Ratios, Rates, and Percents: Chapter Test

The greatest common factor of $52$ and $39$ is $13$, and it must be divided out of both parts together.

$52 : 39 = (52 \div 13) : (39 \div 13) = 4 : 3$

Since $4$ and $3$ share no factor above $1$, this is lowest terms.

The whole is the sum of the parts, so add the two counts and put each count over that total.

$52 + 39 = 91, \qquad \frac{52}{91} = \frac{4}{7}, \qquad \frac{39}{91} = \frac{3}{7}$

The two fractions add to $1$, which is the check that between them they account for the whole orchestra.

What the ratio fixes. Scaling both parts by one factor leaves a ratio unchanged, so $4 : 3$, $8 : 6$ and $52 : 39$ are all the same comparison. The reduced form therefore records only the relationship: four string players for every three wind players. It cannot record the size of the orchestra, because every scaled version of it says the same thing.

$52 : 39 = 4 : 3, \qquad 52 \ne 4, \qquad 39 \ne 3$

Where the $7$ comes from. A part-to-whole fraction needs the whole underneath, and the whole is the sum of the parts, $4 + 3 = 7$. So the $7$ counts a complete group of players, not an entry on the register; the register's own version of the same fraction is $\frac{52}{91}$, and $91 = 13 \times 7$.

The whole is the run of $3400$, so the faulty count goes on top and the run underneath. Setting that ratio equal to a count out of $100$ turns it into a percent.

$\frac{884}{3400} = \frac{p}{100} \quad\Longrightarrow\quad p = \frac{884 \times 100}{3400} = 26$

Dividing gives the same thing: $884 \div 3400 = 0.26 = 26\%$.

The percent sign is an instruction to divide by $100$, which both writes the fraction and moves the point two places left.

$26\% = \frac{26}{100} = \frac{13}{50} = 0.26$

The rest of the run is what is left of the whole, $100\% - 26\% = 74\%$, and the same two conversions apply to it.

$74\% = \frac{74}{100} = \frac{37}{50} = 0.74$

The two decimals add to $1$, and the two fractions add to $\frac{50}{50}$, which is the check that between them they cover the run.

What a percent settles. A percent is a ratio with its second term fixed at $100$, so it says how the faulty leaflets compare with the run, and nothing about how big the run is. Applying it to another run means multiplying that run's own size by it.

$0.26 \times 3400 = 884, \qquad 0.26 \times 5000 = 1300$

The supervisor is right that the percent transfers and wrong that it delivers a count on its own: the second run's size has to be supplied.

Why three forms but only two kinds of information. The percent, the fraction and the decimal are the one ratio written three ways, so each of them scales to any run. The $884$ is not a ratio at all, it is a count, and a count is already attached to the run it was taken from.

"Per tile" puts the tile count underneath, so divide each price by the number of tiles in its pack.

$57.60 \div 24 = 2.40, \qquad 81.20 \div 35 = 2.32$

Both figures are now dollars for a single tile, so they sit on the same scale.

A price per tile scales to any number of tiles by multiplication.

$420 \times 2.32 = 974.40, \qquad 420 \times 2.40 = 1008.00$

The difference between the two totals is the saving.

$1008.00 - 974.40 = 33.60$

Why the buyer's observation cannot discriminate. In these two packs the dearer ticket happens to belong to the bigger pack, so "more money buys more tiles" is true here whatever the large pack costs, as long as it is the dearer of the two. Suppose it cost $105$ dollars for its $35$ tiles: the bigger outlay still buys more tiles, yet the price per tile would then be $3.00$ dollars against $2.40$.

$105 \div 35 = 3.00 > 2.40$

The observation would have reported the same thing in both cases, so it separates nothing. It is not a general law either: at $40$ dollars for $35$ tiles the larger outlay would buy fewer tiles.

What the part A test does. Dividing by the tile count reduces both packs to the same single tile, which is what makes them comparable at all, and it is capable of ranking them either way. The reciprocal figure, tiles per dollar, compares them just as well, with the larger rate winning instead of the smaller.

When a ticket price would be enough. Only if both packs held the same number of tiles, since the tile counts would then divide out and the cheaper ticket would be the cheaper tile.

Put width on top and height underneath on both sides, with the blank in the height position.

$\frac{15}{9} = \frac{35}{h} \quad\Longrightarrow\quad h = \frac{9 \times 35}{15} = \frac{315}{15} = 21$

The known diagonal is $9$ and $35$, and the number diagonally opposite the blank is $15$.

The blank is a width this time, so it sits on top, and the number diagonally opposite it is the $9$.

$\frac{15}{9} = \frac{w}{24} \quad\Longrightarrow\quad w = \frac{15 \times 24}{9} = \frac{360}{9} = 40$

Reducing both ratios of the finished line confirms it, since $15 : 9$ divides by $3$ and $40 : 24$ divides by $8$.

$15 : 9 = 5 : 3, \qquad 40 : 24 = 5 : 3$

What went in where. The left side is width over height. The $24$ is a height, so putting it on top makes the right side height over width, and the two sides are built in opposite orders.

$\frac{15 \text{ wide}}{9 \text{ down}} = \frac{24 \text{ down}}{w} \quad\Longrightarrow\quad w = \frac{9 \times 24}{15} = 14.4$

What it does answer. The $14.4$ is the height that belongs with a width of $24$ centimetres, which is a genuine enlargement, just not the one asked for. A photograph $24$ centimetres across and $14.4$ down is in the same $5 : 3$ shape.

Why it looks fine. Cross-multiplying is arithmetic on four numbers; it has no way to know that one of them is a height. The wrongness lives entirely in the setup, which is why the size of the answer is the only warning available: an enlargement asked to be $24$ centimetres down cannot be $14.4$ centimetres across, since the original is wider than it is tall.

A rise keeps the whole original and adds the extra on top, so the multiplier is $1 + \frac{45}{100}$.

$1.45 \times 28 = 40.60$

The new fare is $40.60$ dollars, which is $145\%$ of the old one.

The vehicle fare's multiplier is $1.06$, and each amount added is the new fare less the old one.

$1.06 \times 260 = 275.60$

$40.60 - 28 = 12.60, \qquad 275.60 - 260 = 15.60$

So the smaller percent has added the larger number of dollars.

What each traveller measures. The percents weigh each rise against its own starting fare; the dollar amounts weigh the rises against each other.

$\frac{12.60}{28} = 45\%, \qquad \frac{15.60}{260} = 6\%$

Which comparison the percent makes. A percent change divides by the original, so it answers how big a change is relative to the thing it happened to. That is why $45\%$ and $6\%$ can be set beside each other at all, while $12.60$ dollars and $15.60$ dollars come from fares nearly ten times apart.

What neither settles alone. A percent alone never says what anyone pays: $6\%$ of $260$ dollars outweighs $45\%$ of $28$ dollars in cash. A dollar amount alone never says how heavy the rise was for that ticket. The pair together does both.

The greatest common factor of $18$ and $300$ is $6$, so divide both parts by it.

$18 : 300 = (18 \div 6) : (300 \div 6) = 3 : 50$

The finished jug is the sum of the two, so the part-to-whole fraction has that sum underneath.

$\frac{18}{18 + 300} = \frac{18}{318} = \frac{3}{53}$

Keep concentrate on top and water underneath on both sides, with the blank in the concentrate position.

$\frac{3}{50} = \frac{c}{750} \quad\Longrightarrow\quad c = \frac{3 \times 750}{50} = 45$

The finished jug is both amounts together.

$45 + 750 = 795$

Make the shared quantity agree. Concentrate appears in both ratios, as a $3$ in one and a $2$ in the other. The least common multiple of $3$ and $2$ is $6$, so scale each ratio, both parts together, until the concentrate reads $6$.

$3 : 50 = 6 : 100, \qquad 2 : 5 = 6 : 15$

Read off the new comparison. Now one description covers all three quantities: $6$ millilitres of concentrate goes with $15$ of syrup base and $100$ of water, so syrup base to water is

$15 : 100 = 3 : 20.$

Why the matching was needed. A ratio fixes a relationship, not an amount, so its numbers only mean anything relative to each other. In $3 : 50$ one part of concentrate is a $3$; in $2 : 5$ it is a $2$. Putting $50$ beside $5$ before those agree would compare a quantity measured in one size of part against a quantity measured in another, which is why the shared term has to be brought to a common value first.

The whole is the number of trains that ran, and the percent is taken of it.

$0.87 \times 1200 = 1044$

A check by pieces: $10\%$ of $1200$ is $120$, so $80\%$ is $960$, and $7\%$ is $84$, giving $1044$.

Here the part and the whole are both known and the percent is the blank.

$\frac{744}{960} = \frac{p}{100} \quad\Longrightarrow\quad p = \frac{744 \times 100}{960} = 77.5$

Dividing agrees: $744 \div 960 = 0.775 = 77.5\%$. A percent is free to carry a decimal.

The third line. The part and the percent are given and the whole is the blank.

$\frac{741}{w} = \frac{78}{100} \quad\Longrightarrow\quad w = \frac{741 \times 100}{78} = 950$

The recovered whole is larger than the part, as it must be for a share below $100\%$. Checking forward: $0.78 \times 950 = 741$.

One relationship, three calculations. Every line is the same statement with a different blank.

$\frac{\text{on time}}{\text{trains run}} = \frac{p}{100}$

The coast line had the part missing, the valley line the percent, the moorland line the whole. Where the blank sits decides whether the finish is a multiplication or a division, but the relationship is one thing, not three rules.

Which number is the whole. The whole is whatever the percent is a percent OF, which here is always the trains run: each report names its percent of the trains that line ran, never of the trains that were punctual. The on-time count is a part of that, which is also why every one of these percents should come out at or below $100\%$.

"Per kilogram" puts the kilograms underneath, so divide the price by the weight, then scale back up.

$19.20 \div 8 = 2.40, \qquad 22 \times 2.40 = 52.80$

The charge for one kilogram is the bridge: divide to reach it, multiply to leave it.

A rise of $12.5\%$ multiplies by $1.125$, and it applies to the charge for one kilogram.

$2.40 \times 1.125 = 2.70, \qquad 22 \times 2.70 = 59.40$

The extra is the difference between this year's price and last year's.

$59.40 - 52.80 = 6.60$

Write both routes as products. Raising the charge and then scaling gives

$22 \times (2.40 \times 1.125) = 22 \times 2.70 = 59.40,$

while scaling and then raising gives

$(22 \times 2.40) \times 1.125 = 52.80 \times 1.125 = 59.40.$

Why the agreement is guaranteed. The two lines contain the very same three factors, $22$, $2.40$ and $1.125$; only the bracketing differs, and multiplication gives the same product however its factors are grouped. Matching answers here are therefore not a coincidence to be checked case by case.

How far it reaches. Nothing in the argument used the particular numbers, so any weight with any percent rise behaves the same way. What would break it is a change that is not a multiplication, such as a flat surcharge added per parcel: adding $3$ dollars to the parcel is not the same as adding $3$ dollars to every kilogram of it.