In 2007, the examination questions in the senior high school entrance examination were sorted-inequality and inequality group I, multiple choice question 1, the solution set of inequality (Jinhua, Zhejiang, 2007) was correctly expressed on the number axis, and the solution set of inequality (Neijiang, Sichuan, 2007) should be expressed on the number axis as (D3) and inequality-1 (Yueyang, Hunan, 2007) The positive integer solution of 5-2x has the solution set of () B (A) 1 (B)2 (C)3 (D)4 5, and (2007 Fuzhou, Fujian) is shown on the number axis as 1. The unequal group is () da.b.c.d.6, (2007). BA, 0 B, -3 c, -2 d,-1 solution: x≤, inequality solution: x ≤- 1, so =- 1, solution: a =-3. 7. The solution set of (2007 Yunnan Shuangbai) inequality is () ca.b.c.d.8 and (2007 Dongying, Shandong) inequality 2x-7 2 14. (2007 Ningbo, Zhejiang) represents the solution set of inequality group on the number axis, and the correct line is () c1. X > 23, the solution of inequality group (Yichang, Hubei, 2007) is. < x < 44, and the integer solution of the inequality group (Xianning, Hubei, 2007) is _ _ _ _ _ _ _ _ _ _ _ _ _. Solution: the solution of the inequality group is-1 < x ≤ 2, the integer solution is: 0, 1 25, and the integer solution of the inequality group (Dezhou, Shandong, 2007) is .26. It is known that the integer solution of the inequality group (Tianmen, Hubei, 2007) has six integer solutions * *, so the integer solution of A. Solution: the solution of the inequality group is: a < x 0；; From ≤4-x，x ≤ 3。 ∴ The solution set of the original inequality group is 0.

, solve the inequality group and get ≤ x ≤

That is to say, buy at least 34 sets and at most 39 sets. The store has six purchase plans.

(2) After the store is sold, the profit is Y yuan.

y =(2000- 1800)x+( 1600- 1500)( 100-x)= 100 x+ 10000。

∫ 100 > 0, ∴ When x is the largest, the value of y is also the largest.

That is, when x = 39, the maximum profit of the store is 13900 yuan 1 1. (In 2007, RoyceWong, a fruit grower in Baiyu Village, Jiangyou, a "national civilized village" in Mianyang City, Sichuan Province, harvested 20 tons of loquat and 0/2 tons of peach/kloc. How many schemes are there? (2) If the transportation cost of each class A truck is 300 yuan and the transportation cost of each class B truck is 240 yuan, which scheme does the fruit farmer choose to have the smallest transportation cost? What's the lowest freight? Solution: (1) If X Class A trucks are arranged, then (8-x) Class B trucks are arranged. According to the meaning of the question, 4x+2 (8-x) ≥ 20 and x+2 (8-x) ≥ 12. Solve this inequality group and get x≥2. The acceptable values of ∴ x are 2, 3 and 4. Therefore, there are three schemes for arranging two kinds of trucks: A-type truck, B-type truck, scheme 1, 2,6 cars, scheme 2,3 cars, scheme 3,4 cars, and 4 cars (2) scheme 1, and the required freight is 300×2+240×6 = 2040 yuan; The freight required for the second scheme is 300×3+240×5 = 2 100 yuan; The freight required for the third scheme is 300×4+240×4 = 2 160 yuan. Therefore, RoyceWong should choose the first scheme with the least freight, with the lowest freight of 2040 yuan. 65,438+02, (Huaihua, Hunan, 2007) In preparation for the 20th anniversary celebration of a county in our city, the garden department decided to use the existing 3,490 pots of A flowers and 2,950 pots of B flowers to put them on both sides of Yingbin Avenue. It is known that 80 potted flowers of A, 40 potted flowers of B, 50 potted flowers of A and 90 potted flowers of B need to be matched with one style. (1) Grade 9 of a school (1) Please help me design it. (2) If the cost of matching a model is 800 yuan and the cost of matching a model is 960 yuan, please explain which scheme in (1) has the lowest cost? How much is the minimum? Solution: If you build a matching model, then this model is one. According to the meaning of the question, you can get:, and you can get: by solving this inequality group, which is an integer. Three collocation schemes can be designed: ① a gardening model, ② a gardening model, ③ a gardening model and ③ a gardening model. (2) Method 1: Because the cost of each model is higher than that of each model, the fewer models, the more the cost. The lowest cost is: (yuan) Method 2: Scheme ① required cost: (yuan) Scheme ② required cost: (yuan) Scheme ③ required cost: Yuan should choose Scheme ③ 13 yuan with the lowest cost. (In 2007, Hebei Province), a mobile phone dealer plans to purchase 60 models A, B and C of a certain brand. And just used up the payment of 6 1 000 yuan. Suppose you buy the X part of the A-type mobile phone and the Y part of the B-type mobile phone. The purchase price and pre-sale price of the three mobile phones are as follows: the purchase price (unit: RMB/set) of type A mobile phone, type B mobile phone and type C mobile phone is 9001200100 pre-sale price (unit: RMB/set) (2) Find the functional relationship between Y and X; (3) Assuming that all the purchased mobile phones have been sold, the mobile phone dealers will spend various fees *** 1500 yuan in the process of buying and selling these mobile phones. ① Find the functional relationship between the estimated profit P (yuan) and X (department); (Note: Estimated profit p = total pre-sale amount-purchase price-various expenses) ② Seek the maximum estimated profit and write down the quantity of each of the three mobile phones purchased at this time. Solution: (1) 60-x-y; (2) According to the meaning of the question, 900x+1200y+1100 (60-x-y) = 61000, and Y = 2x-50. (3) ① According to the meaning of the question, p =1200x+1600y+1300 (60-x-y)-61000-1500, p = 500. ② The number of mobile phones purchased is 29≤x≤34. The value range of ∴x is 29 ≤ x ≤ 34, and x is an integer. (Note: it is not pointed out that X is an integer) ∵P is a linear function of X, k = 500 > 0, ∴P increases with the increase of X, ∴ When X takes the maximum value of 30,