NYJC/2021/JC1/Promo/P1/Q10

Mrs Tan wants to build a wooden fence using vertical planks of equal width and thickness but different lengths. In her plan, the length of the first wooden plank is 2 metres and the length of the planks forms a geometric progression. The length of the 10 th plank is $1.5$ metres.
(i) If the cost of wooden plank is $\$ 18$ per metre, show that the cost of the fence will never exceed $\$ 1200$.
Mrs Tan realised that her plan is not feasible and now wants to build her fence using several identical panels. Each panel comprises 10 vertical planks with identical dimensions to the first 10 planks in her original plan.

[4]

(ii) Find the total length, $T$ metres, of the planks to be used in each panel.

[2]

(iii) She hires a contractor to install the fence. The contractor misunderstands her instructions and uses 10 planks to construct a panel so that the lengths form an arithmetic progression with common difference $d$ metres. If the total length of the planks to be used for one panel is still $T$ metres and the length of the first plank is still 2 metres, find the value of $d$.

[2]

The contractor offers to paint the fence for Mrs Tan. He buys a 5-litre can of paint to do the paint job. To save costs, he fills the can to the 5-litre mark with turpentine to form a uniform mixture when the level of paint in the can falls to the 4-litre mark. He then repeats this process whenever the level of the mixture falls to the 4-litre mark.
(iv) State, in terms of $n$, an expression for the volume of paint remaining in the mixture after the $n$th refill.

[2]

(v) Find the minimum number of refills taken before the mixture is more than $80 \%$ turpentine.

[2]

(i) $u_{10}=1.5$
$a r^9=1.5, a=2$
$r^9=0.75$
$r=(0.75)^{\frac{1}{9}} \approx 0.96854$
$S_{\infty}=\frac{2}{1-(0.75)^{\frac{1}{9}}} \approx 63.574$ or $S_{\infty}=\frac{2}{1-0.96854} \approx 63.573$

Maximum amount to pay $<18 \times S_{\infty} \approx \$ 1144.33$ or $\$ 1144.31$ Hence the cost will not exceed \$1200.

(ii)

$T=S_{10}=\frac{2\left(1-0.75^{\frac{10}{9}}\right)}{1-0.75^{\frac{1}{9}}} \approx 17.394 \approx \approx 17.4 \mathrm{~m}$
or
$T=S_{10}=\frac{2\left(1-968.54^{10}\right)}{1 – 0.96854} \approx 17.394 \approx 17.4 \mathrm{~m}$

(iii) $\frac{10}{2}[2(2)+(10-1) d]=17.394$
$d \approx-0.0579 \mathrm{~m}$

(iv) Original volume of paint $=5$ litres
Let $u_n$ denotes the volume of paint in litres remaining after the $n$th refill.
Before the $1^{\text {st }}$ refill, the quantityofpaint is reduced by a factor of $\frac{4}{5}$.
$\therefore$ the volume of paint is reduced to $u_1= \frac{4}{5} \times 5$

$u_2=\frac{4}{5} \times u_1=\left(\frac{4}{5}\right)^2 \times 5$

$\therefore u_n=4\left(\frac{4}{5}\right)^{n-1}$

(v) If the mixture is more than $80 \%$ turpentine, then it is less than $20 \%$ paint, ie $0.2 \times 5=1$ litre of paint. $u_n=5\left(\frac{4}{5}\right)^n<1 \quad$ or $\quad u_n=4\left(\frac{4}{5}\right)^{n-1}<1$
Using $\mathrm{GC}$, minimum $n=8$
The minimum number of refills taken before the mixture is more than $80 \%$ turpentine is 8 .