Time : 50 minutes
Problem 1
Problem 2
(2) I mix water and 32 g of salt to make 16% of salt solutions.
Problem 2
(2) I mix water and 32 g of salt to make 16% of salt solutions.
Problem 1
(1) Calculation
15 × 6 - 35 / (3 + 2) × 3 =
(2) Find P.
(P + 3/8) / 0.75 - 1/3 = 1/2
(3) Calculation
1.62 × 10/3 - 5/13 × 1.56 =
(4) Find X and Y. X should be integer.
36.65 / 3.012 = X remainder Y
Problem 2
(1) In integers from 1 to 100, find sum total of such numbers as when it is divided by 3, remainder is 2?
(2) I mix water and 32 g of salt to make 16% of salt solutions.
Find the weight of water.
(3) The average of four people of Taro, Jiro, Hanako and Fujiko was 59 points on a certain test.
(4) Hanako has finished reading one book in four days.
(5) There are five sheets of cards of 0, 1, 1, 2, 3.
(6) What cm2 is the shadow area of the figure below?
Problem 3
Problem 4
Problem 5
(3) The average of four people of Taro, Jiro, Hanako and Fujiko was 59 points on a certain test.
In addition, the average of two people of Jiro and Fujiko was 58 points.
The average of three people of Taro, Jiro and Hanako was 60 points.
Find a score of Jiro.
(4) Hanako has finished reading one book in four days.
She read 92 pages on the first day and read 1/3 of the book on the second day.
In addition, she read 1/7 of the book on the 3rd day and read 1/4 of the book on the 4th day.
Find the total pages of this book.
(5) There are five sheets of cards of 0, 1, 1, 2, 3.
I take out two sheets out of five and make two digits of integers.
How many different integers are there in all?
(6) What cm2 is the shadow area of the figure below?
2 people of Taro and Jiro did footrace.
When both of 2 people ran through 100m, Jiro did a goal 4 seconds later than Taro.
Next when Taro started from the point 25m after Jiro, 2 people made the goal at the same time.
Noted that 2 people start at the same time and run with a fixed speed respectively.
(1) Express the speed ratio of Taro and Jiro in the ratio of the least integer.
(2) How many m per second was the speed of Taro?
Problem 4
Rotate the figure below around the straight line A as a rotation axis and make a solid.
Pi is assumed to be 3.14.
(1) Find the area of the cut surface when you cut this solid by the plane including the axis.
(2) Find the volume of this solid.
Problem 5
According to a regulation like a figure below, coins are arranged as the 1st Fig., 2nd Fig., 3rd Fig. ------ .
Answer the following questions.
(1) Find the number of the coin arranged in the 15th Fig.
(2) How many times are there that the total number of the coin becomes a multiple of 90 from the 1st Fig. to the 50th Fig.?
Answer
Problem 1
(1) Calculation
15 × 6 - 35 / (3 + 2) × 3 =
(2) Find P.
(P + 3/8) / 0.75 - 1/3 = 1/2
(3) Calculation
1.62 × 10/3 - 5/13 × 1.56 =
(4) Find X and Y. X should be integer.
36.65 / 3.012 = X remainder Y
Answer
(1) 69
(2) 1/4
(3) 24/5
(4) X = 12, Y = 0.506
Problem 2
(1) In integers from 1 to 100, find sum total of such numbers as when it is divided by 3, remainder is 2?
Answer
1650
Solution
2, 5, 8, 11, ------ are numbers when they are divided by 3, remainder is 2.
The last number is 98.
According to 2 + 3 × 32 = 98, there are 33 numbers until 100.
Therefor the sum total of those numbers is (2 + 98) × 33 / 2 = 1650.
(2) I mix water and 32 g of salt to make 16% of salt solutions.
Find the weight of water.
Answer
168 g
Solution
Because 16% of salt solution is salt of 32 g, the weight of the salt solution is 32 / 0.16 = 200 g.
Therefore water weighs 200 - 32 = 168 g.
(3) The average of four people of Taro, Jiro, Hanako and Fujiko was 59 points on a certain test.
(4) Hanako has finished reading one book in four days.
Answer
(3) The average of four people of Taro, Jiro, Hanako and Fujiko was 59 points on a certain test.
In addition, the average of two people of Jiro and Fujiko was 58 points.
The average of three people of Taro, Jiro and Hanako was 60 points.
Find a score of Jiro.
Answer
60 points
Solution
Total score of Taro, Jiro, Hanako and Fujiko was 59 × 4 = 236 points.
Total score of Taro, Jiro and Hanako was 60 × 3 = 180 points.
Thus the score of Fujiko was 236 - 180 = 56 points.
Since total score of Jiro and Fujiko was 58 × 2 = 116 points, Jiro's score was 116 - 56 = 60 points.
(4) Hanako has finished reading one book in four days.
She read 92 pages on the first day and read 1/3 of the book on the second day.
In addition, she read 1/7 of the book on the 3rd day and read 1/4 of the book on the 4th day.
Find the total pages of this book.
336 pages
Solution
Hanako read 1/3 + 1/4 + 1/7 = 61/84 of this book from the second day to the fourth day.
Remaining 1 - 61/84 = 23/84 is equivalent to 92 pages.
Therefore, the total pages of this book are 92 / 23/84 = 336 pages.
(5) There are five sheets of cards of 0, 1, 1, 2, 3.
(6) What cm2 is the shadow area of the figure below?
Problem 3
Problem 4
Problem 5
(5) There are five sheets of cards of 0, 1, 1, 2, 3.
I take out two sheets out of five and make two digits of integers.
How many different integers are there in all?
Answer
10
Solution
When I use two pieces of 1, the integer to be able to make is 11.
Consider about cases to use 1 one time or zero.
As for tens digit, there are three pieces of numbers of 1, 2 or 3.
As for ones digit, there are three pieces of numbers which are not used in tens digit and 0.
Thus, there are 3 × 3 = 9 ways.
Accordingly there are 1 + 9 = 10 in all.
(6) What cm2 is the shadow area of the figure below?
Answer
24 cm2
Solution
(Shadow area) = (Triangle DBC) - (Triangle HBE) - (Triangle DGF).
△ AHD and△ HBE are homothetic and the homothetic ratio is 12 : 6 = 2 : 1. Since the base of △ HBE = 6 cm and the height = 6 × 1/(1 + 2) = 2 cm, the area is 6 × 2/2 = 6 cm2.
△ ABG and△ DGF are homothetic and the homothetic ratio is 6 : 3 = 2 : 1. Since the base of △DGF = 3 cm and the height= 12 × 1/(1 + 2) = 4 cm, the area is 3 × 4 / 2 = 6 cm2.
Therefore the shadow area = 12 × 6 / 2 - 6 - 6 = 36 - 12 = 24 cm2.
2 people of Taro and Jiro did footrace.
When both of 2 people ran through 100m, Jiro did a goal 4 seconds later than Taro.
Next when Taro started from the point 25m after Jiro, 2 people made the goal at the same time.
Noted that 2 people start at the same time and run with a fixed speed respectively.
(1) Express the speed ratio of Taro and Jiro in the ratio of the least integer.
(2) How many m per second was the speed of Taro?
Answer
(1) 5 : 4
(2) 6.25 m/s
Solution
(1) Because Taro runs 125 m and Jiro runs 100 m in the same time, the speed ratio of Taro : Jiro = 125 : 100 = 5 : 4.
(2) Because the time ratio which two persons run 100 m is to be inverse ratio of the speed ratio, it'll be Taro : Jiro = 4 : 5. 5 - 4 = 1 which is the difference of the ratio is equivalent to 4 seconds.
Time when it takes for Taro to run 100 m is 4 × 4 = 16 seconds.
Therefore speed of Taro is 100/16 = 6.25 m/s.
Problem 4
Rotate the figure below around the straight line A as a rotation axis and make a solid.
Pi is assumed to be 3.14.
(1) Find the area of the cut surface when you cut this solid by the plane including the axis.
(2) Find the volume of this solid.
Answer
(1) 93 cm2
(2) 499.26 cm3
Solution
(1) The area of the cut surface is twice as much as the area of the figure below.
P + Q + R = 2 × 2 +(2+5) ×5 / 2 + 5 ×5 = 4 + 35/2 + 25 = 93/2cm2.
Therefore, an area to find is 93/2 × 2 = 93cm2.
(2) The volume of the solid by the rotation of P is 2 × 2 × 3.14 × 2 = 8 × 3.14.
The volume of the solid by the rotation of R is 5 × 5 × 3.14 × 5 = 125 × 3.14.
△ BCD and △ BEF are homothetic and the homothetic ratio is 5 : 2.
Because BC : BE = 5 : 2, BE = 2cm × 2/(5-2) = 4/3 cm.
The volume of the solid by the rotation of Q is 5 × 5 × 3.14 × (4/3+2) × 1/3 - 2 × 2 × 3.14 × 4/3 × 1/3 = 250/9 × 3.14 - 16/9 × 3.14 = 26 × 3.14.
Therefore the volume to find is 8 × 3.14 + 125 × 3.14 + 26 × 3.14 = 159 × 3.14 = 499.26 cm3.
Problem 5
According to a regulation like a figure below, coins are arranged as the 1st Fig., 2nd Fig., 3rd Fig. ------ .
Answer the following questions.
(1) Find the number of the coin arranged in the 15th Fig.
(2) How many times are there that the total number of the coin becomes a multiple of 90 from the 1st Fig. to the 50th Fig.?
Answer
(1) 1800 pieces
(2) 3 times
Solution
(1) 8 times of the number of coins in a red square of the figure below is the total number of coins arranged in one Fig.
The number of coins in the red square is increased as 1 × 1, 2 × 2, 3 × 3 ----.
Therefore the number of the coin of the 15th figure is 15 × 15 × 8 = 1800.
(2) 90 = 2 × 3 × 3 ×5 and 8 is a multiple of 2.
The number of coins in the square of Ath Fig. is A × A.
It's necessary that A × A becomes a multiple of 45 for the number of coins in the Fig. to become a multiple of 90.
Because it's 45 = 3 × 3 × 5, A has 3 and 5 as divisors and is a multiple of 15.
Therefore it's three times of the 15th Fig., the 30th and the 45th that the total number of coins in Fig. becomes a multiple of 90.
