Common Puzzles Questions

What are the common puzzles asked in an interview?

Some of the common puzzles asked:

Four People on a Rickety Bridge

Question: Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?

Answer: The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.
1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)
Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

A Box of Defective Balls

Question: You have 10 boxes of balls (each ball weighing exactly 10 gm) with one box with defective balls (each one of the defective balls weigh 9 gm). You are given an electronic weighing machine and only one chance at it. How will find out which box has the defective balls?

Answer: For convenience sake, let’s name the boxes from 1 to 10. In order to solve this problem, you have to leverage the fact that you know exactly what each good ball is supposed to weigh and what each defective ball is supposed to weigh. Many of us instinctively will take one ball out of each box and try to find a way to make it work but the trick to take different number of balls from each box.

The number of balls you pick from each bag is equal to the box number. For example, pick 1 ball from box 1, 2 balls from box 2 and so on. In total you will have 55 balls. If all of the boxes have good balls, then the total weight of these balls would be 550 gm.

If box 1 has defective balls, then the total weight should be 1 gm less than expected (only one ball weighing 9 gm). If box 2 has defective balls, then the total weight should be 2 gm less than expected (two balls weighing 9 gm). So once you weigh the set of chosen balls, find out the difference between the total weight and the expected weight. That number represents the box number which contains the defective balls.

The Ant Problem

Question: Three ants are sitting at the three corners of an equilateral triangle. Each ant starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the ants collide?

Answer: So let’s think this through. The ants can only avoid a collision if they all decide to move in the same direction (either clockwise or anti-clockwise). If the ants do not pick the same direction, there will definitely be a collision. Each ant has the option to either move clockwise or anti-clockwise. There is a one in two chance that an ant decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision.

P(No collision) = P(All ants go in a clockwise direction) + P( All ants go in an anti-clockwise direction) = 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5 = 0.25

Clock Hands

Question: How many times a day do the minute and hour hands of a clock overlap?

Answer: Did you think the answer was 24 times? Well if you did, it’s time you think again. Let’s do some math.

In T hours, the minute hand completes T laps. In the same amount of time, the hour hand completes T/12 laps.

The first time the minute and hour hands overlap, the minute hand would have completed 1 lap more than the hour hand. So we have T = T/12 + 1. This implies that the first overlap happens after T = 12/11 hours (~1:05 am)

Similarly, the second time they overlap, the minute hand would have completed two more laps than the hour hand. So for N overlaps, we have T = T/12 + N.
Since we have 24 hours in a day, we can solve the above equation for N
24 = 24/12 + N
24 = 2 + N
N = 22

Thus, the hands of a clock overlap 22 times a day. Thus the hands of the clock overlap at 12:00, ~1:05, ~2:10, ~3:15, ~4:20, ~5:25, ~6:30, ~7:35, ~8:40, ~9:45, ~10:50. Note that there is no ~11:55. This becomes 12:00.

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