Ricardo has two types of assignments for his class. The number of mini assignments, $m$ , he has is $1$ fewer than twice the number of long assignments, $l$ , he has. If he has $46$ assignments in total, which of the following systems of equations can be used to correctly solve for $m$ and $l$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $m=2 l-1$ $\newline$ $m+l=46$ $\newline$ (B) $m=2 l-1$ $\newline$ $m=l+46$ $\newline$ (C) $l=2 m-1$ $\newline$ $m+l=46$

Full solution

Q. Ricardo has two types of assignments for his class. The number of mini assignments,

m

, he has is

1

fewer than twice the number of long assignments,

l

, he has. If he has

46

assignments in total, which of the following systems of equations can be used to correctly solve for

m

and

l

\newline

Choose

1

answer:

\newline

(A)

m=2 l-1

\newline

m+l=46

\newline

(B)

m=2 l-1

\newline

m=l+46

\newline

(C)

l=2 m-1

\newline

m+l=46

Translate into mathematical expressions: Let's translate the word problem into mathematical expressions. We are told that the number of mini assignments, $m$ , is $1$ fewer than twice the number of long assignments, $l$ . This can be written as: $\newline$ $m = 2l - 1$
Total number of assignments: Next, we are told that the total number of assignments is $46$ . This means that the sum of the mini assignments and long assignments is $46$ , which can be written as: $\newline$ $m + l = 46$
Check given choices: Now, we need to find which of the given choices matches the two equations we have derived from the problem statement. Let's check each option: $\newline$ (A) $m = 2l - 1$ and $m + l = 46$ $\newline$ (B) $m = 2l - 1$ and $m = l + 46$ $\newline$ (C) $l = 2m - 1$ and $m + l = 46$ $\newline$ Option (A) matches the equations we derived from the problem statement.
Match with equations: We can disregard options (B) and (C) because they do not match the equations we derived. Option (B) incorrectly states that $m$ equals $l$ plus $46$ , and option (C) incorrectly states that $l$ equals $2m$ minus $1$ .