Solution: Suppose there is a student.
According to the meaning of the question
3a+8-5(a- 1)& lt； 3( 1)
3a+8-5(a- 1)>0(2)
By (1)
3a+8-5a+5 & lt； three
2a & gt 10
A> v
By (2)
3a+8-5a+5 & gt； 0
2a & lt 13
a & lt6.5
Then the value range of a is 5.
Then a=6
There are 6 students, 3×6+8=26 books.
9. The planned construction area of aquatic product market management department is 2,400 m? Market. There are 80 A-type and B-type stores in the shed. The average area of each type A storefront is 28m? Monthly fee 400 yuan; The average area of each B-store is 20m? Monthly fee 360 yuan. The construction area of all storefronts shall not be less than 80% of the total greenhouse area and shall not exceed 85% of the total greenhouse area. Try to make sure that there are several schemes to establish two types of stores, A and B.
Solution: Let Type A store be Room A and Type B store be Room 80-A.
According to the meaning of the question
28a+20(80-a)≥2400×80%( 1)
28a+20(80-a)≤2400×85%(2)
By (1)
28a+ 1600-20a≥ 1920
8a≥320
a≥40
By (2)
28a+ 1600-20a≤2040
8a≤440
a≤55
40≤a≤55
Scheme: A B
40 40
4 1 39
……
55 25
A * * * is the scheme of 55-40+ 1= 16.
X tables and chairs sold in a furniture store, the unit price is one in 300 yuan and one in 60 yuan. The furniture store has made two preferential schemes: (1) buy a table and get two chairs; (2) Pay 87.5% of the total price. A company needs to buy five tables and several chairs (not less than 10). If it is known that X chairs are to be purchased and the company purchases the same number of chairs, which scheme is more economical?
Suppose you need to buy x(x≥ 10) chairs, and the total cost is y.
The first scheme:
y = 300 X5+60×(x- 10)= 1500+60x-600 = 900+60x
The second plan:
y =(300 X5+60x)×87.5% = 13 12.5+52.5 x
If the two schemes cost the same amount of money.
900+60x = 13 12.5+52.5 x
7.5x=4 12.5
x=55
When buying 55 chairs, the two schemes cost the same money.
When it is greater than 55, choose the second scheme.
When it is less than 55, choose the first scheme.
Eleven, a beverage factory developed two new drinks, A and B. The main raw materials are both A and B. The contents of A and B in each bottle are shown in the following table. At present, 2800g of raw materials A and B are produced in trial, and it is planned to produce *** 100 bottles of A and B, with X bottles of drinks. Answer the following questions:
Jiayi
A 20 g 40 g.
30 grams 20 grams
(1) How many production schemes are there? Write the solution process;
(2) If the cost per bottle of beverage A is 2.60 yuan and the cost per bottle of beverage B is 2.80 yuan, and the total cost of these two beverages is Y yuan, please write the relationship between Y and X, and explain what value of X will make the total cost the lowest.
Solution: (1) Suppose that X bottles are needed to produce type A beverage and 100-x bottles are needed for type B beverage.
According to the meaning of the question
20x+30( 100-x)≤2800( 1)
40x+20( 100-x)≤2800(2)
By (1)
20x+3000-30x≤2800
10x≥200
x≥20
By (2)
40x+2000-20x≤2800
20x≤800
x≤40
So the value range of x is 20≤x≤40.
Therefore, the scheme is as follows
Production A B
20 80
2 1 79
……
40 60
A * * * is 40-20+ 1=2 1 scheme.
(2)y = 2.6x+2.8×( 100-x)= 2.6x+280-2.8x = 280-0.2x
Y is a linear function at this time, because 20≤x≤40.
Then when x=40, the cost is the lowest, and at this time the cost y=272 yuan.
12. A real estate development company plans to build 80 single apartments of two types, A and B. Each apartment costs 550,000 yuan and the price is 600,000 yuan. B Each apartment costs 580,000 yuan and costs 640,000 yuan. Set up a development company to build apartment X of A.
(1) Complete the following table according to the given conditions.
A b
Number of groups X 80-x
Single set profit 5 6
Profit 5x 480-6x
If the total profit from selling the house is y million yuan, please write the resolution function of y about x.
y=5x+480-6x=480-x
(2) The funds raised by the company are not less than 44.9 million yuan, but not more than 44.96 million yuan, all of which are used for building houses. What kind of architectural scheme does the company have for these two types of houses? Which scheme is the most profitable?
Solution: according to the meaning of the problem
55x+58(80-x)≥4490( 1)
55x+58(80-x)≤4496(2)
By (1)
55x+4640-58x≥4490
3x≤ 150
x≤50
By (2)
55x+4640-58x≤4496
3x≥ 144
x≥48
48≤x≤50
Therefore, there are three sets of housing plans:
Type a 48 49 50
Type b 32 3 1 30
Y=480-x is a linear function. When x=48, the maximum value of Y = 480-48 = 4.32 million yuan.
(3) In order to meet the market demand, the company has built a number of Class C apartments with the total number of apartments unchanged. Now it is known that each C-apartment costs 530,000 yuan and costs 570,000 yuan. The company plans to use up all the raised funds of 44.9 million yuan. When x= the number of units, the total profit of the houses sold by the company is the largest.
Solution: Set Type B to build Z sleeve, and Set Type C to build 80-x-z sleeve.
55x+58z+53(80-x-z)=4490
55x+58z+4240-53x-53z=4490
2x+5z=250
5z=250-2x
z=50-2/5x
X and z are positive integers, x+z.
50-2/5x+x & lt； 80
3/5 times & lt30
x & lt50
So x can only be a multiple of 5.
x=5，z=48
x= 10.z=46
x= 15，z=44
x=20，z=42
……
x=45，z=32
Profit y=5x+6(50-2/5x)+4(80-x-50+2/5x)
= 5x+300- 12/5x+ 120- 12/5x = 420+ 1/5x
When x=45, the maximum value of y = 420-1/5× 45 = 4.29 million.
Thirteen, a shopping mall spent 36,000 yuan to buy two products, A and B. After the sale, * * * made a profit of 6000 yuan. After knowing the purchase price of product A 120 yuan, product B 138 yuan, and purchase price 120 yuan, sell it.
(1) How many items A and B were bought in this mall;
(2) The mall bought two kinds of goods, A and B, at the original price for the second time. The number of items purchased for B remains unchanged, while the number of items purchased for A is twice that of the first time, and A is sold at the original price. If these two items are sold, you need to earn a profit of not less than 8400 yuan for the second business activity. What's the lowest selling price of B?
Solution: (1)B Commodity price =120× (1+20%) =144 yuan.
Profit of commodity A =138-120 =18 yuan.
Profit of commodity B = 144- 120=24 yuan.
I * * * bought 36,000/120 = 300 A and B products.
Let's buy a commodity A and a commodity B.
a+b=300( 1)
18a+24b=6000(2)
(2)-( 1)× 18
6b=6000-5400
6b=600
b= 100
a=300- 100=200
So I bought 200 A goods, 100 B goods.
(2) according to the meaning of the question
Purchase 100 pieces of Class B goods, 200×2=400 pieces of Class A goods.
The profit of commodity A remains unchanged at 18 yuan.
Suppose the lowest selling price of commodity B is X yuan.
18×400+ 100(x- 120)≥8400
7200+ 100 x- 12000≥8400
100x≥ 13200
x≥ 132
Therefore, the lowest price of commodity B is 132 yuan/piece.
14. Several workers in Workshop A and Workshop B produce the same parts. One person in Workshop A only produces 6 pieces per day, and the others produce 1 1 piece per day. One person in Workshop B only produces 7 pieces per day, and the others produce 10 pieces per day. It is known that the total number of parts produced by two workshops is equal every day, and the total number of parts produced by each workshop is not less than 100 and not more than 200. How many people are there in Workshop A and Workshop B respectively?
Solution: Set one person in Workshop A and one person in Workshop B..
100≤ 1 1(a- 1)+6≤200( 1)
100≤ 10(b- 1)+7≤200(2)
1 1(a- 1)+6 = 10(b- 1)+7(3)
By (3)
1 1a- 1 1+6 = 10 b- 10+7
1 1a- 10b=2
a =( 10 b+2)/ 1 1(4)
By (1)
100≤ 1 1a-5≤200
105≤ 1 1a≤205
105/ 1 1≤a≤205/ 1 1
9 5/11≤ a ≤18 and 7/ 1 1
By (2)
100≤ 10 b- 10+7≤200
103≤ 10b≤203
10.3≤b≤20.3
Because b is a positive integer, b= 1 1, 12, 13, 14, 15, 16, ..., 20.
Substitute (4)
Only b= 13 and a= 12 meet the requirements.
So there are 2 people in Workshop A and 0/3 people in Workshop B/KLOC.
15. A factory has 360 kilograms of raw materials and 290 kilograms of raw materials, and plans to produce 50 AB products with these two raw materials. It is known that 9 kg of raw material A and 3 kg of raw material B are needed to produce a product A, which can make a profit in 700 yuan; To produce a product B, raw material A 4 kg and raw material B 10 kg are needed, and the profit can be 1.200 yuan.
(1) How many schemes are there to arrange the production quantity of AB products as required? Please design it.
Set to produce one product A and 50 products B.
9a+4(50-a)≤360( 1)
3a+ 10(50-a)≤290(2)
By (1)
9a+200-4a≤360
5a≤ 160
a≤32
By (2)
3a+500- 10a≤290
7a≥2 10
a≥30
So 30≤a≤32
One * * * is three schemes.
Produce 30 A products and 20 B products.
Production of A product 3 1 piece, B product 19 pieces.
32 pieces of A products and 8 pieces of B products/KLOC-0 were produced.
(2) Let the production of AB products earn a profit of Y yuan, and the number of pieces produced by one of them is X. Try to write the relationship between Y and X, and point out which scheme gains the most profit.
Set to produce x products.
y = 700 x+ 1200(50-x)= 60000-500 x
As a linear function, y increases with the decrease of x.
So when x=30, the maximum value of y =60000-500×30= 45000 yuan.
16.2009, I was celebrating the 20th anniversary of a county, and the garden department decided to put 3490 pots of A-type flowers and 2950 pots of B-type flowers with AB gardening shapes on both sides of Yingbin Avenue respectively. It is known that 80 pots of A-type flowers and 40 pots of B-type flowers are needed to match an A-type model; To use a type B, you need 50 pots of type A flowers and 90 pots of type B flowers.
(1) A company undertook the design of this supporting scheme for horticultural modeling. How many matching schemes are there? Please help design it.
(2) If the cost of matching a model is 800 yuan and the cost of matching a model is 960 yuan, which scheme in (1) has the lowest cost? How much is the lowest cost plan?
Solution: Suppose Type A has 50-a model, then Type B has 50-a model.
According to the meaning of the question
80a+50(50-a)≤3490( 1)
40a+90(50-a)≤2950(2)
By (1)
80a+2500-50a≤3490
30a≤990
a≤33
By (2)
40a+4500-90a≤2950
50a≥ 1550
a≥3 1
So the range of the group of A is 3 1≤a≤33.
Scheme:
Type a has 3 1 and type b has 19.
There are 32 a and 18 b.
There are 33 type A and 7 type B/KLOC-0.
(2)
Suppose the cost is y yuan.
y = 800 a+960(50-a)= 48000- 160 a
This is a linear function, y decreases with the increase of a, and the required cost is the lowest, so when a=33, the cost is the lowest. At this time, the cost y = 48,000-160× 33 = 42,720 yuan.
Seventeen, a ***25 questions, ask students to choose the correct answer, choose 4 points for each question, and deduct 2 points for not choosing or choosing wrong. If the student scores at least 60 points in this competition, how many questions can he answer correctly at least?
Solution: Set the answer to question A correctly.
According to the meaning of the question
4a-2×(25-a)≥60
4a-50+2a≥60
6a≥ 1 10
A≥55/3= 18 and 1/3.
Answer at least 19 questions correctly.
involve