**Q1- You've got a 10 x 10 x 10 cube that's built up of smaller cubes that are 1 x 1 x 1. The outside of the larger cube has been entirely painted in red to make it stand out. Which of the smaller cubes has red paint on it, and how many of them?**

First and foremost, keep in mind that the larger cube is composed of 1000 smaller cubes. The most straightforward approach to think about this is to consider how many cubes are not painted. The interior cubes of the 8 x 8 x 8 structure are not painted, for a total of 512 cubes. As a result, there are 488 cubes with some paint out of 1000 total. To put it another way, we can say that we painted two 10 × 10 sides (200), two 10 x 8 sides (160), and two 8 x 8 sides (240) and then added them all together (128). 200 + 160 + 128 = 488.

**Answer: **You've got a 10 x 10 x 10 cube that's built up of smaller cubes that are 1 x 1 x 1. The outside of the larger cube has been entirely painted in red to make it stand out. Which of the smaller cubes has red paint on it, and how many of them?

First and foremost, keep in mind that the larger cube is composed of 1000 smaller cubes. The most straightforward approach to think about this is to consider how many cubes are not painted. The interior cubes of the 8 x 8 x 8 structure are not painted, for a total of 512 cubes. As a result, there are 488 cubes with some paint out of 1000 total. To put it another way, we can say that we painted two 10 × 10 sides (200), two 10 x 8 sides (160), and two 8 x 8 sides (240) and then added them all together (128). 200 + 160 + 128 = 488.

**Q2- When travelling at an average speed of 30 miles per hour, a car can cover a distance of 60 miles. How fast would the car have to travel the same 60-mile route back home in order to maintain an average speed of 60 mph throughout the trip?**

**Answer:** The majority of individuals will answer 90 mph, but this is a trick question! The first stage of the journey is 60 miles long and takes an average speed of 30 miles per hour. As a result, the car travelled for a total of 2 hours (60/30). In order for the car to cover 120 miles at an average speed of 60 mph, it would need to travel for exactly 2 hours (120/60). Because the car has already travelled for two hours, it will be impossible for it to maintain an average speed of 60 miles per hour for the duration of the trip.

**Q3- You're given 12 balls and a scale to work with. There are 12 balls total, 11 of which are similar and one which weighs somewhat more. How do you determine which ball is heavier when you only use the scale three times?**

**Answer: **First, compare the weights of five balls versus five balls (1st Use of Scale). If the scales are equal, then discard the first ten balls and weigh the remaining two balls against each other on the remaining scales (Second Use of Scale). The ball with a higher density is the one you're looking for.

If one group is heavier than the other on the first weighing (5 versus 5), then weigh 2 against 2 of the heavier group (2nd Use of Scale). The fifth ball from the heavier group (the one that hasn't been weighed) is the one you're looking for if they're both equal in weight. If one of the groups of two balls is heavier than the other, then take the heavier group of two balls and weigh them against the other group of two balls (Third Use of Scale). The ball with a higher density is the one you're looking for.

You're given 12 balls and a scale to work with. One of the 12 balls weighs slightly more or less than the other 11 balls. The other 11 balls are identical. How do you identify the ball that is different from the others while just using the scale three times AND determining whether it is heavier or lighter than the others?

This question is significantly more difficult than the last one! Weigh 4 vs 4 and compare the results (1st Weighing). If they are all identical, you can be confident that all eight of these balls are "normal." Take three "regular" balls and weigh them against three of the unweighed balls to see which is heavier (2nd Weighing). If the first two balls are identical, the third ball is "different." Take 1 "regular" ball and weigh it against 1 "strange" ball to see which is heavier (3rd Weighing). You should now be able to determine if the "different" ball is heavier or lighter.

If the scales are unequal on the second weighing, you now know whether the "different" ball is heavier (if the three non-normal balls were heavier) or lighter (if the three non-normal balls were lighter) (if the 3 non-normal balls were lighter). Take one of the three "abnormal" balls and weigh it against the other two (3rd Weighing). Assuming they are identical in weight, the third ball that has not been weighed is the "different" one. If they are not equal, then either the heavier or lighter ball is "different," depending on whether the three "non-normal" balls were heavier or lighter in the second weighing, and so "different."

If the balls were not equal on the first weighing, at the very least you would know that the four balls that were not weighed are "normal." After that, take 3 of the "regular balls" and 1 from the heavier group and weigh them against the 1 ball from the lighter group plus the 3 balls you just replaced from the heavier group that you just replaced with the "normal balls" (2nd Weighing). If they are identical in weight, you can tell that the "different" ball is lighter and is one of the three that has not been weighed yet. One of these three balls is weighed against the other three balls (3rd Weighing). If one is lighter, it is the "different" ball; otherwise, the ball that has not been weighed is the "different" and lighter ball.

Assuming that the original heavier group (which contained three "regular" balls) is still heavier on the second weighing from the prior paragraph, then either one of the two balls that were not changed is considered to be "different." Consider taking the one from the heavier side and weighing it against a standard ball (3rd Weighing). Otherwise, the ball that has not been weighed is "different" and heavier; otherwise, the ball that has not been weighed is "different" and lighter. If the initial lighter side becomes heavier on the second weighing, we know that one of the three balls we replaced is "different." Compare and contrast one of these with the other (3rd Weighing). If they are equal, the unweighed ball is "different" and heavier than the weighed ball. Other than that, the "different" ball is the heavier ball (and is heavier).

If you get this correct and are able to answer all of the questions within the 30 minutes allotted for the interview, you are likely to obtain the job.

**Q-4 Three lightbulbs are installed in a room with no windows. You're standing outside the room, in front of three switches, each of which controls one of the lightbulbs. So, if you only have one chance to enter the room, how are you supposed to figure out which switch controls which lightbulb?**

**Answer: **Two switches (designated as A and B) should be turned on for a few minutes and then turned off. Then, using switch B, turn one of them off and walk into the room. Switch A is in charge of controlling the brightness of the light. Make contact with the other two bulbs (they should be off). Switch B is in charge of controlling the one that is still warm. Switch C is responsible for controlling the third bulb (which is both off and chilly).

**Q-5 Four investment bankers must cross a bridge in the middle of the night in order to go to a meeting. They only have one flashlight and only 17 minutes to get there before it gets too dark. The bridge can only sustain two bankers at a time and must be traversed with the flashlight to be effective. One minute for the analyst, two minutes for the associate, five minutes for the vice president and 10 minutes for the medical director to cross the bridge. How are they going to get everyone to the meeting on time?**

**Answer: **In the beginning, the Analyst takes the flashlight and walks across the bridge with the Associate. This procedure takes 2 minutes. After then, the Analyst returns across the bridge with the flashlight, which takes another minute (3 minutes passed so far). The Analyst hands over the flashlight to the Vice President, and the Vice President and the MD cross together, taking a total of 10 minutes (13 minutes passed so far). The Vice President hands the flashlight to the Associate, who returns to the other side of the bridge in 2 minutes (15 minutes passed so far). The Analyst and Associate will now cross the bridge jointly, which will take another 2 minutes. The meeting will begin exactly 17 minutes after everyone has crossed the bridge. It's worth noting that, rather than investment bankers, you'll frequently see the same question asked to members of musical bands (usually either the Beatles or U2).

**Q6- An interviewer places three envelopes in front of you and asks you to choose one of them. One of the envelopes includes a job offer, but the other two are filled with rejection letters. You select an envelope from the pile. The interviewer then proceeds to show you the contents of one of the other envelopes, which turns out to be a letter of rejection. The interviewer now provides you with the opportunity to change your envelope selections. Should you make the switch?**

**Answer: **Yes, it is correct. Let's pretend your first choice was envelope A. Originally, you had a 1/3 chance of receiving the offer letter in envelope A if you opened it. There was a 2/3 probability that the offer letter would be in either envelope B or C, according to the odds. If you stick with envelope A, your chances of winning remain at a third of a chance. Now, the interviewer has rejected one of the envelopes (let's say, envelope B), which carried a rejection letter, from consideration further. Consequently, by switching to envelope C, you have a 2/3 probability of receiving an offer, thereby doubling your chances of receiving an offer.

This question will frequently be asked, but with reference to playing cards (as in 3-Card Monte) or doorways (as in Monte Hall/Make Let's A Deal) instead of envelopes, so be prepared to see it more often.

**Q7- You have a total of 100 balls (50 black balls and 50 white balls) and two buckets at your disposal. The question is, how do you split the balls into the two buckets in such a way that the probability of getting a black ball is maximized if one ball is chosen at random from one of the buckets?**

**Answer: **Please understand that you are expecting that one of the two buckets is chosen at random, and then one of the balls from that bucket is chosen at random, in order to be completely clear. Put one black ball in one of the buckets and all of the other 99 balls in the other bucket if you want to win the game. With this strategy, you have a slightly less than 75 percent chance of getting the black ball in the lottery. Following is an explanation of how the arithmetic works: There is a 50 percent probability of selecting the bucket holding one ball, with a 100 percent chance of selecting a black ball from that bucket when that bucket is selected. Furthermore, there is a 50 percent probability of selecting the bucket holding 99 balls, with a 49.5 percent (49/99) chance of selecting a black ball from that bucket. (50% * 100 %) + (50% * 49.5%) = 74.7 percent is the total probability of selecting a black ball when all other factors are equal.

**Q8- At 3:15 p.m., what is the angle formed by the hour hand and the minute hand of a clock?**

**Answer: **With a quarter past the hour approaching, 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 on the clock. As a result, 1/4 times 1/12 equals 1/48 of the clock. With a clock of 360 degrees, 360/48 = 7.5 degrees is the angle measured.

**Q9- Approximately how many quarters would it take to build a stack from the floor of this room all the way to the ceiling? For simplify, let us assume that the room is 10 feet high and that 12 quarters are 1 inch tall.**

**Answer: **To find the answer, imagine that the room is ten feet high and that twelve quarters are one inch tall.

Osvaldo Zoom 12 quarters every inch multiplied by 12 inches each foot multiplied by 10 feet space equals 1440 quarters

**Q10- Why Are Manhole Covers Round?**

**Answer: **Manhole covers are rounded in order to prevent them from falling into the manholes.

Other possible explanations include: • Because manholes are round; • Because they are easier to transport (simply roll them); and • Because they are easier and less costly to build (smaller surface area than a square cover)

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