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Q.

What is the value of x in $\frac{x + 2}{6} - \frac{11 - x}{3} - \frac{1}{4} = \frac{3 x - 4}{12}$ ?

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a

12

b

13

c

14

d

15

answer is B.

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Detailed Solution

Given,

\frac{x + 2}{6} - \frac{11 - x}{3} - \frac{1}{4} = \frac{3 x - 4}{12}

.
Solve the equation for x.
The prime factorisation of 4 is

2 \times 2

.
The prime factorisation of 6 is

2 \times 3

.
The prime factorisation of 3 is 3.
The LCM between 6, 4 and 3 is 12.

\frac{x + 2}{6} - \frac{11 - x}{3} - \frac{1}{4} = \frac{3 x - 4}{12}

\frac{2 (x + 2) - 4 (11 - x) - 3 (1)}{12} = \frac{3 x - 4}{12}

\frac{2 x + 4 - 44 + 4 x - 3}{12} = \frac{3 x - 4}{12}

6 x - 43 = 3 x - 4

6 x - 3 x = 43 - 4

3x = 39
x = 13
Check whether the solution is correct or not by substitution method.
LHS =

\frac{x + 2}{6} - \frac{11 - x}{3} - \frac{1}{4}

=

\frac{13 + 2}{6} - \frac{11 - 13}{3} - \frac{1}{4}

= \frac{35}{12}

RHS =

\frac{3 x - 4}{12}

= \frac{3 \times 13 - 4}{12}

= \frac{35}{12}

LHS = RHS
So, the value of x is 13.
Hence, option 2 is correct.

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