Latest CTS Aptitude Question SOLUTION: There are 9 balls of equal size and same weight(they look similar) except 1. How may weighs required to find the dissimilar ball using a weighing balance?

Attemt 1:
first of all take 6 balls put it on weighing machine 3 on one side and 3 on another side. both are equal they heavy ball is in remaining 3 balls which are kept in aside starting.
Attempt 2:
same put one ball in one side and another ball another side ,put remaining ball aside.if both are equal ,heavy ball is aside ball.

Minimum 1 weighs to find dissimilar ball.
Expln : Put one ball aside and put the other 8 on the scale - 4 on each side.
If the scale is balanced then - the one you put aside is the heavy ball.

It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.
1. Take 8 coins and weigh 4 against 4.
o If both are not equal, goto step 2
o If both are equal, goto step 3
2. One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.
o If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
 If both are equal, L4 is the odd coin and is lighter.
 If L2 is light, L2 is the odd coin and is lighter.
 If L3 is light, L3 is the odd coin and is lighter.
o If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2
 If both are equal, there is some error.
 If H1 is heavy, H1 is the odd coin and is heavier.
 If H2 is heavy, H2 is the odd coin and is heavier.
o If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4
 If both are equal, L1 is the odd coin and is lighter.
 If H3 is heavy, H3 is the odd coin and is heavier.
 If H4 is heavy, H4 is the odd coin and is heavier.

please any 1 give correct answer with explanation

Minimum no of attempts =1
Explanation- put 8 balls on the weighing machine, 4 on each side and keep 1 ball aside.
If the balance comes out to be equal, then the ball kept aside is the heavy ball.

Maximum no of attempts= 3
Explanation: put 8 balls on the weighing machine, 4 on each side and keep 1 ball aside (ATTEMPT=1). if the scale is not equal, take the 4 balls from the heavy side.
Again put these balls on the weighing machine, 2 on each side (ATTEMPT=2), if the scale is not equal again, Take out the 2 balls from the heavy side
Again put these balls on the weighing machine, 1 on each side (ATTEMPT=3). Now we can easily get the heavy ball out of the 2 balls in the weighing machine.