+0  
 
0
91
2
avatar

Tyace needs at least 40 hot dogs and 40 buns for a cookout. Packages of hotdogs cost $3.50 and contains 10 hotdogs. Packages of buns cost $2.50 and contains 8 buns. The maximum amount of money Tyace can spend on hot dogs and buns is $70. What are possible combinations of packages that Tyace can buy?

 Jan 29, 2020
 #1
avatar+24388 
+2

Tyace needs at least 40 hot dogs and 40 buns for a cookout.
Packages of hotdogs cost $3.50 and contains 10 hotdogs.
Packages of buns cost $2.50 and contains 8 buns.
The maximum amount of money Tyace can spend on hot dogs and buns is $70.
What are possible combinations of packages that Tyace can buy?

 

I assume:


\(\text{Let packages of hotdogs $=x$ and $x_{min} = 4$ packages } \\ \text{Let packages of buns $=y$ and $y_{min} = 5$ packages }\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{$3.50}{10~hotdogs} * 10x + \dfrac{$2.50}{8~buns} * 8y} &=& \mathbf{$70} \\\\ \dfrac{$3.50}{ hotdogs} * x + \dfrac{$2.50}{ buns} * y &=& $ 70 \\\\ 3.50x + 2.50y &=& 70 \quad | \quad * 2 \\\\ \mathbf{7x + 5y} &=& \mathbf{140} \\ \hline \end{array}\)

 

Euler :

\(\begin{array}{|rclrclrcl|} \hline \mathbf{7x + 5y} &=& \mathbf{140} \\ 5y &=& 140 - 7x \\ y &=& \dfrac{140 - 7x}{5} \\ y &=& \dfrac{140 - 5x-2x}{5} \\ y &=& 28 - x - \underbrace{\dfrac{2x}{5}}_{=a} \\ y &=& 28 - x - a & a&=&\dfrac{2x}{5} \\ & & & 5a&=&2x \\ & & & 2x&=&5a \\ & & & x&=&\dfrac{5a}{2} \\ & & & x&=&\dfrac{4a+a}{2} \\ & & & x&=&2a+\underbrace{\dfrac{a}{2}}_{=b} \\ & & & x&=&2a+b & b&=& \dfrac{a}{2} \\ & & & & & & 2b&=& a \\ & & & & & & \mathbf{a}&=& \mathbf{2b} \\ & & & x&=&2(2b)+b \\ & & & \mathbf{x}&=&\mathbf{5b} \\ y &=& 28 - 5b - 2b \\ \mathbf{y} &=& \mathbf{28 - 7b} \\ \hline \end{array}\)

 

The possible combinations of packages:

\(\begin{array}{|c|r|r|} \hline b & x=5b & y = 28 - 7b \\ \hline 1 & 5 & 21 \\ 2 & 10 & 14 \\ 3 & 15 & 7 \\ \hline \end{array}\)

 

1.   5 packages of hotdogs and 21 packages of buns

2. 10 packages of hotdogs and 14 packages of buns

3. 15 packages of hotdogs and 7 packages of buns

 

laugh

 Jan 29, 2020
 #2
avatar+21931 
+3

He needs 40 buns and 40 dogs ...so he HAS to spend

40 dogs/10dogs/pkg   x   $ 3.50/pkg = $14       (4 packages dogs)

40 buns/8buns/pkg     x    $ 2.50/pkg = 12.50    (5 packages buns)       14 + 12.50 = 26.50

     He has     70 - 26.50 = 43.50 left to spend

 

Now any combo that spends 43.50 is allowed as he has met his 40 dog-bun requirement

 

3.50 y  + 2.50 x = 43.50    

I used a graphical approach to see where x,y pairs are integer values (since you cannot split packages) you will see that

x, y   =   2,11    9,6    and   16, 1     all fit the bill   ....remember you have to add the obligatory red packages above !

 

y is dogs   x is buns      so    7 b   15 d            14 b  10 d       and   21 b    5 d     all would spend the total $70

 

 Jan 29, 2020
edited by ElectricPavlov  Jan 29, 2020

25 Online Users

avatar
avatar