CAT 2025 SLOT-3 QA Modern Math Question

1. Concept Used

Topic: Inequalities and Optimization with Constraints
Formula: $$0.3q \leq p \leq 0.7q, \quad p + q + r = 900, \quad 150 < r < 500, \quad r \text{ is a perfect square}$$

2. Calculation

We are given that $p + q + r = 900$, where $r$ is a perfect square lying strictly between 150 and 500. The perfect squares in this range are: $169\ (13^2),\ 196\ (14^2),\ 225\ (15^2),\ 256\ (16^2),\ 289\ (17^2),\ 324\ (18^2),\ 361\ (19^2),\ 400\ (20^2),\ 441\ (21^2),\ 484\ (22^2)$. So the minimum value of $r$ is $169$ and the maximum value of $r$ is $484$.

We are also given that $0.3q \leq p \leq 0.7q$. Adding $q$ throughout:

$q + 0.3q \leq p + q \leq q + 0.7q$

$1.3q \leq p + q \leq 1.7q$

Since $p + q = 900 - r$, we substitute:

$1.3q \leq 900 - r \leq 1.7q$

Finding Minimum value of p:

The minimum value of $p = 0.3q$ is achieved when $q$ is minimized. From the left inequality $1.3q \leq 900 - r$, we get $q \leq \frac{900 - r}{1.3}$. To minimize $q$, we use the boundary $1.3q = 900 - r$ and maximize $r$. With $r = 484$:

$1.3q = 900 - 484 = 416$

$q = \frac{416}{1.3} = 320$

Minimum value of $p = 0.3 \times 320 = 96$.

Verification: $p + q + r = 96 + 320 + 484 = 900$ ✓

Finding Maximum value of p:

The maximum value of $p = 0.7q$ is achieved when $q$ is maximized. From the right inequality $900 - r \leq 1.7q$, we get $q \geq \frac{900 - r}{1.7}$. To maximize $q$, we use the boundary $1.7q = 900 - r$ and minimize $r$. With $r = 169$:

$1.7q = 900 - 169 = 731$

$q = \frac{731}{1.7} = 430$

Maximum value of $p = 0.7 \times 430 = 301$.

Verification: $p + q + r = 301 + 430 + 169 = 900$ ✓

Sum of maximum and minimum values of p:

$p_{\text{max}} + p_{\text{min}} = 301 + 96 = 397$

3. Solution

Answer = 397 ✅

The sum of the maximum and minimum possible values of $p$ is 397.

CAT 2025Slot 3QAQuestion & SolutionQues & Sol

Question

Solution

1. Concept Used

2. Calculation

3. Solution