IMAT 2015 Q20 [Butterfly Cases]

A local museum wishes to exhibit a collection of butterflies which is mounted in nine narrow (only 0.2 m wide) display cases, each 1.5 m long. The museum wants to arrange four tables, each 2 m long and 1 m wide, in such a way that all the display cases can be placed around the edges. The room for the exhibition is 6 m by 6 m and there must be at least 1 m of clear floor space around the outside of the tables.

Which one of the five arrangements shown would be satisfactory?

Simple steps to solve word problems:

  • Underline key information
  • Determine what they are trying to ask, and what you will need to solve it
    • Eliminate any non-essential information
  • Draw a picture, graph, or equation
    • In moments of high stress like exam taking, always work with the paper they give you to avoid careless mistakes.
  • Solve.

This is a question that requires you to find the perimeter of these tables. We require a perimeter large enough to fit all of the display cases without them overlapping, so all we need to do is find the arrangement that: 1) fits the display cases; 2) Fits in the room; 3) leave an area of 1m surrounding the room.
Approach: the easiest approach is to draw on your exam paper in this case and just label out all the dimensions.

Limits:

  • (9 display cases)(1.5m long) = 13.5m
    • So we need a perimeter of at least 13.5m on tables. Note that the display cases are 0.2m wide, so we may need to consider this depending on how the cases are arranged.
  • The room is 6m by 6m, but we need a space of 1m all around so the constraint we have is that the tables must fit withing a 4m by 4m space. This means we can have up to 2 horizontally facing tables and 1 vertically facing table.

A) This is going to be too small. There will not be enough place for all 9 display cases. Also, notice how each side is 3m? It will not even fit 2 display cases aside because we need to factor in width. Each side would need to be an extra 0.4m (0.2m on each side) to accommodate for that extra case room. Therefore A is incorrect.

B) Easiest option to eliminate. Does not fit within the room size limit. Therefore B is incorrect.

C) If you haven’t eliminated all the other options first, let’s plan out the case arrangement. 2 cases fit on the top side (4m side), and this also gives a bit extra room, so the widths of the other display cases on other sides will not be a problem. Next, we can add 2 cases a side on the right and left. This brings us to 6 total so far. Finally, we can add a case to each interior side (all $2$m), which gives us a grand total of 9 cases. Therefore C is correct.

D) We can only fit a max of 2 display cases on the 4m sides and 1 case on the 2m side, giving us only a total of 6 cases. Therefore D is incorrect.

E) Also does not fit within room size limit because we need walking space to go around. Therefore E is also incorrect.

\fcolorbox{red}{grey!30}{Therefore C is the correct answer.}
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