Practice · Match fraction forms to compare sizes · Sensim Math

Variant 1 easy answer: 7

How many whole numbers can go in the $\star$ ?

$1\frac{3}{5} < \frac{\star}{5} < 3\frac{1}{5}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{5}$ sits strictly between the mixed numbers 1 and $\frac{3}{5}$ and 3 and $\frac{1}{5}$ .

Givens

The inequality is 1 $\frac{3}{5}$ < $\frac{\star}{5}$ < 3 $\frac{1}{5}$ .
The middle quantity is $\frac{\star}{5}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 5.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{5}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 5 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 1

\frac{3}{5}

to 5ths: 1 wholes is 5 5ths, plus 3 5ths.

1\tfrac{3}{5} = \frac{1\times 5 + 3}{5} = \frac{8}{5}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 3

\frac{1}{5}

to 5ths: 3 wholes is 15 5ths, plus 1 5ths.

3\tfrac{1}{5} = \frac{3\times 5 + 1}{5} = \frac{16}{5}

Now all three quantities are 5ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{8}{5}

<

\frac{\star}{5}

<

\frac{16}{5}

, so

\star

must be a whole number strictly between 8 and 16: that is 9,10,11,12,13,14,15.

8 < \star < 16 \Rightarrow \star \in \{9,10,11,12,13,14,15\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 7 whole numbers in the list.

\{9,10,11,12,13,14,15\} \Rightarrow 7

Just count the items in the systematic list.

Answer: 7

4 · Reviewdoes it hold up?

The endpoints 8 and 16 are 8 apart; excluding both ends leaves the 7 whole numbers in between, which matches the count. Each candidate lands strictly between 1 $\frac{3}{5}$ and 3 $\frac{1}{5}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{8}{5}$ and $\frac{16}{5}$ on a line of 5ths; the whole-number numerators strictly between them give the same 7 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 5ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 5ths (1 x 5, 3 x 5) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 2 easy answer: 9

How many whole numbers can go in the $\star$ ?

$3\frac{1}{6} < \frac{\star}{6} < 4\frac{5}{6}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{6}$ sits strictly between the mixed numbers 3 and $\frac{1}{6}$ and 4 and $\frac{5}{6}$ .

Givens

The inequality is 3 $\frac{1}{6}$ < $\frac{\star}{6}$ < 4 $\frac{5}{6}$ .
The middle quantity is $\frac{\star}{6}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 6.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{6}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 6 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 3

\frac{1}{6}

to 6ths: 3 wholes is 18 6ths, plus 1 6ths.

3\tfrac{1}{6} = \frac{3\times 6 + 1}{6} = \frac{19}{6}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 4

\frac{5}{6}

to 6ths: 4 wholes is 24 6ths, plus 5 6ths.

4\tfrac{5}{6} = \frac{4\times 6 + 5}{6} = \frac{29}{6}

Now all three quantities are 6ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{19}{6}

<

\frac{\star}{6}

<

\frac{29}{6}

, so

\star

must be a whole number strictly between 19 and 29: that is 20,21,22,23,24,25,26,27,28.

19 < \star < 29 \Rightarrow \star \in \{20,21,22,23,24,25,26,27,28\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 9 whole numbers in the list.

\{20,21,22,23,24,25,26,27,28\} \Rightarrow 9

Just count the items in the systematic list.

Answer: 9

4 · Reviewdoes it hold up?

The endpoints 19 and 29 are 10 apart; excluding both ends leaves the 9 whole numbers in between, which matches the count. Each candidate lands strictly between 3 $\frac{1}{6}$ and 4 $\frac{5}{6}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{19}{6}$ and $\frac{29}{6}$ on a line of 6ths; the whole-number numerators strictly between them give the same 9 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 6ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 6ths (3 x 6, 4 x 6) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 3 easy answer: 6

How many whole numbers can go in the $\star$ ?

$2\frac{1}{6} < \frac{\star}{6} < 3\frac{2}{6}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{6}$ sits strictly between the mixed numbers 2 and $\frac{1}{6}$ and 3 and $\frac{2}{6}$ .

Givens

The inequality is 2 $\frac{1}{6}$ < $\frac{\star}{6}$ < 3 $\frac{2}{6}$ .
The middle quantity is $\frac{\star}{6}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 6.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{6}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 6 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 2

\frac{1}{6}

to 6ths: 2 wholes is 12 6ths, plus 1 6ths.

2\tfrac{1}{6} = \frac{2\times 6 + 1}{6} = \frac{13}{6}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 3

\frac{2}{6}

to 6ths: 3 wholes is 18 6ths, plus 2 6ths.

3\tfrac{2}{6} = \frac{3\times 6 + 2}{6} = \frac{20}{6}

Now all three quantities are 6ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{13}{6}

<

\frac{\star}{6}

<

\frac{20}{6}

, so

\star

must be a whole number strictly between 13 and 20: that is 14,15,16,17,18,19.

13 < \star < 20 \Rightarrow \star \in \{14,15,16,17,18,19\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 6 whole numbers in the list.

\{14,15,16,17,18,19\} \Rightarrow 6

Just count the items in the systematic list.

Answer: 6

4 · Reviewdoes it hold up?

The endpoints 13 and 20 are 7 apart; excluding both ends leaves the 6 whole numbers in between, which matches the count. Each candidate lands strictly between 2 $\frac{1}{6}$ and 3 $\frac{2}{6}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{13}{6}$ and $\frac{20}{6}$ on a line of 6ths; the whole-number numerators strictly between them give the same 6 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 6ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 6ths (2 x 6, 3 x 6) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 4 easy answer: 7

How many whole numbers can go in the $\star$ ?

$3\frac{2}{7} < \frac{\star}{7} < 4\frac{3}{7}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{7}$ sits strictly between the mixed numbers 3 and $\frac{2}{7}$ and 4 and $\frac{3}{7}$ .

Givens

The inequality is 3 $\frac{2}{7}$ < $\frac{\star}{7}$ < 4 $\frac{3}{7}$ .
The middle quantity is $\frac{\star}{7}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 7.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{7}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 7 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 3

\frac{2}{7}

to 7ths: 3 wholes is 21 7ths, plus 2 7ths.

3\tfrac{2}{7} = \frac{3\times 7 + 2}{7} = \frac{23}{7}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 4

\frac{3}{7}

to 7ths: 4 wholes is 28 7ths, plus 3 7ths.

4\tfrac{3}{7} = \frac{4\times 7 + 3}{7} = \frac{31}{7}

Now all three quantities are 7ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{23}{7}

<

\frac{\star}{7}

<

\frac{31}{7}

, so

\star

must be a whole number strictly between 23 and 31: that is 24,25,26,27,28,29,30.

23 < \star < 31 \Rightarrow \star \in \{24,25,26,27,28,29,30\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 7 whole numbers in the list.

\{24,25,26,27,28,29,30\} \Rightarrow 7

Just count the items in the systematic list.

Answer: 7

4 · Reviewdoes it hold up?

The endpoints 23 and 31 are 8 apart; excluding both ends leaves the 7 whole numbers in between, which matches the count. Each candidate lands strictly between 3 $\frac{2}{7}$ and 4 $\frac{3}{7}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{23}{7}$ and $\frac{31}{7}$ on a line of 7ths; the whole-number numerators strictly between them give the same 7 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 7ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 7ths (3 x 7, 4 x 7) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 5 medium answer: 16

How many whole numbers can go in the $\star$ ?

$5\frac{2}{8} < \frac{\star}{8} < 7\frac{3}{8}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{8}$ sits strictly between the mixed numbers 5 and $\frac{2}{8}$ and 7 and $\frac{3}{8}$ .

Givens

The inequality is 5 $\frac{2}{8}$ < $\frac{\star}{8}$ < 7 $\frac{3}{8}$ .
The middle quantity is $\frac{\star}{8}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 8.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{8}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 8 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 5

\frac{2}{8}

to 8ths: 5 wholes is 40 8ths, plus 2 8ths.

5\tfrac{2}{8} = \frac{5\times 8 + 2}{8} = \frac{42}{8}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 7

\frac{3}{8}

to 8ths: 7 wholes is 56 8ths, plus 3 8ths.

7\tfrac{3}{8} = \frac{7\times 8 + 3}{8} = \frac{59}{8}

Now all three quantities are 8ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{42}{8}

<

\frac{\star}{8}

<

\frac{59}{8}

, so

\star

must be a whole number strictly between 42 and 59: that is 43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58.

42 < \star < 59 \Rightarrow \star \in \{43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 16 whole numbers in the list.

\{43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58\} \Rightarrow 16

Just count the items in the systematic list.

Answer: 16

4 · Reviewdoes it hold up?

The endpoints 42 and 59 are 17 apart; excluding both ends leaves the 16 whole numbers in between, which matches the count. Each candidate lands strictly between 5 $\frac{2}{8}$ and 7 $\frac{3}{8}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{42}{8}$ and $\frac{59}{8}$ on a line of 8ths; the whole-number numerators strictly between them give the same 16 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 8ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 8ths (5 x 8, 7 x 8) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 6 medium answer: 5

How many whole numbers can go in the $\star$ ?

$4\frac{3}{8} < \frac{\star}{8} < 5\frac{1}{8}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{8}$ sits strictly between the mixed numbers 4 and $\frac{3}{8}$ and 5 and $\frac{1}{8}$ .

Givens

The inequality is 4 $\frac{3}{8}$ < $\frac{\star}{8}$ < 5 $\frac{1}{8}$ .
The middle quantity is $\frac{\star}{8}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 8.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{8}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 8 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 4

\frac{3}{8}

to 8ths: 4 wholes is 32 8ths, plus 3 8ths.

4\tfrac{3}{8} = \frac{4\times 8 + 3}{8} = \frac{35}{8}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 5

\frac{1}{8}

to 8ths: 5 wholes is 40 8ths, plus 1 8ths.

5\tfrac{1}{8} = \frac{5\times 8 + 1}{8} = \frac{41}{8}

Now all three quantities are 8ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{35}{8}

<

\frac{\star}{8}

<

\frac{41}{8}

, so

\star

must be a whole number strictly between 35 and 41: that is 36,37,38,39,40.

35 < \star < 41 \Rightarrow \star \in \{36,37,38,39,40\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 5 whole numbers in the list.

\{36,37,38,39,40\} \Rightarrow 5

Just count the items in the systematic list.

Answer: 5

4 · Reviewdoes it hold up?

The endpoints 35 and 41 are 6 apart; excluding both ends leaves the 5 whole numbers in between, which matches the count. Each candidate lands strictly between 4 $\frac{3}{8}$ and 5 $\frac{1}{8}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{35}{8}$ and $\frac{41}{8}$ on a line of 8ths; the whole-number numerators strictly between them give the same 5 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 8ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 8ths (4 x 8, 5 x 8) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 7 medium answer: 6

How many whole numbers can go in the $\star$ ?

$5\frac{4}{9} < \frac{\star}{9} < 6\frac{2}{9}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{9}$ sits strictly between the mixed numbers 5 and $\frac{4}{9}$ and 6 and $\frac{2}{9}$ .

Givens

The inequality is 5 $\frac{4}{9}$ < $\frac{\star}{9}$ < 6 $\frac{2}{9}$ .
The middle quantity is $\frac{\star}{9}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 9.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{9}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 9 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 5

\frac{4}{9}

to 9ths: 5 wholes is 45 9ths, plus 4 9ths.

5\tfrac{4}{9} = \frac{5\times 9 + 4}{9} = \frac{49}{9}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 6

\frac{2}{9}

to 9ths: 6 wholes is 54 9ths, plus 2 9ths.

6\tfrac{2}{9} = \frac{6\times 9 + 2}{9} = \frac{56}{9}

Now all three quantities are 9ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{49}{9}

<

\frac{\star}{9}

<

\frac{56}{9}

, so

\star

must be a whole number strictly between 49 and 56: that is 50,51,52,53,54,55.

49 < \star < 56 \Rightarrow \star \in \{50,51,52,53,54,55\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 6 whole numbers in the list.

\{50,51,52,53,54,55\} \Rightarrow 6

Just count the items in the systematic list.

Answer: 6

4 · Reviewdoes it hold up?

The endpoints 49 and 56 are 7 apart; excluding both ends leaves the 6 whole numbers in between, which matches the count. Each candidate lands strictly between 5 $\frac{4}{9}$ and 6 $\frac{2}{9}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{49}{9}$ and $\frac{56}{9}$ on a line of 9ths; the whole-number numerators strictly between them give the same 6 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 9ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 9ths (5 x 9, 6 x 9) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 8 medium answer: 12

How many whole numbers can go in the $\star$ ?

$4\frac{2}{9} < \frac{\star}{9} < 5\frac{6}{9}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{9}$ sits strictly between the mixed numbers 4 and $\frac{2}{9}$ and 5 and $\frac{6}{9}$ .

Givens

The inequality is 4 $\frac{2}{9}$ < $\frac{\star}{9}$ < 5 $\frac{6}{9}$ .
The middle quantity is $\frac{\star}{9}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 9.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{9}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 9 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 4

\frac{2}{9}

to 9ths: 4 wholes is 36 9ths, plus 2 9ths.

4\tfrac{2}{9} = \frac{4\times 9 + 2}{9} = \frac{38}{9}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 5

\frac{6}{9}

to 9ths: 5 wholes is 45 9ths, plus 6 9ths.

5\tfrac{6}{9} = \frac{5\times 9 + 6}{9} = \frac{51}{9}

Now all three quantities are 9ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{38}{9}

<

\frac{\star}{9}

<

\frac{51}{9}

, so

\star

must be a whole number strictly between 38 and 51: that is 39,40,41,42,43,44,45,46,47,48,49,50.

38 < \star < 51 \Rightarrow \star \in \{39,40,41,42,43,44,45,46,47,48,49,50\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 12 whole numbers in the list.

\{39,40,41,42,43,44,45,46,47,48,49,50\} \Rightarrow 12

Just count the items in the systematic list.

Answer: 12

4 · Reviewdoes it hold up?

The endpoints 38 and 51 are 13 apart; excluding both ends leaves the 12 whole numbers in between, which matches the count. Each candidate lands strictly between 4 $\frac{2}{9}$ and 5 $\frac{6}{9}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{38}{9}$ and $\frac{51}{9}$ on a line of 9ths; the whole-number numerators strictly between them give the same 12 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 9ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 9ths (4 x 9, 5 x 9) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 9 hard answer: 8

How many whole numbers can go in the $\star$ ?

$6\frac{5}{10} < \frac{\star}{10} < 7\frac{4}{10}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{10}$ sits strictly between the mixed numbers 6 and $\frac{5}{10}$ and 7 and $\frac{4}{10}$ .

Givens

The inequality is 6 $\frac{5}{10}$ < $\frac{\star}{10}$ < 7 $\frac{4}{10}$ .
The middle quantity is $\frac{\star}{10}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 10.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{10}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 10 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 6

\frac{5}{10}

to 10ths: 6 wholes is 60 10ths, plus 5 10ths.

6\tfrac{5}{10} = \frac{6\times 10 + 5}{10} = \frac{65}{10}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 7

\frac{4}{10}

to 10ths: 7 wholes is 70 10ths, plus 4 10ths.

7\tfrac{4}{10} = \frac{7\times 10 + 4}{10} = \frac{74}{10}

Now all three quantities are 10ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{65}{10}

<

\frac{\star}{10}

<

\frac{74}{10}

, so

\star

must be a whole number strictly between 65 and 74: that is 66,67,68,69,70,71,72,73.

65 < \star < 74 \Rightarrow \star \in \{66,67,68,69,70,71,72,73\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 8 whole numbers in the list.

\{66,67,68,69,70,71,72,73\} \Rightarrow 8

Just count the items in the systematic list.

Answer: 8

4 · Reviewdoes it hold up?

The endpoints 65 and 74 are 9 apart; excluding both ends leaves the 8 whole numbers in between, which matches the count. Each candidate lands strictly between 6 $\frac{5}{10}$ and 7 $\frac{4}{10}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{65}{10}$ and $\frac{74}{10}$ on a line of 10ths; the whole-number numerators strictly between them give the same 8 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 10ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 10ths (6 x 10, 7 x 10) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 10 hard answer: 15

How many whole numbers can go in the $\star$ ?

$7\frac{3}{11} < \frac{\star}{11} < 8\frac{8}{11}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{11}$ sits strictly between the mixed numbers 7 and $\frac{3}{11}$ and 8 and $\frac{8}{11}$ .

Givens

The inequality is 7 $\frac{3}{11}$ < $\frac{\star}{11}$ < 8 $\frac{8}{11}$ .
The middle quantity is $\frac{\star}{11}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 11.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{11}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 11 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 7

\frac{3}{11}

to 11ths: 7 wholes is 77 11ths, plus 3 11ths.

7\tfrac{3}{11} = \frac{7\times 11 + 3}{11} = \frac{80}{11}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 8

\frac{8}{11}

to 11ths: 8 wholes is 88 11ths, plus 8 11ths.

8\tfrac{8}{11} = \frac{8\times 11 + 8}{11} = \frac{96}{11}

Now all three quantities are 11ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{80}{11}

<

\frac{\star}{11}

<

\frac{96}{11}

, so

\star

must be a whole number strictly between 80 and 96: that is 81,82,83,84,85,86,87,88,89,90,91,92,93,94,95.

80 < \star < 96 \Rightarrow \star \in \{81,82,83,84,85,86,87,88,89,90,91,92,93,94,95\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 15 whole numbers in the list.

\{81,82,83,84,85,86,87,88,89,90,91,92,93,94,95\} \Rightarrow 15

Just count the items in the systematic list.

Answer: 15

4 · Reviewdoes it hold up?

The endpoints 80 and 96 are 16 apart; excluding both ends leaves the 15 whole numbers in between, which matches the count. Each candidate lands strictly between 7 $\frac{3}{11}$ and 8 $\frac{8}{11}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{80}{11}$ and $\frac{96}{11}$ on a line of 11ths; the whole-number numerators strictly between them give the same 15 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 11ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 11ths (7 x 11, 8 x 11) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 11 hard answer: 18

How many whole numbers can go in the $\star$ ?

$2\frac{7}{12} < \frac{\star}{12} < 4\frac{2}{12}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{12}$ sits strictly between the mixed numbers 2 and $\frac{7}{12}$ and 4 and $\frac{2}{12}$ .

Givens

The inequality is 2 $\frac{7}{12}$ < $\frac{\star}{12}$ < 4 $\frac{2}{12}$ .
The middle quantity is $\frac{\star}{12}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 12.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{12}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 12 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 2

\frac{7}{12}

to 12ths: 2 wholes is 24 12ths, plus 7 12ths.

2\tfrac{7}{12} = \frac{2\times 12 + 7}{12} = \frac{31}{12}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 4

\frac{2}{12}

to 12ths: 4 wholes is 48 12ths, plus 2 12ths.

4\tfrac{2}{12} = \frac{4\times 12 + 2}{12} = \frac{50}{12}

Now all three quantities are 12ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{31}{12}

<

\frac{\star}{12}

<

\frac{50}{12}

, so

\star

must be a whole number strictly between 31 and 50: that is 32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49.

31 < \star < 50 \Rightarrow \star \in \{32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 18 whole numbers in the list.

\{32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49\} \Rightarrow 18

Just count the items in the systematic list.

Answer: 18

4 · Reviewdoes it hold up?

The endpoints 31 and 50 are 19 apart; excluding both ends leaves the 18 whole numbers in between, which matches the count. Each candidate lands strictly between 2 $\frac{7}{12}$ and 4 $\frac{2}{12}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{31}{12}$ and $\frac{50}{12}$ on a line of 12ths; the whole-number numerators strictly between them give the same 18 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 12ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 12ths (2 x 12, 4 x 12) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Variant 12 hard answer: 17

How many whole numbers can go in the $\star$ ?

$2\frac{5}{14} < \frac{\star}{14} < 3\frac{9}{14}$

Show solution

1 · Understandwhat's really being asked

I need to count how many whole numbers can replace the $\star$ so that the improper fraction $\frac{\star}{14}$ sits strictly between the mixed numbers 2 and $\frac{5}{14}$ and 3 and $\frac{9}{14}$ .

Givens

The inequality is 2 $\frac{5}{14}$ < $\frac{\star}{14}$ < 3 $\frac{9}{14}$ .
The middle quantity is $\frac{\star}{14}$ , where $\star$ is a whole number.
Both ends are mixed numbers with denominator 14.

Unknowns

How many whole numbers $\star$ make the inequality true.

Constraints

$\star$ must be a whole number.
The inequality is strict, so $\frac{\star}{14}$ cannot equal either endpoint.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List

To compare fairly, rewrite both mixed-number endpoints as improper fractions over 14 so every quantity has the same denominator. Then the numerators can be compared directly and the valid whole numbers listed.

3 · Execute4 carry out the plan

1Rewrite the left endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 2

\frac{5}{14}

to 14ths: 2 wholes is 28 14ths, plus 5 14ths.

2\tfrac{5}{14} = \frac{2\times 14 + 5}{14} = \frac{33}{14}

Matching denominators lets us compare fractions just by their numerators.

2Rewrite the right endpoint

#15 Organize Information in More Ways 3.NF.A.3

Convert 3

\frac{9}{14}

to 14ths: 3 wholes is 42 14ths, plus 9 14ths.

3\tfrac{9}{14} = \frac{3\times 14 + 9}{14} = \frac{51}{14}

Now all three quantities are 14ths, so only the top numbers matter.

3Compare numerators and list

#2 Make a Systematic List 3.NF.A.3

The inequality becomes

\frac{33}{14}

<

\frac{\star}{14}

<

\frac{51}{14}

, so

\star

must be a whole number strictly between 33 and 51: that is 34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50.

33 < \star < 51 \Rightarrow \star \in \{34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50\}

With equal denominators, comparing fractions is the same as comparing whole-number numerators.

4Count them

#2 Make a Systematic List 3.OA.A.1

There are 17 whole numbers in the list.

\{34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50\} \Rightarrow 17

Just count the items in the systematic list.

Answer: 17

4 · Reviewdoes it hold up?

The endpoints 33 and 51 are 18 apart; excluding both ends leaves the 17 whole numbers in between, which matches the count. Each candidate lands strictly between 2 $\frac{5}{14}$ and 3 $\frac{9}{14}$ .

Another way: Number-line reasoning (tool 1): mark $\frac{33}{14}$ and $\frac{51}{14}$ on a line of 14ths; the whole-number numerators strictly between them give the same 17 values.

Standardsmin grade 3

3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning — Converting mixed numbers to 14ths so the fractions can be compared by numerator.
3.OA.A.1 Interpret products of whole numbers as total number of objects in groups — Converting whole parts to 14ths (2 x 14, 3 x 14) and counting the valid values.

💡Takeaway. This only needs Grade 3 fraction sense: make the bottoms match, then just compare the top numbers!

Match fraction forms to compare sizes · 12 practice problems

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