某同城速递平台给予骑手每单a元基础收益（a为整数），且该平台在鼓励骑手多劳多得的同时，还鼓励其关注个...

Question

某同城速递平台给予骑手每单a元基础收益（a为整数），且该平台在鼓励骑手多劳多得的同时，还鼓励其关注个人健康，在完成一定数量的订单后及时休息，并为此设定了每日基础接单量和健康临界接单量。骑手在完成基础接单量后，超出的订单每单收益平台补贴1.5元（a+1.5）；但若持续接单，超出健康临界接单量的订单每单收益平台扣除1元（a-1）。已知某骑手某日完成30单，共获得247.5元；第二日完成40单，共获得315元。若该骑手仅第二日超出了健康临界接单量，则扣除后的每单收益为（）元

Thought

Observing pattern shifts

I’m thinking through the delivery platform's pay structure. The base income is supplemented or subtracted based on order volume, encouraging efficiency while allowing for breaks after reaching certain thresholds.

Crunching numbers

I'm analyzing the rider's earnings over two days, factoring in both basic and health order thresholds. This involves calculating the number of orders in each bracket to determine the total.

Piecing together the puzzle

I'm noting that day 1's order earnings involve sums at different rates up to the basic threshold. Day 2's earnings exceed the health threshold, with some orders costing less.

Determining deduction

I’m working to identify the order deduction (a - 1) on day 2 by analyzing the total earnings and considering the feasibility of the threshold T1 being 30 or less.

Adjusting my approach

Simplifying day 1's earnings calculation, I find the equation for day 1 total as 30a + 1.5*(30 - T1). Day 2's total, 40 orders at T2 or more, is 315.

Revisiting day 2's equation

I’m working through day 2's total by reworking the order earnings, resulting in $40a - 40 + 2.5T2 - 1.5T1 = 315$ . This helps refine the calculation.

Solving the equations

I’m working with two equations to find integer solutions for $a$ , $T1$ , and $T2$ . The problem constraints and assumptions guide me in narrowing down the possibilities.

Adjusting my approach

I’m simplifying equation (1) by eliminating decimals to ensure integer solutions for $a$ , $T1$ , and $T2$ , while carefully ensuring the calculations are accurate and integers.

Reworking equations

I’m refining the system by clearing decimals and organizing terms. The rearranged equations now work towards pinpointing integer solutions for $a$ , $T1$ , and $T2$ .

Refining constraints

I’m substituting $T1$ into the simplified equation, leading to $T2 = 61 - 4a$ . Adjusting for $a < 8.25$ , I refine the constraints for $a$ .

Evaluating integer solutions

I'm noting that $a$ should be between 7 and 8. For $a = 7$ , both $T1$ and $T2$ meet the conditions, making it a valid solution.

Confirming calculations

I checked that for $a = 7$ and $a = 8$ , both $T1$ and $T2$ satisfied the conditions, and the total calculations for days 1 and 2 aligned exactly with the expected values.

Confirming logical fit

I’m noting that $a = 7$ , $T1 = 5$ , and $T2 = 33$ fit the requirement that day 1's 30 orders did not surpass $T2$ .

Reasoned about income calculation and parameters for 52 seconds

Answer

题意可简化为：设

$a$ = 每单基础收益（整数），
$T_1$ = 每日「基础接单量」阈值，
$T_2$ = 「健康临界」阈值（超过此阈值的订单单价要扣1元）。

那么对任意一天，如果接单数为 $n$ ，则总收益为：

前 $T_1$ 单：每单收入 $a$ ；
超过 $T_1$ 而不超过 $T_2$ 的部分：每单收入 $a + 1.5$ ；
超过 $T_2$ 的部分：每单收入 $a - 1$ 。

根据题目给出的信息

第一天接了 30 单，总收入 247.5 元。题中说「仅第二日」超过健康临界阈值，所以第一天并未超过 $T_2$ ，即
$30 \leq T_2.$
由于接单数达 30 单，大概率已超过基础接单量 $T_1$ ，故第一天的收入计算式为
$\underbrace{T_1 \times a}_{\text{前 }T_1\text{ 单}} \;+\; \underbrace{(30 - T_1)\times (a + 1.5)}_{\text{超过 }T_1\text{ 但未超 }T_2}.$
给出的总和是 247.5，因此有：
$T_1 \cdot a \;+\; (30 - T_1)\,(a + 1.5)\;=\;247.5. \tag{1}$
第二天接了 40 单，总收入 315 元。题目明确说第二天「超出了健康临界接单量 $T_2$ 」，所以 40 单中有一部分要扣 1 元/单，对应计算式：
$\underbrace{T_1 \times a}_{\text{前 }T_1\text{ 单}} \;+\; \underbrace{(T_2 - T_1)\,(a + 1.5)}_{\text{超过 }T_1\text{ 但未超 }T_2} \;+\;\underbrace{(40 - T_2)\,(a - 1)}_{\text{超过 }T_2\text{ 要扣 1 元}} \;=\;315. \tag{2}$
并且这一天之所以「超过了 $T_2$ 」，必然有
$40 > T_2.$

解方程

先将 (1) 式展开并整理：

\begin{aligned} T_1 \cdot a + (30 - T_1)\,(a + 1.5) &= T_1 a + (30 - T_1)a + (30 - T_1)\times 1.5 \\ &= 30a + 1.5 \,(30 - T_1) \\ &= 247.5. \end{aligned}

故

30a + 45 - 1.5\,T_1 = 247.5 \quad\Longrightarrow\quad 30a - 1.5T_1 = 202.5. \tag{1'}

再看 (2) 式：

\begin{aligned} T_1 a &\;+\; (T_2 - T_1)\,(a + 1.5) \;+\; (40 - T_2)\,(a - 1) = 315. \end{aligned}

把各部分展开并合并同类项，可得到

\begin{aligned} T_1 a &\;+\; (T_2 - T_1)\,a + (T_2 - T_1)\times1.5 \;+\; (40 - T_2)\,a - (40 - T_2)\times1\\ &= T_1 a + T_2 a - T_1 a + 40a - T_2 a \;+\;1.5(T_2 - T_1) \;-\; (40 - T_2)\\ &= 40a \;+\; 1.5\,(T_2 - T_1) \;-\; 40 \;+\; T_2\\ &= 40a - 40 \;+\; T_2 \;+\; 1.5 T_2 \;-\; 1.5 T_1\\ &= 40a - 40 \;+\; 2.5\,T_2 \;-\; 1.5\,T_1 \;=\; 315. \end{aligned}

即

40a - 40 + 2.5\,T_2 - 1.5\,T_1 \;=\; 315. \tag{2'}

于是我们获得方程组：

\begin{cases} 30a \;-\; 1.5\,T_1 = 202.5 \\ 40a \;-\; 40 + 2.5\,T_2 \;-\; 1.5\,T_1 = 315 \end{cases}

为去小数，可以把它们分别乘以 2：

$60a \;-\; 3\,T_1 = 405.$
$80a \;-\; 80 + 5\,T_2 \;-\; 3\,T_1 = 630.$ （注意题中原式 315 乘 2 得到 630）

不过通常做法是先用第一式解出 $T_1$ 并带入第二式。由

60a - 3\,T_1 = 405 \;\Longrightarrow\; 20a - T_1 = 135 \;\Longrightarrow\; T_1 = 20a \;-\; 135.

带入第二式(较简便的那种整理好的形式也行)后，结合「第一天未超 $T_2$ ⇒ $30 \le T_2$ 」「第二天超 $T_2$ ⇒ $40 > T_2$ 」等条件，即可搜索满足整数解的 $a,T_1,T_2$ 。

最终能同时满足的是：

$a = 7$
$T_1 = 5$
$T_2 = 33$

并且符合「第一天没有超 $T_2=33$ 」「第二天确实超了 33 单」。代入可验算：

第一天（30 单未超健康阈 33）： $5\times 7 + (30 - 5)\times (7 + 1.5) = 35 + 25\times 8.5 = 35 + 212.5 = 247.5.$
第二天（40 单，已超健康阈 33）： $5\times 7 \;+\; (33-5)\times(7+1.5) \;+\;(40-33)\times(7-1) = 35 + 28\times 8.5 + 7\times 6 = 35 + 238 + 42 = 315.$

完全符合题意。

最后回答

题目问：「若该骑手仅第二日超出了健康临界接单量，则扣除后的每单收益为多少？」

根据解出的 $a=7$ ，超出健康阈值时平台每单要扣 1 元，故每单收益为

\boxed{\,7 - 1 = 6\,}.

Question

Thought

Answer

根据题目给出的信息

解方程

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