Xavier Aptitude Test 2023 Paper

Question : 67 of 102

Marks: +1, -0

Raju and Sarita play a number game. First, each one of them chooses a positive integer independently. Separately, they both multiply their chosen integers by 2, and then subtract 20 from their resultant numbers. Now, each of them has a new number. Then, they divide their respective new numbers by 5 . Finally, they added their results and found that the sum is 16 . What can be the maximum possible difference between the positive integers chosen by Raju and Sarita?

67
58
49
40
None of the above

Validate

Solution:

Let the positive integers Raju & Sarita chose be 'r' & 's'.
After doing the given operations, the final numbers they get are

\;{2 r-20}/{5} & \;{2 s-20}/{5}

Adding results we get,

\;{2 r-20}/{5}+\;{2 s-20}/{5}=16

\;2 r-20+2 s-20=80

\;2 r+2 s=80+40+40=120

\;r+s=60

For maximum value of

∣ r

-s|, one of '

r

' & 's' has to be maximum and other has to be minimum.
As

{\r} & {\ s}

are positive integers minimum value they can take is " 1 ".
If

r=1

, then

s=59

|r-s|=|1-59|=58

s=1

, then

r=59

|r-s|=|59-1|=58

Hence maximum value of

|r-s|

is 58 . Option B is correct.

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