Question
Four framed prints are labelled with their measured widths: 18 cm, 21 cm, 24 cm and 15 cm. A conservator wants them side by side on an exactly 80 cm shelf, with no gaps or overlaps. Each label was produced by rounding the true width to the nearest whole centimetre; an exact half-centimetre is rounded up.
The labels add to 78 cm, but none of the original measurements survives. Can the conservator guarantee that all four prints will fit? Would the same guarantee hold if the shelf were only 79 cm long? Justify both answers without assuming the rounding errors cancel.
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Hint 1
Every labelled width permits a range of true widths, not just its printed integer.
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Solution
A width labelled \(k\) cm can be any value in the half-open interval \([k-0.5,k+0.5)\) cm: for instance, a width of exactly 18.5 cm would be labelled 19, not 18. The four true widths therefore lie respectively in
\[
[17.5,18.5),\quad [20.5,21.5),\quad [23.5,24.5),\quad [14.5,15.5).
\]
Their total \(T\) lies in \([76,80)\) cm. All four rounding errors can point in the same direction, so adding only the displayed widths would not give a guarantee. Nevertheless, the upper endpoint 80 is excluded by the stated tie rule. Thus \(T<80\) for every admissible set of widths, and the 80 cm shelf is guaranteed to work.
The shorter 79 cm shelf is not guaranteed: widths 18.4, 21.4, 24.4 and 15.4 cm all round to their respective labels, yet total \(79.6\) cm. Conversely, the labels themselves could be exact true widths, totalling 78 cm, which would fit. Hence the 79 cm fit is genuinely undetermined by the measurements. The tie rule determines whether the total can equal 80 cm; either way, the 80 cm shelf fits the frames.
Assumptions
The shelf length is exact, and the frames need no gaps.
Only the four printed widths are known; an exact half-centimetre rounds up.
Final Answer
All four prints are guaranteed to fit on the 80 cm shelf, but not on a 79 cm shelf. Their possible combined width is in \([76, 80)\) cm.