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在一幅比例尺是1:4000000的地图上，量得A、B两地相距12 cm。甲、乙两辆汽车同时从两地出发...

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प्रश्न

在一幅比例尺是1:4000000的地图上，量得A、B两地相距12 cm。甲、乙两辆汽车同时从两地出发，相向而行。某一时刻两车相遇，相遇后甲车每小时比相遇前多行驶20 km，乙车每小时比相遇前少行驶20 km，4小时后两车同时到达B、A两地。相遇前两车每小时分别行驶多少千米？使用小学生视角, 用最简单的方法解决问题

प्रश्न

在一幅比例尺是1:4000000的地图上，量得A、B两地相距12 cm。甲、乙两辆汽车同时从两地出发，相向而行。某一时刻两车相遇，相遇后甲车每小时比相遇前多行驶20 km，乙车每小时比相遇前少行驶20 km，4小时后两车同时到达B、A两地。相遇前两车每小时分别行驶多少千米？使用小学生视角, 用最简单的方法解决问题

विचार

Solving car speeds before meeting

Okay, so the map scale tells us 1 cm on the map equals 40 km in reality. The distance between A and B is 12 cm, so that's 480 km. Two cars start from points A and B and head towards each other until they meet. After meeting, their speeds change: car A speeds up by 20 km/h and car B slows down by 20 km/h. Four hours after they meet, they both reach their destinations. I’ll calculate their speeds before the meeting!

Solving for car speeds

I’ve set up three equations to solve for the speeds. From the relationships between the cars' speeds and the time traveled before and after meeting, I can solve these equations.

By manipulating the equations, I find that the time before they meet is 4 hours. With that, the speed of car A (v_1) is 50 km/h, and the speed of car B (v_2) is 70 km/h.

Thought for 17 seconds

उत्तर

一步步像小学生那样想：

1 把地图上的厘米换成真正的千米

比例尺 1:4 0000001:4\,000 000
1 cm=40 km1\text{ cm}=40\text{ km}

12 cm×40 km/cm=480 km12\text{ cm}\times 40\text{ km/cm}=480\text{ km}

所以 AA、BB 两地相距 480 km。

2 用“设数”法画条小算式

  • 先假设相遇前
    • 甲车每小时跑 xx km
    • 乙车每小时跑 yy km
  • 从两头同时出发，相遇用了 tt 小时。
    相遇前跑的总路程就是整段 480 km：
(x+y)×t=480①(x+y)\times t=480 \quad\text{①}
  • 相遇以后两车还要再跑 4 小时才正好开到对方的城市。
    • 甲车提速 20 km/h → 变成 (x+20)(x+20) km/h
    • 乙车降速 20 km/h → 变成 (y−20)(y-20) km/h

相遇点到终点的那一小段路程：

{(x+20)×4=480−xt②(y−20)×4=480−yt③\begin{cases} (x+20)\times 4=480-xt \quad\text{②}\\[4pt] (y-20)\times 4=480-yt \quad\text{③} \end{cases}

3 把式子化简（只用最简单的代数）

把 ②、③ 展开并整理：

4(x+20)=480−xt    ⇒  x(t+4)=4004(y−20)=480−yt    ⇒  y(t+4)=560\begin{aligned} 4(x+20) &=480-xt \;\;\Rightarrow\; x(t+4)=400 \\[4pt] 4(y-20) &=480-yt \;\;\Rightarrow\; y(t+4)=560 \end{aligned}

于是

x=400t+4,y=560t+4x=\frac{400}{t+4},\qquad y=\frac{560}{t+4}

把它们代回 ①：

t(400t+4+560t+4)=480  ⟹  960tt+4=480t\left(\frac{400}{t+4}+\frac{560}{t+4}\right)=480 \;\Longrightarrow\; \frac{960t}{t+4}=480

解得 t=4t=4 小时。

再把 t=4t=4 代回求 x,yx,y：

x=4004+4=50,y=5604+4=70x=\frac{400}{4+4}=50,\qquad y=\frac{560}{4+4}=70

4 答案

汽车相遇前原来的速度
甲车50 km/h
乙车70 km/h

（相遇后，甲车提速到 70 km/h，乙车降速到 50 km/h，正好再跑 4 小时各到终点。）

प्रश्न

4年级小学生的思维, 更简单一些

विचार

Simplifying the solution

The user wants a super simple explanation, so I have to break it down without complex math. I can use a narrative where I show that after the two vehicles meet, they travel for the same amount of time (4 hours), but with flipped speeds. Rather than focusing on algebra, I can explain that the extra distance they travel is the same, because of their swapped speeds. I'll keep it in simple terms, maybe with a simple example, so a 4th grader could understand!

Simplifying the approach

I should explain that the cars meet after traveling for the same amount of time, but after that, they travel at swapped speeds. I don’t want to overwhelm with too much math, so I’ll focus on the simple idea: The extra distance Car A travels matches the distance Car B loses. For a 4th grader, I can use a picture showing how their paths split into two parts after meeting. Simple steps, like showing how after 4 hours, one car covers more distance than the other at swapped speeds and with clear numbers, could make this easy to understand.

Breaking down the steps

After the cars meet, they each travel for 4 hours at adjusted speeds, so we can set up the equation based on that. Car A's remaining distance is determined by its increased speed, while Car B's distance is shortened by 20 km due to its reduced speed, meaning the sum of both distances equals 480 km. After some simple math, I find that Car A's speed (x) is 50 km/h, and Car B's speed (y) is 70 km/h. I’ll explain this simply for a 4th grader, using little pictures and basic words.

Thought for 38 seconds

उत्तर

像四年级小朋友排“算式”那样做：

① 先把路程换出来

比例尺 1 : 4 000 000
1 cm=40 km1\text{ cm}=40\text{ km}

12 cm×40=480 km12\text{ cm}\times 40=480\text{ km}

② 用 4 小时这条线索列第一行算式

相遇以后，两辆车再跑 4 小时就到终点。

甲车乙车合起来
4 小时跑 4(原速+20)4(\color{blue}{\text{原速}}+20)4 小时跑 4(原速−20)4(\color{green}{\text{原速}}-20)加起来正好 480 km

把两边都除以 4 就得到

(甲原速+20)+(乙原速−20)=120(\color{blue}{\text{甲原速}}+20)+(\color{green}{\text{乙原速}}-20)=120

所以

  甲原速+乙原速=120  （①）\boxed{\;\text{甲原速}+\text{乙原速}=120\;}\qquad\text{（①）}

③ 用“谁先跑谁后跑”列第二行算式

相遇点把路分成两段：

  • 乙车在相遇前跑的那一段（用原速跑了 4 小时）。
  • 甲车在相遇后跑的那一段（提速后跑了 4 小时）。

这两段其实 一样长，于是

4(乙原速)=4(甲原速+20)4\bigl(\color{green}{\text{乙原速}}\bigr)=4\bigl(\color{blue}{\text{甲原速}}+20\bigr)

同样除以 4：

  乙原速=甲原速+20  （②）\boxed{\;\text{乙原速}=\text{甲原速}+20\;}\qquad\text{（②）}

④ 只用加减就能算出速度

把（②）放到（①）里：

甲原速+(甲原速+20)=120\text{甲原速}+(\text{甲原速}+20)=120 2×甲原速=100 ⟹ 甲原速=502\times\text{甲原速}=100\ \Longrightarrow\ \text{甲原速}=50 乙原速=50+20=70\text{乙原速}=50+20=70

答案

  • 甲车相遇前的速度 50 千米/小时
  • 乙车相遇前的速度 70 千米/小时

（全程只用到了加、减、乘、除这四种最基本的运算，完全符合四年级的“口算+列式”水平！）

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