Practice · Track the salt through each pour · Sensim Math

Variant 1 easy

Beaker A holds $400$ g of $a\%$ saltwater and beaker B holds $400$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $15\%$ salt and the saltwater in beaker B is $19\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 400 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 15% in A and 19% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $400$ g of $a\%$ saltwater; beaker B: $400$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $15\%$ salt and B is $19\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

400

g is

4a

g of salt, and

b\%

of

400

g is

4b

g.

\frac{a}{100} \times 400 = 4a,\quad \frac{b}{100} \times 400 = 4b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

200

g carrying

2a

g of salt. A keeps

200

g with

2a

g of salt; B now has

600

g with

2a + 4b

g of salt.

\text{A}: 200 \text{ g},\ 2a \text{ g salt};\qquad \text{B}: 600 \text{ g},\ (2a + 4b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

300

g carrying half its salt,

a + 2b

g. A ends with

500

g and

2a + a + 2b = 3a + 2b

g of salt; B ends with

300

g and

a + 2b

g.

\text{A}: 500 \text{ g},\ (3a + 2b) \text{ g salt};\qquad \text{B}: 300 \text{ g},\ (a + 2b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

15\%

of

500

g is

75

g and

19\%

of

300

g is

57

g.

3a + 2b = 75;\qquad a + 2b = 57

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 18,\quad a = 9;\qquad 9 + 2b = 57,\quad b = 24

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 9

,

b = 24

4 · Reviewdoes it hold up?

A starts with $36$ g of salt and B with $96$ g. After the first pour A has $18$ g in $200$ g and B has $114$ g in $600$ g. Half of B, $300$ g with $57$ g of salt, goes to A: A has $75$ g in $500$ g, which is $15\%$ ; B keeps $57$ g in $300$ g, which is $19\%$ .

Another way: Track only beaker B's final $300$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 4a + \frac{1}{2} \times 4b = 57$ . Total salt is conserved, so $4a + 4b = 75 + 57 = 132$ , giving $a + b = 33$ ; with $a + 2b = 57$ this gives $b = 24$ and $a = 9$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 2 easy

Beaker A holds $400$ g of $a\%$ saltwater and beaker B holds $400$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $14\%$ salt and the saltwater in beaker B is $10\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 400 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 14% in A and 10% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $400$ g of $a\%$ saltwater; beaker B: $400$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $14\%$ salt and B is $10\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

400

g is

4a

g of salt, and

b\%

of

400

g is

4b

g.

\frac{a}{100} \times 400 = 4a,\quad \frac{b}{100} \times 400 = 4b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

200

g carrying

2a

g of salt. A keeps

200

g with

2a

g of salt; B now has

600

g with

2a + 4b

g of salt.

\text{A}: 200 \text{ g},\ 2a \text{ g salt};\qquad \text{B}: 600 \text{ g},\ (2a + 4b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

300

g carrying half its salt,

a + 2b

g. A ends with

500

g and

2a + a + 2b = 3a + 2b

g of salt; B ends with

300

g and

a + 2b

g.

\text{A}: 500 \text{ g},\ (3a + 2b) \text{ g salt};\qquad \text{B}: 300 \text{ g},\ (a + 2b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

14\%

of

500

g is

70

g and

10\%

of

300

g is

30

g.

3a + 2b = 70;\qquad a + 2b = 30

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 40,\quad a = 20;\qquad 20 + 2b = 30,\quad b = 5

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 20

,

b = 5

4 · Reviewdoes it hold up?

A starts with $80$ g of salt and B with $20$ g. After the first pour A has $40$ g in $200$ g and B has $60$ g in $600$ g. Half of B, $300$ g with $30$ g of salt, goes to A: A has $70$ g in $500$ g, which is $14\%$ ; B keeps $30$ g in $300$ g, which is $10\%$ .

Another way: Track only beaker B's final $300$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 4a + \frac{1}{2} \times 4b = 30$ . Total salt is conserved, so $4a + 4b = 70 + 30 = 100$ , giving $a + b = 25$ ; with $a + 2b = 30$ this gives $b = 5$ and $a = 20$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 3 easy

Beaker A holds $400$ g of $a\%$ saltwater and beaker B holds $400$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $10\%$ salt and the saltwater in beaker B is $6\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 400 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 10% in A and 6% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $400$ g of $a\%$ saltwater; beaker B: $400$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $10\%$ salt and B is $6\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

400

g is

4a

g of salt, and

b\%

of

400

g is

4b

g.

\frac{a}{100} \times 400 = 4a,\quad \frac{b}{100} \times 400 = 4b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

200

g carrying

2a

g of salt. A keeps

200

g with

2a

g of salt; B now has

600

g with

2a + 4b

g of salt.

\text{A}: 200 \text{ g},\ 2a \text{ g salt};\qquad \text{B}: 600 \text{ g},\ (2a + 4b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

300

g carrying half its salt,

a + 2b

g. A ends with

500

g and

2a + a + 2b = 3a + 2b

g of salt; B ends with

300

g and

a + 2b

g.

\text{A}: 500 \text{ g},\ (3a + 2b) \text{ g salt};\qquad \text{B}: 300 \text{ g},\ (a + 2b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

10\%

of

500

g is

50

g and

6\%

of

300

g is

18

g.

3a + 2b = 50;\qquad a + 2b = 18

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 32,\quad a = 16;\qquad 16 + 2b = 18,\quad b = 1

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 16

,

b = 1

4 · Reviewdoes it hold up?

A starts with $64$ g of salt and B with $4$ g. After the first pour A has $32$ g in $200$ g and B has $36$ g in $600$ g. Half of B, $300$ g with $18$ g of salt, goes to A: A has $50$ g in $500$ g, which is $10\%$ ; B keeps $18$ g in $300$ g, which is $6\%$ .

Another way: Track only beaker B's final $300$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 4a + \frac{1}{2} \times 4b = 18$ . Total salt is conserved, so $4a + 4b = 50 + 18 = 68$ , giving $a + b = 17$ ; with $a + 2b = 18$ this gives $b = 1$ and $a = 16$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 4 easy

Beaker A holds $800$ g of $a\%$ saltwater and beaker B holds $800$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $12\%$ salt and the saltwater in beaker B is $8\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 800 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 12% in A and 8% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $800$ g of $a\%$ saltwater; beaker B: $800$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $12\%$ salt and B is $8\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

800

g is

8a

g of salt, and

b\%

of

800

g is

8b

g.

\frac{a}{100} \times 800 = 8a,\quad \frac{b}{100} \times 800 = 8b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

400

g carrying

4a

g of salt. A keeps

400

g with

4a

g of salt; B now has

1200

g with

4a + 8b

g of salt.

\text{A}: 400 \text{ g},\ 4a \text{ g salt};\qquad \text{B}: 1200 \text{ g},\ (4a + 8b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

600

g carrying half its salt,

2a + 4b

g. A ends with

1000

g and

4a + 2a + 4b = 6a + 4b

g of salt; B ends with

600

g and

2a + 4b

g.

\text{A}: 1000 \text{ g},\ (6a + 4b) \text{ g salt};\qquad \text{B}: 600 \text{ g},\ (2a + 4b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

12\%

of

1000

g is

120

g and

8\%

of

600

g is

48

g.

6a + 4b = 120,\quad 3a + 2b = 60;\qquad 2a + 4b = 48,\quad a + 2b = 24

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 36,\quad a = 18;\qquad 18 + 2b = 24,\quad b = 3

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 18

,

b = 3

4 · Reviewdoes it hold up?

A starts with $144$ g of salt and B with $24$ g. After the first pour A has $72$ g in $400$ g and B has $96$ g in $1200$ g. Half of B, $600$ g with $48$ g of salt, goes to A: A has $120$ g in $1000$ g, which is $12\%$ ; B keeps $48$ g in $600$ g, which is $8\%$ .

Another way: Track only beaker B's final $600$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 8a + \frac{1}{2} \times 8b = 48$ . Total salt is conserved, so $8a + 8b = 120 + 48 = 168$ , giving $a + b = 21$ ; with $a + 2b = 24$ this gives $b = 3$ and $a = 18$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 5 medium

Beaker A holds $800$ g of $a\%$ saltwater and beaker B holds $800$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $12\%$ salt and the saltwater in beaker B is $16\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 800 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 12% in A and 16% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $800$ g of $a\%$ saltwater; beaker B: $800$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $12\%$ salt and B is $16\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

800

g is

8a

g of salt, and

b\%

of

800

g is

8b

g.

\frac{a}{100} \times 800 = 8a,\quad \frac{b}{100} \times 800 = 8b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

400

g carrying

4a

g of salt. A keeps

400

g with

4a

g of salt; B now has

1200

g with

4a + 8b

g of salt.

\text{A}: 400 \text{ g},\ 4a \text{ g salt};\qquad \text{B}: 1200 \text{ g},\ (4a + 8b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

600

g carrying half its salt,

2a + 4b

g. A ends with

1000

g and

4a + 2a + 4b = 6a + 4b

g of salt; B ends with

600

g and

2a + 4b

g.

\text{A}: 1000 \text{ g},\ (6a + 4b) \text{ g salt};\qquad \text{B}: 600 \text{ g},\ (2a + 4b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

12\%

of

1000

g is

120

g and

16\%

of

600

g is

96

g.

6a + 4b = 120,\quad 3a + 2b = 60;\qquad 2a + 4b = 96,\quad a + 2b = 48

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 12,\quad a = 6;\qquad 6 + 2b = 48,\quad b = 21

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 6

,

b = 21

4 · Reviewdoes it hold up?

A starts with $48$ g of salt and B with $168$ g. After the first pour A has $24$ g in $400$ g and B has $192$ g in $1200$ g. Half of B, $600$ g with $96$ g of salt, goes to A: A has $120$ g in $1000$ g, which is $12\%$ ; B keeps $96$ g in $600$ g, which is $16\%$ .

Another way: Track only beaker B's final $600$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 8a + \frac{1}{2} \times 8b = 96$ . Total salt is conserved, so $8a + 8b = 120 + 96 = 216$ , giving $a + b = 27$ ; with $a + 2b = 48$ this gives $b = 21$ and $a = 6$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 6 medium

Beaker A holds $800$ g of $a\%$ saltwater and beaker B holds $800$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $8\%$ salt and the saltwater in beaker B is $12\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 800 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 8% in A and 12% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $800$ g of $a\%$ saltwater; beaker B: $800$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $8\%$ salt and B is $12\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

800

g is

8a

g of salt, and

b\%

of

800

g is

8b

g.

\frac{a}{100} \times 800 = 8a,\quad \frac{b}{100} \times 800 = 8b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

400

g carrying

4a

g of salt. A keeps

400

g with

4a

g of salt; B now has

1200

g with

4a + 8b

g of salt.

\text{A}: 400 \text{ g},\ 4a \text{ g salt};\qquad \text{B}: 1200 \text{ g},\ (4a + 8b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

600

g carrying half its salt,

2a + 4b

g. A ends with

1000

g and

4a + 2a + 4b = 6a + 4b

g of salt; B ends with

600

g and

2a + 4b

g.

\text{A}: 1000 \text{ g},\ (6a + 4b) \text{ g salt};\qquad \text{B}: 600 \text{ g},\ (2a + 4b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

8\%

of

1000

g is

80

g and

12\%

of

600

g is

72

g.

6a + 4b = 80,\quad 3a + 2b = 40;\qquad 2a + 4b = 72,\quad a + 2b = 36

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 4,\quad a = 2;\qquad 2 + 2b = 36,\quad b = 17

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 2

,

b = 17

4 · Reviewdoes it hold up?

A starts with $16$ g of salt and B with $136$ g. After the first pour A has $8$ g in $400$ g and B has $144$ g in $1200$ g. Half of B, $600$ g with $72$ g of salt, goes to A: A has $80$ g in $1000$ g, which is $8\%$ ; B keeps $72$ g in $600$ g, which is $12\%$ .

Another way: Track only beaker B's final $600$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 8a + \frac{1}{2} \times 8b = 72$ . Total salt is conserved, so $8a + 8b = 80 + 72 = 152$ , giving $a + b = 19$ ; with $a + 2b = 36$ this gives $b = 17$ and $a = 2$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 7 hard

Beaker A holds $1200$ g of $a\%$ saltwater and beaker B holds $1200$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $18\%$ salt and the saltwater in beaker B is $14\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 1200 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 18% in A and 14% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $1200$ g of $a\%$ saltwater; beaker B: $1200$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $18\%$ salt and B is $14\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

1200

g is

12a

g of salt, and

b\%

of

1200

g is

12b

g.

\frac{a}{100} \times 1200 = 12a,\quad \frac{b}{100} \times 1200 = 12b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

600

g carrying

6a

g of salt. A keeps

600

g with

6a

g of salt; B now has

1800

g with

6a + 12b

g of salt.

\text{A}: 600 \text{ g},\ 6a \text{ g salt};\qquad \text{B}: 1800 \text{ g},\ (6a + 12b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

900

g carrying half its salt,

3a + 6b

g. A ends with

1500

g and

6a + 3a + 6b = 9a + 6b

g of salt; B ends with

900

g and

3a + 6b

g.

\text{A}: 1500 \text{ g},\ (9a + 6b) \text{ g salt};\qquad \text{B}: 900 \text{ g},\ (3a + 6b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

18\%

of

1500

g is

270

g and

14\%

of

900

g is

126

g.

9a + 6b = 270,\quad 3a + 2b = 90;\qquad 3a + 6b = 126,\quad a + 2b = 42

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 48,\quad a = 24;\qquad 24 + 2b = 42,\quad b = 9

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 24

,

b = 9

4 · Reviewdoes it hold up?

A starts with $288$ g of salt and B with $108$ g. After the first pour A has $144$ g in $600$ g and B has $252$ g in $1800$ g. Half of B, $900$ g with $126$ g of salt, goes to A: A has $270$ g in $1500$ g, which is $18\%$ ; B keeps $126$ g in $900$ g, which is $14\%$ .

Another way: Track only beaker B's final $900$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 12a + \frac{1}{2} \times 12b = 126$ . Total salt is conserved, so $12a + 12b = 270 + 126 = 396$ , giving $a + b = 33$ ; with $a + 2b = 42$ this gives $b = 9$ and $a = 24$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 8 medium

Beaker A holds $1200$ g of $a\%$ saltwater and beaker B holds $1200$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $10\%$ salt and the saltwater in beaker B is $14\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 1200 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 10% in A and 14% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $1200$ g of $a\%$ saltwater; beaker B: $1200$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $10\%$ salt and B is $14\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

1200

g is

12a

g of salt, and

b\%

of

1200

g is

12b

g.

\frac{a}{100} \times 1200 = 12a,\quad \frac{b}{100} \times 1200 = 12b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

600

g carrying

6a

g of salt. A keeps

600

g with

6a

g of salt; B now has

1800

g with

6a + 12b

g of salt.

\text{A}: 600 \text{ g},\ 6a \text{ g salt};\qquad \text{B}: 1800 \text{ g},\ (6a + 12b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

900

g carrying half its salt,

3a + 6b

g. A ends with

1500

g and

6a + 3a + 6b = 9a + 6b

g of salt; B ends with

900

g and

3a + 6b

g.

\text{A}: 1500 \text{ g},\ (9a + 6b) \text{ g salt};\qquad \text{B}: 900 \text{ g},\ (3a + 6b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

10\%

of

1500

g is

150

g and

14\%

of

900

g is

126

g.

9a + 6b = 150,\quad 3a + 2b = 50;\qquad 3a + 6b = 126,\quad a + 2b = 42

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 8,\quad a = 4;\qquad 4 + 2b = 42,\quad b = 19

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 4

,

b = 19

4 · Reviewdoes it hold up?

A starts with $48$ g of salt and B with $228$ g. After the first pour A has $24$ g in $600$ g and B has $252$ g in $1800$ g. Half of B, $900$ g with $126$ g of salt, goes to A: A has $150$ g in $1500$ g, which is $10\%$ ; B keeps $126$ g in $900$ g, which is $14\%$ .

Another way: Track only beaker B's final $900$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 12a + \frac{1}{2} \times 12b = 126$ . Total salt is conserved, so $12a + 12b = 150 + 126 = 276$ , giving $a + b = 23$ ; with $a + 2b = 42$ this gives $b = 19$ and $a = 4$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 9 medium

Beaker A holds $1200$ g of $a\%$ saltwater and beaker B holds $1200$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $19\%$ salt and the saltwater in beaker B is $15\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 1200 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 19% in A and 15% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $1200$ g of $a\%$ saltwater; beaker B: $1200$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $19\%$ salt and B is $15\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

1200

g is

12a

g of salt, and

b\%

of

1200

g is

12b

g.

\frac{a}{100} \times 1200 = 12a,\quad \frac{b}{100} \times 1200 = 12b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

600

g carrying

6a

g of salt. A keeps

600

g with

6a

g of salt; B now has

1800

g with

6a + 12b

g of salt.

\text{A}: 600 \text{ g},\ 6a \text{ g salt};\qquad \text{B}: 1800 \text{ g},\ (6a + 12b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

900

g carrying half its salt,

3a + 6b

g. A ends with

1500

g and

6a + 3a + 6b = 9a + 6b

g of salt; B ends with

900

g and

3a + 6b

g.

\text{A}: 1500 \text{ g},\ (9a + 6b) \text{ g salt};\qquad \text{B}: 900 \text{ g},\ (3a + 6b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

19\%

of

1500

g is

285

g and

15\%

of

900

g is

135

g.

9a + 6b = 285,\quad 3a + 2b = 95;\qquad 3a + 6b = 135,\quad a + 2b = 45

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 50,\quad a = 25;\qquad 25 + 2b = 45,\quad b = 10

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 25

,

b = 10

4 · Reviewdoes it hold up?

A starts with $300$ g of salt and B with $120$ g. After the first pour A has $150$ g in $600$ g and B has $270$ g in $1800$ g. Half of B, $900$ g with $135$ g of salt, goes to A: A has $285$ g in $1500$ g, which is $19\%$ ; B keeps $135$ g in $900$ g, which is $15\%$ .

Another way: Track only beaker B's final $900$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 12a + \frac{1}{2} \times 12b = 135$ . Total salt is conserved, so $12a + 12b = 285 + 135 = 420$ , giving $a + b = 35$ ; with $a + 2b = 45$ this gives $b = 10$ and $a = 25$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 10 hard

Beaker A holds $1600$ g of $a\%$ saltwater and beaker B holds $1600$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $19\%$ salt and the saltwater in beaker B is $23\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 1600 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 19% in A and 23% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $1600$ g of $a\%$ saltwater; beaker B: $1600$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $19\%$ salt and B is $23\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

1600

g is

16a

g of salt, and

b\%

of

1600

g is

16b

g.

\frac{a}{100} \times 1600 = 16a,\quad \frac{b}{100} \times 1600 = 16b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

800

g carrying

8a

g of salt. A keeps

800

g with

8a

g of salt; B now has

2400

g with

8a + 16b

g of salt.

\text{A}: 800 \text{ g},\ 8a \text{ g salt};\qquad \text{B}: 2400 \text{ g},\ (8a + 16b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

1200

g carrying half its salt,

4a + 8b

g. A ends with

2000

g and

8a + 4a + 8b = 12a + 8b

g of salt; B ends with

1200

g and

4a + 8b

g.

\text{A}: 2000 \text{ g},\ (12a + 8b) \text{ g salt};\qquad \text{B}: 1200 \text{ g},\ (4a + 8b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

19\%

of

2000

g is

380

g and

23\%

of

1200

g is

276

g.

12a + 8b = 380,\quad 3a + 2b = 95;\qquad 4a + 8b = 276,\quad a + 2b = 69

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 26,\quad a = 13;\qquad 13 + 2b = 69,\quad b = 28

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 13

,

b = 28

4 · Reviewdoes it hold up?

A starts with $208$ g of salt and B with $448$ g. After the first pour A has $104$ g in $800$ g and B has $552$ g in $2400$ g. Half of B, $1200$ g with $276$ g of salt, goes to A: A has $380$ g in $2000$ g, which is $19\%$ ; B keeps $276$ g in $1200$ g, which is $23\%$ .

Another way: Track only beaker B's final $1200$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 16a + \frac{1}{2} \times 16b = 276$ . Total salt is conserved, so $16a + 16b = 380 + 276 = 656$ , giving $a + b = 41$ ; with $a + 2b = 69$ this gives $b = 28$ and $a = 13$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 11 hard

Beaker A holds $1600$ g of $a\%$ saltwater and beaker B holds $1600$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $16\%$ salt and the saltwater in beaker B is $12\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 1600 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 16% in A and 12% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $1600$ g of $a\%$ saltwater; beaker B: $1600$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $16\%$ salt and B is $12\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

1600

g is

16a

g of salt, and

b\%

of

1600

g is

16b

g.

\frac{a}{100} \times 1600 = 16a,\quad \frac{b}{100} \times 1600 = 16b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

800

g carrying

8a

g of salt. A keeps

800

g with

8a

g of salt; B now has

2400

g with

8a + 16b

g of salt.

\text{A}: 800 \text{ g},\ 8a \text{ g salt};\qquad \text{B}: 2400 \text{ g},\ (8a + 16b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

1200

g carrying half its salt,

4a + 8b

g. A ends with

2000

g and

8a + 4a + 8b = 12a + 8b

g of salt; B ends with

1200

g and

4a + 8b

g.

\text{A}: 2000 \text{ g},\ (12a + 8b) \text{ g salt};\qquad \text{B}: 1200 \text{ g},\ (4a + 8b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

16\%

of

2000

g is

320

g and

12\%

of

1200

g is

144

g.

12a + 8b = 320,\quad 3a + 2b = 80;\qquad 4a + 8b = 144,\quad a + 2b = 36

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 44,\quad a = 22;\qquad 22 + 2b = 36,\quad b = 7

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 22

,

b = 7

4 · Reviewdoes it hold up?

A starts with $352$ g of salt and B with $112$ g. After the first pour A has $176$ g in $800$ g and B has $288$ g in $2400$ g. Half of B, $1200$ g with $144$ g of salt, goes to A: A has $320$ g in $2000$ g, which is $16\%$ ; B keeps $144$ g in $1200$ g, which is $12\%$ .

Another way: Track only beaker B's final $1200$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 16a + \frac{1}{2} \times 16b = 144$ . Total salt is conserved, so $16a + 16b = 320 + 144 = 464$ , giving $a + b = 29$ ; with $a + 2b = 36$ this gives $b = 7$ and $a = 22$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Variant 12 hard

Beaker A holds $2000$ g of $a\%$ saltwater and beaker B holds $2000$ g of $b\%$ saltwater. Half of the saltwater in beaker A is poured into beaker B and mixed well. Then half of the saltwater now in beaker B is poured into beaker A and mixed well. Now the saltwater in beaker A is $21\%$ salt and the saltwater in beaker B is $17\%$ salt. Find the values of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two beakers of 2000 g saltwater exchange half their contents, first A to B, then B to A. The final strengths are 21% in A and 17% in B. Find the starting strengths a% and b%.

Givens

Beaker A: $2000$ g of $a\%$ saltwater; beaker B: $2000$ g of $b\%$ saltwater.
Half of A is poured into B, then half of B is poured into A.
At the end A is $21\%$ salt and B is $17\%$ salt.

Unknowns

The values of $a$ and $b$ .

Constraints

Pouring moves saltwater; it never creates or destroys salt.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Break the process into its two pours and, after each one, record both the total saltwater and the grams of salt in each beaker. The final concentrations then turn into two equations in $a$ and $b$ .

3 · Execute5 carry out the plan

1Start with grams of salt

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

a\%

of

2000

g is

20a

g of salt, and

b\%

of

2000

g is

20b

g.

\frac{a}{100} \times 2000 = 20a,\quad \frac{b}{100} \times 2000 = 20b

Concentrations do not add, but grams of salt do.

2First pour: A to B

#7 Identify Subproblems 7.RP.A.3

Half of A is

1000

g carrying

10a

g of salt. A keeps

1000

g with

10a

g of salt; B now has

3000

g with

10a + 20b

g of salt.

\text{A}: 1000 \text{ g},\ 10a \text{ g salt};\qquad \text{B}: 3000 \text{ g},\ (10a + 20b) \text{ g salt}

The poured half carries half of A's salt.

3Second pour: B to A

#7 Identify Subproblems 7.RP.A.3

Half of B is

1500

g carrying half its salt,

5a + 10b

g. A ends with

2500

g and

10a + 5a + 10b = 15a + 10b

g of salt; B ends with

1500

g and

5a + 10b

g.

\text{A}: 2500 \text{ g},\ (15a + 10b) \text{ g salt};\qquad \text{B}: 1500 \text{ g},\ (5a + 10b) \text{ g salt}

Well mixed means the half that leaves carries exactly half the salt.

4Turn the final strengths into equations

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

21\%

of

2500

g is

525

g and

17\%

of

1500

g is

255

g.

15a + 10b = 525,\quad 3a + 2b = 105;\qquad 5a + 10b = 255,\quad a + 2b = 51

Each beaker gives one equation about grams of salt.

5Solve

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

Subtract the second equation from the first, then substitute.

2a = 54,\quad a = 27;\qquad 27 + 2b = 51,\quad b = 12

The

2b

terms match, so one subtraction isolates

a

.

Answer:

a = 27

,

b = 12

4 · Reviewdoes it hold up?

A starts with $540$ g of salt and B with $240$ g. After the first pour A has $270$ g in $1000$ g and B has $510$ g in $3000$ g. Half of B, $1500$ g with $255$ g of salt, goes to A: A has $525$ g in $2500$ g, which is $21\%$ ; B keeps $255$ g in $1500$ g, which is $17\%$ .

Another way: Track only beaker B's final $1500$ g: it holds a quarter of A's original salt plus half of B's, so $\frac{1}{4} \times 20a + \frac{1}{2} \times 20b = 255$ . Total salt is conserved, so $20a + 20b = 525 + 255 = 780$ , giving $a + b = 39$ ; with $a + 2b = 51$ this gives $b = 12$ and $a = 27$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the two salt equations together.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting percent concentrations into grams of salt.

💡Takeaway. With mixtures, follow the grams of salt step by step -- the percentages only come back at the very end.

Track the salt through each pour · 12 practice problems

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