2018 H2 Maths Paper 1 Q11 — Worked Solution

Part (i)

Let $r = 1 + \frac{a}{100}$ be the factor by which the total in the account is multiplied at the end of each month, so that with $a = 0.2$ , $r = 1.002$ .

Part (a)

The $100 invested on 1 January 2016 is increased by $0.2\%$ at the end of each of the $12$ months from January to December 2016, so its value at the end of 31 December 2016 is

100r^{12} = 100(1.002)^{12} = \$ 102.43\ \ \text{(to the nearest cent).}

$\therefore$ the $100 is worth $\$ 102.43$ at the end of 31 December 2016.

Part (b)

Approach

The deposit made on 1st of month $n$ (with January $\text{=}1$ ) sits in the account through the growth applied at the end of months $n,n + 1,\ldots,12$ , so it has been multiplied by $r$ a total of $13 - n$ times by the end of December. Summing the $12$ deposits gives a geometric series in $r$ .

Since the deposit made on the 1st of month $n$ is worth $100r^{13 - n}$ at the end of December, the total at the end of 31 December 2016 is

\sum_{n = 1}^{12}100r^{13 - n} = 100\left( r + r^{2} + \cdots + r^{12} \right).

This is a geometric series with first term $r$ , common ratio $r$ and $12$ terms, so

\text{Total} = \frac{100r\left( r^{12} - 1 \right)}{r - 1}.

By GC,

\text{Total} = \frac{100(1.002)\left\lbrack (1.002)^{12} - 1 \right\rbrack}{0.002} = \$ 1215.71\ \ \text{(to the nearest cent).}

$\therefore$ the total in the account at the end of 31 December 2016 is $\$ 1215.71$ .

Part (c)

Approach

With deposits continuing every month, the total at the end of the $n$ th month of investing is $T_{n} = \frac{100r\left( r^{n} - 1 \right)}{r - 1}$ , the same series as in (b) with $12$ replaced by $n$ . We use the GC to find the smallest $n$ for which $T_{n}$ exceeds $3000$ . To decide whether this happens on the first or last day of that month, we also check the amount right after the deposit is made (before that month’s growth is applied) against $3000$ : if that alone already exceeds $3000$ , the crossing happens on the first day; otherwise it only happens once that month’s growth is added, on the last day.

By GC, with $T_{n} = \frac{100r\left( r^{n} - 1 \right)}{r - 1}$ ,

T_{29} = 2988.65\ldots < 3000,\quad\quad T_{30} = 3094.82\ldots > 3000.

So the total first exceeds $\$ 3000$ during the $30$ th month of investing, which (counting January 2016 as month $1$ ) is June 2018.

To decide first day or last day, we check the amount immediately after the deposit on 1 June 2018, before that month’s growth is applied. This is the balance at the end of May 2018, $T_{29}$ , plus the new $\$ 100$ deposit:

T_{29} + 100 = 2988.65\ldots + 100 = 3088.65\ldots > 3000.

Note: this already exceeds $\$ 3000$ before June’s growth is credited.

$\therefore$ the total first exceeds $\$ 3000$ in June 2018, and it does so on the first day of the month, immediately after the deposit is made (before that month’s $0.2\%$ growth is applied).

Part (ii)

Part (a)

The $100 invested on 1 January 2016 earns a fixed bonus of $ $b$ at the end of each of the $12$ months it is held in 2016 (January to December), and these bonuses do not themselves earn further bonus, so its value at the end of 31 December 2016 is

\$(100 + 12b).

(b)

Approach: The $100 deposited on the 1st of month $i$ (with January 2016 $\text{=}1,\ldots,$ December 2017 $\text{=}24$ ) is held from month $i$ to month $24$ inclusive, earning a bonus at the end of each of those $25 - i$ months. Summing over $i = 1,\ldots,24$ gives the total bonus as $b$ times an arithmetic series, and adding the $24$ principal payments of $100 gives the total value.

The total bonus earned by the deposit made in month $i$ is $(25 - i)b$ , so the total bonus over all $24$ deposits is

\sum_{i = 1}^{24}(25 - i)b = b\sum_{k = 1}^{24}k = b \cdot \frac{24(25)}{2} = 300b,

using $k = 25 - i$ . The total principal invested is $24(100) = 2400$ , so the total value at the end of 31 December 2017 is $2400 + 300b$ . Setting this equal to $2800$ ,

2400 + 300b = 2800

300b = 400

b = \frac{4}{3}.

$\therefore$ $b = \frac{4}{3}$ , i.e. $\$ 1.33$ (to the nearest cent).

(iii)

Approach: With $a = 1$ for plan $P$ , the total after $60$ months of investing is the same GP sum as in (i)(b)/(i)(c) with $r = 1.01$ and $n = 60$ . For plan $Q$ , the same argument as in (ii)(b) applies with $24$ replaced by $60$ : the total bonus is $b$ times $\sum_{k = 1}^{60}k$ . Equating the two totals gives $b$ .

For plan $P$ with $a = 1$ , $r = 1.01$ , the total at the end of the $60$ th month is

T_{P} = \frac{100(1.01)\left\lbrack (1.01)^{60} - 1 \right\rbrack}{0.01}.

By GC, $T_{P} = 8248.64$ (to the nearest cent).

For plan $Q$ , by the same reasoning as (ii)(b) with $24$ replaced by $60$ , the total bonus is $b\sum_{k = 1}^{60}k = b \cdot \frac{60(61)}{2} = 1830b$ , and the total principal is $60(100) = 6000$ , so the total value at the end of the $60$ th month is $6000 + 1830b$ . Setting the two plans equal,

6000 + 1830b = 8248.64\ldots

1830b = 2248.64\ldots

b = 1.2288\ldots

$\therefore$ $b = \$ 1.23$ (to the nearest cent).

2018 H2 Maths Paper 1, Question 11

Part (i)

Part (a)

Part (b)

Part (c)

Part (ii)

Part (a)

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