Miscellaneous Logical Puzzles Questions and Answers

Answer: B

The minimum number of guards is 3.

The walkways form one closed loop of five walkways joining five junctions in a ring. A guard at a junction covers exactly the two walkways meeting there, so two guards can cover at most four walkways. Five walkways therefore need at least three guards.

Three are enough. Number the junctions 1 to 5 round the loop and place guards at 1, 3 and 4: walkway 1-2 is watched from 1, walkways 2-3 and 3-4 from 3, walkway 4-5 from 4, and walkway 5-1 from 1. Any choice of three junctions that leaves no walkway with both ends unguarded works.

So 2 is too few, and 4 or 5 guards would work but are not the minimum.

Answer: C

The minimum number of steps is 9.

Each step swaps two neighbouring coins, so it changes the relative order of exactly one pair of coins. In the starting arrangement H H H T T T every H stands to the left of every T; in the final arrangement T T T H H H every T stands to the left of every H. Each of the 3 × 3 = 9 head-tail pairs has to change its order, and one step can reverse only one pair, so at least 9 steps are needed.

Nine steps are also enough: move the first T leftwards past three H coins (3 steps), then the second T past the three H coins (3 steps), then the third (3 steps).

3 and 6 are too few because they reverse only some of the nine pairs, and 12 is more than the minimum.

Answer: C

The pattern that cannot be built is (C), the 3 by 5 rectangle.

Both building blocks contain exactly 4 tiles, and pieces may not overlap, so any pattern made only from them must contain a multiple of 4 tiles. Count the tiles in each option:

• (A) 2 × 4 = 8 tiles: two 2 × 2 squares side by side, or two lines of 4 stacked.
• (B) 3 × 4 = 12 tiles: three horizontal lines of 4.
• (C) 3 × 5 = 15 tiles: not a multiple of 4.
• (D) 4 × 5 = 20 tiles: five vertical lines of 4.

Since 15 is not divisible by 4, no combination of the two patterns can cover the 3 by 5 rectangle exactly, while each of the others is built as shown.

Answer: A

The minimum is 3 cuts.

Make one straight cut through the middle parallel to each pair of opposite faces: the first halves the cube, the second (at right angles) gives four quarters, and the third (perpendicular to both) gives eight equal smaller cubes.

Fewer cuts cannot work, because each full cut through the solid at most doubles the number of pieces, so two cuts give at most 4 pieces. Four, seven or eight cuts would produce more pieces than 8 or are more cuts than necessary.

Answer: A

The four numbers in the last row add up to 11.

The last row has no crosses, so each of its cells counts its non-cross neighbours in the third row (3, X, 6, X) and beside it in the fourth row.

1. First cell: 3 above, X diagonally, its right neighbour: 2.
2. Second cell: 3 and 6 diagonally, X above, its two row neighbours: 4.
3. Third cell: 6 above, two X's diagonally, its two row neighbours: 3.
4. Fourth cell: 6 diagonally, X above, its left neighbour: 2.

So the sum is \(2 + 4 + 3 + 2 = 11\). Answers of 9, 10 or 12 come from counting a cross as a free cell or missing a diagonal neighbour.

Answer: C

The correct choice is bridges on Q, R, T, and V.

Each river segment separates two zones, so a bridge on it joins those two zones:

• P: Z1-Z2; R: Z2-Z3; Q: Z1-Z3; S: Z1-Z5;
• T: Z3-Z5; V: Z3-Z4; U: Z4-Z5.

Connecting 5 zones needs at least 4 bridges, the edges of a spanning tree. Q, R, T and V join Z1, Z2, Z5 and Z4 each directly to Z3, so all five zones are connected with 4 bridges.

Three bridges (P, Q, T) cannot connect five zones. P, Q, S and T connect Z1, Z2, Z3 and Z5 but leave Z4 cut off, because neither U nor V is bridged. P, Q, S, U and V connects everything but uses 5 bridges, one more than needed.

Answer: A

Such a trip is possible for (i), but not for (ii).

The trip is a Hamiltonian cycle: a closed route through every city exactly once.

In (i), a 4 by 4 grid of 16 cities, such a cycle exists. Number rows 1 to 4 from the top and columns 1 to 4 from the left. One route: along row 1 from column 1 to column 4; down to row 2 and left to column 2; down to row 3 and right to column 4; down to row 4 and left to column 1; then up column 1 through rows 3 and 2 back to the start. It passes through all 16 cities once.

In (ii) there are five cities. Three of them have only two highways each, and a cycle through a city of degree two must use both of its highways. Those forced highways already include three edges at the top-left city (to the top-right city, the middle city and the left city). A cycle can use only two edges at any city, so no Hamiltonian cycle exists. That rules out the options claiming a trip for (ii) and the one denying it for (i).

Answer: B

The correct grid is (ii).

The rows must contain 3, 1, 2 and 2 shaded cells from top to bottom, and every column exactly 2. Reading grid (ii) row by row (1 = shaded): 1011, 0100, 0011, 1100. The row counts are 3, 1, 2, 2 and the column counts are 2, 2, 2, 2, as required.

The others fail somewhere:

• (i): 1011, 0101, 0010, 1100. The second row has 2 shaded cells and the third only 1.
• (iii): 1011, 0110, 0010, 1100. The second row has 2, and the third column has 3.
• (iv): 1011, 0100, 0010, 1101. The third row has 1 and the bottom row 3.

Answer: B

The plates in option (B), R, A and R, complete the palindrome RADAR.

With the given A and D, option (B) supplies an R, an upside-down A and an upside-down R. Since plates may be rotated in their plane, turning the last two through 180 degrees gives a normal A and R, and the five letters R, A, D, A, R read the same both ways.

A five-letter palindrome needs two matching pairs and one middle letter. Option (A) gives A, D, D, D, J, with both A and J unpaired. Option (C) gives A, D, Z, E, D, leaving A, Z and E unpaired. Option (D) gives A, D, I, L, Y, five different letters. None of these can be arranged into a palindrome.

Answer: D

The least number of additional ropes is 3.

Number the poles from (0, 0) at the bottom left to (6, 6) at the top right. The land is a 6 by 6 square with two opposite corners cut off; R1 runs from (0, 2) to (2, 2) and R2 from (4, 6) down to (4, 4). Three ropes complete four similar plots:

1. from (2, 2) down to (2, 0), through three poles;
2. from (4, 4) right to (6, 4), through three poles;
3. the diagonal from (2, 2) to (4, 4), through (3, 3).

Each plot is then a square with one corner cut off along a line joining the mid-points of the two sides at that corner: two small ones of side 2 at the bottom left and top right, and two large ones of side 4 at the top left and bottom right. All four have the same shape at two scales, so they are similar, and the areas add up to the 35 square units of the land. Both fixed ropes end up as sides of plots, so nothing is wasted, and 3 is the official least number; the options 4 and 5 use more ropes than this construction needs.