Question

Three times the first of three consecutive odd integers is $3$ more than twice the third. The third integer is:

• A. $9$
• B. $11$
• C. $13$
• D. $15$

Answer 1

The correct option is D $15$

Let the three consecutive odd integers be $x$, $x+2$, $x+4$
Three times the first of three consecutive odd integers is $3$ more than twice the third.
$\therefore$ $3x=2(x+4)+3$
$\therefore$ $x=11$
Hence, the third integer is $11+4=15$.

Hence, option D is correct.