Answer / swati

In order to explain better, give each ball a number, said from 1 to 12. We also use ball zero to indicate a standard ball if that ball is identified as a standard ball.
Step 1. Take 8 balls to weight, ball 1, 2, 3, and 4 on the left, ball 5, 6, 7, and 8 on the right.
If they are balanced, go to Step 2-1.
If they are not balanced, go to Step 2-2.
Step 2-1.
At this time, you know the ball 1 to 8 are OK. Take ball 9, 10 and 11, plus a standard ball - ball zero. Just assume you have ball zero and 9 on the left and ball 10 and ball 11 on the right.
If they are balanced, go to Step 3-1.
If left side is heavier, go to step 3-2.
If right side is heavier, go to Step 3-3.
Step 2-2.
At this time, you know the ball 9 to ball 12 are standard balls. Assume the left side is heavier. Now you need to take 6 balls to weight the second time. You need to choose a standard ball - ball zero, ball 6 which might be lighter, and ball 3 which might be heavier, put them on the left side. Also you choose ball 2 which might be heavier, ball 5 and 7 which might be lighter. Notice that we change sides for ball 2 and ball 6.
If they are balanced, go to step 3-4.
If the left side is heavier, go to step 3-5.
If the right side is heavier, go to step 3-6.
Step 3-1.
From step 2-1, we know ball 9, 10 and 11 are standard balls. We just need to determine the ball 12 is heavier or lighter. We put a standard ball on the left, put ball 12 on the right.
If right side is heavier, the ball 12 is heavier.
If the right side is lighter, the ball 12 is lighter
Step 3-2.
From step 2-1, you know that either ball 9 is heavier, or one of statements should be true: the ball 10 is lighter or the ball 11 is lighter. We can put the ball 10 on the left, the ball 11 on the right.
If they are balanced, they should be OK, the ball 9 is heavier.
If the ball 10 is heavier, the ball 10 is Ok, ball 11 is lighter.
If the ball 10 is lighter, then the ball 11 is OK.
Step 3-3.
From Step 2-1, you know that either ball 9 is lighter, or one of statements should be true: the ball 10 is heavier or the ball 11 is heavier. We can put the ball 10 on the left, the ball 11 on the right.
If they are balanced, they should be OK, the ball 9 is lighter.
If the ball 10 is lighter, the ball 10 is Ok, ball 11 is Heavier.
If the ball 10 is heavier, then the ball 11 is OK.
Step 3-4.
From Step 2-2, you know ball 2, 3, 5, 6, 7 are standard balls. You can determine the ball 1 or ball 4 is heavier, or ball 8 is lighter. We can put ball 1 on the left and the ball 4 on the right.
If they are balanced, we are sure the ball 8 is lighter.
If the ball 1 is lighter, the ball 1 is Ok and ball 4 is heavier.
If the ball 4 is lighter, the ball 4 is OK, the ball 1 is heavier.
Step 3-5.
From Step 2-2, you know ball 2 and ball 6 are OK, because the change side does not affect the consequence. You can determine the ball 3 is heavier, or one of the statements is true: ball 5 is lighter or the ball 7 is lighter. We can put ball 5 on the left and the ball 7 on the right.
If they are balanced, we are sure the ball 3 is heavier.
If the ball 5 is heavier, the ball 5 is Ok and ball 7 is lighter.
If the ball 7 is heavier, the ball 7 is OK, the ball 5 is lighter.
Step 3-6.
From step 2-2, you know that the weighting result is changed because the ball 2 and ball 6 are exchanged. It really means that either ball 2 is heavier or ball 6 is lighter. You still have one more weight, I am sure you know how to do it!

Answer / amit

Make 4 groups of 3 balls each (G1, G2, G3, G4. Picturing a decision tree structure may help as I explain below.

1st Weighing: Place G1 and G2
If Equal then: defective ball in G3 or G4 (Consequently G1 & G2 contain balls all having the same weight i.e. normal balls or control group)
If Not Equal then: defective ball in G1 or G2 (Consequently G3 & G4 contain balls all having the same weight i.e. normal balls or control group)

Assume 1st Weighing is Equal.

2nd Weighing: Place G3 (test) and G1 (control)on pans
If Equal then: defective ball is in G4
If Not Equal then: defective ball in G3. Now if G3 is heavier (pan goes down) that means the abnormal ball is heavier than all others. If G3 is lighter (pan goes up) then that means the abnormal ball is lighter than all others.

Assume 2nd Weighing is Not Equal and G3 group is heavier.

Now we name the 3 balls in G3 group as G3-1, G3-2, G3-3.

3rd Weighing: Place G3-1 and G3-2 on pans.
If Equal then: The G3-3 is the defective (and heavier ball).
If Not Equal then: Then ball in the pan that goes down (i.e. heavier ball) is the defective one.

So 1st Weighing - Narrowing the test group
2nd Weighing - Determining heavy/light status
3rd Weighing - Determining abnormal ball

At any place above depending upon the outcome (Equal / Not Equal) you will have to change the Group in the next weighing accordingly. (E.g. Assume 2nd Weighing was equal then defective ball in G4 - so 3rd weighing you follow same procedure described above with the G4 balls).

I was stumped when I first came across this. And I found no perfect answers online. But some attempts by others helped me figure out part of the solution. Finally I found the missing part (2nd weighing finding heavy/light is crucial and 3rd Weighing with individual balls).

Since many people may be struggling with this - I thought I should help by posting the solution. Test it and vote up if its right!

Cheers!

Answer / wareagle

1. take and put 6 ball on each side of the scale
2. One side will be heavier Try #1
3. Take the heavier side and put 3 balls on each side of
side of the scale. one side is heavier Try #2
4. you now have 3 balls. Take and place one ball on each
side of the scale. If one is the heavier ball the scale
will show it. If the scale measures the same then to ball
in your hand is the heavier. Try #3

Answer / anand

first take 3 balls on one side of the scale and another 3
balls on the other side of the scale. If they don’t weigh
equal then the defective ball is here in this bunch.if they
weigh equal then defective ball is in the set of other 6
balls.Thus we have 6 balls in which we need to find out the
defective one.
now the bunch of other six balls are equal weighted..take 3
balls of from them and weigh them with the bunch of 3 balls
of the above weighted, if they are equal then defective ball
is in the other bunch of 3 balls and if they are not equal
defective ball is in this 3 ball set.(if this set weighs
heavier than the ideal group of 3 balls {that we have
selected from the remaining set of 6 balls} than defective
ball is heavier and vice versa.
now we have 3 balls..take any 2 and weigh them with the
other equal set of 2 balls..if they weigh equal then the
remaining ball is defective..and if they don’t weigh equal
then the ball which is heavier(as we have assumed the set if
3 is heavier than ideal set in the second step.) is
defective in the scale..

Answer / neelesh patil

First take 6 balls on one side of the scale and another 6
balls on the other side of the scale. The defective ball will be there in one of this bunch. Consider the bunch which is Imbalanced. Again weight that 6 balls in the Bunch of 3 balls on one side of the scale and another 3 balls on the other side of the scale. Again Considering the imbalanced weight.weight the ball Single-Single if it is balanced then defective is remaining ball.
in these way we can complete in 3 steps.

Ali Baba had four sons, to whom he bequeathed his 39 camels, with the proviso that the legacy be divided in the following way : The oldest son was to receive one half the property, the next a quarter, the third an eighth and the youngest one tenth. The four brothers were at a loss as how to divide the inheritance among themselves without cutting up a camel, until a stranger appeared upon the scene. Dismounting from his camel, he asked if he might help, for he knew just what to do. The brothers gratefully accepted his offer. Adding his own camel to Ali Baba's 39, he divided the 40 as per the will. The oldest son received 20, the next 10, the third 5 and the youngest 4. One camel remained : this was his, which he mounted and rode away. Scratching their heads in amazement, they started calculating. The oldest thought : is not 20 greater than the half of 39? Someone must have received less than his proper share ! But each brother discovered that he had received more than his due. How is it possible?

2 Answers  

Ek nadi ka Name Ek Ladki ka Name Ek PHOOL ka Name 3 sawalon ka jawab ek hi hona chahiye wo bhi Ek film ka Name hai...........

46 Answers   Ansor, Eagle, FGG, Ganga, Godavari, Infosys, JPSC Jharkhand Public Service Commission, Nilkamal, Parimal, Shani, Sola, SSC, Ugar,

3 boys ,a,b,c . have 3 chocklets .. a ate the chocklet.b thrown away the chocklet and c ate half chocklet .now how many choclect each have in their hand.

33 Answers   TCS,

There are 4 identical tabulates 2 for fever and 2 for cough, you have to take 1 tabulate of fever and 1 tabulate of cough, if you consume 2 same type of tabulates you will die. then how to consume tabulates???

9 Answers   Perfect Pac, Persistent,

Given 3 jabs A,B,C with capacities 10L 8L 5 L You can get water from tap or pour away the water in the jar 1. How do you get 9L in jar A? 2. How do you get 7L in jar B?

4 Answers  

consider 3 by 3 matrix & take 1-9 numbers but not repeat the numbers add each row and column then each row and column has the answer 13 how?

4 Answers  

The son of a rich bullion merchant left home on the death of his father. All he had with him was a gold chain that consisted of 138 links. He rented a place in the city center with a shop at the lower level and an apartment at the upper level. He was required to pay every week one link of the gold chain as rent for the place. The landlady told him that she wanted one link of the gold chain at the end of one week, two gold links at the end of two weeks, three gold links at the end of three weeks and so on. The son realized that he had to cut the links of the gold chain to pay the weekly rent. If the son wished to rent the place for 138 weeks, what would be the minimum number of links he would need to cut?

3 Answers   Zoho,

How many possible combinations are there in a 3x3x3 rubics cube? In other words, if you wanted to solve the rubics cube by trying different combinations, how many might it take you (worst case senerio)? How many for a 4x4x4 cube?

1 Answers  

Theere is a firm by name MR Bean and co. and in the firm there are 20 machines which produce a ball of 1000gms. The balls are packed and are delivered to another company by name royal PVT ltd .The production manager of the Royal company complained that a set of balls produced by a machine of the 20 machines of MR bean and co are defective..There is a loss of 100gm in the balls produced by a machine.... You are provided with a weighing machine ...You are to make use of the machine only once and find the defective machine.

6 Answers   Infosys,

how would u find d exact number of white maruti cars in mumbai???

15 Answers   Bhel, Infosys, RR, TCS,

There are 3 societies A, B, C. A lent cars to B and C as many as they had already. After some time B gave as many tractors to A and C as many as they have. After sometime c did the same thing. At the end of this transaction each one of them had 24. Find the cars each orginally had.

4 Answers   Infosys, TCS, Zoho,

justify 1-1=11...

1 Answers