CAT 2025 Slot1 Quant: The number of distinct pairs of integers
February 20, 2026 2026-02-20 22:50CAT 2025 Slot1 Quant: The number of distinct pairs of integers
AZUCATION • CAT Quant 2025
TITA Question
#Algebra
#Inequalities
Easy-Medium
The number of distinct pairs of integers $(x, y)$ satisfying the inequalities $x > y \ge 3$ and $x + y < 14$ is:
Type your answer below:
⚡ AZUCATION SHORTCUT:
List $y$ values from $3$ upwards. For each $y$, $x$ must be in the range $y < x < 14-y$.
Count the possible values of $x$ for each $y$ until $y$ can no longer be less than $x$.
List $y$ values from $3$ upwards. For each $y$, $x$ must be in the range $y < x < 14-y$.
Count the possible values of $x$ for each $y$ until $y$ can no longer be less than $x$.
Detailed Step-by-Step Solution
Step 1: Understand the Constraints
We need to find integer pairs $(x, y)$ such that:
1. $y \ge 3$
2. $x > y$
3. $x + y < 14 \implies x < 14 - y$
Step 2: Case-by-case Analysis for $y$
Since $x > y$, the smallest possible value for $x$ is $y + 1$.
The largest possible value for $x$ is $13 - y$.
Therefore, for a given $y$, $x$ can take values from $y+1$ to $13-y$.
- If $y = 3$: $x$ can be from $4$ to $(13-3=10)$. Values: $\{4, 5, 6, 7, 8, 9, 10\}$. Count = 7.
- If $y = 4$: $x$ can be from $5$ to $(13-4=9)$. Values: $\{5, 6, 7, 8, 9\}$. Count = 5.
- If $y = 5$: $x$ can be from $6$ to $(13-5=8)$. Values: $\{6, 7, 8\}$. Count = 3.
- If $y = 6$: $x$ can be from $7$ to $(13-6=7)$. Value: $\{7\}$. Count = 1.
- If $y = 7$: $x$ must be $> 7$ and $< (14-7=7)$, which is impossible.
Step 3: Total Count
Total pairs $= 7 + 5 + 3 + 1 = \mathbf{16}$.