The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u = ((x+y))/2 and v = ((y+z))/2 . If x ≥ z, then the minimum possible value of x is (TITA)

CAT 2018 Slot 2 QA Q29 Ratio, Mixtures & Averages

Question 29 of the Quantitative Ability (QA) section, CAT 2018 Slot 2, with CAT's official answer.

The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u = \(\frac{(x+y)}{2}\) and v = \(\frac{(y+z)}{2}\). If x ≥ z, then the minimum possible value of x is (TITA)

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Answer: 105 Ratio, Mixtures & Averages TITA

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CAT 2018 Slot 2 QA Q29 Ratio, Mixtures & Averages

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