 Find the budget.
Answer
(1)-1 15 persons
(1)-2 8640 yen
Solution
(1)-1
Ratio of the price of juice : sandwich : cake = 1/90 : 1/36 : 1/40 = 4 : 10 : 9.
The budget is expressed as 4 × 90 = 360.
360 / (4 + 10 + 9) = 15 persons remainder 15.
(1)-2
Remainder 15 is equivalent to 360 yen.
(2)
(3)
Problem 3
Three persons, Taro, Jiro, and Hanako carry out bagging work of a certain product.
In order to all products are bagged, it will take 150 minutes by two persons, Taro and Jiro, and it will take 125 minutes by two persons, Taro and Hanako and it will take 600 minutes by Jiro.
Find the time of bagging all products done by Taro.
When Taro and Jiro start bagging first and Hanako will join afterwards, find the minimum time when Hanako must work in order to finish bagging all within 120 minutes.
Answer should be rounded out the decimal point and calculated to the unit of minute.
Answer
200 minutes
67 minutes
Solution
All works are set to be LCM of 150, and 125 and 600, and it is 3000.
Sum of workload per minute of A+B = 3000 / 150 = 20.
Sum of workload of A+C = 3000 / 125 = 24.
Workload of B = 3000 / 600 = 5.
A = 20 - 5 = 15.
3000 / 15 = 200 minutes
A+B in 120 minutes = 20 × 120 = 2400.
C = 24 - 15 = 9
2400 / 9 = 66.6--- minutes
Taro and Jiro participated in the triathlon carrying out 1.5 km of swimming, 40 km of bicycle, and 10 km of running in order.
(1) The time of swimming of Taro was 20 minutes and 50 seconds.
Find the speed per minute of swimming of Taro.
The speed ratio of swimming and pedaling a bicycle of Taro was 3 : 20.
Find the time of the bicycle of Taro.
(2) It was 5 minutes and 20 seconds later than Taro for Jiro to start running.
Since the speed of running of Jiro was quicker than Taro by 32 m/m, Jiro caught up with Taro 34 minutes and 40 seconds after Jiro started running.
Find the ratio of the speed per minute of running of Taro and Jiro.
Find the speed per minute of running of Jiro.
Find the distance from the point where Jiro caught up with the Taro to the goal.
(3) Find the sum total time of three events of Jiro.
Answer
(1) 72 m/m
one hour and 23 minutes and 20 seconds
(2) 13 : 15
240 m/m
1680 m
(3) Two hours and 31 minutes and 10 seconds
Solution
(1)
1500 / 135/6 = 72 m/m
Bicycle = 72 × 20/3 = 480 m/m
40000 / 480 = 250/3 minutes = 1:23 and 20 seconds.
(2)
Time ratio of Taro : Jiro = (34 minutes and 40 seconds + 5 minutes and 20 seconds) : (34 minutes and 40 seconds) = 15 : 13.
Speed ratio of Taro : Jiro = 13 : 15.
15 - 13 = 2 is equivalent to 32 m/m.
15 = 32 × 15/2 = 240 m/m.
240 × 34 minutes and 40 seconds = 8320 m
10000 - 8320 = 1680 m
(3)
20 minutes and 50 seconds + 1 hour 3 minutes and 20 seconds + 34 minutes and 40 seconds + 1680 / 240
= 2 hours 31 minutes and 10 seconds
Problem 5
There are four kinds of iron balls A, B, C and D and each volume is different.
When balls were completely sunk into the vessels X and Y which are same form and with same amount of water poured in, and the height of the water surface was compared with them, it became as shown in (1), (2), and (3).
Arrange the ball of A~D in order with small volume.
Answer
C, A, B, D
Solution
According to (1) and (2), C < B.
According to (3), A + A = B + C which means C < A < B.
Considering (1), C < A < B < D.
Problem 6
Solution
(1)
The time to be full in the tank is 12 minutes × (40 / 5) = 96 minutes.
5 cm × (96 / 8) = 60 cm.
(2)
The height of the third tank is 5 cm × (95 / 16) = 30 cm.
Problem 7
Find X
0.384 / 0.24 + 25/27 × 9/10 - (5 - 5/4 + 27/8) / X = 8/15
Answer
15/4
Problem 2
(1)
I became a manager of the party. I can buy 90 juice exactly with the amount of the budget.
If it is sandwiches, I can buy 36 pieces exactly and if it is cakes, I can buy 40 pieces exactly.
When one juice, one sandwiches, and one cake are made into 1 set for one person, I can buy them for all the members within the budget and 360 yen will remain.
The budget is insufficient if there are more members.
(1)-1 Find the number of participants of the party.
(1)-2 Find the budget.
Answer
(1)-1 15 persons
(1)-2 8640 yen
Solution
(1)-1
Ratio of the price of juice : sandwich : cake = 1/90 : 1/36 : 1/40 = 4 : 10 : 9.
The budget is expressed as 4 × 90 = 360.
360 / (4 + 10 + 9) = 15 persons remainder 15.
(1)-2
Remainder 15 is equivalent to 360 yen.
As budget is 360, 360 / 15 × 360 = 8640 yen.
(2)
Find the degree of angle X, Y, and Z in a figure.
In Fig. 1, quadrangle ABCD is a square and curve is a part of circle with a vertex of the square as a center.
In Fig. 2, triangle ABC is an equilateral triangle and pentagon ADEFG is an equilateral pentagon.
In Fig. 1, quadrangle ABCD is a square and curve is a part of circle with a vertex of the square as a center.
In Fig. 2, triangle ABC is an equilateral triangle and pentagon ADEFG is an equilateral pentagon.
Answer
X = 30 degrees
Y = 15 degrees
Z = 48 degrees
Solution
Y = 60 - 45 = 15 degrees
90 - 60 - 30 degrees
(180 - 30) / 2 = 75 degrees
X = 75 - 45 = 30 degrees
Z = 108 - 60 = 48 degrees
X = 30 degrees
Y = 15 degrees
Z = 48 degrees
Solution
Y = 60 - 45 = 15 degrees
90 - 60 - 30 degrees
(180 - 30) / 2 = 75 degrees
X = 75 - 45 = 30 degrees
Z = 108 - 60 = 48 degrees
(3)
In the figure, triangle ABC is an isosceles triangle and a quadrangle ACDE is a trapezoid.
Find the area of X and Y.
Find the area of X and Y.
Answer
X = 36 cm2
Y = 90/13 cm2
Y = 90/13 cm2
Solution
X = 12 × (12 × 1/2) / 2 = 36 cm2
△ACE = 5 × 12 / 2 = 30 cm2
As AE and CD are parallel, △ACE : △CDE = 13 : 3.
Y = 30 × 3/13 = 90/13 cm2
Problem 3
In order to all products are bagged, it will take 150 minutes by two persons, Taro and Jiro, and it will take 125 minutes by two persons, Taro and Hanako and it will take 600 minutes by Jiro.
Find the time of bagging all products done by Taro.
When Taro and Jiro start bagging first and Hanako will join afterwards, find the minimum time when Hanako must work in order to finish bagging all within 120 minutes.
Answer should be rounded out the decimal point and calculated to the unit of minute.
Answer
200 minutes
67 minutes
Solution
All works are set to be LCM of 150, and 125 and 600, and it is 3000.
Sum of workload per minute of A+B = 3000 / 150 = 20.
Sum of workload of A+C = 3000 / 125 = 24.
Workload of B = 3000 / 600 = 5.
A = 20 - 5 = 15.
3000 / 15 = 200 minutes
A+B in 120 minutes = 20 × 120 = 2400.
C = 24 - 15 = 9
2400 / 9 = 66.6--- minutes
---------------------------------------------------------------------------------------------
Problem 4Taro and Jiro participated in the triathlon carrying out 1.5 km of swimming, 40 km of bicycle, and 10 km of running in order.
(1) The time of swimming of Taro was 20 minutes and 50 seconds.
Find the speed per minute of swimming of Taro.
The speed ratio of swimming and pedaling a bicycle of Taro was 3 : 20.
Find the time of the bicycle of Taro.
(2) It was 5 minutes and 20 seconds later than Taro for Jiro to start running.
Since the speed of running of Jiro was quicker than Taro by 32 m/m, Jiro caught up with Taro 34 minutes and 40 seconds after Jiro started running.
Find the ratio of the speed per minute of running of Taro and Jiro.
Find the speed per minute of running of Jiro.
Find the distance from the point where Jiro caught up with the Taro to the goal.
(3) Find the sum total time of three events of Jiro.
Answer
(1) 72 m/m
one hour and 23 minutes and 20 seconds
(2) 13 : 15
240 m/m
1680 m
(3) Two hours and 31 minutes and 10 seconds
Solution
(1)
1500 / 135/6 = 72 m/m
Bicycle = 72 × 20/3 = 480 m/m
40000 / 480 = 250/3 minutes = 1:23 and 20 seconds.
(2)
Time ratio of Taro : Jiro = (34 minutes and 40 seconds + 5 minutes and 20 seconds) : (34 minutes and 40 seconds) = 15 : 13.
Speed ratio of Taro : Jiro = 13 : 15.
15 - 13 = 2 is equivalent to 32 m/m.
15 = 32 × 15/2 = 240 m/m.
240 × 34 minutes and 40 seconds = 8320 m
10000 - 8320 = 1680 m
(3)
20 minutes and 50 seconds + 1 hour 3 minutes and 20 seconds + 34 minutes and 40 seconds + 1680 / 240
= 2 hours 31 minutes and 10 seconds
Problem 5
When balls were completely sunk into the vessels X and Y which are same form and with same amount of water poured in, and the height of the water surface was compared with them, it became as shown in (1), (2), and (3).
Arrange the ball of A~D in order with small volume.
Answer
C, A, B, D
Solution
According to (1) and (2), C < B.
According to (3), A + A = B + C which means C < A < B.
Considering (1), C < A < B < D.
Problem 6
There are three kinds of tanks with equal capacity which are the rectangular prism as shown in the figure.
The side of the tank on which there is the same mark expresses the same length.
The graph expresses a relation with the depth of the water and time when the same amount of water per minute is put in three tanks.
The height of the tank represented by the graph of B is 40 cm.
(1) Find the height of the tank represented by the graph of A.
(2) Find the amount of the water put in per minute.
The side of the tank on which there is the same mark expresses the same length.
The graph expresses a relation with the depth of the water and time when the same amount of water per minute is put in three tanks.
The height of the tank represented by the graph of B is 40 cm.
(1) Find the height of the tank represented by the graph of A.
(2) Find the amount of the water put in per minute.
Answer
(1) 60 cm
(2) 75 cm3
(2) 75 cm3
(1)
The time to be full in the tank is 12 minutes × (40 / 5) = 96 minutes.
5 cm × (96 / 8) = 60 cm.
(2)
The height of the third tank is 5 cm × (95 / 16) = 30 cm.
The volume of the tank is 60 × 40 × 30 = 72000 cm3.
72000 / 96 = 75 cm3.
Problem 7
Find the area of the shaded portion in the rectangle of the figure.
A and B are the centers of each circle.
Pi is assumed to be 3.14 and the answer should round off the 2nd decimal place.
A and B are the centers of each circle.
Pi is assumed to be 3.14 and the answer should round off the 2nd decimal place.
Answer
Solution
Draw auxiliary lines as shown in Fig.1.
1474.3 cm2
Draw auxiliary lines as shown in Fig.1.
Red, green and blue portions can be transformed as shown in Fig.2.
The area of rectangle CTUS = 20 × 37.3 = 746 cm2.
The area of rectangle PQFC = 20 × (37.3 - 2.7) = 692 cm2.
The area of the sector AQF = 20 × 20 × 3.14 × 60/360 = 628/3 cm2.
The height of △AQF = 34.6 / 2 = 17.3 cm
The area of △AQF = 20 × 17 / 2 = 173 cm2.
628/3 - 173 = 109/3 cm2.
In total 746 + 692 + 109/3 = 1474.33----- cm2.
