Practice · Track grams of salt through two pourings · Sensim Math

Variant 1 easy

Beaker A holds $150$ g of salt water that is $a\%$ salt, and beaker B holds $100$ g of salt water that is $b\%$ salt. You pour $100$ g of the salt water from beaker A into beaker B and stir well. Then you pour $50$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $150$ g of $a\%$ salt water.
B: $100$ g of $b\%$ salt water.
Pour $100$ g from A into B and mix; then pour $50$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

100

g of A's water at

a\%

carries

a

g of salt. B already has

b

g of salt in its

100

g.

\frac{a}{100}\times100=a,\quad \frac{b}{100}\times100=b

a\%

of

100

g is

a

g.

2Beaker B after mixing

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

B now has

200

g of water containing

a+b

g of salt.

\frac{a+b}{200}\times100=\frac{a+b}{2}\ (\%)

B's strength is the average of the two, since equal amounts were mixed.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

50

g of B's mixture carries

\frac{1}{4}

of B's salt, because

50

g is

\frac{1}{4}

of

200

g.

\frac{a+b}{2}\times\frac{50}{100}=\frac{a+b}{4}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

50

g at

a\%

, which holds

\frac{a}{2}

g of salt, and receives

50

g more.

\text{salt}=\frac{a}{2}+\frac{a+b}{4}=\frac{3a+b}{4},\quad \text{water}=50+50=100

Put both over a common denominator to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{3a+b}{4}\div100\times100=\frac{3a+b}{4}

\frac{100}{4\times100}=\frac{1}{4}

.

Answer:

\frac{3a+b}{4}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{3a+1a}{4}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $100$ g came from: $50$ g of original A water, plus $25$ g of A water and $25$ g of B water inside the $50$ g poured back. That is $75$ g at $a\%$ and $25$ g at $b\%$ , so the concentration is $\frac{75a+25b}{100}=\frac{3a+b}{4}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $\frac{a}{2}$ and $\frac{a+b}{4}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{3a+b}{4}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 2 easy

Beaker A holds $200$ g of salt water that is $a\%$ salt, and beaker B holds $100$ g of salt water that is $b\%$ salt. You pour $100$ g of the salt water from beaker A into beaker B and stir well. Then you pour $150$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $200$ g of $a\%$ salt water.
B: $100$ g of $b\%$ salt water.
Pour $100$ g from A into B and mix; then pour $150$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

100

g of A's water at

a\%

carries

a

g of salt. B already has

b

g of salt in its

100

g.

\frac{a}{100}\times100=a,\quad \frac{b}{100}\times100=b

a\%

of

100

g is

a

g.

2Beaker B after mixing

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

B now has

200

g of water containing

a+b

g of salt.

\frac{a+b}{200}\times100=\frac{a+b}{2}\ (\%)

B's strength is the average of the two, since equal amounts were mixed.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

150

g of B's mixture carries

\frac{3}{4}

of B's salt, because

150

g is

\frac{3}{4}

of

200

g.

\frac{a+b}{2}\times\frac{150}{100}=\frac{3a+3b}{4}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

100

g at

a\%

, which holds

a

g of salt, and receives

150

g more.

\text{salt}=a+\frac{3a+3b}{4}=\frac{7a+3b}{4},\quad \text{water}=100+150=250

Write

a

as

\frac{4a}{4}

to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{7a+3b}{4}\div250\times100=\frac{7a+3b}{10}

\frac{100}{4\times250}=\frac{1}{10}

.

Answer:

\frac{7a+3b}{10}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{7a+3a}{10}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $250$ g came from: $100$ g of original A water, plus $75$ g of A water and $75$ g of B water inside the $150$ g poured back. That is $175$ g at $a\%$ and $75$ g at $b\%$ , so the concentration is $\frac{175a+75b}{250}=\frac{7a+3b}{10}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $a$ and $\frac{3a+3b}{4}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{7a+3b}{10}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 3 easy

Beaker A holds $200$ g of salt water that is $a\%$ salt, and beaker B holds $150$ g of salt water that is $b\%$ salt. You pour $50$ g of the salt water from beaker A into beaker B and stir well. Then you pour $100$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $200$ g of $a\%$ salt water.
B: $150$ g of $b\%$ salt water.
Pour $50$ g from A into B and mix; then pour $100$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

50

g of A's water at

a\%

carries

\frac{a}{2}

g of salt. B already has

\frac{3b}{2}

g of salt in its

150

g.

\frac{a}{100}\times50=\frac{a}{2},\quad \frac{b}{100}\times150=\frac{3b}{2}

a\%

of

50

g is

\frac{a}{2}

g.

2Beaker B after mixing

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

B now has

200

g of water containing

\frac{a+3b}{2}

g of salt.

\frac{a+3b}{2}\div200\times100=\frac{a+3b}{4}\ (\%)

B's strength is a weighted average of

a

and

b

, weighted by

50

g and

150

g.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

100

g of B's mixture carries

\frac{1}{2}

of B's salt, because

100

g is

\frac{1}{2}

of

200

g.

\frac{a+3b}{4}\times\frac{100}{100}=\frac{a+3b}{4}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

150

g at

a\%

, which holds

\frac{3a}{2}

g of salt, and receives

100

g more.

\text{salt}=\frac{3a}{2}+\frac{a+3b}{4}=\frac{7a+3b}{4},\quad \text{water}=150+100=250

Put both over a common denominator to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{7a+3b}{4}\div250\times100=\frac{7a+3b}{10}

\frac{100}{4\times250}=\frac{1}{10}

.

Answer:

\frac{7a+3b}{10}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{7a+3a}{10}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $250$ g came from: $150$ g of original A water, plus $25$ g of A water and $75$ g of B water inside the $100$ g poured back. That is $175$ g at $a\%$ and $75$ g at $b\%$ , so the concentration is $\frac{175a+75b}{250}=\frac{7a+3b}{10}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $\frac{3a}{2}$ and $\frac{a+3b}{4}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{7a+3b}{10}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 4 easy

Beaker A holds $250$ g of salt water that is $a\%$ salt, and beaker B holds $150$ g of salt water that is $b\%$ salt. You pour $50$ g of the salt water from beaker A into beaker B and stir well. Then you pour $100$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $250$ g of $a\%$ salt water.
B: $150$ g of $b\%$ salt water.
Pour $50$ g from A into B and mix; then pour $100$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

50

g of A's water at

a\%

carries

\frac{a}{2}

g of salt. B already has

\frac{3b}{2}

g of salt in its

150

g.

\frac{a}{100}\times50=\frac{a}{2},\quad \frac{b}{100}\times150=\frac{3b}{2}

a\%

of

50

g is

\frac{a}{2}

g.

2Beaker B after mixing

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

B now has

200

g of water containing

\frac{a+3b}{2}

g of salt.

\frac{a+3b}{2}\div200\times100=\frac{a+3b}{4}\ (\%)

B's strength is a weighted average of

a

and

b

, weighted by

50

g and

150

g.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

100

g of B's mixture carries

\frac{1}{2}

of B's salt, because

100

g is

\frac{1}{2}

of

200

g.

\frac{a+3b}{4}\times\frac{100}{100}=\frac{a+3b}{4}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

200

g at

a\%

, which holds

2a

g of salt, and receives

100

g more.

\text{salt}=2a+\frac{a+3b}{4}=\frac{9a+3b}{4},\quad \text{water}=200+100=300

Write

2a

as

\frac{8a}{4}

to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{9a+3b}{4}\div300\times100=\frac{3a+b}{4}

\frac{100}{4\times300}=\frac{1}{12}

.

Answer:

\frac{3a+b}{4}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{3a+1a}{4}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $300$ g came from: $200$ g of original A water, plus $25$ g of A water and $75$ g of B water inside the $100$ g poured back. That is $225$ g at $a\%$ and $75$ g at $b\%$ , so the concentration is $\frac{225a+75b}{300}=\frac{3a+b}{4}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $2a$ and $\frac{a+3b}{4}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{3a+b}{4}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 5 medium

Beaker A holds $300$ g of salt water that is $a\%$ salt, and beaker B holds $100$ g of salt water that is $b\%$ salt. You pour $100$ g of the salt water from beaker A into beaker B and stir well. Then you pour $50$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $300$ g of $a\%$ salt water.
B: $100$ g of $b\%$ salt water.
Pour $100$ g from A into B and mix; then pour $50$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

100

g of A's water at

a\%

carries

a

g of salt. B already has

b

g of salt in its

100

g.

\frac{a}{100}\times100=a,\quad \frac{b}{100}\times100=b

a\%

of

100

g is

a

g.

2Beaker B after mixing

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

B now has

200

g of water containing

a+b

g of salt.

\frac{a+b}{200}\times100=\frac{a+b}{2}\ (\%)

B's strength is the average of the two, since equal amounts were mixed.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

50

g of B's mixture carries

\frac{1}{4}

of B's salt, because

50

g is

\frac{1}{4}

of

200

g.

\frac{a+b}{2}\times\frac{50}{100}=\frac{a+b}{4}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

200

g at

a\%

, which holds

2a

g of salt, and receives

50

g more.

\text{salt}=2a+\frac{a+b}{4}=\frac{9a+b}{4},\quad \text{water}=200+50=250

Write

2a

as

\frac{8a}{4}

to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{9a+b}{4}\div250\times100=\frac{9a+b}{10}

\frac{100}{4\times250}=\frac{1}{10}

.

Answer:

\frac{9a+b}{10}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{9a+1a}{10}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $250$ g came from: $200$ g of original A water, plus $25$ g of A water and $25$ g of B water inside the $50$ g poured back. That is $225$ g at $a\%$ and $25$ g at $b\%$ , so the concentration is $\frac{225a+25b}{250}=\frac{9a+b}{10}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $2a$ and $\frac{a+b}{4}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{9a+b}{10}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 6 medium

Beaker A holds $350$ g of salt water that is $a\%$ salt, and beaker B holds $200$ g of salt water that is $b\%$ salt. You pour $50$ g of the salt water from beaker A into beaker B and stir well. Then you pour $75$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $350$ g of $a\%$ salt water.
B: $200$ g of $b\%$ salt water.
Pour $50$ g from A into B and mix; then pour $75$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

50

g of A's water at

a\%

carries

\frac{a}{2}

g of salt. B already has

2b

g of salt in its

200

g.

\frac{a}{100}\times50=\frac{a}{2},\quad \frac{b}{100}\times200=2b

a\%

of

50

g is

\frac{a}{2}

g.

2Beaker B after mixing

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

B now has

250

g of water containing

\frac{a+4b}{2}

g of salt.

\frac{a+4b}{2}\div250\times100=\frac{a+4b}{5}\ (\%)

B's strength is a weighted average of

a

and

b

, weighted by

50

g and

200

g.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

75

g of B's mixture carries

\frac{3}{10}

of B's salt, because

75

g is

\frac{3}{10}

of

250

g.

\frac{a+4b}{5}\times\frac{75}{100}=\frac{3a+12b}{20}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

300

g at

a\%

, which holds

3a

g of salt, and receives

75

g more.

\text{salt}=3a+\frac{3a+12b}{20}=\frac{63a+12b}{20},\quad \text{water}=300+75=375

Write

3a

as

\frac{60a}{20}

to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{63a+12b}{20}\div375\times100=\frac{21a+4b}{25}

\frac{100}{20\times375}=\frac{1}{75}

.

Answer:

\frac{21a+4b}{25}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{21a+4a}{25}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $375$ g came from: $300$ g of original A water, plus $15$ g of A water and $60$ g of B water inside the $75$ g poured back. That is $315$ g at $a\%$ and $60$ g at $b\%$ , so the concentration is $\frac{315a+60b}{375}=\frac{21a+4b}{25}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $3a$ and $\frac{3a+12b}{20}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{21a+4b}{25}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 7 medium

Beaker A holds $400$ g of salt water that is $a\%$ salt, and beaker B holds $100$ g of salt water that is $b\%$ salt. You pour $300$ g of the salt water from beaker A into beaker B and stir well. Then you pour $100$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $400$ g of $a\%$ salt water.
B: $100$ g of $b\%$ salt water.
Pour $300$ g from A into B and mix; then pour $100$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

300

g of A's water at

a\%

carries

3a

g of salt. B already has

b

g of salt in its

100

g.

\frac{a}{100}\times300=3a,\quad \frac{b}{100}\times100=b

a\%

of

300

g is

3a

g.

2Beaker B after mixing

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

B now has

400

g of water containing

3a+b

g of salt.

\frac{3a+b}{400}\times100=\frac{3a+b}{4}\ (\%)

B's strength is a weighted average of

a

and

b

, weighted by

300

g and

100

g.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

100

g of B's mixture carries

\frac{1}{4}

of B's salt, because

100

g is

\frac{1}{4}

of

400

g.

\frac{3a+b}{4}\times\frac{100}{100}=\frac{3a+b}{4}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

100

g at

a\%

, which holds

a

g of salt, and receives

100

g more.

\text{salt}=a+\frac{3a+b}{4}=\frac{7a+b}{4},\quad \text{water}=100+100=200

Write

a

as

\frac{4a}{4}

to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{7a+b}{4}\div200\times100=\frac{7a+b}{8}

\frac{100}{4\times200}=\frac{1}{8}

.

Answer:

\frac{7a+b}{8}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{7a+1a}{8}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $200$ g came from: $100$ g of original A water, plus $75$ g of A water and $25$ g of B water inside the $100$ g poured back. That is $175$ g at $a\%$ and $25$ g at $b\%$ , so the concentration is $\frac{175a+25b}{200}=\frac{7a+b}{8}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $a$ and $\frac{3a+b}{4}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{7a+b}{8}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 8 medium

Beaker A holds $400$ g of salt water that is $a\%$ salt, and beaker B holds $200$ g of salt water that is $b\%$ salt. You pour $200$ g of the salt water from beaker A into beaker B and stir well. Then you pour $100$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $400$ g of $a\%$ salt water.
B: $200$ g of $b\%$ salt water.
Pour $200$ g from A into B and mix; then pour $100$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

200

g of A's water at

a\%

carries

2a

g of salt. B already has

2b

g of salt in its

200

g.

\frac{a}{100}\times200=2a,\quad \frac{b}{100}\times200=2b

a\%

of

200

g is

2a

g.

2Beaker B after mixing

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

B now has

400

g of water containing

2a+2b

g of salt.

\frac{2a+2b}{400}\times100=\frac{a+b}{2}\ (\%)

B's strength is the average of the two, since equal amounts were mixed.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

100

g of B's mixture carries

\frac{1}{4}

of B's salt, because

100

g is

\frac{1}{4}

of

400

g.

\frac{a+b}{2}\times\frac{100}{100}=\frac{a+b}{2}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

200

g at

a\%

, which holds

2a

g of salt, and receives

100

g more.

\text{salt}=2a+\frac{a+b}{2}=\frac{5a+b}{2},\quad \text{water}=200+100=300

Write

2a

as

\frac{4a}{2}

to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{5a+b}{2}\div300\times100=\frac{5a+b}{6}

\frac{100}{2\times300}=\frac{1}{6}

.

Answer:

\frac{5a+b}{6}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{5a+1a}{6}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $300$ g came from: $200$ g of original A water, plus $50$ g of A water and $50$ g of B water inside the $100$ g poured back. That is $250$ g at $a\%$ and $50$ g at $b\%$ , so the concentration is $\frac{250a+50b}{300}=\frac{5a+b}{6}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $2a$ and $\frac{a+b}{2}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{5a+b}{6}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 9 hard

Beaker A holds $500$ g of salt water that is $a\%$ salt, and beaker B holds $300$ g of salt water that is $b\%$ salt. You pour $200$ g of the salt water from beaker A into beaker B and stir well. Then you pour $250$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $500$ g of $a\%$ salt water.
B: $300$ g of $b\%$ salt water.
Pour $200$ g from A into B and mix; then pour $250$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

200

g of A's water at

a\%

carries

2a

g of salt. B already has

3b

g of salt in its

300

g.

\frac{a}{100}\times200=2a,\quad \frac{b}{100}\times300=3b

a\%

of

200

g is

2a

g.

2Beaker B after mixing

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

B now has

500

g of water containing

2a+3b

g of salt.

\frac{2a+3b}{500}\times100=\frac{2a+3b}{5}\ (\%)

B's strength is a weighted average of

a

and

b

, weighted by

200

g and

300

g.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

250

g of B's mixture carries

\frac{1}{2}

of B's salt, because

250

g is

\frac{1}{2}

of

500

g.

\frac{2a+3b}{5}\times\frac{250}{100}=\frac{2a+3b}{2}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

300

g at

a\%

, which holds

3a

g of salt, and receives

250

g more.

\text{salt}=3a+\frac{2a+3b}{2}=\frac{8a+3b}{2},\quad \text{water}=300+250=550

Write

3a

as

\frac{6a}{2}

to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{8a+3b}{2}\div550\times100=\frac{8a+3b}{11}

\frac{100}{2\times550}=\frac{1}{11}

.

Answer:

\frac{8a+3b}{11}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{8a+3a}{11}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $550$ g came from: $300$ g of original A water, plus $100$ g of A water and $150$ g of B water inside the $250$ g poured back. That is $400$ g at $a\%$ and $150$ g at $b\%$ , so the concentration is $\frac{400a+150b}{550}=\frac{8a+3b}{11}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $3a$ and $\frac{2a+3b}{2}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{8a+3b}{11}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 10 hard

Beaker A holds $500$ g of salt water that is $a\%$ salt, and beaker B holds $100$ g of salt water that is $b\%$ salt. You pour $400$ g of the salt water from beaker A into beaker B and stir well. Then you pour $250$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $500$ g of $a\%$ salt water.
B: $100$ g of $b\%$ salt water.
Pour $400$ g from A into B and mix; then pour $250$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

400

g of A's water at

a\%

carries

4a

g of salt. B already has

b

g of salt in its

100

g.

\frac{a}{100}\times400=4a,\quad \frac{b}{100}\times100=b

a\%

of

400

g is

4a

g.

2Beaker B after mixing

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

B now has

500

g of water containing

4a+b

g of salt.

\frac{4a+b}{500}\times100=\frac{4a+b}{5}\ (\%)

B's strength is a weighted average of

a

and

b

, weighted by

400

g and

100

g.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

250

g of B's mixture carries

\frac{1}{2}

of B's salt, because

250

g is

\frac{1}{2}

of

500

g.

\frac{4a+b}{5}\times\frac{250}{100}=\frac{4a+b}{2}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

100

g at

a\%

, which holds

a

g of salt, and receives

250

g more.

\text{salt}=a+\frac{4a+b}{2}=\frac{6a+b}{2},\quad \text{water}=100+250=350

Write

a

as

\frac{2a}{2}

to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{6a+b}{2}\div350\times100=\frac{6a+b}{7}

\frac{100}{2\times350}=\frac{1}{7}

.

Answer:

\frac{6a+b}{7}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{6a+1a}{7}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $350$ g came from: $100$ g of original A water, plus $200$ g of A water and $50$ g of B water inside the $250$ g poured back. That is $300$ g at $a\%$ and $50$ g at $b\%$ , so the concentration is $\frac{300a+50b}{350}=\frac{6a+b}{7}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $a$ and $\frac{4a+b}{2}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{6a+b}{7}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 11 hard

Beaker A holds $600$ g of salt water that is $a\%$ salt, and beaker B holds $100$ g of salt water that is $b\%$ salt. You pour $200$ g of the salt water from beaker A into beaker B and stir well. Then you pour $150$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $600$ g of $a\%$ salt water.
B: $100$ g of $b\%$ salt water.
Pour $200$ g from A into B and mix; then pour $150$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

200

g of A's water at

a\%

carries

2a

g of salt. B already has

b

g of salt in its

100

g.

\frac{a}{100}\times200=2a,\quad \frac{b}{100}\times100=b

a\%

of

200

g is

2a

g.

2Beaker B after mixing

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

B now has

300

g of water containing

2a+b

g of salt.

\frac{2a+b}{300}\times100=\frac{2a+b}{3}\ (\%)

B's strength is a weighted average of

a

and

b

, weighted by

200

g and

100

g.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

150

g of B's mixture carries

\frac{1}{2}

of B's salt, because

150

g is

\frac{1}{2}

of

300

g.

\frac{2a+b}{3}\times\frac{150}{100}=\frac{2a+b}{2}

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

400

g at

a\%

, which holds

4a

g of salt, and receives

150

g more.

\text{salt}=4a+\frac{2a+b}{2}=\frac{10a+b}{2},\quad \text{water}=400+150=550

Write

4a

as

\frac{8a}{2}

to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

\frac{10a+b}{2}\div550\times100=\frac{10a+b}{11}

\frac{100}{2\times550}=\frac{1}{11}

.

Answer:

\frac{10a+b}{11}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{10a+1a}{11}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $550$ g came from: $400$ g of original A water, plus $100$ g of A water and $50$ g of B water inside the $150$ g poured back. That is $500$ g at $a\%$ and $50$ g at $b\%$ , so the concentration is $\frac{500a+50b}{550}=\frac{10a+b}{11}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $4a$ and $\frac{2a+b}{2}$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{10a+b}{11}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Variant 12 hard

Beaker A holds $600$ g of salt water that is $a\%$ salt, and beaker B holds $300$ g of salt water that is $b\%$ salt. You pour $300$ g of the salt water from beaker A into beaker B and stir well. Then you pour $200$ g of the salt water from beaker B into beaker A. Write the concentration of the salt water in beaker A in terms of $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Salt water is poured from A to B, mixed, and some is poured back from B to A. Find A's final salt percentage using $a$ and $b$ .

Givens

A: $600$ g of $a\%$ salt water.
B: $300$ g of $b\%$ salt water.
Pour $300$ g from A into B and mix; then pour $200$ g from B into A.

Unknowns

The final concentration of beaker A, in percent, in terms of $a$ and $b$ .

Constraints

Concentration $=$ (grams of salt) $\div$ (grams of salt water) $\times100$ .
After stirring, every gram of B has the same concentration.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #8 Analyze the Units#13 Convert to Algebra

Percentages can't simply be added, but grams of salt can. So follow the grams of salt and the grams of water through each pour, and only turn them back into a percentage at the end.

3 · Execute5 carry out the plan

1Salt in the first pour

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

300

g of A's water at

a\%

carries

3a

g of salt. B already has

3b

g of salt in its

300

g.

\frac{a}{100}\times300=3a,\quad \frac{b}{100}\times300=3b

a\%

of

300

g is

3a

g.

2Beaker B after mixing

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

B now has

600

g of water containing

3a+3b

g of salt.

\frac{3a+3b}{600}\times100=\frac{a+b}{2}\ (\%)

B's strength is the average of the two, since equal amounts were mixed.

3Salt in the pour back

#7 Identify Subproblems 7.RP.A.3

200

g of B's mixture carries

\frac{1}{3}

of B's salt, because

200

g is

\frac{1}{3}

of

600

g.

\frac{a+b}{2}\times\frac{200}{100}=a+b

Same concentration throughout B, so salt splits in proportion to weight.

4Beaker A's totals

#13 Convert to Algebra 7.EE.A.1

A kept

300

g at

a\%

, which holds

3a

g of salt, and receives

200

g more.

\text{salt}=3a+a+b=4a+b,\quad \text{water}=300+200=500

Put both over a common denominator to combine.

5Back to a percentage

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

Divide salt by total weight and multiply by

100

.

(4a+b)\div500\times100=\frac{4a+b}{5}

\frac{100}{1\times500}=\frac{1}{5}

.

Answer:

\frac{4a+b}{5}\%

4 · Reviewdoes it hold up?

If $a=b$ , nothing should change, and indeed $\frac{4a+1a}{5}=a$ . The answer is a weighted average of $a$ and $b$ with most weight on $a$ , which fits A being mostly its own water.

Another way: Count where A's $500$ g came from: $300$ g of original A water, plus $100$ g of A water and $100$ g of B water inside the $200$ g poured back. That is $400$ g at $a\%$ and $100$ g at $b\%$ , so the concentration is $\frac{400a+100b}{500}=\frac{4a+b}{5}$ .

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Converting between percent concentration and grams of salt.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Adding the salt expressions $3a$ and $a+b$.
7.EE.A.2 Rewrite an expression in different forms to shed light on the problem — Rewriting the final ratio as $\frac{4a+b}{5}$.

💡Takeaway. Percents don't add, but grams of salt do: follow the salt.

Track grams of salt through two pourings · 12 practice problems

Generated variants — 12

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