If the positions of the first and the fifth digits of the number 14278359 are interchanged, similarly the positions of the second and the sixth digits are interchanged and so on, then which of the following will be the second digit from the right end after the rearrangement ?

Correct Answer: +ve

Explanation:

We know the Mathematical rules that

 

$$\begin{gathered} + + = + \\ + - = - \\ - + = - \\ - - = + \end{gathered}$$

Correct Answer: 7, 0, 1

Explanation:

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

 

Correct Answer: Indeterminate

Explanation:

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::

• Let us think by this analogy that,

0 / 1 = 0

0 / 2 = 0

????

???..

we can say 0 / n = 0 from this we can write, 0/0 = 0

• Again, another analogy,(from limiting concept of 1/0 = infinity)

1 / 0 = +? or -?

2 /0 = +? or -?

???

.??.

we can say 0 / 0 = ?

• Again, another analogy,

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.

Correct Answer: 749

Correct Answer: 10

Correct Answer: 8

Explanation:

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

Correct Answer: 0

Explanation:

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.

Correct Answer: All of the above

Explanation:

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.

Correct Answer: 19

Correct Answer: 1

Explanation:

'0' is called OFF State in Binary...