IMC 2004 Q19
19. Three rectangular-shaped holes have been drilled passing all the way through a solid 3 × 4 × 5 cuboid. The diagrams show the front, side and top views of the resulting block.
What fraction of the original cuboid remains?
- A \(\dfrac{13}{30}\)
- B \(\dfrac{7}{15}\)
- C \(\dfrac{1}{2}\)
- D \(\dfrac{8}{15}\)
- E \(\dfrac{17}{30}\)
Intermediate Mathematical Challenge 2004, Question 19. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D \(\dfrac{8}{15}\)
Consider the cuboid to be made up of 60 unit cubes. The front and side views show that top and bottom layers consist of the same number of cubes and from the top view we see that this number is 14. The front and side views indicate that only the 4 corner cubes remain in the middle layer, so the total number of cubes remaining is 2 × 14 + 4 = 32. The required fraction, therefore, is \(\dfrac{32}{60} = \dfrac{8}{15}\).
Solution © UK Mathematics Trust, from the IMC 2004 solutions.