← Make each side a single number before comparing anything · Inequality Range Membership

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

6.NS.A.16.EE.B.8

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 2, 3, 4, 5, 6

Find every natural number that ■\blacksquare can be.

823÷1812<18÷6■<218\frac{2}{3} \div 1\frac{8}{12} < 18 \div \frac{6}{\blacksquare} < 21

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.
823÷1812=2658\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÷6■=3×■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.
11115<■<71\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.
■=2,3,4,5,6\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<76 < 7 < 7 at the top, and 1<11115<21 < 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 answer: 1, 2, 3, 4, 5

Find every natural number that ■\blacksquare can be.

11210÷4810<20÷4■<2611\frac{2}{10} \div 4\frac{8}{10} < 20 \div \frac{4}{\blacksquare} < 26

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.
11210÷4810=7311\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÷4■=5×■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.
715<■<515\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.
■=1,2,3,4,5\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<515<65 < 5\frac{1}{5} < 6 at the top, and 0<715<10 < \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 answer: 4, 5, 6, 7, 8

Find every natural number that ■\blacksquare can be.

1235÷136<28÷10■<2412\frac{3}{5} \div 1\frac{3}{6} < 28 \div \frac{10}{\blacksquare} < 24

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.
1235÷136=42512\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÷10■=245×■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<■<8473 < \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.
■=4,5,6,7,8\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<847<98 < 8\frac{4}{7} < 9 at the top, and 3<3<43 < 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 answer: 1, 2, 3, 4

Find every natural number that ■\blacksquare can be.

968÷328<30÷5■<279\frac{6}{8} \div 3\frac{2}{8} < 30 \div \frac{5}{\blacksquare} < 27

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.
968÷328=39\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÷5■=6×■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.
12<■<412\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.
■=1,2,3,4\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<412<54 < 4\frac{1}{2} < 5 at the top, and 0<12<10 < \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 answer: 1, 2, 3

Find every natural number that ■\blacksquare can be.

1056÷314<30÷4■<3010\frac{5}{6} \div 3\frac{1}{4} < 30 \div \frac{4}{\blacksquare} < 30

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.
1056÷314=10310\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÷4■=712×■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.
49<■<4\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.
■=1,2,3\blacksquare = 1, 2, 3
3 values work.
Answer: 1, 2, 3
4 · Reviewdoes it hold up?

The list stops where it should: 3<4<43 < 4 < 4 at the top, and 0<49<10 < \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 answer: 1, 2, 3, 4, 5, 6

Find every natural number that ■\blacksquare can be.

556÷334<28÷6■<305\frac{5}{6} \div 3\frac{3}{4} < 28 \div \frac{6}{\blacksquare} < 30

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.
556÷334=1495\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÷6■=423×■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.
13<■<637\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.
■=1,2,3,4,5,6\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<637<76 < 6\frac{3}{7} < 7 at the top, and 0<13<10 < \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 answer: 2, 3, 4, 5

Find every natural number that ■\blacksquare can be.

945÷1410<33÷6■<329\frac{4}{5} \div 1\frac{4}{10} < 33 \div \frac{6}{\blacksquare} < 32

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.
945÷1410=79\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÷6■=512×■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.
1311<■<59111\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.
■=2,3,4,5\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<5911<65 < 5\frac{9}{11} < 6 at the top, and 1<1311<21 < 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 answer: 2, 3, 4, 5, 6

Find every natural number that ■\blacksquare can be.

736÷1412<20÷4■<357\frac{3}{6} \div 1\frac{4}{12} < 20 \div \frac{4}{\blacksquare} < 35

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.
736÷1412=4587\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÷4■=5×■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.
118<■<71\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.
■=2,3,4,5,6\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<76 < 7 < 7 at the top, and 1<118<21 < 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 answer: 1, 2, 3, 4, 5

Find every natural number that ■\blacksquare can be.

8110÷245<36÷7■<308\frac{1}{10} \div 2\frac{4}{5} < 36 \div \frac{7}{\blacksquare} < 30

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.
8110÷245=81288\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÷7■=517×■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.
916<■<556\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.
■=1,2,3,4,5\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<556<65 < 5\frac{5}{6} < 6 at the top, and 0<916<10 < \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 answer: 1, 2, 3, 4, 5, 6

Find every natural number that ■\blacksquare can be.

1012÷448<40÷8■<3510\frac{1}{2} \div 4\frac{4}{8} < 40 \div \frac{8}{\blacksquare} < 35

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.
1012÷448=7310\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÷8■=5×■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.
715<■<7\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.
■=1,2,3,4,5,6\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<76 < 7 < 7 at the top, and 0<715<10 < \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 answer: 1, 2, 3, 4

Find every natural number that ■\blacksquare can be.

916÷234<42÷7■<289\frac{1}{6} \div 2\frac{3}{4} < 42 \div \frac{7}{\blacksquare} < 28

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.
916÷234=1039\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÷7■=6×■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.
59<■<423\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.
■=1,2,3,4\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<423<54 < 4\frac{2}{3} < 5 at the top, and 0<59<10 < \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 answer: 1, 2

Find every natural number that ■\blacksquare can be.

10510÷213<45÷4■<2410\frac{5}{10} \div 2\frac{1}{3} < 45 \div \frac{4}{\blacksquare} < 24

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.
10510÷213=9210\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÷4■=1114×■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.
25<■<2215\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.
■=1,2\blacksquare = 1, 2
2 values work.
Answer: 1, 2
4 · Reviewdoes it hold up?

The list stops where it should: 2<2215<32 < 2\frac{2}{15} < 3 at the top, and 0<25<10 < \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.