Investment Banking Brain Teasers
Just to be perfectly clear, you are assuming that one of the two buckets is chosen at random and then one of the balls from that bucket is chosen at random. You want to put 1 black ball in 1 of the buckets and all of the other 99 balls in the other bucket. This gives you just slightly less than a 75% change of having a black ball chosen. The math works as follows: There’s a 50% chance of selecting the bucket containing 1 ball with a 100% chance of selecting a black ball from that bucket. And a 50% chance of selecting the bucket containing 99 balls with a ~49.5% (49/99) chance of selecting a black ball from that bucket. Total probability of selecting a black ball is (50% % 100%) + (50% * 49.5%) = 74.7%.
Most people say 90 mph but this is actually a trick question! The first leg of the trip covers 60 miles at an average speed of 30 mph. So, this means the car traveled for 2 hours (60/30). In order for the car to average 60 mph over 120 miles, it would have to travel for exactly 2 hours (120/60). Since the car has already traveled for 2 hours, it is impossible for it to average 60 mph over the entire trip.
The answer is yes. Say your original pick was envelope A. Originally, you had a 1/3 chance that envelope A contained the offer letter. There was a 2/3 chance that the offer letter was either in envelope B or C. If you stick with envelope A, you still have the same 1/3 chance. Now, the interviewer eliminated one of the envelopes (say, envelope B), which contained a rejection letter. So, by switching to envelope C, you now have a 2/3 chance of getting the offer and you’ve doubled your chances.
Note that you will often get this same question but referring to playing cards (as in 3-Card Monte) or doors (as in Monte Hall/Let’s Make a Deal) instead of envelopes.
Turn on two switches (call them A and B) on and leave them on for a few minutes. Then turn one of them off (switch B) and enter the room. The bulb that is lit is controlled by switch A. Touch the other two bulbs (they should be off). The one that is still warm is controlled by switch B. The third bulb (off and cold) is controlled by switch C.
The trick here is that you have 50 pairs which each sum to 101 (e.g. 1+100, 2+99, 3+98, etc.). So, 50 times 101 = 5050.
Significantly harder than the last question! Weigh 4 vs 4 (1st Weighing). If they are identical then you know that all of 8 of these are “normal” balls. Take 3 “normal” balls and weigh them against 3 of the unweighed balls (2nd Weighing). If they are identical, then the last ball is “different.” Take 1 “normal” ball and weigh against the “different” one (3rd Weighing). Now you know if the “different” ball is heavier or lighter.
If, on the 2nd weighing, the scales are unequal then you now know if the “different” ball is heavier (if the 3 non-normal balls were heavier) or lighter (if the 3 non-normal balls were lighter). Take the 3 “non-normal” balls and weigh 1 against the other (3rd Weighing). If they are equal then the third ball not weighed is the “different” one. If they are not equal then either the heavier or lighter ball is “different” depending on if the 3 “non-normal” balls were heavier or lighter in the 2nd Weighing.
If, on the 1st Weighing, the balls were not equal then at least you know that the 4 balls not weighed are “normal.” Next, take 3 of the “normal balls” and 1 from the heavier group and weigh against the 1 ball from the lighter group plus the 3 balls you just replaced from the heavier group (2nd Weighing). If they are equal then you know that the “different” ball is lighter and is 1 of the 3 not weighed. Of these 3, weigh 1 against 1 (3rd Weighing) If one is lighter, that is the “different” ball, otherwise, the ball not weighed is “different” and lighter.
If, on the 2nd weighing from the preceding paragraph, the original heavier group (containing 3 “normal” balls) is still heavier, then either one of the two balls that were NOT replaced are “different.” Take the one from the heavier side and weigh against a normal ball (3rd Weighing). If it is heavier, it is “different, ” and heavier otherwise the ball not weighed is “different” and lighter. If, on the 2nd weighing, the original lighter side is now heavier, then we know that one of the 3 balls we replaced is “different.” Weigh one of these against the other (3rd Weighing). If they are equal, the ball not weighed is “different” and heavier. Otherwise, the heavier ball is the “different” one (and is heavier).
If you get this right and can answer within the 30 minutes alloted for the interview, then you probably do deserve the job.
First, weigh 5 balls against 5 balls (1st Use of Scale). If the scale is equal, then discard those 10 balls and weigh the remaining 2 balls against each other (Second Use of Scale). The heavier ball is the one you are looking for.
If on the first weighing (5 vs 5), one group is heavier, then of the heavier group weigh 2 against 2 (2nd Use of Scale). If they are equal, then the 5th ball from the heavier group (the one not weighed) is the one you are looking for. If one of the groups of 2 balls is heaver, then take the heaver group of 2 balls and weigh them against each other (Third Use of Scale). The heavier ball is the one you are looking for.
At quarter past the hour, the minute-hand is exactly at 3:00 but the hour-hand has moved 1/4 of the way between 3:00 and 4:00. Therefore 1/4 times 1/12 = 1/48 of the clock. With the clock having 360 degrees, 360/48 = 7.5 degrees.