Question

Mai, Abdul, and Chris served a total of 111 orders Monday at the school cafeteria. Abdul served 3 times as many orders as Chris. Mai served 6 more ordersthan Chris. How many orders did they each serve?Number of orders Mal served:Number of orders Abdul served:X5?000Number of orders Chris served:

Answer 1

Let the number of orders served by Mai, Abdul, and Chris be represented by M, A, and C respectively.

A total of 111 orders were served. Thus,

`M\text{ + A +C = 111 ---- equation 1}`

Abdul served 3 times as many orders as Chris. Thus,

`\begin{gathered} A\text{ = 3}* C \\ \Rightarrow A\text{ = 3C ------ equation 2} \end{gathered}`

Mai served 6 more orders than Chris. Thus,

`M\text{ = (6 }+\text{ C) ----- equation 3}`

Substitute equations 2 and 3 into equation 1. Thus, we have

`\begin{gathered} M\text{ + A +C = 111} \\ \Rightarrow(6+C)\text{ + 3C+ C=111} \\ \end{gathered}`

Simplify by opening the brackets

`6\text{ + C + 3C + C = 111}`

Collect like terms, we have

`\begin{gathered} C\text{ + 3C + C = 111-6} \\ \Rightarrow5C\text{ = 105} \end{gathered}`

Solve for C dividing both sides of the equation by the coefficient of C.

The coefficient of C is 5. Thus,

`\begin{gathered} (5C)/(5)=(105)/(5) \\ \Rightarrow C=21 \end{gathered}`

Solve for A by substituting the value of 21 for C in equation 2.

`\begin{gathered} \text{From equation 2,} \\ A\text{ = 3C} \\ \text{= 3}*21 \\ \Rightarrow A=63 \end{gathered}`

Solve for M by substituting the value of 21 for C in equation 3.

`\begin{gathered} \text{From equation 3,} \\ M\text{ = (6 }+\text{ C)} \\ =6\text{ + 21} \\ \Rightarrow M\text{ = 27} \end{gathered}`

Hence,

number of orders served by Mai = 27

number of orders served by Abdul = 63

number of orders served by Chris = 21