Solution: Let's assume that the original speeds of Party A and Party B are A km and B km per day respectively.
According to the meaning of the question
(a+b)×50=200( 1)
10×(a+0.6)+40a+30b+ 10×(b+0.4)= 200(2)
simplify
a+b=4(3)
a+0.6+4a+3b+b+0.4=20
5a+4b= 19(4)
(4)-(3)×4
A =19-4× 4 = 3km.
B = 4-3 = 1 km
A repairs 3 kilometers every day, and B repairs 1 kilometer every day.
A It was originally planned to build 3× 50 = 150km.
B 1× 50 = 50km as originally planned.
2. Xiaohua bought four mechanical pencil and two pens, and paid 14 yuan; Xiaolan bought the same 1 mechanical pencil and two pens, and paid 1 1 yuan. Find the unit price of mechanical pencil and pen.
Solution: Suppose mechanical pencil has a pen with X yuan and a pen with Y yuan.
4X+2Y= 14
X+2Y= 1 1
The solution is X= 1.
Y=5
In mechanical pencil, the unit price is 1 yuan.
Pen unit price 5 yuan
22. For a photo with a length of 29cm and a width of 22cm, it is required that the four sides of the photo frame have the same width, and the area occupied by the photo frame is one quarter of that of the photo frame. What is the width of the photo frame?
Solution: Let the width be one centimeter.
According to the meaning of the question
(29+2a)×a×2+22×a×2 = 1/4×29×22
4a2+58a+44a=3 19/2
8a2+204a-3 19=0
a=(-5 1 √3239)/4
A=(-5 1-√3239)/4 (omitted)
therefore
a =(-5 1+√3239)/4≈ 1.48cm
23. A farmer grows peanuts, and the original peanuts yield 200 kg per mu, and the oil yield is 50% (that is, 50 kg of peanut oil can be processed per 100 kg of peanuts). Now, after planting new peanut varieties, the harvested peanuts per mu can be processed into peanut oil 132 kg, in which the growth rate of peanut oil output is half of that per mu. Seeking the growth rate of yield per mu of new peanut varieties.
Solution: We assume that the yield growth rate per mu of new peanut varieties is a.
Then the oil yield is 1/2a.
200×( 1+a)×50%×( 1+ 1/2a)= 132
(a+ 1)(a+2)= 132/50
a2+3a+2-2.64=0
a2+3a-0.64=0
simplify
25a2+75a- 16=0
a=(-75 85)/50
A=-3.2 (truncated) or a=0.2=20%
Therefore, the yield per mu of new peanut varieties increased by 20%.
Four, a city produces an average of 700 tons of domestic garbage every day, all of which are treated by two garbage factories, A and B. It is known that Factory A treats 55 tons of garbage every hour at a cost of 550 yuan; Factory B handles 45 tons of garbage per hour at a cost of 495 yuan. If it is stipulated that the cost of garbage disposal in this city does not exceed 7370 yuan per day, how many hours does Factory A need to dispose of garbage at least every day?
Solution: A yard should dispose of garbage for at least one hour.
550a+(700-55a)÷45×495≤7370
550 a+(700-55a)× 1 1≤7370
550a+7700-605a≤7370
330≤55a
a≥6
A yard has to deal with garbage for at least 6 hours.
The school allocated several dormitories to the girls in Class 1, Grade 7. It is understood that there are less than 35 girls in this class. If there are five people in each room, the remaining five people have nowhere to live; If there are 8 people in each room, one room will be vacant and one room will be unsatisfactory. How many dorms and girls are there?
Solution: Use dormitory A, and the number of girls is 5a+5.
According to the meaning of the question
a & gt0( 1)
0 & lt5a+5 & lt； 35(2)
0 & lt5a+5-[8(a-2)]& lt； 8(3)
Derived from (2)
-5 & lt； 5a & lt30
- 1 & lt； A< six
By (3)
0 & lt5a+5-8a+ 16 & lt； eight
-2 1 & lt； -3a & lt； - 13
13/3 & lt； A< seven
From this, we determine the value range of a.
4 1/3
A is a positive integer, so a=5.
Then there are five dormitories with 5×5+5=30 girls.