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Two teachers, Mr. Donahue and Mrs. Ortiz, decide to buy mechanical pencilsand rulers for their classrooms. Mrs. Ortizbuys 50 mechanical pencils and 30 rulers for $55. Mr. Donahue buys 30 of each for $45. What was the price per mechanical pencil?What was the price per ruler?a.

Sagot :

jacob193

Answer:

Price per mechanical pencil: [tex]\$ 0.5[/tex].

Price per ruler: [tex]\$ 1[/tex].

Step-by-step explanation:

Let the cost per mechanical pencil be [tex]\$ p[/tex], and let the cost per ruler be [tex]\$ r[/tex].

Set up two equations based on the amount of money that each teacher paid.

The cost of [tex]50[/tex] mechanical pencils ([tex]\$ p[/tex] each) and [tex]30[/tex] rulers ([tex]\$ r[/tex] each) would be [tex](50 \, p + 30\, r)[/tex]. That should be equal to the amount of money that Mrs. Ortiz paid. In other words, [tex]50 \, p + 30\, r = 55[/tex].

Similarly, the cost of [tex]30[/tex] mechanical pencils ([tex]\$ p[/tex] each) and [tex]30 \![/tex] rulers ([tex]\$ r[/tex] each) would be [tex](50 \, p + 30\, r)[/tex]. That should be equal to the amount of money that Mr. Donahue paid. In other words, [tex]30 \, p + 30\, r = 45[/tex].

That gives a system of two equations:

[tex]\left\lbrace \begin{aligned}& 50 \, p + 30\, r = 55 \\ & 30 \, p + 30\, r = 45 \end{aligned}\right.[/tex].

Subtract the second equation for the first:

[tex]20\, p = 10[/tex].

[tex]\displaystyle p = \frac{1}{2} = 0.5[/tex].

In other words, the cost of one mechanical pencil would be [tex]\$0.5[/tex].

Substitute [tex]p = 0.5[/tex] back to either equation and solve for [tex]r[/tex], the cost of one ruler.

[tex]50 \times 0.5 + 30\, r = 55[/tex].

[tex]30\, r = 55 - 25 = 30[/tex].

[tex]r = 1[/tex].

In other words, the cost of one ruler would be [tex]\$ 1[/tex].

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