Positive ribbon lengths and a bounded GP

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? marks

After she cuts that number, Bala continues cutting from the remaining long roll, first 160cm and each successive piece pp times the preceding length, where 0<p<10<p<1.

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

(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. marks

[2022/NJC/Prelim/I/7]

Total marks: 8

Solution

(a)

Amy’s positive-length pieces

Find the maximum number of positive-length pieces Amy can cut.

Hint

  1. Use ℓn>0\ell_n>0, not ℓn≥0\ell_n\ge0.

Answer · (a)

2020 pieces.

(b)

Bala’s ratio and piece count

Find the largest pp, then the maximum pieces Bala can cut before his total exceeds 9.5 m.

Explore this idea

Hint

  1. Ten metres is 1000 cm, not 10 cm.

  2. “Before exceeding” gives an upper bound on n.

Answer · (b)

Largest p=0.84p=0.84; most pieces 1717.

Explore the idea

The last permitted ribbon piece

Use p = 0.84 and measure Bala’s total in centimetres.

Piece countn = 17
Try an example
Bala total (cm) at n = 17
948.3883
Key idea

Seventeen pieces total 948.388 cm; eighteen would total 956.646 cm.

Ideas to try
Compare the values on either side of the boundary. What makes one count qualify?
(c)

Original roll length

Find the original roll length, including the 4.5 cm left over.

Hint

  1. The original length includes the uncut 4.5 cm.

Answer · (c)

2633 cm2633\,\mathrm{cm}.

Takeaway

Use a term condition for lengths and a sum condition for totals

Both people cut from the same roll, so account for all used ribbon plus the 4.5 cm left.