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A children's toy factory recruitment said; 'Working hours of workers in this factory: 8 hours a day, 25 days a month, salary: skilled workers

Analysis: According to the second and third rows of statistics, it takes 15 (=85-70) minutes to make a puppy, and the salary is 0.75 (=5.05-4.30) yuan

< p>According to the first row, it can be seen that it takes 20 (=35-15) minutes to make a car, and the salary is 1.40 (=2.15-0.75) yuan

The monthly working hours are: 60× 8×25=12000 minutes. Assume that X cars are made per month, and the puppy can make (12000-20X)/15. The salary is: 1.4X+0.75×(12000-20X)/15 yuan

< p>To make the piece rate wage not less than 800 yuan, then 1.4X+0.75×(12000-20X)/15≥800

Solution: X≥500, the puppy produced at this time is 133.33 That is to say, at least 501 small cars must be made every month, and the salary must not be less than 800. In this way, the maximum number of puppies can only be 132, which is far less than the number of cars. It is not in line with the factory’s plan and will definitely not be paid. to estimated salary. Therefore, the manufacturer's advertising is fraudulent.

: (1) Assume that the time required to produce each puppy is m minutes, and the time required to produce each car is n minutes.

It can be seen from the meaning of the question:

m+n=35 3m+2n=85,

The solution is: m=15 n=20,

Suppose the piece-rate wage for each puppy produced is a Yuan, the piece-rate wage for producing each car is b Yuan. From the meaning of the question:

a+b=2.8 3a+2b=6.6 The solution is: a=1 b=1.8,

< p>Answer: The time required to produce each car is 20 minutes, and the piece rate wage is 1.8 yuan;

(2) W=x+1.8y+100

The meaning of the question It can be seen that: 15x+20y=8×25×60,

Simplified: y=-3 4 x+600

∴W=-7 20 x+1180;

(3) From the meaning of the question, we can know: x≥2y,

That is, x≥2? (-3 4 x+600),

The solution is x≥480 ,

∵W is a linear function of x, and W decreases as x increases,

When x=480, the maximum W is =1012<1100,

∴Manufacturer recruitment advertisements are fraudulent.