Chapter 13

According to the golden ratio, ideally the ratio of a person’s height to the length from their belly button to the floor is approximately 1.618.

Convert 6 ft 2 in to inches.

6 ft is 72 inches.

6 ft 2 inches is 72 inches + 2 inches = 74 inches

The approximate length from their belly button to the floor is 46 inches.

In the Fibonacci sequence, each number is the sum of the two previous numbers.

1, 1, 2, 3, 5, 8, 13, 21, …

The next number in the sequence is 13 + 21 = 34 scales.

The ratio of $\frac{46,368}{28,657}\approx <!--\; \approx \; -->1.618033988\approx <!--\; \approx \; -->1.618$.

This is very similar to $\mathrm{\varphi <!--\; \varphi \; -->}$.

You can use conversion factors to help.

Gallons used if leave water on:

$\frac{4\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\cancel{\mathrm{b}\mathrm{r}\mathrm{u}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}}}\left(\frac{2\cancel{\mathrm{b}\mathrm{r}\mathrm{u}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{s}}}{\mathrm{d}\mathrm{a}\mathrm{y}}\right)\left(\frac{7\cancel{\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}}}{\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}}}\right)\left(\frac{4\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}\mathrm{s}}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}\right)=224\frac{\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}$

Gallons used if turn water off:

$\frac{0.25\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\cancel{\mathrm{b}\mathrm{r}\mathrm{u}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}}}\left(\frac{2\cancel{\mathrm{b}\mathrm{r}\mathrm{u}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{s}}}{\mathrm{d}\mathrm{a}\mathrm{y}}\right)\left(\frac{7\cancel{\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}}}{\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}}}\right)\left(\frac{4\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}\mathrm{s}}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}\right)=14\frac{\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}$

Find the difference: 224 – 14 = 210.

You will save 210 gallons in a month.

Find one-third of a town’s population (15,000 people).

$\frac{1}{3}\left(15,000\right)=5,000\mathrm{p}\mathrm{e}\mathrm{o}\mathrm{p}\mathrm{l}\mathrm{e}$

One person saves 210 gallons per month. Use this as a conversion factor.

$5,000\cancel{\mathrm{p}\mathrm{e}\mathrm{o}\mathrm{p}\mathrm{l}\mathrm{e}}\left(\frac{210\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{1\cancel{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{o}\mathrm{n}}}\right)=1,050,000\mathrm{p}\mathrm{e}\mathrm{o}\mathrm{p}\mathrm{l}\mathrm{e}$

The town would save 1,050,000 gallons per month.

Find one-sixth of the population of Florida (21.6 million people).

$\frac{1}{6}\left(21,600,000\right)=3,600,000\mathrm{p}\mathrm{e}\mathrm{o}\mathrm{p}\mathrm{l}\mathrm{e}$

Find the savings for one person in a year.

Gallons used if leave water on:

$\frac{4\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\cancel{\mathrm{b}\mathrm{r}\mathrm{u}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}}}\left(\frac{2\cancel{\mathrm{b}\mathrm{r}\mathrm{u}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{s}}}{\mathrm{d}\mathrm{a}\mathrm{y}}\right)\left(\frac{7\cancel{\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}}}{\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}}}\right)\left(\frac{52\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}\mathrm{s}}}{\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}}\right)=2,912\frac{\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}}$

Gallons used if turn water off:

$\frac{0.25\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\cancel{\mathrm{b}\mathrm{r}\mathrm{u}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}}}\left(\frac{2\cancel{\mathrm{b}\mathrm{r}\mathrm{u}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{s}}}{\mathrm{d}\mathrm{a}\mathrm{y}}\right)\left(\frac{7\cancel{\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}}}{\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}}}\right)\left(52\frac{\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}\mathrm{s}}}{\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}}\right)=364\frac{\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}}$

Find the difference: 2,912 – 364 = 2,548 gallons saved per year per person.

$3,600,000\cancel{\mathrm{p}\mathrm{e}\mathrm{o}\mathrm{p}\mathrm{l}\mathrm{e}}\left(\frac{2,548\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{1\cancel{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{o}\mathrm{n}}}\right)=9,172,800,000\mathrm{p}\mathrm{e}\mathrm{o}\mathrm{p}\mathrm{l}\mathrm{e}$

The town would save around 9.828 million gallons per year.

$\frac{1,000}{175\mathrm{w}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{s}}=5.714\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{u}\mathrm{s}\mathrm{e}1\mathrm{k}\mathrm{W}$

How many hours is the TV used in a month?

The TV is on 3 hours per day.

Use 30 days in a month as a conversion factor.

$\frac{3\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}}{\cancel{\mathrm{d}\mathrm{a}\mathrm{y}}}\left(\frac{30\cancel{\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}\right)=90\frac{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}$

Use the knowledge that it takes about 5.714 hours to use 1 kilowatt.

$90\frac{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}\left(\frac{1\mathrm{k}\mathrm{W}}{5.714\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}}\right)\approx <!--\; \approx \; -->15.8\mathrm{k}\mathrm{W}$

The TV will use 15.8 kW in a month on average if it is on three hours a day.

$\frac{1,000}{375\mathrm{w}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{s}}=2.667\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{u}\mathrm{s}\mathrm{e}1\mathrm{k}\mathrm{W}$

Hours of use per day: 9

Use 30 days in a month as a conversion factor.

$\frac{9\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}}{\cancel{\mathrm{d}\mathrm{a}\mathrm{y}}}\left(\frac{30\cancel{\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}\right)=270\frac{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}$

Use the knowledge that it takes about 2.667 hours to use 1 kilowatt and that it costs 14 cents per kilowatt-hour.

$270\frac{\cancel{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}\left(\frac{1\cancel{\mathrm{k}\mathrm{W}}}{2.667\cancel{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}}\right)\left(\frac{\mathrm{\$<!--\; \$\; -->}0.14}{1\cancel{\mathrm{k}\mathrm{W}}}\right)\approx <!--\; \approx \; -->\mathrm{\$<!--\; \$\; -->}14.18\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}$

The refrigerator will cost $14.18 for one month.

$\frac{1,000}{2000\mathrm{w}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{s}}=0.5\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{u}\mathrm{s}\mathrm{e}1\mathrm{k}\mathrm{W}$

Time in use: 15 minutes

$15\cancel{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{s}}\left(\frac{1\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{60\cancel{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{s}}}\right)=0.25\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}$

Use the knowledge that it takes 0.25 hours and it costs 12 cents per kilowatt-hour.

$0.25\cancel{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}}\left(\frac{1\cancel{\mathrm{k}\mathrm{W}}}{0.5\cancel{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}}\right)\left(\frac{\mathrm{\$<!--\; \$\; -->}0.12}{1\cancel{\mathrm{k}\mathrm{W}}}\right)\approx <!--\; \approx \; -->\mathrm{\$<!--\; \$\; -->}0.06$

The refrigerator will cost $0.06 for 15 minutes.

Use 30 days in a month as a conversion factor. Convert the percentage to a decimal.

$\frac{12\mathrm{k}\mathrm{W}}{\cancel{\mathrm{d}\mathrm{a}\mathrm{y}}}\left(\frac{30\cancel{\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{s}}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}\right)\left(0.80\right)=288\frac{\mathrm{k}\mathrm{W}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}$

You will need a 288-kW solar system.

Make a conversion factor out of the fact that one pill is 8.5 mg of the drug.

$25.5\cancel{\mathrm{m}\mathrm{g}}\left(\frac{1\mathrm{p}\mathrm{i}\mathrm{l}\mathrm{l}}{4.25\cancel{\mathrm{m}\mathrm{g}}}\right)=6\mathrm{p}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{s}$

The patient would need 6 pills.

Find the total amount of medicine needed per day.

$60\cancel{\mathrm{k}\mathrm{g}}\left(\frac{1.4\mathrm{m}\mathrm{g}}{1\cancel{\mathrm{k}\mathrm{g}}}\right)=84\mathrm{m}\mathrm{g}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{c}\mathrm{i}\mathrm{n}\mathrm{e}$

Divide this amount into 2 doses.

$\frac{84\mathrm{m}\mathrm{g}}{2}=42\mathrm{m}\mathrm{g}$

The patient needs 42 mg per dose.

Find the total amount of medicine needed per day.

$52\cancel{\mathrm{k}\mathrm{g}}\left(\frac{2.5\mathrm{m}\mathrm{g}}{1\cancel{\mathrm{k}\mathrm{g}}}\right)=130\mathrm{m}\mathrm{g}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{c}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{d}\mathrm{a}\mathrm{y}$

Divide this amount into 2 doses.

$\frac{130\mathrm{m}\mathrm{g}}{2}=65\mathrm{m}\mathrm{g}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{c}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{d}\mathrm{o}\mathrm{s}\mathrm{e}$

Find how many milliliters of the drug to administer.

$65\cancel{\mathrm{m}\mathrm{g}}\left(\frac{1\mathrm{m}\mathrm{L}}{5\cancel{\mathrm{m}\mathrm{g}}}\right)=13\mathrm{m}\mathrm{L}$

Administer 13 mL of the medicine per dose.

The path length is the distance between two people on the shortest path. Count the number of steps it takes to get from one person to another. When you get to 3, you have found a name with a path length of three.

Uma – 1(Vega) – 2(Lev) – 3(Jeffrey). Jeffrey has a path length of three from Uma.

Uma – 1(Vega) – 2(Lev) – 3(Yara)

Yara has a path length of three from Uma.

Uma – 1(Kalani) – 2(Rohan) – 3(Jeffrey)

This is Jeffrey again, so do not list him again.

No one else has a path length of three from Uma, so only Yara and Jeffrey have a path length of three from Uma.

It will help you to remember a few key benchmarks.

A private conversation is around 50 dB. An outboard motor is around 100dB. Firecrackers are around 150 dB.

Birds chirping are around 40dB.

It will help you to remember a few key benchmarks.

A private conversation is around 50 dB. An outboard motor is around 100dB. Firecrackers are around 150 dB.

A window air conditioner is around 60 dB.

It will help you to remember a few key benchmarks.

A private conversation is around 50 dB. An outboard motor is around 100dB. Firecrackers are around 150 dB.

A garbage disposal is around 80 dB.

It will help you to remember a few key benchmarks.

A private conversation is around 50 dB. An outboard motor is around 100dB. Firecrackers are around 150 dB.

A live outdoor concert is around 100 dB.

One half-step up or down is one key up or down. Black and white keys count the same.

One key up from F# is G.

One key down from F# is F.

One half-step up or down is one key up or down. Black and white keys count the same.

One key up from B is C.

One key down from B is B♭. (Alternative answer to down: A#)

One half-step up or down is one key up or down. Black and white keys count the same.

One key up from G♭ is G.

One key down from G♭ is F.

One whole step up or down is two keys up or down. Black and white keys count the same.

Two keys up from D♭ is E♭. (Alternative answer to up: D#)

Two keys down from D♭ is B.

One whole step up or down is two keys up or down. Black and white keys count the same.

Two keys up from C# is D#. (Alternative answer to up: E♭)

Two keys down from C# is B.

One whole step up or down is two keys up or down. Black and white keys count the same.

Two keys up from E is F#. (Alternative answer: G♭)

Two keys down from E is D.

Begin with one note and find the next eleven consecutive half-notes to form an octave.

Start from G and count one key up each time until you have 12 total notes.

G, G#, A, A#, B, C, C#, D, D#, E, F, F#

The frequency of the same note one octave higher is doubled. Since this is the same frequency two octaves lower, divide in two twice.

$\frac{330}{2}=165$ Hz

$\frac{165}{2}=82.5\approx <!--\; \approx \; -->83$ Hz

The approximate frequency is 83 Hz.

The number of games in a single round-robin tournament with n teams is $\frac{n\left(n-<!--\; -\; -->1\right)}{2}$.

n: 5 teams

$\frac{n\left(n-<!--\; -\; -->1\right)}{2}=\frac{5\left(5-<!--\; -\; -->1\right)}{2}=\frac{5\left(4\right)}{2}=\frac{20}{2}=10$

There are 10 games.

The number of games in a single round-robin tournament with n teams is $\frac{n\left(n-<!--\; -\; -->1\right)}{2}$.

n: 12 teams

$\frac{n\left(n-<!--\; -\; -->1\right)}{2}=\frac{12\left(12-<!--\; -\; -->1\right)}{2}=\frac{12\left(11\right)}{2}=\frac{132}{2}=66$

There are 66 games.

The number of games in a single round-robin tournament with n teams is $\frac{n\left(n-<!--\; -\; -->1\right)}{2}$.

n: 25 teams

$\frac{n\left(n-<!--\; -\; -->1\right)}{2}=\frac{25\left(25-<!--\; -\; -->1\right)}{2}=\frac{25\left(24\right)}{2}=\frac{600}{2}=300$

There are 300 games.

The number of games in a single-elimination tournament with n teams is n – 1.

n: 5 teams

Number of games: n – 1 = 5 – 1 = 4

There are 4 games.

The number of games in a single-elimination tournament with n teams is n – 1.

n: 12 teams

Number of games: n – 1 = 12 – 1 = 11

There are 11 games.

The number of games in a single-elimination tournament with n teams is n – 1.

n: 25 teams

Number of games: n – 1 = 25 – 1 = 24

There are 24 games.

In the Fibonacci sequence, each number is the sum of the two previous numbers.

1, 1, 2, 3, 5, 8, 13, 21, 34, 55

13 + 21 = 34

21 + 34 = 55

Find how much water is used by a five-minute shower that uses 2 gallons per minute.

$5\cancel{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}}\left(\frac{2\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\cancel{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}}}\right)=10\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}$

Find how much water is used by a shower a week for a month.

$\frac{10\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\cancel{\mathrm{s}\mathrm{h}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}}}\left(\frac{1\cancel{\mathrm{s}\mathrm{h}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}}}{\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}}}\right)\left(\frac{4\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}\mathrm{s}}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}\right)=\frac{40\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}$

Find how many gallons used by a bath a week for a month.

$\frac{35\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\cancel{\mathrm{b}\mathrm{a}\mathrm{t}\mathrm{h}}}\left(\frac{1\cancel{\mathrm{b}\mathrm{a}\mathrm{t}\mathrm{h}}}{\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}}}\right)\left(\frac{4\cancel{\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{k}\mathrm{s}}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}\right)=\frac{140\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}}{\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{h}}$

Find the difference.

140 – 40 = 100

You save 100 gallons a month by using the water-saving showerhead and taking a five-minute shower once a week.