Scale problems

A (this answer not optimal, better use 4/4/4)

You are given 12 balls and a scale. Of the 12 balls, 11 are identical and 1 weighs slightly more. How do you find the heavier ball using the scale only three times?

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.

A (this answer not optimal, better use 4/4/4)

You are given 12 balls and a scale. Of the 12 balls, 11 are identical and 1 weighs slightly more. How do you find the heavier ball using the scale only three times?

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.

B

You are given a 3-gallon jug and a 5-gallon jug. How do you use them to get 4 gallons of liquid?

Fill the 5-gallon jug completely. Pour the contents of the 5-gallon jug into the 3-gallon jug, leaving 2 gallons of liquid in the 5-gallon jug. Next, dump out the contents of the 3-gallon jug and pour the contents of the 5-gallon jug into the 3-gallon jug. At this point, there are 2 gallons in the 3-gallon jug. Fill up the 5-gallon jug and then pour the contents of the 5-gallon jug into the 3-gallon jug until the 3-gallon jug is full. You will have poured 1 gallon, leaving 4 gallons in the 5-gallon jug.

You are given a 3-gallon jug and a 5-gallon jug. How do you use them to get 4 gallons of liquid?

Fill the 5-gallon jug completely. Pour the contents of the 5-gallon jug into the 3-gallon jug, leaving 2 gallons of liquid in the 5-gallon jug. Next, dump out the contents of the 3-gallon jug and pour the contents of the 5-gallon jug into the 3-gallon jug. At this point, there are 2 gallons in the 3-gallon jug. Fill up the 5-gallon jug and then pour the contents of the 5-gallon jug into the 3-gallon jug until the 3-gallon jug is full. You will have poured 1 gallon, leaving 4 gallons in the 5-gallon jug.

C

You are given 12 balls and a scale. Of the 12 balls, 11 are identical and 1 weighs EITHER slightly more or less. How do you find the ball that is different using the scale only three times AND tell if it is heavier or lighter than the others?

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.

You are given 12 balls and a scale. Of the 12 balls, 11 are identical and 1 weighs EITHER slightly more or less. How do you find the ball that is different using the scale only three times AND tell if it is heavier or lighter than the others?

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.

D

Three envelopes are presented in front of you by an interviewer. One contains a job offer, the other two contain rejection letters. You pick one of the envelopes. The interviewer then shows you the contents of one of the other envelopes, which is a rejection letter. The interviewer now gives you the opportunity to switch envelope choices. Should you switch?

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.

Three envelopes are presented in front of you by an interviewer. One contains a job offer, the other two contain rejection letters. You pick one of the envelopes. The interviewer then shows you the contents of one of the other envelopes, which is a rejection letter. The interviewer now gives you the opportunity to switch envelope choices. Should you switch?

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.

E

You have 100 balls (50 black balls and 50 white balls) and 2 buckets. How do you divide the balls into the two buckets so as to maximize the probability of selecting a black ball if 1 ball is chosen from 1 of the buckets at random? (this answer is soooo stupid to me who designed it?????)

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%.

You have 100 balls (50 black balls and 50 white balls) and 2 buckets. How do you divide the balls into the two buckets so as to maximize the probability of selecting a black ball if 1 ball is chosen from 1 of the buckets at random? (this answer is soooo stupid to me who designed it?????)

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%.

F

You’ve got a 10 x 10 x 10 cube made up of 1 x 1 x 1 smaller cubes. The outside of the larger cube is completely painted red. On how many of the smaller cubes is there any red paint?

First, note that the larger cube is made up of 1000 smaller cubes. The easiest way to think about this is how many cubes are NOT painted? 8 x 8 x 8 inner cubes are not painted which equals 512 cubes. Therefore, 1000 – 512 = 488 cubes that have some paint. Alternatively, we can calculate this by saying that two 10 x 10 sides are painted (200) plus two 10 x 8 sides (160) plus two 8 x 8 sides (128). 200 + 160 + 128 = 488.

You’ve got a 10 x 10 x 10 cube made up of 1 x 1 x 1 smaller cubes. The outside of the larger cube is completely painted red. On how many of the smaller cubes is there any red paint?

First, note that the larger cube is made up of 1000 smaller cubes. The easiest way to think about this is how many cubes are NOT painted? 8 x 8 x 8 inner cubes are not painted which equals 512 cubes. Therefore, 1000 – 512 = 488 cubes that have some paint. Alternatively, we can calculate this by saying that two 10 x 10 sides are painted (200) plus two 10 x 8 sides (160) plus two 8 x 8 sides (128). 200 + 160 + 128 = 488.