Practice · Swapping mixtures shrinks the concentration gap · Sensim Math

Variant 1 easy

Containers A and B each hold $300\ \text{g}$ of salt water, at different concentrations. $100\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $100\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $300\ \text{g}$ each, trade $100\ \text{g}$ one way and then $100\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $300\ \text{g}$ of salt water at different concentrations.
$100\ \text{g}$ goes from A to B and is mixed.
Then $100\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

200\ \text{g}

at

x\%

. B now has

400\ \text{g}

holding

3y

grams of salt from itself and

x

grams from A.

\text{B} = \frac{3y + x}{400} \times 100 = \frac{x + 3y}{4}\ (\%)

y\%

of

300\ \text{g}

is

3y

grams, and

x\%

of

100\ \text{g}

is

x

grams.

2Pour B back into A

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

Taking

100\ \text{g}

out of B does not change B's concentration. That

100\ \text{g}

carries

\frac{x + 3y}{4}

grams of salt into A, which already has

2x

grams in

200\ \text{g}

.

\text{A} = \frac{2x + \frac{x + 3y}{4}}{300} \times 100 = \frac{8x + x + 3y}{12} = \frac{3x + y}{4}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{3x + y}{4} - \frac{x + 3y}{4} = \frac{2x - 2y}{4} = \frac{1}{2}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{1}{2}(x-y) : (x-y) = 1 : 2

Every swap of this size keeps

\frac{1}{2}

of the gap.

Answer:

1:2

4 · Reviewdoes it hold up?

Try A at $8\%$ and B at $4\%$ : B becomes $\frac{12 + 8}{400} = 5\%$ , and A becomes $\frac{16 + 5}{300} = 7\%$ . The gap goes from $4$ to $2$ , and $2 : 4 = 1 : 2$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $1:2$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 2 easy

Containers A and B each hold $400\ \text{g}$ of salt water, at different concentrations. $100\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $100\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $400\ \text{g}$ each, trade $100\ \text{g}$ one way and then $100\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $400\ \text{g}$ of salt water at different concentrations.
$100\ \text{g}$ goes from A to B and is mixed.
Then $100\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

300\ \text{g}

at

x\%

. B now has

500\ \text{g}

holding

4y

grams of salt from itself and

x

grams from A.

\text{B} = \frac{4y + x}{500} \times 100 = \frac{x + 4y}{5}\ (\%)

y\%

of

400\ \text{g}

is

4y

grams, and

x\%

of

100\ \text{g}

is

x

grams.

2Pour B back into A

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

Taking

100\ \text{g}

out of B does not change B's concentration. That

100\ \text{g}

carries

\frac{x + 4y}{5}

grams of salt into A, which already has

3x

grams in

300\ \text{g}

.

\text{A} = \frac{3x + \frac{x + 4y}{5}}{400} \times 100 = \frac{15x + x + 4y}{20} = \frac{4x + y}{5}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{4x + y}{5} - \frac{x + 4y}{5} = \frac{3x - 3y}{5} = \frac{3}{5}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{3}{5}(x-y) : (x-y) = 3 : 5

Every swap of this size keeps

\frac{3}{5}

of the gap.

Answer:

3:5

4 · Reviewdoes it hold up?

Try A at $10\%$ and B at $5\%$ : B becomes $\frac{20 + 10}{500} = 6\%$ , and A becomes $\frac{30 + 6}{400} = 9\%$ . The gap goes from $5$ to $3$ , and $3 : 5 = 3 : 5$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $3:5$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 3 easy

Containers A and B each hold $500\ \text{g}$ of salt water, at different concentrations. $200\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $200\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $500\ \text{g}$ each, trade $200\ \text{g}$ one way and then $200\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $500\ \text{g}$ of salt water at different concentrations.
$200\ \text{g}$ goes from A to B and is mixed.
Then $200\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

300\ \text{g}

at

x\%

. B now has

700\ \text{g}

holding

5y

grams of salt from itself and

2x

grams from A.

\text{B} = \frac{5y + 2x}{700} \times 100 = \frac{2x + 5y}{7}\ (\%)

y\%

of

500\ \text{g}

is

5y

grams, and

x\%

of

200\ \text{g}

is

2x

grams.

2Pour B back into A

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

Taking

200\ \text{g}

out of B does not change B's concentration. That

200\ \text{g}

carries

2 \times \frac{2x + 5y}{7}

grams of salt into A, which already has

3x

grams in

300\ \text{g}

.

\text{A} = \frac{3x + 2 \cdot \frac{2x + 5y}{7}}{500} \times 100 = \frac{21x + 4x + 10y}{35} = \frac{5x + 2y}{7}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{5x + 2y}{7} - \frac{2x + 5y}{7} = \frac{3x - 3y}{7} = \frac{3}{7}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{3}{7}(x-y) : (x-y) = 3 : 7

Every swap of this size keeps

\frac{3}{7}

of the gap.

Answer:

3:7

4 · Reviewdoes it hold up?

Try A at $14\%$ and B at $7\%$ : B becomes $\frac{35 + 28}{700} = 9\%$ , and A becomes $\frac{42 + 18}{500} = 12\%$ . The gap goes from $7$ to $3$ , and $3 : 7 = 3 : 7$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $3:7$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 4 easy

Containers A and B each hold $500\ \text{g}$ of salt water, at different concentrations. $100\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $100\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $500\ \text{g}$ each, trade $100\ \text{g}$ one way and then $100\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $500\ \text{g}$ of salt water at different concentrations.
$100\ \text{g}$ goes from A to B and is mixed.
Then $100\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

400\ \text{g}

at

x\%

. B now has

600\ \text{g}

holding

5y

grams of salt from itself and

x

grams from A.

\text{B} = \frac{5y + x}{600} \times 100 = \frac{x + 5y}{6}\ (\%)

y\%

of

500\ \text{g}

is

5y

grams, and

x\%

of

100\ \text{g}

is

x

grams.

2Pour B back into A

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

Taking

100\ \text{g}

out of B does not change B's concentration. That

100\ \text{g}

carries

\frac{x + 5y}{6}

grams of salt into A, which already has

4x

grams in

400\ \text{g}

.

\text{A} = \frac{4x + \frac{x + 5y}{6}}{500} \times 100 = \frac{24x + x + 5y}{30} = \frac{5x + y}{6}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{5x + y}{6} - \frac{x + 5y}{6} = \frac{4x - 4y}{6} = \frac{2}{3}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{2}{3}(x-y) : (x-y) = 2 : 3

Every swap of this size keeps

\frac{2}{3}

of the gap.

Answer:

2:3

4 · Reviewdoes it hold up?

Try A at $12\%$ and B at $6\%$ : B becomes $\frac{30 + 12}{600} = 7\%$ , and A becomes $\frac{48 + 7}{500} = 11\%$ . The gap goes from $6$ to $4$ , and $4 : 6 = 2 : 3$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $2:3$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 5 medium

Containers A and B each hold $500\ \text{g}$ of salt water, at different concentrations. $300\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $300\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $500\ \text{g}$ each, trade $300\ \text{g}$ one way and then $300\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $500\ \text{g}$ of salt water at different concentrations.
$300\ \text{g}$ goes from A to B and is mixed.
Then $300\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

200\ \text{g}

at

x\%

. B now has

800\ \text{g}

holding

5y

grams of salt from itself and

3x

grams from A.

\text{B} = \frac{5y + 3x}{800} \times 100 = \frac{3x + 5y}{8}\ (\%)

y\%

of

500\ \text{g}

is

5y

grams, and

x\%

of

300\ \text{g}

is

3x

grams.

2Pour B back into A

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

Taking

300\ \text{g}

out of B does not change B's concentration. That

300\ \text{g}

carries

3 \times \frac{3x + 5y}{8}

grams of salt into A, which already has

2x

grams in

200\ \text{g}

.

\text{A} = \frac{2x + 3 \cdot \frac{3x + 5y}{8}}{500} \times 100 = \frac{16x + 9x + 15y}{40} = \frac{5x + 3y}{8}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{5x + 3y}{8} - \frac{3x + 5y}{8} = \frac{2x - 2y}{8} = \frac{1}{4}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{1}{4}(x-y) : (x-y) = 1 : 4

Every swap of this size keeps

\frac{1}{4}

of the gap.

Answer:

1:4

4 · Reviewdoes it hold up?

Try A at $16\%$ and B at $8\%$ : B becomes $\frac{40 + 48}{800} = 11\%$ , and A becomes $\frac{32 + 33}{500} = 13\%$ . The gap goes from $8$ to $2$ , and $2 : 8 = 1 : 4$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $1:4$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 6 medium

Containers A and B each hold $600\ \text{g}$ of salt water, at different concentrations. $100\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $100\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $600\ \text{g}$ each, trade $100\ \text{g}$ one way and then $100\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $600\ \text{g}$ of salt water at different concentrations.
$100\ \text{g}$ goes from A to B and is mixed.
Then $100\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

500\ \text{g}

at

x\%

. B now has

700\ \text{g}

holding

6y

grams of salt from itself and

x

grams from A.

\text{B} = \frac{6y + x}{700} \times 100 = \frac{x + 6y}{7}\ (\%)

y\%

of

600\ \text{g}

is

6y

grams, and

x\%

of

100\ \text{g}

is

x

grams.

2Pour B back into A

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

Taking

100\ \text{g}

out of B does not change B's concentration. That

100\ \text{g}

carries

\frac{x + 6y}{7}

grams of salt into A, which already has

5x

grams in

500\ \text{g}

.

\text{A} = \frac{5x + \frac{x + 6y}{7}}{600} \times 100 = \frac{35x + x + 6y}{42} = \frac{6x + y}{7}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{6x + y}{7} - \frac{x + 6y}{7} = \frac{5x - 5y}{7} = \frac{5}{7}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{5}{7}(x-y) : (x-y) = 5 : 7

Every swap of this size keeps

\frac{5}{7}

of the gap.

Answer:

5:7

4 · Reviewdoes it hold up?

Try A at $14\%$ and B at $7\%$ : B becomes $\frac{42 + 14}{700} = 8\%$ , and A becomes $\frac{70 + 8}{600} = 13\%$ . The gap goes from $7$ to $5$ , and $5 : 7 = 5 : 7$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $5:7$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 7 hard

Containers A and B each hold $700\ \text{g}$ of salt water, at different concentrations. $300\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $300\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $700\ \text{g}$ each, trade $300\ \text{g}$ one way and then $300\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $700\ \text{g}$ of salt water at different concentrations.
$300\ \text{g}$ goes from A to B and is mixed.
Then $300\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

400\ \text{g}

at

x\%

. B now has

1000\ \text{g}

holding

7y

grams of salt from itself and

3x

grams from A.

\text{B} = \frac{7y + 3x}{1000} \times 100 = \frac{3x + 7y}{10}\ (\%)

y\%

of

700\ \text{g}

is

7y

grams, and

x\%

of

300\ \text{g}

is

3x

grams.

2Pour B back into A

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

Taking

300\ \text{g}

out of B does not change B's concentration. That

300\ \text{g}

carries

3 \times \frac{3x + 7y}{10}

grams of salt into A, which already has

4x

grams in

400\ \text{g}

.

\text{A} = \frac{4x + 3 \cdot \frac{3x + 7y}{10}}{700} \times 100 = \frac{40x + 9x + 21y}{70} = \frac{7x + 3y}{10}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{7x + 3y}{10} - \frac{3x + 7y}{10} = \frac{4x - 4y}{10} = \frac{2}{5}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{2}{5}(x-y) : (x-y) = 2 : 5

Every swap of this size keeps

\frac{2}{5}

of the gap.

Answer:

2:5

4 · Reviewdoes it hold up?

Try A at $20\%$ and B at $10\%$ : B becomes $\frac{70 + 60}{1000} = 13\%$ , and A becomes $\frac{80 + 39}{700} = 17\%$ . The gap goes from $10$ to $4$ , and $4 : 10 = 2 : 5$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $2:5$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 8 medium

Containers A and B each hold $700\ \text{g}$ of salt water, at different concentrations. $100\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $100\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $700\ \text{g}$ each, trade $100\ \text{g}$ one way and then $100\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $700\ \text{g}$ of salt water at different concentrations.
$100\ \text{g}$ goes from A to B and is mixed.
Then $100\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

600\ \text{g}

at

x\%

. B now has

800\ \text{g}

holding

7y

grams of salt from itself and

x

grams from A.

\text{B} = \frac{7y + x}{800} \times 100 = \frac{x + 7y}{8}\ (\%)

y\%

of

700\ \text{g}

is

7y

grams, and

x\%

of

100\ \text{g}

is

x

grams.

2Pour B back into A

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

Taking

100\ \text{g}

out of B does not change B's concentration. That

100\ \text{g}

carries

\frac{x + 7y}{8}

grams of salt into A, which already has

6x

grams in

600\ \text{g}

.

\text{A} = \frac{6x + \frac{x + 7y}{8}}{700} \times 100 = \frac{48x + x + 7y}{56} = \frac{7x + y}{8}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{7x + y}{8} - \frac{x + 7y}{8} = \frac{6x - 6y}{8} = \frac{3}{4}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{3}{4}(x-y) : (x-y) = 3 : 4

Every swap of this size keeps

\frac{3}{4}

of the gap.

Answer:

3:4

4 · Reviewdoes it hold up?

Try A at $16\%$ and B at $8\%$ : B becomes $\frac{56 + 16}{800} = 9\%$ , and A becomes $\frac{96 + 9}{700} = 15\%$ . The gap goes from $8$ to $6$ , and $6 : 8 = 3 : 4$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $3:4$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 9 medium

Containers A and B each hold $700\ \text{g}$ of salt water, at different concentrations. $200\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $200\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $700\ \text{g}$ each, trade $200\ \text{g}$ one way and then $200\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $700\ \text{g}$ of salt water at different concentrations.
$200\ \text{g}$ goes from A to B and is mixed.
Then $200\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

500\ \text{g}

at

x\%

. B now has

900\ \text{g}

holding

7y

grams of salt from itself and

2x

grams from A.

\text{B} = \frac{7y + 2x}{900} \times 100 = \frac{2x + 7y}{9}\ (\%)

y\%

of

700\ \text{g}

is

7y

grams, and

x\%

of

200\ \text{g}

is

2x

grams.

2Pour B back into A

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

Taking

200\ \text{g}

out of B does not change B's concentration. That

200\ \text{g}

carries

2 \times \frac{2x + 7y}{9}

grams of salt into A, which already has

5x

grams in

500\ \text{g}

.

\text{A} = \frac{5x + 2 \cdot \frac{2x + 7y}{9}}{700} \times 100 = \frac{45x + 4x + 14y}{63} = \frac{7x + 2y}{9}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{7x + 2y}{9} - \frac{2x + 7y}{9} = \frac{5x - 5y}{9} = \frac{5}{9}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{5}{9}(x-y) : (x-y) = 5 : 9

Every swap of this size keeps

\frac{5}{9}

of the gap.

Answer:

5:9

4 · Reviewdoes it hold up?

Try A at $18\%$ and B at $9\%$ : B becomes $\frac{63 + 36}{900} = 11\%$ , and A becomes $\frac{90 + 22}{700} = 16\%$ . The gap goes from $9$ to $5$ , and $5 : 9 = 5 : 9$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $5:9$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 10 hard

Containers A and B each hold $800\ \text{g}$ of salt water, at different concentrations. $100\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $100\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $800\ \text{g}$ each, trade $100\ \text{g}$ one way and then $100\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $800\ \text{g}$ of salt water at different concentrations.
$100\ \text{g}$ goes from A to B and is mixed.
Then $100\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

700\ \text{g}

at

x\%

. B now has

900\ \text{g}

holding

8y

grams of salt from itself and

x

grams from A.

\text{B} = \frac{8y + x}{900} \times 100 = \frac{x + 8y}{9}\ (\%)

y\%

of

800\ \text{g}

is

8y

grams, and

x\%

of

100\ \text{g}

is

x

grams.

2Pour B back into A

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

Taking

100\ \text{g}

out of B does not change B's concentration. That

100\ \text{g}

carries

\frac{x + 8y}{9}

grams of salt into A, which already has

7x

grams in

700\ \text{g}

.

\text{A} = \frac{7x + \frac{x + 8y}{9}}{800} \times 100 = \frac{63x + x + 8y}{72} = \frac{8x + y}{9}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{8x + y}{9} - \frac{x + 8y}{9} = \frac{7x - 7y}{9} = \frac{7}{9}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{7}{9}(x-y) : (x-y) = 7 : 9

Every swap of this size keeps

\frac{7}{9}

of the gap.

Answer:

7:9

4 · Reviewdoes it hold up?

Try A at $18\%$ and B at $9\%$ : B becomes $\frac{72 + 18}{900} = 10\%$ , and A becomes $\frac{126 + 10}{800} = 17\%$ . The gap goes from $9$ to $7$ , and $7 : 9 = 7 : 9$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $7:9$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 11 hard

Containers A and B each hold $800\ \text{g}$ of salt water, at different concentrations. $300\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $300\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $800\ \text{g}$ each, trade $300\ \text{g}$ one way and then $300\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $800\ \text{g}$ of salt water at different concentrations.
$300\ \text{g}$ goes from A to B and is mixed.
Then $300\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

500\ \text{g}

at

x\%

. B now has

1100\ \text{g}

holding

8y

grams of salt from itself and

3x

grams from A.

\text{B} = \frac{8y + 3x}{1100} \times 100 = \frac{3x + 8y}{11}\ (\%)

y\%

of

800\ \text{g}

is

8y

grams, and

x\%

of

300\ \text{g}

is

3x

grams.

2Pour B back into A

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

Taking

300\ \text{g}

out of B does not change B's concentration. That

300\ \text{g}

carries

3 \times \frac{3x + 8y}{11}

grams of salt into A, which already has

5x

grams in

500\ \text{g}

.

\text{A} = \frac{5x + 3 \cdot \frac{3x + 8y}{11}}{800} \times 100 = \frac{55x + 9x + 24y}{88} = \frac{8x + 3y}{11}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{8x + 3y}{11} - \frac{3x + 8y}{11} = \frac{5x - 5y}{11} = \frac{5}{11}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{5}{11}(x-y) : (x-y) = 5 : 11

Every swap of this size keeps

\frac{5}{11}

of the gap.

Answer:

5:11

4 · Reviewdoes it hold up?

Try A at $22\%$ and B at $11\%$ : B becomes $\frac{88 + 66}{1100} = 14\%$ , and A becomes $\frac{110 + 42}{800} = 19\%$ . The gap goes from $11$ to $5$ , and $5 : 11 = 5 : 11$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $5:11$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Variant 12 hard

Containers A and B each hold $900\ \text{g}$ of salt water, at different concentrations. $200\ \text{g}$ of the salt water in A is poured into B and mixed well, and then $200\ \text{g}$ of the salt water in B is poured back into A. Find the ratio of the difference between the final concentrations of A and B to the difference between their original concentrations, in simplest whole numbers.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two containers of salt water, $900\ \text{g}$ each, trade $200\ \text{g}$ one way and then $200\ \text{g}$ back. Compare the final gap in concentration with the original gap.

Givens

A and B each hold $900\ \text{g}$ of salt water at different concentrations.
$200\ \text{g}$ goes from A to B and is mixed.
Then $200\ \text{g}$ goes from B back to A.

Unknowns

(final gap) : (original gap), in simplest whole numbers.

Constraints

The original concentrations are not given, so the answer must not depend on them.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Call the original concentrations $x\%$ and $y\%$ . Track grams of salt, not percents: salt adds up when solutions are mixed, percents do not. Write each final concentration in $x$ and $y$ , subtract, and compare with $x - y$ .

3 · Execute4 carry out the plan

1Pour A into B

#8 Analyze the Units 7.RP.A.3

A keeps

700\ \text{g}

at

x\%

. B now has

1100\ \text{g}

holding

9y

grams of salt from itself and

2x

grams from A.

\text{B} = \frac{9y + 2x}{1100} \times 100 = \frac{2x + 9y}{11}\ (\%)

y\%

of

900\ \text{g}

is

9y

grams, and

x\%

of

200\ \text{g}

is

2x

grams.

2Pour B back into A

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

Taking

200\ \text{g}

out of B does not change B's concentration. That

200\ \text{g}

carries

2 \times \frac{2x + 9y}{11}

grams of salt into A, which already has

7x

grams in

700\ \text{g}

.

\text{A} = \frac{7x + 2 \cdot \frac{2x + 9y}{11}}{900} \times 100 = \frac{77x + 4x + 18y}{99} = \frac{9x + 2y}{11}\ (\%)

Combining over one denominator shows A is mostly its old self plus a little of B.

3Subtract

#13 Convert to Algebra 7.EE.B.4

The final gap is A minus B.

\frac{9x + 2y}{11} - \frac{2x + 9y}{11} = \frac{7x - 7y}{11} = \frac{7}{11}(x - y)

The unknowns survive only as

x - y

, which is exactly the original gap.

4Form the ratio

#13 Convert to Algebra 7.RP.A.3

Divide by the original gap

x - y

, which is not zero because the concentrations differ.

\frac{7}{11}(x-y) : (x-y) = 7 : 11

Every swap of this size keeps

\frac{7}{11}

of the gap.

Answer:

7:11

4 · Reviewdoes it hold up?

Try A at $22\%$ and B at $11\%$ : B becomes $\frac{99 + 44}{1100} = 13\%$ , and A becomes $\frac{154 + 26}{900} = 20\%$ . The gap goes from $11$ to $7$ , and $7 : 11 = 7 : 11$ .

Another way: Solve an easier related problem: pick any two concentrations, run the two pours with numbers, and see the ratio is $7:11$ every time; the algebra above explains why.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between grams of salt and percent concentration.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the final gap in terms of the two unknown concentrations.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the nested fraction for A as one simple expression.

💡Takeaway. Track grams of salt, not percents, and the unknown concentrations cancel out.

Swapping mixtures shrinks the concentration gap · 12 practice problems

Generated variants — 12

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