Operations interview questions shared by candidates
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 there15 Answers
should be 90
Forgot 90 Milk Shakes
"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
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.
If one person is lefty?
What are your weekensses?4 Answers
I am very talkative and a poor listener.
i would like to do only interested subjects , once i get interest in the subject, i never leave, but its difficult to do which i am not intersted : thats my weakness.
Ever never I loose my temper when anybody cheat me acoording to work
My only weakness is I try to do things genuinely and with perfection, which sometimes creates a problem because it takes Time to do things with perfection. I prefer quality over quantity.
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 point11 Answers
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.
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.
There are 27 balls with one ball having additional weight. Total how many attempts you have to make to check all balls using a sisaw10 Answers
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.
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 -
Estimate number of red cars in Delhi3 Answers
Total population of Delhi - 2 crores On an average every household has 0.75 cars therefore total cars in Delhi 1.5 crores Now out of the total colors sold suppose the combination of all cars is as follows Black = 33% White = 22% Blue = 15% Red = 10% Other shades = 20% Therefore 10% of 1.5 crore will be 15 lakh cars in Delhi are Red
total population of Delhi= 2 crores (approx) avg. family size is 4 members therefore number of households in delhi will be around 50 lakhs ( 2 cr. divided by 4) out of these 70% families will have the required annual income to afford a car, or will even require a car (estimated) thus 70% of 50 lakhs is 35 lakh households assumuing each household has atleast 1 car, we get 35 lakh cars on Delhi roads Now the probability of a car being red is 1/10 that is, 1/(7 colors + black + white + Grey ) thus number of red cars in Delhi is 3500000x1/10 350000 cars
Population of Delhi= 2 crores Avg. family size = 4 No of families in Delhi=2.2 Cr /4 = 55 lakhs No of families who can afford a car in Delhi = 50%(approx) no of families having more than 1 car will be a very small number as compared to 55 lakhs, so ignoring that So no of cars in Delhi = 0.5*55 lakhs = 27.5 lakhs Preference of red colour in car is approximately 5% so probability of a red colour car =5/100 = 0.05 so No of red coloured cars will be 0.05 * 27.5 lakhs = 137500
Would you think this organization is suitable for you?5 Answers
Precisely, thats why I am here.
YES, Becouse i am hard worker and able to this job.
Precisely,that's why I am here. Because I am hard worker and able to this job.
Difficult questions would be from accounts or share market.2 Answers
Be strong with basics of accounts and share market.
can u tell the questions asked about that..
Suppose that 10% of the products sold online were incorrectly priced. What could be a possible solution to this problem and detail out the approach you'll follow.2 Answers
It depends upon the situation,in both the cases what could we do is if cost of the product were priced higher than what expected than we can simply requested to the customer that next perches item by them will be adjusted accordingly and visa versa if cost is less...
I guessed that they were looking for a preventive approach as a solution. So I told them that there are two things that can be done. One is to perform an external benchmarking with the direct competitor on similar product portfolios. This will involve a detailed analysis on how they are able to sell the same product at a better price value. A possible hypothesis is that the competitor has a larger pricing band owing to the number of suppliers in his kitty. Second thing is to attract customers by giving out offers and discounts. I atleast thought this answer went well them :-)
describe an instance when u had topay attention to details1 Answer
please share the answer for this q . I am not able to understand this question.
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