Time : 50 minutes
Passing marks : 70%
Answer : End of the problem
Problem 1
(2)
As shown in a figure, I placed the right-angled isosceles triangle so that each vertex was on the three parallel straight lines X, Y, and Z.
Angle B is right-angled.
(1)
There is a clock which delays 6 minutes in 1 hour. I made this clock fit to the correct time at 1:00 in the afternoon. Find the correct time when the long hand and the hour hand of this clock overlap first.
(2)
Angle B is right-angled.
In addition, the width of X and Y is 5 cm and the width of Y and Z is 3 cm.
The intersection of Y and the side AC is set to D.
Find the length of BD.
Problem 2
Problem 3
Problem 4
(2)
As shown in a figure, I placed the right-angled isosceles triangle so that each vertex was on the three parallel straight lines X, Y, and Z.
Angle B is right-angled.
Problem 2
The salt solution is contained in the three vessels A, B, and C, respectively.
If the salt solution of A and the salt solution of B are mixed, 600g of salt solution will be made with concentration of 25/3 %.
If B and C are mixed, 450g of salt solution will be made with concentration of 50/9 %.
If C and A are mixed, 650g of salt solution will be made with concentration of 110/13%. Answer the following questions.
(1) Find the weight of the salt solution in A, B, and C.
(2) Find the concentration of the salt solution in A, B, and C.
Problem 3
The floor of the rectangular room whose length of the horizontal side is 6 m is covered with square tiles.
This floor is divided into two rectangles by a straight line parallel to the horizontal side (Case 1).
When one rectangle is covered with the tiles whose length of the horizontal side is 50 cm and another rectangle is covered with tiles whose length of the horizontal side is 30 cm, all of the floor will be covered over.
At this time, there are 228 tiles to be used in total.
Moreover, a straight line is shifted while it has been parallel to the horizontal side and a floor is divided into two rectangles different from Case 1 (Case 2).
When one rectangle is covered with the tiles whose length of the horizontal side is 40 cm and another rectangle is covered with tiles whose length of the horizontal side is 30 cm, all of the floor will be covered over.
There are 220 tiles used at this time in total.
Find the length of the vertical side of the room.
Problem 4
There are three kinds of balls A, B, and C.
The same kind of ball is the same weight and the heavier order is A, B, and C among three kinds.
Total of five balls was put into the bag so that one or more balls of every kind might be included and weight of the bag was weighed.
However, weight of a bag is not considered.
When all the cases of combination to be considered were investigated, it turned out that the weight was five kinds.
When I put three balls of A, one of B and one of C into the bag, it shows by the code of (3, 1, 1).
Answer the following questions.
(1) Although combination of three kinds of balls differs, two bags come to become the same weight. Answer the two combinations using a sign ( ).
(2) The weight of the 2nd heaviest bag was 71g and the weight of the 4th heaviest one was 79g among five kinds of weights.
Find the weight of one ball of A, B, and C, respectively.
<Answer>
Problem 1
(1)
There is a clock which delays 6 minutes in 1 hour.
I made this clock fit to the correct time at 1:00 in the afternoon.
Find the correct time when the long hand and the hour hand of this clock overlap first.
Answer
1: 200/33 (1 O'clock and 200/33 minutes)
Solution
The long hand moves 6 degrees per minute.
The hour hand moves 0.5 degrees per minute.
The angle of two hand at 1:00 is 30 degrees.
30 / (6 - 0.5) = 60/11 minutes
60/11 × 60/(60-6) = 200/33 minutes
(2)
Angle B is right-angled.
In addition, the width of X and Y is 5 cm and the width of Y and Z is 3 cm.
The intersection of Y and the side AC is set to D.
Find the length of BD.
Problem 2
Problem 3
Problem 4
Answer
4.25 cm
Solution
△AEB and △BFC are congruent.
△AGD and △AHC are homothetic and the homothetic ratio is 5 : (5+3) = 5 : 8.
GD = (5 - 3) / 8 × 5 = 1.25 cm
BD = 3 + 1.25 = 4.25 cm
Problem 2
The salt solution is contained in the three vessels A, B, and C, respectively.
If the salt solution of A and the salt solution of B are mixed, 600g of salt solution will be made with concentration of 25/3 %.
If B and C are mixed, 450g of salt solution will be made with concentration of 50/9 %.
If C and A are mixed, 650g of salt solution will be made with concentration of 110/13%. Answer the following questions.
(1) Find the weight of the salt solution in A, B, and C.
(2) Find the concentration of the salt solution in A, B, and C.
Answer
(1) A : 400g, B : 200g, C : 250g b
(2) A : 10%, B : 5%, C : 6%
Solution
(1)
Total weight of salt solution is (600 +450 + 650) / 2 = 850g.
A = 850 - 450=400g. B= 850 - 650 = 200g. C = 850 - 600 = 250g.
(2)
The weight of salt in A+B is 600 / 100 × 25/3 = 50g.
The weight of salt in B+C is 450 / 100 × 50/9 = 25g.
The weight of salt in A+B is 650 / 100 × 110/13 = 55g.
Total weight of salt in A, B and C is (50 +25 + 55) / 2 = 65g.
Salt in A = 65 - 25 =40g. B = 65 - 55 = 10g. C= 65 - 50 =15g.
Concentration of A, B , and C is 40 / 100 × 100 = 10%, 10 / 200 × 100 = 5%,
15 / 250 × 100 = 6%.
Problem 3
The floor of the rectangular room whose length of the horizontal side is 6 m is covered with square tiles.
This floor is divided into two rectangles by a straight line parallel to the horizontal side (Case 1).
When one rectangle is covered with the tiles whose length of the horizontal side is 50 cm and another rectangle is covered with tiles whose length of the horizontal side is 30 cm, all of the floor will be covered over.
At this time, there are 228 tiles to be used in total.
Moreover, a straight line is shifted while it has been parallel to the horizontal side and a floor is divided into two rectangles different from Case 1 (Case 2).
When one rectangle is covered with the tiles whose length of the horizontal side is 40 cm and another rectangle is covered with tiles whose length of the horizontal side is 30 cm, all of the floor will be covered over.
There are 220 tiles used at this time in total.
Find the length of the vertical side of the room.
<Example>
Answer
4.7 m
Solution
600 / 50 =12 pieces, 600 / 30 = 20 pieces, 600 / 40 = 15 pieces.
12 × A + 15 × B = 228
(A,B)=(4,9),(9,6),(14,3)
The length of vertical side = 470 cm, 630cm, 790 cm.
15 × C + 12 × D = 220
(C,D)=(4,8),(8,5),(12,2)
The length of vertical side = 400 cm, 470cm, 540 cm.
Therefore it is 470cm = 4.7m.
Problem 4
There are three kinds of balls A, B, and C.
The same kind of ball is the same weight and the heavier order is A, B, and C among three kinds.
Total of five balls was put into the bag so that one or more balls of every kind might be included and weight of the bag was weighed.
However, weight of a bag is not considered.
When all the cases of combination to be considered were investigated, it turned out that the weight was five kinds.
When I put three balls of A, one of B and one of C into the bag, it shows by the code of (3, 1, 1).
Answer the following questions.
(1) Although combination of three kinds of balls differs, two bags come to become the same weight. Answer the two combinations using a sign ( ).
(2) The weight of the 2nd heaviest bag was 71g and the weight of the 4th heaviest one was 79g among five kinds of weights.
Find the weight of one ball of A, B, and C, respectively.
Answer
(1) (2, 1, 2) and (1, 3, 1)
(2) A : 19g, B : 15g, C : 11g
Solution
(1)
Considering combination of two balls, it is found heaviest is A+A, 2nd is A+B and the fifth is C+C.
Then the forth one is B+C.
Since there are five kinds of weights, it is found that A+C = B+B.
Therefore (2,1,2)=(1,3,1)
(2)
The 2nd heaviest combination is (2,2,1) and the fourth one is (1,2,2).
As B+B = A+C, (3,0,2) = (2,2,1) = 79g and (2,0,3) = (1,2,2) = 71 g.
A+C = (79+71) / (3+2) = 30g.
B = 30 / 2 = 15 g.
A - C = 79 - 71 = 8g.
A = (30 +8) / 2 = 19g. C = 30 - 19 = 11g.
