← Adding salt moves the top and the bottom of the fraction · Ratio, Rate and Percent

Adding salt moves the top and the bottom of the fraction · 12 practice problems

7.RP.A.36.RP.A.3

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 75%

To $110\ \text{g}$ of salt water with a concentration of $50\%$ , another $110\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

110 g of 50 percent salt water gets another 110 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 110 g and is 50 percent salt.
110 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

50 percent of 110 g.

110 \times 0.5 = 55

55 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

55 + 110 = 165

165 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

110 + 110 = 220

220 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

165 \div 220 \times 100 = 75

The new water is 75 percent salt.

Answer: 75%

4 · Reviewdoes it hold up?

75 percent of 220 g is 165 g of salt, which is the 55 we started with plus the 110 added. And the new figure is above 50 percent, as adding pure salt must make it.

Another way: Averaging 50 percent with 100 percent gives 75, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 2 easy answer: 52.5%

To $160\ \text{g}$ of salt water with a concentration of $5\%$ , another $160\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

160 g of 5 percent salt water gets another 160 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 160 g and is 5 percent salt.
160 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

5 percent of 160 g.

160 \times 0.05 = 8

8 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

8 + 160 = 168

168 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

160 + 160 = 320

320 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

168 \div 320 \times 100 = 52.5

The new water is 52.5 percent salt.

Answer: 52.5%

4 · Reviewdoes it hold up?

52.5 percent of 320 g is 168 g of salt, which is the 8 we started with plus the 160 added. And the new figure is above 5 percent, as adding pure salt must make it.

Another way: Averaging 5 percent with 100 percent gives 52.5, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 3 easy answer: 68%

To $120\ \text{g}$ of salt water with a concentration of $20\%$ , another $180\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

120 g of 20 percent salt water gets another 180 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 120 g and is 20 percent salt.
180 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

20 percent of 120 g.

120 \times 0.2 = 24

24 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

24 + 180 = 204

204 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

120 + 180 = 300

300 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

204 \div 300 \times 100 = 68

The new water is 68 percent salt.

Answer: 68%

4 · Reviewdoes it hold up?

68 percent of 300 g is 204 g of salt, which is the 24 we started with plus the 180 added. And the new figure is above 20 percent, as adding pure salt must make it.

Another way: Averaging 20 percent with 100 percent gives 60, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 4 easy answer: 28%

To $180\ \text{g}$ of salt water with a concentration of $20\%$ , another $20\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

180 g of 20 percent salt water gets another 20 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 180 g and is 20 percent salt.
20 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

20 percent of 180 g.

180 \times 0.2 = 36

36 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

36 + 20 = 56

56 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

180 + 20 = 200

200 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

56 \div 200 \times 100 = 28

The new water is 28 percent salt.

Answer: 28%

4 · Reviewdoes it hold up?

28 percent of 200 g is 56 g of salt, which is the 36 we started with plus the 20 added. And the new figure is above 20 percent, as adding pure salt must make it.

Another way: Averaging 20 percent with 100 percent gives 60, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 5 medium answer: 34%

To $210\ \text{g}$ of salt water with a concentration of $12\%$ , another $70\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

210 g of 12 percent salt water gets another 70 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 210 g and is 12 percent salt.
70 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

12 percent of 210 g.

210 \times 0.12 = 25.2

25.2 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

25.2 + 70 = 95.2

95.2 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

210 + 70 = 280

280 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

95.2 \div 280 \times 100 = 34

The new water is 34 percent salt.

Answer: 34%

4 · Reviewdoes it hold up?

34 percent of 280 g is 95.2 g of salt, which is the 25.2 we started with plus the 70 added. And the new figure is above 12 percent, as adding pure salt must make it.

Another way: Averaging 12 percent with 100 percent gives 56, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 6 medium answer: 17.2%

To $230\ \text{g}$ of salt water with a concentration of $10\%$ , another $20\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

230 g of 10 percent salt water gets another 20 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 230 g and is 10 percent salt.
20 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

10 percent of 230 g.

230 \times 0.1 = 23

23 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

23 + 20 = 43

43 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

230 + 20 = 250

250 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

43 \div 250 \times 100 = 17.2

The new water is 17.2 percent salt.

Answer: 17.2%

4 · Reviewdoes it hold up?

17.2 percent of 250 g is 43 g of salt, which is the 23 we started with plus the 20 added. And the new figure is above 10 percent, as adding pure salt must make it.

Another way: Averaging 10 percent with 100 percent gives 55, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 7 medium answer: 41.5%

To $260\ \text{g}$ of salt water with a concentration of $10\%$ , another $140\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

260 g of 10 percent salt water gets another 140 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 260 g and is 10 percent salt.
140 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

10 percent of 260 g.

260 \times 0.1 = 26

26 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

26 + 140 = 166

166 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

260 + 140 = 400

400 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

166 \div 400 \times 100 = 41.5

The new water is 41.5 percent salt.

Answer: 41.5%

4 · Reviewdoes it hold up?

41.5 percent of 400 g is 166 g of salt, which is the 26 we started with plus the 140 added. And the new figure is above 10 percent, as adding pure salt must make it.

Another way: Averaging 10 percent with 100 percent gives 55, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 8 medium answer: 38%

To $310\ \text{g}$ of salt water with a concentration of $12\%$ , another $130\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

310 g of 12 percent salt water gets another 130 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 310 g and is 12 percent salt.
130 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

12 percent of 310 g.

310 \times 0.12 = 37.2

37.2 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

37.2 + 130 = 167.2

167.2 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

310 + 130 = 440

440 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

167.2 \div 440 \times 100 = 38

The new water is 38 percent salt.

Answer: 38%

4 · Reviewdoes it hold up?

38 percent of 440 g is 167.2 g of salt, which is the 37.2 we started with plus the 130 added. And the new figure is above 12 percent, as adding pure salt must make it.

Another way: Averaging 12 percent with 100 percent gives 56, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 9 hard answer: 69%

To $310\ \text{g}$ of salt water with a concentration of $50\%$ , another $190\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

310 g of 50 percent salt water gets another 190 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 310 g and is 50 percent salt.
190 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

50 percent of 310 g.

310 \times 0.5 = 155

155 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

155 + 190 = 345

345 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

310 + 190 = 500

500 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

345 \div 500 \times 100 = 69

The new water is 69 percent salt.

Answer: 69%

4 · Reviewdoes it hold up?

69 percent of 500 g is 345 g of salt, which is the 155 we started with plus the 190 added. And the new figure is above 50 percent, as adding pure salt must make it.

Another way: Averaging 50 percent with 100 percent gives 75, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 10 hard answer: 40%

To $360\ \text{g}$ of salt water with a concentration of $15\%$ , another $150\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

360 g of 15 percent salt water gets another 150 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 360 g and is 15 percent salt.
150 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

15 percent of 360 g.

360 \times 0.15 = 54

54 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

54 + 150 = 204

204 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

360 + 150 = 510

510 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

204 \div 510 \times 100 = 40

The new water is 40 percent salt.

Answer: 40%

4 · Reviewdoes it hold up?

40 percent of 510 g is 204 g of salt, which is the 54 we started with plus the 150 added. And the new figure is above 15 percent, as adding pure salt must make it.

Another way: Averaging 15 percent with 100 percent gives 57.5, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 11 hard answer: 32%

To $360\ \text{g}$ of salt water with a concentration of $15\%$ , another $90\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

360 g of 15 percent salt water gets another 90 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 360 g and is 15 percent salt.
90 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

15 percent of 360 g.

360 \times 0.15 = 54

54 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

54 + 90 = 144

144 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

360 + 90 = 450

450 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

144 \div 450 \times 100 = 32

The new water is 32 percent salt.

Answer: 32%

4 · Reviewdoes it hold up?

32 percent of 450 g is 144 g of salt, which is the 54 we started with plus the 90 added. And the new figure is above 15 percent, as adding pure salt must make it.

Another way: Averaging 15 percent with 100 percent gives 57.5, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.

Variant 12 hard answer: 39.8%

To $430\ \text{g}$ of salt water with a concentration of $30\%$ , another $70\ \text{g}$ of salt is added. What is the concentration of the new salt water, as a percent?

Show solution

1 · Understandwhat's really being asked

430 g of 30 percent salt water gets another 70 g of salt stirred in. We want the new concentration.

Givens

The salt water weighs 430 g and is 30 percent salt.
70 g of dry salt is added.

Unknowns

The concentration of the new salt water, as a percent.

Constraints

What is added is pure salt, so all of it counts as salt and all of it counts as weight.

2 · Planchoose the strategy

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

Percentages cannot be added, because the whole they refer to changes. Work in grams -- how much salt, how much water in total -- and turn the pair back into a percentage only at the very end.

3 · Execute4 carry out the plan

1Find the salt already there

#7 Identify Subproblems 6.RP.A.3

30 percent of 430 g.

430 \times 0.3 = 129

129 g of salt to start with.

2Add the new salt

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

The salt goes up by the whole amount added.

129 + 70 = 199

199 g of salt now.

3Add it to the total as well

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

The salt is part of the mixture, so the whole thing gets heavier too. This is the step that is easy to skip.

430 + 70 = 500

500 g of salt water.

4Turn the pair back into a percent

#8 Analyze the Units 6.RP.A.3

Divide the salt by the total, then multiply by 100.

199 \div 500 \times 100 = 39.8

The new water is 39.8 percent salt.

Answer: 39.8%

4 · Reviewdoes it hold up?

39.8 percent of 500 g is 199 g of salt, which is the 129 we started with plus the 70 added. And the new figure is above 30 percent, as adding pure salt must make it.

Another way: Averaging 30 percent with 100 percent gives 65, which is wrong: the two amounts being mixed are not the same size.

Standardsmin grade 7

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Following a percentage through a change in the whole.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Reading a percentage as a part-to-whole rate.

💡Takeaway. Adding salt makes both the salt and the salt water heavier. Count grams first and work out the percent last.