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Understanding Fractions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A measuring label

A label describes $\frac{9}{16}$ of a liter as nine copies of one smaller amount. What is that smaller amount?
1. Hint 1
  The denominator tells the size of each equal part.
2. Hint 2
  How many equal parts is the whole liter cut into?
Answer

$\frac{1}{16}$ of a liter.

Full solution

The denominator $16$ says that one liter is divided into sixteen equal parts.
Each part is $\frac{1}{16}$ of a liter.
The numerator $9$ counts nine such parts.

The part size is therefore $\frac{1}{16}$ of a liter.
Nine of these parts account for the amount on the label.

Answer

$\frac{1}{16}$ of a liter.

Key idea

The numerator counts copies of the part whose size the denominator names.
Problem 1: Mark complete
Problem 2 Between two and three

A fraction has denominator $7$ and a whole-number numerator, and its value is at least $2$ and less than $3$ . Write every such fraction.
1. Hint 1
  Count how many sevenths make each of the two whole numbers.
2. Hint 2
  Decide from "at least" and "less than" whether each boundary value itself is allowed.
Answer

$\frac{14}7$ , $\frac{15}7$ , $\frac{16}7$ , $\frac{17}7$ , $\frac{18}7$ , $\frac{19}7$ , $\frac{20}7$ .

Full solution

Seven sevenths make one whole, so fourteen make two wholes and twenty-one make three.
$\frac{14}7=2$ $\frac{21}7=3$

The value may equal $2$ but must stay below $3$ , so the numerator must be at least $14$ and less than $21$ .
Its whole-number possibilities are $14$ , $15$ , $16$ , $17$ , $18$ , $19$ , and $20$ , giving the seven fractions listed.
Each is improper because its numerator is at least its denominator.

Answer

$\frac{14}7$ , $\frac{15}7$ , $\frac{16}7$ , $\frac{17}7$ , $\frac{18}7$ , $\frac{19}7$ , $\frac{20}7$ .

Key idea

Counting the parts in the boundary wholes gives the allowed numerators.
Problem 2: Mark complete
Problem 3 A point near one

The segment from $0$ to $1$ on a number line is divided into nine equal steps. A point is two of these steps to the left of $1$ . Name the point as a fraction with denominator $9$ .
1. Hint 1
  The point can be named by the number of steps from $0$ .
2. Hint 2
  Find how many of the nine steps remain after moving two steps back from $1$ .
Answer

$\frac79$ .

Full solution

Nine ninth-sized steps from $0$ reach $1$ .
Moving two steps back leaves
$9-2=7$
steps from $0$ .
The point is therefore $\frac79$ .
It lies between $0$ and $1$ , as expected.

Answer

$\frac79$ .

Key idea

A fraction on a number line counts equal steps from zero even when its location is described from the other end.
Problem 3: Mark complete
Problem 4 A moving marker

A marker starts at $0$ . Each move covers $\frac18$ of a unit. It makes nineteen moves right and then six moves left. Give its final position as a fraction with denominator $8$ , and state whether the fraction is proper or improper.
1. Hint 1
  A right move and a left move of the same size undo each other.
2. Hint 2
  Count the right moves that remain after pairing six of them with the left moves.
3. Hint 3
  Compare that count of eighths with the number of eighths that make one whole.
Answer

$\frac{13}{8}$ ; improper.

Full solution

Six left moves undo six of the nineteen right moves, leaving
$19-6=13$
right moves from $0$ .
Each is one eighth of a unit, so the position is $\frac{13}{8}$ .

One whole takes eight moves, and $13$ is more than $8$ , so the marker has passed $1$ .
The numerator is greater than the denominator, so the fraction is improper.

Answer

$\frac{13}{8}$ ; improper.

Key idea

Equal moves can be counted first and then named using their shared part size.
Problem 4: Mark complete
Problem 5 Three fabric orders

Three clubs order lengths from identical full rolls. Club A orders $\frac58$ of a roll, Club B orders $\frac5{12}$ , and Club C orders $\frac7{12}$ . Who orders more in the comparison of A with B, and who orders more in the comparison of B with C? Explain each decision.
1. Hint 1
  Look for the number the two fractions share in each comparison.
2. Hint 2
  For A and B, the counts match but the parts have different sizes.
3. Hint 3
  For B and C, the parts have the same size, so compare their counts.
Answer

A orders more than B; C orders more than B.

Full solution

A and B both take five parts.
An eighth of a roll is larger than a twelfth of the same-sized roll, so
$\frac58>\frac5{12}.$

B and C both count twelfths.
Seven twelfths is more than five twelfths, so
$\frac7{12}>\frac5{12}.$
These comparisons use identical roll sizes, as the statement specifies.

Answer

A orders more than B; C orders more than B.

Key idea

Equal positive counts call for comparing part sizes, while equal part sizes call for comparing counts.
Problem 5: Mark complete
Problem 6 Sharing a roll of wire

A roll of wire is shared equally among $9$ spools, and each spool receives $\frac{40}{9}$ meters. How long was the roll, and how many complete one-meter pieces can be cut from one spool's wire?
1. Hint 1
  Sharing among spools is a division, and the fraction bar is a division sign.
2. Hint 2
  Write $\frac{40}{9}$ as a division and compare it with the roll's length divided by $9$ .
3. Hint 3
  For the pieces, compare $40$ ninths with the number of ninths in each whole meter.
Answer

$40$ meters; $4$ complete pieces.

Full solution

Each spool receives the roll's length divided by $9$ .
The fraction bar is a division sign, so
$\frac{40}{9}=40\div9.$
The roll's length divided by $9$ is therefore $40$ divided by $9$ , so the roll was $40$ meters long.
As a check, nine shares of $40$ ninths are $9\times40=360$ ninths, and since nine ninths make one meter, $360$ ninths are $360\div9=40$ meters, the roll's length.

One meter is nine ninths.
Four complete meters use $4\times9=36$ ninths, and five would need $45$ , more than the $40$ available.
One spool's wire therefore gives four complete one-meter pieces, with $\frac49$ of a meter left over.

Answer

$40$ meters; $4$ complete pieces.

Key idea

When $b$ equal shares are each $\frac{a}{b}$ , reading the bar as $a\div b$ shows the total shared was $a$ , and every $b$ parts of size $\frac1b$ make one whole.
Problem 6: Mark complete
Problem 7 A card counter

A counter starts at $0$ and advances once for each card packed. It reads $24$ when an empty box has been filled to $\frac47$ of its capacity. That box is then filled completely, and packing continues into a second, identical box until the counter reads $72$ . What fraction of the second box is filled?
1. Hint 1
  The twenty-four cards fill four equal parts of one box.
2. Hint 2
  Find the cards in one seventh and in a full box, then count the cards that reach the second box.
3. Hint 3
  Name the second box's cards as a number of sevenths of a box.
Answer

$\frac57$ of the second box.

Full solution

Four sevenths of a box hold $24$ cards, so one seventh holds
$24\div4=6$
cards.
A full box holds $7\times6=42$ cards.

The counter reads $42$ when the first box is full, so the second box receives
$72-42=30$
cards.
Each seventh holds $6$ cards, so $30$ cards fill $30\div6=5$ sevenths, and the second box is $\frac57$ full.

As a check, the $42$ cards in the first box and the $30$ in the second make $42+30=72$ , the final reading.

Answer

$\frac57$ of the second box.

Key idea

The count in several equal parts gives the count in one part, and that tells you how many parts another count fills.
Problem 7: Mark complete
Problem 8 A claim about thirteenths

Nia says that every proper fraction with denominator $13$ and a whole-number numerator of at least $7$ is closer to $1$ than to $0$ on the number line. Is she correct? Explain.
1. Hint 1
  Each allowed point can be measured in thirteenth-sized steps from $0$ and from $1$ .
2. Hint 2
  Use the meaning of a proper fraction to list the numerators that are allowed.
3. Hint 3
  Compare the steps from $0$ with the steps to $1$ , starting with the smallest allowed numerator.
Answer

Yes.

Full solution

A proper fraction has a numerator smaller than its denominator, so the allowed numerators are the whole numbers from $7$ to $12$ .
Thirteen steps of one thirteenth each reach
$\frac{13}{13}=1.$

The point $\frac7{13}$ is $7$ steps from $0$ and $13-7=6$ steps from $1$ , so it is closer to $1$ .
Each larger allowed numerator is one more step from $0$ and one step nearer to $1$ , so every other allowed point is also closer to $1$ .
Nia is correct.

Answer

Yes.

Key idea

With a fixed denominator, comparing a point's steps from $0$ with its steps to $1$ shows which end it is closer to.
Problem 8: Mark complete
Problem 9 Two station readings

A recorder shows $\frac0{11}$ at one station and $\frac0{17}$ at another. Ben says the second value is smaller since seventeenths are smaller parts. Is his conclusion correct? Explain.
1. Hint 1
  Part size matters only after you know how many parts are being counted.
2. Hint 2
  Read each fraction as a division.
Answer

No; both values are $0$ .

Full solution

Both numerators are zero, so neither reading counts any parts.
Equivalently, both divisions give zero:
$\frac0{11}=0$ $\frac0{17}=0$

Seventeenths are smaller than elevenths of the same whole, but taking none of either gives the same amount.
The rule that the smaller denominator wins needs the shared numerator to be positive.

Answer

No; both values are $0$ .

Key idea

Zero copies of different-sized parts still give the same value, zero.
Problem 9: Mark complete
Problem 10 A proposed location

Leah wants to put the expression $\frac{8-8}{5-5}$ at the point $1$ on a number line since its numerator and denominator match. Does the expression name that point? Explain.
1. Hint 1
  Work out the numerator and denominator before applying a fraction rule.
2. Hint 2
  Division must select one value whose product with the denominator gives the numerator.
3. Hint 3
  Ask how many numbers have product $0$ when multiplied by $0$ .
Answer

No; the expression is undefined.

Full solution

Both parts of the fraction are zero:
$8-8=0$ $5-5=0$
The expression would therefore be $\frac00$ .

A division $0\div0$ would need one chosen number whose product with $0$ is $0$ .
Every number has that product, so division does not determine a single value.
The expression is undefined and has no position on the number line.
Equal numerator and denominator give $1$ when their common value is nonzero.

Answer

No; the expression is undefined.

Key idea

A fraction with matching top and bottom equals one only when the common value is nonzero.
Problem 10: Mark complete

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