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QUESTION

CSAT

Medium

Maths

Prelims 2026

Three variables xx, yy and zz take values 2,3,42, 3, 4 or 55 such that their values are always distinct. If MM and NN denote the largest possible value and the smallest possible value, respectively, for the expression {(x×y)+z}\{(x \times y) + z\}; then MNM - N is

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Explanation

To find the largest and smallest possible values for the expression (x×y)+z(x \times y) + z, we must choose distinct values for xx, yy, and zz from the given set {2,3,4,5}\{2, 3, 4, 5\}.

Step 1: Finding the largest possible value (MM) To maximize the value of (x×y)+z(x \times y) + z, we should maximize the product (x×y)(x \times y) because multiplication of numbers greater than 11 yields a larger result than addition.

  • The two largest numbers available are 44 and 55. Therefore, we assign x=4x = 4 and y=5y = 5 (or vice versa), giving a product of 4×5=204 \times 5 = 20.
  • The remaining numbers are 22 and 33. To maximize the overall sum, we assign the largest remaining number to zz, which is 33.
  • Calculating the maximum value: M=(4×5)+3=20+3=23M = (4 \times 5) + 3 = 20 + 3 = 23

Step 2: Finding the smallest possible value (NN) To minimize the value of (x×y)+z(x \times y) + z, we should minimize the product (x×y)(x \times y).

  • The two smallest numbers available are 22 and 33. Therefore, we assign x=2x = 2 and y=3y = 3 (or vice versa), giving a product of 2×3=62 \times 3 = 6.
  • The remaining numbers are 44 and 55. To minimize the overall sum, we assign the smallest remaining number to zz, which is 44.
  • Calculating the minimum value: N=(2×3)+4=6+4=10N = (2 \times 3) + 4 = 6 + 4 = 10

(Self-check: If we tried x=2,y=4,z=3x=2, y=4, z=3, the result would be (2×4)+3=11(2 \times 4) + 3 = 11, which is larger than 1010. Thus, 1010 is indeed the absolute minimum.)

Step 3: Finding the difference (MNM - N) Now, we subtract the smallest value from the largest value: MN=2310=13M - N = 23 - 10 = 13

Therefore, the difference between the largest and smallest possible values is 1313.

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