Practice · Group by numerator plus denominator, then count · Sensim Math

Variant 1 easy

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{3}{4}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{3}{4}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{3}{4}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{3}{4}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 3/4

#7 Identify Subproblems 6.NS.C.6

For

\frac{3}{4}

the sum is

7

, so it is in the bracket with sum

7

, the

6th

bracket.

3 + 4 = 7

Brackets with sums

2

to

6

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 5

fractions.

1 + 2 + 3 + \cdots + 5 = 15

15

fractions come before the sum-

7

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

7

bracket starts at denominator

1

and goes up by one each time.

\frac{6}{1}, \frac{5}{2}, \frac{4}{3}, \frac{3}{4}

\frac{3}{4}

has denominator

4

, so it is

4th

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

15 + 4 = 19

\frac{3}{4}

is the

19th

number.

Answer: 19th

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $7$ bracket: through that bracket there are $1 + 2 + \cdots + 6 = 21$ fractions, and $\frac{3}{4}$ is followed by $\frac{2}{5}, \frac{1}{6}$ , so it is number $21 - 2 = 19$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 2 easy

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{4}{5}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{4}{5}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{4}{5}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{4}{5}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 4/5

#7 Identify Subproblems 6.NS.C.6

For

\frac{4}{5}

the sum is

9

, so it is in the bracket with sum

9

, the

8th

bracket.

4 + 5 = 9

Brackets with sums

2

to

8

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 7

fractions.

1 + 2 + 3 + \cdots + 7 = 28

28

fractions come before the sum-

9

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

9

bracket starts at denominator

1

and goes up by one each time.

\frac{8}{1}, \frac{7}{2}, \frac{6}{3}, \frac{5}{4}, \frac{4}{5}

\frac{4}{5}

has denominator

5

, so it is

5th

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

28 + 5 = 33

\frac{4}{5}

is the

33rd

number.

Answer: 33rd

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $9$ bracket: through that bracket there are $1 + 2 + \cdots + 8 = 36$ fractions, and $\frac{4}{5}$ is followed by $\frac{3}{6}, \frac{2}{7}, \frac{1}{8}$ , so it is number $36 - 3 = 33$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 3 easy

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{2}{5}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{2}{5}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{2}{5}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{2}{5}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 2/5

#7 Identify Subproblems 6.NS.C.6

For

\frac{2}{5}

the sum is

7

, so it is in the bracket with sum

7

, the

6th

bracket.

2 + 5 = 7

Brackets with sums

2

to

6

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 5

fractions.

1 + 2 + 3 + \cdots + 5 = 15

15

fractions come before the sum-

7

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

7

bracket starts at denominator

1

and goes up by one each time.

\frac{6}{1}, \frac{5}{2}, \frac{4}{3}, \frac{3}{4}, \frac{2}{5}

\frac{2}{5}

has denominator

5

, so it is

5th

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

15 + 5 = 20

\frac{2}{5}

is the

20th

number.

Answer: 20th

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $7$ bracket: through that bracket there are $1 + 2 + \cdots + 6 = 21$ fractions, and $\frac{2}{5}$ is followed by $\frac{1}{6}$ , so it is number $21 - 1 = 20$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 4 easy

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{5}{7}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{5}{7}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{5}{7}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{5}{7}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 5/7

#7 Identify Subproblems 6.NS.C.6

For

\frac{5}{7}

the sum is

12

, so it is in the bracket with sum

12

, the

11th

bracket.

5 + 7 = 12

Brackets with sums

2

to

11

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 10

fractions.

1 + 2 + 3 + \cdots + 10 = 55

55

fractions come before the sum-

12

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

12

bracket starts at denominator

1

and goes up by one each time.

\frac{11}{1}, \frac{10}{2}, \frac{9}{3}, \frac{8}{4}, \frac{7}{5}, \frac{6}{6}, \frac{5}{7}

\frac{5}{7}

has denominator

7

, so it is

7th

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

55 + 7 = 62

\frac{5}{7}

is the

62nd

number.

Answer: 62nd

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $12$ bracket: through that bracket there are $1 + 2 + \cdots + 11 = 66$ fractions, and $\frac{5}{7}$ is followed by $\frac{4}{8}, \frac{3}{9}, \frac{2}{10}, \frac{1}{11}$ , so it is number $66 - 4 = 62$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 5 medium

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{6}{7}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{6}{7}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{6}{7}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{6}{7}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 6/7

#7 Identify Subproblems 6.NS.C.6

For

\frac{6}{7}

the sum is

13

, so it is in the bracket with sum

13

, the

12th

bracket.

6 + 7 = 13

Brackets with sums

2

to

12

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 11

fractions.

1 + 2 + 3 + \cdots + 11 = 66

66

fractions come before the sum-

13

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

13

bracket starts at denominator

1

and goes up by one each time.

\frac{12}{1}, \frac{11}{2}, \frac{10}{3}, \frac{9}{4}, \frac{8}{5}, \frac{7}{6}, \frac{6}{7}

\frac{6}{7}

has denominator

7

, so it is

7th

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

66 + 7 = 73

\frac{6}{7}

is the

73rd

number.

Answer: 73rd

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $13$ bracket: through that bracket there are $1 + 2 + \cdots + 12 = 78$ fractions, and $\frac{6}{7}$ is followed by $\frac{5}{8}, \frac{4}{9}, \frac{3}{10}, \frac{2}{11}, \frac{1}{12}$ , so it is number $78 - 5 = 73$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 6 medium

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{7}{3}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{7}{3}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{7}{3}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{7}{3}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 7/3

#7 Identify Subproblems 6.NS.C.6

For

\frac{7}{3}

the sum is

10

, so it is in the bracket with sum

10

, the

9th

bracket.

7 + 3 = 10

Brackets with sums

2

to

9

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 8

fractions.

1 + 2 + 3 + \cdots + 8 = 36

36

fractions come before the sum-

10

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

10

bracket starts at denominator

1

and goes up by one each time.

\frac{9}{1}, \frac{8}{2}, \frac{7}{3}

\frac{7}{3}

has denominator

3

, so it is

3rd

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

36 + 3 = 39

\frac{7}{3}

is the

39th

number.

Answer: 39th

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $10$ bracket: through that bracket there are $1 + 2 + \cdots + 9 = 45$ fractions, and $\frac{7}{3}$ is followed by $\frac{6}{4}, \frac{5}{5}, \frac{4}{6}, \frac{3}{7}, \frac{2}{8}, \frac{1}{9}$ , so it is number $45 - 6 = 39$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 7 medium

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{5}{8}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{5}{8}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{5}{8}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{5}{8}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 5/8

#7 Identify Subproblems 6.NS.C.6

For

\frac{5}{8}

the sum is

13

, so it is in the bracket with sum

13

, the

12th

bracket.

5 + 8 = 13

Brackets with sums

2

to

12

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 11

fractions.

1 + 2 + 3 + \cdots + 11 = 66

66

fractions come before the sum-

13

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

13

bracket starts at denominator

1

and goes up by one each time.

\frac{12}{1}, \frac{11}{2}, \frac{10}{3}, \frac{9}{4}, \frac{8}{5}, \frac{7}{6}, \frac{6}{7}, \frac{5}{8}

\frac{5}{8}

has denominator

8

, so it is

8th

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

66 + 8 = 74

\frac{5}{8}

is the

74th

number.

Answer: 74th

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $13$ bracket: through that bracket there are $1 + 2 + \cdots + 12 = 78$ fractions, and $\frac{5}{8}$ is followed by $\frac{4}{9}, \frac{3}{10}, \frac{2}{11}, \frac{1}{12}$ , so it is number $78 - 4 = 74$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 8 medium

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{3}{8}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{3}{8}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{3}{8}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{3}{8}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 3/8

#7 Identify Subproblems 6.NS.C.6

For

\frac{3}{8}

the sum is

11

, so it is in the bracket with sum

11

, the

10th

bracket.

3 + 8 = 11

Brackets with sums

2

to

10

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 9

fractions.

1 + 2 + 3 + \cdots + 9 = 45

45

fractions come before the sum-

11

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

11

bracket starts at denominator

1

and goes up by one each time.

\frac{10}{1}, \frac{9}{2}, \frac{8}{3}, \frac{7}{4}, \frac{6}{5}, \frac{5}{6}, \frac{4}{7}, \frac{3}{8}

\frac{3}{8}

has denominator

8

, so it is

8th

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

45 + 8 = 53

\frac{3}{8}

is the

53rd

number.

Answer: 53rd

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $11$ bracket: through that bracket there are $1 + 2 + \cdots + 10 = 55$ fractions, and $\frac{3}{8}$ is followed by $\frac{2}{9}, \frac{1}{10}$ , so it is number $55 - 2 = 53$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 9 hard

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{8}{3}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{8}{3}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{8}{3}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{8}{3}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 8/3

#7 Identify Subproblems 6.NS.C.6

For

\frac{8}{3}

the sum is

11

, so it is in the bracket with sum

11

, the

10th

bracket.

8 + 3 = 11

Brackets with sums

2

to

10

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 9

fractions.

1 + 2 + 3 + \cdots + 9 = 45

45

fractions come before the sum-

11

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

11

bracket starts at denominator

1

and goes up by one each time.

\frac{10}{1}, \frac{9}{2}, \frac{8}{3}

\frac{8}{3}

has denominator

3

, so it is

3rd

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

45 + 3 = 48

\frac{8}{3}

is the

48th

number.

Answer: 48th

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $11$ bracket: through that bracket there are $1 + 2 + \cdots + 10 = 55$ fractions, and $\frac{8}{3}$ is followed by $\frac{7}{4}, \frac{6}{5}, \frac{5}{6}, \frac{4}{7}, \frac{3}{8}, \frac{2}{9}, \frac{1}{10}$ , so it is number $55 - 7 = 48$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 10 hard

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{4}{9}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{4}{9}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{4}{9}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{4}{9}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 4/9

#7 Identify Subproblems 6.NS.C.6

For

\frac{4}{9}

the sum is

13

, so it is in the bracket with sum

13

, the

12th

bracket.

4 + 9 = 13

Brackets with sums

2

to

12

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 11

fractions.

1 + 2 + 3 + \cdots + 11 = 66

66

fractions come before the sum-

13

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

13

bracket starts at denominator

1

and goes up by one each time.

\frac{12}{1}, \frac{11}{2}, \frac{10}{3}, \frac{9}{4}, \frac{8}{5}, \frac{7}{6}, \frac{6}{7}, \frac{5}{8}, \frac{4}{9}

\frac{4}{9}

has denominator

9

, so it is

9th

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

66 + 9 = 75

\frac{4}{9}

is the

75th

number.

Answer: 75th

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $13$ bracket: through that bracket there are $1 + 2 + \cdots + 12 = 78$ fractions, and $\frac{4}{9}$ is followed by $\frac{3}{10}, \frac{2}{11}, \frac{1}{12}$ , so it is number $78 - 3 = 75$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 11 hard

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{9}{4}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{9}{4}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{9}{4}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{9}{4}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 9/4

#7 Identify Subproblems 6.NS.C.6

For

\frac{9}{4}

the sum is

13

, so it is in the bracket with sum

13

, the

12th

bracket.

9 + 4 = 13

Brackets with sums

2

to

12

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 11

fractions.

1 + 2 + 3 + \cdots + 11 = 66

66

fractions come before the sum-

13

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

13

bracket starts at denominator

1

and goes up by one each time.

\frac{12}{1}, \frac{11}{2}, \frac{10}{3}, \frac{9}{4}

\frac{9}{4}

has denominator

4

, so it is

4th

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

66 + 4 = 70

\frac{9}{4}

is the

70th

number.

Answer: 70th

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $13$ bracket: through that bracket there are $1 + 2 + \cdots + 12 = 78$ fractions, and $\frac{9}{4}$ is followed by $\frac{8}{5}, \frac{7}{6}, \frac{6}{7}, \frac{5}{8}, \frac{4}{9}, \frac{3}{10}, \frac{2}{11}, \frac{1}{12}$ , so it is number $78 - 8 = 70$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Variant 12 hard

The positive rational numbers are listed, without reducing any fraction, in the following order:

$\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \cdots$

In which position in the list is $\frac{7}{9}$ ? (For example, $\frac{1}{3}$ is the $6$ th number.)

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed group by group: each group collects the fractions whose numerator and denominator add to the same number, with denominators $1, 2, 3, \dots$ inside the group. Find where $\frac{7}{9}$ appears.

Givens

The list begins $\left(\frac{1}{1}\right), \left(\frac{2}{1}, \frac{1}{2}\right), \left(\frac{3}{1}, \frac{2}{2}, \frac{1}{3}\right), \dots$
Fractions are not reduced.
$\frac{1}{3}$ is the $6$ th number.

Unknowns

The position of $\frac{7}{9}$ .

Constraints

Because nothing is reduced, $\frac{2}{2}$ and $\frac{1}{1}$ are different entries.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #7 Identify Subproblems#2 Make a Systematic List

First find the rule behind the brackets: the $k$ th bracket holds the $k$ fractions whose numerator plus denominator is $k + 1$ . Then the position of $\frac{7}{9}$ is the number of fractions in all earlier brackets plus its place within its own bracket.

3 · Execute5 carry out the plan

1Find the rule

#5 Look for a Pattern 4.OA.C.5

In each bracket, numerator plus denominator is constant:

2

, then

3

, then

4

. Inside a bracket the denominator goes

1, 2, 3, \dots

while the numerator counts down.

1 + 1 = 2,\quad 2 + 1 = 1 + 2 = 3,\quad 3 + 1 = 2 + 2 = 1 + 3 = 4

The bracket with sum

s

has

s - 1

fractions.

2Locate the bracket of 7/9

#7 Identify Subproblems 6.NS.C.6

For

\frac{7}{9}

the sum is

16

, so it is in the bracket with sum

16

, the

15th

bracket.

7 + 9 = 16

Brackets with sums

2

to

15

all come before it.

3Count the earlier fractions

#5 Look for a Pattern 4.OA.C.5

Those brackets have

1, 2, \dots, 14

fractions.

1 + 2 + 3 + \cdots + 14 = 105

105

fractions come before the sum-

16

bracket.

4Position inside its bracket

#2 Make a Systematic List 6.NS.C.6

The sum-

16

bracket starts at denominator

1

and goes up by one each time.

\frac{15}{1}, \frac{14}{2}, \frac{13}{3}, \frac{12}{4}, \frac{11}{5}, \frac{10}{6}, \frac{9}{7}, \frac{8}{8}, \frac{7}{9}

\frac{7}{9}

has denominator

9

, so it is

9th

in its bracket.

5Add

#7 Identify Subproblems 4.OA.C.5

Add the earlier count and the place in the bracket.

105 + 9 = 114

\frac{7}{9}

is the

114th

number.

Answer: 114th

4 · Reviewdoes it hold up?

Test the method on the given example: $\frac{1}{3}$ has sum $4$ ; brackets with sums $2$ and $3$ hold $1 + 2 = 3$ fractions, and $\frac{1}{3}$ has denominator $3$ , so it is number $3 + 3 = 6$ , matching the problem.

Another way: Count backwards from the end of the sum- $16$ bracket: through that bracket there are $1 + 2 + \cdots + 15 = 120$ fractions, and $\frac{7}{9}$ is followed by $\frac{6}{10}, \frac{5}{11}, \frac{4}{12}, \frac{3}{13}, \frac{2}{14}, \frac{1}{15}$ , so it is number $120 - 6 = 114$ .

Standardsmin grade 6

4.OA.C.5 Generate a number or shape pattern following a given rule — Finding and extending the rule that builds the list.
6.NS.C.6 Rational numbers on the number line and coordinate plane — Treating unreduced fractions as separate entries of positive rational numbers.

💡Takeaway. Big lists are easy to search once you know what each group has in common.

Group by numerator plus denominator, then count · 12 practice problems

Generated variants — 12

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Where next?