Practice · Two thresholds on one mixture give a range · Sensim Math

Variant 1 easy

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $500$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $450$ L, and its BOD was at least $0.9$ mg/mL. Find the range of the amount of milk, in mL, in the $500$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$3$	$0.06$
Ramen broth	$0.5$	$1.26$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $500$ mL in all, go down the drain. Per mL, milk needs $3$ L of water to purify and has BOD $0.06$ mg/mL; broth needs $0.5$ L and has BOD $1.26$ mg/mL. The total water needed is more than $450$ L and the BOD of the whole $500$ mL is at least $0.9$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $500$ mL of milk and ramen broth.
Per $1$ mL: milk needs $3$ L of water, broth needs $0.5$ L.
Per $1$ mL: milk has BOD $0.06$ mg, broth has BOD $1.26$ mg.
Total water needed is more than $450$ L.
BOD of the mixture is at least $0.9$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $500$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

500 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 3x + 0.5(500 - x) \text{ L}, \qquad \text{oxygen demand} = 0.06x + 1.26(500 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

450

L.

3x + 250 - 0.5x > 450 \;\Rightarrow\; 2.5x > 200 \;\Rightarrow\; x > 80

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

500

mL. It must be at least

0.9

.

\frac{0.06x + 1.26(500 - x)}{500} \ge 0.9 \;\Rightarrow\; 630 - 1.2x \ge 450 \;\Rightarrow\; 1.2x \le 180 \;\Rightarrow\; x \le 150

Milk has a much lower BOD than broth, so too much milk drags the average below

0.9

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

80 < x \le 150

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

80

mL and at most

150

mL (

80 < x \le 150

)

4 · Reviewdoes it hold up?

At $x = 150$ : water $= 450 + 175 = 625$ L, more than $450$ ; BOD $= \frac{9 + 441}{500} = 0.9$ , allowed. At $x = 80$ : water $= 240 + 210 = 450$ L, not more than $450$ , so $80$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $3 - 0.5 = 2.5$ L of water and removes $1.26 - 0.06 = 1.2$ mg of oxygen demand. All broth needs $250$ L, so it takes more than $200 \div 2.5 = 80$ mL of milk to pass $450$ L. All broth gives $630$ mg, and the limit is $450$ mg, so at most $180 \div 1.2 = 150$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 2 easy

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $500$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $400$ L, and its BOD was at least $0.89$ mg/mL. Find the range of the amount of milk, in mL, in the $500$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$2$	$0.05$
Ramen broth	$0.5$	$1.25$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $500$ mL in all, go down the drain. Per mL, milk needs $2$ L of water to purify and has BOD $0.05$ mg/mL; broth needs $0.5$ L and has BOD $1.25$ mg/mL. The total water needed is more than $400$ L and the BOD of the whole $500$ mL is at least $0.89$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $500$ mL of milk and ramen broth.
Per $1$ mL: milk needs $2$ L of water, broth needs $0.5$ L.
Per $1$ mL: milk has BOD $0.05$ mg, broth has BOD $1.25$ mg.
Total water needed is more than $400$ L.
BOD of the mixture is at least $0.89$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $500$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

500 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 2x + 0.5(500 - x) \text{ L}, \qquad \text{oxygen demand} = 0.05x + 1.25(500 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

400

L.

2x + 250 - 0.5x > 400 \;\Rightarrow\; 1.5x > 150 \;\Rightarrow\; x > 100

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

500

mL. It must be at least

0.89

.

\frac{0.05x + 1.25(500 - x)}{500} \ge 0.89 \;\Rightarrow\; 625 - 1.2x \ge 445 \;\Rightarrow\; 1.2x \le 180 \;\Rightarrow\; x \le 150

Milk has a much lower BOD than broth, so too much milk drags the average below

0.89

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

100 < x \le 150

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

100

mL and at most

150

mL (

100 < x \le 150

)

4 · Reviewdoes it hold up?

At $x = 150$ : water $= 300 + 175 = 475$ L, more than $400$ ; BOD $= \frac{7.5 + 437.5}{500} = 0.89$ , allowed. At $x = 100$ : water $= 200 + 200 = 400$ L, not more than $400$ , so $100$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $2 - 0.5 = 1.5$ L of water and removes $1.25 - 0.05 = 1.2$ mg of oxygen demand. All broth needs $250$ L, so it takes more than $150 \div 1.5 = 100$ mL of milk to pass $400$ L. All broth gives $625$ mg, and the limit is $445$ mg, so at most $180 \div 1.2 = 150$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 3 medium

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $1000$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $840$ L, and its BOD was at least $0.98$ mg/mL. Find the range of the amount of milk, in mL, in the $1000$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$1.8$	$0.08$
Ramen broth	$0.6$	$1.28$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $1000$ mL in all, go down the drain. Per mL, milk needs $1.8$ L of water to purify and has BOD $0.08$ mg/mL; broth needs $0.6$ L and has BOD $1.28$ mg/mL. The total water needed is more than $840$ L and the BOD of the whole $1000$ mL is at least $0.98$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $1000$ mL of milk and ramen broth.
Per $1$ mL: milk needs $1.8$ L of water, broth needs $0.6$ L.
Per $1$ mL: milk has BOD $0.08$ mg, broth has BOD $1.28$ mg.
Total water needed is more than $840$ L.
BOD of the mixture is at least $0.98$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $1000$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

1000 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 1.8x + 0.6(1000 - x) \text{ L}, \qquad \text{oxygen demand} = 0.08x + 1.28(1000 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

840

L.

1.8x + 600 - 0.6x > 840 \;\Rightarrow\; 1.2x > 240 \;\Rightarrow\; x > 200

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

1000

mL. It must be at least

0.98

.

\frac{0.08x + 1.28(1000 - x)}{1000} \ge 0.98 \;\Rightarrow\; 1280 - 1.2x \ge 980 \;\Rightarrow\; 1.2x \le 300 \;\Rightarrow\; x \le 250

Milk has a much lower BOD than broth, so too much milk drags the average below

0.98

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

200 < x \le 250

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

200

mL and at most

250

mL (

200 < x \le 250

)

4 · Reviewdoes it hold up?

At $x = 250$ : water $= 450 + 450 = 900$ L, more than $840$ ; BOD $= \frac{20 + 960}{1000} = 0.98$ , allowed. At $x = 200$ : water $= 360 + 480 = 840$ L, not more than $840$ , so $200$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $1.8 - 0.6 = 1.2$ L of water and removes $1.28 - 0.08 = 1.2$ mg of oxygen demand. All broth needs $600$ L, so it takes more than $240 \div 1.2 = 200$ mL of milk to pass $840$ L. All broth gives $1280$ mg, and the limit is $980$ mg, so at most $300 \div 1.2 = 250$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 4 medium

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $1000$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $900$ L, and its BOD was at least $0.75$ mg/mL. Find the range of the amount of milk, in mL, in the $1000$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$2.5$	$0.05$
Ramen broth	$0.5$	$1.05$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $1000$ mL in all, go down the drain. Per mL, milk needs $2.5$ L of water to purify and has BOD $0.05$ mg/mL; broth needs $0.5$ L and has BOD $1.05$ mg/mL. The total water needed is more than $900$ L and the BOD of the whole $1000$ mL is at least $0.75$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $1000$ mL of milk and ramen broth.
Per $1$ mL: milk needs $2.5$ L of water, broth needs $0.5$ L.
Per $1$ mL: milk has BOD $0.05$ mg, broth has BOD $1.05$ mg.
Total water needed is more than $900$ L.
BOD of the mixture is at least $0.75$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $1000$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

1000 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 2.5x + 0.5(1000 - x) \text{ L}, \qquad \text{oxygen demand} = 0.05x + 1.05(1000 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

900

L.

2.5x + 500 - 0.5x > 900 \;\Rightarrow\; 2x > 400 \;\Rightarrow\; x > 200

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

1000

mL. It must be at least

0.75

.

\frac{0.05x + 1.05(1000 - x)}{1000} \ge 0.75 \;\Rightarrow\; 1050 - x \ge 750 \;\Rightarrow\; x \le 300 \;\Rightarrow\; x \le 300

Milk has a much lower BOD than broth, so too much milk drags the average below

0.75

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

200 < x \le 300

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

200

mL and at most

300

mL (

200 < x \le 300

)

4 · Reviewdoes it hold up?

At $x = 300$ : water $= 750 + 350 = 1100$ L, more than $900$ ; BOD $= \frac{15 + 735}{1000} = 0.75$ , allowed. At $x = 200$ : water $= 500 + 400 = 900$ L, not more than $900$ , so $200$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $2.5 - 0.5 = 2$ L of water and removes $1.05 - 0.05 = 1$ mg of oxygen demand. All broth needs $500$ L, so it takes more than $400 \div 2 = 200$ mL of milk to pass $900$ L. All broth gives $1050$ mg, and the limit is $750$ mg, so at most $300 \div 1 = 300$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 5 easy

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $1000$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $800$ L, and its BOD was at least $0.95$ mg/mL. Find the range of the amount of milk, in mL, in the $1000$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$2$	$0.05$
Ramen broth	$0.5$	$1.25$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $1000$ mL in all, go down the drain. Per mL, milk needs $2$ L of water to purify and has BOD $0.05$ mg/mL; broth needs $0.5$ L and has BOD $1.25$ mg/mL. The total water needed is more than $800$ L and the BOD of the whole $1000$ mL is at least $0.95$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $1000$ mL of milk and ramen broth.
Per $1$ mL: milk needs $2$ L of water, broth needs $0.5$ L.
Per $1$ mL: milk has BOD $0.05$ mg, broth has BOD $1.25$ mg.
Total water needed is more than $800$ L.
BOD of the mixture is at least $0.95$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $1000$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

1000 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 2x + 0.5(1000 - x) \text{ L}, \qquad \text{oxygen demand} = 0.05x + 1.25(1000 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

800

L.

2x + 500 - 0.5x > 800 \;\Rightarrow\; 1.5x > 300 \;\Rightarrow\; x > 200

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

1000

mL. It must be at least

0.95

.

\frac{0.05x + 1.25(1000 - x)}{1000} \ge 0.95 \;\Rightarrow\; 1250 - 1.2x \ge 950 \;\Rightarrow\; 1.2x \le 300 \;\Rightarrow\; x \le 250

Milk has a much lower BOD than broth, so too much milk drags the average below

0.95

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

200 < x \le 250

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

200

mL and at most

250

mL (

200 < x \le 250

)

4 · Reviewdoes it hold up?

At $x = 250$ : water $= 500 + 375 = 875$ L, more than $800$ ; BOD $= \frac{12.5 + 937.5}{1000} = 0.95$ , allowed. At $x = 200$ : water $= 400 + 400 = 800$ L, not more than $800$ , so $200$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $2 - 0.5 = 1.5$ L of water and removes $1.25 - 0.05 = 1.2$ mg of oxygen demand. All broth needs $500$ L, so it takes more than $300 \div 1.5 = 200$ mL of milk to pass $800$ L. All broth gives $1250$ mg, and the limit is $950$ mg, so at most $300 \div 1.2 = 250$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 6 easy

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $1000$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $650$ L, and its BOD was at least $0.89$ mg/mL. Find the range of the amount of milk, in mL, in the $1000$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$2$	$0.05$
Ramen broth	$0.5$	$1.25$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $1000$ mL in all, go down the drain. Per mL, milk needs $2$ L of water to purify and has BOD $0.05$ mg/mL; broth needs $0.5$ L and has BOD $1.25$ mg/mL. The total water needed is more than $650$ L and the BOD of the whole $1000$ mL is at least $0.89$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $1000$ mL of milk and ramen broth.
Per $1$ mL: milk needs $2$ L of water, broth needs $0.5$ L.
Per $1$ mL: milk has BOD $0.05$ mg, broth has BOD $1.25$ mg.
Total water needed is more than $650$ L.
BOD of the mixture is at least $0.89$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $1000$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

1000 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 2x + 0.5(1000 - x) \text{ L}, \qquad \text{oxygen demand} = 0.05x + 1.25(1000 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

650

L.

2x + 500 - 0.5x > 650 \;\Rightarrow\; 1.5x > 150 \;\Rightarrow\; x > 100

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

1000

mL. It must be at least

0.89

.

\frac{0.05x + 1.25(1000 - x)}{1000} \ge 0.89 \;\Rightarrow\; 1250 - 1.2x \ge 890 \;\Rightarrow\; 1.2x \le 360 \;\Rightarrow\; x \le 300

Milk has a much lower BOD than broth, so too much milk drags the average below

0.89

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

100 < x \le 300

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

100

mL and at most

300

mL (

100 < x \le 300

)

4 · Reviewdoes it hold up?

At $x = 300$ : water $= 600 + 350 = 950$ L, more than $650$ ; BOD $= \frac{15 + 875}{1000} = 0.89$ , allowed. At $x = 100$ : water $= 200 + 450 = 650$ L, not more than $650$ , so $100$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $2 - 0.5 = 1.5$ L of water and removes $1.25 - 0.05 = 1.2$ mg of oxygen demand. All broth needs $500$ L, so it takes more than $150 \div 1.5 = 100$ mL of milk to pass $650$ L. All broth gives $1250$ mg, and the limit is $890$ mg, so at most $360 \div 1.2 = 300$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 7 medium

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $1000$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $1300$ L, and its BOD was at least $0.76$ mg/mL. Find the range of the amount of milk, in mL, in the $1000$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$2$	$0.1$
Ramen broth	$1$	$1.3$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $1000$ mL in all, go down the drain. Per mL, milk needs $2$ L of water to purify and has BOD $0.1$ mg/mL; broth needs $1$ L and has BOD $1.3$ mg/mL. The total water needed is more than $1300$ L and the BOD of the whole $1000$ mL is at least $0.76$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $1000$ mL of milk and ramen broth.
Per $1$ mL: milk needs $2$ L of water, broth needs $1$ L.
Per $1$ mL: milk has BOD $0.1$ mg, broth has BOD $1.3$ mg.
Total water needed is more than $1300$ L.
BOD of the mixture is at least $0.76$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $1000$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

1000 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 2x + (1000 - x) \text{ L}, \qquad \text{oxygen demand} = 0.1x + 1.3(1000 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

1300

L.

2x + 1000 - x > 1300 \;\Rightarrow\; x > 300 \;\Rightarrow\; x > 300

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

1000

mL. It must be at least

0.76

.

\frac{0.1x + 1.3(1000 - x)}{1000} \ge 0.76 \;\Rightarrow\; 1300 - 1.2x \ge 760 \;\Rightarrow\; 1.2x \le 540 \;\Rightarrow\; x \le 450

Milk has a much lower BOD than broth, so too much milk drags the average below

0.76

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

300 < x \le 450

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

300

mL and at most

450

mL (

300 < x \le 450

)

4 · Reviewdoes it hold up?

At $x = 450$ : water $= 900 + 550 = 1450$ L, more than $1300$ ; BOD $= \frac{45 + 715}{1000} = 0.76$ , allowed. At $x = 300$ : water $= 600 + 700 = 1300$ L, not more than $1300$ , so $300$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $2 - 1 = 1$ L of water and removes $1.3 - 0.1 = 1.2$ mg of oxygen demand. All broth needs $1000$ L, so it takes more than $300 \div 1 = 300$ mL of milk to pass $1300$ L. All broth gives $1300$ mg, and the limit is $760$ mg, so at most $540 \div 1.2 = 450$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 8 medium

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $1000$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $1300$ L, and its BOD was at least $0.85$ mg/mL. Find the range of the amount of milk, in mL, in the $1000$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$2.2$	$0.1$
Ramen broth	$0.7$	$1.6$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $1000$ mL in all, go down the drain. Per mL, milk needs $2.2$ L of water to purify and has BOD $0.1$ mg/mL; broth needs $0.7$ L and has BOD $1.6$ mg/mL. The total water needed is more than $1300$ L and the BOD of the whole $1000$ mL is at least $0.85$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $1000$ mL of milk and ramen broth.
Per $1$ mL: milk needs $2.2$ L of water, broth needs $0.7$ L.
Per $1$ mL: milk has BOD $0.1$ mg, broth has BOD $1.6$ mg.
Total water needed is more than $1300$ L.
BOD of the mixture is at least $0.85$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $1000$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

1000 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 2.2x + 0.7(1000 - x) \text{ L}, \qquad \text{oxygen demand} = 0.1x + 1.6(1000 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

1300

L.

2.2x + 700 - 0.7x > 1300 \;\Rightarrow\; 1.5x > 600 \;\Rightarrow\; x > 400

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

1000

mL. It must be at least

0.85

.

\frac{0.1x + 1.6(1000 - x)}{1000} \ge 0.85 \;\Rightarrow\; 1600 - 1.5x \ge 850 \;\Rightarrow\; 1.5x \le 750 \;\Rightarrow\; x \le 500

Milk has a much lower BOD than broth, so too much milk drags the average below

0.85

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

400 < x \le 500

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

400

mL and at most

500

mL (

400 < x \le 500

)

4 · Reviewdoes it hold up?

At $x = 500$ : water $= 1100 + 350 = 1450$ L, more than $1300$ ; BOD $= \frac{50 + 800}{1000} = 0.85$ , allowed. At $x = 400$ : water $= 880 + 420 = 1300$ L, not more than $1300$ , so $400$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $2.2 - 0.7 = 1.5$ L of water and removes $1.6 - 0.1 = 1.5$ mg of oxygen demand. All broth needs $700$ L, so it takes more than $600 \div 1.5 = 400$ mL of milk to pass $1300$ L. All broth gives $1600$ mg, and the limit is $850$ mg, so at most $750 \div 1.5 = 500$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 9 hard

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $1500$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $1200$ L, and its BOD was at least $0.85$ mg/mL. Find the range of the amount of milk, in mL, in the $1500$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$2$	$0.05$
Ramen broth	$0.5$	$1.25$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $1500$ mL in all, go down the drain. Per mL, milk needs $2$ L of water to purify and has BOD $0.05$ mg/mL; broth needs $0.5$ L and has BOD $1.25$ mg/mL. The total water needed is more than $1200$ L and the BOD of the whole $1500$ mL is at least $0.85$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $1500$ mL of milk and ramen broth.
Per $1$ mL: milk needs $2$ L of water, broth needs $0.5$ L.
Per $1$ mL: milk has BOD $0.05$ mg, broth has BOD $1.25$ mg.
Total water needed is more than $1200$ L.
BOD of the mixture is at least $0.85$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $1500$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

1500 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 2x + 0.5(1500 - x) \text{ L}, \qquad \text{oxygen demand} = 0.05x + 1.25(1500 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

1200

L.

2x + 750 - 0.5x > 1200 \;\Rightarrow\; 1.5x > 450 \;\Rightarrow\; x > 300

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

1500

mL. It must be at least

0.85

.

\frac{0.05x + 1.25(1500 - x)}{1500} \ge 0.85 \;\Rightarrow\; 1875 - 1.2x \ge 1275 \;\Rightarrow\; 1.2x \le 600 \;\Rightarrow\; x \le 500

Milk has a much lower BOD than broth, so too much milk drags the average below

0.85

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

300 < x \le 500

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

300

mL and at most

500

mL (

300 < x \le 500

)

4 · Reviewdoes it hold up?

At $x = 500$ : water $= 1000 + 500 = 1500$ L, more than $1200$ ; BOD $= \frac{25 + 1250}{1500} = 0.85$ , allowed. At $x = 300$ : water $= 600 + 600 = 1200$ L, not more than $1200$ , so $300$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $2 - 0.5 = 1.5$ L of water and removes $1.25 - 0.05 = 1.2$ mg of oxygen demand. All broth needs $750$ L, so it takes more than $450 \div 1.5 = 300$ mL of milk to pass $1200$ L. All broth gives $1875$ mg, and the limit is $1275$ mg, so at most $600 \div 1.2 = 500$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 10 hard

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $1000$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $1600$ L, and its BOD was at least $0.94$ mg/mL. Find the range of the amount of milk, in mL, in the $1000$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$3$	$0.1$
Ramen broth	$1$	$1.5$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $1000$ mL in all, go down the drain. Per mL, milk needs $3$ L of water to purify and has BOD $0.1$ mg/mL; broth needs $1$ L and has BOD $1.5$ mg/mL. The total water needed is more than $1600$ L and the BOD of the whole $1000$ mL is at least $0.94$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $1000$ mL of milk and ramen broth.
Per $1$ mL: milk needs $3$ L of water, broth needs $1$ L.
Per $1$ mL: milk has BOD $0.1$ mg, broth has BOD $1.5$ mg.
Total water needed is more than $1600$ L.
BOD of the mixture is at least $0.94$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $1000$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

1000 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 3x + (1000 - x) \text{ L}, \qquad \text{oxygen demand} = 0.1x + 1.5(1000 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

1600

L.

3x + 1000 - x > 1600 \;\Rightarrow\; 2x > 600 \;\Rightarrow\; x > 300

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

1000

mL. It must be at least

0.94

.

\frac{0.1x + 1.5(1000 - x)}{1000} \ge 0.94 \;\Rightarrow\; 1500 - 1.4x \ge 940 \;\Rightarrow\; 1.4x \le 560 \;\Rightarrow\; x \le 400

Milk has a much lower BOD than broth, so too much milk drags the average below

0.94

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

300 < x \le 400

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

300

mL and at most

400

mL (

300 < x \le 400

)

4 · Reviewdoes it hold up?

At $x = 400$ : water $= 1200 + 600 = 1800$ L, more than $1600$ ; BOD $= \frac{40 + 900}{1000} = 0.94$ , allowed. At $x = 300$ : water $= 900 + 700 = 1600$ L, not more than $1600$ , so $300$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $3 - 1 = 2$ L of water and removes $1.5 - 0.1 = 1.4$ mg of oxygen demand. All broth needs $1000$ L, so it takes more than $600 \div 2 = 300$ mL of milk to pass $1600$ L. All broth gives $1500$ mg, and the limit is $940$ mg, so at most $560 \div 1.4 = 400$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 11 hard

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $2000$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $1400$ L, and its BOD was at least $0.852$ mg/mL. Find the range of the amount of milk, in mL, in the $2000$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$1.5$	$0.04$
Ramen broth	$0.5$	$1.2$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $2000$ mL in all, go down the drain. Per mL, milk needs $1.5$ L of water to purify and has BOD $0.04$ mg/mL; broth needs $0.5$ L and has BOD $1.2$ mg/mL. The total water needed is more than $1400$ L and the BOD of the whole $2000$ mL is at least $0.852$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $2000$ mL of milk and ramen broth.
Per $1$ mL: milk needs $1.5$ L of water, broth needs $0.5$ L.
Per $1$ mL: milk has BOD $0.04$ mg, broth has BOD $1.2$ mg.
Total water needed is more than $1400$ L.
BOD of the mixture is at least $0.852$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $2000$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

2000 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 1.5x + 0.5(2000 - x) \text{ L}, \qquad \text{oxygen demand} = 0.04x + 1.2(2000 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

1400

L.

1.5x + 1000 - 0.5x > 1400 \;\Rightarrow\; x > 400 \;\Rightarrow\; x > 400

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

2000

mL. It must be at least

0.852

.

\frac{0.04x + 1.2(2000 - x)}{2000} \ge 0.852 \;\Rightarrow\; 2400 - 1.16x \ge 1704 \;\Rightarrow\; 1.16x \le 696 \;\Rightarrow\; x \le 600

Milk has a much lower BOD than broth, so too much milk drags the average below

0.852

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

400 < x \le 600

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

400

mL and at most

600

mL (

400 < x \le 600

)

4 · Reviewdoes it hold up?

At $x = 600$ : water $= 900 + 700 = 1600$ L, more than $1400$ ; BOD $= \frac{24 + 1680}{2000} = 0.852$ , allowed. At $x = 400$ : water $= 600 + 800 = 1400$ L, not more than $1400$ , so $400$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $1.5 - 0.5 = 1$ L of water and removes $1.2 - 0.04 = 1.16$ mg of oxygen demand. All broth needs $1000$ L, so it takes more than $400 \div 1 = 400$ mL of milk to pass $1400$ L. All broth gives $2400$ mg, and the limit is $1704$ mg, so at most $696 \div 1.16 = 600$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Variant 12 hard

The table shows, for $1$ mL of each kind of food waste poured down the drain, the amount of water needed to purify it and its BOD (biochemical oxygen demand). A total of $2000$ mL of milk and ramen broth together was poured down the drain. The water needed to purify it was more than $1440$ L, and its BOD was at least $0.95$ mg/mL. Find the range of the amount of milk, in mL, in the $2000$ mL that was poured.

Food waste	Water needed to purify (L)	BOD (mg/mL)
Milk	$2$	$0.05$
Ramen broth	$0.4$	$1.25$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Milk and ramen broth, $2000$ mL in all, go down the drain. Per mL, milk needs $2$ L of water to purify and has BOD $0.05$ mg/mL; broth needs $0.4$ L and has BOD $1.25$ mg/mL. The total water needed is more than $1440$ L and the BOD of the whole $2000$ mL is at least $0.95$ mg/mL. Find the range of the amount of milk.

Givens

Total poured: $2000$ mL of milk and ramen broth.
Per $1$ mL: milk needs $2$ L of water, broth needs $0.4$ L.
Per $1$ mL: milk has BOD $0.05$ mg, broth has BOD $1.25$ mg.
Total water needed is more than $1440$ L.
BOD of the mixture is at least $0.95$ mg/mL.

Unknowns

The range of the amount of milk, in mL.

Constraints

Water needed adds up over all the waste; BOD is an amount per mL, so the mixture's BOD is total oxygen demand divided by $2000$ mL.
Both conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#8 Analyze the Units

One unknown, the amount of milk, controls both quantities. Extend the table to totals, write one inequality for the water and one for the BOD, and keep the values that satisfy both.

3 · Execute4 carry out the plan

1Choose the unknown and extend the table

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

Let the milk be

x

mL, so the broth is

2000 - x

mL. Multiply each per-mL value by the amount to get totals.

\text{water} = 2x + 0.4(2000 - x) \text{ L}, \qquad \text{oxygen demand} = 0.05x + 1.25(2000 - x) \text{ mg}

Each row of the table times its amount gives that food's share of the total.

2Water condition

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

The total water must be more than

1440

L.

2x + 800 - 0.4x > 1440 \;\Rightarrow\; 1.6x > 640 \;\Rightarrow\; x > 400

Milk needs more water than broth, so more milk means more water: the milk has a lower limit.

3BOD condition

#8 Analyze the Units 7.EE.B.4

BOD is milligrams per mL, so the mixture's BOD is the total oxygen demand in mg divided by

2000

mL. It must be at least

0.95

.

\frac{0.05x + 1.25(2000 - x)}{2000} \ge 0.95 \;\Rightarrow\; 2500 - 1.2x \ge 1900 \;\Rightarrow\; 1.2x \le 600 \;\Rightarrow\; x \le 500

Milk has a much lower BOD than broth, so too much milk drags the average below

0.95

: the milk has an upper limit.

4Put the two limits together

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

Both conditions must hold, so take the overlap.

400 < x \le 500

The water condition pushes the milk up, the BOD condition holds it down.

Answer: more than

400

mL and at most

500

mL (

400 < x \le 500

)

4 · Reviewdoes it hold up?

At $x = 500$ : water $= 1000 + 600 = 1600$ L, more than $1440$ ; BOD $= \frac{25 + 1875}{2000} = 0.95$ , allowed. At $x = 400$ : water $= 800 + 640 = 1440$ L, not more than $1440$ , so $400$ itself is excluded.

Another way: Think of each mL of milk swapped in for broth: it adds $2 - 0.4 = 1.6$ L of water and removes $1.25 - 0.05 = 1.2$ mg of oxygen demand. All broth needs $800$ L, so it takes more than $640 \div 1.6 = 400$ mL of milk to pass $1440$ L. All broth gives $2500$ mg, and the limit is $1900$ mg, so at most $600 \div 1.2 = 500$ mL of milk is allowed.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the water and BOD conditions as inequalities in the amount of milk, solving each, and combining them.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Scaling per-mL water and oxygen demand to totals for each amount.

💡Takeaway. One unknown can face two tests at once -- solve each test on its own, then keep only the values that pass both.

Two thresholds on one mixture give a range · 12 practice problems

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