Puzzle Test Questions and Answers

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.

Answer: C

The order of arrival works out uniquely as:

Why: Rajiv's couple has the Garments couple just before it and Rupu's couple just after it, so Rajiv is second or third. Ashutosh's couple is followed by Rinku's, which is followed by the Jewellery couple, so Ashutosh is first or second. Prem comes just before Rosy, so Prem cannot be last.

If Rajiv were second, Ashutosh would be first, Rinku would be Rajiv's wife and the third couple would deal in Jewellery. Prem would then be third and Pappu fourth with Books, leaving Furniture for Rajiv and Rinku, which clue (iv) forbids. So Rajiv is third and Rupu's couple fourth. Ashutosh cannot then be second, because Rinku would be third and the fourth couple, Pappu's, would have to deal in Jewellery though Pappu deals in Books. So Ashutosh is first, Rinku's couple second (Garments), Rajiv's third (Jewellery) and Pappu's fourth (Books). Prem must be second, with Rosy in the third couple as Rajiv's wife, and Nitu is Ashutosh's wife.

The arrival right after Ashutosh's couple was Prem and Rinku, as clue (iv) states directly. Rajiv and Rosy came third and Pappu fourth, so none of them came immediately after Ashutosh.

Answer: C

The tasks run as follows: A from 9 to 11 a.m. (2 h); B after A, from 11 a.m. to 12 noon (1 h); C after A and B, from 12 noon to 1 p.m. (1 h); D alongside B and C, so from 11 a.m., for 3 h, to 2 p.m.; and E only after A, B, C and D are all done.

So the earliest completion of C is 1 p.m. 11 a.m. is when A ends, and 12 noon is when B ends and C can begin. 2 p.m. is when D ends; C does not have to wait for D, because D runs alongside it.

Answer: C

The integer assigned to N is 6.

P is 4. T differs from P by 5, so T is 9 or −1; only 9 lies between 1 and 9, so T = 9.

N differs from T by 3, so N is 6 or 12. Only 6 is in range, so N = 6.

4 is already taken by P, so it cannot also belong to N. 5 and 7 differ from T = 9 by 4 and 2, not 3.