CAT 2025

Slot 1

QA

Question & Solution

1. Concept Used

• Topic: Linear Equations — expressing a target expression as a linear combination of given equations
• Formula: $$ \text{If } E_3 = x \cdot E_1 + y \cdot E_2, \text{ then the value} = x \cdot (\text{RHS}_1) + y \cdot (\text{RHS}_2) $$

2. Calculation

We are given two equations:

$$E_1: a - 6b + 6c = 4$$

$$E_2: 6a + 3b - 3c = 50$$

We want to find the value of the expression $2a + 3b - 3c$. The key insight is to express this target expression as a linear combination of $E_1$ and $E_2$. Suppose:

$$x(a - 6b + 6c) + y(6a + 3b - 3c) = 2a + 3b - 3c$$

Matching coefficients on both sides for $a$, $b$, and $c$:

Coefficient of $a$: $x + 6y = 2$ $\quad \cdots (3)$

Coefficient of $b$: $-6x + 3y = 3 \Rightarrow -2x + y = 1$ $\quad \cdots (4)$

Coefficient of $c$: $6x - 3y = -3 \Rightarrow -2x + y = 1$ $\quad$ (same as equation (4) ✓)

Now solving equations (3) and (4):

From (4): $y = 1 + 2x$

Substituting into (3): $x + 6(1 + 2x) = 2 \Rightarrow x + 6 + 12x = 2 \Rightarrow 13x = -4 \Rightarrow x = -\dfrac{4}{13}$

Then: $y = 1 + 2\left(-\dfrac{4}{13}\right) = 1 - \dfrac{8}{13} = \dfrac{5}{13}$

Now the value of the target expression is:

$$x \cdot (\text{RHS}_1) + y \cdot (\text{RHS}_2) = \left(-\dfrac{4}{13}\right)(4) + \left(\dfrac{5}{13}\right)(50)$$

$$= -\dfrac{16}{13} + \dfrac{250}{13} = \dfrac{234}{13} = 18$$

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3. Solution

Answer = Option C (18) ✅

The final calculated value of $2a + 3b - 3c$ is 18.