Proportions: Free Response

Take each bucket on its own and divide both parts of its ratio by everything the two numbers share.

For the first bucket, $16 = 2 \times 2 \times 2 \times 2$ and $44 = 2 \times 2 \times 11$, so they share two copies of $2$ and $\operatorname{GCF}(16, 44) = 4$:

$\frac{16}{44} = \frac{16 \div 4}{44 \div 4} = \frac{4}{11}.$

For the second, $28 = 2 \times 2 \times 7$ and $77 = 7 \times 11$, so the only prime they share is $7$ and $\operatorname{GCF}(28, 77) = 7$:

$\frac{28}{77} = \frac{28 \div 7}{77 \div 7} = \frac{4}{11}.$

Both ratios reduce to $\frac{4}{11}$, which reads as $4$ grams of powder for every $11$ milliliters of water. Reducing a ratio never changes the comparison it makes, so each bucket makes that same comparison and

$\frac{16}{44} = \frac{28}{77}$

is a true proportion. The two buckets hold different amounts of glaze; what they share is the mix.

Test the helper's bucket exactly as any pair is tested. Set the first bucket's ratio beside it, powder over water on both sides, and multiply along each diagonal:

$16 \times 52 = 832, \qquad 44 \times 24 = 1056.$

The two products are different, so the two ratios are not equal and the helper's bucket is not the studio's glaze.

The fault is in the move itself. A ratio is held by scaling, that is by multiplying both parts by one and the same number, and the helper added the same amount to both parts instead. Adding treats the two numbers as though the gap between them were what mattered, when what matters is how many times one goes into the other.

Now find the water that does keep the mix. Write the studio's ratio against a bucket with $24$ grams of powder, letting $w$ stand for the milliliters of water:

$\frac{16 \text{ g}}{44 \text{ mL}} = \frac{24 \text{ g}}{w \text{ mL}}.$

The diagonal that is fully known is $44$ and $24$, so multiply those two:

$44 \times 24 = 1056.$

That is the value of the other diagonal product, $16 \times w$, so divide by the $16$ facing the blank:

$w = 1056 \div 16 = 66.$

The bucket needs $66$ milliliters of water. Check it by reducing: $\frac{24}{66}$ divides by $6$ to give $\frac{4}{11}$, the studio's mix. The helper's $52$ milliliters was $14$ short of it, so the glaze would have come out thick.

The test has to be earned from something already trusted, and the trustworthy fact is about fractions with a shared denominator: two of them are equal exactly when the numbers on top are equal.

So give the two ratios a shared denominator. Any pair of ratios, written $\frac{a}{b}$ and $\frac{c}{d}$, can both be written over $b \times d$, by multiplying the first above and below by $d$ and the second above and below by $b$:

$\frac{a}{b} = \frac{a \times d}{b \times d}, \qquad \frac{c}{d} = \frac{c \times b}{d \times b}.$

Neither rewriting moved a value, since each multiplied a top and a bottom by the same number, and $d \times b$ is the same product as $b \times d$. The two now sit over one denominator, so they are equal exactly when $a \times d$ and $b \times c$ are equal, and those are precisely the diagonal products.

On this question's numbers, the shared denominator is $44 \times 52$ and the two numerators are

$16 \times 52 = 832, \qquad 24 \times 44 = 1056,$

so the studio's ratio and the helper's become $\frac{832}{2288}$ and $\frac{1056}{2288}$, which count different numbers of the same size of piece and therefore cannot be equal.

The same reasoning judges the helper's move. Add some amount $k$ to both parts of $\frac{a}{b}$ and compare the result with the original: the diagonal products are $a \times (b + k)$ and $b \times (a + k)$, which the distributive property opens out as

$a \times b + a \times k \qquad \text{and} \qquad a \times b + b \times k.$

They agree only when $a \times k$ and $b \times k$ agree, that is only when nothing was added or when the two parts were equal to begin with. A one-to-one mix survives the addition; the studio's $16$ to $44$ does not, which is why the helper's bucket came out wrong.

Write the shop's ratio against the batch, copper over zinc on both sides, with $c$ for the kilograms of copper:

$\frac{5 \text{ copper}}{2 \text{ zinc}} = \frac{c \text{ copper}}{26 \text{ zinc}}.$

The pair fully known is the zinc, $2$ on one side and $26$ on the other, so read the factor off it:

$26 \div 2 = 13.$

A proportion stays true when both parts of a ratio are multiplied by the same number, so the copper takes that same factor of $13$:

$c = 5 \times 13 = 65.$

The batch takes $65$ kilograms of copper. Check by reducing the finished pair: $\frac{65}{26}$ divides by $13$ to give $\frac{5}{2}$, which is the shop's ratio.

Set the batch against the shop's ratio again, copper over zinc on both sides, with $z$ for the kilograms of zinc:

$\frac{5}{2} = \frac{33}{z}.$

Scaling is awkward here because the copper pair does not scale by a whole number: $33 \div 5 = 6.6$, so the neat factor of part A is not available.

Cross-multiplying does not care. The diagonal that is fully known is $2$ and $33$, so multiply them:

$2 \times 33 = 66.$

In a true proportion the two diagonal products are equal, so $5 \times z$ is $66$, and dividing by the $5$ facing the blank gives

$z = 66 \div 5 = 13.2.$

The batch takes $13.2$ kilograms of zinc. There is no reason to round it: a weight of metal need not be a whole number of kilograms, and rounding to $13$ would put the batch off the shop's ratio. Check it by the diagonals: $5 \times 13.2 = 66$ and $2 \times 33 = 66$ agree.

Put the bar's own ratio beside the shop's, copper over zinc in both, and multiply along the diagonals:

$5 \times 20 = 100, \qquad 2 \times 45 = 90.$

The products are not equal, so $\frac{5}{2}$ and $\frac{45}{20}$ are not the same ratio and the bar was not mixed to the shop's brass.

How far off is it? That has two honest answers, depending on which of the bar's two weights is treated as correct.

Taking the copper as correct, the shop's ratio at $45$ kilograms of copper calls for

$2 \times 45 = 90, \qquad 90 \div 5 = 18$

kilograms of zinc. The bar holds $20$, so it carries $2$ kilograms of zinc too many.

Taking the zinc as correct, the ratio at $20$ kilograms of zinc calls for

$5 \times 20 = 100, \qquad 100 \div 2 = 50$

kilograms of copper. The bar holds $45$, so it is $5$ kilograms of copper short.

Both descriptions are of the same single bar, and they suggest different repairs: melt in $5$ kilograms more copper, or start again with $2$ kilograms less zinc. Which one the shop wants depends on what it has to hand, but the arithmetic is the same proportion read from either end.

There are two kinds of quantity here, a count of coins and a weight in grams, and two situations, the reference tray and the sack. A proportion says the two situations compare those quantities in the same way.

Name the unknown first, so the line means something: let $n$ be the number of coins in the sack. Then write coins over grams on the left for the tray and coins over grams on the right for the sack:

$\frac{24 \text{ coins}}{90 \text{ grams}} = \frac{n \text{ coins}}{525 \text{ grams}}.$

The order is the whole of the work. Counts sit on top on both sides and weights sit underneath on both sides, so each diagonal pairs a count with a weight. Writing coins over grams on one side and grams over coins on the other would set down a comparison nobody intended, and the number that came out would answer a different question.

Writing it the other way up throughout, $\frac{90}{24} = \frac{525}{n}$, is just as good, because both sides still list the same kind of quantity in the same place. What is not allowed is mixing the two orders within one line, coins over grams on one side and grams over coins on the other.

There is a third true line hiding in these numbers, $\frac{24}{n} = \frac{90}{525}$, which compares the two situations instead of the two quantities: the tray's coins against the sack's coins, and the tray's grams against the sack's grams. It is a genuine proportion and it leads to the same count. It is not what this part asks for, though, because its sides no longer hold matching kinds of quantity in matching positions.

The diagonal fully known is $24$ and $525$, so multiply those two:

$24 \times 525 = 12600.$

In a true proportion the other diagonal product matches it, and that product is $90 \times n$. Divide by the $90$ facing the blank:

$n = 12600 \div 90 = 140.$

The sack holds $140$ coins.

Now check the finished pair, which is the habit worth keeping: if $\frac{24}{90} = \frac{140}{525}$ is true, both sides reduce to the same lowest terms. On the left, $\operatorname{GCF}(24, 90) = 6$; on the right, $\operatorname{GCF}(140, 525) = 35$:

$\frac{24}{90} = \frac{4}{15}, \qquad \frac{140}{525} = \frac{4}{15}.$

They agree, so the count is consistent with the reference tray. The shared form says the same thing in the smallest whole numbers available: $4$ of these coins weigh $15$ grams.

A proportion between a count and a weight says the two grow together in a fixed comparison: double the count and the weight doubles with it. The sack's own weight breaks that, because it is there whether or not any coins are.

See it on the numbers. The $675$ gram sack carries $675 - 30 = 645$ grams of coins. A sack with twice as many coins would carry $1290$ grams of coins, so it would weigh

$1290 + 30 = 1320 \text{ grams in total,}$

while twice $675$ is $1350$. Twice the coins does not give twice the total, so total weight and coin count are not in proportion, and the empty end settles it just as plainly: a sack with no coins in it still tips the scale at $30$ grams, whereas no coins weigh nothing at all.

What is in proportion is the weight of the coins themselves, which is the quantity the reference tray was measured on. So subtract the sack first and use the proportion on what is left:

$\frac{24 \text{ coins}}{90 \text{ grams}} = \frac{n \text{ coins}}{645 \text{ grams}}.$

Multiply the known diagonal and divide by the number facing the blank:

$24 \times 645 = 15480, \qquad n = 15480 \div 90 = 172.$

The sack holds $172$ coins. The check works as before: $\frac{172}{645}$ divides by $43$ to give $\frac{4}{15}$, the same lowest terms as the reference tray.

Write the two bouquets as ratios of roses to lilies and multiply along each diagonal. The top left times the bottom right is

$27 \times 55 = 1485,$

and the bottom left times the top right is

$45 \times 33 = 1485.$

The two products are equal, so $\frac{27}{45} = \frac{33}{55}$ is a true proportion and the bouquets compare their flowers in the same way.

Reducing agrees, and it names the comparison: $\frac{27}{45}$ divides by $9$ and $\frac{33}{55}$ divides by $11$, and both land on $\frac{3}{5}$. The shop is really working to $3$ roses for every $5$ lilies, and both bouquets obey it.

This line compares the same flower across the two bouquets, so test it the same way. Multiplying along its diagonals:

$27 \times 55 = 1485, \qquad 33 \times 45 = 1485.$

Equal, so $\frac{27}{33} = \frac{45}{55}$ is true as well. Both sides reduce to $\frac{9}{11}$, since $\operatorname{GCF}(27, 33) = 3$ and $\operatorname{GCF}(45, 55) = 5$.

Read that back at the flowers. The roses stand in the ratio $9$ to $11$ between the small bouquet and the large, and so do the lilies. One and the same scaling carried both counts from the small bouquet to the large, which is exactly what it means for the large to be the small one built at a bigger size. Had the two ratios differed, the shop would have grown one flower more than the other and produced a different bouquet, not a bigger one.

There is a reason this line came out true rather than by luck. Its diagonal products are $27 \times 55$ and $33 \times 45$: the same two products as in part A, merely sitting on different diagonals. Any line that keeps those numbers paired the way part A paired them makes the same demand, and part A already showed that demand is met.

One test settles the helper's claim, because a claim that any pairing works is beaten by a single pairing that does not. Multiply along the diagonals of the offered line:

$27 \times 45 = 1215, \qquad 55 \times 33 = 1815.$

Those are different numbers, so $\frac{27}{55} = \frac{33}{45}$ is false and the four numbers may not be paired up any way at all.

What separates it from the two lines that did hold is which numbers meet on a diagonal. In part A the diagonals gave $27 \times 55$ and $45 \times 33$. In part B they gave $27 \times 55$ and $33 \times 45$, the very same two products. The helper's line instead pairs $27$ with $45$ and $55$ with $33$, so it asks a question about two products nobody has shown to be equal, and they are not.

There is one further way of pairing the four numbers, and it is worth checking rather than assuming, because checking it is what makes the rule an exact one. Pairing $27$ with $33$ and $45$ with $55$, which is what the line $\frac{27}{45} = \frac{55}{33}$ does, gives

$27 \times 33 = 891, \qquad 45 \times 55 = 2475,$

and those disagree as well. Four numbers can be split into two pairs in exactly three ways, so with two of the three ruled out, the pairing part A began with is the only one that survives.

So the rule is about the pairing, not about the arrangement on the page. Turning both sides upside down, for instance, gives $\frac{45}{27} = \frac{55}{33}$, whose diagonals are $45 \times 33$ and $27 \times 55$: the original pair again, so that line is true as well. Any rearrangement that leaves $27$ facing $55$ and $45$ facing $33$ across the diagonals survives, and any rearrangement that breaks those pairings is asking something new.

The flowers make the same point in words. The helper's line compares roses in the small bouquet against lilies in the large, which mixes the two kinds of comparison the earlier lines kept apart.

Write the market's trade on the left and the farmer's trade on the right, grain over salt in both, with $s$ for the kilograms of salt:

$\frac{8 \text{ grain}}{5 \text{ salt}} = \frac{56 \text{ grain}}{s \text{ salt}}.$

The diagonal fully known is $5$ and $56$, so multiply those:

$5 \times 56 = 280.$

That is the value of the other diagonal product, $8 \times s$, so divide by the $8$ facing the blank:

$s = 280 \div 8 = 35.$

The farmer receives $35$ kilograms of salt. Scaling gives the same thing here, since the grain pair happens to be friendly: $56 \div 8 = 7$, and $5 \times 7 = 35$.

The second clerk writes salt over grain in both places:

$\frac{5 \text{ salt}}{8 \text{ grain}} = \frac{s \text{ salt}}{56 \text{ grain}}.$

The diagonal fully known is $5$ and $56$ once more, so

$5 \times 56 = 280, \qquad s = 280 \div 8 = 35.$

The same $35$ kilograms, from the same two multiplications. That is not a coincidence: both clerks' lines set $8 \times s$ against $5 \times 56$, because turning both sides of a proportion upside down leaves the same two numbers facing each other across the diagonals. A proportion is a statement that two comparisons agree, and swapping which quantity you count first does not change whether they agree.

A third clerk who wrote grain over salt on the left and salt over grain on the right would have something else entirely:

$\frac{8 \text{ grain}}{5 \text{ salt}} = \frac{s \text{ salt}}{56 \text{ grain}}.$

Here one diagonal is $8 \times 56$, grain against grain, and the other is $5 \times s$, salt against salt, so no diagonal compares the two goods at all. Following it through gives

$8 \times 56 = 448, \qquad s = 448 \div 5 = 89.6,$

which claims the farmer walks away with more salt than the grain he brought, on a market where salt is the scarcer good. The order rule is not bookkeeping fussiness: it is what keeps each diagonal comparing one good against the other.

Settle the case in front of you first. For $36$ kilograms of grain, the known diagonal is $5$ and $36$:

$5 \times 36 = 180, \qquad 180 \div 8 = 22.5.$

That trade calls for $22.5$ kilograms of salt, which is not a whole number, so the market master will not settle it.

Now the general question, which the method itself answers. Whatever weight of grain is brought, the salt comes out of the same two steps: multiply the grain by $5$, then divide by $8$. So the trade settles exactly when $8$ divides the product of $5$ and the grain.

Look at what those two numbers are made of. Since $8 = 2 \times 2 \times 2$, the division needs three factors of $2$, and $5$ is prime and brings none of them. Every one of them must therefore come from the grain itself, which is to say that $8$ has to divide the grain.

Both directions hold, which is what makes the description exact rather than merely necessary. If the grain is $8$ times some whole number, the salt is $5$ times that same whole number, so the trade settles; and if the grain is not a multiple of $8$, the product of $5$ and the grain is not one either, so the division leaves a remainder. The family is exactly

$8, \; 16, \; 24, \; 32, \; 40, \; \ldots$

and $36$ is not among them, which is another way of seeing the first part. Nothing is wrong with the arithmetic of a $36$ kilogram trade; what rules it out is the market master's insistence on whole kilograms.