Answer: Solution: There are X first-class seats and Y second-class seats.
Then it is 300x+200y=5025.
At this time, neither x nor y is a positive integer and should be discarded.
There are x first-class seats and y third-class seats.
Then there are 300x+ 125y=5025, 12x+5y=20 1.
Then x=3 and y=33.
There are x second-class seats and y third-class seats.
Then there are 200x+ 125y=5025 and 8x+5y=20 1.
Then x=7 and y=29.
A: There are two schemes: 3 first-class seats and 33 third-class seats; There are 7 second-class seats and 29 third-class seats.
2. It is known that the linear function y=-x+2 intersects the X axis and the Y axis at point A and point B respectively, and O is the origin of the coordinate system, and the area of △AOB is found.
Answer: solution: when x=0 and Y = 2; When y=0, x=2.
∴AO=2,BO=2
∴S△AOB= 12×2×2=2
A: The area of delta delta △AOB is 2.
3. In the construction of new socialist countryside, a township decided to rebuild a section of highway. It is understood that this project will take 40 days and be completed by the A engineering team alone. If Team B works alone for 65,438+00 days, the remaining projects need two teams to work together for 20 days.
(1) Find the number of days required for Team B to complete the project alone;
(2) Find the number of days required for two teams to complete the project together.
Answer: Solution: (1) It will take X days to form a B engineering team to complete this project alone.
According to the meaning of the question:10x+(1x+140) × 20 =1,
Solution: x=60,
It is verified that x=60 is the solution of the original equation.
Answer: It will take 60 days for Team B to complete this project alone.
Solution: (2) It takes two teams y days to complete this project.
According to the meaning of the question: (140+160) y =1,
Solution: y=24,
Answer: It will take 24 days for two teams to complete this project together.
4. if A(x 1, y 1), B(x2, y2) and C(x3, y3) are all points on the image, with the inverse scaling function y=- 1x and x 1 < 0 < x2 < x3.
Answer: solution: ∵ k =- 1 < 0. The image is located in the second and fourth quadrants respectively. In each quadrant, y increases with the increase of x, and the points on B(x2, y2) and C(x3, y3) are 0 < x2 < x3.
∴B(x2, y2), C(x3, y3) in the fourth quadrant, y2 < y3,
∵ x 1 < 0, ∴A(x 1, y 1) is in the second quadrant, y 1 > 0,
∴y 1, y2 and y3 are arranged in descending order as y2 < y3 < y 1.
Yiwu is a city on wheels. By the end of 2007, the number of cars in the city was 1 14508. It is understood that by the end of 2005, the number of cars in this city was 72,983. Please answer the following questions:
(1) The average annual growth rate of car ownership in our city from the end of 2005 to the end of 2007? (The result is accurate to 0. 1%)
(2) In order to protect the urban environment, it is required that the number of cars in our city should not exceed 158000 by the end of 2009. It is estimated that the number of scrapped cars from the end of 2007 is 4% of the car ownership at the end of last year, so how many cars will be added at most every year? (Assuming that the number of new cars is the same every year, the result is accurate to single digits)
Answer: Solution: (1) Let the average annual growth rate be x, according to the meaning of the question.
72893( 1+x)2 = 1 14508
The solution is x 1≈0.2526, x2≈-2.2526 (irrelevant, omitted).
The average annual growth rate is about 25.3%.
(2) According to the meaning of the question, it is assumed that the number of new cars is X each year.
[ 1 14508( 1-4%)+x]( 1-4%)+x≤ 158000
X≤26770. 12 is obtained.
No more than 26,770 new cars are added each year.
6. For all real numbers x and y, if the function y=f(x) satisfies f(xy)=f(x)f(y) and f(0)≠0, then f(2009)=?
Answer: Solution: Because f(xy)=f(x)f(y),
So f (0) = f (0) f (0);
∫f(0)≠0，
∴f(0)= 1，
∴f(2009× 1)=f(2009)f( 1)，
∴f( 1)= 1.
Similarly, f (2009) =1;
7. A pile of apples on the beach is the property of three monkeys. The first monkey came and divided the apples into three piles, which was 1 apple. Then, he threw the extra apples into the sea and took a bunch of them himself. The second monkey came and divided the remaining apples into three piles and one pile. It also threw the extra apples into the sea and took a pile. The third monkey did the same. So how many apples are there?
Answer: Solution: The third monkey took 1 apple, so the third monkey had 4 apples.
That is, the second monkey left two piles, so the second monkey has two piles, so the second monkey has seven piles left, which can't be the two piles left by the first monkey, so the third monkey can't get one pile.
Taking two is not valid, so the third monkey takes three, the second monkey gets 16, and the first monkey gets 25.
So there are at least 25.