Practice · Multiplying every part by the same positive number keeps the order · Sensim Math

Variant 1 easy

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

$13 \div 10 < \blacksquare \div 12 < 15 \div 10$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 12 has to sit strictly between 13 over 10 and 15 over 10. We want every natural number that works.

Givens

The left end is 13 divided by 10.
The right end is 15 divided by 10.
The middle is the box divided by 12.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 12 leaves the inequality pointing the same way because 12 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

13 \div 10 = 1.3,\quad 15 \div 10 = 1.5

The middle sits between 1.3 and 1.5.

2Multiply every part by 12

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

12 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

1.3 \times 12 < \blacksquare < 1.5 \times 12

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

15.6 < \blacksquare < 18

Strictly between 15.6 and 18.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 16, 17

2 values work.

Answer: 16, 17

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 16 sits above 15.6 and 17 below 18, while 15 and 18 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 2 easy

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

$10 \div 8 < \blacksquare \div 6 < 20 \div 10$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 6 has to sit strictly between 10 over 8 and 20 over 10. We want every natural number that works.

Givens

The left end is 10 divided by 8.
The right end is 20 divided by 10.
The middle is the box divided by 6.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 6 leaves the inequality pointing the same way because 6 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

10 \div 8 = 1.25,\quad 20 \div 10 = 2

The middle sits between 1.25 and 2.

2Multiply every part by 6

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

6 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

1.25 \times 6 < \blacksquare < 2 \times 6

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

7.5 < \blacksquare < 12

Strictly between 7.5 and 12.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 8, 9, 10, 11

4 values work.

Answer: 8, 9, 10, 11

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 8 sits above 7.5 and 11 below 12, while 7 and 12 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 3 easy

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

$13 \div 5 < \blacksquare \div 14 < 22 \div 8$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 14 has to sit strictly between 13 over 5 and 22 over 8. We want every natural number that works.

Givens

The left end is 13 divided by 5.
The right end is 22 divided by 8.
The middle is the box divided by 14.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 14 leaves the inequality pointing the same way because 14 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

13 \div 5 = 2.6,\quad 22 \div 8 = 2.75

The middle sits between 2.6 and 2.75.

2Multiply every part by 14

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

14 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

2.6 \times 14 < \blacksquare < 2.75 \times 14

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

36.4 < \blacksquare < 38.5

Strictly between 36.4 and 38.5.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 37, 38

2 values work.

Answer: 37, 38

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 37 sits above 36.4 and 38 below 38.5, while 36 and 39 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 4 easy

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

$22 \div 20 < \blacksquare \div 6 < 6 \div 4$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 6 has to sit strictly between 22 over 20 and 6 over 4. We want every natural number that works.

Givens

The left end is 22 divided by 20.
The right end is 6 divided by 4.
The middle is the box divided by 6.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 6 leaves the inequality pointing the same way because 6 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

22 \div 20 = 1.1,\quad 6 \div 4 = 1.5

The middle sits between 1.1 and 1.5.

2Multiply every part by 6

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

6 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

1.1 \times 6 < \blacksquare < 1.5 \times 6

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

6.6 < \blacksquare < 9

Strictly between 6.6 and 9.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 7, 8

2 values work.

Answer: 7, 8

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 7 sits above 6.6 and 8 below 9, while 6 and 9 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 5 medium

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

$24 \div 10 < \blacksquare \div 24 < 5 \div 2$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 24 has to sit strictly between 24 over 10 and 5 over 2. We want every natural number that works.

Givens

The left end is 24 divided by 10.
The right end is 5 divided by 2.
The middle is the box divided by 24.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 24 leaves the inequality pointing the same way because 24 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

24 \div 10 = 2.4,\quad 5 \div 2 = 2.5

The middle sits between 2.4 and 2.5.

2Multiply every part by 24

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

24 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

2.4 \times 24 < \blacksquare < 2.5 \times 24

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

57.6 < \blacksquare < 60

Strictly between 57.6 and 60.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 58, 59

2 values work.

Answer: 58, 59

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 58 sits above 57.6 and 59 below 60, while 57 and 60 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 6 medium

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

$23 \div 10 < \blacksquare \div 32 < 5 \div 2$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 32 has to sit strictly between 23 over 10 and 5 over 2. We want every natural number that works.

Givens

The left end is 23 divided by 10.
The right end is 5 divided by 2.
The middle is the box divided by 32.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 32 leaves the inequality pointing the same way because 32 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

23 \div 10 = 2.3,\quad 5 \div 2 = 2.5

The middle sits between 2.3 and 2.5.

2Multiply every part by 32

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

32 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

2.3 \times 32 < \blacksquare < 2.5 \times 32

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

73.6 < \blacksquare < 80

Strictly between 73.6 and 80.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 74, 75, 76, 77, 78, 79

6 values work.

Answer: 74, 75, 76, 77, 78, 79

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 74 sits above 73.6 and 79 below 80, while 73 and 80 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 7 medium

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

$37 \div 5 < \blacksquare \div 24 < 15 \div 2$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 24 has to sit strictly between 37 over 5 and 15 over 2. We want every natural number that works.

Givens

The left end is 37 divided by 5.
The right end is 15 divided by 2.
The middle is the box divided by 24.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 24 leaves the inequality pointing the same way because 24 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

37 \div 5 = 7.4,\quad 15 \div 2 = 7.5

The middle sits between 7.4 and 7.5.

2Multiply every part by 24

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

24 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

7.4 \times 24 < \blacksquare < 7.5 \times 24

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

177.6 < \blacksquare < 180

Strictly between 177.6 and 180.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 178, 179

2 values work.

Answer: 178, 179

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 178 sits above 177.6 and 179 below 180, while 177 and 180 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 8 medium

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

$7 \div 4 < \blacksquare \div 24 < 39 \div 20$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 24 has to sit strictly between 7 over 4 and 39 over 20. We want every natural number that works.

Givens

The left end is 7 divided by 4.
The right end is 39 divided by 20.
The middle is the box divided by 24.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 24 leaves the inequality pointing the same way because 24 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

7 \div 4 = 1.75,\quad 39 \div 20 = 1.95

The middle sits between 1.75 and 1.95.

2Multiply every part by 24

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

24 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

1.75 \times 24 < \blacksquare < 1.95 \times 24

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

42 < \blacksquare < 46.8

Strictly between 42 and 46.8.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 43, 44, 45, 46

4 values work.

Answer: 43, 44, 45, 46

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 43 sits above 42 and 46 below 46.8, while 42 and 47 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 9 hard

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

$9 \div 5 < \blacksquare \div 12 < 40 \div 20$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 12 has to sit strictly between 9 over 5 and 40 over 20. We want every natural number that works.

Givens

The left end is 9 divided by 5.
The right end is 40 divided by 20.
The middle is the box divided by 12.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 12 leaves the inequality pointing the same way because 12 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

9 \div 5 = 1.8,\quad 40 \div 20 = 2

The middle sits between 1.8 and 2.

2Multiply every part by 12

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

12 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

1.8 \times 12 < \blacksquare < 2 \times 12

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

21.6 < \blacksquare < 24

Strictly between 21.6 and 24.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 22, 23

2 values work.

Answer: 22, 23

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 22 sits above 21.6 and 23 below 24, while 21 and 24 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 10 hard

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

$26 \div 8 < \blacksquare \div 14 < 42 \div 12$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 14 has to sit strictly between 26 over 8 and 42 over 12. We want every natural number that works.

Givens

The left end is 26 divided by 8.
The right end is 42 divided by 12.
The middle is the box divided by 14.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 14 leaves the inequality pointing the same way because 14 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

26 \div 8 = 3.25,\quad 42 \div 12 = 3.5

The middle sits between 3.25 and 3.5.

2Multiply every part by 14

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

14 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

3.25 \times 14 < \blacksquare < 3.5 \times 14

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

45.5 < \blacksquare < 49

Strictly between 45.5 and 49.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 46, 47, 48

3 values work.

Answer: 46, 47, 48

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 46 sits above 45.5 and 48 below 49, while 45 and 49 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 11 hard

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

$43 \div 10 < \blacksquare \div 16 < 44 \div 10$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 16 has to sit strictly between 43 over 10 and 44 over 10. We want every natural number that works.

Givens

The left end is 43 divided by 10.
The right end is 44 divided by 10.
The middle is the box divided by 16.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 16 leaves the inequality pointing the same way because 16 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

43 \div 10 = 4.3,\quad 44 \div 10 = 4.4

The middle sits between 4.3 and 4.4.

2Multiply every part by 16

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

16 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

4.3 \times 16 < \blacksquare < 4.4 \times 16

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

68.8 < \blacksquare < 70.4

Strictly between 68.8 and 70.4.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 69, 70

2 values work.

Answer: 69, 70

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 69 sits above 68.8 and 70 below 70.4, while 68 and 71 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Variant 12 hard

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

$16 \div 8 < \blacksquare \div 15 < 45 \div 20$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The box divided by 15 has to sit strictly between 16 over 8 and 45 over 20. We want every natural number that works.

Givens

The left end is 16 divided by 8.
The right end is 45 divided by 20.
The middle is the box divided by 15.

Unknowns

Every natural number the box can hold.

Constraints

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

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #3 Eliminate Possibilities#8 Analyze the Units

The box is buried under a division, so clear it: multiplying all three parts by 15 leaves the inequality pointing the same way because 15 is positive. Then read off the whole numbers strictly inside the range.

3 · Execute4 carry out the plan

1Work out the two ends

#8 Analyze the Units 6.NS.B.3

Do the divisions on the outside first.

16 \div 8 = 2,\quad 45 \div 20 = 2.25

The middle sits between 2 and 2.25.

2Multiply every part by 15

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

15 is positive, so the two signs keep pointing the same way, and the box comes out on its own.

2 \times 15 < \blacksquare < 2.25 \times 15

Now the box stands alone.

3Read the range

#3 Eliminate Possibilities 6.EE.B.5

Both ends are excluded, so a whole number sitting exactly on an end does not count.

30 < \blacksquare < 33.75

Strictly between 30 and 33.75.

4List the natural numbers inside

#3 Eliminate Possibilities 6.EE.B.5

Count up from the first whole number past the left end.

\blacksquare = 31, 32, 33

3 values work.

Answer: 31, 32, 33

4 · Reviewdoes it hold up?

Checking the ends of the list against the range: 31 sits above 30 and 33 below 33.75, while 30 and 34 fall outside it.

Another way: Trying whole numbers one at a time also works, but there is no natural place to stop until the range is known.

Standardsmin grade 6

6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Multiplying a chained inequality through by a positive number.
6.EE.B.5 Solve equations/inequalities by finding values that make them true — Deciding which whole numbers satisfy the strict inequality.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Turning each end into a decimal.

💡Takeaway. You may multiply all three parts of an inequality by the same positive number. The signs keep pointing the way they did.

Multiplying every part by the same positive number keeps the order · 12 practice problems

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