### Brain Teasers

Each week, we ask a brain teaser in The Commons Digest newsletter. Here, you'll find all of the answers!

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#### Week 12

Question: You have got someone working for you for five days and a gold bar to pay him. You must give them a piece of gold at the end of every day. What are the fewest number of cuts to the bar of gold that will allow you to pay him 1/5th each day?

*Answer: *Two cuts. Say you make two cuts to the gold bar and end up with three pieces:Â

- One that is 1/5th of the bar (we'll call that one payment unit)
- Two that are 2/5ths of the bar (two payment units)

Here's how you would pay out the worker:

- Day one, the worker starts with 0 bars and you pay him the first segment of payment unit
- On day two, he now already has one payment unit. You give him the segment with two payment units and ask him to give you back the one payment unit.
- On day three, he now has one block worth of two payment units. You give him the one payment unit back to pay him for day three. You don't ask for anything in return.
- On day four, you pay him your final segment of two payment units and ask for the one payment unit in return. He now has two segments, each worth two payment units (totalling four payment units).
- On day five, you give him the segment with one payment unit. He now has all of the segments and was paid 1/5th of the bar each day!

#### Week 11

Question: You have 9 chairs, 7 clients and me. What are the chances you're sitting next to me?

*Answer:Â *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 (i.e. the 2 chairs on either side of them).

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 a specific seating arrangement.

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#### Week 10

Question: You are blindfolded and are sitting in front of a table. There are 100 coins on the table, 50 of them face up and 50 of them face down. How do you split them into 2 groups where both groups have the same number of coins face up and face down? You cannot feel the coin faces.

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*Answer:Â Split the coins into two groups of 50 coins each, by counting them out at random. Flip one of the groups over so that the coins that were facing down are now facing up, and vice versa for the coins that were facing up. Both groups should now have the same number of couns face up and face down, since there were an even number of each. Say for example, when you split the coins originally, the left pile had 40 up and 10 down. That means that the right pile had to have 40 down and 10 up (since there are even amounts of each). Flipping the right pile would mean you now have 40 up and 10 down, matching the left pile.*

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#### Week 9

Question:Â There are three boxes of eggs. In each box there is either a set of big eggs, small eggs or big and small eggs mixed. The boxes are labelled (shock) LARGE, SMALL and MIXED but each box is labelled incorrectly. What is the least number of boxes you can open to know which eggs are in which box and why?

*Answer: *One!

You need to remember that *all* boxes are labelled incorrectly.

Here's an example.

- You open the box marked BIG and say it's filled of the small eggs. You now know that one boxes must contain the big and mixed eggs. Which box is which? Well, one of the boxes was labelled SMALL, but you opened that box with #1. That means that the third box must have been labelled MIXED, since it's the only one left. You also know that they're all labelled incorrectly, so the MIXEDÂ label cannot be the mixed eggs. That means those eggs must be big.

#### Week 8

Question: At exactly noon each day a scientist puts a bacteria in a petri dish. Every minute the bacteria divides in two. When it's at 1pm, the dish is full. What was the time when the dish was half full?

*Answer: *This brain teaser is a good reminder that you don't always have to overcomplicate things!

After all, the answer is quite simple. If every minute the bacteria doubles, and itâs full at 1pm. It would have been half full a minute earlier at 12:59pm.

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#### Week 7

Question:

There are three boxes, one contains only apples, one contains only oranges, and one contains both apples and oranges. The boxes have been incorrectly labeled such that no label identifies the actual contents of its box. You can pick one box to open (knowing what it has on the label). Without looking in the box, you take out one piece of fruit. By looking at the fruit, how can you immediately label all of the boxes correctly?

*Answer: *

You know all 3 boxes are incorrectly labeled. You want to avoid pulling fruit out of the box with both apples and oranges, since you won't know if it's the both with just the one fruit or both fruits.

So, open the box labeled âApples + Oranges.â Since the label is wrong, it must have one or the other not both.

Whichever fruit comes out is what that box contains. If you took out an apple, the box has only apples. If you took out an orange, the box has only oranges.

Let's say you pull out an apples. You can then move the Apples label to this box. The box that had the apples label has either oranges or both fruits. The third box will have whatever is left (again, either oranges or both fruits). The third box must have had the "oranges" label. And since all boxes are labeled incorrectly, we know that oranges can't be in box #3. So move the oranges label to box #2 and the Apples + Oranges label to box #3.

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#### Week 6

Question:

Four investment bankers need to cross a bridge at night to get to a meeting. They have only one flashlight and 17 minutes to get there. The bridge must be crossed with the flashlight and can only support two bankers at a time.

The Analyst can cross in one minute, the Associate can cross in two minutes, the VP can cross in five minutes, and the MD takes 10 minutes to cross.

How can they all make it to the meeting in time?

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*Answer*:

First crossing: the Analyst takes the flashlight and crosses the bridge with the Associate = 2 minutes

Return crossing: the Analyst returns with the flashlight = 1 minute

Second crossing: The Analyst gives the flashlight to the VP and the VP and MD cross together = 10 minutes

Return crossing: The VP gives the flashlight to the Associate (since the Associate only needs 2 minutes) = 2 minutes

Third crossing: The Analyst and Associate cross the bridge again together = 2 Â minutes.

Total Time = 17 minutes

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#### Week 5

Question:

Letâs say that you have 25 horses, and you want to pick the fastest 3 horses out of those 25. In each race, only 5 horses can run at the same time because there are only 5 tracks. What is the minimum number of races required to find the 3 fastest horses without using a stopwatch?For the sake of clarity, letâs assume that each horseâs speed doesnât change regardless of race conditions, fatigue or good olâ fashioned determination.

You can consider the horses as mechanical, and programmed to run the same speed each time they race. With that said, the horses will come in the same place every race. You can never know their times, only what place they finish in.

How many races will it take to find the top three horses?

*Answer: Check out **Matt's article** for great visualizations. Â *

Step 1: Race every horse at least once. Why?Â Any horse left out could be one of the top three. Since there are 25 horses, and 5 horses per race (25/5)Â that means **we need to run 5 races to start. **

Step 2: You now know which horse was the fastest in each race. But that doesn't mean that they're all in the top three. For example, there could be up to three horses from one race that were faster than all other horses - even the lead horse - in the other races. So, we need to race the winning horses to understand how each of the races stacks up against each other. **Therefore, we're now at race 6.**

Step 3:Â We can now order each of the original five races, base don the placement of each of the lead horses in race 6. We still have the predicament that there could be multiple horses in one of the original five races that makes the top 6. So how do we unpack that? Let's order the races as follows:

- Group 1 = horses from the first group of 5 races that had the fastest horse from race 6
- Group 2 = horses from the first group of 5 races that had the second fastest horse from race 6, etc.
- Group 3 = etc.
- Group 4 = etc.
- Group 5 = horses from the first group of 5 races that had the slowest horse from race 6

Since we're only looking for the top 3 horses, we can eliminate a number of horses based on these facts:

- We
*know*that all of group 4 and all of group 5 were slower than the leading horses from group 1, 2 and 3. Therefore all of group 4 and 5 can be eliminated (we don't need to race them again) - We
*know*that the rest of group 3 (outside the lead horse) will not be in the top 3 either - We
*know*that for groups 1 and 2, only the 3 fastest horses will be in the running (meaning we can eliminate the horses that run in positions 4 and 5 from those groups)

That leaves us with 6 horses:

- The lead horse from group 1
- The 2nd and 3rd place horses from group 1
- The lead horse from group 2
- The 2nd and 3rd place horses from group 2
- The lead horse from group 3

However, we can only run a race with 5 horses. So what do we do? We can eliminate the lead horse from group 1 since we already know it's the fastest horse. At this stage, we're only looking for the second and third placed horses.

**We now run our seventh race with the remaining 5 horses. **

Final Answer:Â 7 races!

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#### Week 4

Question: How many regular golf balls can you fit into a plane?

â*Answer:Â Well, I am sure you guessed, there's no specific answer here. Instead, the interviewer is looking to understand how you think and how you tackle an ambiguous problem. For example:*

*How do you estimate how large the plane is (do you estimate based on what you believe to be the most common plane size? If so, how do you determine that?)**Do you remember to flag, and remove space, for fixtures like seats?*- Do you consider the passenger area, cockpit, overhead and stowaway compartments?
*What sort of educated guesses are you making along the way?**Are you able to tackle this effectively without getting too granular (80/20?**Tip: think out loud the entire time!!*

#### Week 3

Question: What is the next number in the following sequence: 0 0 1 2 2 4 3 6 4 8 5?

â*Answer:Â 10. There are two alternating sequences: 0, 1, 2, 3, 4, 5 and 0, 2, 4, 6, 8.*

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#### Week 2

Question:Â You need to measure out four gallons of water, but you only have a three-gallon jug and a five-gallon jug. How do you measure out four gallons exactly? Â

*Answer: Fill the three-gallon jug and pour all of the water into the five-gallon jug. Then, fill the three-gallon jug again and use it to continue filling the five-gallon jug. Since the five-gallon jug already had three-gallons of water, there are only two-gallons of space remaining in the five-gallon jug. Therefore, two gallons will be poured out of the three-gallon jug, leaving one gallon in the three-gallon jug. Now, dump all of the contents out of the five-gallon jug. Pour the one gallon that is left in the three-gallon jug into the five-gallon jug, and then fill the three-gallon jug again and pour it into the five-gallon jug. This will give you four gallons.*

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#### Week 1

Question: You are in a room that has three switches and a closed door. The switches control three light bulbs on the other side of the door. Once you open the door, you may never touch the switches again. How can you definitively tell which switch is connected to each of the light bulbs?

*Answer:Â Switch on switches 1 & 2, wait a few minutes. Then switch off number 2. Enter the room. Whichever bulb is on is wired to switch 1, whichever bulb is off and hot is wired to switch number 2, and the third is wired to switch 3 (this bulb should be off and cold).*

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