Ratios: Free Response

Lowest terms means dividing both parts by their greatest common factor. Break the two counts down: $42 = 2 \times 3 \times 7$ and $56 = 2 \times 2 \times 2 \times 7$, so the largest factor they share is $2 \times 7 = 14$.

Divide both parts by it, never one of them alone:

$42 : 56 = (42 \div 14) : (56 \div 14) = 3 : 4.$

The parts $3$ and $4$ share no factor beyond $1$, so this is as far as it reduces.

For the second ratio nothing about the shelf has changed, only which kind of book is named first. The same factor of $14$ comes out:

$56 : 42 = (56 \div 14) : (42 \div 14) = 4 : 3.$

So the shelf is $3 : 4$ hardcover to paperback and $4 : 3$ paperback to hardcover. Both statements are true of it, and neither may be written in the other's place: the pair of numbers carries no meaning until the order of the two kinds is fixed.

Nothing else stands on the shelf, so the whole is the sum of the two parts:

$42 + 56 = 98 \text{ books.}$

Now compare one kind with that whole. The greatest common factor of $42$ and $98$ is $14$, since $98 = 2 \times 7 \times 7$:

$42 : 98 = (42 \div 14) : (98 \div 14) = 3 : 7.$

A part-to-whole ratio written as a fraction puts the part on top and the whole underneath, so the hardcovers are $\frac{3}{7}$ of the shelf.

Watch what the denominator is doing. In part A the second number was the other kind of book; here it is the two kinds added together, and $3 + 4 = 7$ is where the $7$ comes from. The check is that the two part-to-whole fractions account for the shelf exactly once: the paperbacks are $56 : 98 = 4 : 7$, and

$\frac{3}{7} + \frac{4}{7} = 1.$

Comparing $42$ with $27$ decides nothing, because the shelves are different sizes. Put the second shelf into lowest terms first. Here $27 = 3 \times 3 \times 3$ and $36 = 2 \times 2 \times 3 \times 3$, so the greatest common factor is $9$:

$27 : 36 = (27 \div 9) : (36 \div 9) = 3 : 4.$

The first shelf reduced to $3 : 4$ as well. Two ratios are equivalent exactly when they reduce to the same lowest terms, so the two shelves do compare their books in the same way.

Reading the same fact upwards makes it concrete. Start from $3 : 4$ and scale both parts by $14$ and you get the first shelf; scale both parts by $9$ and you get the second:

$3 : 4 = 42 : 56, \qquad 3 : 4 = 27 : 36.$

Each shelf is a batch of the same pattern, four paperbacks alongside every three hardcovers, taken $14$ times over on one shelf and $9$ times over on the other. The totals differ, $98$ books against $27 + 36 = 63$, and that difference is invisible to the ratio, because a ratio was never a record of how much there is.

The step from $42 : 56$ to $3 : 4$ divided both counts by $14$, and dividing both parts by the same number leaves the comparison untouched. So the comparison is exactly what survives: on this shelf there are four paperbacks alongside every three hardcovers.

What does not survive is the size. The two shelves in this question reduce to the same thing:

$42 : 56 = 3 : 4, \qquad 27 : 36 = 3 : 4,$

yet one holds $98$ books and the other $63$. So $3 : 4$ cannot tell them apart, and it makes no claim that any shelf holds three hardcovers. It says how the two kinds compare and stops there. To pin the counts down you need one more piece of information, either a total or one of the two counts, and then the scale factor recovers the rest.

Order is the other thing the notation carries, and it is carried entirely by the naming. The pairs $3 : 4$ and $4 : 3$ are built from the same two numbers, and both are true of this shelf, but of different questions about it. Written down with no statement of which kind comes first, the pair says nothing: read as paperbacks to hardcovers, $3 : 4$ would describe a shelf holding more hardcover books than paperback books, and on this shelf the hardcovers are the smaller group.

The three numbers in the ratio are parts, not chairs. Count them first:

$5 + 6 + 9 = 20 \text{ parts.}$

The delivery is cut into that many equal parts, so a single part is worth

$240 \div 20 = 12 \text{ chairs.}$

Now give each hall its own number of parts, at $12$ chairs a part:

$5 \times 12 = 60, \qquad 6 \times 12 = 72, \qquad 9 \times 12 = 108.$

The check is to add the three shares:

$60 + 72 + 108 = 240,$

which is the whole delivery with nothing left over and nothing invented. It is worth doing every time, because it catches the commonest slip in a three-way split, dividing by the number of halls instead of by the number of parts. Dividing $240$ by $3$ would have given $80$ chairs each, a split in the ratio $1 : 1 : 1$ rather than the one that was agreed.

The parts are all the same size, so a hall's share of the delivery is settled by how many parts it holds out of all of them. The third hall holds $9$, and the delivery is $5 + 6 + 9$ parts in all:

$\frac{9}{5 + 6 + 9} = \frac{9}{20}.$

The denominator is the total number of parts. It is not $6$, which counts another hall's parts, and the difference between those two denominators is the difference between a part-to-part comparison and a part-to-whole one.

The same answer comes out of the counts, which is the check. The third hall takes $108$ chairs out of $240$, and the greatest common factor of $108$ and $240$ is $12$:

$108 : 240 = (108 \div 12) : (240 \div 12) = 9 : 20.$

So the fraction and the reduced part-to-whole ratio carry the same information written two ways, and $\frac{9}{20}$ of $240$ is $9 \times 12 = 108$, as it must be.

One half of the parent's statement is right and the other half is a different kind of comparison wearing the same numbers.

Sound: the first two halls really do stand in the ratio $5 : 6$. Every hall's share is its own number of parts, and all the parts are the same size, so the first two compare as $5$ parts against $6$ parts. Ignoring the third hall does not disturb that, because dropping a hall changes what the whole is without changing how any two shares compare with each other.

Faulty: $5 : 6$ is part-to-part. Turning it into a fraction of what the pair receives needs the whole that the pair makes, and that whole is the sum of their parts:

$5 + 6 = 11 \text{ parts,}$

so the first hall's share of the pair's chairs is $\frac{5}{11}$, not $\frac{5}{6}$.

The error can be exposed without counting a single chair. Suppose the first hall did take $\frac{5}{6}$ of the pair's chairs. Then the second would take what is left, $\frac{1}{6}$, and the two would stand in the ratio

$5 : 1,$

which contradicts the $5 : 6$ the parent started from. A rule that destroys its own premise is wrong wherever the numbers came from.

The counts confirm it. The first two halls hold $60 + 72 = 132$ chairs, and

$\frac{5}{11} \times 132 = 5 \times 12 = 60,$

which is the first hall exactly, while $\frac{5}{6}$ of $132$ is $110$, a number no hall in this school receives.

Same method, larger delivery. The ratio still makes

$5 + 6 + 9 = 20 \text{ parts,}$

and at the second school a part is worth

$400 \div 20 = 20 \text{ chairs,}$

so the three halls receive $5 \times 20 = 100$, $6 \times 20 = 120$ and $9 \times 20 = 180$ chairs, which add to $400$.

Now put the two schools side by side. Reduce each set of counts by the factor its own school used:

$60 : 72 : 108 = 5 : 6 : 9,$

$100 : 120 : 180 = 5 : 6 : 9.$

Both return to the same three numbers, because both were built from those three numbers in the first place. That is the whole reason: multiplying every part by the same factor scales each count but leaves each comparison between counts alone, since the factor sits in both sides of every comparison and cancels out of it. The size of a delivery and the ratio it is split in are independent pieces of information, and only the second is what the ratio records.

It is worth being clear about what this does not say. The halls are not receiving equal treatment across the two schools in any other sense: the smallest hall at the second school takes $100$ chairs, more than the middle hall at the first school takes. The ratio is untroubled by that, because it never spoke about how many chairs there are.

Write each comparison as black over white:

$\text{batch A} = \frac{7}{12}, \qquad \text{batch B} = \frac{9}{16}.$

The denominators differ, so the two cannot be read against each other as they stand. A number that both $12$ and $16$ divide into is $48$, since $12 \times 4 = 48$ and $16 \times 3 = 48$. Rebuild each fraction over it, scaling top and bottom by the same factor so that the value is untouched:

$\frac{7}{12} = \frac{7 \times 4}{12 \times 4} = \frac{28}{48},$

$\frac{9}{16} = \frac{9 \times 3}{16 \times 3} = \frac{27}{48}.$

Now the two fractions are cut into pieces of one size, so the numerators decide:

$\frac{28}{48} > \frac{27}{48},$

and batch A is the darker of the two, by one part in $48$.

That margin is the point of doing the work. Batch B holds more black paint than batch A, $9$ cups against $7$, and is still the lighter mix, because it holds so much more white to go with it. A ratio is a comparison, and the size of one of its parts settles nothing on its own.

The batches hold $7$ and $9$ cups of black, and $63$ is a multiple of both, since $7 \times 9 = 63$ and $9 \times 7 = 63$. Scale each batch by whatever its own black needs, and scale the white by the same factor, or the mix would change:

$7 : 12 = (7 \times 9) : (12 \times 9) = 63 : 108,$

$9 : 16 = (9 \times 7) : (16 \times 7) = 63 : 112.$

Neither step changed a shade. Multiplying both parts of a ratio by the same number gives an equivalent ratio, so $63 : 108$ is the same grey as $7 : 12$, and $63 : 112$ is the same grey as $9 : 16$. There is simply more of each.

With the black amounts level, the white amounts decide, and now they can be read directly. Batch B spreads its $63$ cups of black through $112$ cups of white; batch A spreads the same $63$ cups through only $108$, four cups fewer:

$112 - 108 = 4.$

Less white for the same black is the darker mix, so batch A is darker, agreeing with the fractions.

The two routes are one idea. Two comparisons cannot be read against each other while both sides differ, so you level one side and let the other speak. Matching the white does it by scaling the denominators together; matching the black does it by scaling the numerators together.

The shortcut reports a gap, so test the gap on a mix whose shade is known not to have changed. Double batch A, multiplying both parts by $2$:

$7 : 12 = (7 \times 2) : (12 \times 2) = 14 : 24.$

Both parts were scaled by the same number, so $14 : 24$ is the same grey as $7 : 12$, just twice as much of it. The gaps, though, are

$12 - 7 = 5 \qquad \text{and} \qquad 24 - 14 = 10.$

One shade, two different readings. The shortcut would call $7 : 12$ the darker of the pair, and there is nothing there to be darker than.

It fails for a reason worth keeping. A ratio compares by division, and division survives scaling: $7$ compared with $12$ and $14$ compared with $24$ are the same comparison, because the factor $2$ cancels out of it. Subtraction does not survive scaling, because doubling both parts doubles their difference. Any honest test of shade has to be built from the operation the ratio itself is built from.

The shortcut can also give a plainly wrong ranking, not merely an unstable one. Compare $2 : 6$ with $9 : 14$. The gaps are $4$ and $5$, so the shortcut picks $2 : 6$ as the darker, while the fractions over a common denominator say otherwise:

$\frac{2}{6} = \frac{1}{3} = \frac{14}{42}, \qquad \frac{9}{14} = \frac{27}{42}.$

Here $9 : 14$ is very much the darker mix, and the shortcut has it backwards.

The greatest common factor of $150$ and $250$ is $50$, because $150 = 50 \times 3$ and $250 = 50 \times 5$, and $3$ and $5$ share no factor beyond $1$.

$150 : 250 = (150 \div 50) : (250 \div 50) = 3 : 5.$

The agreement is now easy to state and easy to use: out of every $8$ dollars of profit, $3$ follow the first contribution and $5$ follow the second. Note what the reduced ratio has dropped. It no longer records how much either person actually put in, only how the two contributions compare, and comparison is the only thing an agreement about sharing needs.

A profit is cut into as many equal parts as the ratio has between its two sides:

$3 + 5 = 8 \text{ parts.}$

So one part of this profit is worth

$96 \div 8 = 12 \text{ dollars,}$

and each person takes their own number of parts:

$3 \times 12 = 36 \qquad \text{and} \qquad 5 \times 12 = 60.$

The shares are $36$ dollars and $60$ dollars.

Check them twice, because the two checks catch different faults. First add: $36 + 60 = 96$, the whole profit, so nothing has been created or lost. Second, reduce the shares and compare with the agreement:

$36 : 60 = (36 \div 12) : (60 \div 12) = 3 : 5,$

which is the agreed ratio. A split can add up correctly and still be in the wrong ratio, as $48$ and $48$ would be, and it can be in the right ratio and not add up, as $30$ and $50$ would be, so neither check on its own is enough.

Whatever the profit, the first person takes $3$ of the $8$ parts, so on this day those $3$ parts are worth $81$ dollars between them. That fixes a single part:

$81 \div 3 = 27 \text{ dollars.}$

The scale factor from the agreed ratio to this day's shares is therefore $27$. Multiply both parts of the ratio by it:

$3 : 5 = (3 \times 27) : (5 \times 27) = 81 : 135,$

so the second person receives $135$ dollars. The profit is the whole, which is all eight parts:

$8 \times 27 = 216 \text{ dollars,}$

and $81 + 135 = 216$ confirms it.

Nothing from the earlier day was needed, and that is the useful part. The agreement fixes the ratio, not the profit, so each new day brings a new scale factor while $3 : 5$ stays put. Find the factor from whichever amount you are given, and the rest of the day follows from it.

The agreement fixes a ratio, and a ratio compares by division. The second person's argument is about subtraction, so ask what the difference between the shares actually is.

Out of every $8$ parts she takes $5$ and he takes $3$, so she is ahead by

$5 - 3 = 2 \text{ parts,}$

and two parts out of eight is a quarter. The gap between the shares is therefore a quarter of whatever the stall makes, which is not a fixed amount at all. On the $96$ dollar profit the shares were $36$ and $60$, a gap of $24$, and $24$ is a quarter of $96$. On the $216$ dollar profit they were $81$ and $135$, a gap of $54$, and $54$ is a quarter of $216$. Her argument needs that gap to sit at $100$ dollars every time, and it does not sit anywhere.

There is exactly one profit that gives her what she asks for. Two parts must be worth $100$ dollars, so one part is worth

$100 \div 2 = 50 \text{ dollars,}$

and the profit is all eight parts:

$8 \times 50 = 400 \text{ dollars.}$

The shares are then $3 \times 50 = 150$ and $5 \times 50 = 250$, which do differ by $100$. Those two figures are the two contributions themselves, and that is no accident: the contributions stand in the ratio $3 : 5$ and add to $400$, so splitting $400$ in that ratio has to hand each person back exactly what they put in.

The general point is the one the agreement was chosen for. An equal ratio holds the comparison between the shares fixed while the shares themselves grow and shrink, and the difference between them is free to move, and does.

Take each pair on its own. The greatest common factor of $24$ and $60$ is $12$, since $24 = 12 \times 2$ and $60 = 12 \times 5$:

$24 : 60 = (24 \div 12) : (60 \div 12) = 2 : 5.$

The greatest common factor of $18$ and $45$ is $9$, since $18 = 9 \times 2$ and $45 = 9 \times 5$:

$18 : 45 = (18 \div 9) : (45 \div 9) = 2 : 5.$

Both land on $2 : 5$. Two ratios are equivalent exactly when they reduce to the same lowest terms, so these two say the same thing.

Read upwards it is just as clear. Starting from $2 : 5$ and multiplying both parts by $12$ gives the first pair, and multiplying both parts by $9$ gives the second, so both are batches of one pattern, taken $12$ times over and $9$ times over.

Reduce the new pair first. The greatest common factor of $20$ and $52$ is $4$:

$20 : 52 = (20 \div 4) : (52 \div 4) = 5 : 13.$

That is not $2 : 5$, so the two pairs do not reduce to the same lowest terms and are not equivalent. Notice how close they are, which is exactly why the eye is no help here.

To rank them, write each reduced form as a fraction and rebuild both over one denominator. A number that both $5$ and $13$ divide into is $65$:

$\frac{2}{5} = \frac{2 \times 13}{5 \times 13} = \frac{26}{65},$

$\frac{5}{13} = \frac{5 \times 5}{13 \times 5} = \frac{25}{65}.$

Since $26 > 25$, the comparison $24 : 60$ is the stronger, by one part in $65$. Reducing before comparing was worth the trouble: run the same comparison on $24 : 60$ against $20 : 52$ untouched and the common denominator needed is $780$ rather than $65$.

A claim of the form "exactly when" makes two promises, and each has to be tested on its own.

The first promise: if one ratio is reached from another by multiplying both parts by a whole number other than zero, are the two equivalent? Yes. Writing the ratio $a : b$ as $\frac{a}{b}$ and scaling both parts by $k$ gives $\frac{ka}{kb}$, and the factor $k$ sits in both the top and the bottom, so it cancels:

$\frac{ka}{kb} = \frac{a}{b}.$

That cancelling is where the exclusion of zero earns its keep. Taking $k = 0$ would collapse both parts to $0 : 0$, leaving no factor to cancel and no comparison of any kind, which is why the scaling rule has always been stated for a nonzero number. With zero ruled out, this half of the test is sound, and it is the rule the whole idea of an equivalent ratio rests on.

The second promise: if two ratios are equivalent, must one be reachable from the other in that way? Here the test breaks, and the pairs already in front of you break it. The pairs $24 : 60$ and $18 : 45$ are equivalent, since both reduce to $2 : 5$. But

$24 \div 18 = \frac{4}{3}, \qquad 18 \div 24 = \frac{3}{4},$

and neither is a whole number, so neither pair can be reached from the other by multiplying both parts by a whole number, and excluding zero does nothing to rescue it.

The lowest terms show what is going on. Both pairs are built from $2 : 5$, one by scaling by $12$ and the other by scaling by $9$, and $12$ and $9$ are not multiples of each other. The student's test looks for a ladder running from one ratio straight to the other; equivalence asks only that the two stand on the same ladder, and they can do that at rungs neither of which is above the other.

So the verdict is that the test holds one way and fails the other, which means it is not a test of equivalence. The repair is the method used in part A: reduce both and see whether the lowest terms agree. That one does work in both directions, which is why simplifying alone was enough to settle part A.