Fractions: Chapter Test

The denominator records how many equal steps the whole was cut into, and the numerator counts how many of those steps you take. Seven steps make the whole, and you take three of them.

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

So the third mark is three sevenths, and it lands before $1$ because $3$ is smaller than $7$.

Both fractions count elevenths, so the pieces are already the same size. Count them: four elevenths and five more elevenths is nine elevenths, and the size of a piece never changed.

$\frac{4}{11} + \frac{5}{11} = \frac{4 + 5}{11} = \frac{9}{11}$

Since $9$ and $11$ share no factor above $1$, this is already in lowest terms.

The bar is a division, so divide the numerator by the denominator and read the two results. Seven sevenths make one whole, so the quotient counts complete wholes and the remainder counts the sevenths left over.

$50 \div 7 = 7 \text{ remainder } 1, \qquad \frac{50}{7} = 7\tfrac{1}{7}$

The denominator stays $7$, because the pieces were never re-cut, and the remainder $1$ is below $7$, so the fraction part is proper.

Dividing the numerator and the denominator by their greatest common factor reaches lowest terms in a single division. The common factors of $36$ and $60$ are $1, 2, 3, 4, 6$ and $12$, and the greatest is $12$.

$\frac{36}{60} = \frac{36 \div 12}{60 \div 12} = \frac{3}{5}$

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

Multiplying needs no common denominator: the tops multiply to count the pieces you keep and the bottoms multiply to count the pieces in the new whole.

$\frac{4}{9} \times \frac{6}{7} = \frac{4 \times 6}{9 \times 7} = \frac{24}{63} = \frac{8}{21}$

The last step divides top and bottom by their greatest common factor $3$. Canceling the $6$ against the $9$ first gives $\frac{4}{3} \times \frac{2}{7}$ and the same answer with smaller numbers.

Read each one as a division. A denominator of $0$ asks what number times $0$ gives $7$, and nothing times $0$ gives anything but $0$, so there is no answer.

$\frac{0}{7} = 0, \qquad \frac{7}{7} = 1, \qquad \frac{7}{1} = 7, \qquad \frac{7}{0} \text{ is undefined}$

A numerator of $0$ is allowed; a denominator of $0$ is not, because you cannot cut a whole into zero parts.

The two fractions share neither number, so rewrite both over a common denominator. The least common multiple of $11$ and $10$ is $110$.

$\frac{8}{11} = \frac{80}{110}, \qquad \frac{7}{10} = \frac{77}{110}$

With the pieces now the same size, the counts decide, and $80 > 77$. The cross products give the same verdict: $8 \times 10 = 80$ against $11 \times 7 = 77$.

Count in ninths. Nine ninths reach $1$ and eighteen ninths reach $2$, so a fraction of ninths lies between $1$ and $2$ exactly when its numerator is between $9$ and $18$.

$\frac{9}{9} = 1, \qquad \frac{13}{9} \text{ lies between}, \qquad \frac{18}{9} = 2$

Thirteen is between $9$ and $18$, so $\frac{13}{9}$ is the one. It is improper, which is exactly what a value past $1$ has to be.

Convert both factors first, since the multiplication rule expects a single numerator over a single denominator. Multiply the whole number by the denominator and add the numerator.

$2\tfrac{2}{5} = \frac{12}{5}, \qquad 1\tfrac{7}{8} = \frac{15}{8}, \qquad \frac{12}{5} \times \frac{15}{8} = \frac{180}{40} = \frac{9}{2} = 4\tfrac{1}{2}$

Canceling first is quicker: the $12$ pairs with the $8$ and the $15$ pairs with the $5$, leaving $\frac{3}{1} \times \frac{3}{2}$.

Find what the old denominator was multiplied by to reach the new one, then do the same to the numerator, since cutting every part into the same number of pieces leaves the amount alone.

$8 \times 7 = 56, \qquad \frac{5}{8} = \frac{5 \times 7}{8 \times 7} = \frac{35}{56}$

The multiplier has to hit both numbers; changing only one of them would change the value.

Tenths and eighths are different sizes, so rebuild both over a shared size first. The least common multiple of $10$ and $8$ is $40$.

$\frac{7}{10} = \frac{28}{40}, \qquad \frac{3}{8} = \frac{15}{40}, \qquad \frac{28}{40} + \frac{15}{40} = \frac{43}{40}$

Each rebuild multiplied top and bottom by the same number, so nothing moved. Since $43$ and $40$ share no factor above $1$, the improper answer $\frac{43}{40}$ is finished.

Dividing by a fraction is multiplying by its reciprocal, because a fraction times its reciprocal is $1$. Leave the first fraction alone and flip only the divisor.

$\frac{5}{9} \div \frac{10}{27} = \frac{5}{9} \times \frac{27}{10} = \frac{135}{90} = \frac{3}{2}$

Canceling first is lighter: the $27$ pairs with the $9$ and the $5$ pairs with the $10$. The answer is above $1$, which fits, since $\frac{10}{27}$ is smaller than $\frac{5}{9}$ and so fits into it more than once.

No two of the three share a number, so neither glance-shortcut applies and each pair has to be rebuilt over a shared denominator. For $\frac{5}{7}$ against $\frac{8}{11}$ that denominator is $77$.

$\frac{5}{7} = \frac{55}{77}, \qquad \frac{8}{11} = \frac{56}{77}, \qquad \text{so} \qquad \frac{5}{7} < \frac{8}{11}$

The cross products are that same rebuild with the shared denominator left unwritten: $55$ and $56$ are exactly the products $5 \times 11$ and $7 \times 8$. So they settle the other two pairs in one line each.

$3 \times 7 = 21 > 4 \times 5 = 20, \qquad 3 \times 11 = 33 > 4 \times 8 = 32$

Those give $\frac{3}{4} > \frac{5}{7}$ and $\frac{3}{4} > \frac{8}{11}$, so the order is $\frac{5}{7}, \frac{8}{11}, \frac{3}{4}$.

Counting in sevenths, every seven steps completes one whole, so divide the numerator by the denominator and read off both results.

$38 \div 7 = 5 \text{ remainder } 3, \qquad \frac{38}{7} = 5\tfrac{3}{7}$

Five complete wholes use up $35$ sevenths and three sevenths remain, so the point lies past $5$ but short of $6$, because three more sevenths do not reach a whole.

Write the whole number over $1$ and multiply by the reciprocal of the divisor, canceling the shared $7$ before multiplying.

$14 \div \frac{7}{9} = \frac{14}{1} \times \frac{9}{7} = \frac{2}{1} \times \frac{9}{1} = 18$

The counting reading agrees. One whole holds $\frac{9}{7}$ of a piece of size $\frac{7}{9}$, so $14$ wholes hold $14 \times \frac{9}{7} = 18$ such pieces. The count is larger than $14$, which is what dividing by a number below $1$ has to give.

Put the fraction parts over their least common denominator $12$, then look at the fraction column: $\tfrac{3}{12}$ is not enough to take $\tfrac{10}{12}$ from, so borrow one whole and add it as $\tfrac{12}{12}$.

$6\tfrac{3}{12} - 2\tfrac{10}{12} = 5\tfrac{15}{12} - 2\tfrac{10}{12} = 3\tfrac{5}{12}$

Converting instead gives the same thing: $\frac{25}{4} - \frac{17}{6} = \frac{75}{12} - \frac{34}{12} = \frac{41}{12}$.

Work each one out and compare it with $\frac{5}{8}$.

$\frac{5}{8} \times \frac{9}{10} = \frac{9}{16}, \quad \frac{5}{8} \div \frac{4}{3} = \frac{15}{32}, \quad \frac{5}{8} \times \frac{8}{9} = \frac{5}{9}, \quad \frac{5}{8} \div \frac{5}{6} = \frac{3}{4}$

Over a shared denominator of $288$ those are $162$, $135$, $160$ and $216$, against $180$ for $\frac{5}{8}$, so only the last is above it.

The pattern behind that: for a positive starting number, multiplying by a fraction below $1$ shrinks it and dividing by a fraction below $1$ grows it. Three of the four second fractions here sit below $1$, and only one of those three is a division.

Convert both mixed numbers first, then flip only the divisor.

$5\tfrac{1}{4} = \frac{21}{4}, \qquad 1\tfrac{5}{9} = \frac{14}{9}, \qquad \frac{21}{4} \div \frac{14}{9} = \frac{21}{4} \times \frac{9}{14}$

Cancel the $7$ shared by $21$ and $14$, leaving $\frac{3}{4} \times \frac{9}{2} = \frac{27}{8}$, and $27 \div 8 = 3$ remainder $3$, so the answer is $3\tfrac{3}{8}$.

A whole number joins the calculation as a fraction over $1$, rebuilt over the common denominator. The least common multiple of $6$ and $4$ is $12$, and $3 = \frac{36}{12}$.

$3 - \frac{5}{6} - \frac{3}{4} = \frac{36}{12} - \frac{10}{12} - \frac{9}{12} = \frac{17}{12}$

Since $17$ and $12$ share no factor above $1$, the answer is finished, and it is above $1$, which fits: the two fractions taken away come to less than $2$.

"How many of these fit into that" is a division, so convert the mixed number and flip the divisor.

$5\tfrac{5}{6} = \frac{35}{6}, \qquad \frac{35}{6} \div \frac{7}{12} = \frac{35}{6} \times \frac{12}{7} = 5 \times 2 = 10$

The $7$ cancels into the $35$ and the $6$ into the $12$. The count is larger than the number of liters, which is what dividing by an amount below one liter has to give.