Time : 50 minutes
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Answer : End of the Problem
Problem 1
Calculation
(1) 17/10 - 5/6 + 4/5
(2) 3.5 ÷ (2.2 - 0.8) + 0.6 × 4.5
(3) (0.2 - 1/8 × 16/15) ÷ (7/6 - 0.75)
Problem 2
(1) Length of both side of a square is made 1 cm longer each and a new square is made.
The area of new square is 19 cm2 bigger than the original square.
Find the original square area.
(2) Three people of Taro, Jiro, Hanako took a test of Math.
The total score of three people was 228 points.
Taro's score was 18 points higher than Jiro's score.
The average score of Taro and Jiro was the same as a score of Hanako.
Find Taro's score.
(3) The price of pork 150g and beef 300g is 1500 yen.
The price of pork 250g and beef 150g is 1240 yen.
Find the price of pork 100g and beef 100g respectively.
Problem 3
There are P point and Q point 720m away.
Taro and Jiro left P at the same time and made a round trip between P and Q.
Jiro reached P three minutes later after Taro reached P.
The ratio of speed of Taro and Jiro is 3 : 2.
(1) Find the speed of Taro in meter per minute.
(2) Find the time of minutes and seconds when two people passed each other after they left P.
Problem 4
Some pieces of apple and pear were sold respectively at a fruit parlor.
On Day 1, 50% of the number of the whole apple, 40% of the number of the whole pear sold.
On Day 2, remaining apples and pears were sold.
All pears were sold.
As for the apple, only the same number was sold as pear sold on Day 1 and 42 remained.
As for the total number of apple and pear sold on Day 2, 18 decreased from the total number of two fruits sold on Day 1.
Find the each number of apple and pear in this parlor first.
Problem 5
There is a trapezoid ABCD whose side AD and BC are parallel as showing in the figure below.
E is a point on the side BC. AB and DE are parallel.
This trapezoid is divided into five triangle P, Q, R, S, and T with AC, DE, and BF.
The area of P is 32 cm2.
The area of Q is larger than the area of S by 8 cm2.
(1) Answer triangles whose area is equal among P, Q, R, S, and T.
(2) When the height of trapezoid ABCD is 16cm and an area is 280cm2, find the length of the EC.
Problem 6
There are some number of two kinds of trapezoidal-shaped tile A and B shown as below respectively.
I will arrange A and B by connecting sides of A and B with 1cm in length and make figures such as Fig.1 and Fig.2.
For instance, as for Fig.1, it is a figure where three A tiles were arranged and Fig.2 is a figure where two A and two B were arranged.
Answer the following questions.
(1) I make a figure by arranging two A and six B.
Find the area ratio between the figure surrounded by inner sides of this figure and the figure surrounded by inner sides of Fig.2.
(2) I make a figure by arranging four A and two B.
Find the length of surroundings outside of the figure.
Problem 7
There is a bar of length 180cm.
I mark points on the bar where it is divided M equally and N equally respectively.
M and N are integers and M is bigger than 2 and is smaller than N.
For example when M is 4 and N is 6, it is marked respectively.
As the mark at one point is overlapped as shown in the lower figure, number of point becomes seven.
Answer the following questions.
(1) In case M is 5 and N is 6, find the number of point.
Find the shortest length of interval between two points.
(2) In case M is 8 and N is 12, find the number of point.
Find the shortest length of interval between two points.
(3) There are two cases of combination of M and N which make the number of point eight.
Answer these combinations of M and N.
(4) Answer all combinations of M and N that the shortest length becomes 12cm among the length of the interval between points.
Problem 1
Calculation
(1) 17/10 - 5/6 + 4/5
(2) 3.5 ÷ (2.2 - 0.8) + 0.6 × 4.5
(3) (0.2 - 1/8 × 16/15) ÷ (7/6 - 0.75)
Answer
(1) 5/3
(2) 5.2
(3) 4/25
Problem 2
(1) Length of both side of a square is made 1 cm longer each and a new square is made.
The area of new square is 19 cm2 bigger than the original square.
Find the original square area.
Answer
81 cm2
Solution
1 × 1 = 1 cm2
19 - 1 = 18 cm2
18 / 2 = 9 cm2
X = 9 / 1 = 9 cm
Therefore 9 × 9 = 81 cm2.
(2) Three people of Taro, Jiro, Hanako took a test of Math.
The total score of three people was 228 points.
Taro's score was 18 points higher than Jiro's score.
The average score of Taro and Jiro was the same as a score of Hanako.
Find Taro's score.
Answer
85 points
Solution
Accordning to Taro + Jiro + Hanako = 228 and Taro + Jiro = Hanako × 2, Hanako × 3 = 228.
Thus Hanako = 228 / 3 = 76.
Taro + Jiro = 76 × 2 = 152
Taro - Jiro = 18
Therefore Taro = (152 + 18) / 2 = 85 points.
(3) The price of pork 150g and beef 300g is 1500 yen.
The price of pork 250g and beef 150g is 1240 yen.
Find the price of pork 100g and beef 100g respectively.
Answer
Pork : 280 yen, Beef : 360 yen
Solution
According to pork250g + beef150g = 1240 yen, pork500g + beef 300g = 2480 yen.
As pork 150g + beef 300g = 1500g, pork350g = 2480 - 1500 = 980 yen.
Thus pork 100g = 980 / 350 × 100 = 280 yen.
Beef 300g = 2480 - 280 × 5 = 1080.
Therefore beef 100g = 1080 / 3 = 360g.
Problem 3
There are P point and Q point 720m away.
Taro and Jiro left P at the same time and made a round trip between P and Q.
Jiro reached P three minutes later after Taro reached P.
The ratio of speed of Taro and Jiro is 3 : 2.
(1) Find the speed of Taro in meter per minute.
(2) Find the time of minutes and seconds when two people passed each other after they left P.
Answer
(1) 240 m/m
(2) 3 minutes 36 seconds
Solution
(1) The ratio of distance is 3 : 2 which is the same as the speed ratio of Taro and Jiro.
Jiro moved the distance of 3 -2 = 1 in 3 minutes.
Jiro took 3 minutes × 3 = 9 minutes to make a round trip between P and Q.
Taro took 9 - 3 = 6 minutes.
Thus the spped of Taro is 720 × 2 / 6 = 240 m/m.
(2) The speed of Jiro is 240 × 2/3 = 160 m/m.
720 × 2 / (240 + 160) = 3.6 minutes = 3 minutes 36 seconds.
Problem 4
Some pieces of apple and pear were sold respectively at a fruit parlor.
On Day 1, 50% of the number of the whole apple, 40% of the number of the whole pear sold.
On Day 2, remaining apples and pears were sold.
All pears were sold.
As for the apple, only the same number was sold as pear sold on Day 1 and 42 remained.
As for the total number of apple and pear sold on Day 2, 18 decreased from the total number of two fruits sold on Day 1.
Find the each number of apple and pear in this parlor first.
Answer
Apple : 180 pieces, Pear : 120 pieces
Solution
Each number of apple and pear first in the parlor is set to A and P respectively.
The number sold in Day1 is A × 0.5 apples and P × 0.4 pears.
The number sold of pear in Day2 is P × 0.6.
The number sold of apple in Day 2 is P × 0.4.
As 42 apples remained, A × 0,5 - P × 0.4 = 42.
Furthermore A × 0.5 - P × 0.6 = 18.
Thus P × 0.2 = 42 - 18 = 24.
P = 24 / 0.2 = 120.
P × 0.4 = 120 × 0.4 = 48
A × 0.5 = 48 + 42 = 90
Therefore A = 90 / 0.5= 180.
Problem 5
There is a trapezoid ABCD whose side AD and BC are parallel as showing in the figure below.
E is a point on the side BC. AB and DE are parallel.
This trapezoid is divided into five triangle P, Q, R, S, and T with AC, DE, and BF.
The area of P is 32 cm2.
The area of Q is larger than the area of S by 8 cm2.
(1) Answer triangles whose area is equal among P, Q, R, S, and T.
(2) When the height of trapezoid ABCD is 16cm and an area is 280cm2, find the length of the EC.
Answer
(1) R, T
(2) 15 cm
Solution
(1) As quadrangle ADEB is parallelogram, P + R = Q. As triangle ACD = ABD, P + T = Q.
Thus R = T.
(2) According to 280 × 2 / 16 = 35, AD + BC = 35 cm.
P = 32 (cm2),
As P +R = Q, 32 + R = Q.
R = Q - 32.
As Q = S + 8 and Q = S - 8, R = S - 24 and T = R = S - 24.
Thus, 32 + S + 8 + S - 24 + S -24 = 280.
4S = 288
S = 288/4 = 72
R = 72 - 24 = 48
BE : CE = 48 : 72 = 2 : 3
AD : BE : EC = 2 : 2 : 3
Thus EC = 35 × 3/ (2+2+3) = 15 cm.
Problem 6
There are some number of two kinds of trapezoidal-shaped tile A and B shown as below respectively.
I will arrange A and B by connecting sides of A and B with 1cm in length and make figures such as Fig.1 and Fig.2.
For instance, as for Fig.1, it is a figure where three A tiles were arranged and Fig.2 is a figure where two A and two B were arranged.
Answer the following questions.
(1) I make a figure by arranging two A and six B.
Find the area ratio between the figure surrounded by inner sides of this figure and the figure surrounded by inner sides of Fig.2.
(2) I make a figure by arranging four A and two B.
Find the length of surroundings outside of the figure.
Answer
(1) 7：2
(2) 16 cm
Solution
(1) The figure is as shown in Fig.4 below.
Inside of this figure is rectangle and inside of Fig.2 is also rectangle.
The vertical length of both rectangles is same.
The horizontal length of the rectangle inside of Fig.4 is 2 + 3 + 2 = 7(cm).
The horizontal length of the rectangle inside of Fig.2 is 2cm.
Thus area ratio is also 7 : 2.
(2) The figure is as shown in Fig.5 below.
As you can see, the length of surroundings outside of the figure is 3 × 4 + 2 × 2 = 12 + 4 = 16 cm.
Problem 7
There is a bar of length 180cm.
I mark points on the bar where it is divided M equally and N equally respectively.
M and N are integers and M is bigger than 2 and is smaller than N.
For example when M is 4 and N is 6, it is marked respectively.
As the mark at one point is overlapped as shown in the lower figure, number of point becomes seven.
Answer the following questions.
(1) In case M is 5 and N is 6, find the number of point.
Find the shortest length of interval between two points.
(2) In case M is 8 and N is 12, find the number of point.
Find the shortest length of interval between two points.
(3) There are two cases of combination of M and N which make the number of point eight.
Answer these combinations of M and N.
(4) Answer all combinations of M and N that the shortest length becomes 12cm among the length of the interval between points.
Answer
(1) 9 pieces, 6 cm
(2) 15 pieces, 7.5 cm
(3) (3,7),(3,9)
(4) (3,5),(3,15),(5,15)
Solution
(1) 180 / 5 = 36 (cm) 0 36 72 108 144 180
180 / 6 = 30 (cm) 0 30 60 90 120 150 180
Number of mark is nine.
The shortest is 36 - 30 = 6 cm.
(2) 180 / 8 = 22.5 (cm) 0 22.5 45 67.5 90 112.5 135 157.5 180
180 / 12 =15(cm) 0 15 30 45 60 75 90 105 120 135 150 165 180
Number of mark is 15.
The shortest is 22.5 - 15 = 7.5 cm.
(3)
In case M is 3 and N is 7, there is no mark overlapped.
Thus number of mark is 8.
In case M is 3 and N is 9, there are two overlapping points.
Thus number of mark is 2 + 8 -2 = 8.
(4) According to the table below, (M,N) = (3,5),(3,15),(5,15).
In case of (3,5), 36 × 2 -60 = 12.
In case of (3,15), 60- 12 × 4 = 12.
In case of (5,15), 36 - 12 × 2 = 12.
