Practice · Meshed gears share teeth times turns · Sensim Math

Variant 1 easy

Gears P and Q, whose numbers of teeth are in the ratio $1:4$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $3:4$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=1:4$ .
Teeth of Q : teeth of R $=3:4$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

4

parts in the first ratio and

3

parts in the second. Rescale both so Q is

12

parts.

\text{P}:\text{Q}=1:4=3:12,\quad \text{Q}:\text{R}=3:4=12:16 \Rightarrow \text{P}:\text{Q}:\text{R}=3:12:16

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

3a

,

12a

and

16a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

3a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

16a \times y

teeth.

3a \times x = 16a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

16a

.

y=\frac{3a}{16a}x=\frac{3}{16}x

R has more teeth than P, so R turns more slowly; the factor

\frac{3}{16}

is less than

1

.

Answer:

y=\frac{3}{16}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=16$ : P passes $3 \times 16=48$ teeth a minute. Q, with $12$ teeth, turns $4$ times and passes $48$ teeth. R, with $16$ teeth, turns $3$ times: $16 \times 3=48$ . The formula gives $y=\frac{3}{16}\times16=3$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{1}{4}x$ times and R turns $\frac{3}{4}$ of Q's turns: $y=\frac{3}{4}\times\frac{1}{4}x=\frac{3}{16}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $3:12:16$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{3}{16}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 2 easy

Gears P and Q, whose numbers of teeth are in the ratio $2:3$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $4:3$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=2:3$ .
Teeth of Q : teeth of R $=4:3$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

3

parts in the first ratio and

4

parts in the second. Rescale both so Q is

12

parts.

\text{P}:\text{Q}=2:3=8:12,\quad \text{Q}:\text{R}=4:3=12:9 \Rightarrow \text{P}:\text{Q}:\text{R}=8:12:9

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

8a

,

12a

and

9a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

8a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

9a \times y

teeth.

8a \times x = 9a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

9a

.

y=\frac{8a}{9a}x=\frac{8}{9}x

R has more teeth than P, so R turns more slowly; the factor

\frac{8}{9}

is less than

1

.

Answer:

y=\frac{8}{9}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=9$ : P passes $8 \times 9=72$ teeth a minute. Q, with $12$ teeth, turns $6$ times and passes $72$ teeth. R, with $9$ teeth, turns $8$ times: $9 \times 8=72$ . The formula gives $y=\frac{8}{9}\times9=8$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{2}{3}x$ times and R turns $\frac{4}{3}$ of Q's turns: $y=\frac{4}{3}\times\frac{2}{3}x=\frac{8}{9}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $8:12:9$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{8}{9}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 3 easy

Gears P and Q, whose numbers of teeth are in the ratio $1:2$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $3:4$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=1:2$ .
Teeth of Q : teeth of R $=3:4$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

2

parts in the first ratio and

3

parts in the second. Rescale both so Q is

6

parts.

\text{P}:\text{Q}=1:2=3:6,\quad \text{Q}:\text{R}=3:4=6:8 \Rightarrow \text{P}:\text{Q}:\text{R}=3:6:8

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

3a

,

6a

and

8a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

3a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

8a \times y

teeth.

3a \times x = 8a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

8a

.

y=\frac{3a}{8a}x=\frac{3}{8}x

R has more teeth than P, so R turns more slowly; the factor

\frac{3}{8}

is less than

1

.

Answer:

y=\frac{3}{8}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=8$ : P passes $3 \times 8=24$ teeth a minute. Q, with $6$ teeth, turns $4$ times and passes $24$ teeth. R, with $8$ teeth, turns $3$ times: $8 \times 3=24$ . The formula gives $y=\frac{3}{8}\times8=3$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{1}{2}x$ times and R turns $\frac{3}{4}$ of Q's turns: $y=\frac{3}{4}\times\frac{1}{2}x=\frac{3}{8}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $3:6:8$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{3}{8}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 4 medium

Gears P and Q, whose numbers of teeth are in the ratio $3:5$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $3:4$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=3:5$ .
Teeth of Q : teeth of R $=3:4$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

5

parts in the first ratio and

3

parts in the second. Rescale both so Q is

15

parts.

\text{P}:\text{Q}=3:5=9:15,\quad \text{Q}:\text{R}=3:4=15:20 \Rightarrow \text{P}:\text{Q}:\text{R}=9:15:20

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

9a

,

15a

and

20a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

9a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

20a \times y

teeth.

9a \times x = 20a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

20a

.

y=\frac{9a}{20a}x=\frac{9}{20}x

R has more teeth than P, so R turns more slowly; the factor

\frac{9}{20}

is less than

1

.

Answer:

y=\frac{9}{20}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=20$ : P passes $9 \times 20=180$ teeth a minute. Q, with $15$ teeth, turns $12$ times and passes $180$ teeth. R, with $20$ teeth, turns $9$ times: $20 \times 9=180$ . The formula gives $y=\frac{9}{20}\times20=9$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{3}{5}x$ times and R turns $\frac{3}{4}$ of Q's turns: $y=\frac{3}{4}\times\frac{3}{5}x=\frac{9}{20}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $9:15:20$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{9}{20}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 5 hard

Gears P and Q, whose numbers of teeth are in the ratio $5:3$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $2:5$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=5:3$ .
Teeth of Q : teeth of R $=2:5$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

3

parts in the first ratio and

2

parts in the second. Rescale both so Q is

6

parts.

\text{P}:\text{Q}=5:3=10:6,\quad \text{Q}:\text{R}=2:5=6:15 \Rightarrow \text{P}:\text{Q}:\text{R}=10:6:15

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

10a

,

6a

and

15a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

10a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

15a \times y

teeth.

10a \times x = 15a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

15a

.

y=\frac{10a}{15a}x=\frac{10}{15}x=\frac{2}{3}x

R has more teeth than P, so R turns more slowly; the factor

\frac{2}{3}

is less than

1

.

Answer:

y=\frac{2}{3}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=15$ : P passes $10 \times 15=150$ teeth a minute. Q, with $6$ teeth, turns $25$ times and passes $150$ teeth. R, with $15$ teeth, turns $10$ times: $15 \times 10=150$ . The formula gives $y=\frac{2}{3}\times15=10$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{5}{3}x$ times and R turns $\frac{2}{5}$ of Q's turns: $y=\frac{2}{5}\times\frac{5}{3}x=\frac{2}{3}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $10:6:15$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{2}{3}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 6 medium

Gears P and Q, whose numbers of teeth are in the ratio $1:5$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $4:3$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=1:5$ .
Teeth of Q : teeth of R $=4:3$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

5

parts in the first ratio and

4

parts in the second. Rescale both so Q is

20

parts.

\text{P}:\text{Q}=1:5=4:20,\quad \text{Q}:\text{R}=4:3=20:15 \Rightarrow \text{P}:\text{Q}:\text{R}=4:20:15

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

4a

,

20a

and

15a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

4a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

15a \times y

teeth.

4a \times x = 15a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

15a

.

y=\frac{4a}{15a}x=\frac{4}{15}x

R has more teeth than P, so R turns more slowly; the factor

\frac{4}{15}

is less than

1

.

Answer:

y=\frac{4}{15}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=15$ : P passes $4 \times 15=60$ teeth a minute. Q, with $20$ teeth, turns $3$ times and passes $60$ teeth. R, with $15$ teeth, turns $4$ times: $15 \times 4=60$ . The formula gives $y=\frac{4}{15}\times15=4$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{1}{5}x$ times and R turns $\frac{4}{3}$ of Q's turns: $y=\frac{4}{3}\times\frac{1}{5}x=\frac{4}{15}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $4:20:15$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{4}{15}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 7 easy

Gears P and Q, whose numbers of teeth are in the ratio $2:3$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $2:5$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=2:3$ .
Teeth of Q : teeth of R $=2:5$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

3

parts in the first ratio and

2

parts in the second. Rescale both so Q is

6

parts.

\text{P}:\text{Q}=2:3=4:6,\quad \text{Q}:\text{R}=2:5=6:15 \Rightarrow \text{P}:\text{Q}:\text{R}=4:6:15

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

4a

,

6a

and

15a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

4a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

15a \times y

teeth.

4a \times x = 15a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

15a

.

y=\frac{4a}{15a}x=\frac{4}{15}x

R has more teeth than P, so R turns more slowly; the factor

\frac{4}{15}

is less than

1

.

Answer:

y=\frac{4}{15}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=15$ : P passes $4 \times 15=60$ teeth a minute. Q, with $6$ teeth, turns $10$ times and passes $60$ teeth. R, with $15$ teeth, turns $4$ times: $15 \times 4=60$ . The formula gives $y=\frac{4}{15}\times15=4$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{2}{3}x$ times and R turns $\frac{2}{5}$ of Q's turns: $y=\frac{2}{5}\times\frac{2}{3}x=\frac{4}{15}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $4:6:15$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{4}{15}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 8 medium

Gears P and Q, whose numbers of teeth are in the ratio $3:5$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $3:2$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=3:5$ .
Teeth of Q : teeth of R $=3:2$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

5

parts in the first ratio and

3

parts in the second. Rescale both so Q is

15

parts.

\text{P}:\text{Q}=3:5=9:15,\quad \text{Q}:\text{R}=3:2=15:10 \Rightarrow \text{P}:\text{Q}:\text{R}=9:15:10

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

9a

,

15a

and

10a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

9a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

10a \times y

teeth.

9a \times x = 10a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

10a

.

y=\frac{9a}{10a}x=\frac{9}{10}x

R has more teeth than P, so R turns more slowly; the factor

\frac{9}{10}

is less than

1

.

Answer:

y=\frac{9}{10}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=10$ : P passes $9 \times 10=90$ teeth a minute. Q, with $15$ teeth, turns $6$ times and passes $90$ teeth. R, with $10$ teeth, turns $9$ times: $10 \times 9=90$ . The formula gives $y=\frac{9}{10}\times10=9$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{3}{5}x$ times and R turns $\frac{3}{2}$ of Q's turns: $y=\frac{3}{2}\times\frac{3}{5}x=\frac{9}{10}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $9:15:10$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{9}{10}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 9 medium

Gears P and Q, whose numbers of teeth are in the ratio $1:5$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $2:1$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=1:5$ .
Teeth of Q : teeth of R $=2:1$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

5

parts in the first ratio and

2

parts in the second. Rescale both so Q is

10

parts.

\text{P}:\text{Q}=1:5=2:10,\quad \text{Q}:\text{R}=2:1=10:5 \Rightarrow \text{P}:\text{Q}:\text{R}=2:10:5

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

2a

,

10a

and

5a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

2a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

5a \times y

teeth.

2a \times x = 5a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

5a

.

y=\frac{2a}{5a}x=\frac{2}{5}x

R has more teeth than P, so R turns more slowly; the factor

\frac{2}{5}

is less than

1

.

Answer:

y=\frac{2}{5}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=5$ : P passes $2 \times 5=10$ teeth a minute. Q, with $10$ teeth, turns $1$ times and passes $10$ teeth. R, with $5$ teeth, turns $2$ times: $5 \times 2=10$ . The formula gives $y=\frac{2}{5}\times5=2$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{1}{5}x$ times and R turns $2$ of Q's turns: $y=2\times\frac{1}{5}x=\frac{2}{5}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $2:10:5$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{2}{5}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 10 hard

Gears P and Q, whose numbers of teeth are in the ratio $2:3$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $5:6$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=2:3$ .
Teeth of Q : teeth of R $=5:6$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

3

parts in the first ratio and

5

parts in the second. Rescale both so Q is

15

parts.

\text{P}:\text{Q}=2:3=10:15,\quad \text{Q}:\text{R}=5:6=15:18 \Rightarrow \text{P}:\text{Q}:\text{R}=10:15:18

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

10a

,

15a

and

18a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

10a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

18a \times y

teeth.

10a \times x = 18a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

18a

.

y=\frac{10a}{18a}x=\frac{10}{18}x=\frac{5}{9}x

R has more teeth than P, so R turns more slowly; the factor

\frac{5}{9}

is less than

1

.

Answer:

y=\frac{5}{9}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=18$ : P passes $10 \times 18=180$ teeth a minute. Q, with $15$ teeth, turns $12$ times and passes $180$ teeth. R, with $18$ teeth, turns $10$ times: $18 \times 10=180$ . The formula gives $y=\frac{5}{9}\times18=10$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{2}{3}x$ times and R turns $\frac{5}{6}$ of Q's turns: $y=\frac{5}{6}\times\frac{2}{3}x=\frac{5}{9}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $10:15:18$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{5}{9}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 11 hard

Gears P and Q, whose numbers of teeth are in the ratio $2:5$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $6:5$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=2:5$ .
Teeth of Q : teeth of R $=6:5$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

5

parts in the first ratio and

6

parts in the second. Rescale both so Q is

30

parts.

\text{P}:\text{Q}=2:5=12:30,\quad \text{Q}:\text{R}=6:5=30:25 \Rightarrow \text{P}:\text{Q}:\text{R}=12:30:25

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

12a

,

30a

and

25a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

12a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

25a \times y

teeth.

12a \times x = 25a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

25a

.

y=\frac{12a}{25a}x=\frac{12}{25}x

R has more teeth than P, so R turns more slowly; the factor

\frac{12}{25}

is less than

1

.

Answer:

y=\frac{12}{25}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=25$ : P passes $12 \times 25=300$ teeth a minute. Q, with $30$ teeth, turns $10$ times and passes $300$ teeth. R, with $25$ teeth, turns $12$ times: $25 \times 12=300$ . The formula gives $y=\frac{12}{25}\times25=12$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{2}{5}x$ times and R turns $\frac{6}{5}$ of Q's turns: $y=\frac{6}{5}\times\frac{2}{5}x=\frac{12}{25}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $12:30:25$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{12}{25}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Variant 12 hard

Gears P and Q, whose numbers of teeth are in the ratio $1:6$ , mesh with each other, and gears Q and R, whose numbers of teeth are in the ratio $5:2$ , mesh with each other, so all three gears turn together. Gear P makes $x$ turns per minute and gear R makes $y$ turns per minute. Write an equation that gives $y$ in terms of $x$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three gears are chained P-Q-R. We know two teeth ratios that share gear Q. Relate P's turns per minute, $x$ , to R's turns per minute, $y$ .

Givens

Teeth of P : teeth of Q $=1:6$ .
Teeth of Q : teeth of R $=5:2$ .
P meshes with Q, and Q meshes with R.

Unknowns

An equation giving $y$ in terms of $x$ .

Constraints

When two gears mesh, the same number of teeth pass the contact point on each gear every minute.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #15 Organize Information in More Ways#13 Convert to Algebra

Think in the unit 'teeth per minute': for meshed gears it is equal, and it equals (teeth) $\times$ (turns per minute). First reorganize the two ratios so Q has the same number in both, giving one ratio P:Q:R, then write the equal-teeth equation.

3 · Execute4 carry out the plan

1Line up the two ratios

#15 Organize Information in More Ways 6.RP.A.3

Q is

6

parts in the first ratio and

5

parts in the second. Rescale both so Q is

30

parts.

\text{P}:\text{Q}=1:6=5:30,\quad \text{Q}:\text{R}=5:2=30:12 \Rightarrow \text{P}:\text{Q}:\text{R}=5:30:12

A shared middle term has to be written with the same number before the ratios can be joined.

2Name the teeth

#13 Convert to Algebra 6.RP.A.3

Let the gears have

5a

,

30a

and

12a

teeth for some natural number

a

.

Only the ratio is known, so a common multiplier

a

stands in for the actual size.

3Teeth per minute are equal

#8 Analyze the Units 8.F.B.4

P turns

x

times a minute, so

5a \times x

teeth pass the contact point each minute. That number passes on Q too, and Q passes the same number on to R. R turns

y

times a minute, so it moves

12a \times y

teeth.

5a \times x = 12a \times y

Q's own speed does not matter; it just hands the teeth count from P to R.

4Solve for y

#13 Convert to Algebra 8.F.B.4

Divide both sides by

12a

.

y=\frac{5a}{12a}x=\frac{5}{12}x

R has more teeth than P, so R turns more slowly; the factor

\frac{5}{12}

is less than

1

.

Answer:

y=\frac{5}{12}x

4 · Reviewdoes it hold up?

Take $a=1$ and $x=12$ : P passes $5 \times 12=60$ teeth a minute. Q, with $30$ teeth, turns $2$ times and passes $60$ teeth. R, with $12$ teeth, turns $5$ times: $12 \times 5=60$ . The formula gives $y=\frac{5}{12}\times12=5$ .

Another way: Go one mesh at a time. Turns are inversely proportional to teeth, so Q turns $\frac{1}{6}x$ times and R turns $\frac{5}{2}$ of Q's turns: $y=\frac{5}{2}\times\frac{1}{6}x=\frac{5}{12}x$ .

Standardsmin grade 8

6.RP.A.3 Use ratio and rate reasoning to solve problems — Rescaling two ratios to a common middle term to get $5:30:12$.
8.F.B.4 Construct a function to model a linear relationship between two quantities — Writing the function rule $y=\frac{5}{12}x$ that relates the two rates.

💡Takeaway. For meshed gears, multiply teeth by turns: that product is the same on every gear in the chain.

Meshed gears share teeth times turns · 12 practice problems

Generated variants — 12

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