As shown in the analysis, the cruise ship finally reached the third place, with a linear distance of 50 kilometers from A. Because the diameter of the circular artificial lake is 50 kilometers, it can be concluded that the linear distance between A and C is the diameter of the circle. According to the fillet theorem, if the fillets ∠ A, B and C opposite to the diameter are right angles, then △ A, B and C are right triangles.
Since Party A and Party B = 30km and Party A, Party B and Party C = 50km, according to Pythagorean theorem, Party A, Party B and Party C = 40km. Excluding the stay time of 36 minutes, the cruise from A to B to C takes * * * 120-36 = 84 minutes. Therefore, the speed of the cruise ship = (30+40)/84 = 5/6km/min, so the straight-line driving time from A to C = distance/speed = 50 \.
So the correct answer is B.
Test site itinerary problem
62 answer a
According to the analysis of the valve stem condition,
A+B = C+D...①;
A-B =240, that is, A = B+240...②;
D-C = 160, that is, D = C+ 160...③;
Substituting Formula ② and Formula ③ into Formula ①, we can get: 2B+240 = 2B+160, which is simplified to: 40 pieces of C-B, that is, 40 pieces of B are less than C.
So the correct answer is a.
Test Center and Difference Ratio Problem
63 answer c
According to the meaning of the question, the total cost of flying includes air ticket price (excluding tax), transportation fee and air ticket tax. Suppose the full-price air ticket price (excluding tax) from city A to city B is X yuan. According to the relationship of 1.4 times of two discounts, it can be expressed as: 0.6x+90+60 =1.4x (0.4x+90+60), and the solution is x= 1500 yuan.
So the correct answer is C.
Economic profit of the proving ground
64 answer a
65 answer b
analyse
Suppose that there were X people who scored well in the examination last year and 1.2x this year. According to the meaning of the question, the total number of people in these two years is 100, so the number of people who scored well or below in the examination this year is less than last year100×15% = 100. If the number of outstanding people in two years is the least, the number of outstanding people in two years is taken as the minimum number 0, and the number of outstanding people in two years = 75+90- 100 =
So the correct answer is B.
Test Center and Difference Ratio Problem
Answer c
The analysis shows that 27% of the patents applied by enterprise A are invention patents, that is, invention patents/applied patents =27/ 100, so the number of patents applied by enterprise A is an integer multiple of 100. It is also known that the number of patents applied by enterprise B is less than that of enterprise A, and the sum of them is more than 300, so enterprise A applied for 200 or 300 patents. Classification discussion:
If enterprise A applies for 200 patents, the invention patents of enterprise A at this time are 200×27%=54. The number of invention patents applied by enterprise B is more than that of enterprise A, and the invention/non-invention of enterprise B is 8/ 13, which means that the invention patents of enterprise B are integer multiples of 8 and more than 54, so there are at least 56 invention patents of enterprise B, and the non-invention patents of enterprise B are 13×56/8=9 1, so two enterprises.
If enterprise A applies for 300 patents, the invention patent of enterprise A is 300×27%=8 1. According to enterprise B, the number of invention patents applied for is more than that of enterprise A and enterprise B, and the invention/non-invention ratio is 8/13, which means that the invention patents of enterprise B are an integer multiple of 8 and greater than 8 1, so there are at least 88 invention patents of enterprise B, and the corresponding total number of patents of enterprise B is 88+13× 88/8 = 23.
All in all, the two companies applied for at least 237 non-invention patents.
So the correct answer is C.
Remarks: In the actual exam, only 237 items need to be calculated in the first case, and 237 is the least result among the options, so there can be no smaller situation, so there is no need to discuss the second case.
Maximum value problem of test point
67. Answer A
According to the analysis, "the output per unit time of production line A is five times that of production line B", the efficiency of A is 5, and the efficiency of B is 1. According to the meaning of the question, production line A works 1 hour and rests for 3 hours, while production line B continues to work, and the following table can be obtained:
Observing the table, we can see that every 4 hours is a change cycle.
First look at the changing trend: the output difference in the first 1 hour of each cycle is increasing, and the output difference in the last 3 hours is decreasing, excluding C;
Look at the time corresponding to the inflection point (horizontal axis T): the ratio of the time of the rising process corresponding to each cycle to the time of the falling process is 1:3, excluding b;
Finally, look at the corresponding height (vertical axis L): every cycle is up 4 and then down 3, and the downward value is more than half of the upward value, excluding D.
So the correct answer is a.
Test site others
Answer c
So the correct answer is C.
Test site probability problem
Answer c
According to the meaning of the question, the efficiency of a, b and c satisfies the following relationship:
2× B = A+C … ①
3× (A+B) +7× (B+C) =7× (A+B+C)...②
So the correct answer is C.
Test site engineering problems
Answer c
From the analysis of the meaning of the question, it can be seen that car A transported 2×35=70 boxes of goods twice before, and after car B joined, it was fully loaded 10 times to complete the task. At this time, car B transported 10 more boxes of goods than car A, so it can be obtained that the total amount of 10 B-( 10×35) is 70+(35+43)× 10=850 boxes, 850/43.
So the correct answer is C.
Test Center and Difference Ratio Problem
7 1 answer b
The age of the test center
72 answer d
According to the meaning of the question, if the number of days when the temperature exceeds 30 degrees Celsius at noon in July is as few as possible, then two conditions must be met at the same time:
(1) The whole sky above 30 degrees Celsius appears for three consecutive days;
(2) All the days not exceeding 30 degrees Celsius appear in the form of 120÷24= 5 consecutive days.
Ask "at least" and substitute from the smallest option.
Item C: If the number of days with the temperature exceeding 30 degrees Celsius is 12, the number of days with the temperature not exceeding 30 degrees Celsius is 3 1- 12= 19. According to rule ②, water 12 days = 4 times; According to rule ③, 19 ÷ 5 = 3 ... 4, that is,19 times a day. 4+3=7 times, which contradicts "8 times of irrigation" and is excluded;
Item D: If the number of days exceeding 30 degrees Celsius is 15 days, the number of days not exceeding 30 degrees Celsius is 3 1- 15= 16 days. According to rule ②, water 15÷3 = 15 times a day; According to rule ③, 16 ÷ 5 = 3... 1, that is, 16 drinks three times a day, and 5+3=8 times. Satisfy the meaning of the problem
Maximum value problem of test point
73 answer a
Test site probability problem
74 answer a
Arrangement and combination of test sites
75 answer d
Geometric problems of test sites