Miscellaneous Logical Puzzles Questions and Answers

Answer: A

The view from Y is (A).

Seen from direction Y, the column holding the upper cube appears as two squares stacked on the right, and the other column as a single square on the left. The small cube rests on the lower layer directly in front of the upper cube, on the side nearer the viewer, so it hides part of the upper cube's face and shows as a small square at the bottom of the upper square.

In Figure A the small cube sits against the half of the upper cube that is on the right as seen from Y (Figure B confirms it occupies only a corner of its quarter), so the small square sits at the lower right of the upper square, as in (A). (C) puts it on the wrong side, (D) puts it outside the upper cube, and (B) shows a full-height dividing line that no edge of the assembly produces.

Answer: A

The minimum angle is 120°.

A body diagonal of a cube passes through two opposite vertices. At each of those vertices three edges and three faces meet, arranged symmetrically around the diagonal. A rotation of one third of a turn, 120°, carries each of these edges and faces onto the next, so the cube looks the same.

A 60° rotation would need six-fold symmetry about the diagonal, which the three edges at a vertex do not have. 90° is the symmetry about an axis through the centres of opposite faces, and 180° is not the smallest such rotation.

Answer: B

The minimum number of guards is 3.

The walkways form a single loop of 5 junctions and 5 walkways. A guard at a junction watches exactly the 2 walkways meeting there, so 2 guards can watch at most 4 walkways, which is not enough for 5. At least 3 guards are needed.

Three are enough. Label the roof apex A, the eave corners B and C, and the floor corners D and E. Guards at A, D and E watch AB and AC (from A), BD and DE (from D), and CE (from E): all five walkways. So 2 is too few, and 4 or 5 are more than the minimum.

Answer: A

The correct view is drawing (A).

Looking along the arrow, at the end of the long lower block, that block shows as a plain rectangle: its end face is flat, and the upper block sits on top of it, so no line of the upper block can cross this rectangle. The upper block appears above it in its trapezoidal cross-section. In the picture its face towards the viewer has a sloping edge on the near-left side and an almost vertical edge on the far side, so in the view the trapezoid has its sloping side on the left and its vertical side on the right, standing towards the right of the rectangle.

Drawings (B) and (D) carry a sloping line across the lower rectangle, which the flat end face cannot show. Drawing (C) puts the sloping side of the trapezoid on the right, which is the mirror image of the correct view.

Answer: A

The minimum time is 6 hours 30 minutes.

The arrows leave little choice. From R the only route is to Q; from Q the only route is to P; from P the person can go straight to S; and from S there is a direct route back to R. So the journey is R, Q, P, S, R, and the time is set by the waits for connecting trains:

1. R: first train at 7 AM, leave 7:00, reach Q at 8:00.
2. Q: trains at 5, 7, 9 AM (every 120 min); wait for 9:00, reach P at 10:00.
3. P: trains at 8:00, 9:30, 11:00 (every 90 min); wait for 11:00, reach S at 12:00.
4. S: trains from 8:00 every 45 min, so 11:45 then 12:30; leave 12:30, reach R at 1:30 PM.

From 7:00 AM to 1:30 PM is 6 hours 30 minutes. The shorter options count only the four hours of travel plus some of the waits; every connection here forces a wait, and none can be skipped because each city has only one useful outgoing route.

Answer: B

The shadow that is NOT possible is Q.

A parallel beam casts a shadow that is the parallel projection of the solid onto the screen. A pyramid is a convex solid, and the projection of a convex solid is always a convex region: any straight segment between two points of the shadow stays inside the shadow. So every possible shadow of the pyramid is a convex polygon.

Shadow Q has a notch cut into its upper edge, a re-entrant corner, so it is not convex and cannot be cast. The others can: looking straight along the axis gives the square S, and tilted orientations give other convex outlines, five-sided like P or four-sided like R, depending on which vertices end up on the boundary of the shadow.

Answer: B

The minimum number of guards is 3.

The walkways form a single closed loop of 5 junctions and 5 walkways. Each guard covers at most the 2 walkways that meet at his junction, so 2 guards can watch at most 4 walkways, which is not enough for 5.

Three guards suffice: number the junctions 1 to 5 round the loop and place guards at 1, 3 and 4. Guard 1 covers walkways 5-1 and 1-2, guard 3 covers 2-3 and 3-4, and guard 4 covers 4-5.

4 and 5 guards also work, but they are not the minimum.

Answer: D

There are 28 triangles in the figure.

Take each square as 2 units wide, so the whole figure is 4 by 2. Count by size:

\[ 8 + 4 + 4 + 2 + 8 + 2 = 28 \]

Counting only the small pieces gives 12, and missing the triangles that span both squares gives 18 or less.

Answer: C

The ring returns to child-1 after 30 passes.

The k-th pass moves the ring k places (1 to 2 is one place, 2 to 4 is two, 4 to 7 is three). After n passes it has moved \(1 + 2 + \dots + n = n(n+1)/2\) places, and it is back with child-1 when that total is a multiple of 31.

\[ \begin{aligned} \frac{n(n+1)}{2} &\equiv 0 \pmod{31} \\ n = 30 &: \frac{30 \times 31}{2} = 465 = 15 \times 31 \end{aligned} \]

Since 31 is prime, it must divide n or n + 1, so the smallest n is 30. n = 31 also works (31 × 32/2 = 496 = 16 × 31), but 30 comes first, and 28, 29 and 32 do not give a multiple of 31.

Answer: D

The figure holds 28 triangles.

Each of the three cells is a rectangle cut by both diagonals. Inside one cell there are 4 small triangles (the four parts the diagonals make) and 4 larger ones, each made of two small parts, with a side of the cell as the base. That is 8 per cell.

\[ 3 \times 8 = 24 \]

Triangles also span two neighbouring cells. Take the top edge of two adjacent cells as the base, with the apex at the bottom point where the diagonals meet on the shared line. The same works with the bottom edge and the apex at the top. That is 2 for each pair of neighbouring cells, and there are 2 such pairs.

\[ 24 + 2 \times 2 = 28 \]

No triangle spans all three cells, since the outer diagonals do not meet over a common apex.