Three variables x, y and z take values 2, 3, 4 or 5 such that their values are always distinct. If M and N denote the largest possible value and the smallest possible value, respectively, for the expression {(x × y) + z}; then M - N is

Correct Option: (C) 13

This question tests the arithmetic principles of combinatorial optimization, specifically maximizing and minimizing algebraic expressions. It aligns perfectly with the quantitative aptitude framework standard to the UPSC Civil Services Aptitude Test (CSAT).

To evaluate the expression E = (x × y) + z using distinct values from the set {2, 3, 4, 5}, we must systematically find the maximum (M) and minimum (N) possible values.

Finding the Largest Value (M): To maximize a sum containing a product, assign the largest available integers to the multiplicands. The largest numbers in our set are 4 and 5. Let x = 4 and y = 5, giving x × y = 20. The remaining values for z are {2, 3}. To maximize the overall expression, we select the largest remaining value, z = 3. Thus, M = 20 + 3 = 23.

Finding the Smallest Value (N): Conversely, to minimize the expression, assign the smallest available integers to the multiplicands. The smallest numbers in the set are 2 and 3. Let x = 2 and y = 3, giving x × y = 6. The remaining values for z are {4, 5}. To minimize the total sum, we pick the smallest remaining value, z = 4. Thus, N = 6 + 4 = 10.

Calculating the difference: M - N = 23 - 10 = 13. Option C is correct.

Why the other options are incorrect:

• Option A (11): Incorrect. This typically results from a simple arithmetic error during subtraction or a failure to correctly identify the true upper or lower bounding values.
• Option B (12): Incorrect. This occurs if an aspirant incorrectly calculates N = 11 by misassigning variables (x=2, y=4 ⇒ 8 + 3 = 11), leading to 23 - 11 = 12.
• Option D (14): Incorrect. This stems from calculating N = 9 by violating the "always distinct" constraint and reusing the number 3 (2 × 3 + 3 = 9), resulting in 23 - 9 = 14.

Takeaway: In optimization problems involving sums and products, multiplication scales faster than addition. Always assign extreme values (largest or smallest) to the multiplied variables first to accurately establish your limits.