Practice · Read each runner's rate from one graph point · Sensim Math

Variant 1 easy

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $100$ m race at the same moment. Runner A reaches $30$ m after $5$ seconds, and runner B reaches $30$ m after $6$ seconds. How many seconds apart are their finishing times for the $100$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $100$ m time and subtract.

Givens

Runner A's line passes through $(5, 30)$ .
Runner B's line passes through $(6, 30)$ .
Both start together at the origin and run $100$ m.

Unknowns

The difference between their $100$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $100$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

30

m in

5

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{30}{5} = 6$, so $y = 6x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

6

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

30

m but needs

6

s.

k_B = \frac{30}{6} = 5$, so $y = 5x

B is slower:

5

meters each second.

3Time each runner over 100 m

#8 Analyze the Units 7.RP.A.2

Put

y = 100

into each equation and solve for

x

.

6x = 100 \Rightarrow x = \frac{50}{3}, \quad 5x = 100 \Rightarrow x = 20

Time

=

distance

\div

speed:

100 \div 6

and

100 \div 5

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

20 - \frac{50}{3} = \frac{60}{3} - \frac{50}{3} = \frac{10}{3}

Rewrite both times over the denominator

3

, then subtract.

Answer:

\frac{10}{3}

seconds

4 · Reviewdoes it hold up?

A needs about 16.7 s and B 20 s, a gap of about 3.3 s, matching $\frac{10}{3}$ . A gap of a few seconds over $100$ m between $6$ m/s and $5$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{6}$ s per meter and B takes $\frac{1}{5}$ s, a gap of $\frac{1}{5} - \frac{1}{6} = \frac{1}{30}$ s per meter, so over $100$ m the gap is $100 \times \frac{1}{30} = \frac{10}{3}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 2 easy

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $100$ m race at the same moment. Runner A reaches $36$ m after $4$ seconds, and runner B reaches $36$ m after $6$ seconds. How many seconds apart are their finishing times for the $100$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $100$ m time and subtract.

Givens

Runner A's line passes through $(4, 36)$ .
Runner B's line passes through $(6, 36)$ .
Both start together at the origin and run $100$ m.

Unknowns

The difference between their $100$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $100$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

36

m in

4

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{36}{4} = 9$, so $y = 9x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

9

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

36

m but needs

6

s.

k_B = \frac{36}{6} = 6$, so $y = 6x

B is slower:

6

meters each second.

3Time each runner over 100 m

#8 Analyze the Units 7.RP.A.2

Put

y = 100

into each equation and solve for

x

.

9x = 100 \Rightarrow x = \frac{100}{9}, \quad 6x = 100 \Rightarrow x = \frac{50}{3}

Time

=

distance

\div

speed:

100 \div 9

and

100 \div 6

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

\frac{50}{3} - \frac{100}{9} = \frac{150}{9} - \frac{100}{9} = \frac{50}{9}

Rewrite both times over the denominator

9

, then subtract.

Answer:

\frac{50}{9}

seconds

4 · Reviewdoes it hold up?

A needs about 11.1 s and B about 16.7 s, a gap of about 5.6 s, matching $\frac{50}{9}$ . A gap of a few seconds over $100$ m between $9$ m/s and $6$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{9}$ s per meter and B takes $\frac{1}{6}$ s, a gap of $\frac{1}{6} - \frac{1}{9} = \frac{1}{18}$ s per meter, so over $100$ m the gap is $100 \times \frac{1}{18} = \frac{50}{9}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 3 easy

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $100$ m race at the same moment. Runner A reaches $42$ m after $6$ seconds, and runner B reaches $42$ m after $7$ seconds. How many seconds apart are their finishing times for the $100$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $100$ m time and subtract.

Givens

Runner A's line passes through $(6, 42)$ .
Runner B's line passes through $(7, 42)$ .
Both start together at the origin and run $100$ m.

Unknowns

The difference between their $100$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $100$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

42

m in

6

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{42}{6} = 7$, so $y = 7x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

7

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

42

m but needs

7

s.

k_B = \frac{42}{7} = 6$, so $y = 6x

B is slower:

6

meters each second.

3Time each runner over 100 m

#8 Analyze the Units 7.RP.A.2

Put

y = 100

into each equation and solve for

x

.

7x = 100 \Rightarrow x = \frac{100}{7}, \quad 6x = 100 \Rightarrow x = \frac{50}{3}

Time

=

distance

\div

speed:

100 \div 7

and

100 \div 6

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

\frac{50}{3} - \frac{100}{7} = \frac{350}{21} - \frac{300}{21} = \frac{50}{21}

Rewrite both times over the denominator

21

, then subtract.

Answer:

\frac{50}{21}

seconds

4 · Reviewdoes it hold up?

A needs about 14.3 s and B about 16.7 s, a gap of about 2.4 s, matching $\frac{50}{21}$ . A gap of a few seconds over $100$ m between $7$ m/s and $6$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{7}$ s per meter and B takes $\frac{1}{6}$ s, a gap of $\frac{1}{6} - \frac{1}{7} = \frac{1}{42}$ s per meter, so over $100$ m the gap is $100 \times \frac{1}{42} = \frac{50}{21}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 4 hard

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $100$ m race at the same moment. Runner A reaches $40$ m after $5$ seconds, and runner B reaches $40$ m after $8$ seconds. How many seconds apart are their finishing times for the $100$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $100$ m time and subtract.

Givens

Runner A's line passes through $(5, 40)$ .
Runner B's line passes through $(8, 40)$ .
Both start together at the origin and run $100$ m.

Unknowns

The difference between their $100$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $100$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

40

m in

5

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{40}{5} = 8$, so $y = 8x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

8

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

40

m but needs

8

s.

k_B = \frac{40}{8} = 5$, so $y = 5x

B is slower:

5

meters each second.

3Time each runner over 100 m

#8 Analyze the Units 7.RP.A.2

Put

y = 100

into each equation and solve for

x

.

8x = 100 \Rightarrow x = \frac{25}{2}, \quad 5x = 100 \Rightarrow x = 20

Time

=

distance

\div

speed:

100 \div 8

and

100 \div 5

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

20 - \frac{25}{2} = \frac{40}{2} - \frac{25}{2} = \frac{15}{2}

Rewrite both times over the denominator

2

, then subtract.

Answer:

\frac{15}{2}

seconds

4 · Reviewdoes it hold up?

A needs 12.5 s and B 20 s, a gap of 7.5 s, matching $\frac{15}{2}$ . A gap of a few seconds over $100$ m between $8$ m/s and $5$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{8}$ s per meter and B takes $\frac{1}{5}$ s, a gap of $\frac{1}{5} - \frac{1}{8} = \frac{3}{40}$ s per meter, so over $100$ m the gap is $100 \times \frac{3}{40} = \frac{15}{2}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 5 hard

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $100$ m race at the same moment. Runner A reaches $45$ m after $5$ seconds, and runner B reaches $45$ m after $9$ seconds. How many seconds apart are their finishing times for the $100$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $100$ m time and subtract.

Givens

Runner A's line passes through $(5, 45)$ .
Runner B's line passes through $(9, 45)$ .
Both start together at the origin and run $100$ m.

Unknowns

The difference between their $100$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $100$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

45

m in

5

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{45}{5} = 9$, so $y = 9x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

9

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

45

m but needs

9

s.

k_B = \frac{45}{9} = 5$, so $y = 5x

B is slower:

5

meters each second.

3Time each runner over 100 m

#8 Analyze the Units 7.RP.A.2

Put

y = 100

into each equation and solve for

x

.

9x = 100 \Rightarrow x = \frac{100}{9}, \quad 5x = 100 \Rightarrow x = 20

Time

=

distance

\div

speed:

100 \div 9

and

100 \div 5

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

20 - \frac{100}{9} = \frac{180}{9} - \frac{100}{9} = \frac{80}{9}

Rewrite both times over the denominator

9

, then subtract.

Answer:

\frac{80}{9}

seconds

4 · Reviewdoes it hold up?

A needs about 11.1 s and B 20 s, a gap of about 8.9 s, matching $\frac{80}{9}$ . A gap of a few seconds over $100$ m between $9$ m/s and $5$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{9}$ s per meter and B takes $\frac{1}{5}$ s, a gap of $\frac{1}{5} - \frac{1}{9} = \frac{4}{45}$ s per meter, so over $100$ m the gap is $100 \times \frac{4}{45} = \frac{80}{9}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 6 easy

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $100$ m race at the same moment. Runner A reaches $24$ m after $3$ seconds, and runner B reaches $24$ m after $4$ seconds. How many seconds apart are their finishing times for the $100$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $100$ m time and subtract.

Givens

Runner A's line passes through $(3, 24)$ .
Runner B's line passes through $(4, 24)$ .
Both start together at the origin and run $100$ m.

Unknowns

The difference between their $100$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $100$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

24

m in

3

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{24}{3} = 8$, so $y = 8x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

8

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

24

m but needs

4

s.

k_B = \frac{24}{4} = 6$, so $y = 6x

B is slower:

6

meters each second.

3Time each runner over 100 m

#8 Analyze the Units 7.RP.A.2

Put

y = 100

into each equation and solve for

x

.

8x = 100 \Rightarrow x = \frac{25}{2}, \quad 6x = 100 \Rightarrow x = \frac{50}{3}

Time

=

distance

\div

speed:

100 \div 8

and

100 \div 6

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

\frac{50}{3} - \frac{25}{2} = \frac{100}{6} - \frac{75}{6} = \frac{25}{6}

Rewrite both times over the denominator

6

, then subtract.

Answer:

\frac{25}{6}

seconds

4 · Reviewdoes it hold up?

A needs 12.5 s and B about 16.7 s, a gap of about 4.2 s, matching $\frac{25}{6}$ . A gap of a few seconds over $100$ m between $8$ m/s and $6$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{8}$ s per meter and B takes $\frac{1}{6}$ s, a gap of $\frac{1}{6} - \frac{1}{8} = \frac{1}{24}$ s per meter, so over $100$ m the gap is $100 \times \frac{1}{24} = \frac{25}{6}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 7 medium

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $100$ m race at the same moment. Runner A reaches $35$ m after $5$ seconds, and runner B reaches $35$ m after $7$ seconds. How many seconds apart are their finishing times for the $100$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $100$ m time and subtract.

Givens

Runner A's line passes through $(5, 35)$ .
Runner B's line passes through $(7, 35)$ .
Both start together at the origin and run $100$ m.

Unknowns

The difference between their $100$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $100$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

35

m in

5

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{35}{5} = 7$, so $y = 7x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

7

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

35

m but needs

7

s.

k_B = \frac{35}{7} = 5$, so $y = 5x

B is slower:

5

meters each second.

3Time each runner over 100 m

#8 Analyze the Units 7.RP.A.2

Put

y = 100

into each equation and solve for

x

.

7x = 100 \Rightarrow x = \frac{100}{7}, \quad 5x = 100 \Rightarrow x = 20

Time

=

distance

\div

speed:

100 \div 7

and

100 \div 5

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

20 - \frac{100}{7} = \frac{140}{7} - \frac{100}{7} = \frac{40}{7}

Rewrite both times over the denominator

7

, then subtract.

Answer:

\frac{40}{7}

seconds

4 · Reviewdoes it hold up?

A needs about 14.3 s and B 20 s, a gap of about 5.7 s, matching $\frac{40}{7}$ . A gap of a few seconds over $100$ m between $7$ m/s and $5$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{7}$ s per meter and B takes $\frac{1}{5}$ s, a gap of $\frac{1}{5} - \frac{1}{7} = \frac{2}{35}$ s per meter, so over $100$ m the gap is $100 \times \frac{2}{35} = \frac{40}{7}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 8 medium

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $100$ m race at the same moment. Runner A reaches $28$ m after $4$ seconds, and runner B reaches $28$ m after $7$ seconds. How many seconds apart are their finishing times for the $100$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $100$ m time and subtract.

Givens

Runner A's line passes through $(4, 28)$ .
Runner B's line passes through $(7, 28)$ .
Both start together at the origin and run $100$ m.

Unknowns

The difference between their $100$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $100$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

28

m in

4

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{28}{4} = 7$, so $y = 7x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

7

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

28

m but needs

7

s.

k_B = \frac{28}{7} = 4$, so $y = 4x

B is slower:

4

meters each second.

3Time each runner over 100 m

#8 Analyze the Units 7.RP.A.2

Put

y = 100

into each equation and solve for

x

.

7x = 100 \Rightarrow x = \frac{100}{7}, \quad 4x = 100 \Rightarrow x = 25

Time

=

distance

\div

speed:

100 \div 7

and

100 \div 4

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

25 - \frac{100}{7} = \frac{175}{7} - \frac{100}{7} = \frac{75}{7}

Rewrite both times over the denominator

7

, then subtract.

Answer:

\frac{75}{7}

seconds

4 · Reviewdoes it hold up?

A needs about 14.3 s and B 25 s, a gap of about 10.7 s, matching $\frac{75}{7}$ . A gap of a few seconds over $100$ m between $7$ m/s and $4$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{7}$ s per meter and B takes $\frac{1}{4}$ s, a gap of $\frac{1}{4} - \frac{1}{7} = \frac{3}{28}$ s per meter, so over $100$ m the gap is $100 \times \frac{3}{28} = \frac{75}{7}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 9 medium

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $100$ m race at the same moment. Runner A reaches $24$ m after $4$ seconds, and runner B reaches $24$ m after $6$ seconds. How many seconds apart are their finishing times for the $100$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $100$ m time and subtract.

Givens

Runner A's line passes through $(4, 24)$ .
Runner B's line passes through $(6, 24)$ .
Both start together at the origin and run $100$ m.

Unknowns

The difference between their $100$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $100$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

24

m in

4

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{24}{4} = 6$, so $y = 6x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

6

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

24

m but needs

6

s.

k_B = \frac{24}{6} = 4$, so $y = 4x

B is slower:

4

meters each second.

3Time each runner over 100 m

#8 Analyze the Units 7.RP.A.2

Put

y = 100

into each equation and solve for

x

.

6x = 100 \Rightarrow x = \frac{50}{3}, \quad 4x = 100 \Rightarrow x = 25

Time

=

distance

\div

speed:

100 \div 6

and

100 \div 4

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

25 - \frac{50}{3} = \frac{75}{3} - \frac{50}{3} = \frac{25}{3}

Rewrite both times over the denominator

3

, then subtract.

Answer:

\frac{25}{3}

seconds

4 · Reviewdoes it hold up?

A needs about 16.7 s and B 25 s, a gap of about 8.3 s, matching $\frac{25}{3}$ . A gap of a few seconds over $100$ m between $6$ m/s and $4$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{6}$ s per meter and B takes $\frac{1}{4}$ s, a gap of $\frac{1}{4} - \frac{1}{6} = \frac{1}{12}$ s per meter, so over $100$ m the gap is $100 \times \frac{1}{12} = \frac{25}{3}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 10 medium

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $100$ m race at the same moment. Runner A reaches $32$ m after $4$ seconds, and runner B reaches $32$ m after $8$ seconds. How many seconds apart are their finishing times for the $100$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $100$ m time and subtract.

Givens

Runner A's line passes through $(4, 32)$ .
Runner B's line passes through $(8, 32)$ .
Both start together at the origin and run $100$ m.

Unknowns

The difference between their $100$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $100$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

32

m in

4

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{32}{4} = 8$, so $y = 8x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

8

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

32

m but needs

8

s.

k_B = \frac{32}{8} = 4$, so $y = 4x

B is slower:

4

meters each second.

3Time each runner over 100 m

#8 Analyze the Units 7.RP.A.2

Put

y = 100

into each equation and solve for

x

.

8x = 100 \Rightarrow x = \frac{25}{2}, \quad 4x = 100 \Rightarrow x = 25

Time

=

distance

\div

speed:

100 \div 8

and

100 \div 4

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

25 - \frac{25}{2} = \frac{50}{2} - \frac{25}{2} = \frac{25}{2}

Rewrite both times over the denominator

2

, then subtract.

Answer:

\frac{25}{2}

seconds

4 · Reviewdoes it hold up?

A needs 12.5 s and B 25 s, a gap of 12.5 s, matching $\frac{25}{2}$ . A gap of a few seconds over $100$ m between $8$ m/s and $4$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{8}$ s per meter and B takes $\frac{1}{4}$ s, a gap of $\frac{1}{4} - \frac{1}{8} = \frac{1}{8}$ s per meter, so over $100$ m the gap is $100 \times \frac{1}{8} = \frac{25}{2}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 11 hard

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $200$ m race at the same moment. Runner A reaches $48$ m after $6$ seconds, and runner B reaches $48$ m after $8$ seconds. How many seconds apart are their finishing times for the $200$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $200$ m time and subtract.

Givens

Runner A's line passes through $(6, 48)$ .
Runner B's line passes through $(8, 48)$ .
Both start together at the origin and run $200$ m.

Unknowns

The difference between their $200$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $200$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

48

m in

6

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{48}{6} = 8$, so $y = 8x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

8

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

48

m but needs

8

s.

k_B = \frac{48}{8} = 6$, so $y = 6x

B is slower:

6

meters each second.

3Time each runner over 200 m

#8 Analyze the Units 7.RP.A.2

Put

y = 200

into each equation and solve for

x

.

8x = 200 \Rightarrow x = 25, \quad 6x = 200 \Rightarrow x = \frac{100}{3}

Time

=

distance

\div

speed:

200 \div 8

and

200 \div 6

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

\frac{100}{3} - 25 = \frac{100}{3} - \frac{75}{3} = \frac{25}{3}

Rewrite both times over the denominator

3

, then subtract.

Answer:

\frac{25}{3}

seconds

4 · Reviewdoes it hold up?

A needs 25 s and B about 33.3 s, a gap of about 8.3 s, matching $\frac{25}{3}$ . A gap of a few seconds over $200$ m between $8$ m/s and $6$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{8}$ s per meter and B takes $\frac{1}{6}$ s, a gap of $\frac{1}{6} - \frac{1}{8} = \frac{1}{24}$ s per meter, so over $200$ m the gap is $200 \times \frac{1}{24} = \frac{25}{3}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Variant 12 hard

The graph shows part of the relationship between the time $x$ seconds and the distance $y$ m run by two runners, A and B, who start a $200$ m race at the same moment. Runner A reaches $60$ m after $6$ seconds, and runner B reaches $60$ m after $10$ seconds. How many seconds apart are their finishing times for the $200$ m?

Your answer

Show solution

1 · Understandwhat's really being asked

Both runners' distance is proportional to time, so each graph is a line $y=kx$ through the origin. One marked point on each line gives its constant $k$ (the speed). Use the speeds to find each runner's $200$ m time and subtract.

Givens

Runner A's line passes through $(6, 60)$ .
Runner B's line passes through $(10, 60)$ .
Both start together at the origin and run $200$ m.

Unknowns

The difference between their $200$ m times, in seconds.

Constraints

Each graph is a straight line through the origin, so the speed stays constant.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#8 Analyze the Units

Split the job: first read each runner's constant from one point, then turn $200$ m into a time for each runner, then compare. The units (meters per second) say the constant is a speed.

3 · Execute4 carry out the plan

1Read A's speed

#13 Convert to Algebra 7.RP.A.2

A covers

60

m in

6

s, so

\frac{y}{x}

is the same everywhere on A's line.

k_A = \frac{60}{6} = 10$, so $y = 10x

On a line through the origin,

\frac{y}{x}

is the constant of proportionality:

10

meters each second.

2Read B's speed

#13 Convert to Algebra 7.RP.A.2

B covers the same

60

m but needs

10

s.

k_B = \frac{60}{10} = 6$, so $y = 6x

B is slower:

6

meters each second.

3Time each runner over 200 m

#8 Analyze the Units 7.RP.A.2

Put

y = 200

into each equation and solve for

x

.

10x = 200 \Rightarrow x = 20, \quad 6x = 200 \Rightarrow x = \frac{100}{3}

Time

=

distance

\div

speed:

200 \div 10

and

200 \div 6

seconds.

4Subtract the times

#7 Identify Subproblems 7.NS.A.1

B finishes later, so subtract A's time from B's using a common denominator.

\frac{100}{3} - 20 = \frac{100}{3} - \frac{60}{3} = \frac{40}{3}

Rewrite both times over the denominator

3

, then subtract.

Answer:

\frac{40}{3}

seconds

4 · Reviewdoes it hold up?

A needs 20 s and B about 33.3 s, a gap of about 13.3 s, matching $\frac{40}{3}$ . A gap of a few seconds over $200$ m between $10$ m/s and $6$ m/s runners is reasonable.

Another way: Compare per meter instead: A takes $\frac{1}{10}$ s per meter and B takes $\frac{1}{6}$ s, a gap of $\frac{1}{6} - \frac{1}{10} = \frac{1}{15}$ s per meter, so over $200$ m the gap is $200 \times \frac{1}{15} = \frac{40}{3}$ s.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading each runner's constant of proportionality (speed) from a point on the graph.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Subtracting the two fractional times.

💡Takeaway. One point on a line through the origin tells you the whole line: divide

y

by

x

.

Read each runner's rate from one graph point · 12 practice problems

Generated variants — 12

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