Applied Mathematics Study Guide for the WorkKeys

Skills Needed for Level 7 Questions

Problems in Level 7 may be presented in a way that you’re not used to. It’s very likely the information given will seem incomplete, but can be used to come up with other facts that will help you solve the problem. In other words, problems will take at least two or three logical steps to complete. Solid geometric figures like spheres, cylinders, and cones will show up at this question level.

Fractions in Problems

Problems with ratios, rates, or proportions can have given information in the form of fractions as well as whole numbers. You should be able to set up and solve these, rearranging if need be.

Ratio Problems

We know that ratios are fractions, but ratios may also contain fractions, which doesn’t change the concept, but does create more necessary steps.

Example:

A recipe calls for $\dfrac{1}{2}$ cup of raisins and $2\dfrac{1}{4}$ cups of flour. What is the ratio of raisins to flour?

We’re looking for the ratio $\dfrac{\text{raisins}}{\text{flour}}$.

Putting in the amounts given gives us this: $\dfrac{\frac{1}{2}}{2\frac{1}{4}}$

Change the denominator to an improper fraction: $\dfrac{\frac{1}{2}}{\frac{9}{4}}$

To divide fractions, invert the denominator and multiply by the numerator.

\[\dfrac{1}{2} \times \dfrac{4}{9}\] \[\dfrac{4}{18}\] \[\dfrac{2}{9}\]

Rate Problems

In Level 7, the rate problems you see will still use what you already know about rates, but will be a bit more work. Example:

Claire earned $\$540$ for a $40$-hour work week and decided to put in a few hours of overtime on Saturday to earn a little extra. The rate for overtime work is $1.5$ times the usual wage. If Claire worked $4.5$ hours on Saturday, how much extra money did she earn?

Find Claire’s normal hourly wage using a ratio of dollars to hours.

\[\dfrac{\$ 540}{40}=\$ 13.50\]

Find Claire’s overtime wage.

\[\$13.50 \times 1.5 = \$20.25\]

Find the amount she earned on Saturday.

\[\$ 20.25 \times 4.5=\$ 91.125 \approx \$ 91.13\]

Proportion Problems

A proportion is usually used to solve for an unknown quantity and will look like this:

\[\dfrac {3}{8}=\dfrac {x}{32}\]

Solve by multiplying both sides by $32$ and reducing.

\[32\times\dfrac {3}{8}=\require{cancel}\cancel{32}\times \dfrac {x}{\cancel{32}}\] \[\dfrac{96}{8}=x\] \[12=x\]

Finding the Reason for an Error

Many math errors are due to simple calculation mistakes, like $3\times 2=8$. Other errors are due to faulty reasoning and are a little harder to spot. Here’s an example of one of those.

Test Question: A soap maker mixes $80$ g of lye pellets with water to make $300$ g of solution. What is the ratio of lye to water?

A student writes $\dfrac{80}{300}$ and simplifies it to $\dfrac{4}{15}$. The teacher gives only a small amount of partial credit for that answer. What was the student’s flaw in thinking that led to the wrong answer?

Answer: The student found the ratio of lye to solution, not lye to water. He should have subtracted $80$ g from $300$ g to get the mass of water: $220$ g. This would have given a ratio of $\dfrac{80}{220}$, which would simplify to $\dfrac{4}{11}$.

More Complicated Unit Conversions

Unit conversions can be done with any type of number, and here are some examples using different types.

With Fractions

An American farm implement lost a $\frac{9}{16}$ inch diameter bolt and nut, but the German farmer had only metric replacement bolts and nuts. What is the biggest metric bolt that will fit in the $\frac{9}{16}$ inch hole? (Metric bolts come in whole number mm sizes.)

Change $\frac{9}{16}$ inch to mm. ($1$ inch $= 25.4$ mm)

\[\dfrac{9}{16} \text{ in.} \times \dfrac {25.4 \text { mm}}{1 \text{ in.}} =14.28 \text{ mm}\]

The hole size is $14.28$ mm, so the largest metric bolt that will fit is a $14$ mm bolt.

With Mixed Numbers

Suppose a casual walking pace is about $2\dfrac{1}{3} \frac{\text { mi}}{\text { hour}}$. What would this speed be in $\dfrac{ \text{ft.}}{ \text{sec.}}$?

Use $1$ mi. $= 5280$ ft. and $1$ hour = $3600$ sec. to convert.

Change $2\dfrac{1}{3}$ to an improper fraction and multiply by two conversion factors.

\[\require{cancel}\dfrac{7 \cancel{\text{ mi.}}}{3 \cancel{\text { hour}}} \times \dfrac{1 \cancel {\text{ hour}}}{3600 \text{ sec.}} \times \dfrac{5280 \text{ ft.}}{1 \cancel{\text { mi.}}} = 3.4 \dfrac{ \text { ft.}}{\text { sec.} }\]

With Decimals

A board needs to be cut to a length of $3\frac{1}{4} \text{ ft.}$ but the French carpenter has a tape measure marked in centimeters. How long, to the nearest cm, should he make the board? ($1$ inch $= 2.54$ cm)

Change $\frac{1}{4}$ to a decimal and convert ft. to in.:

\[3.25 \text{ ft.} \times \dfrac{12 \text{ in.}}{1 \text{ ft.}}=39 \text { in.}\]

Convert in. to cm:

\[39 \text { in.} \times \dfrac{2.54 \text{ cm}}{1 \text{ in.}}=99.06 \approx 99 \text { cm}\]

With Percentages

I walked $5 \%$ of a mile yesterday. What percent of a foot would this be?

Change $5 \%$ to $0.05$ and convert.

\[0.05 \text{ mi.} \times \dfrac{5280 \text{ ft.}}{1 \text{ mi.}}=264 \text { ft.}\]

Convert $264$ to $26,400 \%$ of a foot.

Additional Volume Calculations

On the formula sheet that goes with this test, you’ll have volume formulas for cylinders, cones, and spheres. Here are three sample problems to give you some experience using the formulas.

Sphere

Which has a larger total volume, a sphere with radius $2r$ cm or two spheres, each with radius $r$ cm?

Make up any number for $r$. Let’s make it $1$ to stay simple. So $2r$ will be $2$. Now put those into the formula $V=\dfrac{4}{3} \pi r^{3}$

Using $r=1$: $V=\dfrac{4}{3} \times 3.14 \times 1^{3} \approx 4.2 \text{ cm}^{3}$

Double that number because there are two spheres this size.

\[2\times 4.2=8.4 \text{ cm}^{3}\]

Using $r = 2$:

\[V=\dfrac{4}{3} \times 3.14 \times 2^{3} \approx 33.5 \text{ cm}^{3}\]

The single sphere wins. It actually has a volume $4$ times larger than the two smaller spheres.

Cylinder

A cylindrical tank sitting on its base has a diameter of $3$ ft. and a height of $4$ ft. It has water inside that is $30$ inches deep. What is the volume of water?

\[V=\pi r^{2}h\]

The radius of the tank is $1.5$ ft and the height (depth) of the water is $2.5$ ft.

$V=3.14 \times 1.5^{2} \times {2.5}$ $V=17.7 \text{ft}^{3}$

Cone

There is a cylinder with a radius of $3$ and a height of $5$ and a cone with a radius of $3$ and a height of $5$. What is the ratio of the volume of the cylinder to the volume of the cone?

Volume of a cylinder = $V=\pi r^{2}h$

Volume of a cone = $V=\frac{1}{3}\pi r^{2}h$

You could put the given values into the two formulas and calculate the ratio from the two results.

Or you could cleverly notice that because of the $\frac{1}{3}$ in its formula, the volume of a cone will be $\frac{1}{3}$ that of the cylinder if $r$ and $h$ are the same.

That makes the cylinder’s volume $3$ times larger than that of the cone.

The ratio is $3\text{:}1$.

Using Volume Calculations

Volume calculations can be combined with unit conversions and the volume formulas in these problems may need to be rearranged or used more than once.

Example:

A cubic tank with edges of $1$ m is to be filled with wax by melting a number of spherical wax balls. If the radius of each ball is $20$ cm, how many balls need to be melted to fill the tank? Assume that the volume of wax doesn’t change when it is melted and that you can melt a portion of a ball if you need to. Answer to the nearest tenth of a ball.

First, change $1$ m to $100$ cm and calculate volumes.

Tank: $V= (100 \text{ cm})^3 = 1,000,000 \text { cm}^{3}$

Ball: $V=\dfrac{4}{3} \times 3.14 \times (20 \text{ cm})^{3} = 33,510 \text{ cm}^{3}$

Now convert $\text{cm}^{3}$ to balls

\[1,000,000 \text{ cm}^{3} \times \dfrac{1 \text { ball}}{ 33,510 \text{ cm}^{3}} = 29.8 \text { balls}\]

Finding Better Economic Value

Everyone wants to get the best deal when they buy something. There are different ways to help choose the better way to go.

Using Graphics

Let’s say you were going to rent a car from either Acme or Standard rental agencies, and you wanted to get the best deal. The graph above shows the pricing structure for each company. For an $8$-day rental, which company gives the best deal?

At $8$ days the Standard graph (dotted line) is lower than the Acme graph, so Standard is cheaper.

How long a rental period would give the same price for each company?

The crossover point where both graphs have the same value is $12$ days.

The graph shows you that for fewer than $12$ days, Standard is cheaper, but for more than $12$ days, Acme is cheaper.

Using Percentage Difference

Suppose there are two identical new cars for sale at different prices. Honest Charlie is selling one for $\$29,200$ and Dapper Dan is selling one for $\$28,900$. Clearly, Dapper Dan has the lowest price, but his dealership is two hours away. How do you decide? It might help if you calculated something called percent difference. It’s a way of helping judge the relative amount of difference between numbers. A three-hundred dollar difference is nothing to ignore, but is it really all that much compared to the price?

Percent difference = $\dfrac{\text {Price 1}-\text {Price 2}}{\text{Average of Prices}} \times 100 \%$

The average price is $\dfrac{29,200-28,900}{2}=29,050$

Percent difference = $\dfrac{29,200-28,900}{29,050} \times 100 \%$

Percent difference = $\dfrac {300}{29,050} \times 100 \%$

Percent difference = $1.03 \%$

So you’re talking about a $1\%$ difference, which doesn’t sound like much. Still, three hundred bucks is three hundred bucks.

Using Unit Cost

This is a good way to compare deals if you are buying something that comes in different packaged quantities, like corn flakes or bags of concrete mix. We did an example previously in Level 6. The method is to divide the cost by the amount, getting units like

$\dfrac{ \text{ cents}}{\text {oz.}}$ or $\dfrac{ \text{ dollars}}{\text {lb.}}$

Example:

Chocolates are selling in one store for $\$7.95$ per pound or in a different store for $59$ cents per ounce. Which price is the best value?

Make sure you are comparing the same units when you divide, so change the units of pounds to ounces first. Then divide.

\[\dfrac{\$7.95}{1 \text{ lb.}}\] \[\dfrac{\$7.95}{16 \text{ oz.}}\] \[\dfrac {\$0.50}{\text { oz.}}\]

We can see that $\dfrac {\$0.50}{\text { oz.}}$ is a better value than $\dfrac {\$0.59}{\text { oz.}}$.

Using Statistics

You should remember a few statistical calculations, as listed below, and be able to find and interpret measures of central tendency, spread, and tolerance.

Finding the Weighted Mean

A weighted mean is done a little differently than a regular mean. To show the difference, imagine that $13$ students took a history test. $12$ of them did very well, and each got an A $(4.0)$. The other didn’t study and got a D $(1.0)$. What is the class average (mean)?

One way (the wrong way) to do this might be to say there are two different grades, $(4.0)$ and $(1.0)$, so add them and divide by $2$. That would give a class mean of $2.5$, which seems way off because almost all the students got four points. The mean should be a lot closer to $4.0$.

The right way to do this is to add the points earned by all $13$ of them and divide the sum by $13$.

\[12 (4.0) + 1 (1.0) = 49.0\] \[\dfrac{49}{13}=3.8\]

This is called a weighted average; the score $4.0$ is given a lot more weight in the total than the score $1.0$ because many more people had a $4.0$.

Measures of Central Tendency

Central tendency is the idea of finding a single number to represent a set of numbers (data). It means finding a number that the data seem to be clustered around. This was covered in Level 4.

Mean—A common way to calculate central tendency is to simply find the average (called the mean) of the group. Add all the numbers and divide by how many there are.

Median—A second way is to arrange the data from lowest to highest and pick the value in the exact middle. This is called the median. If there are an even number of values, the median is the mean of the two middle terms.

Mode—A third way is to pick the data value that shows up most often. This is the mode. If no values are repeated, then there is no mode.

As an example, suppose this is a set of seven winter temperatures in Detroit.

\[{30, 31, 29, 32, 29, 26, 54}\]

The mean is $\dfrac{30+31+29+32+29+26+54}{7}=33$

The median is $\require{enclose}26, 29, 29, \enclose{circle}{30}, 31, 32, 54$

The mode is $29$ because it shows up most often (twice).

Measure of Spread

Some sets of numbers are all very close to each other, not very spread out. Other sets are really spread out. One measure of this spread is called the range of the data. It is simply the difference between the largest and smallest values.

The range of ${ 66, 72, 79, 88, 92, 99}$ is $99-66=33$.

A more sophisticated measure of spread is called standard deviation. Calculating this is beyond the scope of this WorkKeys Test.

Tolerance

In manufacturing, some parts have to be made very accurately, while others are not as critical. Even with very sophisticated equipment, It’s impossible to make two parts exactly the same size. In very accurate machining of, say, a piston for an engine, there will still be a very tiny difference from one piston to the next. The amount of difference allowed is called the tolerance for that part. Engineers specify that when they design the parts. The size of a main bearing, for example, may be specified as $2.4500 \pm 0.0002 \text { inches}$.

The tolerance is $\pm0.0002 \text{ inch}$, meaning the bearing is allowed to be $0.0002 \text{ inch}$ bigger or smaller than $2.4500 \text{ inches}$.

Another way to say this is that the allowed measurements are from $2.4498 \text{ inches}$ to $2.4502 \text{ inches}$. The allowed range is $0.0004 \text{ inch}$.