Miscellaneous Logical Puzzles Questions and Answers

Answer: C

The opposite face shows 3.

On a standard die, opposite faces always add up to 7: 1 is opposite 6, 2 opposite 5, and 3 opposite 4. So when 4 is on top, 3 is on the bottom.

1, 2 and 6 are opposite 6, 5 and 1 respectively, and all of those are on the side faces when 4 is on top.

Answer: A

The statement that must be true is C is adjacent to E.

Each face of a cube touches four others and is opposite one. A touches B, D and F, so its fourth neighbour and its opposite face are C and E in some order. Whichever of C and E is opposite A, the other is one of A's neighbours, and a face opposite A touches all four of A's neighbours. So C and E are always adjacent.

'C is opposite E' is therefore impossible. 'B is opposite F' and 'A is adjacent to C' may or may not hold, depending on how the remaining faces are placed.

Answer: B

The number opposite 3 is 4.

Numbers seen together on a picture are adjacent, never opposite.

• 6 is seen with 3, 2, 4 (I, II) and 5 (III), so 6 is opposite 1.
• 2 is seen with 6, 3, 4 (I, II) and 1 (IV), so 2 is opposite 5.

The remaining pair, 3 and 4, must be opposite each other. 1 is opposite 6 and 5 opposite 2, and the data are sufficient.

A die has three pairs of opposite faces; once two pairs are found, the third is whatever is left.

Answer: A

The circle will have 16 radii.

Each diameter is made of two radii, and dividing a circle into 16 equal sectors needs 8 diameters, each passing through the centre. Those 8 diameters give \(8 \times 2 = 16\) radii, one bounding line for each of the 16 parts, just as a cake cut into 16 slices has 16 cut lines from the centre.

32 doubles the count by treating each part as having two radii of its own, forgetting that neighbouring parts share them. 64 and 24 match no construction.

Answer: C

The rope must be cut 11 times.

On a single straight rope, \(n\) cuts give \(n + 1\) pieces. When the rope is folded double, every cut goes through two strands, so each cut adds two pieces:

\[ \begin{aligned} 2n + 1 &= 23 \\ n &= 11 \end{aligned} \]

22 cuts would be needed on a single rope, so 23 confuses pieces with cuts. 10 cuts give only 21 pieces, and 9 cuts only 19.

Answer: C

The figure holds 17 triangles.

Call the rectangle PQRS (P top left, Q top right, R bottom right, S bottom left), the left tip L, the midpoint of PS as M, the middle of QR as N and the point where the diagonals meet O. Count by kind:

Triangles with O as apex on a side of the rectangle: POQ, QOR, ROS, SOP, which is 4.

Triangles cut by one diagonal: PQR, PQS, SQR, SRP, which is 4.

Triangles split by the horizontal line MN: QON, RON, MOP, MOS, which is 4.

Triangles using the left tip: LMP, LMS, LPS, LPO, LSO, which is 5.

\[ 4 + 4 + 4 + 5 = 17 \]

The listing has 17 distinct triangles with none counted twice, so 15, 16 and 18 are miscounts.

Answer: B

The figure holds 20 rectangles, counting squares and tilted ones.

The central square is split into four small squares by a horizontal and a vertical line. Upright rectangles: 4 small squares, 4 made of two small squares, and the whole central square, which is 9.

Tilted rectangles: the four large tilted squares at the corners, and the tilted square joining the midpoints of the central square's sides, which is 5. Chaining these tilted squares along a diagonal direction gives rectangles two squares long (four of them) and three squares long (two of them), which is 6 more.

\[ 9 + 5 + 6 = 20 \]

Counting only the upright ones gives 9, and adding the four corner squares gives 13. Both miss the tilted ones.

Answer: A

The figure holds 21 triangles.

There are three triangles of the same shape, with apexes at the top of the centre line and at two lower points on it. Count the triangles on each apex:

Top apex: the whole triangle and its two halves cut by the centre line, which is 3.

Middle apex: the whole triangle with two halves, and the smaller triangle above the first horizontal line with two halves, which is 6.

Lowest apex: the whole triangle with two halves, and the smaller triangle above the next horizontal line with two halves, which is 6.

Then the small triangles that sit against the left and right sides of the rectangle between a slanting line and the side: 3 on each side, which is 6.

\[ 3 + 6 + 6 + 6 = 21 \]