Practice · Make each side a single number before comparing anything · Sensim Math

Variant 1 easy

Find every natural number that $\blacksquare$ can be.

$8\frac{2}{3} \div 1\frac{8}{12} < 18 \div \frac{6}{\blacksquare} < 21$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 21: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 8 and 2/3 divided by 1 and 8/12.
The middle is 18 divided by 6 over the box.
The right end is 21.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

8\frac{2}{3} \div 1\frac{8}{12} = \frac{26}{5}

The left end is 5 and 1/5.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 6 over the box multiplies by the box over 6, so the box comes out as a plain factor.

18 \div \frac{6}{\blacksquare} = 3 \times \blacksquare

The middle is 3 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

3 is positive, so the signs keep pointing the same way and the box stands alone.

1\frac{11}{15} < \blacksquare < 7

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 2, 3, 4, 5, 6

5 values work.

Answer: 2, 3, 4, 5, 6

4 · Reviewdoes it hold up?

The list stops where it should: $6 < 7 < 7$ at the top, and $1 < 1\frac{11}{15} < 2$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 2 easy

Find every natural number that $\blacksquare$ can be.

$11\frac{2}{10} \div 4\frac{8}{10} < 20 \div \frac{4}{\blacksquare} < 26$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 26: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 11 and 2/10 divided by 4 and 8/10.
The middle is 20 divided by 4 over the box.
The right end is 26.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

11\frac{2}{10} \div 4\frac{8}{10} = \frac{7}{3}

The left end is 2 and 1/3.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 4 over the box multiplies by the box over 4, so the box comes out as a plain factor.

20 \div \frac{4}{\blacksquare} = 5 \times \blacksquare

The middle is 5 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

5 is positive, so the signs keep pointing the same way and the box stands alone.

\frac{7}{15} < \blacksquare < 5\frac{1}{5}

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 1, 2, 3, 4, 5

5 values work.

Answer: 1, 2, 3, 4, 5

4 · Reviewdoes it hold up?

The list stops where it should: $5 < 5\frac{1}{5} < 6$ at the top, and $0 < \frac{7}{15} < 1$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 3 easy

Find every natural number that $\blacksquare$ can be.

$12\frac{3}{5} \div 1\frac{3}{6} < 28 \div \frac{10}{\blacksquare} < 24$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 24: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 12 and 3/5 divided by 1 and 3/6.
The middle is 28 divided by 10 over the box.
The right end is 24.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

12\frac{3}{5} \div 1\frac{3}{6} = \frac{42}{5}

The left end is 8 and 2/5.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 10 over the box multiplies by the box over 10, so the box comes out as a plain factor.

28 \div \frac{10}{\blacksquare} = 2\frac{4}{5} \times \blacksquare

The middle is 2 and 4/5 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

2 and 4/5 is positive, so the signs keep pointing the same way and the box stands alone.

3 < \blacksquare < 8\frac{4}{7}

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 4, 5, 6, 7, 8

5 values work.

Answer: 4, 5, 6, 7, 8

4 · Reviewdoes it hold up?

The list stops where it should: $8 < 8\frac{4}{7} < 9$ at the top, and $3 < 3 < 4$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 4 medium

Find every natural number that $\blacksquare$ can be.

$9\frac{6}{8} \div 3\frac{2}{8} < 30 \div \frac{5}{\blacksquare} < 27$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 27: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 9 and 6/8 divided by 3 and 2/8.
The middle is 30 divided by 5 over the box.
The right end is 27.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

9\frac{6}{8} \div 3\frac{2}{8} = 3

The left end is 3.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 5 over the box multiplies by the box over 5, so the box comes out as a plain factor.

30 \div \frac{5}{\blacksquare} = 6 \times \blacksquare

The middle is 6 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

6 is positive, so the signs keep pointing the same way and the box stands alone.

\frac{1}{2} < \blacksquare < 4\frac{1}{2}

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 1, 2, 3, 4

4 values work.

Answer: 1, 2, 3, 4

4 · Reviewdoes it hold up?

The list stops where it should: $4 < 4\frac{1}{2} < 5$ at the top, and $0 < \frac{1}{2} < 1$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 5 easy

Find every natural number that $\blacksquare$ can be.

$10\frac{5}{6} \div 3\frac{1}{4} < 30 \div \frac{4}{\blacksquare} < 30$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 30: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 10 and 5/6 divided by 3 and 1/4.
The middle is 30 divided by 4 over the box.
The right end is 30.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

10\frac{5}{6} \div 3\frac{1}{4} = \frac{10}{3}

The left end is 3 and 1/3.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 4 over the box multiplies by the box over 4, so the box comes out as a plain factor.

30 \div \frac{4}{\blacksquare} = 7\frac{1}{2} \times \blacksquare

The middle is 7 and 1/2 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

7 and 1/2 is positive, so the signs keep pointing the same way and the box stands alone.

\frac{4}{9} < \blacksquare < 4

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 1, 2, 3

3 values work.

Answer: 1, 2, 3

4 · Reviewdoes it hold up?

The list stops where it should: $3 < 4 < 4$ at the top, and $0 < \frac{4}{9} < 1$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 6 medium

Find every natural number that $\blacksquare$ can be.

$5\frac{5}{6} \div 3\frac{3}{4} < 28 \div \frac{6}{\blacksquare} < 30$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 30: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 5 and 5/6 divided by 3 and 3/4.
The middle is 28 divided by 6 over the box.
The right end is 30.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

5\frac{5}{6} \div 3\frac{3}{4} = \frac{14}{9}

The left end is 1 and 5/9.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 6 over the box multiplies by the box over 6, so the box comes out as a plain factor.

28 \div \frac{6}{\blacksquare} = 4\frac{2}{3} \times \blacksquare

The middle is 4 and 2/3 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

4 and 2/3 is positive, so the signs keep pointing the same way and the box stands alone.

\frac{1}{3} < \blacksquare < 6\frac{3}{7}

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 1, 2, 3, 4, 5, 6

6 values work.

Answer: 1, 2, 3, 4, 5, 6

4 · Reviewdoes it hold up?

The list stops where it should: $6 < 6\frac{3}{7} < 7$ at the top, and $0 < \frac{1}{3} < 1$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 7 medium

Find every natural number that $\blacksquare$ can be.

$9\frac{4}{5} \div 1\frac{4}{10} < 33 \div \frac{6}{\blacksquare} < 32$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 32: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 9 and 4/5 divided by 1 and 4/10.
The middle is 33 divided by 6 over the box.
The right end is 32.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

9\frac{4}{5} \div 1\frac{4}{10} = 7

The left end is 7.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 6 over the box multiplies by the box over 6, so the box comes out as a plain factor.

33 \div \frac{6}{\blacksquare} = 5\frac{1}{2} \times \blacksquare

The middle is 5 and 1/2 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

5 and 1/2 is positive, so the signs keep pointing the same way and the box stands alone.

1\frac{3}{11} < \blacksquare < 5\frac{9}{11}

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 2, 3, 4, 5

4 values work.

Answer: 2, 3, 4, 5

4 · Reviewdoes it hold up?

The list stops where it should: $5 < 5\frac{9}{11} < 6$ at the top, and $1 < 1\frac{3}{11} < 2$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 8 medium

Find every natural number that $\blacksquare$ can be.

$7\frac{3}{6} \div 1\frac{4}{12} < 20 \div \frac{4}{\blacksquare} < 35$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 35: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 7 and 3/6 divided by 1 and 4/12.
The middle is 20 divided by 4 over the box.
The right end is 35.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

7\frac{3}{6} \div 1\frac{4}{12} = \frac{45}{8}

The left end is 5 and 5/8.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 4 over the box multiplies by the box over 4, so the box comes out as a plain factor.

20 \div \frac{4}{\blacksquare} = 5 \times \blacksquare

The middle is 5 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

5 is positive, so the signs keep pointing the same way and the box stands alone.

1\frac{1}{8} < \blacksquare < 7

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 2, 3, 4, 5, 6

5 values work.

Answer: 2, 3, 4, 5, 6

4 · Reviewdoes it hold up?

The list stops where it should: $6 < 7 < 7$ at the top, and $1 < 1\frac{1}{8} < 2$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 9 hard

Find every natural number that $\blacksquare$ can be.

$8\frac{1}{10} \div 2\frac{4}{5} < 36 \div \frac{7}{\blacksquare} < 30$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 30: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 8 and 1/10 divided by 2 and 4/5.
The middle is 36 divided by 7 over the box.
The right end is 30.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

8\frac{1}{10} \div 2\frac{4}{5} = \frac{81}{28}

The left end is 2 and 25/28.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 7 over the box multiplies by the box over 7, so the box comes out as a plain factor.

36 \div \frac{7}{\blacksquare} = 5\frac{1}{7} \times \blacksquare

The middle is 5 and 1/7 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

5 and 1/7 is positive, so the signs keep pointing the same way and the box stands alone.

\frac{9}{16} < \blacksquare < 5\frac{5}{6}

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 1, 2, 3, 4, 5

5 values work.

Answer: 1, 2, 3, 4, 5

4 · Reviewdoes it hold up?

The list stops where it should: $5 < 5\frac{5}{6} < 6$ at the top, and $0 < \frac{9}{16} < 1$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 10 hard

Find every natural number that $\blacksquare$ can be.

$10\frac{1}{2} \div 4\frac{4}{8} < 40 \div \frac{8}{\blacksquare} < 35$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 35: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 10 and 1/2 divided by 4 and 4/8.
The middle is 40 divided by 8 over the box.
The right end is 35.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

10\frac{1}{2} \div 4\frac{4}{8} = \frac{7}{3}

The left end is 2 and 1/3.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 8 over the box multiplies by the box over 8, so the box comes out as a plain factor.

40 \div \frac{8}{\blacksquare} = 5 \times \blacksquare

The middle is 5 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

5 is positive, so the signs keep pointing the same way and the box stands alone.

\frac{7}{15} < \blacksquare < 7

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 1, 2, 3, 4, 5, 6

6 values work.

Answer: 1, 2, 3, 4, 5, 6

4 · Reviewdoes it hold up?

The list stops where it should: $6 < 7 < 7$ at the top, and $0 < \frac{7}{15} < 1$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 11 hard

Find every natural number that $\blacksquare$ can be.

$9\frac{1}{6} \div 2\frac{3}{4} < 42 \div \frac{7}{\blacksquare} < 28$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 28: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 9 and 1/6 divided by 2 and 3/4.
The middle is 42 divided by 7 over the box.
The right end is 28.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

9\frac{1}{6} \div 2\frac{3}{4} = \frac{10}{3}

The left end is 3 and 1/3.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 7 over the box multiplies by the box over 7, so the box comes out as a plain factor.

42 \div \frac{7}{\blacksquare} = 6 \times \blacksquare

The middle is 6 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

6 is positive, so the signs keep pointing the same way and the box stands alone.

\frac{5}{9} < \blacksquare < 4\frac{2}{3}

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 1, 2, 3, 4

4 values work.

Answer: 1, 2, 3, 4

4 · Reviewdoes it hold up?

The list stops where it should: $4 < 4\frac{2}{3} < 5$ at the top, and $0 < \frac{5}{9} < 1$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Variant 12 hard

Find every natural number that $\blacksquare$ can be.

$10\frac{5}{10} \div 2\frac{1}{3} < 45 \div \frac{4}{\blacksquare} < 24$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A quotient of two mixed numbers, an expression holding the box, and 24: the middle has to sit strictly between the other two. We want every natural number the box can be.

Givens

The left end is 10 and 5/10 divided by 2 and 1/3.
The middle is 45 divided by 4 over the box.
The right end is 24.

Unknowns

Every natural number the box can hold.

Constraints

Both signs are strict, so neither end value is allowed.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Nothing here is a number yet. Simplify each part in turn -- the left into one fraction, the middle into a multiple of the box -- and only then is there an inequality to read.

3 · Execute4 carry out the plan

1Simplify the left end

#7 Identify Subproblems 6.NS.A.1

Two mixed numbers divided: write both as improper fractions and multiply by the reciprocal.

10\frac{5}{10} \div 2\frac{1}{3} = \frac{9}{2}

The left end is 4 and 1/2.

2Simplify the middle

#13 Convert to Algebra 6.NS.A.1

Dividing by 4 over the box multiplies by the box over 4, so the box comes out as a plain factor.

45 \div \frac{4}{\blacksquare} = 11\frac{1}{4} \times \blacksquare

The middle is 11 and 1/4 times the box.

3Divide every part by that factor

#13 Convert to Algebra 6.EE.B.8

11 and 1/4 is positive, so the signs keep pointing the same way and the box stands alone.

\frac{2}{5} < \blacksquare < 2\frac{2}{15}

Now it is a plain range.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.8

Both ends are excluded, so take only what falls strictly between them.

\blacksquare = 1, 2

2 values work.

Answer: 1, 2

4 · Reviewdoes it hold up?

The list stops where it should: $2 < 2\frac{2}{15} < 3$ at the top, and $0 < \frac{2}{5} < 1$ at the bottom.

Another way: Substituting 1, 2, 3 and so on into the middle expression and testing each against both ends gives the same list, with no way of knowing when to stop.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Simplifying each division of fractions.
6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Reducing the chained inequality until the box stands alone.

💡Takeaway. You cannot compare things that are still sums and quotients. Turn each one into a number first.

Make each side a single number before comparing anything · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?