NinetynineCAT 2025 Slot 2 QA Q1 Number Systems
Question 1 of the Quantitative Ability (QA) section, CAT 2025 Slot 2, with CAT's official answer and a worked solution.
If \( m \) and \( n \) are integers such that \( ( m + 2 n ) ( 2 m + n ) = 27 \), then the maximum possible value of \( 2 m - 3 n \) is
TITA: type in the answer. There are no options and no negative marking.
Answer: 17 Number Systems TITA
Solution
- Let \(u = m+2n\), \(v = 2m+n\), so \(uv = 27\). Then \(m = \frac{2v-u}{3}\), \(n = \frac{2u-v}{3}\), and \(m+n = \frac{u+v}{3}\) must be an integer.
- Factor pairs with \(u+v\) divisible by 3: \((3,9), (9,3), (-3,-9), (-9,-3)\); pairs like \((1,27)\) give \(u+v = 28\), not allowed.
- \(2m - 3n = \frac{2(2v-u) - 3(2u-v)}{3} = \frac{7v - 8u}{3}\).
- Values: \((3,9) \to 13\), \((9,3) \to -17\), \((-3,-9) \to -13\), \((-9,-3) \to \frac{-21+72}{3} = 17\).
- Maximum is \(17\) (at \(m = 1, n = -5\): \((1-10)(2-5) = 27\)).
Shortcut. Change variables to the two factors; \(2m-3n\) becomes linear in \(u, v\), so just test the four valid factor pairs.