# Operations Manager Interview Questions

Operations manager interview questions shared by candidates

## Top Interview Questions

### 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 15 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? |

### 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 |

### 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 |

Would you think this organization is suitable for you? 5 AnswersPrecisely, thats why I am here. YES, Becouse i am hard worker and able to this job. Yes Show more responses Yes Precisely,that's why I am here. Because I am hard worker and able to this job. |

### Operation Manager at Technomine was asked...

What if better offer from other company ? 1 AnswerMoney is a secondary thing Mainly Job satisfaction if its getting here then NO for other offers |

### Manager Operations at Matrix ComSec was asked...

Why you want to leave schneider 1 AnswerGetting opportunity from asst. manager production to factory head |

### Operations Manager at DuPont was asked...

career aspiration for next 10 years? 1 Answerbelief in self & organisational growth ,career flows with business strategy . |

where do you see this company after 5 year 1 Answerstill planning to see that |

Menu costing staff hiring 1 AnswerOverall ok |

Something any myself 1 AnswerWhich places worked |

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