Positive ribbon lengths and a bounded GP · source reference

Printed question

Amy cuts pieces from a long roll of ribbon. The first is 160cm and each successive piece is 8cm shorter than the preceding one.

(a) What is the maximum number of pieces she can cut? [2]

After she cuts that number, Bala continues cutting from the remaining long roll, first 160cm and each successive piece $p$ times the preceding length, where $0<p<1$.

(b) Given that Bala’s total can never exceed 10m regardless of the number cut, find the largest $p$. Taking that value and solving an appropriate inequality, find the maximum number he can cut before his total exceeds 9.5m. [4]

(c) After Amy cuts the number in (a), and Bala the number in (b), 4.5cm remains. Determine the original roll length, to the nearest cm. [2]

Printed working

(a) $160+(n-1)(-8)>0\Rightarrow168-8n>0\Rightarrow n<21$. At most 20 pieces.

(b) $S_{\infty}\le1000\Rightarrow\frac{160}{1-p}\le1000\Rightarrow160\le1000(1-p)\Rightarrow0<p\le0.84$. Largest $p=0.84$.

For total not exceeding 9.5m,

$$\frac{160(1-0.84^n)}{1-0.84}\le950\Rightarrow160(1-0.84^n)\le152\Rightarrow0.84^n\ge0.05,$$ $$n\ln0.84\ge\ln0.05\Rightarrow n\le\frac{\ln0.05}{\ln0.84}\Rightarrow0<n\le17.18.$$

At most 17 pieces, since $n$ is integer.

(c) Total roll length

$$\frac{20}2[2(160)+19(-8)]+\frac{160(1-0.84^{17})}{1-0.84}+4.5=1680+948.388+4.5=2632.888\mathrm{cm}\approx2633\mathrm{cm}.$$

Teacher notes

Amy: AP, first term 160, common difference −8; $u_n=a+(n-1)d$ is the $n$th length and must be positive.

Bala: GP, first term 160, ratio $p$, infinite sum $a/(1-r)$.

Change the sign when dividing/multiplying an inequality by a negative value; $\ln0.84<0$.

Total ribbon length = 20 Amy pieces + 17 Bala pieces + 4.5cm remainder.

Printed answer key

(a) 20; (b) 0.84; 17; (c) 2633cm (nearest cm)

Mathematical review and qualifications

(a) $160-8(n-1)>0$ gives $n<21$, so at most 20 pieces. Their total is $20(160+8)/2=1680$ cm.

(b) $160/(1-p)\le1000$ gives $p\le0.84$, so maximum $p=0.84$. The condition $1000(1-0.84^n)\le950$ gives $0.84^n\ge0.05$, hence $n\le\ln0.05/\ln0.84\approx17.18$. Maximum $n=17$.

(c) The roll length is $1680+1000(1-0.84^{17})+4.5\approx2632.888$ cm, so 2633 cm.

A zero-length twenty-first piece is not a ribbon piece; convert metres to centimetres before combining totals.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2022/NJC/Prelim/I/7] Question PDF pages 4; solution PDF pages 17. The original files remain private QA records outside this native content.