Puzzle Test Questions and Answers

Answer: B

The correct statement is S cannot be placed at a corner.

Among the six points, two lines are parallel only when one joins the mid-points of two sides and the other lies along the third side. Suppose S were a corner. Then QS lies along a side through S, so P and R must be the mid-points of the other two sides, one of which is the side opposite S. U must not share a side with S, and the only point not on a side through S is the mid-point of the side opposite S, which is already P or R. So there is no place for U, and S cannot be a corner.

The other options are possible. Take corners X, Y, Z with T = X, Q and S the mid-points of XY and XZ, and P = Y, R = Z, U the mid-point of YZ. Every condition holds, yet P and R sit at corners and U at a mid-point, so options A, C and D are false.

Answer: A

The correct statement is The line segment joining R and T is parallel to the line segment joining Q and S.

In a regular hexagon there are three lengths: side, short diagonal and long diagonal. A long diagonal is perpendicular only to short diagonals, and a short diagonal is perpendicular to sides and long diagonals. Trying the possibilities, RS longer than and perpendicular to PQ, together with RU parallel to TQ, can only be met with RS a long diagonal and PQ a short diagonal.

Up to rotation and reflection the order round the hexagon is then forced: P, S, Q, U, R, T. Check: R and S are opposite; P and Q sit either side of S, so PQ is a short diagonal perpendicular to RS; RU is a side and TQ is the long diagonal parallel to it.

In this order RT and QS are opposite sides, so they are parallel. TQ and PU are two different long diagonals, so they meet at the centre and are not parallel. RP is a short diagonal, and UQ and QS are sides that each meet its direction at 30 degrees, not 90, so the two perpendicularity claims fail.

Answer: B

The hardest pair is learnt by Q.

Chain the comparisons from easiest to hardest: Spanish is easier than French, French easier than Dutch, Dutch easier than Japanese, and Japanese easier than Chinese:

Spanish \(\lt\) French \(\lt\) Dutch \(\lt\) Japanese \(\lt\) Chinese

Q studies Chinese and Japanese, the two hardest languages, so Q's pair is the most difficult.

S studies Dutch and Japanese, the next-hardest pair; P has French and Dutch; R has Spanish and French, the easiest pair of all.

Answer: B

The box takes ✱ (the star).

Number the rows and columns 1 to 4 from the top left; '?' is in row 1, column 2.

1. In the top-right 2×2 square, row 2, column 4 cannot be ○ (row 2 already has ○) and cannot be △ or ✱ (both are already in that square), so it is □.
2. That leaves ○ for row 1, column 3, the last empty box of the top-right square.
3. Row 1 now has ○ and △, so its two empty boxes take ✱ and □. Column 1 already has ✱ in row 4, so row 1, column 1 is □ and '?' is ✱.

○ and △ are already in row 1, so they cannot repeat there, and □ is forced into column 1, leaving the star as the only symbol for the marked box.

Answer: B

The best orator is Meritorius' brother.

The sibling pairs are Meritorius and his brother, and the son and the daughter. Condition (ii) puts the son diagonally across the sibling of the worst orator.

If the worst orator were the brother, the son would sit diagonally across Meritorius, leaving the daughter and the brother in the other two boxes, which lie in different columns; this breaks (i). If the worst were Meritorius, the son sits diagonally across the brother, so the daughter must take the box in the brother's column, and Meritorius sits in the brother's row. If the worst were the son, the son sits diagonally across his sister, and by (i) the brother takes the other box in her column, which is in the son's row.

In both possible cases, (iii) puts the best orator in the worst orator's row, and that person is the brother. So Meritorius, the son and the daughter are not the best.

Answer: A

The correct pie chart is V 10%, R 20%, Y 30%, G 40%.

Check each chart against the four conditions, taking the percentages as counts out of 100:

• (i) Y + 2V = 30 + 20 = 50
• (ii) V + G = 10 + 40 = 50
• (iii) Y + R = 30 + 20 = 50
• (iv) V + 2R = 10 + 40 = 50

All four hold. Solving directly gives the same: (i) and (iv) give Y + 2V = V + 2R, so Y = 2R - V; with (iii), 3R - V = 50, and with (iv), V = 50 - 2R, so R = 20, V = 10, Y = 30 and G = 40. The total is 100, so the counts are also the percentages.

The other charts fail at once: V 40% and Y 20% give Y + 2V = 100; V 30%, Y 10% gives 70; V 20%, Y 40% gives 80.

Answer: C

P is 3.

Combine the conditions: R < T, T < P, P < S and P < Q. So R and T are smaller than P, and S and Q are larger.

With the digits 1 to 5, exactly two are below P and two above, so P must be the middle value, 3. Then R = 1, T = 2, and S and Q are 4 and 5 in some order (for example 34152).

P = 1 or 2 would leave too few digits below it for R and T, and P = 4 too few above it for S and Q.

Answer: C

The minimum number of steps is 9.

Each step swaps two adjacent coins, so it can change the relative order of only that one pair. In the starting row every one of the 3 heads is to the left of every one of the 3 tails, and in the final row every head is to the right of every tail. So each of the 3 × 3 = 9 head-tail pairs must be reversed, and at least 9 steps are needed.

9 steps are enough: move the leftmost tail three places left (3 steps), then the next tail three places (3 steps), then the last (3 steps). 3 and 6 steps cannot reverse all 9 pairs, and 12 is more than necessary.

Answer: A

The sum is 11.

Each new number counts the neighbouring cells without a cross. The last row has no crosses, and the row above it reads 3, X, 6, X.

• Cell 1: neighbours 3 (no cross), X, and cell 2 of the last row: 2.
• Cell 2: 3, X, 6 above, cells 1 and 3 beside: 4.
• Cell 3: X, 6, X above, cells 2 and 4 beside: 3.
• Cell 4: 6 and X above, cell 3 beside: 2.

\[ 2 + 4 + 3 + 2 = 11 \]

The other totals come from counting a cross as clear or forgetting the diagonal neighbours.

Answer: C

Why: N and O are not industrial, so the three industrial cities are L, M and P. O is the hill station and so has a university. M is a port, and no port has a university, so M has none; the two industrial cities with universities are therefore L and P. That makes three universities (L, O, P). The second port cannot be O (not a port) or L or P (they have universities), so it is N.

The city with industries and a port but no university is M. 'L and O' and 'O and M' both include O, the hill station, which is not industrial and has no port. N is a port but not industrial.