Fractions: Chapter Test

The numerator $3$ says the $9$ tins are three equal eighths, so undo that count first to find one eighth.

$9 \div 3 = 3 \text{ tins in one eighth}$

The denominator $8$ says a full box is eight of those equal parts.

$3 \times 8 = 24 \text{ tins in a full box}$

One eighth of a box is $3$ tins, so count how many eighths the $33$ tins make.

$33 \div 3 = 11 \text{ eighths}, \qquad \frac{11}{8}$

The numerator is larger than the denominator, so the fraction is improper. Eight eighths make $1$ and sixteen eighths make $2$, and $11$ sits between $8$ and $16$, so the value lies between $1$ and $2$. The bar is a division sign, so $\frac{11}{8} = 11 \div 8$.

The denominator does two jobs at once: it counts the pieces the whole was cut into and, by doing so, fixes how big each piece is. That second job only works when the pieces are equal.

$\frac{1}{3} \text{ of a box} = \text{one of } 3 \text{ EQUAL parts of the box}$

Three layers of unequal depth are three pieces, but they are not three equal pieces, so "it is one of the three layers" is not a reason for calling it a third. The packer's argument is what fails.

That is not the same as saying the layer is not a third. Equal layers would settle the name, but they are not the only thing that could. Three layers of $\tfrac{1}{3}$, $\tfrac{1}{4}$ and $\tfrac{5}{12}$ of a box fill it exactly, no two of them are equal, and the first is still exactly one third.

$\tfrac{1}{3} + \tfrac{1}{4} + \tfrac{5}{12} = \tfrac{4}{12} + \tfrac{3}{12} + \tfrac{5}{12} = 1$

So there are two honest ways to earn the name here: repack so that all three layers hold the same amount, $8$ tins each in a box of $24$; or measure this layer and find it holds $8$ of the $24$ tins. What cannot earn it is counting the layers.

Watch out. "One of three" and "one third" are different claims. The first is a count of pieces; the second is a measurement, and a measurement needs a fixed unit.

Ask what the old denominator was multiplied by to reach the new one, then do the same on top.

$9 \times 5 = 45, \qquad \frac{4}{9} = \frac{4 \times 5}{9 \times 5} = \frac{20}{45}$

Watch out. Writing $\frac{4}{45}$ changes the denominator without paying for it on top. That is a different number, one fifth of the one asked for.

One division is enough when the number you divide by is the greatest common factor. The factors shared by $42$ and $70$ are $1$, $2$, $7$ and $14$.

$\operatorname{GCF}(42, 70) = 14, \qquad \frac{42}{70} = \frac{42 \div 14}{70 \div 14} = \frac{3}{5}$

Check that nothing is left: $3$ and $5$ share no factor above $1$, so the fraction is finished.

Settle the equality first, with cross products.

$4 \times 14 = 56 \qquad\text{against}\qquad 9 \times 9 = 81$

The two products differ, so the fractions are not equal. Rewriting both over $126$ says the same thing: $\frac{4}{9} = \frac{56}{126}$ and $\frac{9}{14} = \frac{81}{126}$, so the classmate's fraction is the larger.

The reason the building rule works is that multiplying by $c$ describes an action on the picture: each of the $9$ parts becomes $c$ smaller pieces, so the whole holds $9c$ of them, and each of the $4$ shaded parts becomes $c$ pieces, so $4c$ are shaded. Nothing was added or removed, only cut lines drawn. Adding $5$ to each term describes no such action: it does not cut the parts, it invents five more of them on top and five more underneath, and the shaded share changes.

Watch out. "The same change to both" is not the test. The test is whether the change is one the picture can perform, and only multiplying and dividing are.

The bar is a division, so divide the numerator by the denominator and keep the remainder.

$58 \div 7 = 8 \text{ remainder } 2, \qquad \frac{58}{7} = 8\tfrac{2}{7}$

Seven sevenths fill one whole, so $8$ counts the whole units and $2$ counts the sevenths that could not make another one. A remainder is always below the divisor, so the fraction part is proper.

A mixed number is a sum, so write the whole number over the same denominator and add.

$6\tfrac{5}{9} = \frac{6 \times 9}{9} + \frac{5}{9} = \frac{54 + 5}{9} = \frac{59}{9}$

Running the same shortcut on part A's answer returns the fraction it came from.

$8\tfrac{2}{7} = \frac{8 \times 7 + 2}{7} = \frac{58}{7}$

First line. The shortcut is multiply then add, because each whole is $\frac{9}{9}$ and six wholes are $54$ ninths, not $6$ ninths.

$6\tfrac{5}{9} = \frac{6 \times 9 + 5}{9} = \frac{59}{9}, \qquad \text{not} \qquad \frac{6 + 5}{9} = \frac{11}{9}$

Since $\frac{11}{9}$ is only a little above $1$, the slip has thrown away $\frac{48}{9}$, that is five and a third wholes.

Second line. The remainder counts leftover pieces, and the pieces are sevenths, so it belongs over $7$.

$\frac{58}{7} = 8\tfrac{2}{7}, \qquad \text{not} \qquad 8\tfrac{2}{58}$

Two fifty-eighths is far smaller than two sevenths, so this report falls short of the true value.

Both slips break the same thing. The denominator names the size of one piece, and converting between the two forms only regroups pieces into wholes or breaks wholes back into pieces. No piece is ever re-cut, so the denominator is the one number that must survive both directions untouched.

Twelfths and eighths are different sizes, so rebuild both over a shared size. The least common multiple of $12$ and $8$ is $24$.

$\frac{11}{12} = \frac{22}{24}, \qquad \frac{5}{8} = \frac{15}{24}$

Now every piece is a twenty-fourth, so count them.

$\frac{22}{24} + \frac{15}{24} = \frac{37}{24}$

Since $37$ is prime and does not divide $24$, the fraction is already in lowest terms. It is improper, which fits: the walker has passed one full lap.

Watch out. Adding the denominators too, as $\frac{11 + 5}{12 + 8} = \frac{16}{20} = \frac{4}{5}$, returns less than the first stretch on its own, which no total of two positive lengths can be.

The rebuilds from part A serve here too, since the pieces have to match for a subtraction just as they do for an addition.

$\frac{11}{12} - \frac{5}{8} = \frac{22}{24} - \frac{15}{24} = \frac{7}{24}$

Since $7$ and $24$ share no factor above $1$, the answer is finished.

Rebuild over $24$ again and subtract in the order written.

$\frac{5}{8} - \frac{11}{12} = \frac{15}{24} - \frac{22}{24} = \frac{15 - 22}{24} = -\frac{7}{24}$

The like-denominator rule sends the whole question down to the numerators, and $15 - 22$ is a subtraction of whole numbers that lands to the left of zero.

So the walker is half right. The gap between the two stretches is $\frac{7}{24}$ of a lap either way, and that is what "how far apart" means. But a subtraction reports a signed change, not a bare gap, and reversing the order flips its sign, since $b - a$ is the opposite of $a - b$ for any $a$ and $b$.

Watch out. The two are equal only when the two fractions are equal, in which case both differences are $0$.

"Of" is a multiplication, so the purchase is $\frac{5}{6}$ of $\frac{8}{15}$. Once the product is written as one fraction, any numerator may be cancelled against any denominator, whichever fraction each came from.

Divide the $5$ and the $15$ by $5$, leaving $1$ and $3$. Divide the $8$ and the $6$ by $2$, leaving $4$ and $3$.

$\frac{5}{6} \times \frac{8}{15} = \frac{1}{3} \times \frac{4}{3} = \frac{1 \times 4}{3 \times 3} = \frac{4}{9}$

The long way agrees: multiplying straight across gives $\frac{40}{90}$, and dividing top and bottom by their greatest common factor $10$ gives $\frac{4}{9}$.

"How many of these fit into that" is a division, and dividing by a fraction is multiplying by its reciprocal.

$\frac{8}{15} \div \frac{4}{45} = \frac{8}{15} \times \frac{45}{4}$

Cancel the $4$ into the $8$ and the $15$ into the $45$.

$\frac{2}{1} \times \frac{3}{1} = 6$

The count is above $1$, which is what dividing by a length below one whole reel has to give.

Work the assistant's line out.

$\frac{15}{8} \times \frac{4}{45} = \frac{60}{360} = \frac{1}{6}$

That is the reciprocal of $6$, and it is not an accident of these numbers. Write the two calculations in general letters.

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

The second is the first with its numerator and denominator swapped, so it is always the reciprocal of the correct quotient. Four numbers have to be nonzero for that sentence to mean anything, and each for its own reason: $b$ and $d$ because they are denominators, $c$ because dividing by $\frac{c}{d}$ is dividing by zero otherwise, and $a$ because turning $\frac{a}{b}$ over puts $a$ underneath. That last one is also what gives the correct quotient a reciprocal to be.

Watch out. A reciprocal-shaped answer is a useful alarm. A count of pieces that comes out below $1$, when the pieces are smaller than the amount being cut up, is worth rechecking before it is written down.

Put both fraction parts over the least common denominator $12$ before looking at either cut.

$9\tfrac{5}{12} - 3\tfrac{3}{12} \qquad\text{and}\qquad 9\tfrac{5}{12} - 3\tfrac{8}{12}$

First roll. Five twelfths covers three twelfths, so the columns run as they stand.

$9\tfrac{5}{12} - 3\tfrac{3}{12} = 6\tfrac{2}{12} = 6\tfrac{1}{6}$

Second roll. Five twelfths does not cover eight twelfths, so one whole metre is broken open and counted out as $\tfrac{12}{12}$.

$9\tfrac{5}{12} = 8\tfrac{17}{12}, \qquad 8\tfrac{17}{12} - 3\tfrac{8}{12} = 5\tfrac{9}{12} = 5\tfrac{3}{4}$

So what decides it is a comparison, not the size of the numbers: break a whole open exactly when the piece being removed exceeds the piece already there. Should the two be equal, nothing is broken open and the answer has no fraction part at all.

The fraction part here is bigger than one whole, so the line cannot be a finished length: a mixed number carries a proper fraction. What has happened is that a whole metre has moved across, counted out over the denominator already in use.

$\tfrac{8}{8} = 1$

So undo the move by handing that whole back to the whole-number part.

$3\tfrac{15}{8} = 3 + \tfrac{8}{8} + \tfrac{7}{8} = 4\tfrac{7}{8}$

Check both lines against a single fraction, and they agree.

$3\tfrac{15}{8} = \frac{24 + 15}{8} = \frac{39}{8}, \qquad 4\tfrac{7}{8} = \frac{32 + 7}{8} = \frac{39}{8}$

Through all of it the $8$ never moves. It names the size of one piece, an eighth of a metre, and breaking a whole metre open re-cuts nothing; it only counts that metre out in pieces the calculation was already using.

Check the length before the shape.

$4\tfrac{21}{12} = \frac{4 \times 12 + 21}{12} = \frac{69}{12}, \qquad 5\tfrac{3}{4} = \frac{5 \times 4 + 3}{4} = \frac{23}{4} = \frac{69}{12}$