Math 2000 Solutions 3

Math 2000 - Solutions 3
Spring 2002

Unit 2A 15d. First we convert 5.50 euros to dollars:

$\begin{displaymath} 5.50 {\rm\ euros} = 5.50 {\rm\ euros} \times \frac{ \$0.9544}{1\rm\ euros} =\$5.25. \end{displaymath}$

Hence, gasoline in Germany sells for $5.25 per gallon. 26. The field area is $150 {\rm\ feet} \times 100 {\rm\ feet}= 15,000$ square feet, which we must first convert to square yards:

$\begin{displaymath} 15,000 {\rm\ ft^2} \times \frac{1 \rm\ yd^2}{9\rm\ ft^2} = 1666.67 {\rm\ yd^2}. \end{displaymath}$

Covering the field with turf will cost:

$\begin{displaymath} 1666.67 {\rm\ yd^2} \times \frac{\$7.50}{1\rm\ yd^2} =\$12,500 \end{displaymath}$

Unit 2B 4d. First, note that the area of the field is $100 \rm yd \times 60 \rm yd = 6000 \rm\ yd^2$ . We can now convert the volume of the garbage to cubic yards. Because ${1\rm\ yd^3}=({3\rm\ ft})^3={27\rm\ ft^3}$ , we have

$\begin{displaymath} {500,000 \rm\ ft^3} = {500,000 \rm\ ft^3} \times\frac{1\rm\ yd^s}{27\rm\ ft^3}= 18,520 \mbox{ cubic yards}. \end{displaymath}$

To find the height of the pile, we divide the total volume of the garbage by the area of the field (notice how the units work out):

$\begin{displaymath} \mbox{height of pile} = \frac{18,520\rm\ yd^3}{6000 \rm\ yd^2} = 3.0867 \mbox{ yards} = 9.26 \mbox{ feet}. \end{displaymath}$

The problem could also be worked by converting all units to feet. 8b. Converting from pounds to kilograms using 1 pound = 2.2 kg (or equivalently, 1 kg = 0.5 pounds), we have

$\begin{displaymath} {150\rm\ lb} \times\frac{0.4536\rm\ kg}{1\rm\ lb}, \end{displaymath}$

which is about 68 kilograms. 49a. Solutions will vary, but let's assume you weigh 155 pounds. As shown in class, there are several steps to this solution. a. A 155-pound body contains about

$\begin{displaymath}155{\rm\ lb}\times\frac{1\rm\ kg}{2.205\rm\ lb} \times\frac{... ...es\frac{1\rm\ L}{1000\rm\ mL}= 4.92 \mbox{ liters of blood.} \end{displaymath}$

At a concentration of $0.08\%=0.0008$ , the amount of alcohol in this much blood is

$\begin{displaymath} 0.0008 \times 4.92\rm\ L = 0.0039\rm\ L = 3.9\rm\ mL. \end{displaymath}$

Because the alcohol content of wine is 13%, we know that

$\begin{displaymath} 13\% \times \mbox{ Amount of wine} = 3.9\rm\ mL. \end{displaymath}$

It follows that the amount of wine is $\frac{3.9\rm\ mL}{0.13}= 30$ mL, which is about 1 oz. 49b. At the near-fatal blood-alcohol concentration of 0.5%, the amount of alcohol in the blood is $0.5\% \times 4.92\rm\ L = 0.0246\rm\ L = 24.6\rm\ mL$ . The amount of wine that leads to 24.6 mL of alcohol in the blood $\frac{24.6\rm\ mL}{0.13}= 189.23$ mL, or about 7 oz. Unit 3A 6c. Note that $\frac{2345}{4023}=0.5829 = 58.29\%$ . Thus,

is $58.29\%$ of

. 8. The absolute change is

deaths, and the percentage change is

$\begin{displaymath} -\frac{63000}{1008000}=-0.0625 = -6.25\%. \end{displaymath}$

The number of deaths decreased by $6.25\%$ . The negative changes indicates a decrease. 12. The absolute change is

gallons, and the percentage change is

$\begin{displaymath} \frac{0.53}{0.73}=0.7260= 72.60\%. \end{displaymath}$

Beer consumption increased by $72.60\%$ . Extra credit. The key to this problem (see page 100 of the text) is that a British pint is 20% larger than an American pint. Therefore, an American pint is 16 oz and a British pint is $16\rm\ oz + (20\% \times 16\rm\ oz) = 19.2 \rm\ oz$ . Let's also assume that $\$1 = 0.66 \mbox{ British pounds}$ . We can now see that American beer costs

$\begin{displaymath} \frac{\$2.50}{12 \rm\ oz} = \$0.21 \mbox{ per ounce}. \end{displaymath}$

In the same units, British beer costs

$\begin{displaymath} \frac{1.75 \mbox{ pounds }}{1 \rm\ pint} \times \frac{1 \rm... ... \frac{\$1}{0.66 \mbox{ pounds}} = \$0.14 \mbox{ per ounce}. \end{displaymath}$

We see that the British beer costs less on a per-ounce basis than the American beer.