Problem description:

1. The unit prices of A, B and C in a certain store are 10 yuan respectively. 6 yuan, 4 yuan and Wang Mai bought several pieces of each of these three commodities, and * * * paid 40 yuan. Later, Xiao Wang thought that one of the goods was overbought and wanted to return two, but the clerk only had RMB with 20 yuan face value and no change to return. Xiao Wang had to adjust the purchase quantity of the other two commodities to keep the total price unchanged. How many products did Wang Mai buy?

The distance between a and b is 2400 kilometers. A starts from A and B at the same time and runs back and forth between A and B. A runs 300 meters per minute and B runs 240 meters per minute, and stops exercising after 35 minutes. When did A and B meet for the first time, the closest to A? What's the nearest distance?

Analysis:

The first question:

The analysis shows that Xiao Wang has three commodities, A, B and C, that is, A, B and C are not zero, whether before or after the exchange. In addition, the total value of the three commodities before and after the exchange is 40 yuan.

1. First of all, the extra goods can't be A, otherwise two A's will be returned, and the salesperson only needs to return them to Xiao Wang 20 yuan, and no other change smaller than 20 yuan's face value is needed.

2. The number of known A can only be 1, 2, 3; Because when A=4, in order not to surpass 40 yuan, b and c can only be 0, which does not meet the meaning of the question.

(1) When A= 1, the total value of B and C is 30 yuan, and since both B and C must be natural numbers, B=3, C=3 or B= 1, C=6.

(2) When A=2, the total value of B and C is 20 yuan, so B = 2 and C = 2 (because B and C must be natural numbers, this is the only possible situation).

(3) When A=3, the total value of B and C is 10 yuan, so B = 1 and C = 1.

3. We know from 1 that the extra goods can only be B or C, and it is because Xiao Wang thinks one of them is overbought and wants to return two, that is, the quantity of returned goods must be greater than 2, otherwise it is impossible to return two (even if it is equal to 2, because returning two will be gone). Then in the case of 2 (2) and (2)

4. To sum up, we only need to consider the situation of (1), which can be divided into two types:

(1) Assuming that the extra commodity is C, quitting two Cs means quitting 8 yuan. If you want to keep the total price unchanged by adjusting the purchase quantity of the other two commodities (namely, A and B), you must quit A (because B is 6 yuan, 8-6=2, and the extra 2 yuan is not enough to buy another B, so you can only make up the difference by quitting A), and A is only 65,438.

(2) Assuming that the extra commodity is B, quitting two B's is the extra 12 yuan, which can be used to buy all C's and three B's (similarly, you can't quit A, because there is only one A, and it will be gone if you quit).

5. So the answer can only be A= 1, B=3 and C=3. After returning goods, A= 1, B= 1, and C=6.

Xiao Wang bought 1 product B (the first 3 returned products).

The second question:

A runs 300 meters per minute, and B runs 240 meters per minute. After 35 minutes, A * * * runs10500m, and B * * * runs 8400m. The two add up to 18900 meters, which is 18.9 kilometers. The title says a and B. ......

Can we meet here?

I just automatically think this is the wrong number. A and b are 2400 meters apart.

A can run 300*35/2400=4.375 times between A and B in 35 minutes, and B can run 240*35/2400=3.5 times between A and B in 35 minutes. In other words, when B goes from B to A, then to B, and then to the midpoint between A and B, the time is just right.

1. From B to A

Suppose it took one minute to meet for the first time, and they ran 2400 meters when they met for the first time, then (300+240)* A = 2400, that is, 540a=2400, that is to say, the total length of A and B is converted into 540a meters represented by A, and this conversion result will be used many times in the future.

At this time, their position is 300 meters away from A and 240a meters away.

After that, A continued to run the remaining 240m, which took 240a/300 = 4a/5min. In the meantime, B continues to run to A (4a/5)*240= 192a meters, and the distance from B to A is still 300 a- 192a meters.

2.b from a to B.

A starts to run from b to a, b continues to run to a and then starts to run to b,

When a runs to b, start timing. Suppose they meet B for the second time after B minutes. At this time, they actually ran 2400 meters+108a meters.

When they met, A ran 300 meters and B ran 240 meters, so (300+240)*b=2400+ 108a, that is.

540b=2400+ 108a, and the total length of 2400m can be changed to 540a.

So 540b = 540a+ 108a = 648a, and b = 6a/5.

When they met again, A ran 300b meters and B ran 240b meters, which means A ran 360 meters and B ran 288 meters. At this time, the distance between B and A is 360m, and 288- 108A = 180A.

After that, A continued to run to A, which took 180a/300=3a/5 minutes. During this period of time, B has reached B(3a/5)* 240 = 144 a, and there is still 360a- 144a = 2 144a from B..

3.b from b to a.

When A arrived at A, he started to run to B, and B was still 2 16a meters away from B,

When A runs to A, it starts to time. Suppose it meets B for the third time after C minutes.

Then A ran 300c meters, B ran 240c meters, and the two ran 540c meters.

The 240c meters run by B includes the remaining 2 16a meters of B. That is, when they met for the third time, they ran 2400+2 16a meters.

The equation 540c=2400+2 16a can be obtained, and the total length of 2400 meters can be replaced by 540a meters.

540c=540a+2 16a=756a，c=756a/540，c=7a/5

So when they met for the third time, A ran 300c meters =420a meters and B ran 240c meters =336a meters.

At this time, the meeting place is 420a meters away and b 120 A meters away.

After that, it takes 120a/300=2a/5 minutes for A to continue running, and when A arrives at B, B runs 240*2a/5=96a meters. At this time, B is 420a-96a=324a meters away from A, and B needs 324 A/240 = 27a/ ..

4.B from a to b (note that it's time when b runs between a and b)

Because B can only run to the midpoint of A and B at most, and the midpoint distance A is 540a/2=270a meters, the distance from the venue to A cannot be greater than 270a meters.

When A runs to B, it starts to time. Suppose it meets B for the fourth time after D minutes.

At this time, A ran 300d meters, B ran 240d meters, and the two ran 300d+240d=540d meters.

And this 540d meter also includes the remaining distance of 324a meters from A to B when A runs to B.

Therefore, we get 540d=2400+324a=540a+324a=864a m, and d=864a/540=8a/5.

So when they meet for the fourth time, the distance B =480a is 300d, and the distance a is 540 a-480 a.

5. After that, it takes 60a/300=a/5 minutes for A to run to A, and during this time, B runs to B(A/5)* 240 = 48a meters, and there is still 270a-48a-60a= 162a meters from the midpoint between A and B..

Since the time is up when B runs to the midpoint between A and B, it mainly depends on whether A can catch up with B again before this.

When A runs to A and starts to return, B is meters away from the midpoint of A and B 162a, and it takes162a/240 = 27a/40min to reach the midpoint. At this time, A has just set off from A, 270a meters away from the midpoint between the two places.

During this period, A can run (27a/40)*300=202.5a meters, less than 270a meters, so A will not catch up with B, and there will be no fifth meeting.

6. To sum up, we can know that Party A and Party B can meet four times in 35 minutes:

First time: 300 meters away from a place.

The second time: the distance is a 180 a meters.

The third time: 420a meters away from a place.

The fourth time: 60 meters away from location A.

Because the total length of 2400 meters in A and B can be represented by 540a, 60a = 60 * 2400/540 = 800/3m.

When we met for the fourth time, we were closest to A, and the nearest distance was 800 meters (about 266.67 meters).