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)

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\\[4mm] 11d =&\,\, 3700\\[4mm] d =&\,\, \frac{3700}{11}. \end{align*}

Year 4 is the fourth term:

\begin{align*} u_4 =&\,\, 300+3d\\[4mm] =&\,\, 300+3\left(\frac{3700}{11}\right)\\[4mm] =&\,\, 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 4 项到第 1 项的差距是第 12 项到第 11 项的 $\frac{3}{11}$ 。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)\\[4mm] =&\,\, 300 + \frac{3}{11}(4000-300)\\[4mm] =&\,\, 300 + \frac{3}{11}(3700)\\[4mm] =&\,\, 300 + \frac{11100}{11}\\[4mm] =&\,\, 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=&\,\left(\frac{4000}{300}\right)^{\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

展开

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)\\[4mm] =&\,\, 25800. \end{align*}

For Model B,

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

where

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

\begin{align*} S_B =&\,\, \frac{300(1.2655\ldots^{12}-1)}{1.2655\ldots-1}\\[4mm] =&\,\, 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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We can list the values of wheat produced in each of the 12 years:

Model A (AP): $a = 300, d = \frac{3700}{11} \approx 336.36$
- Year 1: $300$
- Year 2: $636.36$
- …
- Year 12: $4000$
- Sum $S_A = 25800$
Model B (GP): $a = 300, r = \left(\frac{40}{3}\right)^{1/11} \approx 1.2655$
- Year 1: $300$
- Year 2: $379.66$
- Year 3: $480.47$
- …
- Year 12: $4000$
- Sum $S_B = 300 + 379.66 + 480.47 + \dots + 4000 \approx 17935.2$

The total difference is:

\begin{align*} \text{Difference} = 25800 - 17935.2 = 7864.8 \end{align*}

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

IAL 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