XAT 2023 — QA & DI Question 1
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?
Answer & solution
- A
67
58
- C
49
- D
40
- E
None of the above
Let the numbers choosen by Raju and Sarita be r and s respectively.
Multiplying both the numbers with 2, we get the numbers as 2r and 2s respectively.
Subtracting 20 from both the numbers, we get the numbers as 2r - 20 and 2s - 20 respectively.
Dividing both the numbers by 5, we get the numbers as (2r - 20)/5 and (2s - 20)/5 respectively.
∴ (2r - 20)/5 + (2s - 20)/5 = 16
⇒ (2r - 20) + (2s - 20) = 80
⇒ 2r + 2s = 120
⇒ r + s = 60
For maximum difference one number should be least possible and the other maximum possible.
Since r and s both are positive integers, the least value one of them can take is 1 hence the maximum value of other will be (60 - 1 = 59)
⇒ Maximum difference between them = 59 - 1 = 58.
Hence, option (b).