Question

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?

• A. 67
• B. 58
• C. 49
• D. 40
• E. None of the above

Answer 1

The Correct Option is B

Step 1: Let the numbers chosen by Raju and Sarita be:

\[ x \quad \text{and} \quad y \] 

Step 2: Apply the operations as per the problem.

• After doubling: \(2x, \; 2y\)
• Subtract 20: \(2x - 20, \; 2y - 20\)
• Divide by 5: \(\frac{2x - 20}{5}, \; \frac{2y - 20}{5}\)

Step 3: The sum of the results is given as 16.

\[ \frac{2x - 20}{5} + \frac{2y - 20}{5} = 16 \]

Step 4: Simplify the equation.

\[ \frac{2x + 2y - 40}{5} = 16 \]

\[ 2x + 2y - 40 = 80 \]

\[ 2x + 2y = 120 \quad \Rightarrow \quad x + y = 60 \]

Step 5: Find the maximum difference between \(x\) and \(y\).

Since their sum is fixed at 60, the difference will be maximum when one number is as small as possible and the other as large as possible.

Thus:

\[ x = 1, \; y = 59 \quad \text{or} \quad x = 59, \; y = 1 \]

Step 6: Maximum difference.

\[ |x - y| = |59 - 1| = 58 \]

Final Answer:

\[ \boxed{58} \]

Correct Option:

Option B