Practice · Rewrite both ends over one denominator · Sensim Math

Variant 1 easy

Among the numbers between $-\frac{4}{3}$ and $\frac{3}{4}$ , consider those that can be written as a fraction with denominator $6$ (that is, $\frac{k}{6}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{6}$ that lie strictly between $-\frac{4}{3}$ and $\frac{3}{4}$ .

Givens

The lower end is $-\frac{4}{3}$ .
The upper end is $\frac{3}{4}$ .
Each number counted has the form $\frac{k}{6}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{6}{6} = -1$ or $\frac{0}{6} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{6}$ , so rewrite both ends over $6$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 6

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{4}{3}

by

2

.

-\frac{4}{3} = -\frac{8}{6}

The left end is exactly

-\frac{8}{6}

, and it is not included.

2Trap the right end between fractions over 6

#15 Organize Information in More Ways 6.NS.C.7

\frac{3}{4}

is not a whole number over

6

, so find the two fractions over

6

around it by comparing them all over the common denominator

12

.

\frac{4}{6} = \frac{8}{12} < \frac{9}{12} = \frac{3}{4} < \frac{10}{12} = \frac{5}{6}

\frac{3}{4}

sits between

\frac{4}{6}

and

\frac{5}{6}

, so the largest allowed numerator is

4

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-8 < k

and

\frac{k}{6} < \frac{3}{4}

, so

k

runs from

-7

up to

4

.

-8 < k < \frac{9}{2}$, $k = -7, -6, \dots, 4$; $4 - (-7) + 1 = 12

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{6}

is an integer only when

k

is a multiple of

6

. In the range, those are

k = -6

(giving

-1

) and

k = 0

(giving

0

).

12 - 2 = 10

Two of the

12

fractions are really integers, so they drop out.

Answer: 10

4 · Reviewdoes it hold up?

The interval has length $\frac{4}{3} + \frac{3}{4} = 2\frac{1}{12}$ , and steps of $\frac{1}{6}$ start exactly at the left end, so $2\frac{1}{12} \times 6 = 12\frac{1}{2}$ means $12$ of them fit strictly inside. Removing $-1$ and $0$ leaves $10$ .

Another way: List them on a number line in groups: just $-\frac{7}{6}$ ( $1$ ), from $-\frac{5}{6}$ to $-\frac{1}{6}$ ( $5$ ) and from $\frac{1}{6}$ to $\frac{4}{6}$ ( $4$ ); $1 + 5 + 4 = 10$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $6$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{3}{4}$ with nearby fractions over $6$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 2 easy

Among the numbers between $-\frac{3}{2}$ and $\frac{2}{5}$ , consider those that can be written as a fraction with denominator $6$ (that is, $\frac{k}{6}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{6}$ that lie strictly between $-\frac{3}{2}$ and $\frac{2}{5}$ .

Givens

The lower end is $-\frac{3}{2}$ .
The upper end is $\frac{2}{5}$ .
Each number counted has the form $\frac{k}{6}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{6}{6} = -1$ or $\frac{0}{6} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{6}$ , so rewrite both ends over $6$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 6

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{3}{2}

by

3

.

-\frac{3}{2} = -\frac{9}{6}

The left end is exactly

-\frac{9}{6}

, and it is not included.

2Trap the right end between fractions over 6

#15 Organize Information in More Ways 6.NS.C.7

\frac{2}{5}

is not a whole number over

6

, so find the two fractions over

6

around it by comparing them all over the common denominator

30

.

\frac{2}{6} = \frac{10}{30} < \frac{12}{30} = \frac{2}{5} < \frac{15}{30} = \frac{3}{6}

\frac{2}{5}

sits between

\frac{2}{6}

and

\frac{3}{6}

, so the largest allowed numerator is

2

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-9 < k

and

\frac{k}{6} < \frac{2}{5}

, so

k

runs from

-8

up to

2

.

-9 < k < \frac{12}{5}$, $k = -8, -7, \dots, 2$; $2 - (-8) + 1 = 11

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{6}

is an integer only when

k

is a multiple of

6

. In the range, those are

k = -6

(giving

-1

) and

k = 0

(giving

0

).

11 - 2 = 9

Two of the

11

fractions are really integers, so they drop out.

Answer: 9

4 · Reviewdoes it hold up?

The interval has length $\frac{3}{2} + \frac{2}{5} = 1\frac{9}{10}$ , and steps of $\frac{1}{6}$ start exactly at the left end, so $1\frac{9}{10} \times 6 = 11\frac{2}{5}$ means $11$ of them fit strictly inside. Removing $-1$ and $0$ leaves $9$ .

Another way: List them on a number line in groups: from $-\frac{8}{6}$ to $-\frac{7}{6}$ ( $2$ ), from $-\frac{5}{6}$ to $-\frac{1}{6}$ ( $5$ ) and from $\frac{1}{6}$ to $\frac{2}{6}$ ( $2$ ); $2 + 5 + 2 = 9$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $6$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{2}{5}$ with nearby fractions over $6$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 3 easy

Among the numbers between $-\frac{5}{4}$ and $\frac{1}{3}$ , consider those that can be written as a fraction with denominator $8$ (that is, $\frac{k}{8}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{8}$ that lie strictly between $-\frac{5}{4}$ and $\frac{1}{3}$ .

Givens

The lower end is $-\frac{5}{4}$ .
The upper end is $\frac{1}{3}$ .
Each number counted has the form $\frac{k}{8}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{8}{8} = -1$ or $\frac{0}{8} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{8}$ , so rewrite both ends over $8$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 8

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{5}{4}

by

2

.

-\frac{5}{4} = -\frac{10}{8}

The left end is exactly

-\frac{10}{8}

, and it is not included.

2Trap the right end between fractions over 8

#15 Organize Information in More Ways 6.NS.C.7

\frac{1}{3}

is not a whole number over

8

, so find the two fractions over

8

around it by comparing them all over the common denominator

24

.

\frac{2}{8} = \frac{6}{24} < \frac{8}{24} = \frac{1}{3} < \frac{9}{24} = \frac{3}{8}

\frac{1}{3}

sits between

\frac{2}{8}

and

\frac{3}{8}

, so the largest allowed numerator is

2

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-10 < k

and

\frac{k}{8} < \frac{1}{3}

, so

k

runs from

-9

up to

2

.

-10 < k < \frac{8}{3}$, $k = -9, -8, \dots, 2$; $2 - (-9) + 1 = 12

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{8}

is an integer only when

k

is a multiple of

8

. In the range, those are

k = -8

(giving

-1

) and

k = 0

(giving

0

).

12 - 2 = 10

Two of the

12

fractions are really integers, so they drop out.

Answer: 10

4 · Reviewdoes it hold up?

The interval has length $\frac{5}{4} + \frac{1}{3} = 1\frac{7}{12}$ , and steps of $\frac{1}{8}$ start exactly at the left end, so $1\frac{7}{12} \times 8 = 12\frac{2}{3}$ means $12$ of them fit strictly inside. Removing $-1$ and $0$ leaves $10$ .

Another way: List them on a number line in groups: just $-\frac{9}{8}$ ( $1$ ), from $-\frac{7}{8}$ to $-\frac{1}{8}$ ( $7$ ) and from $\frac{1}{8}$ to $\frac{2}{8}$ ( $2$ ); $1 + 7 + 2 = 10$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $8$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{1}{3}$ with nearby fractions over $8$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 4 easy

Among the numbers between $-\frac{7}{4}$ and $\frac{2}{3}$ , consider those that can be written as a fraction with denominator $8$ (that is, $\frac{k}{8}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{8}$ that lie strictly between $-\frac{7}{4}$ and $\frac{2}{3}$ .

Givens

The lower end is $-\frac{7}{4}$ .
The upper end is $\frac{2}{3}$ .
Each number counted has the form $\frac{k}{8}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{8}{8} = -1$ or $\frac{0}{8} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{8}$ , so rewrite both ends over $8$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 8

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{7}{4}

by

2

.

-\frac{7}{4} = -\frac{14}{8}

The left end is exactly

-\frac{14}{8}

, and it is not included.

2Trap the right end between fractions over 8

#15 Organize Information in More Ways 6.NS.C.7

\frac{2}{3}

is not a whole number over

8

, so find the two fractions over

8

around it by comparing them all over the common denominator

24

.

\frac{5}{8} = \frac{15}{24} < \frac{16}{24} = \frac{2}{3} < \frac{18}{24} = \frac{6}{8}

\frac{2}{3}

sits between

\frac{5}{8}

and

\frac{6}{8}

, so the largest allowed numerator is

5

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-14 < k

and

\frac{k}{8} < \frac{2}{3}

, so

k

runs from

-13

up to

5

.

-14 < k < \frac{16}{3}$, $k = -13, -12, \dots, 5$; $5 - (-13) + 1 = 19

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{8}

is an integer only when

k

is a multiple of

8

. In the range, those are

k = -8

(giving

-1

) and

k = 0

(giving

0

).

19 - 2 = 17

Two of the

19

fractions are really integers, so they drop out.

Answer: 17

4 · Reviewdoes it hold up?

The interval has length $\frac{7}{4} + \frac{2}{3} = 2\frac{5}{12}$ , and steps of $\frac{1}{8}$ start exactly at the left end, so $2\frac{5}{12} \times 8 = 19\frac{1}{3}$ means $19$ of them fit strictly inside. Removing $-1$ and $0$ leaves $17$ .

Another way: List them on a number line in groups: from $-\frac{13}{8}$ to $-\frac{9}{8}$ ( $5$ ), from $-\frac{7}{8}$ to $-\frac{1}{8}$ ( $7$ ) and from $\frac{1}{8}$ to $\frac{5}{8}$ ( $5$ ); $5 + 7 + 5 = 17$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $8$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{2}{3}$ with nearby fractions over $8$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 5 medium

Among the numbers between $-\frac{5}{2}$ and $\frac{3}{7}$ , consider those that can be written as a fraction with denominator $10$ (that is, $\frac{k}{10}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{10}$ that lie strictly between $-\frac{5}{2}$ and $\frac{3}{7}$ .

Givens

The lower end is $-\frac{5}{2}$ .
The upper end is $\frac{3}{7}$ .
Each number counted has the form $\frac{k}{10}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{20}{10} = -2$ or $-\frac{10}{10} = -1$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{10}$ , so rewrite both ends over $10$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 10

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{5}{2}

by

5

.

-\frac{5}{2} = -\frac{25}{10}

The left end is exactly

-\frac{25}{10}

, and it is not included.

2Trap the right end between fractions over 10

#15 Organize Information in More Ways 6.NS.C.7

\frac{3}{7}

is not a whole number over

10

, so find the two fractions over

10

around it by comparing them all over the common denominator

70

.

\frac{4}{10} = \frac{28}{70} < \frac{30}{70} = \frac{3}{7} < \frac{35}{70} = \frac{5}{10}

\frac{3}{7}

sits between

\frac{4}{10}

and

\frac{5}{10}

, so the largest allowed numerator is

4

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-25 < k

and

\frac{k}{10} < \frac{3}{7}

, so

k

runs from

-24

up to

4

.

-25 < k < \frac{30}{7}$, $k = -24, -23, \dots, 4$; $4 - (-24) + 1 = 29

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{10}

is an integer only when

k

is a multiple of

10

. In the range, those are

k = -20

(giving

-2

),

k = -10

(giving

-1

) and

k = 0

(giving

0

).

29 - 3 = 26

Three of the

29

fractions are really integers, so they drop out.

Answer: 26

4 · Reviewdoes it hold up?

The interval has length $\frac{5}{2} + \frac{3}{7} = 2\frac{13}{14}$ , and steps of $\frac{1}{10}$ start exactly at the left end, so $2\frac{13}{14} \times 10 = 29\frac{2}{7}$ means $29$ of them fit strictly inside. Removing $-2$ , $-1$ and $0$ leaves $26$ .

Another way: List them on a number line in groups: from $-\frac{24}{10}$ to $-\frac{21}{10}$ ( $4$ ), from $-\frac{19}{10}$ to $-\frac{11}{10}$ ( $9$ ), from $-\frac{9}{10}$ to $-\frac{1}{10}$ ( $9$ ) and from $\frac{1}{10}$ to $\frac{4}{10}$ ( $4$ ); $4 + 9 + 9 + 4 = 26$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $10$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{3}{7}$ with nearby fractions over $10$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 6 medium

Among the numbers between $-\frac{8}{5}$ and $\frac{1}{3}$ , consider those that can be written as a fraction with denominator $10$ (that is, $\frac{k}{10}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{10}$ that lie strictly between $-\frac{8}{5}$ and $\frac{1}{3}$ .

Givens

The lower end is $-\frac{8}{5}$ .
The upper end is $\frac{1}{3}$ .
Each number counted has the form $\frac{k}{10}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{10}{10} = -1$ or $\frac{0}{10} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{10}$ , so rewrite both ends over $10$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 10

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{8}{5}

by

2

.

-\frac{8}{5} = -\frac{16}{10}

The left end is exactly

-\frac{16}{10}

, and it is not included.

2Trap the right end between fractions over 10

#15 Organize Information in More Ways 6.NS.C.7

\frac{1}{3}

is not a whole number over

10

, so find the two fractions over

10

around it by comparing them all over the common denominator

30

.

\frac{3}{10} = \frac{9}{30} < \frac{10}{30} = \frac{1}{3} < \frac{12}{30} = \frac{4}{10}

\frac{1}{3}

sits between

\frac{3}{10}

and

\frac{4}{10}

, so the largest allowed numerator is

3

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-16 < k

and

\frac{k}{10} < \frac{1}{3}

, so

k

runs from

-15

up to

3

.

-16 < k < \frac{10}{3}$, $k = -15, -14, \dots, 3$; $3 - (-15) + 1 = 19

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{10}

is an integer only when

k

is a multiple of

10

. In the range, those are

k = -10

(giving

-1

) and

k = 0

(giving

0

).

19 - 2 = 17

Two of the

19

fractions are really integers, so they drop out.

Answer: 17

4 · Reviewdoes it hold up?

The interval has length $\frac{8}{5} + \frac{1}{3} = 1\frac{14}{15}$ , and steps of $\frac{1}{10}$ start exactly at the left end, so $1\frac{14}{15} \times 10 = 19\frac{1}{3}$ means $19$ of them fit strictly inside. Removing $-1$ and $0$ leaves $17$ .

Another way: List them on a number line in groups: from $-\frac{15}{10}$ to $-\frac{11}{10}$ ( $5$ ), from $-\frac{9}{10}$ to $-\frac{1}{10}$ ( $9$ ) and from $\frac{1}{10}$ to $\frac{3}{10}$ ( $3$ ); $5 + 9 + 3 = 17$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $10$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{1}{3}$ with nearby fractions over $10$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 7 medium

Among the numbers between $-\frac{7}{5}$ and $\frac{1}{4}$ , consider those that can be written as a fraction with denominator $10$ (that is, $\frac{k}{10}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{10}$ that lie strictly between $-\frac{7}{5}$ and $\frac{1}{4}$ .

Givens

The lower end is $-\frac{7}{5}$ .
The upper end is $\frac{1}{4}$ .
Each number counted has the form $\frac{k}{10}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{10}{10} = -1$ or $\frac{0}{10} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{10}$ , so rewrite both ends over $10$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 10

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{7}{5}

by

2

.

-\frac{7}{5} = -\frac{14}{10}

The left end is exactly

-\frac{14}{10}

, and it is not included.

2Trap the right end between fractions over 10

#15 Organize Information in More Ways 6.NS.C.7

\frac{1}{4}

is not a whole number over

10

, so find the two fractions over

10

around it by comparing them all over the common denominator

20

.

\frac{2}{10} = \frac{4}{20} < \frac{5}{20} = \frac{1}{4} < \frac{6}{20} = \frac{3}{10}

\frac{1}{4}

sits between

\frac{2}{10}

and

\frac{3}{10}

, so the largest allowed numerator is

2

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-14 < k

and

\frac{k}{10} < \frac{1}{4}

, so

k

runs from

-13

up to

2

.

-14 < k < \frac{5}{2}$, $k = -13, -12, \dots, 2$; $2 - (-13) + 1 = 16

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{10}

is an integer only when

k

is a multiple of

10

. In the range, those are

k = -10

(giving

-1

) and

k = 0

(giving

0

).

16 - 2 = 14

Two of the

16

fractions are really integers, so they drop out.

Answer: 14

4 · Reviewdoes it hold up?

The interval has length $\frac{7}{5} + \frac{1}{4} = 1\frac{13}{20}$ , and steps of $\frac{1}{10}$ start exactly at the left end, so $1\frac{13}{20} \times 10 = 16\frac{1}{2}$ means $16$ of them fit strictly inside. Removing $-1$ and $0$ leaves $14$ .

Another way: List them on a number line in groups: from $-\frac{13}{10}$ to $-\frac{11}{10}$ ( $3$ ), from $-\frac{9}{10}$ to $-\frac{1}{10}$ ( $9$ ) and from $\frac{1}{10}$ to $\frac{2}{10}$ ( $2$ ); $3 + 9 + 2 = 14$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $10$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{1}{4}$ with nearby fractions over $10$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 8 hard

Among the numbers between $-\frac{11}{6}$ and $\frac{1}{5}$ , consider those that can be written as a fraction with denominator $12$ (that is, $\frac{k}{12}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{12}$ that lie strictly between $-\frac{11}{6}$ and $\frac{1}{5}$ .

Givens

The lower end is $-\frac{11}{6}$ .
The upper end is $\frac{1}{5}$ .
Each number counted has the form $\frac{k}{12}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{12}{12} = -1$ or $\frac{0}{12} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{12}$ , so rewrite both ends over $12$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 12

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{11}{6}

by

2

.

-\frac{11}{6} = -\frac{22}{12}

The left end is exactly

-\frac{22}{12}

, and it is not included.

2Trap the right end between fractions over 12

#15 Organize Information in More Ways 6.NS.C.7

\frac{1}{5}

is not a whole number over

12

, so find the two fractions over

12

around it by comparing them all over the common denominator

60

.

\frac{2}{12} = \frac{10}{60} < \frac{12}{60} = \frac{1}{5} < \frac{15}{60} = \frac{3}{12}

\frac{1}{5}

sits between

\frac{2}{12}

and

\frac{3}{12}

, so the largest allowed numerator is

2

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-22 < k

and

\frac{k}{12} < \frac{1}{5}

, so

k

runs from

-21

up to

2

.

-22 < k < \frac{12}{5}$, $k = -21, -20, \dots, 2$; $2 - (-21) + 1 = 24

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{12}

is an integer only when

k

is a multiple of

12

. In the range, those are

k = -12

(giving

-1

) and

k = 0

(giving

0

).

24 - 2 = 22

Two of the

24

fractions are really integers, so they drop out.

Answer: 22

4 · Reviewdoes it hold up?

The interval has length $\frac{11}{6} + \frac{1}{5} = 2\frac{1}{30}$ , and steps of $\frac{1}{12}$ start exactly at the left end, so $2\frac{1}{30} \times 12 = 24\frac{2}{5}$ means $24$ of them fit strictly inside. Removing $-1$ and $0$ leaves $22$ .

Another way: List them on a number line in groups: from $-\frac{21}{12}$ to $-\frac{13}{12}$ ( $9$ ), from $-\frac{11}{12}$ to $-\frac{1}{12}$ ( $11$ ) and from $\frac{1}{12}$ to $\frac{2}{12}$ ( $2$ ); $9 + 11 + 2 = 22$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $12$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{1}{5}$ with nearby fractions over $12$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 9 medium

Among the numbers between $-\frac{5}{3}$ and $\frac{2}{5}$ , consider those that can be written as a fraction with denominator $12$ (that is, $\frac{k}{12}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{12}$ that lie strictly between $-\frac{5}{3}$ and $\frac{2}{5}$ .

Givens

The lower end is $-\frac{5}{3}$ .
The upper end is $\frac{2}{5}$ .
Each number counted has the form $\frac{k}{12}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{12}{12} = -1$ or $\frac{0}{12} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{12}$ , so rewrite both ends over $12$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 12

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{5}{3}

by

4

.

-\frac{5}{3} = -\frac{20}{12}

The left end is exactly

-\frac{20}{12}

, and it is not included.

2Trap the right end between fractions over 12

#15 Organize Information in More Ways 6.NS.C.7

\frac{2}{5}

is not a whole number over

12

, so find the two fractions over

12

around it by comparing them all over the common denominator

60

.

\frac{4}{12} = \frac{20}{60} < \frac{24}{60} = \frac{2}{5} < \frac{25}{60} = \frac{5}{12}

\frac{2}{5}

sits between

\frac{4}{12}

and

\frac{5}{12}

, so the largest allowed numerator is

4

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-20 < k

and

\frac{k}{12} < \frac{2}{5}

, so

k

runs from

-19

up to

4

.

-20 < k < \frac{24}{5}$, $k = -19, -18, \dots, 4$; $4 - (-19) + 1 = 24

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{12}

is an integer only when

k

is a multiple of

12

. In the range, those are

k = -12

(giving

-1

) and

k = 0

(giving

0

).

24 - 2 = 22

Two of the

24

fractions are really integers, so they drop out.

Answer: 22

4 · Reviewdoes it hold up?

The interval has length $\frac{5}{3} + \frac{2}{5} = 2\frac{1}{15}$ , and steps of $\frac{1}{12}$ start exactly at the left end, so $2\frac{1}{15} \times 12 = 24\frac{4}{5}$ means $24$ of them fit strictly inside. Removing $-1$ and $0$ leaves $22$ .

Another way: List them on a number line in groups: from $-\frac{19}{12}$ to $-\frac{13}{12}$ ( $7$ ), from $-\frac{11}{12}$ to $-\frac{1}{12}$ ( $11$ ) and from $\frac{1}{12}$ to $\frac{4}{12}$ ( $4$ ); $7 + 11 + 4 = 22$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $12$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{2}{5}$ with nearby fractions over $12$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 10 hard

Among the numbers between $-\frac{10}{7}$ and $\frac{3}{5}$ , consider those that can be written as a fraction with denominator $14$ (that is, $\frac{k}{14}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{14}$ that lie strictly between $-\frac{10}{7}$ and $\frac{3}{5}$ .

Givens

The lower end is $-\frac{10}{7}$ .
The upper end is $\frac{3}{5}$ .
Each number counted has the form $\frac{k}{14}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{14}{14} = -1$ or $\frac{0}{14} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{14}$ , so rewrite both ends over $14$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 14

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{10}{7}

by

2

.

-\frac{10}{7} = -\frac{20}{14}

The left end is exactly

-\frac{20}{14}

, and it is not included.

2Trap the right end between fractions over 14

#15 Organize Information in More Ways 6.NS.C.7

\frac{3}{5}

is not a whole number over

14

, so find the two fractions over

14

around it by comparing them all over the common denominator

70

.

\frac{8}{14} = \frac{40}{70} < \frac{42}{70} = \frac{3}{5} < \frac{45}{70} = \frac{9}{14}

\frac{3}{5}

sits between

\frac{8}{14}

and

\frac{9}{14}

, so the largest allowed numerator is

8

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-20 < k

and

\frac{k}{14} < \frac{3}{5}

, so

k

runs from

-19

up to

8

.

-20 < k < \frac{42}{5}$, $k = -19, -18, \dots, 8$; $8 - (-19) + 1 = 28

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{14}

is an integer only when

k

is a multiple of

14

. In the range, those are

k = -14

(giving

-1

) and

k = 0

(giving

0

).

28 - 2 = 26

Two of the

28

fractions are really integers, so they drop out.

Answer: 26

4 · Reviewdoes it hold up?

The interval has length $\frac{10}{7} + \frac{3}{5} = 2\frac{1}{35}$ , and steps of $\frac{1}{14}$ start exactly at the left end, so $2\frac{1}{35} \times 14 = 28\frac{2}{5}$ means $28$ of them fit strictly inside. Removing $-1$ and $0$ leaves $26$ .

Another way: List them on a number line in groups: from $-\frac{19}{14}$ to $-\frac{15}{14}$ ( $5$ ), from $-\frac{13}{14}$ to $-\frac{1}{14}$ ( $13$ ) and from $\frac{1}{14}$ to $\frac{8}{14}$ ( $8$ ); $5 + 13 + 8 = 26$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $14$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{3}{5}$ with nearby fractions over $14$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 11 hard

Among the numbers between $-\frac{9}{7}$ and $\frac{1}{3}$ , consider those that can be written as a fraction with denominator $14$ (that is, $\frac{k}{14}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{14}$ that lie strictly between $-\frac{9}{7}$ and $\frac{1}{3}$ .

Givens

The lower end is $-\frac{9}{7}$ .
The upper end is $\frac{1}{3}$ .
Each number counted has the form $\frac{k}{14}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{14}{14} = -1$ or $\frac{0}{14} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{14}$ , so rewrite both ends over $14$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 14

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{9}{7}

by

2

.

-\frac{9}{7} = -\frac{18}{14}

The left end is exactly

-\frac{18}{14}

, and it is not included.

2Trap the right end between fractions over 14

#15 Organize Information in More Ways 6.NS.C.7

\frac{1}{3}

is not a whole number over

14

, so find the two fractions over

14

around it by comparing them all over the common denominator

42

.

\frac{4}{14} = \frac{12}{42} < \frac{14}{42} = \frac{1}{3} < \frac{15}{42} = \frac{5}{14}

\frac{1}{3}

sits between

\frac{4}{14}

and

\frac{5}{14}

, so the largest allowed numerator is

4

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-18 < k

and

\frac{k}{14} < \frac{1}{3}

, so

k

runs from

-17

up to

4

.

-18 < k < \frac{14}{3}$, $k = -17, -16, \dots, 4$; $4 - (-17) + 1 = 22

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{14}

is an integer only when

k

is a multiple of

14

. In the range, those are

k = -14

(giving

-1

) and

k = 0

(giving

0

).

22 - 2 = 20

Two of the

22

fractions are really integers, so they drop out.

Answer: 20

4 · Reviewdoes it hold up?

The interval has length $\frac{9}{7} + \frac{1}{3} = 1\frac{13}{21}$ , and steps of $\frac{1}{14}$ start exactly at the left end, so $1\frac{13}{21} \times 14 = 22\frac{2}{3}$ means $22$ of them fit strictly inside. Removing $-1$ and $0$ leaves $20$ .

Another way: List them on a number line in groups: from $-\frac{17}{14}$ to $-\frac{15}{14}$ ( $3$ ), from $-\frac{13}{14}$ to $-\frac{1}{14}$ ( $13$ ) and from $\frac{1}{14}$ to $\frac{4}{14}$ ( $4$ ); $3 + 13 + 4 = 20$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $14$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{1}{3}$ with nearby fractions over $14$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Variant 12 hard

Among the numbers between $-\frac{11}{8}$ and $\frac{1}{3}$ , consider those that can be written as a fraction with denominator $16$ (that is, $\frac{k}{16}$ for some integer $k$ ). How many of them are not integers?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the non-integer fractions $\frac{k}{16}$ that lie strictly between $-\frac{11}{8}$ and $\frac{1}{3}$ .

Givens

The lower end is $-\frac{11}{8}$ .
The upper end is $\frac{1}{3}$ .
Each number counted has the form $\frac{k}{16}$ with $k$ an integer.

Unknowns

How many such numbers are not integers.

Constraints

Both ends are excluded ("between").
A number like $-\frac{16}{16} = -1$ or $\frac{0}{16} = 0$ is an integer and must not be counted.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#16 Count the Complement

Every candidate is a whole number of steps of size $\frac{1}{16}$ , so rewrite both ends over $16$ and count the possible numerators $k$ . Then take away the numerators that make an integer.

3 · Execute4 carry out the plan

1Write the left end over 16

#15 Organize Information in More Ways 6.NS.C.6

Multiply top and bottom of

\frac{11}{8}

by

2

.

-\frac{11}{8} = -\frac{22}{16}

The left end is exactly

-\frac{22}{16}

, and it is not included.

2Trap the right end between fractions over 16

#15 Organize Information in More Ways 6.NS.C.7

\frac{1}{3}

is not a whole number over

16

, so find the two fractions over

16

around it by comparing them all over the common denominator

48

.

\frac{5}{16} = \frac{15}{48} < \frac{16}{48} = \frac{1}{3} < \frac{18}{48} = \frac{6}{16}

\frac{1}{3}

sits between

\frac{5}{16}

and

\frac{6}{16}

, so the largest allowed numerator is

5

.

3Count the numerators

#2 Make a Systematic List 6.NS.C.7

The numerator

k

must be an integer with

-22 < k

and

\frac{k}{16} < \frac{1}{3}

, so

k

runs from

-21

up to

5

.

-22 < k < \frac{16}{3}$, $k = -21, -20, \dots, 5$; $5 - (-21) + 1 = 27

Counting from one integer to another, inclusive, is last minus first plus one.

4Remove the integers

#16 Count the Complement 6.NS.C.6

\frac{k}{16}

is an integer only when

k

is a multiple of

16

. In the range, those are

k = -16

(giving

-1

) and

k = 0

(giving

0

).

27 - 2 = 25

Two of the

27

fractions are really integers, so they drop out.

Answer: 25

4 · Reviewdoes it hold up?

The interval has length $\frac{11}{8} + \frac{1}{3} = 1\frac{17}{24}$ , and steps of $\frac{1}{16}$ start exactly at the left end, so $1\frac{17}{24} \times 16 = 27\frac{1}{3}$ means $27$ of them fit strictly inside. Removing $-1$ and $0$ leaves $25$ .

Another way: List them on a number line in groups: from $-\frac{21}{16}$ to $-\frac{17}{16}$ ( $5$ ), from $-\frac{15}{16}$ to $-\frac{1}{16}$ ( $15$ ) and from $\frac{1}{16}$ to $\frac{5}{16}$ ( $5$ ); $5 + 15 + 5 = 25$ .

Standardsmin grade 6

6.NS.C.6 Rational numbers on the number line and coordinate plane — Placing fractions with denominator $16$ on the number line and recognizing which are integers.
6.NS.C.7 Order and absolute value of rational numbers — Comparing $\frac{1}{3}$ with nearby fractions over $16$.

💡Takeaway. When every candidate shares one denominator, turn the question into counting numerators.

Rewrite both ends over one denominator · 12 practice problems

Generated variants — 12

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Where next?