六年级数学　 /　期中试卷 · 第 24 题GRADE 6 MATH　 /　 MIDTERM · QUESTION 24

摩托车能把物品送多远？How far can the motorcycles go?

满油的摩托车一起出发，每箱油能走 120 千米。支援车沿途给同伴分油，再直接返回起点。只有甲送货到终点，每辆车最终都恰好耗尽燃油。Each motorcycle starts with a full tank and can travel 120 km on it. Helpers share fuel on the way out, then ride straight home. Only bike A (甲) delivers the parcel. Every bike finishes with an empty tank.

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逐站看分油Follow the fuel, stop by stop

选择车辆数，再切换折返点，观察这一站的支援车怎样分配一箱油。Choose a fleet size, then select a stop to see how its returning helper uses one tank.

行驶路线　 /　按比例绘制ROUTE MAP　 /　 TO SCALE

油量分配　 /　本段初始满箱FUEL ALLOCATION　 /　 ONE FULL TANK

本段去程用掉Burned on this leg
分给续行的车Given to continuing bikes
留下直接回家Kept for the ride home

查看折返点Inspect a turnaround

五个问题，逐步求解The five questions, worked through

任务 1　 /　问题 1、2TASK 1　 /　 QUESTIONS 1–2

两辆车：乙的一箱油分三份Two bikes share a tank in thirds

乙留下 1/3 箱返程。去、回路程相同，所以去程也用 1/3 箱。折返点距家 120 × 1/3 = 40 千米。B (乙) keeps 1/3 of a tank for the trip home. The outward trip is just as long, so it also burns 1/3. The stop is 120 × 1/3 = 40 km from home.
乙还可给甲 1 − 1/3 − 1/3 = 1/3 箱。甲刚好也用掉了 1/3，收油后又满了。B can give A the remaining 1 − 1/3 − 1/3 = 1/3 of a tank. A has just burned that much, so the gift fits exactly.
甲先走 40 千米，再用满箱油走 120 千米。把甲前后两段的耗油相加。A travels 40 km with B, then 120 km alone on its refilled tank. Add what A burns on both legs.

① 用油 1/3 + 1 = 4/3 箱
② 最远 40 + 120 = 160 千米① Fuel burned　1/3 + 1 = 4/3 tanks
② Distance　40 + 120 = 160 km

任务 2　 /　问题 3TASK 2　 /　 QUESTION 3

三辆车：乙从 B 地怎么回家？Three bikes make two stops

丙在 A 地折返。它的一箱油分为四份：去程、回家、给甲、给乙各 1/4 箱。因此 A 地距起点 30 千米，甲、乙再次满油。C (丙) turns at the first stop (point A), 30 km from home. Its tank covers four quarters: riding out, riding home, topping up A, and topping up B. Both remaining bikes leave full.
乙在 A 地先预留 1/4 箱，用于最后从 A 地回家。余下 3/4 箱分成三份，用于 A 到 B、B 回 A、补给甲。每份 (3/4) × (1/3) = 1/4 箱。B sets aside 1/4 of its refilled tank for the first stop to home. It splits the remaining 3/4 equally between riding to the second stop, riding back to the first stop, and topping up A. Each share is (3/4) × (1/3) = 1/4.
B 地距家 30 + 30 = 60 千米。乙给甲 1/4 箱后，留下 1/2 箱走完全程回家。甲补满后再走 120 千米。The second stop (point B) is 30 + 30 = 60 km from home. B gives A a quarter and keeps half a tank for the whole return. A is full again and rides 120 km alone.

③ 甲实际用油 1/4 + 1/4 + 1 = 3/2 箱
对应路程 30 + 30 + 120 = 180 千米。③ A burns　1/4 + 1/4 + 1 = 3/2 tanks
A’s route measures 30 + 30 + 120 = 180 km.

任务 3　 /　问题 4TASK 3　 /　 QUESTION 4

11 辆车：把一箱油画成 12 格Eleven bikes divide a tank into 12 parts

第一辆支援车要给另外 10 辆各补上第一段所用的油。它自己去、回各用同样的 1 格，合计 1 + 10 + 1 = 12 格。一格能走 120 ÷ 12 = 10 千米。The first helper replaces the fuel that 10 other bikes burned on the first leg. Its own trip out and its ride home each take one more share. That makes 1 + 10 + 1 = 12 shares. One share covers 120 ÷ 12 = 10 km.
第二辆支援车刚在上一站补满。再走 10 千米后，本段用 1 格，给其余 9 辆各 1 格，留 2 格回家。合计仍是 12 格。The second helper starts the next leg full. After another 10 km, it has spent 1 share, gives 9 shares to the others, and keeps 2 to get home. That still totals 12.
第 j 站也一样：本段用 1 格，补给 11 − j 辆各 1 格，留 j 格回家。10 辆车依次折返，最后一次分油在第 10 站。At stop j, the helper spends 1 share on this leg, gives 1 to each of the 11 − j continuing bikes, and keeps j for home. Ten helpers turn in all, so the last refill is at stop 10.

第一站的油量分配。每一格是 1/12 箱，十个绿色格分别给十辆车。The first helper’s tank at departure. Each green block is 1/12 of a tank for one continuing bike.

④ 最远距离 = 10 × 10 + 120 = 220 千米。④ Maximum distance = 10 × 10 + 120 = 220 km.

任务 3　 /　问题 5TASK 3　 /　 QUESTION 5

继续加车，能到 260 千米吗？Could more bikes reach 260 km?

盯住最后一辆分油的支援车。它分油后必须自己回家，而一箱油最多走 120 千米，所以最后一次分油地点离家最多 120 千米。Look at the last helper. After sharing fuel it must ride all the way home, and even a full tank covers only 120 km. Its turnaround is therefore at most 120 km from home.
甲在这里即使加满，也最多再向前走 120 千米。两段加起来至多 240 千米，小于 260 千米。From that stop, A can ride at most another 120 km. Even this generous upper bound gives just 120 + 120 = 240 km, short of 260 km.

⑤ 不能。有限辆车甚至到不了 240 千米。⑤ No. Every finite fleet falls short even of 240 km.

同一把尺上比较最远距离。增加车辆可以逼近 240 千米。Maximum distances on one scale. Adding bikes brings the destination closer to 240 km.

任意辆车的最远距离The maximum for any fleet

有 n 辆车，就把一箱分成 n + 1 格，每格能走 120/(n + 1) 千米。每次分油后，继续前进的车都重新满油。第 j 站的支援车可以这样分配它在上一站加满的一箱油：For n bikes, divide a tank into n + 1 shares. One share takes a bike 120/(n + 1) km. Every continuing bike leaves each stop full. At stop j, the returning helper accounts for its tank at the start of this leg as follows.

本段用 1 格＋给同伴 (n − j) 格＋回家留 j 格＝ n + 1 格1 burned on this leg + (n − j) given to others + j kept for home = n + 1 shares

甲走过 n − 1 段后，还有满箱油可以再走 120 千米。After n − 1 short legs, A has a full tank for a final 120 km.

D_n = 120 + (n − 1) × 120/(n + 1)

也就是 Dₙ = 240 − 240/(n + 1) 千米。每增加车辆，距 240 千米的差距就缩小一点，但只要车辆有限，差距始终大于 0。Equivalently, Dₙ = 240 − 240/(n + 1) km. The gap to 240 km shrinks as the fleet grows, and stays positive for every finite fleet.

提高题：如果把折返点挪一挪，会不会更远？Going further: could unevenly spaced stops do better?

下面允许各段距离任意，也允许续行车辆没有补满。把 120 千米当作 1 个距离单位。由近到远，设第 j 辆支援车在 xⱼ 处折返。Sⱼ 是前 j 个折返距离的和，S₀ = 0。Allow any lengths for the legs and any legal fuel transfers. Measure distance in units of 120 km. Let xⱼ be the distance from home to turnaround j in outward order. Write Sⱼ for the sum of the first j turnaround distances, with S₀ = 0.

到第 j 站分油后，前 j 辆支援车走完全程往返共需 2Sⱼ 箱，其中包含已留下的返程油。其余 n − j 辆各已用掉 xⱼ 箱，且各自最多还装 1 箱。总油量为 n 箱，所以 n ≤ 2Sⱼ + (n − j)xⱼ + (n − j)。At stop j, the first j helpers account for 2Sⱼ tanks for their complete round trips, including the fuel reserved for coming home. Each of the remaining n − j bikes has burned xⱼ tanks and can now hold at most one. Accounting for all n starting tanks gives n ≤ 2Sⱼ + (n − j)xⱼ + (n − j).
代入 xⱼ = Sⱼ − Sⱼ₋₁，整理得 (n − j + 2)Sⱼ ≥ j + (n − j)Sⱼ₋₁。右边随上一项增大而增大。等距方案每次都让续行车辆满油，所以从 S₀ = 0 起，每一步都恰好取等号。逐站比较，其他方案的 Sⱼ 都不会更小。Substitute xⱼ = Sⱼ − Sⱼ₋₁ and rearrange to get (n − j + 2)Sⱼ ≥ j + (n − j)Sⱼ₋₁. A larger previous sum can only raise the lower bound. The equal-leg plan fills every continuing tank, so it gives equality at each step. Starting at S₀ = 0, no other plan can have a smaller sum at any step.
所有油最终耗尽。支援车共耗 2Sₙ₋₁ 箱，所以留给甲全程消耗的油是 n − 2Sₙ₋₁ 箱。等距方案使 Sₙ₋₁ 最小，因而甲走得最远。All starting fuel is burned. The helpers use 2Sₙ₋₁ tanks on their round trips, leaving n − 2Sₙ₋₁ tanks for A’s whole journey. The equal-leg plan minimizes that sum and therefore maximizes A’s distance.

用程序验算Check the solution in code

每一箱用 n + 1 个整数格表示。程序逐辆扣除每段的耗油，逐次补给、留足返程，并检查油箱容量、最终余油和全车队用油总量。可下载的 Python 代码另用分数从油箱容量不等式算出上界，与实际跑出的最远距离核对。The browser counts fuel in integer shares. It deducts each leg bike by bike, performs every top-up, reserves the ride home, then checks capacities, empty tanks and fleet-wide fuel use. The downloadable Python program also derives an independent upper bound from tank capacities using exact fractions and compares it with the simulated route.

下载 Python 验证代码Download the Python verifier