题目

Problem

Wheat is grown on a farm.

In year 1, the farm produced $300$ tonnes of wheat.
In year 12, the farm is predicted to produce $4000$ tonnes of wheat.

Model A assumes that the amount of wheat produced on the farm will increase by the same amount each year.

(a) Using model A, find the amount of wheat produced on the farm in year 4.

Give your answer to the nearest $10$ tonnes.

(3)

Model B assumes that the amount of wheat produced on the farm will increase by the same percentage each year.

(b) Using model B, find the amount of wheat produced on the farm in year 2.

Give your answer to the nearest $10$ tonnes.

(3)

(c) Calculate, according to the two models, the difference between the total amounts of wheat predicted to be produced on the farm from year 1 to year 12 inclusive.

Give your answer to the nearest $10$ tonnes.

(3)

题目中文翻译

某农场种植小麦。

第 1 年，农场生产了 $300$ 吨小麦。
预计第 12 年，农场将生产 $4000$ 吨小麦。

模型 A 假设农场每年的小麦产量增加相同的数量。

(a) 使用模型 A，求第 4 年的小麦产量，答案取最接近的 $10$ 吨。

（3分）

模型 B 假设农场每年的小麦产量增加相同的百分比。

(b) 使用模型 B，求第 2 年的小麦产量，答案取最接近的 $10$ 吨。

（3分）

解答

(a)

解法一

思路

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Model A 是等差数列。year 1 是第 $1$ 项，year 12 是第 $12$ 项，所以从第 $1$ 项到第 $12$ 项一共有 $11$ 个公差。

答题过程

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Let the common difference be $d$ .

Since year 1 is $300$ tonnes and year 12 is $4000$ tonnes,

\begin{align*} 4000 =&\, 300+11d\\ 11d =&\, 3700\\ d =&\, \frac{3700}{11}. \end{align*}

Year 4 is the fourth term:

\begin{align*} u_4 =&\, 300+3d\\ =&\, 300+3\bigg(\frac{3700}{11}\bigg)\\ =&\, 1309.09\ldots \end{align*}

To the nearest $10$ tonnes, the amount is $1310\text{ tonnes}$ .

解法二

思路

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利用线性插值法。第 1 项为 $300$ ，第 12 项为 $4000$ 。第 1 年到第 4 年共有 3 个间隔，整个第 1 年到第 12 年共有 11 个间隔，所以前一增量是总增量的 $\frac{3}{11}$ 。使用公式 $u_4 = u_1 + \frac{3}{11}(u_{12}-u_1)$ 可以直接算出，无需算出公差 $d$ 。

答题过程

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Using the property of arithmetic sequences, the change from year 1 to year 4 is $\frac{3}{11}$ of the total change from year 1 to year 12:

\begin{align*} u_4 =&\, u_1 + \frac{3}{11}(u_{12}-u_1)\\ =&\, 300 + \frac{3}{11}(4000-300)\\ =&\, 300 + \frac{3}{11}(3700)\\ =&\, 300 + \frac{11100}{11}\\ =&\, 1309.09\ldots \end{align*}

To the nearest $10$ tonnes, the amount of wheat is $1310\text{ tonnes}$ .

(b)

解法一

思路

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Model B 是等比数列。year 12 是 $300r^{11}$ ，先求公比 $r$ ，再用 $300r$ 求 year 2。

答题过程

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Let the common ratio be $r$ .

Then

\begin{align*} 4000=&\,300r^{11}\\ r^{11}=&\,\frac{4000}{300}\\ r=&\,\bigg(\frac{4000}{300}\bigg)^{\frac1{11}}\\ =&\,1.2655\ldots \end{align*}

Year 2 is

\begin{align*} 300r=&\,300(1.2655\ldots)\\ =&\,379.65\ldots \end{align*}

To the nearest $10$ tonnes, the amount is

\begin{align*} 380\text{ tonnes}. \end{align*}

(c)

解法一

思路

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分别求两个模型从 year 1 到 year 12 的总和。Model A 用等差数列求和，Model B 用等比数列求和，最后相减。

答题过程

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For Model A,

\begin{align*} S_A =&\, \frac{12}{2}(300+4000)\\ =&\, 25800. \end{align*}

For Model B,

\begin{align*} S_B =&\, \frac{300(r^{12}-1)}{r-1}, \end{align*}

where

\begin{align*} r = \bigg(\frac{4000}{300}\bigg)^{\frac1{11}}. \end{align*}

\begin{align*} S_B =&\, \frac{300(1.2655\ldots^{12}-1)}{1.2655\ldots-1}\\ =&\, 17935.20\ldots \end{align*}

The difference is

\begin{align*} 25800-17935.20\ldots =&\, 7864.79\ldots \end{align*}

To the nearest $10$ tonnes, the difference is $7860\text{ tonnes}$ .

解法二

思路

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列举法（Listing values）。由于总共只有 12 年，我们可以在草稿纸上直接计算两个模型在每一年对应的值并求和： Model A 公差为 $d \approx 336.36$ ，Model B 公比为 $r \approx 1.2655$ 。把 12 年的数据累加，或直接计算两模型差值的总和。

答题过程

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Using the unrounded values $d=\dfrac{3700}{11}$ and $r=(40/3)^{1/11}$ , calculate all twelve years. Values below are displayed to four decimal places; the sums use unrounded values.

Year	Model A (tonnes)	Model B (tonnes)
1	300.0000	300.0000
2	636.3636	379.6544
3	972.7273	480.4582
4	1309.0909	608.0268
5	1645.4545	769.4668
6	1981.8182	973.7715
7	2318.1818	1232.3220
8	2654.5455	1559.5215
9	2990.9091	1973.5972
10	3327.2727	2497.6160
11	3663.6364	3160.7696
12	4000.0000	4000.0000

Adding all twelve terms in each model gives

\begin{align*} S_A =&\,25800,\\ S_B =&\,17935.203872\ldots,\\ S_A-S_B =&\,7864.796127\ldots. \end{align*}

To the nearest $10$ tonnes, the difference is $\boxed{7860\text{ tonnes}}$ .

P2 2024 Jan Q7

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

解法二

思路

答题过程