Split the coins into 2 groups of 50 coins, it does not matter which coins you pick. Flip one of the groups over so that the coins face down will be face up and vice versa. Both groups should now have the same number of coins face up and face down.

Straight flushes go from A-5 to 10-A (technically a royal flush) over four suits. This means 40 combinations or 36 combinations excluding the royal flushes. The tricky part of this question is that there are ostensibly only 13 four of a kinds (Ace to King) when the four cards can be paired with any of the 12 other cards.

He just walks to his horse. Trains were not invented until he was in exile.

This has been a popular question in Asia and Europe. We are aware that it is a stupid question, but the point of the question is to see whether the candidate is rattled and can remain composed and laugh it off. Good articulation of a wrong thinking process is also helpful.

The 58th minute. At the 59th minute it is 50% full, as it is ½ of 100%. At the 58th minute it is ½ of the 59th minute’s waffle content.

She will just walk out of the room with the gold. The candidate will need to know the day’s gold price per ounce, but assuming it continues to be range bound between US$1,200 and US$1,400, and there are 16 ounces in a pound this is around 50 lbs.

Given that the interviewer is sitting in one of the 9 chairs, the other 8 chairs are distributed randomly to the 7 clients and you. Out of the 8 chairs you could possibly sit in, there are 2 combinations where you are sitting next to the interviewers. Thus, the chance that you are sitting next to the interviewer is 2/8 or 25%. Which chair the interviewer sits in is irrelevant in this question, given that every person has an equal probability of sitting in every seat (1/9). How the rest of the clients sit is also irrelevant, as without further information it is logical to assume the clients are identical and have no preference on sitting arrangement.

At 9:45, the hour hand has moved forward from 9 by 3/4th of an hour, or 3/4 * 1/12 of 360 degrees. A quick calculation gets us to 22 1/2 degrees away from 9, where the minute hand is at currently. Different variations of this question should be expected.

80 km/hr is incorrect. When averaging speed, you are not trying to find the arithmetic mean but the harmonic mean. The average speed in this case is 1/((1/100 + 1/60)/2) = 75 km/hr. To verify the answer, you can take the simplest case where distance = 300 km (lowest common denominator). It would take 3 hours to get to the interview and 5 hours to get back, meaning 1200 km was traversed in 8 hours. 600/8 = 75 km/hr.

A 8 x 8 x 8 cube is comprised of 64 x 8 = 512 1 x 1 x 1 cubes.

• 8 cubes are on the corner and are coated on 3 sides, 512 – 8 = 504
• 6 cubes are on each edge, there are 12 edges, 6 x 12 = 72 cubes are coated on 2 sides, 504 – 72 = 432
• 6 x 6 cubes are each face (excluding edges and corners), there are 6 faces, 36 x 6 = 216 cubes are coated on 1 side, 432 – 216 = 216
• 6 x 6 x 6 cubes are not on the faces, edges or corners. 36 x 6 = 216 are coated on 0 sides, which matches up to our answer from elimination

1. Fill up the 5 L bucket. 3L[0], 5L[5]
2. Pouring from the 5 L bucket to the 3 L bucket, you are left with 2 L in the 5 L bucket. 3L[3], 5L[2]
3. Pour out the water in the 3 L bucket. 3L[0], 5L[2]
4. Pour the 2 L of water from the 5 L bucket into the 3 L bucket. 3L[2], 5L[0]
5. Fill up the 5 L bucket. 3L[2], 5L[5]
6. Pour from the 5 L bucket to the 3 L bucket until the 3 L bucket is full. You should have poured out 1 L of water, and 4 L of water remain in the 5 L bucket. 3L[3], 5L[4]

There are at least 2 logical methods to tackle this question, although they may not arrive at the same answer.

1. Start with the (estimated) annual sales of McDonald’s,  divide that figure by 365 to get the average sales per day. Estimate a reasonable figure for COGS per day from that, and then estimate how much of the COGS is ground beef. Divide that figure by a reasonable dollars per pound ground beef figure (maybe half of Costco price to account for economies of scale) to get pounds ground beef per day consumed. Estimate the number of cows it takes to produce that many pounds of ground beef.
2. Start with the (estimated) number of burgers sold in a day at a single McDonald’s, multiply that figure by the (estimated) number of restaurants in the world. Multiply that by an average pound of ground beef per burger (~1/4) to get pound ground beef per day consumed. Estimate the number of cows it takes to produce that many pounds of ground beef.

By breaking it down to (3^2)^3, this question becomes easier. 9^3 = 81*9 = 729.

Given that you are looking at US and Canada, you would start with the US & Canadian population (~300 million and ~30 million respectively). Assuming an average household size of 4 people, you would arrive at 82.5 million households. Given the importance of Thanksgiving in US and Canadian culture, you would expect most households (80% may be a reasonable estimate) to partake in a turkey meal. 82.5 million * 0.8 = 66 million turkeys consumed on Thanksgiving.

The most important fact in the question is that all the boxes are labelled incorrectly, which gives us the following information:

• Box labelled apples -> Must be oranges or apples + oranges
• Box labelled oranges -> Must be apples or apples + oranges
• Box labelled apples + oranges -> Must be apples or oranges

Start with the box labelled apples + oranges, and pick a fruit out of the box. You know that it has to be only apples or only oranges, since the box was labelled incorrectly, so you would label the box with the fruit you ended up picking from the box (apple in this example). Now that you know only oranges and apples + oranges are left, the box labelled oranges would have to be apples + oranges, and you would label it as such. The box labelled apples would be apples + oranges by process of elimination.

To simplify this question, remember 1 + 999 = 2 + 998 … 499 + 501 = 1000. There are 499 pairs of numbers from 1 to 1000 that add to 1000, and their sum is 499 x 1000 = 499,000.  Adding the 2 numbers leftover, 1000 and 500 gets us to 500,500.

There are several methods to tackle this question, one of which is described below:

1. Weigh 4 random balls against 4 other ones.
   1. If the both groups weigh the same, the heavier ball must be in the group of 4 balls.
   2. If one group weighs more than the other, the heavier ball must be in the group that weighs more.
2. You’ve now narrowed it down to a 4 ball group. Weigh 2 random balls from that group against the other 2.
   1. One group must weigh more than the other, the heavier group has the heavier ball.
3. You’ve now narrowed it down to a 2 ball group. Weigh the last 2 against each other.
   1. The heaver ball is the one you are trying to find.

This question is significantly more difficult, as you are solving the same problem under the same constraint with more degrees of freedom. If you can solve this question within the allotted time, you probably deserve the job. The key to solving this problem is to use the information from the previous steps to choose which balls to weigh. There may be more methods to tackle this question, the one we found is described below:

To help you visualize the problem easier, label the balls from 1 to 12: 1   2   3   4   5   6   7   8   9   10   11   12, in which 1 is unique and the rest identical.

• Step 1: You weigh balls Group 1 (1 2 3 4) against Group 2 (5 6 7 8).
  • Outcome 1: Group 1 weighs the same as Group 2 (1 2 3 4 = 5 6 7 8). In this case the unique ball must be balls 9 10 11 or 12, and balls (1 2 3 4 5 6 7 8) must be identical and will be referred to as “o”.
    • Step 2: You weigh you weigh group A (9 10) against group B (11 “0”).
      • Outcome 1: Group A weighs the same as group B(9 10 = 11 “o”). In this case the unique ball must be 12.
        • Step 3: You weigh 12 against “o”. 12 has to weigh more or less than “o”, therefore 12 is the unique ball.
      • Outcome 2: Group A weighs more than group B (9 10 > 11 “o”). In this case if the unique ball is in group A (9 10) it must be heavier than the identical balls, and if the unique ball is in group B (11) it must be lighter than the identical balls.
        • Step 3: you weigh 9 against 10.
          • Outcome 1: 9 weighs the same as 10, 11 must be the unique ball and is lighter.
          • Outcome 2: 9 weights more than 10, 9 is the unique ball and is heavier, consistent with our previous statement.
          • Outcome 3: 9 weights less than 10, 10 is the unique ball and is heavier, consistent with our previous statement.
      • Outcome 3: Group A weighs less than group B (9 10 < 11 “o”). In this case if the unique ball is in Group A (9 10) it must be lighter than the identical balls, and if the unique ball is in group B (11) it must be heavier than the identical balls.
        • Step 3: You weight 9 against 10.
          • Outcome 1: 9 weighs the same as 10, 11 must be the unique ball and is heavier.
          • Outcome 2: 9 weights more than 10, 10 is the unique ball and is lighter, consistent with our previous statement.
          • Outcome 3: 9 weights less than 10, 9 is the unique ball and is lighter, consistent with our previous statement.
  • Outcome 2: Group 1 weighs more than Group 2 (1 2 3 4 > 5 6 7 8). In this case if the unique ball is in group 1 (1 2 3 4) it must be heavier than the identical balls, and if the unique ball is in group 2 (5 6 7 8) it must be lighter than the identical balls.
    • Step 2: You weigh Group A (1 3 5 7) against Group B (2 6 “o” “o”).
      • Outcome 1: Group A weighs the same as Group B (1 3 5 7 = 2 6 “o” “o”). In this case the unique ball must be either 4 or 8.
        • Step 3: You weigh 4 against “o”.
          • Outcome 1: 4 weighs the same as “o”, 8 is the unique ball and is lighter, consistent with our previous statement.
          • Outcome 2: 4 weighs more than “o”, 4 is the unique ball and is heavier, consistent with our previous statement.
      • Outcome 2: Group A weighs more than Group B (1 3 5 7 > 2 6 “o” “o”). In this case the unique ball must be either 1, 3 or 6.
        • Step 3: You weigh 1 against 3.
          • Outcome 1: 1 weighs the same as 3, 6 is the unique ball and is lighter, consistent with our previous statement.
          • Outcome 2: 1 weighs more than 3, 1 is the unique ball and is heavier, consistent with our previous statement.
          • Outcome 1: 1 weighs less than 3, 3 is the unique ball and is heavier, consistent with our previous statement.
      • Outcome 2: Group A weighs less than Group B (1 3 5 7 < 2 6 “o” “o”). In this case the unique ball must be either 5, 7 or 2.
        • Step 3: You weigh 5 against 7.
          • Outcome 1: 5 weighs the same as 7, 2 is the unique ball and is heavier, consistent with our previous statement.
          • Outcome 2: 5 weighs more than 7, 7 is the unique ball and is lighter, consistent with our previous statement.
          • Outcome 1: 5 weighs less than 7, 5 is the unique ball and is lighter, consistent with our previous statement.
  • Outcome 3: Group 2 weighs more than group 1 (1 2 3 4 < 5 6 7 8).
    • Solve as you would solve for outcome 2 above.