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

To solve this problem, let's define the process and steps involved based on the given conditions:

1. Let the integer chosen by Raju be \(x\) and the integer chosen by Sarita be \(y\).
2. Raju and Sarita both multiply their numbers by 2, giving them \(2x\) and \(2y\), respectively.
3. They then subtract 20 from their results, which gives them \(2x - 20\) and \(2y - 20\).
4. Each of these results is divided by 5:
   Raju: \(\frac{2x - 20}{5}\) 
   Sarita: \(\frac{2y - 20}{5}\)
5. According to the problem, the sum of these two results is 16:
   \(\frac{2x - 20}{5} + \frac{2y - 20}{5} = 16\)
6. Combining and simplifying the equation:
   \(\frac{2x - 20 + 2y - 20}{5} = 16\)
   \(2x + 2y - 40 = 80\)
   \(2x + 2y = 120\)
   Dividing everything by 2:
   \(x + y = 60\)
7. To find the maximum possible difference between the numbers chosen by Raju and Sarita, we want to maximize \(|x - y|\).
8. Assume Raju chooses \(x = 1\) (the smallest positive integer). Then, \(y = 60 - 1 = 59\), making the difference:
   \(|x - y| = |1 - 59| = 58\)
9. Assume Sarita chooses \(y = 1\). Then, \(x = 60 - 1 = 59\), making the difference the same:
   \(|x - y| = |59 - 1| = 58\)

In both cases, the maximum possible difference between the integers chosen by Raju and Sarita is 58. Thus, the correct answer is 58.

Answer 2

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