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Question 13 : 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?
Let the number chosen by Raju be ‘r’ and the number chosen by Sarita be ‘s’.
The final numbers that they have are \( \frac { 2 r - 20 } { 5 } \) and \( \frac { 2 s - 20 } { 5 } \).
The sum of these numbers is 16.
\( \frac { 2 r - 20 } { 5 } + \frac { 2 s - 20 } { 5 } = 16 \)
\( 2 ( r + s ) - 40 = 80 \)
\( r + s = 60 \)
r and s are positive integers. For their difference to be as large as possible. r and s should be as far apart as possible.
r = 1; s = 59
The max difference between r and s is 59 - 1 = 58.
Choice B is the correct answer.
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