Chapter 1, Problem 5CA

a.

To determine

To create: A system of equations that models the given situation.

Answer to Problem 5CA

Equations are $x+y+z=7,30x+60y+60z=360andy=2x$ .

Explanation of Solution

Given Information: Comedy episode lasts $30\min utes$ , drama and reality episodes last $60\min utes$ .

Calculation:

Let no. of comedy episodes be $x$ , drama episodes be $y$ and reality episodes be $z$ ,

As per the given situation system turns out to be,

  $\begin{array}{l}x+y+z=7-(i)\\ 30x+60y+60z=360-(ii)\\ y=2x-(iii)\end{array}$

b.

To determine

How many episodes of each type are there in DVD of part $a$ .

Answer to Problem 5CA

  $x=2,y=4andz=1$ .

Explanation of Solution

Given Information: Comedy episode lasts $30\min utes$ , drama and reality episodes last $60\min utes$ . System of equations from previous part.

Calculation:

Substituting $(iii)$ in $(i)$ ,

  $\begin{array}{l}x+2x+z=7\\ 3x+z=7-(iv)\end{array}$

Substituting $(iii)$ in $(ii)$ ,

  $\begin{array}{l}30x+60(2x)+60z=360\\ 30x+120x+60z=360\\ 150x+60z=360-(v)\end{array}$

Multiplying $(iv)$ by $50$ and subtracting it from $(v)$ ,

  $\begin{array}{l}150x+60z=360\\ 150x+50z=350\\ =============\\ 10z=10\\ z=1\end{array}$

Substituting value of $zin(iv)$ ,

  $\begin{array}{l}3x+1=7\\ 3x=7-1\\ x=2\end{array}$

Putting value of $x$ in $(iii)$ ,

  $\begin{array}{l}y=2(2)\\ y=4\end{array}$

Thus, $x=2,y=4andz=1$ .

c.

To determine

The episodes of each type contained in the new DVD made by taking care of given constraints.

Answer to Problem 5CA

Two DVDs have different number of episodes of each category, second DVD has $6$ reality based episodes.

Explanation of Solution

Given Information: Comedy episode lasts $30\min utes$ , drama and reality episodes last $60\min utes$ . System of equations from previous part.

Calculation:

New system of equations is,

  $\begin{array}{l}x+y+z=6-(vi)\\ 30x+60y+60z=360-(vii)\\ y=2x-(viii)\end{array}$

Substituting $(viii)$ in $(vi)$ ,

  $\begin{array}{l}x+2x+z=6\\ 3x+z=6-(ix)\end{array}$

Substituting $(viii)$ in $(vii)$ ,

  $\begin{array}{l}30x+60(2x)+60z=360\\ 30x+120x+60z=360\\ 150x+60z=360-(x)\end{array}$

Multiplying $(ix)$ by $50$ and subtracting it from $(x)$ ,

  $\begin{array}{l}150x+60z=360\\ 150x+50z=300\\ =============\\ 10z=60\\ z=6\end{array}$

  $x=0,y=0andz=6$

Thus, two DVDs have different number of episodes of each category, second DVD has $6$ reality based episodes.