← Rounding cannot be undone, but it can be bounded · Inequality Range Membership

Rounding cannot be undone, but it can be bounded · 12 practice problems

6.EE.B.86.NS.B.35.NBT.A.4

Generated variants — 12

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

Variant 1 easy answer: 9

A number was divided by 99, and the quotient rounded to the nearest whole number came to 77. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 9 rounds to 7. We want how many different numbers it could have been.

Givens
  • The number was divided by 9.
  • The quotient rounded to the nearest whole number is 7.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 6.5 rounds up to 7.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 6.5 up to but not including 7.5 rounds to 7.
6.5≤quotient<7.56.5 \le \text{quotient} < 7.5
Half a unit either side of the rounded value.

2Multiply both bounds by 9

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 9, so multiplying undoes it -- and the bounds move with it.
58.5≤number<67.558.5 \le \text{number} < 67.5
The number lives between 58.5 and 67.5.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 58.5 is 59, and the largest below 67.5 is 67.
59≤number≤6759 \le \text{number} \le 67
From 59 to 67.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
67−59+1=967 - 59 + 1 = 9
9 numbers work.
Answer: 9
4 · Reviewdoes it hold up?

Checking the ends without dividing: 59 is at least 58.5 and below 67.5, while 58 falls short of 58.5 and 68 reaches 67.5 -- so the run really does stop where it stops.

Another way: The window is 9 wide, so it holds 9 or so whole numbers -- 9 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 2 easy answer: 4

A number was divided by 44, and the quotient rounded to the nearest whole number came to 99. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 4 rounds to 9. We want how many different numbers it could have been.

Givens
  • The number was divided by 4.
  • The quotient rounded to the nearest whole number is 9.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 8.5 rounds up to 9.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 8.5 up to but not including 9.5 rounds to 9.
8.5≤quotient<9.58.5 \le \text{quotient} < 9.5
Half a unit either side of the rounded value.

2Multiply both bounds by 4

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 4, so multiplying undoes it -- and the bounds move with it.
34≤number<3834 \le \text{number} < 38
The number lives between 34 and 38.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 34 is 34, and the largest below 38 is 37.
34≤number≤3734 \le \text{number} \le 37
From 34 to 37.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
37−34+1=437 - 34 + 1 = 4
4 numbers work.
Answer: 4
4 · Reviewdoes it hold up?

Checking the ends without dividing: 34 is at least 34 and below 38, while 33 falls short of 34 and 38 reaches 38 -- so the run really does stop where it stops.

Another way: The window is 4 wide, so it holds 4 or so whole numbers -- 4 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 3 easy answer: 6

A number was divided by 66, and the quotient rounded to the nearest whole number came to 1111. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 6 rounds to 11. We want how many different numbers it could have been.

Givens
  • The number was divided by 6.
  • The quotient rounded to the nearest whole number is 11.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 10.5 rounds up to 11.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 10.5 up to but not including 11.5 rounds to 11.
10.5≤quotient<11.510.5 \le \text{quotient} < 11.5
Half a unit either side of the rounded value.

2Multiply both bounds by 6

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 6, so multiplying undoes it -- and the bounds move with it.
63≤number<6963 \le \text{number} < 69
The number lives between 63 and 69.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 63 is 63, and the largest below 69 is 68.
63≤number≤6863 \le \text{number} \le 68
From 63 to 68.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
68−63+1=668 - 63 + 1 = 6
6 numbers work.
Answer: 6
4 · Reviewdoes it hold up?

Checking the ends without dividing: 63 is at least 63 and below 69, while 62 falls short of 63 and 69 reaches 69 -- so the run really does stop where it stops.

Another way: The window is 6 wide, so it holds 6 or so whole numbers -- 6 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 4 easy answer: 11

A number was divided by 1111, and the quotient rounded to the nearest whole number came to 88. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 11 rounds to 8. We want how many different numbers it could have been.

Givens
  • The number was divided by 11.
  • The quotient rounded to the nearest whole number is 8.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 7.5 rounds up to 8.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 7.5 up to but not including 8.5 rounds to 8.
7.5≤quotient<8.57.5 \le \text{quotient} < 8.5
Half a unit either side of the rounded value.

2Multiply both bounds by 11

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 11, so multiplying undoes it -- and the bounds move with it.
82.5≤number<93.582.5 \le \text{number} < 93.5
The number lives between 82.5 and 93.5.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 82.5 is 83, and the largest below 93.5 is 93.
83≤number≤9383 \le \text{number} \le 93
From 83 to 93.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
93−83+1=1193 - 83 + 1 = 11
11 numbers work.
Answer: 11
4 · Reviewdoes it hold up?

Checking the ends without dividing: 83 is at least 82.5 and below 93.5, while 82 falls short of 82.5 and 94 reaches 93.5 -- so the run really does stop where it stops.

Another way: The window is 11 wide, so it holds 11 or so whole numbers -- 11 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 5 medium answer: 12

A number was divided by 1212, and the quotient rounded to the nearest whole number came to 66. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 12 rounds to 6. We want how many different numbers it could have been.

Givens
  • The number was divided by 12.
  • The quotient rounded to the nearest whole number is 6.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 5.5 rounds up to 6.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 5.5 up to but not including 6.5 rounds to 6.
5.5≤quotient<6.55.5 \le \text{quotient} < 6.5
Half a unit either side of the rounded value.

2Multiply both bounds by 12

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 12, so multiplying undoes it -- and the bounds move with it.
66≤number<7866 \le \text{number} < 78
The number lives between 66 and 78.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 66 is 66, and the largest below 78 is 77.
66≤number≤7766 \le \text{number} \le 77
From 66 to 77.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
77−66+1=1277 - 66 + 1 = 12
12 numbers work.
Answer: 12
4 · Reviewdoes it hold up?

Checking the ends without dividing: 66 is at least 66 and below 78, while 65 falls short of 66 and 78 reaches 78 -- so the run really does stop where it stops.

Another way: The window is 12 wide, so it holds 12 or so whole numbers -- 12 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 6 medium answer: 5

A number was divided by 55, and the quotient rounded to the nearest whole number came to 1212. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 5 rounds to 12. We want how many different numbers it could have been.

Givens
  • The number was divided by 5.
  • The quotient rounded to the nearest whole number is 12.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 11.5 rounds up to 12.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 11.5 up to but not including 12.5 rounds to 12.
11.5≤quotient<12.511.5 \le \text{quotient} < 12.5
Half a unit either side of the rounded value.

2Multiply both bounds by 5

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 5, so multiplying undoes it -- and the bounds move with it.
57.5≤number<62.557.5 \le \text{number} < 62.5
The number lives between 57.5 and 62.5.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 57.5 is 58, and the largest below 62.5 is 62.
58≤number≤6258 \le \text{number} \le 62
From 58 to 62.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
62−58+1=562 - 58 + 1 = 5
5 numbers work.
Answer: 5
4 · Reviewdoes it hold up?

Checking the ends without dividing: 58 is at least 57.5 and below 62.5, while 57 falls short of 57.5 and 63 reaches 62.5 -- so the run really does stop where it stops.

Another way: The window is 5 wide, so it holds 5 or so whole numbers -- 5 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 7 medium answer: 7

A number was divided by 77, and the quotient rounded to the nearest whole number came to 1313. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 7 rounds to 13. We want how many different numbers it could have been.

Givens
  • The number was divided by 7.
  • The quotient rounded to the nearest whole number is 13.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 12.5 rounds up to 13.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 12.5 up to but not including 13.5 rounds to 13.
12.5≤quotient<13.512.5 \le \text{quotient} < 13.5
Half a unit either side of the rounded value.

2Multiply both bounds by 7

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 7, so multiplying undoes it -- and the bounds move with it.
87.5≤number<94.587.5 \le \text{number} < 94.5
The number lives between 87.5 and 94.5.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 87.5 is 88, and the largest below 94.5 is 94.
88≤number≤9488 \le \text{number} \le 94
From 88 to 94.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
94−88+1=794 - 88 + 1 = 7
7 numbers work.
Answer: 7
4 · Reviewdoes it hold up?

Checking the ends without dividing: 88 is at least 87.5 and below 94.5, while 87 falls short of 87.5 and 95 reaches 94.5 -- so the run really does stop where it stops.

Another way: The window is 7 wide, so it holds 7 or so whole numbers -- 7 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 8 medium answer: 10

A number was divided by 1010, and the quotient rounded to the nearest whole number came to 1414. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 10 rounds to 14. We want how many different numbers it could have been.

Givens
  • The number was divided by 10.
  • The quotient rounded to the nearest whole number is 14.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 13.5 rounds up to 14.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 13.5 up to but not including 14.5 rounds to 14.
13.5≤quotient<14.513.5 \le \text{quotient} < 14.5
Half a unit either side of the rounded value.

2Multiply both bounds by 10

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 10, so multiplying undoes it -- and the bounds move with it.
135≤number<145135 \le \text{number} < 145
The number lives between 135 and 145.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 135 is 135, and the largest below 145 is 144.
135≤number≤144135 \le \text{number} \le 144
From 135 to 144.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
144−135+1=10144 - 135 + 1 = 10
10 numbers work.
Answer: 10
4 · Reviewdoes it hold up?

Checking the ends without dividing: 135 is at least 135 and below 145, while 134 falls short of 135 and 145 reaches 145 -- so the run really does stop where it stops.

Another way: The window is 10 wide, so it holds 10 or so whole numbers -- 10 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 9 hard answer: 14

A number was divided by 1414, and the quotient rounded to the nearest whole number came to 44. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 14 rounds to 4. We want how many different numbers it could have been.

Givens
  • The number was divided by 14.
  • The quotient rounded to the nearest whole number is 4.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 3.5 rounds up to 4.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 3.5 up to but not including 4.5 rounds to 4.
3.5≤quotient<4.53.5 \le \text{quotient} < 4.5
Half a unit either side of the rounded value.

2Multiply both bounds by 14

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 14, so multiplying undoes it -- and the bounds move with it.
49≤number<6349 \le \text{number} < 63
The number lives between 49 and 63.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 49 is 49, and the largest below 63 is 62.
49≤number≤6249 \le \text{number} \le 62
From 49 to 62.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
62−49+1=1462 - 49 + 1 = 14
14 numbers work.
Answer: 14
4 · Reviewdoes it hold up?

Checking the ends without dividing: 49 is at least 49 and below 63, while 48 falls short of 49 and 63 reaches 63 -- so the run really does stop where it stops.

Another way: The window is 14 wide, so it holds 14 or so whole numbers -- 14 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 10 hard answer: 8

A number was divided by 88, and the quotient rounded to the nearest whole number came to 1515. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 8 rounds to 15. We want how many different numbers it could have been.

Givens
  • The number was divided by 8.
  • The quotient rounded to the nearest whole number is 15.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 14.5 rounds up to 15.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 14.5 up to but not including 15.5 rounds to 15.
14.5≤quotient<15.514.5 \le \text{quotient} < 15.5
Half a unit either side of the rounded value.

2Multiply both bounds by 8

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 8, so multiplying undoes it -- and the bounds move with it.
116≤number<124116 \le \text{number} < 124
The number lives between 116 and 124.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 116 is 116, and the largest below 124 is 123.
116≤number≤123116 \le \text{number} \le 123
From 116 to 123.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
123−116+1=8123 - 116 + 1 = 8
8 numbers work.
Answer: 8
4 · Reviewdoes it hold up?

Checking the ends without dividing: 116 is at least 116 and below 124, while 115 falls short of 116 and 124 reaches 124 -- so the run really does stop where it stops.

Another way: The window is 8 wide, so it holds 8 or so whole numbers -- 8 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 11 hard answer: 16

A number was divided by 1616, and the quotient rounded to the nearest whole number came to 55. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 16 rounds to 5. We want how many different numbers it could have been.

Givens
  • The number was divided by 16.
  • The quotient rounded to the nearest whole number is 5.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 4.5 rounds up to 5.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 4.5 up to but not including 5.5 rounds to 5.
4.5≤quotient<5.54.5 \le \text{quotient} < 5.5
Half a unit either side of the rounded value.

2Multiply both bounds by 16

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 16, so multiplying undoes it -- and the bounds move with it.
72≤number<8872 \le \text{number} < 88
The number lives between 72 and 88.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 72 is 72, and the largest below 88 is 87.
72≤number≤8772 \le \text{number} \le 87
From 72 to 87.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
87−72+1=1687 - 72 + 1 = 16
16 numbers work.
Answer: 16
4 · Reviewdoes it hold up?

Checking the ends without dividing: 72 is at least 72 and below 88, while 71 falls short of 72 and 88 reaches 88 -- so the run really does stop where it stops.

Another way: The window is 16 wide, so it holds 16 or so whole numbers -- 16 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.
Variant 12 hard answer: 3

A number was divided by 33, and the quotient rounded to the nearest whole number came to 2020. How many natural numbers could that number be?

Show solution
1 · Understandwhat's really being asked

Some natural number divided by 3 rounds to 20. We want how many different numbers it could have been.

Givens
  • The number was divided by 3.
  • The quotient rounded to the nearest whole number is 20.
Unknowns
  • How many natural numbers fit that description.
Constraints
  • Rounding is to the nearest whole number, so a quotient of exactly 19.5 rounds up to 20.
2 · Planchoose the strategy

#11 Work Backwards · also uses: #3 Eliminate Possibilities#8 Analyze the Units

Rounding loses information, so the original quotient cannot be recovered -- but it can be trapped between two bounds. Find those, multiply them back up, and count the whole numbers that survive.

3 · Execute4 carry out the plan

1Bound the quotient

#11 Work Backwards 5.NBT.A.4
Anything from 19.5 up to but not including 20.5 rounds to 20.
19.5≤quotient<20.519.5 \le \text{quotient} < 20.5
Half a unit either side of the rounded value.

2Multiply both bounds by 3

#8 Analyze the Units 6.EE.B.8
The quotient came from dividing by 3, so multiplying undoes it -- and the bounds move with it.
58.5≤number<61.558.5 \le \text{number} < 61.5
The number lives between 58.5 and 61.5.

3Find the first and last whole number inside

#3 Eliminate Possibilities 6.EE.B.8
The smallest natural number at least 58.5 is 59, and the largest below 61.5 is 61.
59≤number≤6159 \le \text{number} \le 61
From 59 to 61.

4Count them

#3 Eliminate Possibilities 6.NS.B.3
Both ends are included, so subtract and add one.
61−59+1=361 - 59 + 1 = 3
3 numbers work.
Answer: 3
4 · Reviewdoes it hold up?

Checking the ends without dividing: 59 is at least 58.5 and below 61.5, while 58 falls short of 58.5 and 62 reaches 61.5 -- so the run really does stop where it stops.

Another way: The window is 3 wide, so it holds 3 or so whole numbers -- 3 here. That is a quick check on the count without redoing the work.

Standardsmin grade 6
  • 6.EE.B.8 Write and graph inequalities of the form x>c or x<c — Writing the possible values as an inequality and scaling it.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Counting the whole numbers the range contains.
  • 5.NBT.A.4 Round decimals to any place — Reading what 'rounds to the nearest whole number' allows.
💡Takeaway. A rounded number does not tell you what it was. It tells you the smallest and largest it could have been.