# Manager Interview Questions in Hyderabad, India

Manager interview questions shared by candidates

## Top Interview Questions

### Program Manager at Microsoft was asked...

Consider a lift lobby where people are waiting for the lift. Now when lift arrives people get in irrespective of who came first. Thus, the person who came first keeps waiting and others keep getting in. Design a lift system which can solve the problem of this person. 17 Answersincrease the load capacity of lobby by its size andhaving a narrow path in front of lobby so that the person who came first can only enter first, just like we stand for ticket in railway station. a.)If you are talking about the technical customization, HAVE A ELEVATOR INSTEAD OF LIFT. Regards Vishnu 9884578010 install a camera with face recognition near the external button system, system scans the face of the person and stores, when the lift comes, there is a face recognition system too.. the persons can only move in the lift in the sequence of arrival in the lobby and subsequent face scan ... if the person who came first, gets his face scanned first does not get to move in the lift first .. instead an other person moves and tries to get his scan ratified ... system would throw a voice messege ... pls let the person before you in the que come first!!!! Show more responses I don't find any of the answers posted above by Vikram, Vishnu or Deb as appropriate! @Vikram: Can you increase the size of a lift lobby??? FYI a lift lobby is the space u generally have between an array of lifts located on opposite sides of a wall. Having a narrow path and having a long waiting queue?? Imagine such a situation in an office or a shopping mall or an apartment! and will u demolish an existing building to reinstall this design u are suggesting - who will buy ur design? @Vishnu: Elevator - do u mean escalator? Imagine a 30-40 floor building and and escalator going 30 floors up and down! Think of the time u will have to take to reach the top floor. There can be n types of solution, but it is important to evaluate every solution. we are not suggesting just an alternate solution, but addressing the users problem. I think we are increasing his problem by this design. And also consider the technical feasibility of such design in say 50 floor building? and again will u demolish an existing building to reinstall this design u are suggesting - who will buy ur design? @Deb: Again, we are solving the users problem here and not just suggesting solution for the sake of it. Imagine such a lift system installed in your apartment complex! The camera would know that you are no. 7, how would you know that? not knowing this you will try to enter a lift and then u say that the lift will stop u!!! and there may be 6 lifts in the lobby. Say at a particular instance of time 2 lifts - lift 1 (going down) and lift 4 (going up) arrive. you will not let in person no 10 who wants to go down while all other 9 who came before and wants to go up. And how would the camera know which floor the person wants to go?? I am just giving pointers and this is the way the options will be struck down, if they do not address the actual problem/ or you do not consider the feasibility of implementation. The solution that i gave was this: When one enters a lift lobby first thing he does is to press up or down button to call a lift going up or down respectively. Once we enter a lift we have the keyboard to press the floor number inside the lift. What i suggested is to have a similar keyboard somewhere at the entrance of the lift lobby along with a small embedded display screen. So that when i am entering the lobby i press say 11 (to go to the 11th floor). While i press that, instantly the lobby shows me the lift number (say Lift 1, 2, ..6) i need to get into. So i know which lift i need to get into and i will obviously stand in front of that lift. At any time, the same lift number will not be issued to more than X people (where X is a predefined number of persons who can go in a lift at a time). Here the design is not in reinventing the entire lift system. Nobody will incur so much real estate cost to scrap the existing lift system and build a totally new designed one! The actually design here would be to build the backend logic that the lift system would use to assign the right lift number to every arriver in the most efficient way. e.g. Assume a simple start state (for a 2 lift lobby) when all lifts are stationary at the ground floor. When P1 (going to F10) comes System assigns him L1. For P2 (F5) assigns L2. Now to P3 (F12) assign L1, P4(F11) assign L1, P5(F4) assign L2 and so on. This is a simple situation. All we need is an algorithm which can address this lift assignment in the most efficient way. I WAS SELECTED :) The lift system you described is one they use in a large hospital close to where I live. I think they did it so that you would not have to press buttons inside the lift and not for the particular problem described in the question. Have a two way door opening in the lift. But always entry should be in one side and exit other side. Sorry for my previous response. Wrongly understood the question @Anonymus (One who posted this question): I see two major flaws in this solution. 1) If a group of people wants to go to the same floor together and only person among the group presses the button, then the lift under this system would assume that only one person would enter and allot only 1 person's capacity to the entire group, thus exceeding the capacity frequently. Under this system, every individual would have to press the button even if they come together which is a bit of an inconvenience I feel. 2) Also, if someone presses the button, is allocated a lift and then doesn't enter, then the lift wont know and will not 'free' the capacity leading to it running under-capacity when there are other people waiting to use the lift. IMO there should be a token vending machine just outside the lift where as and when people come, take out token. Now these tokens are embedded with a single chip with a kind of dynamic number saved on them. The vending machine would draw out tokens in a numeric sequence, once a person enters a life he would have to punch/drop his token into a reading device which will open the gate for the lift in a particular order. Once all are done, the machine would be reset and again the procedure can be repeated for the next set of people. This is something similar to Metro Train Tokens. Create a queuing system with barricades, just like at airport ticketing counters. What's the problem here i can't find it ? increase the no of lifts to serve for peak periods. Decrease the width of lift doors. enhance the lift algorithm to serve better. Try to control the flow, by analyzing the causes. Keep a notice board or appoint a security , if required. It was not mentioned i the question that there are several lifts in th lobby. Hence, it a very simple solution for the first person to enter first is to arrange a baricade- Q system. Thats it....age old and very simple, price less solution. Show more responses One way can be to have a id card scanner installed in the lift area. Since it is an office it is safe to assume that people will be having their id cards. When people come to the lift they scan their id cards, there can be a tv screen next to the lift which can show the order in which people have arrived. In my view , This is not any technical issue we are trying to solve, problem is rooted through people discipline. Install cc cam and publish few lift lobby disciplines, make sure violation of rules will be consider as HR escalation. I am sure this problem will be corrected without any huge technology investment. With this approach two benefit I can see : Huge R&D investment can be saved. If we could correct People behaviours that will be a permanent fix for the many other initiatives. If the question is asked to measure a person's reasoning and interviewer expecting a solution through some logical technology initiation , selected candidate has given a reasonable answer of course a complex algorithm need to be develop with all permutations and combinations along with pros and cons, still it is possible through embedded technology. It’s simple guys, keep one notice FCFS 👍🏻 Form a que..so that the person arrive first will be standing in front and enters the lift first |

### Operations Manager at Amazon was asked...

If in a room there are 10 persons and if each person has to shake hands with all other then total how many hand shakes will be there 16 Answers10C2=45 should be 90 Forgot 90 Milk Shakes Show more responses 10(10-1)/2 "Each other" leaves this very open-ended; that depends on if A shakes with B or A shakes with B & C, OR if A shakes with all the other nine, etc. I would say the answer would have to be one of two: 10 or 100. If each person chooses only one to shake with, it would be ten. IF each person shakes with everyone there, all ten, it would be 100. Since this question is pretty vague, Some people may come to the conclusion that the answer is Either 90 assuming everybody stayed to shake hands with each other meaning the first person shook hands with 9 people and the 2nd person did the same etc etc bringing it to the conclusion that you got 90 handshakes. Another answer towards for people would be 45 being that the first person gave a hand shake to 9 people and then left and then the 2nd person gave a handshake to 8 people n then left etc and etc making it 9+8+7+6+5+4+3+2+1=45. simply formulas: n*(n-1)/2 (n-1)*(n/2) Lock in 90 thanks 45 5 n(n-1)/2 = 10(10-1)/2 = 45 If there are 10 persons, the first shakes hands nine times, the second eight times (already shook with first person), and so on. In other words, 9+8+7+6+5+4... 1. n(n-1)/2 2. 10*9/2 3 90/2 4. 45 If you say 90 you are including the repetetions. If first person shakes 9 times, the next one requires only 8 shakes. So it goes like, 9, 8 ,7 ,6...1. Can use the formula n*(n-1)/2/. So we get only 45 shake hands. It is similar to no of squares in the chess board. There squares will apply. 9 Show more responses 45 If one person is lefty? 45 |

### Operations Manager at Amazon was asked...

If u are running at 10km/hr and anothet person is running st 20km/hr then in which round u both will be at same point 11 Answers4 4 is wrong answer to this question. Its answer for next question. Posted here by mistake No matter what size track they run on, they will meet at the starting line every two laps. Show more responses With slower person's perspective, they will meet at starting point each round. Person running 10 km/hr, will complete one round, while the other person will complete two round during same time. You are assuming that the track is exactly 1KM in length. Most tracks are not that, what are they 440? Yes, with reference to the slower runner, they'll meet every round and with reference to the faster runner, they'll meet every alternate round. There is not enough infomation here. You really do need to know the length of the track to determine future meeting points. every round for the slower runner. at the end of second round (w.r.t to the person with 20km/hr speed). From 20kmph perspective Every alternative round they will meet at starting point , From 10kmph perspective every round they will meet at starting point. 2nd round |

### Program Manager at Google was asked...

what is angle between hour hand and minute hand in clock at 4:20 ? what is biggest conflict management you have handled in your work place 13 Answers4:20 - ANGLE WOULD BE 120 CIRCLE IS 360 DEGREES , EVERY 5 MINUTES = 30 DEGREES IF min's hand rotates for 360 deg, hr's hand rotates for 30 deg - so for 60 min - hr's hand rotates for 30 deg. For 1 min it rotates for 0.5 degree. Hence, for 20 min it rotates for 10 degrees. The angle between hr's hand and min's hand should be 10 deg. have you ppl looked at an analog clock recently it is 0 or 360 degrees Show more responses The angle between hour and minute hand in 4:20 is 10 degrees. For a minute, the hour hand rotates by 30/60 = 1/2 degrees. hence, for 20 minutes it rotates by an angle of 20*1/2 = 10 degrees. The formula is 180 - | 180 - | m * 6 - (h * 30 + m * 0.5) | Heres the complete formula.. for better understanding.. ABS is absolute value 180 - ABS (180 - ABS ( m * 6 - (h * 30 + m * 0.5) ) ) An analogue clock is degrees. 360/12=30 degrees per hour 360/60=6 degrees per minute So for 4 hours - 30*4=120 degrees For 20 mins - 20*6=120 degrees therefore, Answer is zero 20 minutes is 1/3 (20/60) of the way around the dial, 4 o'clock is also 1/3 of the way around a clock (4/12), so they are perfectly aligned and the angle between them must be zero. There are two approaches to this question: - If you assume the hour hand stays put and only moves every hour o'clock (which is a simplification, but a reasonable one), then the angle is 0º. - If you assume the hour hand rotates at a constant angular speed, then the answer is 10º. Since there's 360º / 12 = 30º between hours: + The minute hand points at 4. That's 30º x 4 = 120º. + The hour points slightly past 4. Since a third of the hour has passed (20' = 60' / 3), then the minute hour has moved a third of the way between hours. Hence the hand is at 120º + 30º / 3 = 130º. The difference between the two hands is 130º - 120º = 10º So I guess you get some poits for saying 0º, more points for saying 10º. My question is, did you get to use pen and paper, or you had to answer on the spot? in every 60 minutes hour hand moves 30 degrees. so at 14:20 ( it means that 20 minutes have past since last turn) hour hand moves 20*30/60 = 10 degrees so the answer is 10 degrees... 360/(12*3) = 10 0 10 degrees |

### Operations Manager at Amazon was asked...

There are 27 balls with one ball having additional weight. Total how many attempts you have to make to check all balls using a sisaw 10 Answers4 3. group the 27 balls into3 groups of nine, weigh 2 of the groups. If one is heavier, continue with that group. If they are equal, continue with the unweighed group. Break those nine into 3 groups of 3, weigh 2 groups. If one is heavier, continue with that group, otherwise use the unweighed group. With the 3 left, weigh any two. If one is heavier, you have it. If they are the same, then it is the unweighed ball that is heaviest. You have used the scale 3 times friend what sort question they may ask to electrical student.amazon is coming to my college.please reply. Show more responses Break it into 3 groups of 9, based on worse case scenario 1 test A & B group, if same go to step 2 if different go to step 3 2 test A & C group, take the heaviest group and split into 3 groups of 3 3 test A & B group if same go to step 4 if different go to step 5 4 test A & C group, take heaviest group and split into single balls 5 test A & B ball, if same go to step 6 if different you have the answer 6 test A and C ball Worse case is 6 times you will need to use the scale It’s actually FIVE different weightings. Try a visual tree with the breakdown of structure. Stars indicates each weighing. 27 split 9 vs 9 9 9* vs 9 9* split 3 vs 3 3 3* vs 3 3* split 1 vs 1 ( 1) (1*) If last weight is uneven that item is the heaviest. If it's even then the unmeasured stone is the heaviest. First make three groups of 9. then make three groups of 3 with the heaviest. Finally 1 to 1. So total 8 attempts can pick the odd one. 1, 2, or 4 One attempt in the best case scenario and four attempts in the worst case scenario. Using the method below, your answer will be revealed in the first, second, or fourth attempt using the seesaw. 1. Divide ball in 2 groups of 13 and weigh. Your 27th ball not being weighed. --> If equal, you have the answer on first attempt. The heavy ball is the 27 ball, not weighed. --> If not equal, remove light balls and repeat the comparison with the heavier group of balls. 2. Divide balls in two groups of 6 and weigh. Your 13th ball is not weighed. --> If equal, you have the answer on the second attempt. The heavy ball is the 13th ball, not weighed. --> If not equal, remove the light balls and repeat the comparison with the heavier group of balls. 3. Divide balls in two groups of 3 and weigh. --> Remove the light balls and repeat the comparison with the heavier group of balls. 4. You now have three balls. Weigh two of the balls to get your answer. --> If equal, the heavy ball is the one not weighed. --> If not equal, your heavy ball is identified by the low end of the seesaw. 3 You can guarantee the answer in 3 steps, if required. However, you no longer stand a chance of obtaining the answer in the first 2 weighings as I described immediately above. 1. Divide balls into 3 groups of 9. --> Weigh two of the groups. This will identify the heavy group. 2. Divide the identified heavy group into 3 groups of 3. --> Weigh two of the groups. This will identify the heavy group. 3. Separate the 3 balls in the heavy group and weigh. --> This comparison will identify the one heavy ball from the original group of 27. 3 attempts - attempt 1 - divide 3 sections 9balls 9balls 9balls pick a section which is different in weight, attempt 2 - divide 3 sections 3balls 3balls 3 balls pick a section which is different in weight, Attempt-3- divide 3 sections 1ball 1ball 1ball pick a ball which is in different in weight Attempt 3 - 4-5 attemps |

### Program Manager at Google was asked...

How many ways that you can choose 3 desserts from a menu of 10? 7 AnswersC (10, 3) = 120 - This is very simple Just to clarify: the original reviewer's answer - C(10,3) - is indeed incorrect. More specifically, C(10,3) represents a number of ways to pick 3 DISTINCT desserts. The problem statement does not suggest that the desserts should be unique; moreover, it's very natural to order three identical desserts! I believe the correct answer to be a sum of C(10,3) /* number of ways to pick 3 distinct desserts */ + 10 * 9 /*number of ways to pick 2 matching and 1 distinct dessert */ + 10 /* number of ways to pick 3 matching desserts */ = 220 if repetition is allowed, its 210 ways if repetition is not allowed, C(10,3) = 120 ways Show more responses If order matters (it would to me if I were eating them), and repetition is allowed, the answer would be 1,000 since you can choose any of the 10 for the first, second and third dessert resulting in 10 X 10 X 10 possible dessert sequences. 10x10x10 = 1000 10*9*8 = 720 combinations 45 I suppose |

### Product Manager at Google was asked...

If you host a celebrity website which displays ads and suddenly notice a drop in traffic to your site/clicks on ads, how do you root cause the issue? 5 AnswersKey factors to consider - Fans' profiles, celebrity's profile, profession, products advertized, and macro issues. Drill down on each level to extract potential issues relating to the drop. 1. is the trend all across the site and for all ads? if yes, factors would be content, look & feel of the site, ease of navigation, introduction of new features on the site if no, then every page has to be looked into individually on the above mentioned factors. also compare with those pages which still get more clicks(usually the landing page) 2. is the fan base constant? customization of the site to let the users decide what kind of ads they would like to see A/B testing with Web optimizer. Show more responses First check whether all ads and site is working fine or not. If it's all good then for site, check the least performing channel from where traffic is comparatively low. Check why particular channel has lower traffic. Next for ads, identify which ad's contribution to traffic is lower than usual. Check it's position, visibility. Identify the source of problem to fix it. Check multiple things 1. Is there any patch applied by us recently which is giving a bad user experience 2. Is it a phenomenon across other sites too or only this specific celebrity? 3. Any statements or negative news? 4. Any new celebrity making headlines and so captured all attention 5. Any problems with our web server access or hosting...etc 6. Celebrity profile. User profile , geography....etc 7. Any changes in ad algorithms which might have given unrelated ads.. 8. Website performance or latency metrics |

### Program Manager at Google was asked...

suggest 5 ways of improving google maps & gmail ? 5 AnswersIn Google maps, needs to develop an option which would help the user to simply select a place as FROM ADDRESS and select another place as TO ADDRESS, representing GREEN and RED buttons respectively. Route is shown from Green button to Red button, may add options if selected place is not clear, mean that nearest recognized places. Steps: 1) Select the option of SELECT DIRECTIONS 2) Select/Drag/Click GREEN button and place it on the starting address / from address Ex: 10 Kent Ct, Princeton, NJ 08540, USA. 3) Select/drag/Click RED button and place it on the ending address / to address Ex: Princeton Junction at Vaughn Drive. 4) Needs to show the route with available options. Just like in gmail, get only map view for low speed internet connection. where it needs to show the basic map as requested by the user. As in rural parts of India the phone/internet signal strength will be at its minimum/low. This might be a great help for the user to select basic map view for low speed internet connection. In Gmail, there needs a option like "Show mails which have attachments", this will reduce user's time in searching for the old mails which have attachments. Show more responses While composing a new mail in gmail, icons related to Formatting Options, Attach Files, Insert Files using Drive, Insert Photo, Insert Link etc links needs to be shown to the user at the bottom of the compose pane. In Google Maps just like Notifications icon, need to include and update the status of new mails if linked with the gmail account. In gmail need to add a link to Google Maps. |

1. You were driving a truck, your combined weight(yourself & the truck) is 1 ton & you have a bridge to cross which can sustain max 1 ton weight. When you reach the middle of the bridge, a bird weighing 500 gms sits on the bonnet of the truck. However you cross the bridge safely without it crashing. When the max weight the bridge could take is 1 ton, then why didnt the bridge crash? 4 AnswersSince the bird sat on truck in the middle of journey. torque =r*F(at the moddle r=0 , r will increase when the truck moves toward the other end of the bridge.When the truck will reach at the end the torque will be maximum but at that time truck would have covered the bridge, so the bridge did not crash) by the time the truck reached the center of the bridge, it had burned away enough of its fuel to compensate for the weight of the bird i.e. the truck had burned away more than or equal to 500gms of fuel. The weight of the bird nullifies the normal reaction occured by the weight of the truck and driver. So the bridge didn't break Show more responses What kind of questions do they ask in the 3rd round of the interview? |

### Product Manager at Google was asked...

How would you estimate the number of seniors in India who use gmail? 3 Answerspopulation of india: 1.2 bil approx seniors in india: 100 mil internet reach in india: 10% this leaves about 10 mil seniors with internet access. gmail reach in india: close to 70% would give the answer as 7 million gmail users Defining seniors as people above 55yrs 1.2 bn people in India. India has one of the highest proportion of working age population. So I assume about 30% of the above population to be above 55 = 360mn. The rural-urban ratio is about 60-40, which leaves about 200mn rural and 160mn urban. Safe to assume that there are practically zero seniors from rural areas on gmail due to lack of education, internet penetration, power etc. That leaves about 160mn in urban areas. Breaking this up into tier1 and tier 2,3,4, tier1 cities would have 40% of the seniors and 60% in tier 2,3,4, ie, 60mn in tier 1 and 100 mn in tier 2,3,4 The internet penetration should be about 40% in tier 1 cities and 10% in tier 2,3,4 cities. This leads to about 24mn seniors with internet access in tier 1 cities and about 10 mn in tier 2,3,4 cities. Assuming an adoption rate of 20% in tier-1 and 10% in tier 2,3,4, it leads to about 4.8mn in tier-1 and 1mn in tier 2,3,4. So in all about 6mn gmail accounts by people over 55 in India. Not everyone has gmail account, google launched in 1998 and in 2017 it's reach in India is about 35%, Out of which 30% are youth , so 5% of 1 billion people |

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