Operations analyst interview questions shared by candidates
25 horses, 5 race tracks. How many races you have to run to select top 5 horses.59 Answers
7 Races First race all 25 in groups for 5 to figure out top 3 in each group. (15 left) Call them A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3 Here A1 is faster than A2. A2 is faster than A3. The same applies for the rest of the groups. Now race A1, B1 C1 D1 E1 Lets say the order of the horses according to ranking was A1, B1,C1,D1,E1 So A1 is No 1 Now A1 is faster than B1 and B1 is faster than C1. So we can get rid of the entire D and E groups 9 horses remaining Also A1 is faster than B1 and B1 is faster than C1. So we can get rid of C2 and C3 7 horses remaining A1 is faster than B1. B1 is faster than B2. get rid of B3 6 horses remaining Out of these, We know A1 is the fastest. So now race A2, A3, B1,B2,C1 to figure out No2 and No3 positions
One race is the answer
because, the top 5 are the fastest right ? so you run one race and take the best five times. It's a trick question to see if you'll make a mountain out of a mole hill.
The first question to ask to clarify is: how many horses can fit in one track? Technically, one track can be used for all 25 horses, but this is too crowded for an efficient race. Let's say track size is not a factor, then, you need 1 race.
Without overcomplicating things - the answer is 1 race. All other answers are based on dissecting the problem into, imo, unnecessary details. There is no mention of the capacity of each race track or number of gates available etc. Even if there are less than 25 gates available at a track, I'd say fetch enough number of gates so you can find the outcome with single race.
This isn't a trick question. Normally this question has a track limit of 5 horses. I agree with Anuradha's answer.
Anuradha's solution still has problems. (Even if we go with Anuradha's assumptions that you can only race one horse per track, and also assuming that we don't have a stopwatch and must compare horses placing positions) What if the fastest five horses are A1, B1, C1, D1, and E1 ? In Anuradha's second step, he elminates two of the fastest horses (D1 and E1) . He's assuming that A2, B2, or some of the other horses from the other heats are faster, but he hasn't actually tested to see if that is true.
@anuradha: I think there are 2 problems: 1. How do you know A4 is not faster than B1 .... Suppose A1,A2,A3,A4,A5 are the fastest. 2. Using the logic posted we can get the answer in 6 races, 5 races for finding A1,B1,C1,D1,E1 and 1 race to find the ranking.
Assuming 1 track for 1 horse then it need 5 races to select top 5 horses. Each race for 5 horses and count the time for each race. The top 5 fastest horses are the top 5 horses.
2 max: top 4 from first race, if 5th is a tie, then two, or else just one.
You guys are not doing CS! 10 runs is my answer. 1. randomize 5 groups, each of 5 horses 2. rank them within each group, I will use Anuradha's notation (5 races) 3. pick the best of each group, race to figure the 1st place, call it A1 (1 race) It should be clear, it wins all times, every one lost once. 4. remove it. substitute 2nd best in. repeat 3 (in my eg. A2,B1,C1,D1,E1) now you have second place. keep going, you get the first 5 and ranking! So, 5+5=10 races in total.
The answer is 9. Assuming: - There's no time measuring (stopwatch), just relative places. - The horses perform consistently. - A maximum of 5 horses per race. First we need 5 races (A to E) to get relative scores for all 25 horses. Let's take a worst scenario: the list was already ordered (A1 fastest and E5 slowest), so race A contained the top 5. The 6th race would be the winners of the 5 races (A1, B1, C1, D1, E1), and would give A1 as the fastest of all. This would also mean that some horses can be excluded (only 4 more places to fill): B5 C4, C5 D3, D4, D5 E2, E3, E4, E5 For the 7th race, A2 would replace A1, and A2 would be appointed as the runner-up (of all). We also can exclude some more (only 3 more places to fill): B4 C3 D2 E1 For the 8th race, A3 would replace A2, but as E1 has been excluded, we got a vacancy. Let's add C2 for worst case scenario. The winner would be A3, and we can exclude more horses (only 2 more places to fill): B3 C2 D1 At this point there're only 5 horses who have not yet been classified or excluded, so the winner and runner-up of the 9th race would give 4th and 5th overall.
To Patrick: I had to slightly modify your almost correct solution and got 10 races. You has an excellent idea to remove subset of candidates after final selection of the i-th horse, and my modifications are not an essential ones. Actually it is necessary to fix two small things: 1. get rid off the suggestion about the ordered list To avoid this suggestion I would rename groups from the fastest A to the slowest E with order by 1st horse after final selection of the i-th horse. So after race #6 we have order: A1, B1, C1, D1, E1, but there is no suggestion that all A are better than all B and so on. And after every race #7, #8, #9 we have to rename groups. With this modification your method works fine until final selection of horse #5 2. Without this suggestion you need race #10 to find the last 5th horse. To test both methods you may use 25 randomly selected integers and select the 5 minimal or maximum ones. I can't prove that 10 is the minimum number of required tests, but it looks very convincing. Thank you.
Maybe Ziqi Dai also meant the same method, but it was difficult to understand without a couple of details.
To Leo, I checked some random sequences with a spreadsheet, and all the remaining (not placed or excluded) horses run in the 9th race, so there's no need for a 10th race. Sometimes there're only 4 or even 3 horses left in the 9th race. I didn't check every possibility, but there was no indication a 10th was ever necessary.
This answer is 1 regardless of track. If it can fit all 25 then you only need one race. If you make the assumption that only 5 can race at a time then you put 5 on each track and start the race on each track at the same time. Either way, the fastest 5 horses win. Using complex math on this is pointless since a pragmatic approach is available.
@ Anuradha -- this is a great solution to the wrong problem! The classic problem is to find the THREE fastest, and that's what your solution is. However, the question posed in this thread is (likely incorrectly quoted from the interview) to find the FIVE fastest of the group.
why would you make an assumption that makes this more difficult? I mean given this is an IQ test of sorts, making it harder than stated is........
Attention to details, Anuradha! You're answer would've been correct had the question asked for top 3 horses.
This is like merge sort. In the first 5 races (when we run 5 horses per race) we get 5 sorted sublists. Each sub-list contains 5 horses sorted by their relative speeds within that sub-list. Now just merge these 5 sorted sublists to get your answer. Every "merge" means 1 race to find the fastest horse from the front of these 5 sorted sublists. To summarize: 1. Creating 5 sorted sublists of 5 horses each = 5 races 2. Getting fastest horse = 1st round of merge = 1 race 3. Getting second horse = 2nd round of merge = 1 more race 4. and so on...
Why is this harder than having five races with five horses, recording the times of each horse, and sorting the list of observed values to get the top 3?
it is seven race
7 full explanation; http://www.programmerinterview.com/index.php/puzzles/25-horses-3-fastest-5-races-puzzle/
For 3 fastest horses no doubt the answer is 7 but for 5 fastest I worked it out to be 9. And pretty sure too
1 only if you can do good flow control and horse will start with different time / space out based on best estimate. It is kind of feeding the horse to the gate.
Simply 5 races only, i.e. 5 horses in 5 race tracks per race. Finally, select top 5 horses by best fastest timing at finish line after 5 races.
As a BA, the answer I would give is: We don't yet have enough information to provide a solution. There is critical information missing that could affect the solution. And flushing out all the critical factors of the problem are one of the most important responsibilities of the BA. As a BA, it is very important to NOT start building solutions to problems that you don't yet fully understand. Actually, it's one of the biggest mistakes inexperienced BAs make: solving the wrong problem because they rushed to solutions and did not fully investigate and understand the problem! Once they're committed to the wrong problem, they end up with a beautifully done inadequate solution.
5 races, because you can fit 5 horses in 1 race and there are 25 horses. You just note the timings of each horse and compare them after all the races are done. That way, you won't have to arrange more than 5 races and stopwatch is a common tool for every phone!!!
First of all guys, you guys are making critical mistakes. You assume we have a timer but that would make it too easy, right? I'm just assuming that 5 horses a track and no timer. So I have two potential solutions: 1. You could always just race them all at the same time on 5 tracks...but that'll be too easy. Then there's my solution: --------------------------------------- First, group all the horses into five groups of five. It doesn't matter which horse goes into which group, it honestly doesn't matter. Then, race all the horse in one groups against each other. It helps to name them A-E(Group name) and the number place they came in (ex: 1st place a group would be A1). So now we have A1 to A5 and B1 to B5 and so on until E5. Then race all the 1s against each other. You should get something like A1, C1, B1, E1, D1 and then you should arrange their groups in collumns from left to right based on that race (1st on left, 5th on right). Here is where it starts getting complicated, so I will just assume it is A1 fastest in that race and E1 slowest and so on just for simplicity. So so far this is 6 races. Get E1 and race them against D2-5. You can ignore E2-5 because no matter what they can never be in the top 5 because A1, B1, C1, and D1 is the fastest of their groups and you know that the fastest five can come from those groups but not E2-5 because E1 is the slowest of the 1's and there's no way those horses can be in the top five. SO just throw them away. Since there are four spots available max you race the slowest four of the D which happens to be D2-5. Get the place numbers and arrange them. EX: D1 D2 E1 D3 D4 --- D5 E1 can be anywhere based on E1's speed but you immediately want to discard the D5. You might be saying, "Hey, why can't we discard anything below D2 since A1 and B1 and C1 are all faster than D1 and E1?" But...you could be unlucky and have A2-5 and B2-5 and C2-5 be all dirt slow. So just keep them for now. So the rule now is to keep all the horses that are in the top 5 and discard the rest until you get to A, when you know, "Oh, A1 has to be the fastest horse." 7 races. So we have our list above just for examples. Next you want to race D4 against C5 to see if our group can EVEN pass the C's, cause we might get super unlucky and see that C1-5 is faster than D1...if D4 loses in 5th place we just throw this all away and our new list becomes: C1 C2 C3 C4 C5 But otherwise we'll just assume that our D4 luckily got to 3rd place. It becomes: C1 - Our group can go anywhere inside the lines so race them again C2 minus D4 since its in 3rd place to see where they go. Let's assume we got SUPER lucky and C2 became last place in that race. - C3 C4 C5 8 races. We raced this group at that time and left out D4 because D4's fate was already decided: D1 D2 E1 D3 C2 It turnes into this: C1 - D1 D2 The single dash indicates where our new group was spliced, ofc the triple E1 dash lines show where the cutoff is, so discard everything under the line. D3 -(---) C2 D4 C5 9 races. Kay, so now we have the five fastest in the E, D, and C groups. Now continue with the B groups and let's say that our D3 passed B4 only... It becomes: B1 B2 - With our top five so far group racing against B4 B3 - D3 B4 B5 10 races. And then with another race you get this with lets say...only our top two making it through with B3 getting ironically first place: B1 B2 B3 C1 D1 --- D2 E1 Yep...E1 got kicked out. Aw well. SO NOW our to group D3 is the group above the triple dash marks. B4 B5 11 races. Now we're up against the A group. Assume that they are all wusses and that our D2 beat all of them except for A1. So we got lucky and the new lineup is this: A1 B1 B2 B3 C1 --- It was inevitable that A1 would become first but now this is how you do it! D1 D2 A2 A3 A4 A5 12 races/13 if our B1 didn't beat all of the other A's on the first try. Conclusion: 5 races (initial races)+1 race(for initial group leader ranking)+1 races(E-->D)+2 races(D-->C)+2 races(C-->B)+2 races(B-->A) =13 races/10 races if you get insanely lucky and have all our previous groups beat all the #2s of the groups.
Oh nvm ignore that it was all crushed into one place with no spaces...I'll stick a better one later.
The answer is 8 races assuming that you can fit only one horse on one track. Divide the group into 5 horses. In five races we can find the 5 fastest horses in the 5 groups. The 6th race is among the top 5 fastest horses. The top three fastest horses in the 6th race are faster than the horses in the group of the 4th and the 5th fastest horse. The horse that came third in the 6th race may or may not be faster that the horses in the group of the fastest and the send fastest horse. To determine that, we hold 2 more races. So 6+2 = 8 races is what I think is the correct answer
****************ANSWER -- 3 RACES.************* Given : 25 horses, 5 tracks Assuming : 5 horses to a track Five groups of 5 horses for each track (5 tracks, tracks ABCDE, each track with horses labeled "X#") First race allows you to find out the top 5 horses (however this isnt the true top 5) [A1,B1,C1,D1,E1] Second race the top horse is discovered (lets say its A1) Now you have to find out if B1 is better than A2, B2 is better than C1, etc. (each predecessor of each winner) We know that B1 is better than [(CDE)1] Third race we have A2,B1,B2,C1,D1 Conclusion : If both of the predecessors are NOT last place, then we have our answer at race 3 by taking the top 4 to be the bottom 4 of the top 5. Otherwise, repeat this process one more time to find the 5th placed horse.
Ans is 14
Patrick gets the right number but his logic is not the most efficient. Initially follow the standard procedure and race 5 groups of 5 (races 1-5) and then race the winners (race 6). The winner is the fastest overall. Now name the groups with the horses in the winners group A1,A2,A3,A4,A5, the first four horses in the runner-ups group B2,B3,B4,B5, the first three horses in the third group C3, C4, C5 down to E5 for the winning horse from the slowest group. The numbers indicate the highest possible position a horse could attain (i.e. C4 is definitely slower than C3, B2 and A1 but may be slower than several others). The rest of the horses are eliminated and so don't need to be labeled. Now race (race 7) the horses labeled 2 or 3 (there are 5 - A2,A3, B2, B3, C3) The top two are second and third overall and the loser is eliminated. You can eliminate a total of 5 or 6 horses - those known to be slower than the fourth or fifth place finisher and any two positions slower than the third place finisher in race 7. There aren't that many options here so you can just inventory all the possibilities if you wish. At this point you have a maximum of 7 horses for the remaining two positions You just race any 5 (race 8) to eliminate at least 3 and race the rest (race 9) to get the fourth and fifth fastest overall. If you choose all the potential fourth place finishers in race you may not need race 9. As noted elsewhere Anuradha gets the right answer to a different question than the one asked.
I have stepped among a lot of smart folks here. Maybe even several analyst's. So here it goes. I am looking at this as who the question is coming from and for what. A Business Operations Analyst for Google. Everyone is so focused on answering the question based on assumptions. Ie. That every horse and rider are at their best condition and that every track is exactly the same. That you are running all the races in the same day and that horses, riders, and tracks dont change. Horses and riders get tired, not to mention the possibility of injuries, and track conditions can change so forth and so on. So to me. In order to (analyze) to get the best possible answer. I would start by asking questions and not assuming. The more information you have leads to the best possible answer.
Guys think as a programmer, (mind my language and grammar) algorithm would be of these steps ->group the horses (5 horses each) -> get the best runner of each group (five races) -> store the horses in winning orders to five stacks (a,b,c,d,e) ->until the top 5 are found, repeat ->race all the 5 on top of the stack ->pop the winner from its stack and store in top five that all, now lets see how it will work. lets have 5 horses per group and group them as a,b,c,d,e. 1=a1 a2 a3 a4 a5 a1 (winner) 2=b1 b2 b3 b4 b5 b1 3=c1 c2 c3 c4 c5 c1 4=d1 d2 d3 d4 d5 d1 5=e1 e2 e3 e4 e5 e1 now we have best of each group, let them race 6=a1 b1 c1 d1 e1 a1 for every winner get the best candidate for the group, and race again, until top 5 are found 7=a2 b1 c1 d1 e1 a2 8=a3 b1 c1 d1 e1 a3 9=a4 b1 c1 d1 e1 a4 10.a5 b1 c1 d1 e1 a5 for any scenario just get the so the ten races. its like finding five highest numbers from 25 numbers, when you can only compare 5 at a time.
Answer- 7 Races Give names H1....H25 Divide into 5 groups as 5 can run at a time H1, H6 , H11, H16, H21 are fastest in there groups Now we can remove last two from each groups as we need fastest 3 now we left with 15 horses 6th Race- between all fastest in groups After 6th Race H1> H6 > H11> H16,>H21 Now H21 was fastest in group but is 5th in 6th race so H22 and H23 who already slower than H21 can't be in first 3 Same thing applied for H17 and H18 as H16 was fastest in group race but 4th in 6th race Same applied for H12 and H13 as H11 was fastest in group race but 3rd in 6th race Already three horses are faster than H16 and H21 so they both also can be rules out Now we left with H1, H2, H3,H6, H7, H8 and H11 But H8 is slower than H7 and H7 is slower than H6 in group race and we have H1 on top already so H8 also can't be in fastest 3 Now we left with H1, H2, H3,H6, H7, H11 H1 we know is fastest..so lets 7th race between H2, H3, H6, H7 and H11 and find the other two After 7th race you will get fastest 3
Make a group of 5 horses per track and race them all at once. The winners of each track totals to "the top 5" !
The answer is 7. Assuming: - There's no time measuring (stopwatch), just relative places. - The horses perform consistently. - A maximum of 5 horses per race. Race r1: A1 A2 A3 A4 A5 (By order of winners) Race r2: B1 B2 B3 B4 B5 Race r3: C1 C2 C3 C4 C5 Race r4: D1 D2 D3 D4 D5 Race r5: E1 E2 E3 E4 E5 Race r6: A1 B1 C1 D1 E1 (rename R1 R2 R3 R4 R5) Here comes the tricky race. From race 6, we already know the ranks. As a demonstration, assume R1 = A1, R2 = B1, R3 = C1, R4 = D1, R5 = E1. Then: 1) D1 D2 D3 D4 D5, E1 E2 E3 E4 E5 are out of the contest. Because D1 and E1 are ranked 4th and 5th in a race. 2) C2 C3 C4 C5 are out of the contest. Because two horses already are faster than C1 ==> 3 horses are already faster than C2 C3 C4 C5. 3) B3 B4 B5 are out of the contest. Because one horse already faster than B1 ==> 3 horses already faster than B3 B4 B5. Remaining potential horses are: A1 A2 A3 B1 B2 C1. Here, A1 is redundant because we already know it is fastest. Race 7: A2 A3 B1 B2 C1. The top 2 will be added to A1 to be the fastest three.
Take a look at previous horse race statistics from horse newspapers. Make your analytic and you have your top 5 horses without any new race :)
It will take min 9 race not 7: first five race and find top 3 @ every slot now total race=5; one race between first five, now we find top 3 and take 2nd and 3rd from the race as 1st is declared. race=1; 1 race between 2nd finisher of all slot and take 2 from them , race=1; 1 race between 3rd finisher of all slot and take top finisher, race=1; now we have race between 5 horses(2+2+1) and find 2nd and 3rd; so total no of races= 5+1+1+1+1=9 ;
I have grouped them in 5 groups. See the groups below. I have also ranked them in the order of the wins. So for e.g. in the first group, 1st horse came first, 2nd horse came second and so on In the second group 6th horse came first, 7th horse came second and so on. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 After the first five races, there will be two more races So the sixth race will be among the horses below (who were first in the previous races) and I have ranked them in the same order in the 6th race 1 6 11 16 21 - - So 1st horse came first, 6th horse came second and so on. Now for the 7th race, we will take the winner of the 5th race and make him compete against the first four winners of the 6 the race. So it will be something like this and I ranked them in the same order for the 7th race. 1 6 11 16 26 - So from 7 races it will is very clear that the fastest four horses are 1 6 11 16. Now, only 21 and 26 are the contenders for the fifth position, based on their timings in their respective races (6th and 7th ones), we can decide which one is faster between the two. So in total , we would need 7 races.
The question is not a good one! Let those horses be marked X1~X25. First divide them into 5 groups X1~X5, X6~X10, .... Assume the first X1, X6, X11, X16, X21 are the fast ones in each group after the first round. Also, X2 > X3, X6>X7, during the first round. We then, during the 2nd round, form the first group with those fastest from each group in the first round. Assuming that X1 > X6 > X11 > X16 > X21 after the first try of the 2nd round, we will then have X2, X3, X6, X7, X11 to form a new group and determine which are fastest 2 in this new group. Nevertheless, during the 2nd try of 2nd round, it can happen that X3 > X2 or X11 > X6. The inference fails all the relation set in the previous try. ( that the X2 > X3 ,after the first round, and X6 > X11, after the first try of 2nd round). It would be better to change all those horses to be boxes carrying different integer value. You want to sort those boxes to find which carrying the largest 3 numbers and, at any time, you can only open 5 boxes to inspect . You cannot mark the content of boxes but you can put down the relation between boxes during the inspection. Then, the number of inspections needed to find boxes with the largest 3 number will be 7#
This seems like a simple answer, but I'm pretty sure it's 9. Race 5 groups of 5, then race all the 3rd place, then the winning 3rd races the seconds, then you have either have five firsts and two seconds left or 5 firsts a second and a third left from the original racing groups. Worst case scenario you have one more race of three ones and two twos or two ones and the 2nd and third place to find the fastest there out of that last group. 9 races.
So driving down the road I figured out I should have started from the top after the first five races. Then worst case scenario you run three more races to find out the top three. 11111, 21111, 31111. Damn.
Six races, six groups. First group will have five horses, the 5 others groups will have only 4 horses. The winner of the first group of 5 horses will pass to the second race, then the winner of the second race will pass to the third race and so on, bay the end of the 6 race you will have top 5 horses.
25 horses, 5 race tracks. How many races you have to run to select top 5 horses. Solution: http://codinginterviewquestionsans.blogspot.in/2017/07/25-horses-5-tracks-5-fastest-horses.html Step 1: First, we group the horses into groups of 5 and race each group in the race course. This gives us 5 races. W11 W12 W13 W14 W15 W21 W22 W23 W24 W25 W31 W32 W33 W34 W35 W41 W42 W43 W44 W45 W51 W52 W53 W54 W55 Step 2:we race the 5 level 1 winners(w11,w21,w31,w41,w51) and assume winning order of this race is w11,w21,w31,w41,w51 (THIS IS 6TH RACE) Step 3: BECAUSE WE NEED TOP 5 AND W51 HAS COME 5TH Position that is the reason we don't need to consider W52 W53 W54 W55 now we have W11 W12 W13 W14 W15 W21 W22 W23 W24 W25 W31 W32 W33 W34 W35 W41 W42 W43 W44 W45 W51 Step 4: because we need top 5 then dont need W25 W34 W35 W43 W44 W45 now we have W11 W12 W13 W14 W15 W21 W22 W23 W24 W31 W32 W33 W41 W42 W51 Step 5: top 1 is already achieved which is W11(winner of 6th race) remaining are X W12 W13 W14 W15 W21 W22 W23 W24 W31 W32 W33 W41 W42 W51 Step 6: candidates for 5th position: W51 W42 W33 W24 W15. 1 RACE TO GET 5TH POSITION (this is 7th race) remaining are X W12 W13 W14 W21 W22 W23 W31 W32 W41 Step 7: candidates for 4th position: W41 W32 W23 W14. 1 race to get 4th position ((this is 8th race) X W12 W13 W21 W22 W31 Step 8: candidates for 2nd and 3rd position: W12 W13 W21 W22 W31. 1 race to get 2nd and 3rd position ((this is 9th race) Hence answer is 9 races.
Assuming no limit to how many horses on a track, 1, because the fastest 5 will obviously finish in 1st, second, third, fourth, and fifth place. If there is a limit to the number of horses on a track, then it will take (25/number of horses to a track) rounded up to the next integer races because you can just compare the times of all of the horses and pick out the five fastest times.
There are a lot of wrong answers on here or answers for different questions. Answer for top 5 is 8 races. You can find explanations online for this.
Friends only 5 races are required As mentioned we have 5 tracks and 25 horses. So divide them into set of 5 And winner of each set is fastest so only 5 races are required.
If there in no limit on track one race is enough. But if 5 horses can run at a time we need 6 races. 5 races to find top of each group. And additional 1 race among those top 5 horses to find out top 3.
answer is 8 suppose we have limit of 5 horses per track and all horses run together on 5 track so *-5 races-* conducting on five track, the top one of each track is selected for further race and now race is conducting between the all top 5 horses which is *-6th race-* and the remaining 4 horses will race again for 2nd and 3rd position hence total no. of race are 5(all five races)+1(6th race)+1(7th race)+1(8th race).
So many wrong answers here, first of all the question is about FIVE fastest horses, not three, so all the posts giving 7 as an answer are wrong. All the answers above 9 are also wrong. You need 7 races to find 3 fastest horses, after that by using logic you come down to either 6 or 7 remaining horses as candidates for places 4-5, and you need 2 races to figure out which 2 of them are the fastest. So it's 9 races in total.
6 races: 25 Horses - categorize in 5 races of 5 horses each (if there is a 5 horse limit) merge and sort timings for top 3 of each of the above races. Now race 6--> race 5 horses with top 5 timing --->Winner
Rate yourself on a scale of 1 to 10 how weird you are.16 Answers
just know the company's culture and if they like uniqueness or not
well, do they like uniqueness or not?
about a 5.6719151431
"Extremely weird" since my scale shows 156 lbs.
I am within two standard deviations from the norm...
I am a 10. I am a software engineer. We are all very weird.
please read this as, "hmmm, how will you answer this question?"
My perception of weird and your perception are going to be different, so will the interviewer's. Always consider the audience that you are speaking to and make sure to add some, not a lot, of levity!
One number lower than whoever came up with this question
Maximus- love that answer- halarious.
This is most likely fake, if it was asked it probably was just to amuse the interviewer, not so much of a deciding factor.
This is to assess the candidate's perception, witt, and confidence, as well as ability to present a rational response to an "irrational" question. My anwer would be: Weirdness is a subjective topic, based on a scale of (1-10) one would first have to determine what the reference point is. I live life through my own eyes and judge people and events through my own moral compass. Therefore i would be a "1" and measure everything else with me as the reference point. There is no wrong or right answer, this is about confidence in presenting your argument. I would have a different answer of course if the question was about things other than weirdness, such as talent, uniqueness, difference, etc..
my answer would be "1" because normalcy is relative.
5 because im inbetween both a normal one....^_^
CASE: Cross-selling Credit Insurance to Cardholders direct mail: .50, 1% response rate, avg balance $1000, 5% claim insurance, etc. Profitable? How make more profitable? What if response rate doubled but claims doubled? Make chart of profit curve, what does it mean if..., etc.13 Answers
see cross selling case on CO's website
Hi Sir, Can you confirm if below is the correct answer. I believe its always unprofitable. Assume 100 direct mail offers were sent for card insurance, 1% response rate means, 1 person bought credit insurance Revenue = 1% of avg balance = (1/100)*1000 = 10$ Cost = Insurance claim Cost + Mailing Cost = (Response Count* claim rate * Avg Bal) + (No. of mails sent * Cost per mail) (1* (5/100)*1000) + (100 * 0.5) = 50 + 50 = 100 $ Selling credit insurance in this case is unprofitable even without the mkt cost. For other questions if we assume, response rate = x%, claim rate = y% with the assumption, 100 offers were sent for credit insurance Revenue = x*(1/100)*1000 = 10x Insurance Cost = x*(y/100)*1000 = 10xy Mkt Cost = 100 * 0.5 = 50 Profit = 10x-10xy-50 Assumption is that y% claim rate means, of x people who take the offer, y% of those x, file for claim, which means company has to cover their avg bal of 1000 for people who filed the claim, and hence is a loss for the company. At this point its a loss to sell credit insurance. Let me know if i am doing anything incorrectly.
Great the way you laid it out, very thorough and clearly organizing your thoughts. Remember to think out loud/explain your thinking as you write during the case. Yes, something is incorrect, very small but key, changes the whole answer: you are assuming everything is in the same time unit. When you calculate revenue to be $10, you should realize right away that would be a MONTHLY amount, while claims et all would be annual. Helps to know that credit cards in the US regularly try to entice customers to add little $5-10/month services to their bills, whether insurance, credit report monitoring, etc so that $10 couldn't possibly be yearly, ie insurance less than 0.90/month.
Thanks for your response and guidance. I knew the revenue is monthly, but i thought claim rate is also monthly and hence calculated profit for each credit insurance sold per month. In light of your clarification Profit per card insurance per year = (10*12)-50-50 = 20$ per year If we chose to calculate per month, we will need to consider monthly claim rate as (5/12) and also amortize the marketing expense over next 12 month. Profit per card insurance per month = 10-(50/12)-(50/12)= (20/12)$ per month The profitability equation (per card per year) = 120x-10xy-50 For calculating any of the break even rates (x or y assuming 1 is known), 120x-10xy-50 = 0. For graph, P = 120x-10xy-50 Let me know if my analysis/answer is accurate and up to the mark. Thanks a lot for all your guidance.
Aspirant, why did you take the revenue as 1% of the average balance? I think the average balance is the revenue. Whereas the response rate is 1%. Assuming 100 people are sent the mail: Revenue : (Response rate * 1000) = 1 * 1000 = 1000 Cost : (100 * Mailing cost) + (Response rate * claim rate * 1000) = (100 * 0.5) + (1*(5/100)*1000) = 50 + 50 = 100 Profit = Revenue - Cost = 1000 - 100 = 900 (This is for 100 mails sent) So the profit per mail sent = 900/100 = $9 Now, assuming the response rate is x% (instead of 1% as given) Revenue = 1000x Cost = (100*0.5) + (x*(5/100)*1000) = 50 + 50x Profit = 1000x - (50+50x) = 950x - 50 Profit per person = (950x - 50)/100 = 9.5x - 0.5 To make it more profitable, try to increase response rate x. Now if response rate is doubled and claims doubled, Profit = (Revenue) - (Cost) =(2000) - (50 + 100) = 1850 So profit per person = 1850/100 = $18.5 Now to make a chart of profit curve, i.e profit vs response rate, we use Profit per person = 9.5x - 0.5 plot : y= 9.5x - 0.5 Meaning : To break even, we need 9.5x - 0.5 = 0 i.e x = 0.05 (approximately) So we need a response rate of atleast 0.05% to be profitable (or 1 in 2000 people to respond)
Looks like I left out the price of the insurance: customers would pay 1% of monthly balance for insurance.
@Aspirant 2: You are using response rate as a number and at the same time using claim rate correctly as 5/100. use Response rate as 1/100 wherever applicable. Revenue should be $10 Cost = 50.50 Profit(Loss) = 10-50.50 Profit (Loss) per mail sent = (10-50.50)/100
Why isn't churn rate mentioned in here? Isn't that an important factor? Or were those number given to you without probing? Thank you! =)
@ ghachla: The response rate is 1% that is 1 out of 100. I assumed 100 people are sent the mail. Hence the response rate is 1. As for the claim rate, the people who don't respond can't make claims. So out of the people who respond, the claim rate is 5% (i.e 1 person responds in 100 and out of that 1, 0.05 make the claim); Or to make it more clearer, if we assume 10000 people are sent the mail, 100 respond (because of the 1% response rate) and 5 out of the 100 make claims (because of the 5% claim rate).
@A2: I understand what you are saying. the words were just misleading. What about the insurance rate? That's revenue to the bank, right? "By intrvw candidate: Looks like I left out the price of the insurance: customers would pay 1% of monthly balance for insurance." = monthly avg balance x 10 x # of people who made claims? and then convert to annual numbers or monthly as the case may be? How do you suggest to answer the profit: annual numbers or monthly numbers?
Hello, can someone tell me how to plot claim rate vs response rate. What units goes on the x and y axis. Also how to find response rate for maximum profit?
This discussion was helpful, I have been thinking about this problem for a bit now. Only one thing, shouldnt equation for market cost be 50/x since if the response rate goes up, the cost per customer will go down as a function of the response rate.
for per customer profit calculation, we choose the number of people to whom mails were sent or number of people who responded to the mails ?
You have the choice between using first class or third class mail for a letter you are sending out to potential customers. First class costs $0.50 per piece and reach 100% of potential customers. Third class costs $0.40 per piece and reaches 80% of potential customers. Which do you use?11 Answers
It provokes you to ask questions. You do not nearly have enough information to perform the analysis. You need to know the present value of each customer, how many mailings, and calculate the point of indifference in terms of how many letters you are sending, etc.
First class should be used, as the cost of each delivered letter will be the same, but you will reach all of your intended audience. Example: 1000 pieces to be sent, sending first class costs $500 and reaches 1000 ($.50 per peice), and sending second class costs $400 and reaches 800 ($.50 per piece)
either one is good. For example, you will reach 0.8 person by spending $0.4. If divided 0.4 by 0.8, we get that we will reach 1 person by spending $0.5. So two methods have the same effect.
Need to know the Revenue per customer and no of customers before we can decide on either
@gaurav: don't make it too complicated. they just want to know the costs of reaching the customers. peppermint is right.
peppermint is not right. say you make 1000 dollars off of each customer you reach at probability .5. Assuming you send 1000 letters, first class reaches all 1000 and has total cost 500 dollars. revenue is (1000 customers reached)(.5 prob make 1000 dollars)(1000 dollars) = 500,000 and profit is 500,000-500 = 499,500. third class costs 400 dollars, reaches 800 people, gives revenue of (800)(.5)(1000)= 400,000, profit is 400,000-400=399,600. You can generalize this to a formula to determine which is better given number of letters to be sent, expected money off of customers who accept, and probability that the customers letters reach will accept.
Peppermint is right. Assumption: only when the letter has reached the recipient, he becomes a potential customer. Say they send letters to x people So, Cost per Potential Customer= 0.5/x and 0.4/(0.8x) for respective cases So, cost per potential customer is same in both cases.
I believe that the point of cost is moot; what is the content of the letter and how imperative is it that it be delivered to the potential customer? If it is vital, go for 100% success rate. If it is vital and digital copies can be provided, then go for a scanned copy. Digital signature documents also work. I know most prospects yield their info to cold callers. Why do math when you can just use logic?
The real answer is.... Send the letter through 1st class mail if the price of the product capital one is trying to sell is above $0.90. If not, send the letter through 3rd class mail. Assumptions:- •Potentially customers = actual customers Sample calculation:- Profit = Revenue - Cost Assume all potential customers = 100 Assume Price = $1 1st class mail profit = ($1x100)-($0.5x100) = $50 3rd class mail profit = ($1x80)-($0.4x80) = $48 Assume Price = $0.80 1st class mail profit = ($0.8x100)-($0.5x100) = $30 3rd class mail profit = ($0.8x80)-($0.4x80) = $32 Assume Price = $0.90 1st class mail profit = ($0.9x100)-($0.5x100) = $40 3rd class mail profit = ($0.9x80)-($0.4x80) = $40
Must also consider response of 3rd class versus 1st class-perception
cost per reached customer is the same in both cases - 50cents (.5/100 or .4/80) however you have to send more letters with 3rd mail to reach the same number of customers. There is a cost associated with printing and handling letters then you'd use 1st class.
Pretend 1% of the population has a disease. You have a test that determines if you have that disease, but it's only 80% accurate and 20% of the time you get a false positive, how likely is it you have the disease.12 Answers
divide by reciprocal, multiply by .8% = 1%
 Fact: 1% of the population has the disease (given)  Data: Test is only 80% accurate, and 20% inaccuarate (given): Assume, Population = 10,000 people 1% have the disease = 100 people 99% do not have the disease = 9,900 people Of the 1% who have the disease 80% tested +ve = 80 Of the 99% who don't have the disease 20 tested +ve = 1980  Question: How likely is it that you have the disease? To identify that you have the disease you have to test +ve and actually have the disease = 80 / (80+1980) = 80 / 2060 = 3.88%
Its easier if you draw this out in quadrants, Accuracy = 80% = P(Test shows true and having disease) + P (Test showing false and not having disease) Also, we know false positive rate = P(Test shows true but not having disease) = 20% That leaves us with the fact that P(Test being false and person having disease) = 0% Hence likelihood of having the disease is 1%
You don't need the test. The answer is given... 1% of the population has the disease... Answer: 1%
Very similar question, with a step by step walk through to the solution: http://www.mathsisfun.com/data/probability-false-negatives-positives.html Answer is 3.88%
1% Given: 1. 1% of the population has a disease 2. A test exists to identify disease carriers The questions makes no statement regarding whether or not you have taken the test, the second piece of data has no relevance. You have a 1% chance
If you have test positive, how likely is it you have the disease = All sick with positive result / All positive (sick and healthy). All sick with positive result = 1% = (1% x 80%) accurate positive result + (1% x 20%) non accurate but non false positive result. All positive = all sick with positive result + all healthy with positive resultall = 1% + (99% x 20%) = 1% + 19,8% = 20.8% ANSWER = 1% / 20.8% = 5,050505%
I thinks this question is ambiguous. If it meant the possibility that you have the disease, then 1%; If it meant the possibility that your test result shows you have the disease, then 1%*80%+99%*20%
Sample: 1,000 X=10 P=.8 If the test is only 80% accurate then 8 of the 10 infected will be positive and 2 will walk around unbeknownst, but 20% of the non-infected population will be falsely identified, 1000-10=990 (non-infected) 990*.2=198 (false positive) + 8 (true positive)= 296 (total positive tests) Of this population only 8/296=2.7% were correctly identified. This is the chance you have the disease
I think the question is not complete. It should read: "Pretend 1% of the population has a disease. You have a test that determines if you have that disease, but it's only 80% accurate and 20% of the time you get a false positive, how likely is it you have the disease "if you got tested positive. " Use four quadrants method and imagine total number of people = 1000 + Tested -Tested ------------------- ----------------------------------------------------------------- D | 80% of 10 = 8 | 10 - 2 = 8 | 10 ND | 20 % of 990 = 199 | 990 - 199 = 791 | 990 ---------------------------------------------------------------------------------------- | 207 | | 1000 So if one is tested +, they are in first column. Total of which is 207. However only 8 of them actually have disease. So probability that test came out + and the person has disease will be 8/207% = 3.865 ~ 3.7%
1%. The first sentence gives you all the info you need.
What are the number of new customers needed in order to break even?7 Answers
@ Interview Candidate : How did you even get to 8000? Did you know what was the profit per new applicant?
Can you please explain your question in more detail? Thanks in advance
It's over 9000!
Can you please explain how you calculated the breakeven to be 8000 or 9000??
are you all guessing??!! rev: fees + interest + charge, all dependent on # of customers cost: cash back+bad debt+capital cost, dependent on # of customers; system & operation+other fixed cost, independent you only have a credit limit, what can you do??
My knee-jerk first answer is "How much did you lose in terms of customer revenue? It isn't a question of how many customers you've lost, but of P+L."
Given two different mailing options with different cost rates and effectiveness rates, which option would be best in various scenarios? (For example, unlimited mailing list, unlimited budget, have both be unlimited, specific numbers of each, etc.5 Answers
Do you remember the numbers in the case? Did the let you use a calculator?
They let you use a calculator if you bring one.
For unlimited budget, the price doesn't matter so we choose the mailing option with the higher effectiveness rate. Is this correct? I'm confused with the scenario "specific number of budget". Say we want to optimize the number of successful mailing response. This number N should be = fund * effectiveness rate / cost rate. Then N is actually nothing to do with the fund. Which mailing method can have a bigger N depends on the ratio effectiveness rate / cost rate. Is this the right way to solve the problem?
If you use regression analysis or dynamic range function analysis you are more likely to get an unbounded answer. I would stick with Cartesian power thermos.
It's important to focus on the random forest approach in this example. Simple Lagrange multiplier will not provide the right determination.
If you have two coins, one is a fair coin and other is with two heads. If you get a coin and flip it, then if you know that is head, what is the probability that this coin is the fair one.4 Answers
Using Bayes' theorem: 1/3.
yes its 1/3 using Bayes Rule.
If we undertake a campaign to contact our 30 day overdue group of customers, what is our overall probability of success given the following probabilities: 50% chance of having their e-mail on file 20% chance of a customer paying their bill without us contacting them 30% chance of a customer reading the e-mail we send them 30% chance of a customer making a payment if they read the e-mail etc.5 Answers
Use a decision tree.
this is likely an applied probability question: My answer is to apply total probability formula here.
It's ~25%. .5 x .3 x .3 = .045 or 4.5% success rate from emailing. Add to that the unprompted payment rate (.2 or 20%) and get 24.5%
^That method won't work because it's possible that there are those on the email list that would respond without an email, so there would be some overlap, which means you can't just add the two together. Easiest way to approach it would be to draw some sort of Venn Diagram
20% of customers will pay no matter whether their email is on file. For the remaining 80% of customers who won't pay without contacting, the chance campaign is successful is 0.5*0.3*0.3. Total probability of success = 0.2 + 0.8*0.5*0.3*0.3=23.6%. However, this only works if the email campaign probabilities for the 2 populations (customers will pay anyway and customers won't pay w/o contact) are the same.
What would do with a Facebook user who was having trouble with their account?4 Answers
Was mostly just like some kind of customer service case interview. They're looking to see how you would problem-solve in a dynamic setting, given the possibility of interacting with these real people/users.
ask them whts d prob and then give ur suggestion on that..
Investigate about the problem being faced by the user and provide/assist in providing the best feasible solution.
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