Question

Ross has two types of boxes. The number of mini boxes, $m$ is 1 fewer than twice the number of big boxes, $l$. If he has 46 boxes in total, which of the following systems of equations can be used to correctly solve for $m$ and $l$?

• A. $m=2l-1$
  $m+l=46$
• B. $m=2l-1$
  $m=l+46$
  
• C. $l=2m-1$
  $m+l=46$
  
• D. $l=2m-1$
  $m=l+46$

Answer 1

The correct option is A $m=2l-1$

$m+l=46$
First, let us determine $m$ in terms of $l$. If $m$ is one less than two times $l$, then $m=2l-1$
[by this one can directly eliminate option C and D]
Next, given is that the total number of boxes is 46. Therefore $m+l=46$
The best system for this scenario is
$m=2l-1$
$m+l=46$