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**General Information / Chapter Title:**
第 2 讲 和差倍 (Lecture 2: Sum, Difference, and Multiples)
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**Question 1**
* **Section Title:** 优秀篇 (Excellent Section)
* **Difficulty Annotation:** (本题题目难度为★★至★★★) (Difficulty of this question is from 2 to 3 stars)
* **Question Stem:** 甲水池有水 260 立方米, 乙水池有水 120 立方米, 如果甲水池里的水以每分钟 23 立方米的速度流到乙水池, 那么多少分钟后, 乙水池中的水是甲水池的 4 倍?
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**Question 2**
* **Question Stem:** 某镇上有东西两个公交车站, 东站有客车 84 辆, 西站有客车 56 辆, 每天从东站到西站有 7 辆车开出, 从西站到东站有 11 辆车开出, 几天后, 东站车辆是西站的 4 倍?
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**Question 3**
* **Question Stem:** 维尼熊和跳跳虎去摘苹果。维尼熊爬上树去摘, 跳跳虎在地上跳着摘。维尼熊摘了 80 分钟, 跳跳虎摘了 50 分钟就累了, 不摘了。他们回来后数了数, 一共摘了 480 个苹果, 那么其中维尼熊摘的苹果有 ____ 个。
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Today we'll solve three classic sum, difference, and multiples problems. The first involves water flowing between pools until one pool contains four times the water of another. The second deals with buses moving between stations until one station has four times as many buses. The third is about apple picking with different working times. These problems share a common pattern: quantities change over time until specific mathematical relationships are achieved.
To solve these problems systematically, we follow a four-step framework. First, define a variable, usually time t. Second, express how each quantity changes after time t using linear functions. Third, set up an equation based on the given condition or target relationship. Fourth, solve the equation algebraically. This approach works because these problems involve linear functions, and we're finding the intersection point where a specific ratio is achieved.
Let's solve the water pool problem step by step. First, we define t as time in minutes. After t minutes, Pool A will have 260 minus 23t cubic meters, while Pool B will have 120 plus 23t cubic meters. The condition states that Pool B should equal 4 times Pool A, giving us the equation: 120 plus 23t equals 4 times 260 minus 23t. Expanding this: 120 plus 23t equals 1040 minus 92t. Collecting terms: 115t equals 920. Therefore, t equals 8 minutes.
Now let's solve the bus station problem using the same framework. We define d as time in days. For the East station: it starts with 84 buses, loses 7 per day to West, but gains 11 per day from West, giving us 84 plus 4d buses. For the West station: it starts with 56 buses, gains 7 per day from East, but loses 11 per day to East, giving us 56 minus 4d buses. The condition requires East equals 4 times West: 84 plus 4d equals 4 times 56 minus 4d. Expanding: 84 plus 4d equals 224 minus 16d. Solving: 20d equals 140, so d equals 7 days.
Let's verify our solutions and summarize key strategies. For Problem 1, after 8 minutes: Pool A has 76 cubic meters, Pool B has 304 cubic meters, and 304 equals 4 times 76, confirming our answer. For Problem 2, after 7 days: East station has 112 buses, West station has 28 buses, and 112 equals 4 times 28, verifying our solution. Problem 3 requires additional information about the picking rates to determine individual contributions. The key strategies are: define a time variable, track how quantities change, set up equations based on target relationships, and always verify your solutions by substituting back into the original conditions.