Question

Three consecutive positive even numbers are such that thrice the first number exceeds double the third by $2$, the third number is?

• A. $10$
• B. $14$
• C. $16$
• D. $12$

Answer 1

The correct option is B $14$

Let the three consecutive positive even integers be $x,x+2$ and $x+4$.

It is given that thrice the first number exceeds double the third by $2$, therefore, we have:

$(3\times x)-2=2(x+4)\Rightarrow 3x-2=2x+8\Rightarrow 3x-2x=8+2\Rightarrow x=10$

Therefore, $x=10$ and the other two integers are:

$x+2=10+2=12$ and

$x+4=10+4=14$

Hence, the third number is $14$.