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CAT 2003 Slot 2 — QA Question 39

Number TheoryEasy
Passage / Data

Answer the following question based on the information given below.

A string of three English letters is formed as per the following rules:

  1. The first letter is any vowel.
  2. The second letter is m, n or p.
  3. If the second letter is m, then the third letter is any vowel which is different from the first letter.
  4. If the second letter is n, then the third letter is e or u.
    1. If the second letter is p, then the third letter is the same as the first letter.

If three positive real numbers x, y, z satisfy y – x = z – y and xyz = 4, then what is the minimum possible value of y?

Answer & solution

  • A

    21/3

  • 22/3

  • C

    21/4

  • D

    23/4

Solution

y − x = z − y

2y = x + z                                                  ... (i)

xyz = 4                                                      ... (ii)

It is known that Arithmetic Mean (A.M.) is greater than or equal to Geometric Mean (G.M.)

i.e. A.M. ≥ G.M.

Hence, x+y+z3 ≥ (xyz)1/3 .....(iii)

From (i), (ii) and (iii), we get 3y3 ≥ 41/3

∴ y ≥ 22/3

∴ The minimum value of y is 22/3

Hence, option (b).

CAT 2003 Slot 2 QA Q39: If three positive real numbers x, y, z satisfy y – x = z – y and xyz = 4, then what is the minimum — Solution | TheCATExam