Time : 50 minutes
Passing mark : 70%
Answer : End of the problem
Problem 1
(2)
There are 365 cards piled up on which the date of the non-leap year is written.
It is written to the 1st card as January 1, to the 2nd card as January 2, to the 3rd card as January 3rd and to the 365th card as December 31. Answer the following questions.
(2)-1 Remove even-numbered cards counted from the top.
In this case, the date written on the top of the remaining card is January 1 and the date of the 2nd card is January 3.
Find the date of the 28th card.
(2)-2 As a next step, remove odd-numbered cards counted from the top among remaining cards.
In this case, which number of card counted from the top among remaining cards is written as September 12?
(2)-3 Suppose that January 1 is Monday.
Find the day of the week of the 69th card counted from the top among remaining cards.
Problem 2
(1)
① (26.5 / 2.5 - 17/3) / 37/24 - 0.96 × 11/6 =
② Find X
53/36 - 11/25 / (27/13 - X × 9/65) = 5/4
③ 3 × 5 × 5 × 5 + 4 × 5 × 5 + 2 × 5 + 1
= A × A × A + B × A × A + C × A + 3.
Find A, B and C, respectively.
A, B and C are the integers from 1 to 9.
Find A, B and C, respectively.
A, B and C are the integers from 1 to 9.
(2)
It is written to the 1st card as January 1, to the 2nd card as January 2, to the 3rd card as January 3rd and to the 365th card as December 31. Answer the following questions.
(2)-1 Remove even-numbered cards counted from the top.
In this case, the date written on the top of the remaining card is January 1 and the date of the 2nd card is January 3.
Find the date of the 28th card.
(2)-2 As a next step, remove odd-numbered cards counted from the top among remaining cards.
In this case, which number of card counted from the top among remaining cards is written as September 12?
(2)-3 Suppose that January 1 is Monday.
Find the day of the week of the 69th card counted from the top among remaining cards.
Problem 2
A figure is a circle with a radius of 3 cm and AB is a diameter of the circle.
Problem 3
All ● in a figure is 15 degrees. Pi is assumed to be 3.14.
(1) Find the sum of the area of the slash portion of P and Q.
(2) Find the area of the slash portion of Q.
Problem 3
When the cube of Fig. 1 is opened up by cutting by the bold line of Fig. 2, the development view become Fig. 3.
How should it be cut respectively, in order for the development view of each cube to become as shown in (1) and (2)?
As shown in Fig. 2, write a bold line in the sketch of the cube.
Problem 4
As shown in Fig. 2, write a bold line in the sketch of the cube.
Problem 4
As shown in a figure, the square card was equally divided into four and the integer was written to each portion sequentially from 1.
A lot of these cards were made.
All the sizes of the card are same.
(1) Cards were arranged according to the direction in the figure from left to right sequentially from the 1st card.
Which card is the first card for the sum of the two number of the upper rows of one card to become larger than 120?
(2) The card was piled up in order by direction in the figure from the 1st card to the 50th card.
Find the sum total of 50 numbers which has overlapped with the number 1 of the 1st card including 1 of the 1st card.
(3) Rotate 90 degrees of the 1st card counterclockwise and pile up on the 2nd card.
Next, rotate 90 degrees counterclockwise both of 1st and 2nd cards piled up and 2 cards are piled up on the 3rd card.
Thus, when you pile up 50 cards, find the sum totals of 50 numbers including 3 which has overlapped with the number 3.
Answer the following questions.
Which card is the first card for the sum of the two number of the upper rows of one card to become larger than 120?
(2) The card was piled up in order by direction in the figure from the 1st card to the 50th card.
Find the sum total of 50 numbers which has overlapped with the number 1 of the 1st card including 1 of the 1st card.
(3) Rotate 90 degrees of the 1st card counterclockwise and pile up on the 2nd card.
Next, rotate 90 degrees counterclockwise both of 1st and 2nd cards piled up and 2 cards are piled up on the 3rd card.
Thus, when you pile up 50 cards, find the sum totals of 50 numbers including 3 which has overlapped with the number 3.
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Problem 5
There is a rectangle ABCD with 20 cm in length and 10 cm in width as shown in a figure.
The point P leaves the point A and moves in order of A→B→C→D→A→-------- on the side of this rectangle.
The point Q leaves the point B simultaneously with the point P and moves in order of B→C→D→A→ ----.
The point P moves at the speed of 1 cm/s and the point Q moves 2 cm/s in the first round.
Then, the point P increases the speed per 1 second for every round as 2 cm/s in 2nd round, 3 cm/s in 3rd round.
The point Q also increases the speed per 1 second for every round as 3 cm/s in 2nd round, 4 cm/s in 3rd round.
Answer the following questions.
(1) Find the time after leaving when the point Q overlaps with the point P for the first time.
(2) Regarding from 1st round to 3rd round, write the time when two point P and Q take moving one round of rectangle ABCD, respectively in a lower table.
Let the measure be second.
(3) 100 seconds after leaving, where are two point P and Q?
Write the position of point P in Fig. 1 and write the position of the point Q in Fig. 2. according to the (Example) below.
(Example) When the point P is at the position of 7 cm from A on the side AD in 100 seconds.
(4) In 100 seconds after leaving, find the last time of seconds when the point Q overlaps with the point P.
<Answer>
Problem 1
(1)
① (26.5 / 2.5 - 17/3) / 37/24 - 0.96 × 11/6 =
② Find X
53/36 - 11/25 / (27/13 - X × 9/65) = 5/4
③ 3 × 5 × 5 × 5 + 4 × 5 × 5 + 2 × 5 + 1
= A × A × A + B × A × A + C × A + 3.
Find A, B and C, respectively.
A, B and C are the integers from 1 to 9.
Find A, B and C, respectively.
A, B and C are the integers from 1 to 9.
Answer
① 36/25
② 7/10
③ A = 7, B = 2, C = 6
(2)
Problem 2
Solution
3 × 5 × 5 × 5 + 4 × 5 × 5 + 2 × 5 + 1 = 486
A × A × A + B × A × A + C × A + 3 = A × (A × A + B × A + C) + 3 = 486
As A is divisor of 483 = 3 × 7 × 23, A = 7.
7 × 7 + B × 7 + C = 69
B × 7 + C = 20
The B = 2 and C = 6.
<Another solution>
3 × 5 × 5 × 5 + 4 × 5 × 5 + 2 × 5 + 1 means 3421 of notation system of base 5.
A × A × A + B × A × A + C × A + 3 means 1BC3 of notation system of base A.
As A is divisor of 483 = 3 × 7 × 23, A = 7.
486 is 1263 by notation system of base 7.
(2)
There are 365 cards piled up on which the date of the non-leap year is written.
It is written to the 1st card as January 1, to the 2nd card as January 2, to the 3rd card as January 3rd and to the 365th card as December 31. Answer the following questions.
(2)-1 Remove even-numbered cards counted from the top.
In this case, the date written on the top of the remaining card is January 1 and the date of the 2nd card is January 3.
Find the date of the 28th card.
(2)-2 As a next step, remove odd-numbered cards counted from the top among remaining cards.
In this case, which number of card counted from the top among remaining cards is written as September 12?
(2)-3 Suppose that January 1 is Monday.
Find the day of the week of the 69th card counted from the top among remaining cards.
It is written to the 1st card as January 1, to the 2nd card as January 2, to the 3rd card as January 3rd and to the 365th card as December 31. Answer the following questions.
(2)-1 Remove even-numbered cards counted from the top.
In this case, the date written on the top of the remaining card is January 1 and the date of the 2nd card is January 3.
Find the date of the 28th card.
(2)-2 As a next step, remove odd-numbered cards counted from the top among remaining cards.
In this case, which number of card counted from the top among remaining cards is written as September 12?
(2)-3 Suppose that January 1 is Monday.
Find the day of the week of the 69th card counted from the top among remaining cards.
Answer
(2)-1 February 24
(2)-2 64th card
(2)-3 Tuesday
(2)-2 64th card
(2)-3 Tuesday
Solution
(2)-1
28th number of odd numbers is 28 × 2 - 1 = 55th number of original cards.
28th number of odd numbers is 28 × 2 - 1 = 55th number of original cards.
Jan./55th = Feb./55 - 31 = 24th.
(2)-2
Sept. 12th is 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 12 = 255th of original cards.
Sept. 12th is 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 12 = 255th of original cards.
Remaining cards are 3, 7, 11, 15, 19, ---- .
255 = 3 + 4 × (64 - 1), then Card of Sept. 12th is 64th.
(2)-3
69th is 3 + 4 × (69 - 1) = 275th of original cards.
According to 275 / 7 = 39 remainder 2, it is Tuesday.
Problem 2
A figure is a circle with a radius of 3 cm and AB is a diameter of the circle.
All ● in a figure is 15 degrees. Pi is assumed to be 3.14.
(1) Find the sum of the area of the slash portion of P and Q.
(2) Find the area of the slash portion of Q.
Answer
Problem 3
(1) 11.565 cm2
(2) 6.96 cm2
(2) 6.96 cm2
Solution
(1)
3 × 3 × 3.14 × 1/4 + 3 × 3 / 2 = 11.565 cm2
(2)
3 × 1.5 / 2 + 3 × 3 × 3.14 × 30/360 = 2.25 + 2.355 = 4.605 cm2
11.565 - 4.605 = 6.96 cm2
Problem 3
When the cube of Fig. 1 is opened up by cutting by the bold line of Fig. 2, the development view become Fig. 3.
How should it be cut respectively, in order for the development view of each cube to become as shown in (1) and (2)?
As shown in Fig. 2, write a bold line in the sketch of the cube.
How should it be cut respectively, in order for the development view of each cube to become as shown in (1) and (2)?
As shown in Fig. 2, write a bold line in the sketch of the cube.
Answer
Problem 4
(1)
(2)
Solution
(1)
Replacement of rectangles to be original squares as below.
Then you can find cut line as below.
(2)
Replacement of triangles to be original squares as below.
Then you can find cut line as below.
Problem 4
As shown in a figure, the square card was equally divided into four and the integer was written to each portion sequentially from 1.
A lot of these cards were made.
All the sizes of the card are same.
(1) Cards were arranged according to the direction in the figure from left to right sequentially from the 1st card.
Which card is the first card for the sum of the two number of the upper rows of one card to become larger than 120?
(2) The card was piled up in order by direction in the figure from the 1st card to the 50th card.
Find the sum total of 50 numbers which has overlapped with the number 1 of the 1st card including 1 of the 1st card.
(3) Rotate 90 degrees of the 1st card counterclockwise and pile up on the 2nd card.
Next, rotate 90 degrees counterclockwise both of 1st and 2nd cards piled up and 2 cards are piled up on the 3rd card.
Thus, when you pile up 50 cards, find the sum totals of 50 numbers including 3 which has overlapped with the number 3.
Answer the following questions.
Which card is the first card for the sum of the two number of the upper rows of one card to become larger than 120?
(2) The card was piled up in order by direction in the figure from the 1st card to the 50th card.
Find the sum total of 50 numbers which has overlapped with the number 1 of the 1st card including 1 of the 1st card.
(3) Rotate 90 degrees of the 1st card counterclockwise and pile up on the 2nd card.
Next, rotate 90 degrees counterclockwise both of 1st and 2nd cards piled up and 2 cards are piled up on the 3rd card.
Thus, when you pile up 50 cards, find the sum totals of 50 numbers including 3 which has overlapped with the number 3.
Answer
Problem 5
(1) 16th card
(2) 4950
(3) 5025
(2) 4950
(3) 5025
Solution
(1)
Sum of upper row is 3, 11, 19, 27, 35, ----
Then 3 + 8 × (N - 1) = 8 × N - 5 > 120.
N > 15.6, thus N = 16.
(2)
Overlapped numbers with 1 are 1, 5, 9, 13, 17, ------.
50th number of card is 1 + 4 × (50 -1) = 197.
(1 + 197) × 50 /2 = 4950
(3)
3 → 6 → 9 → 16 can be considered as one cycle.
Sum of first one cycle is 3 + 6 + 9 + 16 = 34.
50 sheets / 4 sheets = 12 cycles remainder 2 sheets
First number of first cycle is 3 and first number of second cycle is 19 which is 16 larger than 3.
Sum of numbers in 2nd cycle is 34 + 16 × 4 = 34 + 64 = 98.
Sum of numbers of 12th cycle can be calculated as 34 + 64 × (12 -1) = 738.
First number of 13th cycle is 3 + 16 × (13-1) = 195 and 2nd is 195 + 3 = 198.
Sum total of 1st ~ 50th cards are (34 + 738) × 12 /2 + 195 + 198 = 4632 + 195 + 198 = 5025.
Problem 5
There is a rectangle ABCD with 20 cm in length and 10 cm in width as shown in a figure.
The point P leaves the point A and moves in order of A→B→C→D→A→-------- on the side of this rectangle.
The point Q leaves the point B simultaneously with the point P and moves in order of B→C→D→A→ ----.
The point P moves at the speed of 1 cm/s and the point Q moves 2 cm/s in the first round.
Then, the point P increases the speed per 1 second for every round as 2 cm/s in 2nd round, 3 cm/s in 3rd round.
The point Q also increases the speed per 1 second for every round as 3 cm/s in 2nd round, 4 cm/s in 3rd round.
Answer the following questions.
(1) Find the time after leaving when the point Q overlaps with the point P for the first time.
(2) Regarding from 1st round to 3rd round, write the time when two point P and Q take moving one round of rectangle ABCD, respectively in a lower table.
Let the measure be second.
(3) 100 seconds after leaving, where are two point P and Q?
Write the position of point P in Fig. 1 and write the position of the point Q in Fig. 2. according to the (Example) below.
(Example) When the point P is at the position of 7 cm from A on the side AD in 100 seconds.
(4) In 100 seconds after leaving, find the last time of seconds when the point Q overlaps with the point P.
Answer
(1) 40 seconds after
(2)
(3)
(4) 3392/35 seconds
Solution
(1)
Q overlaps with P in 2nd round.
60 / 2 = 30 seconds.
When Q is 1st lap, P is at C.
20 / (3 - 1) = 10 seconds.
30 + 10 = 40 seconds.
(2)
As for P, 60 / 1 = 60 seconds, 60 / 2 = 30 seconds and 60 / 3 = 20 seconds.
As for Q, 60 / 2 = 30 seconds, 60 /3 = 20 seconds and 60 / 4 = 15 seconds.
(3)
100 seconds after for P means that 100 - (60 + 30) = 10 seconds in the 3rd round of P.
Since 3 cm/s × 10 sec. = 30 cm which shows that P is at C.
100 seconds after for Q means that 100 - ( 30 + 20 + 15 + 10 + 60/7) = 31/7 seconds in the 7th round of Q.
The speed of Q in the 7th round is 8 cm/s.
8 × 31/7 = 248/7 cm
248/7－(20＋10) = 38/7 cm from D.
(4)
At 100 seconds, the distance between P and Q is 10 + 38/7 = 108/7 cm.
The speed of P is 3 cm/m and P is 8 cm/s.
108/7 / (8 - 3) = 108/35 seconds before 100 seconds.
Thus 100 - 108/35 = 3392/35 seconds.
