Practice · Find the mixing ratio, then the bottleneck · Sensim Math

Variant 1 easy

There are $600$ g of a paint in which white and brown are mixed in the ratio $5:7$ , and $500$ g of a paint in which white and brown are mixed in the ratio $1:2$ . You want to mix these two paints to make paint in which white and brown are in the ratio $2:3$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 5:7 and 1:2), 600 g and 500 g, are mixed to get white:brown = 2:3. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $5:7$ ; there are $600$ g of it.
Paint Q is white and brown in the ratio $1:2$ ; there are $500$ g of it.
The mixture must be white and brown in the ratio $2:3$ .

Unknowns

The largest number of grams of $2:3$ paint that can be made.

Constraints

You cannot use more than $600$ g of paint P or more than $500$ g of paint Q.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{5}{12}

white and

\frac{7}{12}

brown; paint Q is

\frac{1}{3}

white and

\frac{2}{3}

brown.

\text{white} = \frac{5}{12}x + \frac{1}{3}y,\quad \text{brown} = \frac{7}{12}x + \frac{2}{3}y

A ratio

5:7

means

5

parts out of

12

are white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

2:3

, so

3

times the white equals

2

times the brown. Multiply by

12

to clear the fractions.

3\left(\frac{5}{12}x + \frac{1}{3}y\right) = 2\left(\frac{7}{12}x + \frac{2}{3}y\right),\quad 15x + 12y = 14x + 16y,\quad x = 4y

In any batch, paint P must be

4

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 4y

, paint P is used up 4 times as fast. If all of paint Q were used, we would need

2000

g of paint P, which we do not have. So paint P runs out first:

x = 600

forces

y = 150

.

x = 600,\quad y = \frac{600}{4} = 150

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

150

g of paint Q.

600 + 150 = 750

Total paint is

5y

, so it is largest when

y

is.

Answer:

750

g

4 · Reviewdoes it hold up?

White: $\frac{5}{12} \times 600 + \frac{1}{3} \times 150 = 250 + 50 = 300$ g. Brown: $\frac{7}{12} \times 600 + \frac{2}{3} \times 150 = 350 + 100 = 450$ g. $300:450 = 2:3$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $60$ : paint P is $\frac{25}{60}$ white, paint Q is $\frac{20}{60}$ , the target is $\frac{24}{60}$ . The target sits $1$ part from P and $4$ parts from Q, so the amounts of P and Q must be in the ratio $4:1$ -- the same $x = 4y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 4y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 2 easy

There are $600$ g of a paint in which white and brown are mixed in the ratio $1:2$ , and $500$ g of a paint in which white and brown are mixed in the ratio $1:4$ . You want to mix these two paints to make paint in which white and brown are in the ratio $3:7$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 1:2 and 1:4), 600 g and 500 g, are mixed to get white:brown = 3:7. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $1:2$ ; there are $600$ g of it.
Paint Q is white and brown in the ratio $1:4$ ; there are $500$ g of it.
The mixture must be white and brown in the ratio $3:7$ .

Unknowns

The largest number of grams of $3:7$ paint that can be made.

Constraints

You cannot use more than $600$ g of paint P or more than $500$ g of paint Q.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{1}{3}

white and

\frac{2}{3}

brown; paint Q is

\frac{1}{5}

white and

\frac{4}{5}

brown.

\text{white} = \frac{1}{3}x + \frac{1}{5}y,\quad \text{brown} = \frac{2}{3}x + \frac{4}{5}y

A ratio

1:2

means

1

part out of

3

is white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

3:7

, so

7

times the white equals

3

times the brown. Multiply by

15

to clear the fractions.

7\left(\frac{1}{3}x + \frac{1}{5}y\right) = 3\left(\frac{2}{3}x + \frac{4}{5}y\right),\quad 35x + 21y = 30x + 36y,\quad 5x = 15y,\quad x = 3y

In any batch, paint P must be

3

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 3y

, paint P is used up 3 times as fast. If all of paint Q were used, we would need

1500

g of paint P, which we do not have. So paint P runs out first:

x = 600

forces

y = 200

.

x = 600,\quad y = \frac{600}{3} = 200

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

200

g of paint Q.

600 + 200 = 800

Total paint is

4y

, so it is largest when

y

is.

Answer:

800

g

4 · Reviewdoes it hold up?

White: $\frac{1}{3} \times 600 + \frac{1}{5} \times 200 = 200 + 40 = 240$ g. Brown: $\frac{2}{3} \times 600 + \frac{4}{5} \times 200 = 400 + 160 = 560$ g. $240:560 = 3:7$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $30$ : paint P is $\frac{10}{30}$ white, paint Q is $\frac{6}{30}$ , the target is $\frac{9}{30}$ . The target sits $1$ part from P and $3$ parts from Q, so the amounts of P and Q must be in the ratio $3:1$ -- the same $x = 3y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 3y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 3 easy

There are $600$ g of a paint in which white and brown are mixed in the ratio $2:3$ , and $600$ g of a paint in which white and brown are mixed in the ratio $1:4$ . You want to mix these two paints to make paint in which white and brown are in the ratio $1:2$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 2:3 and 1:4), 600 g and 600 g, are mixed to get white:brown = 1:2. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $2:3$ ; there are $600$ g of it.
Paint Q is white and brown in the ratio $1:4$ ; there are $600$ g of it.
The mixture must be white and brown in the ratio $1:2$ .

Unknowns

The largest number of grams of $1:2$ paint that can be made.

Constraints

You cannot use more than $600$ g of either paint.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{2}{5}

white and

\frac{3}{5}

brown; paint Q is

\frac{1}{5}

white and

\frac{4}{5}

brown.

\text{white} = \frac{2}{5}x + \frac{1}{5}y,\quad \text{brown} = \frac{3}{5}x + \frac{4}{5}y

A ratio

2:3

means

2

parts out of

5

are white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

1:2

, so

2

times the white equals

1

times the brown. Multiply by

5

to clear the fractions.

2\left(\frac{2}{5}x + \frac{1}{5}y\right) = 1\left(\frac{3}{5}x + \frac{4}{5}y\right),\quad 4x + 2y = 3x + 4y,\quad x = 2y

In any batch, paint P must be

2

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 2y

, paint P is used up 2 times as fast. If all of paint Q were used, we would need

1200

g of paint P, which we do not have. So paint P runs out first:

x = 600

forces

y = 300

.

x = 600,\quad y = \frac{600}{2} = 300

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

300

g of paint Q.

600 + 300 = 900

Total paint is

3y

, so it is largest when

y

is.

Answer:

900

g

4 · Reviewdoes it hold up?

White: $\frac{2}{5} \times 600 + \frac{1}{5} \times 300 = 240 + 60 = 300$ g. Brown: $\frac{3}{5} \times 600 + \frac{4}{5} \times 300 = 360 + 240 = 600$ g. $300:600 = 1:2$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $15$ : paint P is $\frac{6}{15}$ white, paint Q is $\frac{3}{15}$ , the target is $\frac{5}{15}$ . The target sits $1$ part from P and $2$ parts from Q, so the amounts of P and Q must be in the ratio $2:1$ -- the same $x = 2y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 2y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 4 easy

There are $600$ g of a paint in which white and brown are mixed in the ratio $2:3$ , and $400$ g of a paint in which white and brown are mixed in the ratio $1:3$ . You want to mix these two paints to make paint in which white and brown are in the ratio $3:5$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 2:3 and 1:3), 600 g and 400 g, are mixed to get white:brown = 3:5. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $2:3$ ; there are $600$ g of it.
Paint Q is white and brown in the ratio $1:3$ ; there are $400$ g of it.
The mixture must be white and brown in the ratio $3:5$ .

Unknowns

The largest number of grams of $3:5$ paint that can be made.

Constraints

You cannot use more than $600$ g of paint P or more than $400$ g of paint Q.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{2}{5}

white and

\frac{3}{5}

brown; paint Q is

\frac{1}{4}

white and

\frac{3}{4}

brown.

\text{white} = \frac{2}{5}x + \frac{1}{4}y,\quad \text{brown} = \frac{3}{5}x + \frac{3}{4}y

A ratio

2:3

means

2

parts out of

5

are white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

3:5

, so

5

times the white equals

3

times the brown. Multiply by

20

to clear the fractions.

5\left(\frac{2}{5}x + \frac{1}{4}y\right) = 3\left(\frac{3}{5}x + \frac{3}{4}y\right),\quad 40x + 25y = 36x + 45y,\quad 4x = 20y,\quad x = 5y

In any batch, paint P must be

5

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 5y

, paint P is used up 5 times as fast. If all of paint Q were used, we would need

2000

g of paint P, which we do not have. So paint P runs out first:

x = 600

forces

y = 120

.

x = 600,\quad y = \frac{600}{5} = 120

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

120

g of paint Q.

600 + 120 = 720

Total paint is

6y

, so it is largest when

y

is.

Answer:

720

g

4 · Reviewdoes it hold up?

White: $\frac{2}{5} \times 600 + \frac{1}{4} \times 120 = 240 + 30 = 270$ g. Brown: $\frac{3}{5} \times 600 + \frac{3}{4} \times 120 = 360 + 90 = 450$ g. $270:450 = 3:5$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $40$ : paint P is $\frac{16}{40}$ white, paint Q is $\frac{10}{40}$ , the target is $\frac{15}{40}$ . The target sits $1$ part from P and $5$ parts from Q, so the amounts of P and Q must be in the ratio $5:1$ -- the same $x = 5y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 5y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 5 medium

There are $900$ g of a paint in which white and brown are mixed in the ratio $5:7$ , and $500$ g of a paint in which white and brown are mixed in the ratio $1:3$ . You want to mix these two paints to make paint in which white and brown are in the ratio $3:5$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 5:7 and 1:3), 900 g and 500 g, are mixed to get white:brown = 3:5. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $5:7$ ; there are $900$ g of it.
Paint Q is white and brown in the ratio $1:3$ ; there are $500$ g of it.
The mixture must be white and brown in the ratio $3:5$ .

Unknowns

The largest number of grams of $3:5$ paint that can be made.

Constraints

You cannot use more than $900$ g of paint P or more than $500$ g of paint Q.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{5}{12}

white and

\frac{7}{12}

brown; paint Q is

\frac{1}{4}

white and

\frac{3}{4}

brown.

\text{white} = \frac{5}{12}x + \frac{1}{4}y,\quad \text{brown} = \frac{7}{12}x + \frac{3}{4}y

A ratio

5:7

means

5

parts out of

12

are white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

3:5

, so

5

times the white equals

3

times the brown. Multiply by

12

to clear the fractions.

5\left(\frac{5}{12}x + \frac{1}{4}y\right) = 3\left(\frac{7}{12}x + \frac{3}{4}y\right),\quad 25x + 15y = 21x + 27y,\quad 4x = 12y,\quad x = 3y

In any batch, paint P must be

3

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 3y

, paint P is used up 3 times as fast. If all of paint Q were used, we would need

1500

g of paint P, which we do not have. So paint P runs out first:

x = 900

forces

y = 300

.

x = 900,\quad y = \frac{900}{3} = 300

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

300

g of paint Q.

900 + 300 = 1200

Total paint is

4y

, so it is largest when

y

is.

Answer:

1200

g

4 · Reviewdoes it hold up?

White: $\frac{5}{12} \times 900 + \frac{1}{4} \times 300 = 375 + 75 = 450$ g. Brown: $\frac{7}{12} \times 900 + \frac{3}{4} \times 300 = 525 + 225 = 750$ g. $450:750 = 3:5$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $24$ : paint P is $\frac{10}{24}$ white, paint Q is $\frac{6}{24}$ , the target is $\frac{9}{24}$ . The target sits $1$ part from P and $3$ parts from Q, so the amounts of P and Q must be in the ratio $3:1$ -- the same $x = 3y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 3y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 6 medium

There are $900$ g of a paint in which white and brown are mixed in the ratio $2:7$ , and $500$ g of a paint in which white and brown are mixed in the ratio $1:8$ . You want to mix these two paints to make paint in which white and brown are in the ratio $1:4$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 2:7 and 1:8), 900 g and 500 g, are mixed to get white:brown = 1:4. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $2:7$ ; there are $900$ g of it.
Paint Q is white and brown in the ratio $1:8$ ; there are $500$ g of it.
The mixture must be white and brown in the ratio $1:4$ .

Unknowns

The largest number of grams of $1:4$ paint that can be made.

Constraints

You cannot use more than $900$ g of paint P or more than $500$ g of paint Q.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{2}{9}

white and

\frac{7}{9}

brown; paint Q is

\frac{1}{9}

white and

\frac{8}{9}

brown.

\text{white} = \frac{2}{9}x + \frac{1}{9}y,\quad \text{brown} = \frac{7}{9}x + \frac{8}{9}y

A ratio

2:7

means

2

parts out of

9

are white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

1:4

, so

4

times the white equals

1

times the brown. Multiply by

9

to clear the fractions.

4\left(\frac{2}{9}x + \frac{1}{9}y\right) = 1\left(\frac{7}{9}x + \frac{8}{9}y\right),\quad 8x + 4y = 7x + 8y,\quad x = 4y

In any batch, paint P must be

4

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 4y

, paint P is used up 4 times as fast. If all of paint Q were used, we would need

2000

g of paint P, which we do not have. So paint P runs out first:

x = 900

forces

y = 225

.

x = 900,\quad y = \frac{900}{4} = 225

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

225

g of paint Q.

900 + 225 = 1125

Total paint is

5y

, so it is largest when

y

is.

Answer:

1125

g

4 · Reviewdoes it hold up?

White: $\frac{2}{9} \times 900 + \frac{1}{9} \times 225 = 200 + 25 = 225$ g. Brown: $\frac{7}{9} \times 900 + \frac{8}{9} \times 225 = 700 + 200 = 900$ g. $225:900 = 1:4$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $45$ : paint P is $\frac{10}{45}$ white, paint Q is $\frac{5}{45}$ , the target is $\frac{9}{45}$ . The target sits $1$ part from P and $4$ parts from Q, so the amounts of P and Q must be in the ratio $4:1$ -- the same $x = 4y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 4y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 7 medium

There are $900$ g of a paint in which white and brown are mixed in the ratio $2:3$ , and $900$ g of a paint in which white and brown are mixed in the ratio $3:7$ . You want to mix these two paints to make paint in which white and brown are in the ratio $3:5$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 2:3 and 3:7), 900 g and 900 g, are mixed to get white:brown = 3:5. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $2:3$ ; there are $900$ g of it.
Paint Q is white and brown in the ratio $3:7$ ; there are $900$ g of it.
The mixture must be white and brown in the ratio $3:5$ .

Unknowns

The largest number of grams of $3:5$ paint that can be made.

Constraints

You cannot use more than $900$ g of either paint.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{2}{5}

white and

\frac{3}{5}

brown; paint Q is

\frac{3}{10}

white and

\frac{7}{10}

brown.

\text{white} = \frac{2}{5}x + \frac{3}{10}y,\quad \text{brown} = \frac{3}{5}x + \frac{7}{10}y

A ratio

2:3

means

2

parts out of

5

are white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

3:5

, so

5

times the white equals

3

times the brown. Multiply by

10

to clear the fractions.

5\left(\frac{2}{5}x + \frac{3}{10}y\right) = 3\left(\frac{3}{5}x + \frac{7}{10}y\right),\quad 20x + 15y = 18x + 21y,\quad 2x = 6y,\quad x = 3y

In any batch, paint P must be

3

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 3y

, paint P is used up 3 times as fast. If all of paint Q were used, we would need

2700

g of paint P, which we do not have. So paint P runs out first:

x = 900

forces

y = 300

.

x = 900,\quad y = \frac{900}{3} = 300

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

300

g of paint Q.

900 + 300 = 1200

Total paint is

4y

, so it is largest when

y

is.

Answer:

1200

g

4 · Reviewdoes it hold up?

White: $\frac{2}{5} \times 900 + \frac{3}{10} \times 300 = 360 + 90 = 450$ g. Brown: $\frac{3}{5} \times 900 + \frac{7}{10} \times 300 = 540 + 210 = 750$ g. $450:750 = 3:5$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $40$ : paint P is $\frac{16}{40}$ white, paint Q is $\frac{12}{40}$ , the target is $\frac{15}{40}$ . The target sits $1$ part from P and $3$ parts from Q, so the amounts of P and Q must be in the ratio $3:1$ -- the same $x = 3y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 3y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 8 hard

There are $900$ g of a paint in which white and brown are mixed in the ratio $4:5$ , and $1000$ g of a paint in which white and brown are mixed in the ratio $1:2$ . You want to mix these two paints to make paint in which white and brown are in the ratio $5:7$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 4:5 and 1:2), 900 g and 1000 g, are mixed to get white:brown = 5:7. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $4:5$ ; there are $900$ g of it.
Paint Q is white and brown in the ratio $1:2$ ; there are $1000$ g of it.
The mixture must be white and brown in the ratio $5:7$ .

Unknowns

The largest number of grams of $5:7$ paint that can be made.

Constraints

You cannot use more than $900$ g of paint P or more than $1000$ g of paint Q.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{4}{9}

white and

\frac{5}{9}

brown; paint Q is

\frac{1}{3}

white and

\frac{2}{3}

brown.

\text{white} = \frac{4}{9}x + \frac{1}{3}y,\quad \text{brown} = \frac{5}{9}x + \frac{2}{3}y

A ratio

4:5

means

4

parts out of

9

are white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

5:7

, so

7

times the white equals

5

times the brown. Multiply by

9

to clear the fractions.

7\left(\frac{4}{9}x + \frac{1}{3}y\right) = 5\left(\frac{5}{9}x + \frac{2}{3}y\right),\quad 28x + 21y = 25x + 30y,\quad 3x = 9y,\quad x = 3y

In any batch, paint P must be

3

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 3y

, paint P is used up 3 times as fast. If all of paint Q were used, we would need

3000

g of paint P, which we do not have. So paint P runs out first:

x = 900

forces

y = 300

.

x = 900,\quad y = \frac{900}{3} = 300

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

300

g of paint Q.

900 + 300 = 1200

Total paint is

4y

, so it is largest when

y

is.

Answer:

1200

g

4 · Reviewdoes it hold up?

White: $\frac{4}{9} \times 900 + \frac{1}{3} \times 300 = 400 + 100 = 500$ g. Brown: $\frac{5}{9} \times 900 + \frac{2}{3} \times 300 = 500 + 200 = 700$ g. $500:700 = 5:7$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $36$ : paint P is $\frac{16}{36}$ white, paint Q is $\frac{12}{36}$ , the target is $\frac{15}{36}$ . The target sits $1$ part from P and $3$ parts from Q, so the amounts of P and Q must be in the ratio $3:1$ -- the same $x = 3y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 3y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 9 hard

There are $600$ g of a paint in which white and brown are mixed in the ratio $1:4$ , and $1000$ g of a paint in which white and brown are mixed in the ratio $1:9$ . You want to mix these two paints to make paint in which white and brown are in the ratio $1:5$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 1:4 and 1:9), 600 g and 1000 g, are mixed to get white:brown = 1:5. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $1:4$ ; there are $600$ g of it.
Paint Q is white and brown in the ratio $1:9$ ; there are $1000$ g of it.
The mixture must be white and brown in the ratio $1:5$ .

Unknowns

The largest number of grams of $1:5$ paint that can be made.

Constraints

You cannot use more than $600$ g of paint P or more than $1000$ g of paint Q.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{1}{5}

white and

\frac{4}{5}

brown; paint Q is

\frac{1}{10}

white and

\frac{9}{10}

brown.

\text{white} = \frac{1}{5}x + \frac{1}{10}y,\quad \text{brown} = \frac{4}{5}x + \frac{9}{10}y

A ratio

1:4

means

1

part out of

5

is white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

1:5

, so

5

times the white equals

1

times the brown. Multiply by

10

to clear the fractions.

5\left(\frac{1}{5}x + \frac{1}{10}y\right) = 1\left(\frac{4}{5}x + \frac{9}{10}y\right),\quad 10x + 5y = 8x + 9y,\quad 2x = 4y,\quad x = 2y

In any batch, paint P must be

2

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 2y

, paint P is used up 2 times as fast. If all of paint Q were used, we would need

2000

g of paint P, which we do not have. So paint P runs out first:

x = 600

forces

y = 300

.

x = 600,\quad y = \frac{600}{2} = 300

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

300

g of paint Q.

600 + 300 = 900

Total paint is

3y

, so it is largest when

y

is.

Answer:

900

g

4 · Reviewdoes it hold up?

White: $\frac{1}{5} \times 600 + \frac{1}{10} \times 300 = 120 + 30 = 150$ g. Brown: $\frac{4}{5} \times 600 + \frac{9}{10} \times 300 = 480 + 270 = 750$ g. $150:750 = 1:5$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $30$ : paint P is $\frac{6}{30}$ white, paint Q is $\frac{3}{30}$ , the target is $\frac{5}{30}$ . The target sits $1$ part from P and $2$ parts from Q, so the amounts of P and Q must be in the ratio $2:1$ -- the same $x = 2y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 2y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 10 medium

There are $800$ g of a paint in which white and brown are mixed in the ratio $3:5$ , and $1000$ g of a paint in which white and brown are mixed in the ratio $1:3$ . You want to mix these two paints to make paint in which white and brown are in the ratio $1:2$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 3:5 and 1:3), 800 g and 1000 g, are mixed to get white:brown = 1:2. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $3:5$ ; there are $800$ g of it.
Paint Q is white and brown in the ratio $1:3$ ; there are $1000$ g of it.
The mixture must be white and brown in the ratio $1:2$ .

Unknowns

The largest number of grams of $1:2$ paint that can be made.

Constraints

You cannot use more than $800$ g of paint P or more than $1000$ g of paint Q.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{3}{8}

white and

\frac{5}{8}

brown; paint Q is

\frac{1}{4}

white and

\frac{3}{4}

brown.

\text{white} = \frac{3}{8}x + \frac{1}{4}y,\quad \text{brown} = \frac{5}{8}x + \frac{3}{4}y

A ratio

3:5

means

3

parts out of

8

are white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

1:2

, so

2

times the white equals

1

times the brown. Multiply by

8

to clear the fractions.

2\left(\frac{3}{8}x + \frac{1}{4}y\right) = 1\left(\frac{5}{8}x + \frac{3}{4}y\right),\quad 6x + 4y = 5x + 6y,\quad x = 2y

In any batch, paint P must be

2

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 2y

, paint P is used up 2 times as fast. If all of paint Q were used, we would need

2000

g of paint P, which we do not have. So paint P runs out first:

x = 800

forces

y = 400

.

x = 800,\quad y = \frac{800}{2} = 400

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

400

g of paint Q.

800 + 400 = 1200

Total paint is

3y

, so it is largest when

y

is.

Answer:

1200

g

4 · Reviewdoes it hold up?

White: $\frac{3}{8} \times 800 + \frac{1}{4} \times 400 = 300 + 100 = 400$ g. Brown: $\frac{5}{8} \times 800 + \frac{3}{4} \times 400 = 500 + 300 = 800$ g. $400:800 = 1:2$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $24$ : paint P is $\frac{9}{24}$ white, paint Q is $\frac{6}{24}$ , the target is $\frac{8}{24}$ . The target sits $1$ part from P and $2$ parts from Q, so the amounts of P and Q must be in the ratio $2:1$ -- the same $x = 2y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 2y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 11 hard

There are $1200$ g of a paint in which white and brown are mixed in the ratio $3:7$ , and $1200$ g of a paint in which white and brown are mixed in the ratio $1:4$ . You want to mix these two paints to make paint in which white and brown are in the ratio $2:5$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 3:7 and 1:4), 1200 g and 1200 g, are mixed to get white:brown = 2:5. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $3:7$ ; there are $1200$ g of it.
Paint Q is white and brown in the ratio $1:4$ ; there are $1200$ g of it.
The mixture must be white and brown in the ratio $2:5$ .

Unknowns

The largest number of grams of $2:5$ paint that can be made.

Constraints

You cannot use more than $1200$ g of either paint.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{3}{10}

white and

\frac{7}{10}

brown; paint Q is

\frac{1}{5}

white and

\frac{4}{5}

brown.

\text{white} = \frac{3}{10}x + \frac{1}{5}y,\quad \text{brown} = \frac{7}{10}x + \frac{4}{5}y

A ratio

3:7

means

3

parts out of

10

are white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

2:5

, so

5

times the white equals

2

times the brown. Multiply by

10

to clear the fractions.

5\left(\frac{3}{10}x + \frac{1}{5}y\right) = 2\left(\frac{7}{10}x + \frac{4}{5}y\right),\quad 15x + 10y = 14x + 16y,\quad x = 6y

In any batch, paint P must be

6

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 6y

, paint P is used up 6 times as fast. If all of paint Q were used, we would need

7200

g of paint P, which we do not have. So paint P runs out first:

x = 1200

forces

y = 200

.

x = 1200,\quad y = \frac{1200}{6} = 200

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

200

g of paint Q.

1200 + 200 = 1400

Total paint is

7y

, so it is largest when

y

is.

Answer:

1400

g

4 · Reviewdoes it hold up?

White: $\frac{3}{10} \times 1200 + \frac{1}{5} \times 200 = 360 + 40 = 400$ g. Brown: $\frac{7}{10} \times 1200 + \frac{4}{5} \times 200 = 840 + 160 = 1000$ g. $400:1000 = 2:5$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $70$ : paint P is $\frac{21}{70}$ white, paint Q is $\frac{14}{70}$ , the target is $\frac{20}{70}$ . The target sits $1$ part from P and $6$ parts from Q, so the amounts of P and Q must be in the ratio $6:1$ -- the same $x = 6y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 6y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Variant 12 hard

There are $1200$ g of a paint in which white and brown are mixed in the ratio $3:5$ , and $500$ g of a paint in which white and brown are mixed in the ratio $1:7$ . You want to mix these two paints to make paint in which white and brown are in the ratio $1:2$ . At most how many grams of this paint can you make?

Your answer

Show solution

1 · Understandwhat's really being asked

Two paints (white:brown = 3:5 and 1:7), 1200 g and 500 g, are mixed to get white:brown = 1:2. Find the largest possible amount of the new paint.

Givens

Paint P is white and brown in the ratio $3:5$ ; there are $1200$ g of it.
Paint Q is white and brown in the ratio $1:7$ ; there are $500$ g of it.
The mixture must be white and brown in the ratio $1:2$ .

Unknowns

The largest number of grams of $1:2$ paint that can be made.

Constraints

You cannot use more than $1200$ g of paint P or more than $500$ g of paint Q.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#3 Eliminate Possibilities

Organize the white and brown in a table, use the target ratio to find how the two amounts must compare, and then ask which paint runs out first.

3 · Execute4 carry out the plan

1Split each paint into white and brown

#15 Organize Information in More Ways 7.RP.A.3

Use

x

g of paint P and

y

g of paint Q. Paint P is

\frac{3}{8}

white and

\frac{5}{8}

brown; paint Q is

\frac{1}{8}

white and

\frac{7}{8}

brown.

\text{white} = \frac{3}{8}x + \frac{1}{8}y,\quad \text{brown} = \frac{5}{8}x + \frac{7}{8}y

A ratio

3:5

means

3

parts out of

8

are white.

2Apply the target ratio

#13 Convert to Algebra 8.EE.C.7

White to brown must be

1:2

, so

2

times the white equals

1

times the brown. Multiply by

8

to clear the fractions.

2\left(\frac{3}{8}x + \frac{1}{8}y\right) = 1\left(\frac{5}{8}x + \frac{7}{8}y\right),\quad 6x + 2y = 5x + 7y,\quad x = 5y

In any batch, paint P must be

5

times as heavy as paint Q.

3Find the bottleneck

#3 Eliminate Possibilities 7.RP.A.2

Because

x = 5y

, paint P is used up 5 times as fast. If all of paint Q were used, we would need

2500

g of paint P, which we do not have. So paint P runs out first:

x = 1200

forces

y = 240

.

x = 1200,\quad y = \frac{1200}{5} = 240

The batch grows with

y

, and the most

y

can be is set by the scarcer ingredient.

4Add up

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

The largest batch uses all of paint P and

240

g of paint Q.

1200 + 240 = 1440

Total paint is

6y

, so it is largest when

y

is.

Answer:

1440

g

4 · Reviewdoes it hold up?

White: $\frac{3}{8} \times 1200 + \frac{1}{8} \times 240 = 450 + 30 = 480$ g. Brown: $\frac{5}{8} \times 1200 + \frac{7}{8} \times 240 = 750 + 210 = 960$ g. $480:960 = 1:2$ , and neither paint is overdrawn.

Another way: Compare white fractions over a common denominator of $24$ : paint P is $\frac{9}{24}$ white, paint Q is $\frac{3}{24}$ , the target is $\frac{8}{24}$ . The target sits $1$ part from P and $5$ parts from Q, so the amounts of P and Q must be in the ratio $5:1$ -- the same $x = 5y$ .

Standardsmin grade 8

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Turning each ratio into the fraction of white and brown.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Using the proportional relationship $x = 5y$ to scale the batch.
8.EE.C.7 Solve linear equations in one variable — Solving the linear equation that comes from the target ratio.

💡Takeaway. First find the recipe (how the two ingredients must compare), then let the scarcer ingredient decide how big the batch can be.

Find the mixing ratio, then the bottleneck · 12 practice problems

Generated variants — 12

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