Practice · Rounding to n means a half-open interval · Sensim Math

Variant 1 hard

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $5<\left[\frac{x}{4}+3\right]<8$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{4}+3\right]$ is a whole number strictly between $5$ and $8$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$5<\left[\frac{x}{4}+3\right]<8$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{4}+3\right]

is an integer with

5<\left[\frac{x}{4}+3\right]<8

, so it is

6

or

7

.

\left[\frac{x}{4}+3\right]\in\{6, 7\}

Only two whole numbers fit strictly between

5

and

8

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

6

covers

5.5\le t<6.5

and rounding to

7

covers

6.5\le t<7.5

; together they make one interval.

5.5\le\frac{x}{4}+3<7.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Subtract

3

from every part, then multiply every part by

4

.

2.5\le\frac{x}{4}<4.5 \;\Rightarrow\; 10\le x<18

Subtracting and multiplying by a positive number keep every inequality sign.

Answer:

10\le x<18

4 · Reviewdoes it hold up?

At $x=10$ : $\frac{10}{4}+3=5.5$ , which rounds up to $6$ , allowed. At $x=18$ : $\frac{18}{4}+3=7.5$ , which rounds to $8$ , not allowed. At $x=9.6$ : $5.4$ rounds to $5$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $6$ or $7$ , so $\frac{x}{4}+3$ is in $[5.5, 7.5)$ , so $\frac{x}{4}$ is in $[2.5, 4.5)$ , so $x$ is in $[10, 18)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 2 easy

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $3<\left[\frac{x}{6}-2\right]<6$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{6}-2\right]$ is a whole number strictly between $3$ and $6$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$3<\left[\frac{x}{6}-2\right]<6$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{6}-2\right]

is an integer with

3<\left[\frac{x}{6}-2\right]<6

, so it is

4

or

5

.

\left[\frac{x}{6}-2\right]\in\{4, 5\}

Only two whole numbers fit strictly between

3

and

6

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

4

covers

3.5\le t<4.5

and rounding to

5

covers

4.5\le t<5.5

; together they make one interval.

3.5\le\frac{x}{6}-2<5.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Add

2

to every part, then multiply every part by

6

.

5.5\le\frac{x}{6}<7.5 \;\Rightarrow\; 33\le x<45

Adding and multiplying by a positive number keep every inequality sign.

Answer:

33\le x<45

4 · Reviewdoes it hold up?

At $x=33$ : $\frac{33}{6}-2=3.5$ , which rounds up to $4$ , allowed. At $x=45$ : $\frac{45}{6}-2=5.5$ , which rounds to $6$ , not allowed. At $x=32.4$ : $3.4$ rounds to $3$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $4$ or $5$ , so $\frac{x}{6}-2$ is in $[3.5, 5.5)$ , so $\frac{x}{6}$ is in $[5.5, 7.5)$ , so $x$ is in $[33, 45)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 3 medium

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $2<\left[\frac{x}{6}+1\right]<6$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{6}+1\right]$ is a whole number strictly between $2$ and $6$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$2<\left[\frac{x}{6}+1\right]<6$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{6}+1\right]

is an integer with

2<\left[\frac{x}{6}+1\right]<6

, so it is

3

,

4

or

5

.

\left[\frac{x}{6}+1\right]\in\{3, 4, 5\}

Only three whole numbers fit strictly between

2

and

6

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

3

covers

2.5\le t<3.5

and rounding to

4

covers

3.5\le t<4.5

and rounding to

5

covers

4.5\le t<5.5

; together they make one interval.

2.5\le\frac{x}{6}+1<5.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Subtract

1

from every part, then multiply every part by

6

.

1.5\le\frac{x}{6}<4.5 \;\Rightarrow\; 9\le x<27

Subtracting and multiplying by a positive number keep every inequality sign.

Answer:

9\le x<27

4 · Reviewdoes it hold up?

At $x=9$ : $\frac{9}{6}+1=2.5$ , which rounds up to $3$ , allowed. At $x=27$ : $\frac{27}{6}+1=5.5$ , which rounds to $6$ , not allowed. At $x=8.4$ : $2.4$ rounds to $2$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $3$ , $4$ or $5$ , so $\frac{x}{6}+1$ is in $[2.5, 5.5)$ , so $\frac{x}{6}$ is in $[1.5, 4.5)$ , so $x$ is in $[9, 27)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 4 easy

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $1<\left[\frac{x}{4}+2\right]<5$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{4}+2\right]$ is a whole number strictly between $1$ and $5$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$1<\left[\frac{x}{4}+2\right]<5$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{4}+2\right]

is an integer with

1<\left[\frac{x}{4}+2\right]<5

, so it is

2

,

3

or

4

.

\left[\frac{x}{4}+2\right]\in\{2, 3, 4\}

Only three whole numbers fit strictly between

1

and

5

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

2

covers

1.5\le t<2.5

and rounding to

3

covers

2.5\le t<3.5

and rounding to

4

covers

3.5\le t<4.5

; together they make one interval.

1.5\le\frac{x}{4}+2<4.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Subtract

2

from every part, then multiply every part by

4

.

-0.5\le\frac{x}{4}<2.5 \;\Rightarrow\; -2\le x<10

Subtracting and multiplying by a positive number keep every inequality sign.

Answer:

-2\le x<10

4 · Reviewdoes it hold up?

At $x=-2$ : $\frac{-2}{4}+2=1.5$ , which rounds up to $2$ , allowed. At $x=10$ : $\frac{10}{4}+2=4.5$ , which rounds to $5$ , not allowed. At $x=-2.4$ : $1.4$ rounds to $1$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $2$ , $3$ or $4$ , so $\frac{x}{4}+2$ is in $[1.5, 4.5)$ , so $\frac{x}{4}$ is in $[-0.5, 2.5)$ , so $x$ is in $[-2, 10)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 5 hard

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $3<\left[\frac{x}{2}-5\right]<7$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{2}-5\right]$ is a whole number strictly between $3$ and $7$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$3<\left[\frac{x}{2}-5\right]<7$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{2}-5\right]

is an integer with

3<\left[\frac{x}{2}-5\right]<7

, so it is

4

,

5

or

6

.

\left[\frac{x}{2}-5\right]\in\{4, 5, 6\}

Only three whole numbers fit strictly between

3

and

7

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

4

covers

3.5\le t<4.5

and rounding to

5

covers

4.5\le t<5.5

and rounding to

6

covers

5.5\le t<6.5

; together they make one interval.

3.5\le\frac{x}{2}-5<6.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Add

5

to every part, then multiply every part by

2

.

8.5\le\frac{x}{2}<11.5 \;\Rightarrow\; 17\le x<23

Adding and multiplying by a positive number keep every inequality sign.

Answer:

17\le x<23

4 · Reviewdoes it hold up?

At $x=17$ : $\frac{17}{2}-5=3.5$ , which rounds up to $4$ , allowed. At $x=23$ : $\frac{23}{2}-5=6.5$ , which rounds to $7$ , not allowed. At $x=16.8$ : $3.4$ rounds to $3$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $4$ , $5$ or $6$ , so $\frac{x}{2}-5$ is in $[3.5, 6.5)$ , so $\frac{x}{2}$ is in $[8.5, 11.5)$ , so $x$ is in $[17, 23)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 6 medium

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $4<\left[\frac{x}{2}+1\right]<7$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{2}+1\right]$ is a whole number strictly between $4$ and $7$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$4<\left[\frac{x}{2}+1\right]<7$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{2}+1\right]

is an integer with

4<\left[\frac{x}{2}+1\right]<7

, so it is

5

or

6

.

\left[\frac{x}{2}+1\right]\in\{5, 6\}

Only two whole numbers fit strictly between

4

and

7

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

5

covers

4.5\le t<5.5

and rounding to

6

covers

5.5\le t<6.5

; together they make one interval.

4.5\le\frac{x}{2}+1<6.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Subtract

1

from every part, then multiply every part by

2

.

3.5\le\frac{x}{2}<5.5 \;\Rightarrow\; 7\le x<11

Subtracting and multiplying by a positive number keep every inequality sign.

Answer:

7\le x<11

4 · Reviewdoes it hold up?

At $x=7$ : $\frac{7}{2}+1=4.5$ , which rounds up to $5$ , allowed. At $x=11$ : $\frac{11}{2}+1=6.5$ , which rounds to $7$ , not allowed. At $x=6.8$ : $4.4$ rounds to $4$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $5$ or $6$ , so $\frac{x}{2}+1$ is in $[4.5, 6.5)$ , so $\frac{x}{2}$ is in $[3.5, 5.5)$ , so $x$ is in $[7, 11)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 7 medium

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $0<\left[\frac{x}{8}-1\right]<3$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{8}-1\right]$ is a whole number strictly between $0$ and $3$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$0<\left[\frac{x}{8}-1\right]<3$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{8}-1\right]

is an integer with

0<\left[\frac{x}{8}-1\right]<3

, so it is

1

or

2

.

\left[\frac{x}{8}-1\right]\in\{1, 2\}

Only two whole numbers fit strictly between

0

and

3

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

1

covers

0.5\le t<1.5

and rounding to

2

covers

1.5\le t<2.5

; together they make one interval.

0.5\le\frac{x}{8}-1<2.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Add

1

to every part, then multiply every part by

8

.

1.5\le\frac{x}{8}<3.5 \;\Rightarrow\; 12\le x<28

Adding and multiplying by a positive number keep every inequality sign.

Answer:

12\le x<28

4 · Reviewdoes it hold up?

At $x=12$ : $\frac{12}{8}-1=0.5$ , which rounds up to $1$ , allowed. At $x=28$ : $\frac{28}{8}-1=2.5$ , which rounds to $3$ , not allowed. At $x=11.2$ : $0.4$ rounds to $0$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $1$ or $2$ , so $\frac{x}{8}-1$ is in $[0.5, 2.5)$ , so $\frac{x}{8}$ is in $[1.5, 3.5)$ , so $x$ is in $[12, 28)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 8 easy

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $2<\left[\frac{x}{4}-1\right]<5$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{4}-1\right]$ is a whole number strictly between $2$ and $5$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$2<\left[\frac{x}{4}-1\right]<5$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{4}-1\right]

is an integer with

2<\left[\frac{x}{4}-1\right]<5

, so it is

3

or

4

.

\left[\frac{x}{4}-1\right]\in\{3, 4\}

Only two whole numbers fit strictly between

2

and

5

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

3

covers

2.5\le t<3.5

and rounding to

4

covers

3.5\le t<4.5

; together they make one interval.

2.5\le\frac{x}{4}-1<4.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Add

1

to every part, then multiply every part by

4

.

3.5\le\frac{x}{4}<5.5 \;\Rightarrow\; 14\le x<22

Adding and multiplying by a positive number keep every inequality sign.

Answer:

14\le x<22

4 · Reviewdoes it hold up?

At $x=14$ : $\frac{14}{4}-1=2.5$ , which rounds up to $3$ , allowed. At $x=22$ : $\frac{22}{4}-1=4.5$ , which rounds to $5$ , not allowed. At $x=13.6$ : $2.4$ rounds to $2$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $3$ or $4$ , so $\frac{x}{4}-1$ is in $[2.5, 4.5)$ , so $\frac{x}{4}$ is in $[3.5, 5.5)$ , so $x$ is in $[14, 22)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 9 hard

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $1<\left[\frac{x}{8}+2\right]<4$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{8}+2\right]$ is a whole number strictly between $1$ and $4$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$1<\left[\frac{x}{8}+2\right]<4$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{8}+2\right]

is an integer with

1<\left[\frac{x}{8}+2\right]<4

, so it is

2

or

3

.

\left[\frac{x}{8}+2\right]\in\{2, 3\}

Only two whole numbers fit strictly between

1

and

4

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

2

covers

1.5\le t<2.5

and rounding to

3

covers

2.5\le t<3.5

; together they make one interval.

1.5\le\frac{x}{8}+2<3.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Subtract

2

from every part, then multiply every part by

8

.

-0.5\le\frac{x}{8}<1.5 \;\Rightarrow\; -4\le x<12

Subtracting and multiplying by a positive number keep every inequality sign.

Answer:

-4\le x<12

4 · Reviewdoes it hold up?

At $x=-4$ : $\frac{-4}{8}+2=1.5$ , which rounds up to $2$ , allowed. At $x=12$ : $\frac{12}{8}+2=3.5$ , which rounds to $4$ , not allowed. At $x=-4.8$ : $1.4$ rounds to $1$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $2$ or $3$ , so $\frac{x}{8}+2$ is in $[1.5, 3.5)$ , so $\frac{x}{8}$ is in $[-0.5, 1.5)$ , so $x$ is in $[-4, 12)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 10 easy

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $1<\left[\frac{x}{2}-3\right]<4$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{2}-3\right]$ is a whole number strictly between $1$ and $4$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$1<\left[\frac{x}{2}-3\right]<4$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{2}-3\right]

is an integer with

1<\left[\frac{x}{2}-3\right]<4

, so it is

2

or

3

.

\left[\frac{x}{2}-3\right]\in\{2, 3\}

Only two whole numbers fit strictly between

1

and

4

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

2

covers

1.5\le t<2.5

and rounding to

3

covers

2.5\le t<3.5

; together they make one interval.

1.5\le\frac{x}{2}-3<3.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Add

3

to every part, then multiply every part by

2

.

4.5\le\frac{x}{2}<6.5 \;\Rightarrow\; 9\le x<13

Adding and multiplying by a positive number keep every inequality sign.

Answer:

9\le x<13

4 · Reviewdoes it hold up?

At $x=9$ : $\frac{9}{2}-3=1.5$ , which rounds up to $2$ , allowed. At $x=13$ : $\frac{13}{2}-3=3.5$ , which rounds to $4$ , not allowed. At $x=8.8$ : $1.4$ rounds to $1$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $2$ or $3$ , so $\frac{x}{2}-3$ is in $[1.5, 3.5)$ , so $\frac{x}{2}$ is in $[4.5, 6.5)$ , so $x$ is in $[9, 13)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 11 medium

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $1<\left[\frac{x}{4}-3\right]<4$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{4}-3\right]$ is a whole number strictly between $1$ and $4$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$1<\left[\frac{x}{4}-3\right]<4$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{4}-3\right]

is an integer with

1<\left[\frac{x}{4}-3\right]<4

, so it is

2

or

3

.

\left[\frac{x}{4}-3\right]\in\{2, 3\}

Only two whole numbers fit strictly between

1

and

4

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

2

covers

1.5\le t<2.5

and rounding to

3

covers

2.5\le t<3.5

; together they make one interval.

1.5\le\frac{x}{4}-3<3.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Add

3

to every part, then multiply every part by

4

.

4.5\le\frac{x}{4}<6.5 \;\Rightarrow\; 18\le x<26

Adding and multiplying by a positive number keep every inequality sign.

Answer:

18\le x<26

4 · Reviewdoes it hold up?

At $x=18$ : $\frac{18}{4}-3=1.5$ , which rounds up to $2$ , allowed. At $x=26$ : $\frac{26}{4}-3=3.5$ , which rounds to $4$ , not allowed. At $x=17.6$ : $1.4$ rounds to $1$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $2$ or $3$ , so $\frac{x}{4}-3$ is in $[1.5, 3.5)$ , so $\frac{x}{4}$ is in $[4.5, 6.5)$ , so $x$ is in $[18, 26)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Variant 12 hard

Let $[a]$ be the integer obtained by rounding $a$ at its tenths digit (to the nearest whole number, with halves rounded up), so $[3.49]=3$ and $[7.5]=8$ . Find the range of values of $x$ that satisfy $2<\left[\frac{x}{10}-1\right]<5$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The rounded value $\left[\frac{x}{10}-1\right]$ is a whole number strictly between $2$ and $5$ . Find which whole numbers those are, turn "rounds to $n$ " into an ordinary inequality, and solve for $x$ .

Givens

$[a]$ rounds $a$ at the tenths digit, halves up: $[3.49]=3$ , $[7.5]=8$ .
$2<\left[\frac{x}{10}-1\right]<5$

Unknowns

The range of $x$ .

Constraints

A rounded value is always an integer.
Halves round up, so the lower end of each rounding interval is included and the upper end is not.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #2 Make a Systematic List#1 Draw a Diagram

List the possible rounded values, then rewrite "rounds to $n$ " as $n-0.5\le t<n+0.5$ (picture it on a number line). The rounding disappears and a plain linear inequality is left.

3 · Execute3 carry out the plan

1List the possible rounded values

#2 Make a Systematic List 7.EE.B.4

\left[\frac{x}{10}-1\right]

is an integer with

2<\left[\frac{x}{10}-1\right]<5

, so it is

3

or

4

.

\left[\frac{x}{10}-1\right]\in\{3, 4\}

Only two whole numbers fit strictly between

2

and

5

.

2Undo the rounding

#1 Draw a Diagram 5.NBT.A.4

A number rounds to

n

exactly when it is at least

n-0.5

and less than

n+0.5

. Rounding to

3

covers

2.5\le t<3.5

and rounding to

4

covers

3.5\le t<4.5

; together they make one interval.

2.5\le\frac{x}{10}-1<4.5

On a number line the rounding intervals sit side by side with no gap.

3Solve for $x$

#13 Convert to Algebra 7.EE.B.4

Add

1

to every part, then multiply every part by

10

.

3.5\le\frac{x}{10}<5.5 \;\Rightarrow\; 35\le x<55

Adding and multiplying by a positive number keep every inequality sign.

Answer:

35\le x<55

4 · Reviewdoes it hold up?

At $x=35$ : $\frac{35}{10}-1=2.5$ , which rounds up to $3$ , allowed. At $x=55$ : $\frac{55}{10}-1=4.5$ , which rounds to $5$ , not allowed. At $x=34$ : $2.4$ rounds to $2$ , not allowed. So the endpoints are right.

Another way: Work backwards from the outside in: the rounded value is $3$ or $4$ , so $\frac{x}{10}-1$ is in $[2.5, 4.5)$ , so $\frac{x}{10}$ is in $[3.5, 5.5)$ , so $x$ is in $[35, 55)$ .

Standardsmin grade 7

5.NBT.A.4 Round decimals to any place — Describing exactly which numbers round to a given whole number.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving the resulting linear inequality for $x$.

💡Takeaway. Rounding hides an interval: "rounds to

3

" just means "from

2.5

up to, but not including,

3.5

".

Rounding to n means a half-open interval · 12 practice problems

Generated variants — 12

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