Measurements and Units
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For the Classic ACT exam:
The ACT Mathematics test is a timed exam...60 questions in 60 minutes
This implies that you have to solve each question in one minute.
Each of the first 20 questions (less challenging) will typically take less than a minute a solve.
Each of the next 20 questions (medium challenging) may take about a minute to solve.
Each of the last 20 questions (more challenging) may take more than a minute to solve.
The goal is to maximize your time.
You use the time saved on the questions you solve in less than a minute to solve questions that will take more
than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.
Also: please note that unless specified otherwise, geometric figures are drawn to scale. So, you can figure out
the correct answer by eliminating the incorrect options.
Other suggestions are listed in the solutions/explanations as applicable.
These are the solutions to the ACT past questions on the topics: Measurements and Units.
When applicable, the TI-84 Plus CE calculator (also applicable to TI-84 Plus calculator) solutions are provided
for some questions.
The link to the video solutions will be provided for you. Please
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Please use the conversion equation specified in the questions as applicable.
However, if the conversion equation was not given to you, then it is necessary to use these tables.
It is also important to memorize the prefixes and the multiplication factors for the International System of
Units.
The three tables are:
Table 1
| Prefix | Symbol | Multiplication Factor |
|---|---|---|
| yocto | y | $10^{-24}$ |
| zepto | z | $10^{-21}$ |
| atto | a | $10^{-18}$ |
| femto | f | $10^{-15}$ |
| pico | p | $10^{-12}$ |
| nano | n | $10^{-9}$ |
| micro | $\mu$ | $10^{-6}$ |
| milli | m | $10^{-3}$ |
| centi | c | $10^{-2}$ |
| deci | d | $10^{-1}$ |
| deka | da | $10^1$ |
| hecto | h | $10^2$ |
| kilo | K | $10^3$ |
| mega | M | $10^6$ |
| giga | G | $10^9$ |
| tera | T | $10^{12}$ |
| peta | P | $10^{15}$ |
| exa | E | $10^{18}$ |
| zetta | Z | $10^{21}$ |
| yotta | Y | $10^{24}$ |
Table 2
| Measurement | Customary | Customary | Unit Conversion Factor |
|---|---|---|---|
| Length | inch (in) | foot (ft) | $12\:inches = 1\:ft$ |
| Length | foot (ft) | yard (yd) | $3\:ft = 1\:yd$ |
| Length | yard (yd) | mile (mi) | $1760\:yd = 1\:mi$ |
| Length | foot (ft) | mile (mi) | $5280\:ft = 1\:mi$ |
| Length | rod/pole | yards (yd) | $1\:rod = 5.5\:yd$ |
| Length | furlong | rod | $1\:furlong = 40\;rod$ |
| Length | fathom | feet (ft) | $1\:fathom = 6\;ft$ |
| Length | league/marine | nautical miles | $1\:league = 3\;nautical\;\;miles$ |
| Mass | pound (lb) | ounce (oz) | $1\:lb = 16\:oz$ |
| Mass | short ton (ton) | pound (lb) | $1\:short\:ton = 2000\:lb$ |
| Mass | long ton | pound (lb) | $1\:long\:ton = 2240\:lb$ |
| Mass | stone | pound (lb) | $1\:\:stone = 14\:lb$ |
| Mass | long ton | stone | $1\:long\:ton = 160\:stones$ |
| Area | acre (acre) | square feet ($ft^2$) | $1\:acre = 43560\:ft^2$ |
| Volume | quart (qt) | pint (pt) | $1\:qt = 2\:pt$ |
| Volume | pint (pt) | cup (cup) | $1\:pt = 2\:cups$ |
| Volume | quart (qt) | cup (cup) | $1\:qt = 4\:cups$ |
| Volume | quart (qt) | fluid ounce (fl. oz) | $1\:qt = 32\:fl.\:oz$ |
| Volume | pint (pt) | fluid ounce (fl. oz) | $1\:pt = 16\:fl.\:oz$ |
| Volume | cup (cup) | fluid ounce (fl. oz) | $1\:cup = 8\:fl.\:oz$ |
| Volume | gallon (gal) | quart (qt) | $1\:gal = 4\:qt$ |
| Volume | gallon (gal) | quart (pt) | $1\:gal = 8\:pt$ |
| Volume | gallon (gal) | cup (cup) | $1\:gal = 16\:cups$ |
| Volume | gallon (gal) | fluid ounce (fl. oz) | $1\:gal = 128\:fl.\:oz$ |
| Volume | gallon (gal) | cubic inches ($in^3$) | $1\:gal = 231\:in^3$ |
Table 3
| Measurement | Metric | Customary | Unit Conversion Factor |
|---|---|---|---|
| Length | meter (m) | foot (ft) | $1\:ft = 0.3048\:m$ |
| Length | kilometer (km) | nautical miles | $1\:nautical\;\;miles = 1.852\;km$ |
| Mass | gram (g) | pound (lb) | $1\:lb = 453.59237\:g$ |
| Mass | metric ton (tonne) | kilogram (kg) | $1\:tonne = 1000\:kg$ |
| Volume |
liter or cubic decimeters (L or $dm^3$) |
gallons (gal) | $1\:L = 0.26417205\:gal$ |
On the model, $\overline{BC}$ represents an actual length of 90 feet in the park.
On the model, $\overline{DE}$ represents what actual length, in feet, in the park?
$ F.\;\; 67.5 \\[3ex] G.\;\; 72 \\[3ex] H.\;\; 112.5 \\[3ex] J.\;\; 120 \\[3ex] K.\;\; 150 \\[3ex] $
| inch | feet |
|---|---|
| 4 | 90 |
| 5 | what |
$ \dfrac{what}{90} = \dfrac{5}{4} \\[5ex] what \cdot 4 = 90 \cdot 5 \\[3ex] what = \dfrac{90 \cdot 5}{4} \\[5ex] what = 112.5\;feet $
A scatterplot of the data is shown below.
To the nearest pound, the average weight of these pumpkins was 10 pounds, and the average number of seeds per pumpkin was 444 seeds.
An equation of the regression line of best fit is y = 15x + 294, where x is the weight, in pounds, and y is the number of seeds.
An object that weighs 1 pound on Earth has a mass of 0.45 kilograms.
What is the mass, to the nearest 0.1 kilogram, of a pumpkin that weighs the same as the average weight of the 23 pumpkins?
$ F.\;\; 4.5 \\[3ex] G.\;\; 10.4 \\[3ex] H.\;\; 14.9 \\[3ex] J.\;\; 22.2 \\[3ex] K.\;\; 51.1 \\[3ex] $
$ \underline{\text{Unity Fraction Method}} \\[3ex] 10\;lbs * \dfrac{...kg}{...lbs} \\[5ex] 10\;lbs * \dfrac{0.45\;kg}{1\;lb} \\[5ex] 4.5\;kg $
(Note: No sales tax is charged.)
$ A.\;\; \$4.40 \\[3ex] B.\;\; \$5.40 \\[3ex] C.\;\; \$6.00 \\[3ex] D.\;\; \$6.30 \\[3ex] E.\;\; \$7.20 \\[3ex] $
$ 4\dfrac{4}{5} = 4.8 \\[5ex] \underline{\text{Unity Fraction Method}} \\[3ex] 4.8\;pounds * \dfrac{...\$}{...pound} \\[5ex] 4.8\;pounds * \dfrac{\$1.50}{1\;pound} \\[5ex] \$7.20 $
You cut it into 3 pieces: one is exactly 3 feet 6 inches long and another is exactly 2 feet 3 inches long, as shown below.
If each cut is exactly $\dfrac{1}{8}$ inch wide, how long, in feet and inches, is the third piece?
$ F.\;\; 2\;ft\;3\;in \\[3ex] G.\;\; 2\;ft\;2\dfrac{3}{4}\;in \\[5ex] H.\;\; 2\;ft\;2\dfrac{5}{8}\;in \\[5ex] J.\;\; 2\;ft\;\dfrac{3}{4}\;in \\[5ex] K.\;\; 2\;ft \\[3ex] $
To get the length of the 3rd piece, we need to consider:
1st: the length of the 1st piece
2nd: the length of the 2nd piece
3rd: the width of each cut, also known as the kerf. The kerf is the thickness (width) of the cut made by the blade. It is an important factor to consider when making accurate cuts.
We shall determine the sum of these cuts, and subtract the sum from the length of the lumber to get the length of the third piece.
$ \underline{Total\;\;Length\;\;of\;\;Cuts} \\[3ex] Length\;\;of\;\;1st\;\;Piece = 3\;ft\;6\;in \\[3ex] Length\;\;of\;\;2nd\;\;Piece = 2\;ft\;3\;in \\[3ex] Thickness\;\;of\;\;1st\;\;cut = \dfrac{1}{8} = 0.125\;in \\[5ex] Thickness\;\;of\;\;2nd\;\;cut = \dfrac{1}{8} = 0.125\;in \\[5ex] \begin{array}{c} 3ft\;\;\;6in \\ +\hspace{7em} \\ 2ft\;\;\;3in \\ +\hspace{7em} \\ ~~~~~~~0ft\;\;\;0.125in \\ +\hspace{7em} \\ ~~~~~~~0ft\;\;\;0.125in \\ \hline ~~~~~5ft\;\;\;9.25in \\ \hline \end{array} $
$ \underline{Length\;\;of\;\;3rd\;\;Piece} \\[3ex] \begin{array}{c} 8ft\;\;\;0in \\ -\hspace{7em} \\ ~~~~~5ft\;\;\;9.25in \\ \hline ~~~~~2ft\;\;\;2.75in \\ \hline \end{array} $
$ \text{Length of 3rd Piece} = 2ft\;\;\;2.75in \\[3ex] Convert\;\;decimal\;\;to\;\;fraction \\[3ex] 2.75 = 2 + 0.75 \\[3ex] 0.75 = \dfrac{75}{100} = \dfrac{3}{4} \\[5ex] \implies \\[3ex] \text{Length of 3rd Piece} = 2\;ft\;2\dfrac{3}{4}\;in $
Convert all inches to feet
$ \underline{Unity\;\;Fraction\;\;Method} \\[3ex] ...\;in * \dfrac{...\;ft}{...\;in} \\[5ex] 6\;in = \dfrac{6}{12}\;ft \\[5ex] 3\;in = \dfrac{3}{12}\;ft \\[5ex] \dfrac{1}{8}\;in = \dfrac{1}{8}\;in * \dfrac{1\;ft}{12\;in} = \dfrac{1}{96}\;in \\[5ex] $
$\text{Length of 3rd Piece} = 2\;ft\;2\dfrac{3}{4}\;in$
When checking meters, he travels a rectangular path that starts and ends at the corner of Main Street and 1st Avenue, as shown in the figure below.
Matt travels this path 15 times in a typical day.
In traveling this path 15 times, about how many miles does Matt travel?
(Note: 1 mile = 5,280 feet)
$ A.\;\; 13 \\[3ex] B.\;\; 16 \\[3ex] C.\;\; 31 \\[3ex] D.\;\; 88 \\[3ex] E.\;\; 352 \\[3ex] $
$ \underline{\text{Rectangular Path}} \\[3ex] Length, L = 3500\;ft \\[3ex] Width, W = 2000\;ft \\[3ex] Perimeter, P = 2(L + W) \\[3ex] = 2(3500 + 2000) \\[3ex] = 11000\;ft \\[5ex] \text{Matt travels this path 15 times} \implies 15(11000\;ft) = 165000\;ft \\[5ex] 165000\;ft \;\;to\;\; miles \implies \\[3ex] = 165000\;ft * \dfrac{...miles}{...ft} \\[5ex] = 165000\;ft * \dfrac{1\;mile}{5280\;ft} \\[5ex] = 31.25\;miles \\[3ex] $ Option C. is the correct answer because of the wording: about how many miles
Each element is composed of small squares that are 8 mm wide and 8 mm long.
Each element after the 1st element is a square that is 8 mm wider and 8 mm longer than the previous element.
What is the area, in square centimeters, of the 4th element?
$ A.\;\; 16 \\[3ex] B.\;\; 20 \\[3ex] C.\;\; 25 \\[3ex] D.\;\; 40 \\[3ex] E.\;\; 64 \\[3ex] $
| 1st Element | 2nd Element | 3rd Element | 4th Element | |
|---|---|---|---|---|
| Length (mm) | $8 + 8 = 16$ | $16 + 8 = 24$ | $16 + 8(2) = 32$ | $16 + 8(3) = 40$ |
| Width (mm) | $8 + 8 = 16$ | $16 + 8 = 24$ | $16 + 8(2) = 32$ | $16 + 8(3) = 40$ |
| Area (mm²) | $16 \cdot 16$ | $24 \cdot 24$ | $32 \cdot 32$ | $40 \cdot 40 = 1600$ |
Convert 1600 mm² to cm²
$ \underline{\text{Unity Fraction Method}} \\[3ex] 1600\;mm * mm * \dfrac{...m}{...mm} * \dfrac{...m}{...mm} * \dfrac{...cm}{...m} * \dfrac{...cm}{...m} \\[5ex] 16 \cdot 10^2\;mm * mm * \dfrac{10^{-3}\;m}{1\;mm} * \dfrac{10^{-3}\;m}{1\;mm} * \dfrac{1\;cm}{10^{-2}\;m} * \dfrac{1\;cm}{10^{-2}\;m} \\[6ex] 16 \cdot 10^{2 - 3 - 3 + 2 + 2} \\[4ex] 16 \cdot 10^0 \\[4ex] 16 \cdot 1 \\[3ex] 16\;cm^2 $
