Latest Capgemini Aptitude Question SOLUTION: there are 6561balls are there out of them 1 is heavy,find the minimum no of times the balls have to b weighted for finding out the heavy ball?

as 6561 =3^8

so minimum 8 number of times , balls have to be weighed.

min 8 times....divide 6187 by 3 we get 3 parts=>2187 2187 2187 in minimal form compare any of 2 once take higher quantity..2187 again divided into 3 parts =>729 729 729...these vll be weighted once nd again dividing 729 by 3 parts ....so on

or problems like this, just find the 3rd root of the number. For ex: consider 3 balls, u can find the heavier by just weighing 2 balls, so min = 1. consider 9 balls, divide them into 3 equal groups, now u can weigh 2 groups to find the group in which the heavier ball lies, and u ll be left with 3 and u know how to find out of 3. so its 3 pow 2 = 9, then 3 pow 3 = 27...... 3 pow 8 = 6561. So the ans is 8.

Lets make this simple.

Suppose there are 9 balls

Let us give name to each ball B1 B2 B3 B4 B5 B6 B7 B8 B9

Now we will divide all the balls into 3 groups.

Group1 - B1 B2 B3

Group2 - B4 B5 B6

Group3 - B7 B8 B9

Step1 - Now weigh any two groups. Let's assume we choose Group1 on left side of the scale and Group2 on the right side.

So now when we weigh these two groups we can get 3 outcomes.

Weighing scale tilts on left - Group1 has a heavy ball.
Weighing scale tilts on right - Group2 has a heavy ball.
Weighing scale remains balanced - Group3 has a heavy ball.
Lets assume we got the outcome as 3. i.e Group 3 has a heavy ball.

Step2 - Now weigh any two balls from Group3. Lets assume we keep B7 on left side of the scale and B8 on right side.

So now when we weigh these two balls we can get 3 outcomes.

Weighing scale tilts on left - B7 is the heavy ball.
Weighing scale tilts on right - B8 is the heavy ball.
Weighing scale remains balanced - B9 is the heavy ball.
The conclusion we get from this Problem is that each time weigh. We element 2/3 of the balls.

As we came to conclusion that Group3 has the heavy ball from Step1, we remove 6 balls from the equation i.e (2/3) of 9.

Simillarly we do the ame thing for the Step2.

Now going with this conclusion. We have 6561 balls.

Step - 1

Divided into 3 groups

Group1 - 2187Balls

Group2 - 2187Balls

Group3 - 2187Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 2

Divided into 3 groups

Group1 - 729Balls

Group2 - 729Balls

Group3 - 729Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 3

Divided into 3 groups

Group1 - 243Balls

Group2 - 243Balls

Group3 - 243Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 4

Divided into 3 groups

Group1 - 81Balls

Group2 - 81Balls

Group3 - 81Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 5

Divided into 3 groups

Group1 - 27Balls

Group2 - 27Balls

Group3 - 27Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 6

Divided into 3 groups

Group1 - 9Balls

Group2 - 9Balls

Group3 - 9Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 7

Divided into 3 groups

Group1 - 3Balls

Group2 - 3Balls

Group3 - 3Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 8

So now when we weigh 2 balls out of 3 we can get 3 outcomes.

Weighing scale tilts on left - left side placed is the heavy ball.
Weighing scale tilts on right - right side placed is the heavy ball.
Weighing scale remains balanced - remaining ball is the heavy ball.
So the general answer to this question is, it is always multiple of 3 steps.

For 9 balls. 3^2 = 9. therefore 2 steps

For 6561 balls 3^8 = 6561 therefore 8 steps

those who dint undrstud..check this..
if there r 9 balls..we wud hv divided them into 3 equal parts of 3 balls each..
take two parts weigh them(assume weighing machine as in old kirana shops) ..if they r equal then odd one presnt in other part ,if they r nt then odd one present in one of them...again we take heavier part and divide them into 3 parts...and we get the heavier one..so, for nine balls(9=3^2) we took 2 steps.
so for 6561 balls=3^8,i.e ,8 steps..

plz_explain_

1 is the ans as they are asking minimum no .of times . Minimum you have to do it once

ans 16
6561 is divided into 3 groups and they are weighted like that
it is weighted for 16 times

8 times
First, divide the group of 6561 into 3 groups of 2187. Take two arbitrary groups and weigh them. Either they will be imbalanced (in which case you've identified the group with the heaviest ball), or they are equal, in which case you know the 3rd group has the heaviest ball. Split the heaviest group of 2187 into 3 groups of 729, and continue on - so first test 6561, then 2187, then 729, then 243, then 81, then 9, then 3 balls remain. You can deduce which one is heaviest at this (7th) step in which 3 balls remain.
2187>729>243>81>27>9>3>1

Suppose there are 9 balls
Let us give name to each ball B1 B2 B3 B4 B5 B6 B7 B8 B9
Now we will divide all the balls into 3 groups.
Group1 - B1 B2 B3
Group2 - B4 B5 B6
Group3 - B7 B8 B9
Step1 - Now weigh any two groups. Let's assume we choose Group1 on left side of the scale and Group2 on the right side.
So now when we weigh these two groups we can get 3 outcomes.
Weighing scale tilts on left - Group1 has a heavy ball.
Weighing scale tilts on right - Group2 has a heavy ball.
Weighing scale remains balanced - Group3 has a heavy ball.
Lets assume we got the outcome as 3. i.e Group 3 has a heavy ball.
Step2 - Now weigh any two balls from Group3. Lets assume we keep B7 on left side of the scale and B8 on right side.
So now when we weigh these two balls we can get 3 outcomes.
Weighing scale tilts on left - B7 is the heavy ball.
Weighing scale tilts on right - B8 is the heavy ball.
Weighing scale remains balanced - B9 is the heavy ball.
The conclusion we get from this Problem is that each time weigh. We element 2/3 of the balls.
As we came to conclusion that Group3 has the heavy ball from Step1, we remove 6 balls from the equation i.e (2/3) of 9.
Simillarly we do the ame thing for the Step2.
Now going with this conclusion. We have 6561 balls.
Step - 1
Divided into 3 groups
Group1 - 2187Balls
Group2 - 2187Balls
Group3 - 2187Balls
Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.
Step - 2
Divided into 3 groups
Group1 - 729Balls
Group2 - 729Balls
Group3 - 729Balls
Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.
Step - 3
Divided into 3 groups
Group1 - 243Balls
Group2 - 243Balls
Group3 - 243Balls
Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.
Step - 4
Divided into 3 groups
Group1 - 81Balls
Group2 - 81Balls
Group3 - 81Balls
Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.
Step - 5
Divided into 3 groups
Group1 - 27Balls
Group2 - 27Balls
Group3 - 27Balls
Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.
Step - 6
Divided into 3 groups
Group1 - 9Balls
Group2 - 9Balls
Group3 - 9Balls
Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.
Step - 7
Divided into 3 groups
Group1 - 3Balls
Group2 - 3Balls
Group3 - 3Balls
Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.
Step - 8
So now when we weigh 2 balls out of 3 we can get 3 outcomes.
Weighing scale tilts on left - left side placed is the heavy ball.
Weighing scale tilts on right - right side placed is the heavy ball.
Weighing scale remains balanced - remaining ball is the heavy ball.
So the general answer to this question is, it is always multiple of 3 steps.
For 9 balls small 3^{2}= 9. therefore 2 steps
For 6561 balls small 3^{8} = 6561 therefore 8 steps

6561=6560+1
minimum number of times is 1

it's simple we can use bubble search to find the heaviest ball. and we know know it take LOGm base 2.