Question

The sum of three consecutive numbers is $36$. The smallest number among the three is

• A. 12
• B. 11
• C. 13
• D. 10

Answer 1

The correct option is B

11

Let the first number be $a$. So the three consecutive numbers will be $a,a+1,a+2$
$\text{a +a +1 +a +2 =36}$
$\text{3a +3 = 36}$
$\text{3a = 36 -3 = 33}$
$a=\frac{33}{3}=11$
$\text{So the three consecutive numbers will be 11, 12, 13}$.