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 = 9. therefore 2 steps
For 6561 balls = 6561 therefore 8 steps
Condone and offence have a contradictory relation as an offence cannot be condoned. The same relationship is exhibited by relationship overlook : Aberration.
This options says that aggressiveness is common way for both.
Hence X will face in the end towards South.
The pattern is the letter change their place in cyclic order.
Then, number of legs = 4x + 2 x (x/2) = 5x.
So, 5X = 70 or x = 14.
Grenade and gun are fire arms whereas head and brain are body organs.
All flowers are trees + All trees are jungles = A + A = A = All flowers are jungles ? conversion ? Some jungles are flowers Hence III follows. All trees are jungle + No jungle is hill = A + E = E = No tree is hill. Hence III follows. All flowers are trees + No tree is hill. Hence II follows. All flowers are trees + No tree is hill = A + E = E = No flower is hill. Hence I follows.
According to question,
A's weight is twice as much as of B.
1. A = 2B.
B's weight is four and half times as much as of C.
2. B = 4.5C
C's weight is half times as much as of D.
3. C = D/2.
D's weight is half as much as of E.
4. D = E/2.
E weight is less than A but more than C.
5. A > E > C
from 1 and 2
6. A = 2B = 9C
from 6 and 3
7. A = 2B = 9C = 4.5D
from 7 and 4
8. A = 2B = 9C = 4.5D = 2.25E
so from 8 and 5
A > B > E > D > C
So A is the heaviest in the weight.
Why are some excesses bound to happen? The speaker must be assuming II.
Again, to be satisfied with the success of SOG implies that the speaker must be assuming I also.
The meaningful order would be :
(d) Jungle ? (c) Timber ? (b) Pulp ? (e) Paper ? (a) Book
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