0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
0 | 1 | 4 | 8 | 2 | 6 | 5 | 7 | 3 | 9 |
Here pairs interchanged are
2, 4
3, 8
5, 6
Unchanged are 0, 1, 7, 9
Indeterminate are those,which mathematics can?t determine.
Because we don?t actually know which one is the exact answer for this due to having lots of results for this.
Explanation::
0 / 1 = 0
0 / 2 = 0
????
???..
we can say 0 / n = 0 from this we can write, 0/0 = 0
1 / 0 = +? or -?
2 /0 = +? or -?
???
.??.
we can say 0 / 0 = ?
1 / 1 = 1
2/2 = 1
.??..
???..
we can say, 0/0 = 1
Now, you see it?s confusing. Every analog is mathematically correct then which one is the right answer for 0/0. Therefore Mathematics can?t determine this fact.
So, 0/0 is indeterminate.
The Lami's theorem states that " If three forces coplanar and concurrent forces acting on a particle keep it in equilibrium, then each force is proportional to the sine of the angle between the other two and the constant of proportionality is the same." This theorem is derived from the Sine rule of triangles.
18 tens 20 ones, the given stateent can be split into
18 tens = 18 x 10 = 180
20 ones = 20 x 1 = 20
Then add both => 180 + 20 = 200
Hence, 18 tens 20 ones = 200.
We know the Mathematical rules that
Given number is 14278359
As given 1st and 5th digits, 2nd & 6th, 3rd & 7th and 4th & 8th digits are to to be interchanged.
Therfore, the rearranged number is 83591427
Then, the second digit from the right end is 2.
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
The place value of 0 in 103 is given by the product of digits place in the number and the digit.
Here 0 is in 10's place in 103
=> 0 x 10 = 0.
Hence, The place value of 0 in 103 = 0.
Aptitude tests measures an individual's level of competency to perform a certain type of task and their mental and physical potentials in dealing different situations. These tests include General reasoning, Verbal comprehension, Perceptual speed and numerical operations.
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