Chapter 7, Problem 49RE

To determine

Find the number of each type of commercial spray that gives the desired mixture.

Answer to Problem 49RE

The amount of spray X, Y and Z present in the mixture are as follows

  $\frac{\text{89}}{\text{6}}\text{ gallons, }\frac{\text{13}}{\text{4}}\text{ gallons and }\frac{\text{107}}{\text{12}}\text{ gallons}$

Explanation of Solution

Given:

The commercial spray X contains 1,2 and 2 parts of the chemical A, B and C

Therefore,

  $x_x=a+2b+2c$

Also, the commercial spray Y contains only chemical C, therefore

  $x_y=c$

Again, the commercial spray Z contains the chemicals A, B and C in equal amounts, therefore

  $x_z=a+b+c$

Suppose $x_x,x_y\text{ and }{x}_z$ be the amount of commercial spray X, Y and Z respectively needed for the mixture.

Also, suppose a, b and c be the amount of chemicals A, B and C

Since the commercial spray X contains 1,2 and 2 parts of the chemical A, B and C

Therefore,

  $x_x=a+2b+2c$

Also, the commercial spray Y contains only chemical C, therefore

  $x_y=c$

Again, the commercial spray Z contains the chemicals A, B and C in equal amounts, therefore

  $x_z=a+b+c$

The amount of chemicals A,B and C contains in the spray is

  $\begin{array}{l}\begin{array}{l}{x}_x+x_y+x_z=a+2b+2c+c+a+b+c\hfill \\ =2a+3b+4c\hfill \end{array}\\ \begin{array}{cc}& =a+2b+2c+c+a+b+c\\ & =2a+3b+4c\end{array}\end{array}$

Since the spray contains 6 gallons of chemicals A, 8 gallons of chemical B and 13 gallons of chemical C, therefore

  $\begin{array}{l}2a=6\hfill \\ 3b=8\hfill \\ 4c=13\hfill \end{array}$

Solve to above equation to get

  $a=3,b=\frac{8}{3}andc=\frac{13}{4}$

Put the values of $a,bandc$ in the $x_x,x_y\text{ and }{x}_z$

  $x_x=3+2\times \frac{8}{3}+2\times \frac{13}{4}$

  $=\frac{89}{6}$

  $\begin{array}{l}{x}_y=c\hfill \\ =\frac{13}{4}\hfill \end{array}$

  $x_z=a+b+c$

  $\begin{array}{l} =3+\frac{8}{3}+\frac{13}{4}\hfill \\ =\frac{107}{12}\hfill \end{array}$

Thus, the amount of spray X, Y and Z present in the mixture are as follows

  $\frac{\text{89}}{\text{6}}\text{ gallons, }\frac{\text{13}}{\text{4}}\text{ gallons and }\frac{\text{107}}{\text{12}}\text{ gallons}$