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Arithmetic Reasoning problems

1. 7/9 of the people present in a hall are sitting in 9/13 of the chairs available and the rest are standing. If there are 28 empty 9 sitting chairs, how many chairs would have been still empty if everyone in the hall was sitting?
Options
A. 15
B. 12
C. 18
D. 10

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2. In the given numbers, if in each number first digit is replaced by the third digit, second digit is replaced by the first digit and third digit is replaced by the second digit then which number will be the biggest ? 653 749 872 933 541
Options
A. 933
B. 872
C. 749
D. 653

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4. Rajitha has a calculator, which has 10 digits. She want to clean the calculator by removing all the keys and while reassembling it, she interchanged the locations of 3 pairs of keys. For example, if she placed the key ?3? in the place of the key ?7?, she placed the key ?7? only in the place of the key ?3? and not any other key. All keys that were interchanged corresponded only to numbers. Later, she input some calculations and obtained outputs as provided in the table below. The input is shown in terms of the keys that Kavitapressed, while the output is the output displayed by the calculator. Output: 1. 95 + 84=128 2. 128 + 64=195 3. 912 + 43=942 4. 178 + 19=192 Question : Which of the following pair of digits is unchanged ?
Options
A. 9, 0, 3
B. 1, 3, 8
C. 7, 0, 1
D. 2, 4, 1

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6. 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 ?
Options
A. 5
B. 2
C. 7
D. 4

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7. There are 6561 balls are there out of them 1 is heavy. Find the minimum number of times the balls have to be weighted for finding out the heavy ball ?
Options
A. 2414
B. 204
C. 87
D. 8

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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