If $a - 6b + 6c = 4$ and $6a + 3b - 3c = 50$ , where $a$ , $b$ and $c$ are real numbers, the value of $2a + 3b - 3c$ is

✅ Correct Option: 2

Look at what we're asked to find: $2a + 3b - 3c$ .

Notice that $3b - 3c = 3(b - c)$ . Now glance at the given equations:

$a - 6b + 6c = 4 \implies a - 6(b - c) = 4 \quad \cdots (1)$

$6a + 3b - 3c = 50 \implies 6a + 3(b - c) = 50 \quad \cdots (2)$

See how $b$ and $c$ always appear together as $(b - c)$ ? That means we don't need to find $b$ and $c$ individually — we only need $(b - c)$ and $a$ .

From Equation $(1)$ :

$a = 4 + 6(b - c) \quad \cdots (3)$

Substitute into Equation $(2)$ :

$6[4 + 6(b - c)] + 3(b - c) = 50$

$24 + 36(b - c) + 3(b - c) = 50$

$39(b - c) = 26$

$b - c = \dfrac{26}{39} = \dfrac{2}{3}$

Plug $b - c = \dfrac{2}{3}$ back into Equation $(3)$ :

$a = 4 + 6 \times \dfrac{2}{3} = 4 + 4 = 8$

$2a + 3b - 3c = 2a + 3(b - c)$

$= 2(8) + 3\!\left(\dfrac{2}{3}\right)$

$= 16 + 2$

$= \boxed{18}$

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CAT Past Year Questions (Topic-Wise)

If $a - 6b + 6c = 4$ and $6a + 3b - 3c = 50$ , where $a$ , $b$ and $c$ are real numbers, the value of $2a + 3b - 3c$ is