Time : 50 minutes
Passing mark : 70%
Answer : End o the problem
Problem 1
Problem 2
Problem 3
Problem 4
(1) Find <4,1>
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
{(4.5 - 7/3) × 14/13 - 1.6} / 11/9 =
Problem 2
Find X.
0.5 / 2/5 - (5/4 - X) × 5/24 = 9/8
Problem 3
The fraction lines up according to a certain rule as follows.
Answer the following questions.
(1) Find the 20th fraction counting from the first.
(2) Find the number of fractions that are larger than 1/4.
(3) Which number is the fraction counting from the first that is nearest to 2/9?
(2) Find the number of fractions that are larger than 1/4.
(3) Which number is the fraction counting from the first that is nearest to 2/9?
Problem 4
The sign <a,b> shows the number of ways to arrange when a piece of ○ and b piece of × are arranged in a row.
For instance, <2,1> shows the number of ways to arrange when 2 ○ and 1 × are arranged in a row and it becomes <2,1> = 3 because there are three ways of arrangement of ○○×, ○×○, and ×○○.
Answer the following questions.
Answer the following questions.
(1) Find <4,1>
(2) Find X. <X,1> = 9
(3) Find <2,2>
(3) Find <2,2>
Problem 5
I decided to sell 100 pieces of certain product in three shops of A, B, and C.
50 pieces were sold by the list price at A shop.
25 pieces were sold by 20% discounted of the list price at B shop.
25 pieces were sold by 35% discounted of the list price at C shop.
The total amount of sales of three shops has decreased by 3850 yen comparing to when all of the 100 pieces are sold at the list price.
Find the list price of this product.
Moreover, find the amount of sales in the shop of C.
Problem 6
Hanako took the tests of mathematics(sansue), Japanese, science, and social studies.
As a result, the score of sansue was 83 and the score of social studies was 6 points lower than that of Japanese and the scores of Japanese and science was same.
The average score of sansue, science, and the social studies was 4 points lower than the score of Japanese.
Find the score of Japanese.
Moreover, find the average score of four subjects.
Problem 7
There is a sector OAB whose central angle is 90° and quadrangle OCDE is a rectangle in the figure below.
CE is a diagonal of the rectangle and the sum total of areas of triangle DEF and triangle OCF is 12cm2 and the length of DF is 3cm.
Find the length of CE.
Moreover, find the area for the shaded portion.
Pi is assumed to be 3.14.
Problem 8
Taro ordered total 100 pieces of marbles and super balls together to a certain shop.
Price of a piece of marble is five yen and that of super ball is 25 yen.
However, the shop staff heard wrong by mistake the number of marbles and super ball to be reversed.
Thereby, the ratio of the amount of money Taro ordered and the staff calculated by mistake became 7 : 18.
Find the amount of money Taro ordered.
Moreover, find the number of super ball ordered.
Problem 9
As shown in Fig.1 below, there is a water tank whose shape is a rectangular prism with 200cm2 of bottom area and 12 cm in height and are faucet A that pours the water into the tank and faucet B that drains the water.
This water tank is divided into two parts of P and Q partitioned by a rectangle plate stood vertically to the bottom.
It began to pour water from A into the part of Q with a fixed rate.
When B was opened on the way and it drained the water with a fixed rate, the water tank became full in 18 minutes from the beginning.
The graph of Fig. 2 is showing of the relation of the time after it begins to pour water and the highest surface of the water.
Noted that the thickness of the partition is not considered.
Answer the following questions.
(1) Find the height of the partition.
(2) Find the area of the bottom of P.
(3) Find the number that applies to X of Fig. 2.
(4) Find the volume of the water drained.
<Answer>
Problem 1
Problem 2
Problem 3
Problem 4
(1) Find <4,1>
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
{(4.5 - 7/3) × 14/13 - 1.6} / 11/9 =
Answer
3/5
Problem 2
Find X.
0.5 / 2/5 - (5/4 - X) × 5/24 = 9/8
Answer
13/20
Problem 3
The fraction lines up according to a certain rule as follows.
Answer the following questions.
(1) Find the 20th fraction counting from the first.
(2) Find the number of fractions that are larger than 1/4.
(3) Which number is the fraction counting from the first that is nearest to 2/9?
(2) Find the number of fractions that are larger than 1/4.
(3) Which number is the fraction counting from the first that is nearest to 2/9?
Answer
(1) 4/29
(2) 4 pieces
(3) 5th
Solution
(1)
As the numerator of the fraction is a cycle of 3 and 4, 20th numerator is 4.
20th denominator is calculated as 10 + (20 - 1) = 29.
(2)
After 4/13, the fraction is smaller than 0.25 = 1/4.
Therefore there are 4 fractions larger than 1/4.
(3)
9/2=4.5
10/3=3.33----
11/4=2.75
12/3=4
13/4=3.25
14/3=4.66---
15/4=3.75
16/3=5.33---
17/4=4.25
18/3=6
19/4=4.75
Therefore the fraction nearest to 2/9 is 3/14 which is 5th fraction.
Problem 4
The sign <a,b> shows the number of ways to arrange when a piece of ○ and b piece of × are arranged in a row.
For instance, <2,1> shows the number of ways to arrange when 2 ○ and 1 × are arranged in a row and it becomes <2,1> = 3 because there are three ways of arrangement of ○○×, ○×○, and ×○○.
Answer the following questions.
Answer the following questions.
(1) Find <4,1>
(2) Find X. <X,1> = 9
(3) Find <2,2>
(3) Find <2,2>
Answer
(1) 5
(2) 8
(3) 6
Solution
(1)
5 of ○○○○×，○○○×○，○○×○○，○×○○○，×○○○○.
(2)
According to the result of (1), the answer of <X,1> is to be found by counting the number of place of × can be put in X+1 places.
Thereby it is X + 1 = 9 and X = 9 - 1 = 8.
(3)
The answer is the number of cases to find the number which two × can be placed in 2+2=4 places.
(4 × 3) / (2 × 1) = 6
Problem 5
I decided to sell 100 pieces of certain product in three shops of A, B, and C.
50 pieces were sold by the list price at A shop.
25 pieces were sold by 20% discounted of the list price at B shop.
25 pieces were sold by 35% discounted of the list price at C shop.
The total amount of sales of three shops has decreased by 3850 yen comparing to when all of the 100 pieces are sold at the list price.
Find the list price of this product.
Moreover, find the amount of sales in the shop of C.
Answer
280 yen,
4550 yen
Solution
List price is set to 1.
The sales amount in A shop is 1 × 50 = 50.
The sales amount in B shop is 0.8 × 25 = 20.
The sales amount in C shop is 0.65 × 25 = 16.25.
Total amount is 50 + 20 + 16.25 = 86.25.
1 × 100 - 86.25 = 13.75
13.75 is equivalent to 3850 yen.
Then list price 1 is 3850 / 13.75 = 280 yen.
The sales amount in C shop is 16.25 = 280 × 16.25 = 4550 yen.
Problem 6
Hanako took the tests of mathematics(sansue), Japanese, science, and social studies.
As a result, the score of sansue was 83 and the score of social studies was 6 points lower than that of Japanese and the scores of Japanese and science was same.
The average score of sansue, science, and the social studies was 4 points lower than the score of Japanese.
Find the score of Japanese.
Moreover, find the average score of four subjects.
Answer
Answer
89 point,
86 point
Solution
The score of Japanese is set to □.
The score of social studies = □ - 6 and science = □.
The average score of sansue, science, and the social studies was 4 points lower than the score of Japanese.
Then total score of sansue, science, and the social studies was 4 × 3 = 12 points lower than □ × 3.
83 + □ + □ - 6 + 12 = □ × 3
Thus □ = 89 points.
The average score of four subjects is (89 + 85 × 3) / 4 = 86 points.
Problem 7
There is a sector OAB whose central angle is 90° and quadrangle OCDE is a rectangle in the figure below.
CE is a diagonal of the rectangle and the sum total of areas of triangle DEF and triangle OCF is 12cm2 and the length of DF is 3cm.
Find the length of CE.
Moreover, find the area for the shaded portion.
Pi is assumed to be 3.14.
Answer
8 cm,
26.24 cm2
Solution
△DEF + △OCF = 12 cm2 which is half of the area of rectangle OCDE.
Thus the aea of OCDE = 12 × 2 = 24 cm2.
△CDE = CE × DF / 2 = 24 / 2 = 12 cm2.
As CE × 3 = 24, CE = 24 / 3 = 8 cm.
Radius OA = OD = CE = 8cm.
The area of the shaded portion = 8 × 8 × 3.14 × 90/360 - 24 = 26.24 cm2.
Problem 8
Taro ordered total 100 pieces of marbles and super balls together to a certain shop.
Price of a piece of marble is five yen and that of super ball is 25 yen.
However, the shop staff heard wrong by mistake the number of marbles and super ball to be reversed.
Thereby, the ratio of the amount of money Taro ordered and the staff calculated by mistake became 7 : 18.
Find the amount of money Taro ordered.
Moreover, find the number of super ball ordered.
Answer
840 yen,
17 pieces
Solution
Number of marble Taro ordered is set to X pieces and number of super ball is set to Y pieces.
Total money Taro ordered can be expressed as 5 × X + 25 × Y = 7.
Total money the staff calculated can be expressed as 5 × Y + 25 × X = 18.
Sum total is 5 × X + 25 × Y + 5 × X + 25 × Y = 30 × X + 30 × Y = 7 + 18 = 25.
As X + Y = 100, 25 = 30 × 100 = 3000 yen.
7 = 3000 / 25 × 7 = 840 yen.
Y = (840 - 5 × 100) / (25 - 5) = 17 pieces.
Problem 9
As shown in Fig.1 below, there is a water tank whose shape is a rectangular prism with 200cm2 of bottom area and 12 cm in height and are faucet A that pours the water into the tank and faucet B that drains the water.
This water tank is divided into two parts of P and Q partitioned by a rectangle plate stood vertically to the bottom.
It began to pour water from A into the part of Q with a fixed rate.
When B was opened on the way and it drained the water with a fixed rate, the water tank became full in 18 minutes from the beginning.
The graph of Fig. 2 is showing of the relation of the time after it begins to pour water and the highest surface of the water.
Noted that the thickness of the partition is not considered.
Answer the following questions.
(1) Find the height of the partition.
(2) Find the area of the bottom of P.
(3) Find the number that applies to X of Fig. 2.
(4) Find the volume of the water drained.
Answer
(1) 4 cm
(2) 40 cm2
(3) 10
(4) 480 cm3
Solution
(1)
According to the graph, the height of the partition is 4 cm.
(2)
It took 4 minutes in Q and 5 - 4 = 1 minute in P.
Thus the area ratio of P : Q = 1 : 4.
The area of the bottom of P = 200 × 1/(1+4) = 40 cm2.
(3)
As it took 5 minutes to the height of 4cm, it take double to the height of 8cm.
Thus X is 5 × 2 = 10 minutes.
(4)
A is pouring water 200 × 8 / 10 = 160 cm3 per minute.
In 18 minutes water was poured 160 × 18 = 2880 cm3.
The volume of the water tank is 200 × 12 = 2400cm3.
Thus the volume of the water drained is 2880 - 2400 = 480 cm3.
