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\title{Word Problems and Applications}


\begin{document}
\maketitle
\section{Objective} To continue to build on the techniques for solving word problems.\\
\begin{itemize}
%---------------------------------------------------
\item \textbf{Recall 1:} If a polynomial factors, then it is divisible by any one of  its factors.

For example, the polynomial $$x^4 + 13x^2 + 36$$ factors as $$(x - 2)(x + 2)(x - 3)(x + 3).$$
This means that any one of the factors divides  $x^4 + 13x^2 + 36$ without remainder.
 Indeed, if you divide   $x^4 + 13x^2 + 36$ by $(x - 2)$ you will find the result to be  $x^3 + 2x^2 - 9x - 18,$ exactly. 

\end{itemize}
%---------------------------------------------------
\textbf{Note:} Finding the factors of a polynomial equation is not always easy.  The polynomials encountered in algebra textbooks are generally easier to factor than arbitrary ones . The next recall suggests some tricks for finding the factors.
%---------------------------------------------------

\begin{itemize} 
\item \textbf{Recall 2:} Consider the polynomial
$$2x^3 - 3x^2 - 5x + 6.$$
By letting $x = 1,$ we find that the polynomial becomes 0; that  is:
$$2(1)^3 - 3(1)^2 - 5(1) + 6 = 0.$$
This means that $(x - 1)$ is a factor of the polynomial.  We   can test this by performing long division.
$$\begin{array}{ll}
&\;\;\;\;\;\;\;\;\;\;\;2x^2 \;\;-x - 6\\
x-1&\overline{)2x^3 - 3x^2 - 5x + 6}\\
& \; \underline{2x^3 - 2x^2}\\
&\;\;\;\;\;\;\;\;\;\; -x^2 - 5x\\
&\;\;\;\;\;\;\;\;\;\; \underline{-x^2 + \;x}\\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; -6x + 6\\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \underline{-6x + 6}\\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0
\end{array}
$$
From this, we see that
$$2x^3 - 3x^2 - 5x + 6 = (x - 1)(2x^2 - x - 6).$$
The quadratic at the right side of this last equation factors   as
$$2x^2 - x - 6 = (x - 2)(2x + 3),$$
which means that the original polynomial factors as
$$2x^3 - 3x^2 - 5x + 6 = (x - 1)(x - 2)(2x + 3).$$
%---------------------------------------------------
\item \textbf{Example 2.}\emph{(Percent Problem)}

In an experiment, it is calculated that in administering a new drug for the treatment of ulcers, $85.5 \%$ of patients who received the drug had had significant improvements of their ulcer condition.  If 1881 ulcer patients had significant improvements in their condition, how many patients participated in the experiment?

\emph{Solution:} Remember that \emph{percent} means \emph{hundredths.}  So, if you say $85.5$ you mean $85.5$ hundredths or $.85.5/100$. This means that if you say

85.5 \emph{percent of} $x$,    you mean    $\frac{85.5}{100}\cdot x.$

  Let $x$ be the number of patients participating in the experiment.

  We know that $\frac{85.5}{100}x = 1881.$

  Solving for $x$ we find that
    $$x = \frac{1881}{\frac{85.5}{100}}$$
so
   $$x = 2200 \; \textrm{patients}.$$ 
\end{itemize}
%---------------------------------------------------
\textbf{Note} Remember to check your answer.  Is it true that $85.5 \%$ of 2200 equals 1881?
To find out, multiply  $\frac{85.5}{100}$ times $2200$.  The result should be 1881.  Is it?

%---------------------------------------------------
\begin{itemize}
\item \textbf{Example 3.} \emph{(Percent Problem)}

A division of the Transylvania Company manufactures light bulbs.  Each month $20 \%$ of the bulbs it manufactures do not pass inspection.  Another $12 \%$ are damaged on route to stores.  If the company can sell $18,020$ bulbs, how many bulbs should it manufacture?

 \emph{Solution:} Let $x$ = the number of bulbs that should be manufactured.

  $20\%$ of $x$ is the number of bulbs that do not pass.  
  
  Hence,
  $\frac{20}{100}x =$ the number of bulbs that do not pass.

 So  $12\%$ of $x$ is the number of damaged bulbs.

  $\frac{12}{100}x =$ the number of damaged bulbs.
 

  Use the fact that

``the number of bulbs to be manufactured" - ``the number of bulbs that do not pass''-``the number of damaged bulbs" = 18,020.

  This translates into: $$x - \frac{20}{100}x - \frac{12}{100}x = 18020.$$
  Solving for $x$, we get

    $$x - .20x - .12x = 18020,$$
    or $$x - .32x = 18020,$$
    so $$x(.68) = 18020,$$
    so $$x = 26,500 \; \textrm{light}\; \textrm{bulbs.}$$ 

%---------------------------------------------------


\item \textbf{Example 4.} \emph{(Mixture problem)}

A supermarket contracts an ice cream manufacturer to mix 2000 gallons of Haag ice cream that contains $18 \%$ fat with $1,800$ gallons of Dass ice cream that contains $12\%$ fat to produce an ice cream called Hagg and Dass.  What is the amount of fat content in the Hagg and Dass mixture?

 Solution: Let $x$ be the content of fat in Hagg and Dass.

  We use the fact that
``the~amount~of~fat~in~one~cream'' + ``the~amount~of~fat~in~second~cream'' = ``the~amount~of~fat~in~mixture."

So $18\%$ of 2000 + $12\%$ of 1800 $= x\%$ of $(2000 + 1800),$
the quantity $(2000 + 1800)$ being the new mixture, which contains $(2000 + 1800)$ gallons of ice cream.
Thus, $$\frac{18}{100}2000 + \frac{12}{100}18000 = \frac{x}{100}3800.$$
Or,
$$ 360 + 216 = 38x,$$
so $$576 = 38x,$$
so $$x = 15.16 \%.$$ 
\end{itemize}
%---------------------------------------------------
\textbf{Note:} Check that this makes sense.  If the answer is correct, then $15\%$ of the total amount of the mixture should be approximately equal $18\%$ of 2000 gallons plus $12\%$ of 1800 gallons.

18\% of 2000 = 360,

12\% of 1800 = 216,

15\% of 3800 = 570. 

check that $570 = 360 + 216.$ \\
%---------------------------------------------------

\textbf{Note:}
In the example above, two batches of ice cream are mixed to get a new batch of ice cream, the mixture.  The amount of substance in the mixture is ``the amount of substance in the first batch'' +``the amount of substance in the second batch."  

The substance in the above example is ice cream.  This means that the amount of ice cream in the first mixture plus the amount of ice cream in the second mixture is equal to the amount of ice cream in the mixture.  We used this fact when grouping the 2000 gallons with the 1800 gallons to get the 3800 gallons of mixture.
%---------------------------------------------------
\begin{itemize}
\item \textbf{Example 5.} \emph{(Geometry problem)}

A high rise apartment building is planned for a $20,000$ square foot lot.  Because of strict zoning laws, the number of stories permitted for an apartment building is tied to the number of tenants in the building.  Each tenant must have a parking space. There are 20 tenants to a floor, and each parking space requires 40 square feet of space.  

If the area of the base of the building is 10 times the square of the number of floors, what is the largest number of floors that the zoning law will permit?

\emph{Solution:} Let $x$ = number of floors that the zoning law will permit.

Then the size of the base of the building is $10x^2.$

The number of tenants is $20x.$

The amount of space required for the parking lot is $(20x)40.$

We therefore know that the building space  +  the parking space  $\le$   the size of the lot, so 
$$10x^2            +       (20x)40             \le         20,000.$$
The maximum $x$ will be achieved when the inequality is an equality; in other words, when
$$10x^2 + (20x)40 = 20,000.$$
Solving for $x$, we get $$10x^2 + 800x = 20,000,$$ or $$x^2 + 80x - 2,000 = 0.$$
This quadratic equation factors as
$$(x + 100)(x - 20) = 0.$$
The solutions to this quadratic are $x = -100$ and $x = 20.$  Since $x$  is the number of floors of the high rise, it must be a positive  number.  We therefore reject the negative solution    $x = -100.$  So the answer is $$x=20 \; \textrm{floors.}$$ 
%---------------------------------------------------
\item \textbf{Example 6.} A horse race track is 1.5 miles long.  The winner comes in only five seconds before the second horse.  If the winner averaged 80 miles/hour during the race, how much faster (on average) should the second horse have gone in order to keep up with the first?

 \emph{Solution:} Let $x$ = the velocity of the second horse.

Recall the relation between distance, velocity and time:

    distance = velocity $\cdot$ time.

Since units need to match up, we must convert the 80 miles/hour to miles/second. Since 1~hour~=~60~minutes  and  1~minute~=~60~seconds, 1~hour~=~60~(60)~seconds.

Therefore, the velocity of the winner--converted into miles per second--is

80 $\frac{\textrm{miles}}{\textrm{hour}}= \frac{1}{45} \frac{\textrm{miles}}{\textrm{second}}.$

The total distance of the race is 1.5 miles.  Therefore, using the distance relation above, we get

1.5 miles = $\frac{1}{45}$ $\frac{\textrm{miles}}{\textrm{second}}\cdot t,$

where $t$ is winner's time in finishing the race (in seconds).

We know that the second horse came in 5 seconds after the winner.  Hence, we know that ``the velocity of the second horse" $\cdot(t + 5) = 1.5.$ In other words, if $x$ = the velocity of the second horse, then
$$x *(t + 5) = 1.5.$$
We may find $t$ by solving the equation
$$\frac{1}{45} \frac{\textrm{miles}}{\textrm{second}}\cdot t= 1.5\; \textrm{miles}$$
for $t$.  We find that $t = 67.5$ seconds.  Therefore,
$$x \cdot(67.5 + 5) = 1.5.$$
Solving for $x$ gives $x = 0.021$ miles/second.
The difference between the winners velocity and this velocity is
$$\frac{1}{45}- \frac{21}{1000}  = \frac{55}{45000} \frac{\textrm{miles}}{\textrm{second}}.$$

We can convert this back to a velocity in miles per hour by multiplying by 3600, since there are 3600 seconds to an hour.  We get   $$\frac{55}{45000}\cdot 3600 \frac{\textrm{miles}}{\textrm{hour}}.$$

\end{itemize}
%---------------------------------------------------

\section{Exercises}\mbox{}\\
1. Two numbers have a relationship where one minus eight times the other equals 57.  The average of the two numbers is 6.  Find the numbers.\\
 

 
2. In a certain far away land, a duke was given on his twentieth birthday a large number of gold ingots.  When he reached 30, he fell in love with a beautiful princess whose fancy he could not easily attract.  After many attempts, he fell to drink and gambling.  At 31, he gambled and lost 1/6 of his gold ingots, at 32 he lost 1/5 of the remaining ingots.  At age 34 he had only 10 ingots left.  How many gold ingots did the duke receive for his twentieth birthday?\\
 

 
3.  The editor of a large publishing company reviews algebra manuscripts for possible publication.  If she rejects 98.5\% of the manuscripts that she reads per year, how many manuscripts does she review for publication, if she accepts 15 new manuscripts per year?\\
 

 
4. A chemical plant discharges diluted waste that contains 0.04\% $H_2S$ (hydrogen sulfide) into a $15,000$ gallon holding container that contains $10,000$ gallons of liquid containing 0.01\% $H_2S.$  The mixture is released into a stream when the concentration reaches 0.02\% $H_2S$.  How much waste is needed before the valves of the holding container are opened to allow the mixture at 0.02\% to flow into the stream?\\
 

 
5. The English Channel tunnel was dug by an English team and a French team.  The French team drilled toward England at the rate of 130 feet per week, while the English team, having poorer drilling equipment drilled at a rate of 90 feet per week.  If it took two years to complete the drilling, how long is the tunnel?\\
 

 
6. A group of young teenagers assemble computers in a garage.  They decide to go into business by selling three different computers.  One contains a single floppy drive and costs \$740.00.  Another has one floppy drive and one hard drive and costs \$1080.00.  A third contains two floppy drives plus a hard drive and costs \$1200.00.  If the costs are entirely due to the drives, what does a floppy drive cost?  What does a hard drive cost?\\
 

 
7. A health food store offers three mixtures of Colombian coffee.  The first contains 5\% chicory and costs \$4.50 a pound.  The second has 5\% chicory and 5\% cocoa bean and costs \$4.55 a pound.  The third has 5\% cocoa bean and costs \$4.75 per pound.  If the price of the coffee is entirely due to the chicory and cocoa bean, what is the price per pound of the cocoa bean?\\
 

 
8. During the Ming Dynasty in China cricket fighting was so much in vogue in the Imperial Palace that the people were permitted to pay taxes by sending crickets.  Taxes were based on the amount of crops that were grown and brought to market.  A bushel of rice could be sold at the market for 40 yuan.  The taxes were assessed at 10\% of market value; but one good fighting cricket could be offered in place of 55 yuan worth of taxes.  If a peasant sold 500 bushels of rice in the market during the year, how many crickets would need if he wished to pay 22\% of his taxes in good fighting crickets?\\
 

 
9. Suppose that the value of a good fighting cricket is as it was in exercise 8.  Also suppose that a bushel of rice sells for 40 yaun but requires 10\% tax, a bushel of maize for 25 yaun requiring 12\% tax and a bushel of potatoes for 12 yaun with 15\% tax.  If a peasant sells 200 bushels of rice, 100 bushels of maize and 50 bushels of potatoes, how many good fighting crickets will he need to pay 50\% of the taxes in good fighting crickets?  (According to Ming tax laws any fractional value of a cricket is equivalent to the next larger whole number of crickets.)\\
 

 
10. The Emperor's cricket was entered in a cricket match.  Tickets were sold at 20 yaun, 40 yaun and 80 yaun each.  Ch'ang-an collected tickets and noticed that the number of 40 yaun tickets was three times the number of least expensive tickets and 50 less than the number of most expensive tickets.   The total amount collected was 8,000 yaun.  How many of each class of tickets was sold?\\
 
 

 
11 Two 19-th Century fictional characters are to fight a dual over a beautiful woman.  They choose their weapons, stand back-to-back and walk in opposite directions.  The taller of the two walks at a rate of one feet per second faster than the shorter.  At the end of 8 seconds they are 56 feet apart.  Find the speed of each person.\\
 

 
12 A troop of soldiers is marching in a line that is 400 feet long.  A messenger at the end of the line must deliver a message to the leader of the troop in the front of the line.  If the troops are marching at a rate of 8 feet per second and the messenger is running at a rate of 10 feet per second, how long will it take for the messenger to deliver his message?\\
 

 
13 In a foot race between Achilles and a tortoise, the tortoise was given a head start of 400 feet.  If Achilles runs at a rate of 20 feet per second and the tortoise runs at a rate of 2 feet per second, how long will it take for Achilles to pass the tortoise? 

%---------------------------------------------------
\section{Solutions} \mbox{}

1. 17 and -5\\

 
2.  15 gold ingots\\

 
3.   1,000 manuscripts\\

 
4.  15,000 gals. of mix or 5,000 gals of waste ?? \\
 
5.  22,880 feet\\

 
6.  \$120.00 for the floppy drive;\\

   \$340.00 for the hard drive.\\
 
7.  \$0.80 per pound\\

 
8.  8 crickets\\

 
9.  11 crickets\\

10. 7 tickets at 20 yaun, 21 tickets at 40 yaun and 71 tickets at 80 yaun\\
 

11. One walks at a rate of 3 feet/sec, the other at 4 feet/sec.\\
 

12. 200 seconds\\

13.$\frac{400}{18}$ seconds or  approx. 22.23 secs \\

\end{document}
