High-frequency question type on quantitative relationship in the 2013 national examination: itinerary question.

The 2013 National Examination is coming soon. How can we prepare lessons more effectively? In particular, the quantitative relationship in the Administrative Vocational Aptitude Test often gives many candidates a headache, and it has always been the part with the lowest score rate. Through investigation, the National Civil Service Examination Network found that the root of the problem lies in the fact that everyone does not know the quantitative relationship test questions, but that it is difficult to find a suitable method to solve it in a short time. The large number of test questions makes it difficult to grasp the direction of the test proposition. In response to the above problems, we By decomposing the itinerary problems in the quantitative relationship of the national examination in the past three years, we can give some guidance to the candidates.

74 Questions in the 2012 National Examination

Two people, A and B, plan to walk from point A to point B. B sets off at 7:00 in the morning and walks at a constant speed. A is delayed due to something. Departs at 9:00. In order to catch up with B, A decides to run forward. The running speed is 2.5 times the walking speed, but he needs to rest for half an hour every half hour of running. So when can A catch up with B?

A.10:20 B.12:00 C.14:30 D.16:10

Analytical answer C

Let B’s speed be 12 , then A’s running speed is 30 and his resting speed is 0, substitute option, so A can catch up with B at 14:30, answer C.

2011 National Examination 66 Questions

Xiao Wang’s walking speed is 50% slower than running, and running is 50% slower than cycling. If he rides a bicycle from City A to City B, and then walks back to City A, it will take *** 2 hours. Ask Xiao Wang how long does it take to run from city A to city B?

A. 45 B. 48 C. 56 D. 60

Analytical answer B

This question is a proportional stroke problem. Suppose the walking speed is 1, the running speed is 2, the cycling speed is 4, and the AB distance is L, then L/4+L/1=2, then L/2=48, so choose option B.

53 Questions in the 2010 National Examination Test

A tourism department plans a tourist route from scenic spot A to scenic spot B. After testing, it takes a tourist boat to travel at a constant speed from A to B along the river. 3 hours; it takes 4 hours to drive from B back to A against the current at a constant speed. Assume that the water speed is constant, the distance between A and B is y kilometers, and it takes x hours for a tourist boat to travel y kilometers at a constant speed in still water, then the equation that x satisfies is:

　A.1/3-1/ X=1/X-1/4 B.1/3-1/X=1/X+1/4

C.1/(X+3)=1/4-1/X D.1/(4-X)=1/X+1/3

Analytical answer A

Suppose the distance between A and B is 1, then the speed along the current is 1/3, The speed against water is 1/4 and the speed under still water is 1/X, so 1/3-1/X and 1/X-1/4 are both water speeds. Therefore choose A.

Through the above analysis, it is obvious that the itinerary question is a required question type in the national examination, and the moderate difficulty is one of the question types that candidates must score.

Itinerary problems are generally divided into encounter problems, catching up problems and flowing water problems. The problem-solving techniques include equation method, proportion method, substitution method, drawing method and formula method, etc. The above three real questions reflect the practical application of these skills. The following summarizes the relevant theories that you must know about itinerary problems.

Theoretical summary:

n Basic formula: distance = speed × time;

n Commonly used methods: list equations and solve equations;

n Key to problem-solving: Find the correct stroke process, quickly formulate equations, and accurately solve equations.

Tips on skills:

n Typical model formula:

Encounter problem: encounter distance = (large speed + small speed) × encounter time

< p> Catch-up problem: catch-up distance = (large speed - small speed) × catch-up time

Departure problem: divergence distance = (large speed + small speed) × divergence time

Reverse Movement: Ring circumference = (large speed + small speed) × encounter time.

Movement in the same direction: circular circumference = (large speed - small speed) × meeting time

Downstream distance = downstream speed × downstream time = (ship speed + water speed ) × Downstream Time

Countercurrent Distance = Countercurrent Speed ??× Countercurrent Time = (Ship Speed ??– Water Speed) × Countercurrent Time

n Two people start from both ends and walk towards each other. Return immediately after reaching the end point on the opposite side, and so on, then:

When two people meet head-on for the first, second, third, fourth... time, the sum of the distances traveled by the two people is 1, 3, 5, 7 respectively. …a whole journey;

When the two people catch up and meet each other for the first, second, third, fourth... time, the distance difference between the two is 1, 3, 5, 7... a whole journey respectively.