## Questions d'entretien

Entretien pour Operations Manager

-Hyderâbâd

# 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

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## Réponses aux questions d'entretien

10 réponse(s)

30

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

Tim le

1

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.

Rico le

1

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 -

Utilisateur anonyme le

2

4

Anub le

0

4-5 attemps

Utilisateur anonyme le

1

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.

Dawn le

0

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.

Utilisateur anonyme le

0

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.

Rico le

1