Practice · Count one quadrant, then use point symmetry · Sensim Math

Variant 1 easy

In the figure, the graphs of the direct proportion $y=3x$ and the inverse proportion $y=\frac{3}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{3}x$ and $y=\frac{3}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=3x$ , $y=\frac{1}{3}x$ , and $y=\frac{3}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=3x$ and $y=\frac{1}{3}x$ , curve $y=\frac{3}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{3}x \le y \le 3x$ and $xy \le 3$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

3x = \frac{3}{x} \Rightarrow x = 1,\ A(1, 3); \quad \frac{1}{3}x = \frac{3}{x} \Rightarrow x = 3,\ B(3, 1)

So the first-quadrant region runs from

x = 0

to

x = 3

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

3

,

y

must satisfy

\frac{x}{3} \le y \le 3x

and

y \le \frac{3}{x}

.

\frac{x}{3} \le y \le \min\left(3x, \frac{3}{x}\right)

The lower edge is the line

y=\frac{1}{3}x

; the upper edge is

y=3x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

1 \le y \le 3

,

3

points.

x=2, 3

: only

y=1

,

1

point each.

3 + 1 + 1 = 5

For example at

x=2

:

y \ge \frac{2}{3}

and

y \le 1.5

, so

y = 1

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

5

points. The origin is a corner of both regions and is counted once.

5 \times 2 + 1 = 11

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

11

4 · Reviewdoes it hold up?

Spot-check the curve: $(2, 1)$ gives $xy = 2 \le 3$ (in) but $(2, 2)$ gives $4 > 3$ (out). The corner points A $(1, 3)$ and B $(3, 1)$ are included.

Another way: Count by rows instead: for $y = 1$ to $3$ list the allowed $x$ (from $\frac{y}{3}$ up to $\min(3y, \frac{3}{y})$ ): $3, 1, 1$ -- the same $5$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=3x$ and $y=\frac{1}{3}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 2 easy

In the figure, the graphs of the direct proportion $y=4x$ and the inverse proportion $y=\frac{4}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{4}x$ and $y=\frac{4}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=4x$ , $y=\frac{1}{4}x$ , and $y=\frac{4}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=4x$ and $y=\frac{1}{4}x$ , curve $y=\frac{4}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{4}x \le y \le 4x$ and $xy \le 4$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

4x = \frac{4}{x} \Rightarrow x = 1,\ A(1, 4); \quad \frac{1}{4}x = \frac{4}{x} \Rightarrow x = 4,\ B(4, 1)

So the first-quadrant region runs from

x = 0

to

x = 4

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

4

,

y

must satisfy

\frac{x}{4} \le y \le 4x

and

y \le \frac{4}{x}

.

\frac{x}{4} \le y \le \min\left(4x, \frac{4}{x}\right)

The lower edge is the line

y=\frac{1}{4}x

; the upper edge is

y=4x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

1 \le y \le 4

,

4

points.

x=2

:

y = 1, 2

,

2

points.

x=3, 4

: only

y=1

,

1

point each.

4 + 2 + 1 + 1 = 8

For example at

x=3

:

y \ge 0.75

and

y \le \frac{4}{3}

, so

y = 1

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

8

points. The origin is a corner of both regions and is counted once.

8 \times 2 + 1 = 17

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

17

4 · Reviewdoes it hold up?

Spot-check the curve: $(2, 2)$ gives $xy = 4 \le 4$ (in) but $(2, 3)$ gives $6 > 4$ (out). The corner points A $(1, 4)$ and B $(4, 1)$ are included.

Another way: Count by rows instead: for $y = 1$ to $4$ list the allowed $x$ (from $\frac{y}{4}$ up to $\min(4y, \frac{4}{y})$ ): $4, 2, 1, 1$ -- the same $8$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=4x$ and $y=\frac{1}{4}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 3 easy

In the figure, the graphs of the direct proportion $y=5x$ and the inverse proportion $y=\frac{5}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{5}x$ and $y=\frac{5}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=5x$ , $y=\frac{1}{5}x$ , and $y=\frac{5}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=5x$ and $y=\frac{1}{5}x$ , curve $y=\frac{5}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{5}x \le y \le 5x$ and $xy \le 5$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

5x = \frac{5}{x} \Rightarrow x = 1,\ A(1, 5); \quad \frac{1}{5}x = \frac{5}{x} \Rightarrow x = 5,\ B(5, 1)

So the first-quadrant region runs from

x = 0

to

x = 5

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

5

,

y

must satisfy

\frac{x}{5} \le y \le 5x

and

y \le \frac{5}{x}

.

\frac{x}{5} \le y \le \min\left(5x, \frac{5}{x}\right)

The lower edge is the line

y=\frac{1}{5}x

; the upper edge is

y=5x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

1 \le y \le 5

,

5

points.

x=2

:

y = 1, 2

,

2

points.

x=3, 4, 5

: only

y=1

,

1

point each.

5 + 2 + 1 + 1 + 1 = 10

For example at

x=2

:

y \ge 0.4

and

y \le 2.5

, so

y = 1, 2

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

10

points. The origin is a corner of both regions and is counted once.

10 \times 2 + 1 = 21

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

21

4 · Reviewdoes it hold up?

Spot-check the curve: $(2, 2)$ gives $xy = 4 \le 5$ (in) but $(2, 3)$ gives $6 > 5$ (out). The corner points A $(1, 5)$ and B $(5, 1)$ are included.

Another way: Count by rows instead: for $y = 1$ to $5$ list the allowed $x$ (from $\frac{y}{5}$ up to $\min(5y, \frac{5}{y})$ ): $5, 2, 1, 1, 1$ -- the same $10$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=5x$ and $y=\frac{1}{5}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 4 easy

In the figure, the graphs of the direct proportion $y=6x$ and the inverse proportion $y=\frac{6}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{6}x$ and $y=\frac{6}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=6x$ , $y=\frac{1}{6}x$ , and $y=\frac{6}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=6x$ and $y=\frac{1}{6}x$ , curve $y=\frac{6}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{6}x \le y \le 6x$ and $xy \le 6$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

6x = \frac{6}{x} \Rightarrow x = 1,\ A(1, 6); \quad \frac{1}{6}x = \frac{6}{x} \Rightarrow x = 6,\ B(6, 1)

So the first-quadrant region runs from

x = 0

to

x = 6

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

6

,

y

must satisfy

\frac{x}{6} \le y \le 6x

and

y \le \frac{6}{x}

.

\frac{x}{6} \le y \le \min\left(6x, \frac{6}{x}\right)

The lower edge is the line

y=\frac{1}{6}x

; the upper edge is

y=6x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

1 \le y \le 6

,

6

points.

x=2

:

1 \le y \le 3

,

3

points.

x=3

:

y = 1, 2

,

2

points.

x=4, 5, 6

: only

y=1

,

1

point each.

6 + 3 + 2 + 1 + 1 + 1 = 14

For example at

x=4

:

y \ge \frac{2}{3}

and

y \le 1.5

, so

y = 1

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

14

points. The origin is a corner of both regions and is counted once.

14 \times 2 + 1 = 29

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

29

4 · Reviewdoes it hold up?

Spot-check the curve: $(2, 3)$ gives $xy = 6 \le 6$ (in) but $(2, 4)$ gives $8 > 6$ (out). The corner points A $(1, 6)$ and B $(6, 1)$ are included.

Another way: Count by rows instead: for $y = 1$ to $6$ list the allowed $x$ (from $\frac{y}{6}$ up to $\min(6y, \frac{6}{y})$ ): $6, 3, 2, 1, 1, 1$ -- the same $14$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=6x$ and $y=\frac{1}{6}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 5 medium

In the figure, the graphs of the direct proportion $y=2x$ and the inverse proportion $y=\frac{8}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{2}x$ and $y=\frac{8}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=2x$ , $y=\frac{1}{2}x$ , and $y=\frac{8}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=2x$ and $y=\frac{1}{2}x$ , curve $y=\frac{8}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{2}x \le y \le 2x$ and $xy \le 8$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

2x = \frac{8}{x} \Rightarrow x = 2,\ A(2, 4); \quad \frac{1}{2}x = \frac{8}{x} \Rightarrow x = 4,\ B(4, 2)

So the first-quadrant region runs from

x = 0

to

x = 4

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

4

,

y

must satisfy

\frac{x}{2} \le y \le 2x

and

y \le \frac{8}{x}

.

\frac{x}{2} \le y \le \min\left(2x, \frac{8}{x}\right)

The lower edge is the line

y=\frac{1}{2}x

; the upper edge is

y=2x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

y = 1, 2

,

2

points.

x=2

:

1 \le y \le 4

,

4

points.

x=3, 4

: only

y=2

,

1

point each.

2 + 4 + 1 + 1 = 8

For example at

x=3

:

y \ge 1.5

and

y \le \frac{8}{3}

, so

y = 2

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

8

points. The origin is a corner of both regions and is counted once.

8 \times 2 + 1 = 17

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

17

4 · Reviewdoes it hold up?

Spot-check the curve: $(3, 2)$ gives $xy = 6 \le 8$ (in) but $(3, 3)$ gives $9 > 8$ (out). The corner points A $(2, 4)$ and B $(4, 2)$ are included.

Another way: Count by rows instead: for $y = 1$ to $4$ list the allowed $x$ (from $\frac{y}{2}$ up to $\min(2y, \frac{8}{y})$ ): $2, 4, 1, 1$ -- the same $8$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=2x$ and $y=\frac{1}{2}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 6 medium

In the figure, the graphs of the direct proportion $y=3x$ and the inverse proportion $y=\frac{12}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{3}x$ and $y=\frac{12}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=3x$ , $y=\frac{1}{3}x$ , and $y=\frac{12}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=3x$ and $y=\frac{1}{3}x$ , curve $y=\frac{12}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{3}x \le y \le 3x$ and $xy \le 12$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

3x = \frac{12}{x} \Rightarrow x = 2,\ A(2, 6); \quad \frac{1}{3}x = \frac{12}{x} \Rightarrow x = 6,\ B(6, 2)

So the first-quadrant region runs from

x = 0

to

x = 6

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

6

,

y

must satisfy

\frac{x}{3} \le y \le 3x

and

y \le \frac{12}{x}

.

\frac{x}{3} \le y \le \min\left(3x, \frac{12}{x}\right)

The lower edge is the line

y=\frac{1}{3}x

; the upper edge is

y=3x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

1 \le y \le 3

,

3

points.

x=2

:

1 \le y \le 6

,

6

points.

x=3

:

1 \le y \le 4

,

4

points.

x=4

:

y = 2, 3

,

2

points.

x=5, 6

: only

y=2

,

1

point each.

3 + 6 + 4 + 2 + 1 + 1 = 17

For example at

x=5

:

y \ge \frac{5}{3}

and

y \le 2.4

, so

y = 2

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

17

points. The origin is a corner of both regions and is counted once.

17 \times 2 + 1 = 35

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

35

4 · Reviewdoes it hold up?

Spot-check the curve: $(3, 4)$ gives $xy = 12 \le 12$ (in) but $(3, 5)$ gives $15 > 12$ (out). The corner points A $(2, 6)$ and B $(6, 2)$ are included.

Another way: Count by rows instead: for $y = 1$ to $6$ list the allowed $x$ (from $\frac{y}{3}$ up to $\min(3y, \frac{12}{y})$ ): $3, 6, 4, 2, 1, 1$ -- the same $17$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=3x$ and $y=\frac{1}{3}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 7 medium

In the figure, the graphs of the direct proportion $y=4x$ and the inverse proportion $y=\frac{16}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{4}x$ and $y=\frac{16}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=4x$ , $y=\frac{1}{4}x$ , and $y=\frac{16}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=4x$ and $y=\frac{1}{4}x$ , curve $y=\frac{16}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{4}x \le y \le 4x$ and $xy \le 16$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

4x = \frac{16}{x} \Rightarrow x = 2,\ A(2, 8); \quad \frac{1}{4}x = \frac{16}{x} \Rightarrow x = 8,\ B(8, 2)

So the first-quadrant region runs from

x = 0

to

x = 8

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

8

,

y

must satisfy

\frac{x}{4} \le y \le 4x

and

y \le \frac{16}{x}

.

\frac{x}{4} \le y \le \min\left(4x, \frac{16}{x}\right)

The lower edge is the line

y=\frac{1}{4}x

; the upper edge is

y=4x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

1 \le y \le 4

,

4

points.

x=2

:

1 \le y \le 8

,

8

points.

x=3

:

1 \le y \le 5

,

5

points.

x=4

:

1 \le y \le 4

,

4

points.

x=5

:

y = 2, 3

,

2

points.

x=6, 7, 8

: only

y=2

,

1

point each.

4 + 8 + 5 + 4 + 2 + 1 + 1 + 1 = 26

For example at

x=5

:

y \ge 1.25

and

y \le 3.2

, so

y = 2, 3

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

26

points. The origin is a corner of both regions and is counted once.

26 \times 2 + 1 = 53

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

53

4 · Reviewdoes it hold up?

Spot-check the curve: $(3, 5)$ gives $xy = 15 \le 16$ (in) but $(3, 6)$ gives $18 > 16$ (out). The corner points A $(2, 8)$ and B $(8, 2)$ are included.

Another way: Count by rows instead: for $y = 1$ to $8$ list the allowed $x$ (from $\frac{y}{4}$ up to $\min(4y, \frac{16}{y})$ ): $4, 8, 5, 4, 2, 1, 1, 1$ -- the same $26$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=4x$ and $y=\frac{1}{4}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 8 medium

In the figure, the graphs of the direct proportion $y=2x$ and the inverse proportion $y=\frac{18}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{2}x$ and $y=\frac{18}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=2x$ , $y=\frac{1}{2}x$ , and $y=\frac{18}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=2x$ and $y=\frac{1}{2}x$ , curve $y=\frac{18}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{2}x \le y \le 2x$ and $xy \le 18$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

2x = \frac{18}{x} \Rightarrow x = 3,\ A(3, 6); \quad \frac{1}{2}x = \frac{18}{x} \Rightarrow x = 6,\ B(6, 3)

So the first-quadrant region runs from

x = 0

to

x = 6

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

6

,

y

must satisfy

\frac{x}{2} \le y \le 2x

and

y \le \frac{18}{x}

.

\frac{x}{2} \le y \le \min\left(2x, \frac{18}{x}\right)

The lower edge is the line

y=\frac{1}{2}x

; the upper edge is

y=2x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

y = 1, 2

,

2

points.

x=2

:

1 \le y \le 4

,

4

points.

x=3

:

2 \le y \le 6

,

5

points.

x=4

:

2 \le y \le 4

,

3

points.

x=5, 6

: only

y=3

,

1

point each.

2 + 4 + 5 + 3 + 1 + 1 = 16

For example at

x=5

:

y \ge 2.5

and

y \le 3.6

, so

y = 3

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

16

points. The origin is a corner of both regions and is counted once.

16 \times 2 + 1 = 33

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

33

4 · Reviewdoes it hold up?

Spot-check the curve: $(4, 4)$ gives $xy = 16 \le 18$ (in) but $(4, 5)$ gives $20 > 18$ (out). The corner points A $(3, 6)$ and B $(6, 3)$ are included.

Another way: Count by rows instead: for $y = 1$ to $6$ list the allowed $x$ (from $\frac{y}{2}$ up to $\min(2y, \frac{18}{y})$ ): $2, 4, 5, 3, 1, 1$ -- the same $16$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=2x$ and $y=\frac{1}{2}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 9 hard

In the figure, the graphs of the direct proportion $y=5x$ and the inverse proportion $y=\frac{20}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{5}x$ and $y=\frac{20}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=5x$ , $y=\frac{1}{5}x$ , and $y=\frac{20}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=5x$ and $y=\frac{1}{5}x$ , curve $y=\frac{20}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{5}x \le y \le 5x$ and $xy \le 20$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

5x = \frac{20}{x} \Rightarrow x = 2,\ A(2, 10); \quad \frac{1}{5}x = \frac{20}{x} \Rightarrow x = 10,\ B(10, 2)

So the first-quadrant region runs from

x = 0

to

x = 10

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

10

,

y

must satisfy

\frac{x}{5} \le y \le 5x

and

y \le \frac{20}{x}

.

\frac{x}{5} \le y \le \min\left(5x, \frac{20}{x}\right)

The lower edge is the line

y=\frac{1}{5}x

; the upper edge is

y=5x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

1 \le y \le 5

,

5

points.

x=2

:

1 \le y \le 10

,

10

points.

x=3

:

1 \le y \le 6

,

6

points.

x=4

:

1 \le y \le 5

,

5

points.

x=5

:

1 \le y \le 4

,

4

points.

x=6

:

y = 2, 3

,

2

points.

x=7, 8, 9, 10

: only

y=2

,

1

point each.

5 + 10 + 6 + 5 + 4 + 2 + 1 + 1 + 1 + 1 = 36

For example at

x=8

:

y \ge 1.6

and

y \le 2.5

, so

y = 2

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

36

points. The origin is a corner of both regions and is counted once.

36 \times 2 + 1 = 73

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

73

4 · Reviewdoes it hold up?

Spot-check the curve: $(3, 6)$ gives $xy = 18 \le 20$ (in) but $(3, 7)$ gives $21 > 20$ (out). The corner points A $(2, 10)$ and B $(10, 2)$ are included.

Another way: Count by rows instead: for $y = 1$ to $10$ list the allowed $x$ (from $\frac{y}{5}$ up to $\min(5y, \frac{20}{y})$ ): $5, 10, 6, 5, 4, 2, 1, 1, 1, 1$ -- the same $36$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=5x$ and $y=\frac{1}{5}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 10 hard

In the figure, the graphs of the direct proportion $y=3x$ and the inverse proportion $y=\frac{27}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{3}x$ and $y=\frac{27}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=3x$ , $y=\frac{1}{3}x$ , and $y=\frac{27}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=3x$ and $y=\frac{1}{3}x$ , curve $y=\frac{27}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{3}x \le y \le 3x$ and $xy \le 27$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

3x = \frac{27}{x} \Rightarrow x = 3,\ A(3, 9); \quad \frac{1}{3}x = \frac{27}{x} \Rightarrow x = 9,\ B(9, 3)

So the first-quadrant region runs from

x = 0

to

x = 9

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

9

,

y

must satisfy

\frac{x}{3} \le y \le 3x

and

y \le \frac{27}{x}

.

\frac{x}{3} \le y \le \min\left(3x, \frac{27}{x}\right)

The lower edge is the line

y=\frac{1}{3}x

; the upper edge is

y=3x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

1 \le y \le 3

,

3

points.

x=2

:

1 \le y \le 6

,

6

points.

x=3

:

1 \le y \le 9

,

9

points.

x=4

:

2 \le y \le 6

,

5

points.

x=5

:

2 \le y \le 5

,

4

points.

x=6

:

2 \le y \le 4

,

3

points.

x=7, 8, 9

: only

y=3

,

1

point each.

3 + 6 + 9 + 5 + 4 + 3 + 1 + 1 + 1 = 33

For example at

x=4

:

y \ge \frac{4}{3}

and

y \le 6.75

, so

2 \le y \le 6

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

33

points. The origin is a corner of both regions and is counted once.

33 \times 2 + 1 = 67

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

67

4 · Reviewdoes it hold up?

Spot-check the curve: $(4, 6)$ gives $xy = 24 \le 27$ (in) but $(4, 7)$ gives $28 > 27$ (out). The corner points A $(3, 9)$ and B $(9, 3)$ are included.

Another way: Count by rows instead: for $y = 1$ to $9$ list the allowed $x$ (from $\frac{y}{3}$ up to $\min(3y, \frac{27}{y})$ ): $3, 6, 9, 5, 4, 3, 1, 1, 1$ -- the same $33$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=3x$ and $y=\frac{1}{3}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 11 hard

In the figure, the graphs of the direct proportion $y=2x$ and the inverse proportion $y=\frac{32}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{2}x$ and $y=\frac{32}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=2x$ , $y=\frac{1}{2}x$ , and $y=\frac{32}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=2x$ and $y=\frac{1}{2}x$ , curve $y=\frac{32}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{2}x \le y \le 2x$ and $xy \le 32$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

2x = \frac{32}{x} \Rightarrow x = 4,\ A(4, 8); \quad \frac{1}{2}x = \frac{32}{x} \Rightarrow x = 8,\ B(8, 4)

So the first-quadrant region runs from

x = 0

to

x = 8

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

8

,

y

must satisfy

\frac{x}{2} \le y \le 2x

and

y \le \frac{32}{x}

.

\frac{x}{2} \le y \le \min\left(2x, \frac{32}{x}\right)

The lower edge is the line

y=\frac{1}{2}x

; the upper edge is

y=2x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

y = 1, 2

,

2

points.

x=2

:

1 \le y \le 4

,

4

points.

x=3

:

2 \le y \le 6

,

5

points.

x=4

:

2 \le y \le 8

,

7

points.

x=5

:

3 \le y \le 6

,

4

points.

x=6

:

3 \le y \le 5

,

3

points.

x=7, 8

: only

y=4

,

1

point each.

2 + 4 + 5 + 7 + 4 + 3 + 1 + 1 = 27

For example at

x=5

:

y \ge 2.5

and

y \le 6.4

, so

3 \le y \le 6

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

27

points. The origin is a corner of both regions and is counted once.

27 \times 2 + 1 = 55

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

55

4 · Reviewdoes it hold up?

Spot-check the curve: $(5, 6)$ gives $xy = 30 \le 32$ (in) but $(5, 7)$ gives $35 > 32$ (out). The corner points A $(4, 8)$ and B $(8, 4)$ are included.

Another way: Count by rows instead: for $y = 1$ to $8$ list the allowed $x$ (from $\frac{y}{2}$ up to $\min(2y, \frac{32}{y})$ ): $2, 4, 5, 7, 4, 3, 1, 1$ -- the same $27$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=2x$ and $y=\frac{1}{2}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Variant 12 hard

In the figure, the graphs of the direct proportion $y=4x$ and the inverse proportion $y=\frac{36}{x}$ meet at points A and D, and the graphs of the direct proportion $y=\frac{1}{4}x$ and $y=\frac{36}{x}$ meet at points B and C. Count the points in the shaded regions whose $x$ - and $y$ -coordinates are both integers. (Points on the lines and curves count too.)

Your answer

Show solution

1 · Understandwhat's really being asked

The shaded regions are bounded by $y=4x$ , $y=\frac{1}{4}x$ , and $y=\frac{36}{x}$ . Count the integer points inside or on the boundary. All three graphs are symmetric about the origin, so count the first-quadrant region, double it, and add the origin.

Givens

Lines $y=4x$ and $y=\frac{1}{4}x$ , curve $y=\frac{36}{x}$ .
First-quadrant region: between the two lines and under the curve (corners O, A, B).
Third-quadrant region: its reflection through O (corners O, C, D).
Boundary points count.

Unknowns

The number of points with integer coordinates in the shaded regions.

Constraints

In the first quadrant a point counts when $\frac{1}{4}x \le y \le 4x$ and $xy \le 36$ .

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #17 Visualize Spatial Relationships#7 Identify Subproblems

Use symmetry to cut the work in half, then make a systematic list column by column: for each integer $x$ , find the lowest and highest allowed $y$ .

3 · Execute4 carry out the plan

1Find the corners

#7 Identify Subproblems 7.RP.A.2

Solve each line with the curve in the first quadrant.

4x = \frac{36}{x} \Rightarrow x = 3,\ A(3, 12); \quad \frac{1}{4}x = \frac{36}{x} \Rightarrow x = 12,\ B(12, 3)

So the first-quadrant region runs from

x = 0

to

x = 12

.

2Bounds in each column

#2 Make a Systematic List 7.RP.A.2

For an integer

x

from

1

to

12

,

y

must satisfy

\frac{x}{4} \le y \le 4x

and

y \le \frac{36}{x}

.

\frac{x}{4} \le y \le \min\left(4x, \frac{36}{x}\right)

The lower edge is the line

y=\frac{1}{4}x

; the upper edge is

y=4x

until A, then the curve.

3List the columns

#2 Make a Systematic List 6.NS.C.8

x=1

:

1 \le y \le 4

,

4

points.

x=2

:

1 \le y \le 8

,

8

points.

x=3

:

1 \le y \le 12

,

12

points.

x=4

:

1 \le y \le 9

,

9

points.

x=5

:

2 \le y \le 7

,

6

points.

x=6

:

2 \le y \le 6

,

5

points.

x=7

:

2 \le y \le 5

,

4

points.

x=8

:

2 \le y \le 4

,

3

points.

x=9

:

y = 3, 4

,

2

points.

x=10, 11, 12

: only

y=3

,

1

point each.

4 + 8 + 12 + 9 + 6 + 5 + 4 + 3 + 2 + 1 + 1 + 1 = 56

For example at

x=5

:

y \ge 1.25

and

y \le 7.2

, so

2 \le y \le 7

.

4Use symmetry

#17 Visualize Spatial Relationships 6.NS.C.8

Each graph is symmetric about the origin, so the third-quadrant region holds the mirror images of these

56

points. The origin is a corner of both regions and is counted once.

56 \times 2 + 1 = 113

(x, y)

is in one region exactly when

(-x, -y)

is in the other.

Answer:

113

4 · Reviewdoes it hold up?

Spot-check the curve: $(4, 9)$ gives $xy = 36 \le 36$ (in) but $(4, 10)$ gives $40 > 36$ (out). The corner points A $(3, 12)$ and B $(12, 3)$ are included.

Another way: Count by rows instead: for $y = 1$ to $12$ list the allowed $x$ (from $\frac{y}{4}$ up to $\min(4y, \frac{36}{y})$ ): $4, 8, 12, 9, 6, 5, 4, 3, 2, 1, 1, 1$ -- the same $56$ , because the region is symmetric about $y=x$ too.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Finding where the lines $y=4x$ and $y=\frac{1}{4}x$ meet the curve and bounding each column between them.
6.NS.C.8 Solve problems by graphing points in four quadrants — Listing integer points in the coordinate plane and mirroring them through the origin.

💡Takeaway. Symmetry halves the counting; a column-by-column list keeps it honest.

Count one quadrant, then use point symmetry · 12 practice problems

Generated variants — 12

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Where next?