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CIE 9709 2025 June Paper 15 Q6

A Level / CIE / P1

CIE 9709 2025 June Paper 15 Paper · Question 6

题目

Problem

Each year, on her birthday, Ananya receives some money from each of her parents.

On Ananya’s first birthday, her father gives her $10. Every subsequent year, her father gives her $5 more than he gave her the previous year.

On Ananya’s first birthday, her mother also gives her $10. Every subsequent year, her mother gives her 20%20\% more than she gave her the previous year.

(a) Show that on Ananya’s eleventh birthday she receives more from her mother than from her father.

[3]

(b) Find the total amount of money Ananya receives up to and including her eighteenth birthday.

[5]
题目中文翻译

每年生日时,Ananya 都会从父母双方各收到一些钱。

在 Ananya 的第一个生日,她的父亲给她 $10。之后每一年,她的父亲都比前一年多给她 $5。

在 Ananya 的第一个生日,她的母亲也给她 $10。之后每一年,她的母亲都比前一年多给她 20%20\%。

(a) 证明在 Ananya 的第 11 个生日,她从母亲那里收到的钱比从父亲那里收到的多。

(b) 求 Ananya 到第 18 个生日(含第 18 个生日)为止收到的总金额。

解答

(a)

解法一

思路

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父亲的礼金构成等差数列,母亲的构成等比数列。第十一项相对于首项经过十次递推。

答题过程

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On the eleventh birthday, the father gives 10+10(5)=6010+10(5)=60 dollars. The mother gives 10(1.2)10≈61.9210(1.2)^{10}\approx61.92 dollars. Since 61.92>6061.92>60, the mother gives more on that birthday.

(b)

解法一

思路

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十八次礼金都要累加。分别求两个数列的前十八项和,最后才把金额舍入到分。

答题过程

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The father’s total is

182(2(10)+17(5))=945.\frac{18}{2}\bigl(2(10)+17(5)\bigr)=945.

The mother’s total is 101.218−11.2−1≈1281.16666410\frac{1.2^{18}-1}{1.2-1}\approx1281.166664. Thus the combined total, to the nearest cent, is $2226.17.