CAT 2022 Slot 1 QA Q4 Number Systems

Question 4 of the Quantitative Ability (QA) section, CAT 2022 Slot 1, with CAT's official answer and a worked solution.

For natural numbers \(x, y\), and \(z\), if \(x y+y z=19\) and \(y z+x z=51\), then the minimum possible value of \(x y z\) is

TITA: type in the answer. There are no options and no negative marking.

Answer: 34 Number Systems TITA

Solution

\(y(x+z) = 19\) and 19 is prime; \(x+z \ge 2\), so \(y = 1\) and \(x + z = 19\).
\(z(x+y) = 51\) becomes \(z(x+1) = z(20 - z) = 51\), i.e. \(z^2 - 20z + 51 = 0\), so \(z = 3\) or \(z = 17\).
\(z = 3, x = 16\): \(xyz = 48\). \(z = 17, x = 2\): \(xyz = 34\).
Minimum \(xyz = 34\).