Time : 60 minutes
Problem 5
Passing mark : 70%
Answer : The end of the problem
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Problem 1
Problem 2
Problem 3
Problem 4
As 0, 1, 2, 3, 5, 7, 10, 11, 12, 13, 15, 17, 20, ----, the number made only of 0, 1, 2, 3, 5, 7 is arranged in small order.
Answer the following questions.
(1) Which number in this sequence is 2007?
(2) Find the 100th number in the multiple of 5.
0 is assumed as the multiple of 5.
(3) Find the 100th number in the multiple of 3.
0 is assumed as the multiple of 3.
Problem 2
There is a rectangle ABCD as shown in the figure.
AB = 6 cm and AD = 12 cm. AM : MB = 2 : 1 and DN : NC = 1 : 2.
The point P moves on the side AD.
The intersection of the extension of PM and the extension of BC is set to Q.
The intersection of the extension of PN and the extension of BC is set to R.
Answer the following questions.
(1) When the triangle PQR turns into an isosceles triangle of PQ = PR, find the length of AP.
(2) Find the length of AP at the time of BQ = CR.
(3) When the area of the triangle PQR is 90 cm2, find the area of pentagon PMBCN.
AB = 6 cm and AD = 12 cm. AM : MB = 2 : 1 and DN : NC = 1 : 2.
The point P moves on the side AD.
The intersection of the extension of PM and the extension of BC is set to Q.
The intersection of the extension of PN and the extension of BC is set to R.
Answer the following questions.
(1) When the triangle PQR turns into an isosceles triangle of PQ = PR, find the length of AP.
(2) Find the length of AP at the time of BQ = CR.
(3) When the area of the triangle PQR is 90 cm2, find the area of pentagon PMBCN.
Problem 3
Taro and Jiro respectively decided to make a plan to do homework of the mathematics of the summer vacation.
Taro decided to solve three problems a day from the first day of the summer vacation.
Jiro decided not do its homework for six days of the beginning of the summer vacation and from seventh day to solve [ ] problems a day until he finish the homework.
It was found that several days after Jiro started to solve, the number of problems solved by Taro and Jiro are the same.
It was found that several days after Jiro started to solve, the number of problems solved by Taro and Jiro are the same.
Since the day above Taro solved five problems a day until the day he would finish, but it was two days after the day Jiro finished that Taro finished.
The number of days that Taro solved three problems a day is same as the number of days Jiro solved five days a day.
Answer the following questions.
(1) How many days did Taro take to finish the homework?
(2) How many days did Jiro take to finish the homework?
(3) Find the number applied in [ ].
The number of days that Taro solved three problems a day is same as the number of days Jiro solved five days a day.
Answer the following questions.
(1) How many days did Taro take to finish the homework?
(2) How many days did Jiro take to finish the homework?
(3) Find the number applied in [ ].
Problem 4
Taro and Jiro participated in the competition of the triathlon.
A triathlon is a game which carry out three athletic games in order of swimming, a bicycle (49 km), and marathon (20 km), and all athletes compete for speed there.
In a swimming race, the ratio of the speed of Taro and Jiro is 15 : 13 and there is distance difference of 12 m in one minute.
It was 6 minutes after since Taro begins a bicycle race that Jiro made a goal by swimming.
In the bicycle race, the ratio of the speed of Taro and Jiro was 5 : 7.
Since the tire was flat on the way, Jiro took 12 minutes and 15 seconds for the repair.
It was 9 minutes and 45 seconds after Jiro started marathon game when Taro made a goal by the bicycle.
Answer the following questions.
(1) Find the speed per minute at which Taro swims.
(2) Find the distance of the swimming race.
(3) Find the speed per hour of Jiro in a bicycle race.
(4) In the last marathon, 7.2 km from the start is the uphill course and remaining 12 km is flat course.
Through the uphill course, the speed ratio of Taro and Jiro is 6 : 5.
Taro was able to shorten the time difference with Jiro in the this up hill course for 5 minutes.
Find the speed per minute when Jiro runs in the flat course to make a goal ahead of Taro.
The speed ratio in the uphill and flat course of Taro is 9 : 10.
A triathlon is a game which carry out three athletic games in order of swimming, a bicycle (49 km), and marathon (20 km), and all athletes compete for speed there.
In a swimming race, the ratio of the speed of Taro and Jiro is 15 : 13 and there is distance difference of 12 m in one minute.
It was 6 minutes after since Taro begins a bicycle race that Jiro made a goal by swimming.
In the bicycle race, the ratio of the speed of Taro and Jiro was 5 : 7.
Since the tire was flat on the way, Jiro took 12 minutes and 15 seconds for the repair.
It was 9 minutes and 45 seconds after Jiro started marathon game when Taro made a goal by the bicycle.
Answer the following questions.
(1) Find the speed per minute at which Taro swims.
(2) Find the distance of the swimming race.
(3) Find the speed per hour of Jiro in a bicycle race.
(4) In the last marathon, 7.2 km from the start is the uphill course and remaining 12 km is flat course.
Through the uphill course, the speed ratio of Taro and Jiro is 6 : 5.
Taro was able to shorten the time difference with Jiro in the this up hill course for 5 minutes.
Find the speed per minute when Jiro runs in the flat course to make a goal ahead of Taro.
The speed ratio in the uphill and flat course of Taro is 9 : 10.
Problem 5
There is a square pyramid with a square In the bottom and an isosceles triangle in the side as shown in Fig. 1.
The intersection of AC and BD is set to O.
When O is connected to the vertex P, PO will become vertical to AC and BD, respectively.
The point in the middle of PO is set to M.
The intersection of the extension of CM and the side PA is set to E.
Moreover, the point which divides the side PB into 3 : 1 is set to F.
The intersection of the plane EPC and side PD is set to G.
Three point G, M, and F are located on a straight line.
Answer the following questions at this time.
The ratio should be by the least integer.
The intersection of AC and BD is set to O.
When O is connected to the vertex P, PO will become vertical to AC and BD, respectively.
The point in the middle of PO is set to M.
The intersection of the extension of CM and the side PA is set to E.
Moreover, the point which divides the side PB into 3 : 1 is set to F.
The intersection of the plane EPC and side PD is set to G.
Three point G, M, and F are located on a straight line.
Answer the following questions at this time.
The ratio should be by the least integer.
(1) Find PE : EA with reference to Fig. 2.
(2) Find PG : GD with reference to Fig. 3.
(3) Find the ratio of the volume of triangular pyramid P-ECG to the volume of square pyramid P-ABCD.
(4) Cut the square pyramid P-ABCD with Plane EFCG.
The volume of the upper solid is set to U and the lower solid is set to V.
Find the ratio of U to V at this time.
<Answer>
Problem 1
As for the number of double figures, since numbers of tens digit are five other than 0 and numbers of ones digit are six, there are 5 × 6 = 30 pieces in all.
Similarly, numbers of triple figures are 5 × 6 × 6 = 180 pieces.
The number of 4 figures whose thousands digit is 1 is 1 × 6 × 6 × 6 = 216 pieces.
2000 is the 6 + 30 + 180 + 216 + 1 = 433th number.
Therefore, 2007 is 2001, 2002, 2003, 2005, and 2007, and it is 433 + 5 = 438th number.
(2) The number of single figure is two pieces of 0 and 5.
As for the number of double figures, since number of ones digit are two kinds of 0 and 5, there are 5 × 2 = 10 pieces.
The number of triple figures is 5 × 6 × 2 = 60 pieces.
1000 is the 2 + 10 + 60 + 1 = 73rd number.
Furthermore, number of the number whose thousands digit is 1 and hundreds digit is 0 or 1 is 1 × 2 × 6 × 2 = 24 pieces.
Thus, 1200 is the 72 + 24 + 1 = 97th number.
Since 1205, 1210, and 1215, are following 1200, the 100th number is 1215.
(3) When 36 of single and double figures are classified according to the remainder when it is divided by 3, it is shown as in the table below.
As for triple figures, in case hundreds digit of the triple figures is 1, 2, 3, 5, 7, they will become multiples of 3 when following double figures are C, B, A, B, and C, respectively.
Thus the number of the multiple of 3 of triple figures is 12 × 5 = 60 pieces.
Thus, there are 60 + 12 = 72 pieces of multiple of 3 up to 1000.
As for four figures whose thousands digit is 1 and hundreds digit is 0, number of the multiple of 3 is 12 pieces.
As for four figures whose thousands digit is 1 and hundreds digit is 1, number of the multiple of 3 is 12 pieces as well.
Thus, 1200 becomes the 72 + 12 × 2 + 1 = 97th number.
Since 1203, 1212, and 1215 are following 1200, the 100th number is 1215.
Problem 2
Problem 3
Problem 4
Problem 5
As 0, 1, 2, 3, 5, 7, 10, 11, 12, 13, 15, 17, 20, ----, the number made only of 0, 1, 2, 3, 5, 7 is arranged in small order.
Answer the following questions.
(1) Which number in this sequence is 2007?
(2) Find the 100th number in the multiple of 5.
0 is assumed as the multiple of 5.
(3) Find the 100th number in the multiple of 3.
0 is assumed as the multiple of 3.
Answer
(1) 438th
(2) 1215
(3) 1215
Solution
(1) The number of single figure is six pieces of 0, 1, 2, 3, 5, and 7. As for the number of double figures, since numbers of tens digit are five other than 0 and numbers of ones digit are six, there are 5 × 6 = 30 pieces in all.
Similarly, numbers of triple figures are 5 × 6 × 6 = 180 pieces.
The number of 4 figures whose thousands digit is 1 is 1 × 6 × 6 × 6 = 216 pieces.
2000 is the 6 + 30 + 180 + 216 + 1 = 433th number.
Therefore, 2007 is 2001, 2002, 2003, 2005, and 2007, and it is 433 + 5 = 438th number.
(2) The number of single figure is two pieces of 0 and 5.
As for the number of double figures, since number of ones digit are two kinds of 0 and 5, there are 5 × 2 = 10 pieces.
The number of triple figures is 5 × 6 × 2 = 60 pieces.
1000 is the 2 + 10 + 60 + 1 = 73rd number.
Furthermore, number of the number whose thousands digit is 1 and hundreds digit is 0 or 1 is 1 × 2 × 6 × 2 = 24 pieces.
Thus, 1200 is the 72 + 24 + 1 = 97th number.
Since 1205, 1210, and 1215, are following 1200, the 100th number is 1215.
(3) When 36 of single and double figures are classified according to the remainder when it is divided by 3, it is shown as in the table below.
Thus the number of the multiple of 3 of triple figures is 12 × 5 = 60 pieces.
Thus, there are 60 + 12 = 72 pieces of multiple of 3 up to 1000.
As for four figures whose thousands digit is 1 and hundreds digit is 0, number of the multiple of 3 is 12 pieces.
As for four figures whose thousands digit is 1 and hundreds digit is 1, number of the multiple of 3 is 12 pieces as well.
Thus, 1200 becomes the 72 + 12 × 2 + 1 = 97th number.
Since 1203, 1212, and 1215 are following 1200, the 100th number is 1215.
Problem 2
There is a rectangle ABCD as shown in the figure.
AB = 6 cm and AD = 12 cm. AM : MB = 2 : 1 and DN : NC = 1 : 2.
The point P moves on the side AD.
The intersection of the extension of PM and the extension of BC is set to Q.
The intersection of the extension of PN and the extension of BC is set to R.
Answer the following questions.
(1) When the triangle PQR turns into an isosceles triangle of PQ = PR, find the length of AP.
(2) Find the length of AP at the time of BQ = CR.
(3) When the area of the triangle PQR is 90 cm2, find the area of pentagon PMBCN.
AB = 6 cm and AD = 12 cm. AM : MB = 2 : 1 and DN : NC = 1 : 2.
The point P moves on the side AD.
The intersection of the extension of PM and the extension of BC is set to Q.
The intersection of the extension of PN and the extension of BC is set to R.
Answer the following questions.
(1) When the triangle PQR turns into an isosceles triangle of PQ = PR, find the length of AP.
(2) Find the length of AP at the time of BQ = CR.
(3) When the area of the triangle PQR is 90 cm2, find the area of pentagon PMBCN.
Answer
(1) 8 cm
(2) 9.6 cm
(3) 56 cm2
(2) 9.6 cm
(3) 56 cm2
Solution
(1) Since △PQR is an isosceles triangle, ∠APQ=∠DPN.
Therefore, △APM and △DPN are homothetic.
Since AB = DC and AM : MB = 2 : 1 and DN : NC = 1 : 2, AM : DN = 2 : 1.
The homothetic ratio of △APM and △DPN is also set to 2 : 1.
Thus, AP : DP = 2 : 1.
Therefore, AP = 12 cm × 2/(2+1) = 8 cm.
(2) △MAP and △MBQ are homothetic and homothetic ratio is AM : MB = 2 : 1.
Thus, as shown in the figure, BQ is set to 2, AP = 2 × 2 = 4.
△NDP and △NCR are also homothetic and homothetic ratio is DN : NC = 1 : 2.
Thus, since CR = 2, PD = 2/2 = 1.
AP + PD = 4 + 1 = 5 = 12cm, AP = 4 = 12cm × 4/5 = 9.6 cm.
(3) Since the area of △PQR measures 90 cm2, the length of QR is 90 × 2/ 6 = 30 cm.
BQ is set to 1, AP = 2.
Since AD = 12 cm, PD = 12cm - 2.
CR = PD × 2 = (12cm - 2) × 2 = 24cm - 4.
Then BQ + CR = 1 + 24cm - 4 = 24cm - 3.
Since QR = 30 cm and BC = 12cm, BQ + CR = 30 - 12 = 18cm.
Thus, 24 cm - 3 = 18 cm and 3 is equivalent to 24 - 18 = 6cm.
1 = 6cm / 3 = 2cm.
As mentioned above, since BQ = 2 cm, the area of △MBQ is 2 × 2 / 2 = 2cm2 and CR = 24 cm - 4 = 24 cm - 2cm × 4 = 16cm.
Therefore, he area of △NCR = 16 × 4 / 2 = 32cm2.
The area of pentagon PMBCN =△PQR - △MBQ - △NCR = 90 - 2 - 32 = 56cm2.
Problem 3
Taro and Jiro respectively decided to make a plan to do homework of the mathematics of the summer vacation.
Taro decided to solve three problems a day from the first day of the summer vacation.
Jiro decided not do its homework for six days of the beginning of the summer vacation and from seventh day to solve [ ] problems a day until he finish the homework.
It was found that several days after Jiro started to solve, the number of problems solved by Taro and Jiro are the same.
It was found that several days after Jiro started to solve, the number of problems solved by Taro and Jiro are the same.
Since the day above Taro solved five problems a day until the day he would finish, but it was two days after the day Jiro finished that Taro finished.
The number of days that Taro solved three problems a day is same as the number of days Jiro solved five days a day.
Answer the following questions.
The number of days that Taro solved three problems a day is same as the number of days Jiro solved five days a day.
Answer the following questions.
(1) How many days did Taro take to finish the homework?
(2) How many days did Jiro take to finish the homework?
(3) Find the number applied in [ ].
Answer
In addition, Taro and Jiro solved the number of the same problems at the time of the end in the first half. Based on these statements, it is found that the ratio of the number of the problems solved B in first half and second half is 3 : 5 and the ratio of numbers Jiro solved in first half and second half is also 3 : 5.
As the number of problems Jiro solved a day is same through days, the ratio of days of A : B that is the ratio of days Jiro solved is 3 : 5 in the figure.
As A + 6 = B + 2, the difference between A and B are 6 - 2 = 4 days. 4days is equivalent to 5 -3 =2 which is the differences of the ration. Thus, the ratio 1 is equivalent to 2 days (4days / 2 = 2).
Therefore, the days Jiro solved problems are 2days × (3+5) = 16 days.
(1) 16 days
(2) 24 days
(3) six
(2) 24 days
(3) six
Solution
(1)The number of days (first half) that Taro solved five a day is same as the number of days (second half) that Taro solved five a day.In addition, Taro and Jiro solved the number of the same problems at the time of the end in the first half. Based on these statements, it is found that the ratio of the number of the problems solved B in first half and second half is 3 : 5 and the ratio of numbers Jiro solved in first half and second half is also 3 : 5.
As the number of problems Jiro solved a day is same through days, the ratio of days of A : B that is the ratio of days Jiro solved is 3 : 5 in the figure.
As A + 6 = B + 2, the difference between A and B are 6 - 2 = 4 days. 4days is equivalent to 5 -3 =2 which is the differences of the ration. Thus, the ratio 1 is equivalent to 2 days (4days / 2 = 2).
Therefore, the days Jiro solved problems are 2days × (3+5) = 16 days.
(2) Taro took eight days (6+2=8) more than Jiro.
It is 16 + 8 = 24days.
(3) Total number of problems is = (3 + 5) × 12 days =96 problems.
Since Jiro solved 96 problems 16 days, the number solved per day is 96 / 16 = 6.
Problem 4
Taro and Jiro participated in the competition of the triathlon.
A triathlon is a game which carry out three athletic games in order of swimming, a bicycle (49 km), and marathon (20 km), and all athletes compete for speed there.
In a swimming race, the ratio of the speed of Taro and Jiro is 15 : 13 and there is distance difference of 12 m in one minute.
It was 6 minutes after since Taro begins a bicycle race that Jiro made a goal by swimming.
In the bicycle race, the ratio of the speed of Taro and Jiro was 5 : 7.
Since the tire was flat on the way, Jiro took 12 minutes and 15 seconds for the repair.
It was 9 minutes and 45 seconds after Jiro started marathon game when Taro made a goal by the bicycle.
Answer the following questions.
(1) Find the speed per minute at which Taro swims.
(2) Find the distance of the swimming race.
(3) Find the speed per hour of Jiro in a bicycle race.
(4) In the last marathon, 7.2 km from the start is the uphill course and remaining 12 km is flat course.
Through the uphill course, the speed ratio of Taro and Jiro is 6 : 5.
Taro was able to shorten the time difference with Jiro in the this up hill course for 5 minutes.
Find the speed per minute when Jiro runs in the flat course to make a goal ahead of Taro.
The speed ratio in the uphill and flat course of Taro is 9 : 10.
A triathlon is a game which carry out three athletic games in order of swimming, a bicycle (49 km), and marathon (20 km), and all athletes compete for speed there.
In a swimming race, the ratio of the speed of Taro and Jiro is 15 : 13 and there is distance difference of 12 m in one minute.
It was 6 minutes after since Taro begins a bicycle race that Jiro made a goal by swimming.
In the bicycle race, the ratio of the speed of Taro and Jiro was 5 : 7.
Since the tire was flat on the way, Jiro took 12 minutes and 15 seconds for the repair.
It was 9 minutes and 45 seconds after Jiro started marathon game when Taro made a goal by the bicycle.
Answer the following questions.
(1) Find the speed per minute at which Taro swims.
(2) Find the distance of the swimming race.
(3) Find the speed per hour of Jiro in a bicycle race.
(4) In the last marathon, 7.2 km from the start is the uphill course and remaining 12 km is flat course.
Through the uphill course, the speed ratio of Taro and Jiro is 6 : 5.
Taro was able to shorten the time difference with Jiro in the this up hill course for 5 minutes.
Find the speed per minute when Jiro runs in the flat course to make a goal ahead of Taro.
The speed ratio in the uphill and flat course of Taro is 9 : 10.
Answer
(1) 90 m/m
(2) 3510 m
(3) 42 km/h
(4) 287 m/m
(2) 3510 m
(3) 42 km/h
(4) 287 m/m
Solution
(1) Since the ratio of swimming speed of Taro and Jiro is 15 : 13, the ratio of distance is also 15 : 13.
Since there is a difference of 12 m in every minute, 15 - 13 = 2 which is a difference of the distance ratio is equivalent to 12 m.
Therefore, the speed per minute of Taro is 12m / 2 × 15 = 90m.
(2) Since the ratio of the swimming time of Taro and Jiro is an inverse ratio of the swimming speed ratio and it is 13 : 15.
15 - 13 = 2 which is a difference of this time ratio is equivalent to 6 minutes of an actual time lag.
The time when Taro took for swimming is 6 minutes /2 × 13 = 39 minutes.
Therefore, the distance of a swimming race is 90 m/m × 39 minutes = 3510m.
(3) Jiro started the bicycle race 6 minutes later than Taro.
Although Jiro took 12 minutes and 15 seconds for repair of the blowout in the middle of the race, he made a goal 9 minute and 45 seconds earlier than Taro.
The time lag of the bicycle race becomes 6 minutes + 12 minutes and 15 seconds +9 minutes and 45 seconds = 28 minutes.
Since the ratio of the speed of Taro and Jiro was 5 : 7, the ratio of the time of the bicycle race is an inverse ratio of the speed ratio and it is Taro : Jiro = 7 : 5.
7 - 5 = 2 which is a difference of this time ratio is equivalent to 28 minutes of an actual time lag.
Therefore, the time when Jiro took is 28 minutes /2 × 5 = 70 minutes.
Therefore, the speed of Jiro is 49km/ 70/60 hour = 42km/h.
(4) The distance ratio of the uphill and flat course is 7.2 : 12.8 = 9 : 16.
Since the speed ratio of ascent and descent of Taro is 9 : 10, the time ratio is 9/9 : 16/10 = 1 : 1.6 = 5 : 8.
Since the speed ratio of Taro and Jiro in the uphill course is 6 : 5, the time ratio turns into an inverse ratio of speed ratio and it is 5 : 6.
Since Taro contracted the time difference with Jiro in this uphill course by 5 minutes, 6 - 5 = 1 which is a difference of a time ratio is equivalent to 5 minutes of actual time lag.
Thus, the actual time Taro took in the uphill course is 5 minute /1 × 5 = 25 minutes.
The time in a flat course is 25 minute × 8/5 = 40 minutes.
In order for Jiro to make a goal ahead of Taro, he should run less than 40 minutes + 9 minutes 45 seconds - 5 minutes = 44 minutes 45 seconds in the flat course.
According to the calculation of 12800/444560=286.03 --, the speed of the integer to be found is 287 m/m.
Problem 5
There is a square pyramid with a square In the bottom and an isosceles triangle in the side as shown in Fig. 1.
The intersection of AC and BD is set to O.
When O is connected to the vertex P, PO will become vertical to AC and BD, respectively.
The point in the middle of PO is set to M.
The intersection of the extension of CM and the side PA is set to E.
Moreover, the point which divides the side PB into 3 : 1 is set to F.
The intersection of the plane EPC and side PD is set to G.
Three point G, M, and F are located on a straight line.
Answer the following questions at this time.
The ratio should be by the least integer.
(1) Find PE : EA with reference to Fig. 2.
(2) Find PG : GD with reference to Fig. 3.
(3) Find the ratio of the volume of triangular pyramid P-ECG to the volume of square pyramid P-ABCD.
(4) Cut the square pyramid P-ABCD with Plane EFCG.
The volume of the upper solid is set to U and the lower solid is set to V.
Find the ratio of U to V at this time.
Answer
(1) 1 : 2
(2) 3 : 5
(3) 1 : 16
(4) 3 : 13
(2) 3 : 5
(3) 1 : 16
(4) 3 : 13
Solution
(1)
In the left figure below, as for △HPM and △COM, since HP and OC are parallel and PM = OM, △HPM and △COM are congruent.
HP : CO = 1 : 1.
In the right figure below, as for △HPE and △CAE, since HP and AC are parallel, △HPE and △CAE are homothetic.
Moreover, since HP = OC = 1 and AC is twice of OC, AC = 2.
The homothetic ratio of two triangles is 1 : 2.
Therefore, PE : EA = 1 : 2 as well.
(2)
As for △KPF and △LBF, since KP and BL are parallel, △KPM and △LBF are homothetic.
Since PF : BF = 3 : 1, the homothetic ratio of two triangles is 3 : 1.
Thus, KP : LB = 3 : 1.
As for △KPM and △LOM, since KP and OL are parallel and PM = OM, △KPM and △LOM are congruent.
Thus, KP : LO = 1 : 1.
Since △KPG and △LDG are also homothetic and KP = LO and KP : LB = 3 : 1, LO : LB = 3 : 1.
Thus, BO=LO - LB = 3 - 1 = 2.
Since DO = BO, DL = 2 + 3 = 5.
Since the homothetic ratio of △KPG and △LDG is 3 : 5, PG : GD = 3 : 5.
(3)
As for triangular pyramid P-ECG and P-ACD, when vertex C is assumed as the vertex of two solids, bottom is as shown in the figure below.
The area ratio of △PEG and △PAD = 3 × 1 : 8 × 3 = 1 : 8.
Thus the volume ratio of triangular pyramid P-ECG and P-ACD is also 1 : 8.
The volume of the square pyramid P-ABCD is twice of that of the triangular pyramid P-ACD.
Thus the volume ratio of the triangular pyramid P-ECG and the square pyramid P-ABCD is 1 : 8 × 2 = 1 : 16.
(4) Because PE : PA = 1 : 3 and PF : PB = 3 : 4, the area ratio of △PEF and △PAB is (1 × 3) : (3 × 4) =3 : 12 = 1 : 4.
The volume ratio of the triangular pyramid P-ECF and P-ABC is also 1 : 4.
The volume ratio of triangular pyramid P-ECF and square pyramid P-ABCD is 1 : (4 × 2) = 1 : 8.
The volume of square pyramid P-ABCD is set to 16.
The volume of triangular pyramid P-ECG is 1 and the volume of triangular pyramid P-ECF is 2.
Then U = 1 + 2 = 3.
Then V=16 - U = 16 - 3 = 13.
