Practice · Times give equations in reciprocal speeds · Sensim Math

Variant 1 easy

A fires at a target $680$ m away and hears the bullet hit the target $3$ seconds after firing. B stands $800$ m from A and $1140$ m from the target, and hears the bullet hit the target $2$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $680$ m from A; A hears the hit $3$ seconds after firing.
B is $800$ m from A and $1140$ m from the target.
B hears the hit $2$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 3 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

680

m to the target, then the sound of the hit travels the same

680

m back to A.

\frac{680}{x} + \frac{680}{y} = 3

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 2 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{800}{y}

seconds, and the sound of the hit reaches B after

\frac{680}{x} + \frac{1140}{y}

seconds. The gap between these is

2

seconds.

\frac{680}{x} + \frac{1140}{y} - \frac{800}{y} = 2,\quad \frac{680}{x} + \frac{340}{y} = 2

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{680}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{680}{y} - \frac{340}{y} = 3 - 2,\quad \frac{340}{y} = 1,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{680}{340} = 2

seconds to come back to A, so the bullet took the other

1

second of A's

3

.

\frac{680}{x} = 3 - 2 = 1,\quad x = 680

Once one block is known, the other comes straight out of the first equation.

Answer:

680

m per second

4 · Reviewdoes it hold up?

With $x = 680$ and $y = 340$ : A waits $1 + 2 = 3$ seconds. For B, the hit sound arrives at $1 + \frac{1140}{340}$ seconds and the gunshot at $\frac{800}{340}$ seconds; the gap is $1 + \frac{340}{340} = 1 + 1 = 2$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $680u + 680v = 3$ and $680u + 340v = 2$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{680}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 2 easy

A fires at a target $1020$ m away and hears the bullet hit the target $5$ seconds after firing. B stands $700$ m from A and $1380$ m from the target, and hears the bullet hit the target $4$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $1020$ m from A; A hears the hit $5$ seconds after firing.
B is $700$ m from A and $1380$ m from the target.
B hears the hit $4$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 5 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

1020

m to the target, then the sound of the hit travels the same

1020

m back to A.

\frac{1020}{x} + \frac{1020}{y} = 5

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 4 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{700}{y}

seconds, and the sound of the hit reaches B after

\frac{1020}{x} + \frac{1380}{y}

seconds. The gap between these is

4

seconds.

\frac{1020}{x} + \frac{1380}{y} - \frac{700}{y} = 4,\quad \frac{1020}{x} + \frac{680}{y} = 4

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{1020}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{1020}{y} - \frac{680}{y} = 5 - 4,\quad \frac{340}{y} = 1,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{1020}{340} = 3

seconds to come back to A, so the bullet took the other

2

seconds of A's

5

.

\frac{1020}{x} = 5 - 3 = 2,\quad x = 510

Once one block is known, the other comes straight out of the first equation.

Answer:

510

m per second

4 · Reviewdoes it hold up?

With $x = 510$ and $y = 340$ : A waits $2 + 3 = 5$ seconds. For B, the hit sound arrives at $2 + \frac{1380}{340}$ seconds and the gunshot at $\frac{700}{340}$ seconds; the gap is $2 + \frac{680}{340} = 2 + 2 = 4$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $1020u + 1020v = 5$ and $1020u + 680v = 4$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{510}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 3 easy

A fires at a target $1700$ m away and hears the bullet hit the target $7$ seconds after firing. B stands $980$ m from A and $2000$ m from the target, and hears the bullet hit the target $5$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $1700$ m from A; A hears the hit $7$ seconds after firing.
B is $980$ m from A and $2000$ m from the target.
B hears the hit $5$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 7 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

1700

m to the target, then the sound of the hit travels the same

1700

m back to A.

\frac{1700}{x} + \frac{1700}{y} = 7

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 5 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{980}{y}

seconds, and the sound of the hit reaches B after

\frac{1700}{x} + \frac{2000}{y}

seconds. The gap between these is

5

seconds.

\frac{1700}{x} + \frac{2000}{y} - \frac{980}{y} = 5,\quad \frac{1700}{x} + \frac{1020}{y} = 5

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{1700}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{1700}{y} - \frac{1020}{y} = 7 - 5,\quad \frac{680}{y} = 2,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{1700}{340} = 5

seconds to come back to A, so the bullet took the other

2

seconds of A's

7

.

\frac{1700}{x} = 7 - 5 = 2,\quad x = 850

Once one block is known, the other comes straight out of the first equation.

Answer:

850

m per second

4 · Reviewdoes it hold up?

With $x = 850$ and $y = 340$ : A waits $2 + 5 = 7$ seconds. For B, the hit sound arrives at $2 + \frac{2000}{340}$ seconds and the gunshot at $\frac{980}{340}$ seconds; the gap is $2 + \frac{1020}{340} = 2 + 3 = 5$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $1700u + 1700v = 7$ and $1700u + 1020v = 5$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{850}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 4 easy

A fires at a target $2040$ m away and hears the bullet hit the target $8$ seconds after firing. B stands $1500$ m from A and $1840$ m from the target, and hears the bullet hit the target $3$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $2040$ m from A; A hears the hit $8$ seconds after firing.
B is $1500$ m from A and $1840$ m from the target.
B hears the hit $3$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 8 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

2040

m to the target, then the sound of the hit travels the same

2040

m back to A.

\frac{2040}{x} + \frac{2040}{y} = 8

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 3 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{1500}{y}

seconds, and the sound of the hit reaches B after

\frac{2040}{x} + \frac{1840}{y}

seconds. The gap between these is

3

seconds.

\frac{2040}{x} + \frac{1840}{y} - \frac{1500}{y} = 3,\quad \frac{2040}{x} + \frac{340}{y} = 3

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{2040}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{2040}{y} - \frac{340}{y} = 8 - 3,\quad \frac{1700}{y} = 5,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{2040}{340} = 6

seconds to come back to A, so the bullet took the other

2

seconds of A's

8

.

\frac{2040}{x} = 8 - 6 = 2,\quad x = 1020

Once one block is known, the other comes straight out of the first equation.

Answer:

1020

m per second

4 · Reviewdoes it hold up?

With $x = 1020$ and $y = 340$ : A waits $2 + 6 = 8$ seconds. For B, the hit sound arrives at $2 + \frac{1840}{340}$ seconds and the gunshot at $\frac{1500}{340}$ seconds; the gap is $2 + \frac{340}{340} = 2 + 1 = 3$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $2040u + 2040v = 8$ and $2040u + 340v = 3$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{1020}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 5 medium

A fires at a target $1360$ m away and hears the bullet hit the target $6$ seconds after firing. B stands $1100$ m from A and $2120$ m from the target, and hears the bullet hit the target $5$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $1360$ m from A; A hears the hit $6$ seconds after firing.
B is $1100$ m from A and $2120$ m from the target.
B hears the hit $5$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 6 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

1360

m to the target, then the sound of the hit travels the same

1360

m back to A.

\frac{1360}{x} + \frac{1360}{y} = 6

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 5 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{1100}{y}

seconds, and the sound of the hit reaches B after

\frac{1360}{x} + \frac{2120}{y}

seconds. The gap between these is

5

seconds.

\frac{1360}{x} + \frac{2120}{y} - \frac{1100}{y} = 5,\quad \frac{1360}{x} + \frac{1020}{y} = 5

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{1360}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{1360}{y} - \frac{1020}{y} = 6 - 5,\quad \frac{340}{y} = 1,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{1360}{340} = 4

seconds to come back to A, so the bullet took the other

2

seconds of A's

6

.

\frac{1360}{x} = 6 - 4 = 2,\quad x = 680

Once one block is known, the other comes straight out of the first equation.

Answer:

680

m per second

4 · Reviewdoes it hold up?

With $x = 680$ and $y = 340$ : A waits $2 + 4 = 6$ seconds. For B, the hit sound arrives at $2 + \frac{2120}{340}$ seconds and the gunshot at $\frac{1100}{340}$ seconds; the gap is $2 + \frac{1020}{340} = 2 + 3 = 5$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $1360u + 1360v = 6$ and $1360u + 1020v = 5$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{680}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 6 medium

A fires at a target $2040$ m away and hears the bullet hit the target $10$ seconds after firing. B stands $1300$ m from A and $2660$ m from the target, and hears the bullet hit the target $8$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $2040$ m from A; A hears the hit $10$ seconds after firing.
B is $1300$ m from A and $2660$ m from the target.
B hears the hit $8$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 10 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

2040

m to the target, then the sound of the hit travels the same

2040

m back to A.

\frac{2040}{x} + \frac{2040}{y} = 10

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 8 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{1300}{y}

seconds, and the sound of the hit reaches B after

\frac{2040}{x} + \frac{2660}{y}

seconds. The gap between these is

8

seconds.

\frac{2040}{x} + \frac{2660}{y} - \frac{1300}{y} = 8,\quad \frac{2040}{x} + \frac{1360}{y} = 8

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{2040}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{2040}{y} - \frac{1360}{y} = 10 - 8,\quad \frac{680}{y} = 2,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{2040}{340} = 6

seconds to come back to A, so the bullet took the other

4

seconds of A's

10

.

\frac{2040}{x} = 10 - 6 = 4,\quad x = 510

Once one block is known, the other comes straight out of the first equation.

Answer:

510

m per second

4 · Reviewdoes it hold up?

With $x = 510$ and $y = 340$ : A waits $4 + 6 = 10$ seconds. For B, the hit sound arrives at $4 + \frac{2660}{340}$ seconds and the gunshot at $\frac{1300}{340}$ seconds; the gap is $4 + \frac{1360}{340} = 4 + 4 = 8$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $2040u + 2040v = 10$ and $2040u + 1360v = 8$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{510}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 7 medium

A fires at a target $1700$ m away and hears the bullet hit the target $9$ seconds after firing. B stands $1300$ m from A and $2660$ m from the target, and hears the bullet hit the target $8$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $1700$ m from A; A hears the hit $9$ seconds after firing.
B is $1300$ m from A and $2660$ m from the target.
B hears the hit $8$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 9 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

1700

m to the target, then the sound of the hit travels the same

1700

m back to A.

\frac{1700}{x} + \frac{1700}{y} = 9

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 8 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{1300}{y}

seconds, and the sound of the hit reaches B after

\frac{1700}{x} + \frac{2660}{y}

seconds. The gap between these is

8

seconds.

\frac{1700}{x} + \frac{2660}{y} - \frac{1300}{y} = 8,\quad \frac{1700}{x} + \frac{1360}{y} = 8

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{1700}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{1700}{y} - \frac{1360}{y} = 9 - 8,\quad \frac{340}{y} = 1,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{1700}{340} = 5

seconds to come back to A, so the bullet took the other

4

seconds of A's

9

.

\frac{1700}{x} = 9 - 5 = 4,\quad x = 425

Once one block is known, the other comes straight out of the first equation.

Answer:

425

m per second

4 · Reviewdoes it hold up?

With $x = 425$ and $y = 340$ : A waits $4 + 5 = 9$ seconds. For B, the hit sound arrives at $4 + \frac{2660}{340}$ seconds and the gunshot at $\frac{1300}{340}$ seconds; the gap is $4 + \frac{1360}{340} = 4 + 4 = 8$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $1700u + 1700v = 9$ and $1700u + 1360v = 8$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{425}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 8 medium

A fires at a target $2720$ m away and hears the bullet hit the target $12$ seconds after firing. B stands $1900$ m from A and $2580$ m from the target, and hears the bullet hit the target $6$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $2720$ m from A; A hears the hit $12$ seconds after firing.
B is $1900$ m from A and $2580$ m from the target.
B hears the hit $6$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 12 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

2720

m to the target, then the sound of the hit travels the same

2720

m back to A.

\frac{2720}{x} + \frac{2720}{y} = 12

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 6 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{1900}{y}

seconds, and the sound of the hit reaches B after

\frac{2720}{x} + \frac{2580}{y}

seconds. The gap between these is

6

seconds.

\frac{2720}{x} + \frac{2580}{y} - \frac{1900}{y} = 6,\quad \frac{2720}{x} + \frac{680}{y} = 6

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{2720}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{2720}{y} - \frac{680}{y} = 12 - 6,\quad \frac{2040}{y} = 6,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{2720}{340} = 8

seconds to come back to A, so the bullet took the other

4

seconds of A's

12

.

\frac{2720}{x} = 12 - 8 = 4,\quad x = 680

Once one block is known, the other comes straight out of the first equation.

Answer:

680

m per second

4 · Reviewdoes it hold up?

With $x = 680$ and $y = 340$ : A waits $4 + 8 = 12$ seconds. For B, the hit sound arrives at $4 + \frac{2580}{340}$ seconds and the gunshot at $\frac{1900}{340}$ seconds; the gap is $4 + \frac{680}{340} = 4 + 2 = 6$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $2720u + 2720v = 12$ and $2720u + 680v = 6$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{680}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 9 hard

A fires at a target $2380$ m away and hears the bullet hit the target $9$ seconds after firing. B stands $1300$ m from A and $3000$ m from the target, and hears the bullet hit the target $7$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $2380$ m from A; A hears the hit $9$ seconds after firing.
B is $1300$ m from A and $3000$ m from the target.
B hears the hit $7$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 9 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

2380

m to the target, then the sound of the hit travels the same

2380

m back to A.

\frac{2380}{x} + \frac{2380}{y} = 9

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 7 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{1300}{y}

seconds, and the sound of the hit reaches B after

\frac{2380}{x} + \frac{3000}{y}

seconds. The gap between these is

7

seconds.

\frac{2380}{x} + \frac{3000}{y} - \frac{1300}{y} = 7,\quad \frac{2380}{x} + \frac{1700}{y} = 7

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{2380}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{2380}{y} - \frac{1700}{y} = 9 - 7,\quad \frac{680}{y} = 2,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{2380}{340} = 7

seconds to come back to A, so the bullet took the other

2

seconds of A's

9

.

\frac{2380}{x} = 9 - 7 = 2,\quad x = 1190

Once one block is known, the other comes straight out of the first equation.

Answer:

1190

m per second

4 · Reviewdoes it hold up?

With $x = 1190$ and $y = 340$ : A waits $2 + 7 = 9$ seconds. For B, the hit sound arrives at $2 + \frac{3000}{340}$ seconds and the gunshot at $\frac{1300}{340}$ seconds; the gap is $2 + \frac{1700}{340} = 2 + 5 = 7$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $2380u + 2380v = 9$ and $2380u + 1700v = 7$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{1190}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 10 hard

A fires at a target $3060$ m away and hears the bullet hit the target $13$ seconds after firing. B stands $1500$ m from A and $2520$ m from the target, and hears the bullet hit the target $7$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $3060$ m from A; A hears the hit $13$ seconds after firing.
B is $1500$ m from A and $2520$ m from the target.
B hears the hit $7$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 13 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

3060

m to the target, then the sound of the hit travels the same

3060

m back to A.

\frac{3060}{x} + \frac{3060}{y} = 13

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 7 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{1500}{y}

seconds, and the sound of the hit reaches B after

\frac{3060}{x} + \frac{2520}{y}

seconds. The gap between these is

7

seconds.

\frac{3060}{x} + \frac{2520}{y} - \frac{1500}{y} = 7,\quad \frac{3060}{x} + \frac{1020}{y} = 7

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{3060}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{3060}{y} - \frac{1020}{y} = 13 - 7,\quad \frac{2040}{y} = 6,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{3060}{340} = 9

seconds to come back to A, so the bullet took the other

4

seconds of A's

13

.

\frac{3060}{x} = 13 - 9 = 4,\quad x = 765

Once one block is known, the other comes straight out of the first equation.

Answer:

765

m per second

4 · Reviewdoes it hold up?

With $x = 765$ and $y = 340$ : A waits $4 + 9 = 13$ seconds. For B, the hit sound arrives at $4 + \frac{2520}{340}$ seconds and the gunshot at $\frac{1500}{340}$ seconds; the gap is $4 + \frac{1020}{340} = 4 + 3 = 7$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $3060u + 3060v = 13$ and $3060u + 1020v = 7$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{765}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 11 hard

A fires at a target $2040$ m away and hears the bullet hit the target $9$ seconds after firing. B stands $2100$ m from A and $3120$ m from the target, and hears the bullet hit the target $6$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $2040$ m from A; A hears the hit $9$ seconds after firing.
B is $2100$ m from A and $3120$ m from the target.
B hears the hit $6$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 9 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

2040

m to the target, then the sound of the hit travels the same

2040

m back to A.

\frac{2040}{x} + \frac{2040}{y} = 9

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 6 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{2100}{y}

seconds, and the sound of the hit reaches B after

\frac{2040}{x} + \frac{3120}{y}

seconds. The gap between these is

6

seconds.

\frac{2040}{x} + \frac{3120}{y} - \frac{2100}{y} = 6,\quad \frac{2040}{x} + \frac{1020}{y} = 6

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{2040}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{2040}{y} - \frac{1020}{y} = 9 - 6,\quad \frac{1020}{y} = 3,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{2040}{340} = 6

seconds to come back to A, so the bullet took the other

3

seconds of A's

9

.

\frac{2040}{x} = 9 - 6 = 3,\quad x = 680

Once one block is known, the other comes straight out of the first equation.

Answer:

680

m per second

4 · Reviewdoes it hold up?

With $x = 680$ and $y = 340$ : A waits $3 + 6 = 9$ seconds. For B, the hit sound arrives at $3 + \frac{3120}{340}$ seconds and the gunshot at $\frac{2100}{340}$ seconds; the gap is $3 + \frac{1020}{340} = 3 + 3 = 6$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $2040u + 2040v = 9$ and $2040u + 1020v = 6$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{680}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Variant 12 hard

A fires at a target $3400$ m away and hears the bullet hit the target $14$ seconds after firing. B stands $2600$ m from A and $4300$ m from the target, and hears the bullet hit the target $9$ seconds after hearing A's gunshot. Find the speed of the bullet. (The bullet and the sound each travel at a constant speed.)

Your answer

Show solution

1 · Understandwhat's really being asked

A bullet flies to a target and the sound of the hit flies back. Two listeners time it differently: A from the moment of firing, B from the moment the gunshot reaches B. Find the bullet's speed.

Givens

The target is $3400$ m from A; A hears the hit $14$ seconds after firing.
B is $2600$ m from A and $4300$ m from the target.
B hears the hit $9$ seconds after hearing the gunshot.

Unknowns

The speed of the bullet.

Constraints

The speed of sound is not given, but it is constant, so it is a second unknown.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Each time in the story is a distance divided by either the bullet's speed or the speed of sound. Drawing each listener's timeline gives two equations; both contain the same bullet time, so subtracting them leaves only sound.

3 · Execute5 carry out the plan

1Name the two speeds

#8 Analyze the Units 6.RP.A.3

Let the bullet travel

x

m per second and sound travel

y

m per second. Every duration in the story is a distance divided by one of these speeds.

\text{time} = \frac{\text{distance}}{\text{speed}}

A time of travel is just distance over speed, so the unknown speeds appear in the denominators.

2Follow A's 14 seconds

#1 Draw a Diagram 8.EE.C.8

The bullet flies

3400

m to the target, then the sound of the hit travels the same

3400

m back to A.

\frac{3400}{x} + \frac{3400}{y} = 14

A's waiting time splits into two legs: bullet out, sound back.

3Follow B's 9 seconds

#1 Draw a Diagram 8.EE.C.8

Measured from the shot, the gunshot reaches B after

\frac{2600}{y}

seconds, and the sound of the hit reaches B after

\frac{3400}{x} + \frac{4300}{y}

seconds. The gap between these is

9

seconds.

\frac{3400}{x} + \frac{4300}{y} - \frac{2600}{y} = 9,\quad \frac{3400}{x} + \frac{1700}{y} = 9

Put both sounds on one timeline starting at the shot, then take the difference.

4Subtract to remove the bullet

#13 Convert to Algebra 8.EE.C.8

Both equations contain the same bullet time

\frac{3400}{x}

. Subtracting the second from the first leaves only the speed of sound.

\frac{3400}{y} - \frac{1700}{y} = 14 - 9,\quad \frac{1700}{y} = 5,\quad y = 340

Treat each fraction as one block; the bullet block cancels exactly like a variable would.

5Find the bullet's speed

#13 Convert to Algebra 8.EE.C.8

Sound takes

\frac{3400}{340} = 10

seconds to come back to A, so the bullet took the other

4

seconds of A's

14

.

\frac{3400}{x} = 14 - 10 = 4,\quad x = 850

Once one block is known, the other comes straight out of the first equation.

Answer:

850

m per second

4 · Reviewdoes it hold up?

With $x = 850$ and $y = 340$ : A waits $4 + 10 = 14$ seconds. For B, the hit sound arrives at $4 + \frac{4300}{340}$ seconds and the gunshot at $\frac{2600}{340}$ seconds; the gap is $4 + \frac{1700}{340} = 4 + 5 = 9$ seconds. Both match, and $340$ m per second is the real speed of sound in air.

Another way: Write $u = \frac{1}{x}$ and $v = \frac{1}{y}$ . The equations become the ordinary linear system $3400u + 3400v = 14$ and $3400u + 1700v = 9$ , which elimination solves in the usual way: $v = \frac{1}{340}$ , $u = \frac{1}{850}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Setting up and solving the two timing equations together.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning each distance and speed into a travel time.

💡Takeaway. When the unknowns sit in denominators, treat each whole fraction as the thing to eliminate -- the system is still linear in those blocks.

Times give equations in reciprocal speeds · 12 practice problems

Generated variants — 12

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