Time : 60 minutes
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Passing mark : 70 %
Answer : End of the problem
Problem 1
6 - 121/13 / {39.2 - 189/4 × (2/3 - 1/5) / 0.7} × 4.9 =
Problem 2
There are three kinds of salt solutions A, B, and C weigh 60 g, 120 g, and 100g, respectively.
The salt concentration of A is 3%.
① When 20g of A and 30g of B are mixed, the salt concentration of salt solution is same as C.
② The salt concentration of salt solution D made by mixing A, B, and C altogether is to be 7.5%.
Answer the following questions.
The salt concentration of a salt solution is a ratio of the weight of salt to the weight of a salt solution.
(1) Find the weight of the salt contained in D.
(2) Find the salt concentration (%) of B.
(3) Find the salt concentration (%) of C.
Problem 3
There are A station and B station and another station C is between A and B.
Train X will leave A station at 23:10 and will arrive at intermediate C station at 2:10 on the next day.
After stopping for 20 minutes, train X will leave C station at 2:30 and it will arrive at B station at 6:10.
The speed of this train is 2.675 km per hour faster than the speed which is calculated the distance between A and B is divided by 7 hours which is actual time taken from A to B.
Find the distance of A station and B station.
Find the distance of A station and B station.
Problem 4
Either the sign + of addition or the sign × of multiplication is put in the following [ ] and calculate.
Answer the following questions.
(1) When 1[ ]2[ ]3[ ]4 is calculated, Answer each calculation result to small order.
When there come out same calculation results, write only once.
(2) Answer two kinds of manner putting in + and × applicable to [ ] of the following formula.
1[ ]2[ ]3[ ]4[ ]5 = 2[ ]3[ ]4[ ]5[ ]6
Problem 5
There is a rectangular prism 3 cm in length, 4 cm in width and 5 cm in height.
As for faces of this rectangular prism, the face of the rectangle with 3cm and 4cm side is set to face A, the face with 4cm and 5cm is set to face B and the face with 5cm and 3cm is set to face C.
Answer the following questions.
(1) Make small rectangular prisms by cutting with planes which are parallel to face A, face B and face C, once, once and twice respectively.
① Find the number of small rectangular prism.
① Find the number of small rectangular prism.
② Find the sum total of the surface area of these small rectangular prisms.
Noted that the surface area of a rectangular prism is the sum total of the area of all the faces of the rectangular prism.
Noted that the surface area of a rectangular prism is the sum total of the area of all the faces of the rectangular prism.
(2) This rectangular prism was cut with planes which are parallel to face A, face B and face C, X times, Y times and Z times respectively.
In this case, there are 90 small rectangular prisms made and the sum total of the surface area of these rectangular prisms was 462 cm2.
Find the number applicable to X, Y, and Z.
Problem 6
There is the sector OAB which is 1/4 part a circle.
Noted that the straight lines OA, CD, and EF are parallel.
(2) As shown in Fig. 2, the straight line parallel to OA was drawn from each point which divided the arc AB of the sector OAB into five equally. OA is 5 cm.
Find the sum of the area of two shadow areas.
Pi is assumed to be 3.14.
Answer the following questions.
(1) Find the ratio of the area of a shadow area and the sector OAB in Fig. 1. Noted that the straight lines OA, CD, and EF are parallel.
(2) As shown in Fig. 2, the straight line parallel to OA was drawn from each point which divided the arc AB of the sector OAB into five equally. OA is 5 cm.
Find the sum of the area of two shadow areas.
Pi is assumed to be 3.14.
Answer
Problem 1
6 - 121/13 / {39.2 - 189/4 × (2/3 - 1/5) / 0.7} × 4.9 =
Answer
1/13
Problem 2
There are three kinds of salt solutions A, B, and C weigh 60 g, 120 g, and 100g, respectively.
The salt concentration of A is 3%.
① When 20g of A and 30g of B are mixed, the salt concentration of salt solution is same as C.
② The salt concentration of salt solution D made by mixing A, B, and C altogether is to be 7.5%.
Answer the following questions.
The salt concentration of a salt solution is a ratio of the weight of salt to the weight of a salt solution.
(1) Find the weight of the salt contained in D.
(2) Find the salt concentration (%) of B.
(3) Find the salt concentration (%) of C.
Answer
(1) 21 g
(2) 10 %
(3) 7.2 %
Solution
(1) The weight of D = 60 + 120 + 100 = 280 g.
280 × 0.075 = 21 g.
(2) As A 20g + B 30g are mixed, the salt concentration of salt solution is same as C, in case that A 40g + B 60g are mixed, it is also the same concentration of salt solution is same as C.
Thus A 60g + B 120g + C 100g = A 60g + B 120g + A 40g + B 60g = A 100g + B 180g also become D.
The weight of salt in A 100g = 100 × 3% = 3g.
The weight of salt in B 180g = 21g - 3g = 18g.
Therefore the salt concentration (%) of B = 18 / 180 = 10%.
(3) The weight of salt in A 40g + B 60g = 40 × 3% + 60 × 10% = 7.2g which is the weight of salt in C 100g.
Therefore the salt concentration (%) of C = 7.2 / 100 = 7.2%.
Problem 3
There are A station and B station and another station C is between A and B.
The speed of this train is 2.675 km per hour faster than the speed which is calculated the distance between A and B is divided by 7 hours which is actual time taken from A to B.
Find the distance of A station and B station.
Train X will leave A station at 23:10 and will arrive at intermediate C station at 2:10 on the next day.
After stopping for 20 minutes, train X will leave C station at 2:30 and it will arrive at B station at 6:10. The speed of this train is 2.675 km per hour faster than the speed which is calculated the distance between A and B is divided by 7 hours which is actual time taken from A to B.
Find the distance of A station and B station.
Answer
374.5 km
Solution
The actual time the train ran on the way from A station to B station is calculated by stoppage-time 20 minutes is subtracted from 7 hours.
7 hour - 20 minutes = 20/3 hours.
The ratio of the time is 20/3 : 7 = 20/3 : 21/3 = 20 : 21.
The ratio of the speed is an inverse ratio and is 21 : 20.
Since 21 - 20 = 1 of the difference of this ratio is equivalent to 2.675 km per hour which is a difference of actual speed, the speed of this train is 2.675 km per hour × 21.
Since the running time is 20/3 hours, the distance between A and B is 2.675 × 21 × 20/3 = 374.5 km.
Problem 4
Either the sign + of addition or the sign × of multiplication is put in the following [ ] and calculate.
Answer the following questions.
(1) When 1[ ]2[ ]3[ ]4 is calculated, Answer each calculation result to small order.
When there come out same calculation results, write only once.
(2) Answer two kinds of manner putting in + and × applicable to [ ] of the following formula.
1[ ]2[ ]3[ ]4[ ]5 = 2[ ]3[ ]4[ ]5[ ]6
Answer
(1) 9, 10, 11, 14, 15, 24, 25
(2) 1[+]2[+]3[×]4[+]5 = 2[+]3[+]4[+]5[+]6,
1[×]2[+]3[+]4[×]5 = 2[+]3[×]4[+]5[+]6
Solution
(1)
1 + 2 + 3 + 4 = 10
1 × 2 + 3 + 4 = 9
1 + 2 × 3 + 4 = 11
1 + 2 + 3 × 4 = 15
1 × 2 × 3 + 4 = 10
1 × 2 + 3 × 4 = 14
1 + 2 × 3 × 4 = 25
1 × 2 × 3 × 4 = 24
(2)
As for 1[ ]2[ ]3[ ]4[ ]5, write down until 10th calculation.
1 + 2 + 3 + 4 + 5 = 15
1 × 2 + 3 + 4 + 5 = 14
1 + 2 × 3 + 4 + 5 = 16
1 + 2 + 3 × 4 + 5 = 20
1 + 2 + 3 + 4 × 5 = 26
1 × 2 × 3 + 4 + 5 = 15
1 × 2 + 3 × 4 + 5 = 19
1 × 2 + 3 + 4 × 5 = 25
1 + 2 × 3 × 4 + 5 = 30
1 + 2 × 3 + 4 × 5 = 27
Then move to writing down of the calculation of 2[ ]3[ ]4[ ]5[ ]6
2 + 3 + 4 + 5 + 6 = 20
2 × 3 + 4 + 5 + 6 = 21
2 + 3 × 4 + 5 + 6 = 25
At this calculation, two kinds could be found.
Problem 5
There is a rectangular prism 3 cm in length, 4 cm in width and 5 cm in height.
As for faces of this rectangular prism, the face of the rectangle with 3cm and 4cm side is set to face A, the face with 4cm and 5cm is set to face B and the face with 5cm and 3cm is set to face C.
Answer the following questions.
(1) Make small rectangular prisms by cutting with planes which are parallel to face A, face B and face C, once, once and twice respectively.
① Find the number of small rectangular prism.
① Find the number of small rectangular prism.
② Find the sum total of the surface area of these small rectangular prisms.
Noted that the surface area of a rectangular prism is the sum total of the area of all the faces of the rectangular prism.
Noted that the surface area of a rectangular prism is the sum total of the area of all the faces of the rectangular prism.
(2) This rectangular prism was cut with planes which are parallel to face A, face B and face C, X times, Y times and Z times respectively.
In this case, there are 90 small rectangular prisms made and the sum total of the surface area of these rectangular prisms was 462 cm2.
Find the number applicable to X, Y, and Z.
Answer
(1) ① 12 pieces
② 218 cm2
(2) X = 2, Y = 5, Z = 4
② 218 cm2
(2) X = 2, Y = 5, Z = 4
Solution
(1)
①
According to the figure below, the number of small rectangular prism is 12.
②
Total surface area of the original rectangular prism = (3 × 4 + 3 × 5 + 4 × 5) × 2 = 94 cm2.
When cutting with plane which is parallel to face A once, the total surface is increased by 24 cm2 as the area of cut surface is 12 cm2.
When cutting with planes which are parallel to face B and face C once, the total surface is increased by 20 × 2 = 40 cm2 and 15 × 2 = 30 cm2respectively.
Thus sum of total surface area of small rectangular prisms = 94 + 24 + 40 + 30 × 2 = 218 cm2.
(2)
It is (X + 1) × (Y + 1) × ( Z + 1) = 90.
90 = 1 × 2 × 3 × 3 × 5
Increased total surface area = 462 - 94 = 368 cm2.
Thus 24 × X + 40 × Y + 30 × Z = 368.
As ones digit of 368 is 8, X should be 2 or 7 or 12.
As 2 + 1 = 3, 7 + 1 = 8, 12 + 1 = 13, 3 is the only number to be a divisor of 90.
Thus X = 2
40 × Y + 30 × Z = 368 - 24 × 2 = 320 and (Y + 1) × (Z + 1) = 90 / 3 = 30
As Z should be multiple of 4, Z = 4.
Then Y = 5.
Problem 6
There is the sector OAB which is 1/4 part a circle.
Answer the following questions.
(1) Find the ratio of the area of a shadow area and the sector OAB in Fig. 1.
Noted that the straight lines OA, CD, and EF are parallel.
(2) As shown in Fig. 2, the straight line parallel to OA was drawn from each point which divided the arc AB of the sector OAB into five equally. OA is 5 cm.
Find the sum of the area of two shadow areas.
Pi is assumed to be 3.14.
Answer the following questions.
(1) Find the ratio of the area of a shadow area and the sector OAB in Fig. 1.
Noted that the straight lines OA, CD, and EF are parallel.
(2) As shown in Fig. 2, the straight line parallel to OA was drawn from each point which divided the arc AB of the sector OAB into five equally. OA is 5 cm.
Find the sum of the area of two shadow areas.
Pi is assumed to be 3.14.
Answer
(1) 4 : 9
(2) 7.85 cm2
(2) 7.85 cm2
Solution
(1)
As △OEF and △OCD are congruent, part of shadow area can be transferred equivalently as shown in the figure below.
Then the area ratio of sector ODF : OAB = (90° - 25° × 2) : 90° = 40° : 90° = 4 : 9.
(2)
The area of shadow areas = (The area of shadow area in Fig.4) - (The area of shadow area in Fig.5)
Based on the idea of (1), since △OBG and △OPH are congruent in Fig.4, (The area of shadow area in Fig.4) = (The area of the sector OGH).
Furthermore since △OIQ and △OJR are congruent in Fig.5, (The area of shadow area in Fig.5) = (The area of the sector OIJ).
Thus the area of shadow area in Fig.2 = (The area of the sector OGH) - (The area of the sector OIJ).
Therefore the area to be found = 5 × 5 × 3.14 × 1/4 × (3/5 - 1/5) ＝7.85 cm2.
