CAT 2025 Slot 1 QA Question 22: 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

Question

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

Accepted Answer

The expression required contains $3b - 3c$, so we first isolate that from the second equation.
$6a + 3b - 3c = 50$
$3b - 3c = 50 - 6a$
$2a + 3b - 3c = 2a + 50 - 6a = 50 - 4a$

Now find $a$ by eliminating $b$ and $c$.
$a - 6b + 6c = 4$
$2(6a + 3b - 3c = 50)$
$12a + 6b - 6c = 100$
$13a = 104$
$a = 8$

If a−6b+6c=4a - 6b + 6c = 4a−6b+6c=4 and 6a+3b−3c=506a + 3b - 3c = 506a+3b−3c=50, where aaa, bbb and ccc are real numbers, the value of 2a+3b−3c2a + 3b - 3c2a+3b−3c is

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