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Density and Scale Models: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Read a material label

Two solid tiles each occupy 180 cubic centimeters. Tile A has mass 126 grams and tile B has mass 216 grams.

Find both average mass densities. Explain what the larger density tells you when the tile volumes are equal.

Use the mass and volume printed on each tile label.

Text description of this figure

Two equal-sized label panels list the same volume of 180 cubic centimeters and different masses of 126 grams and 216 grams.
1. Hint 1
  Read grams per cubic centimeter as grams for each one cubic centimeter on average.
2. Hint 2
  Divide each measured mass by the same volume, then compare the quotients.
Answer

Tile A averages 0.7 grams per cubic centimeter. Tile B averages 1.2 grams per cubic centimeter and has more mass in the same volume.

Full solution

Tile A.
Its mass per cubic centimeter is
$126\div180=0.7.$

Tile B.
Its mass per cubic centimeter is
$216\div180=1.2.$

The larger rate means more mass in the same volume.
Here both tiles occupy 180 cubic centimeters, and tile B has the larger mass.

Answer

Tile A averages 0.7 grams per cubic centimeter. Tile B averages 1.2 grams per cubic centimeter and has more mass in the same volume.

Key idea

Dividing by the common volume expresses the masses on a one-cubic-centimeter basis.
Problem 2 Compare two town centers

North Center has 296 residents in 0.8 square kilometers. Market Center has 420 residents in 1,200,000 square meters.

Which center has the greater average population density per square kilometer, and by how much? Explain how the center with fewer residents can have the greater density.
1. Hint 1
  Put both areas in square kilometers. Start with 1 kilometer as 1,000 meters.
2. Hint 2
  A square kilometer contains one million square meters. Divide each population by the area of its own center.
Answer

North Center, by 20 residents per square kilometer. It has fewer residents spread over sufficiently less area.

Full solution

North Center.
Divide 296 by 0.8 to get 370 residents per square kilometer.

Market Center.
Its area is 1.2 square kilometers.
Divide 420 by 1.2 to get 350 residents per square kilometer.

North Center has 20 more residents per square kilometer.
Its population is smaller, but its area is smaller by a larger proportion.

Answer

North Center, by 20 residents per square kilometer. It has fewer residents spread over sufficiently less area.

Key idea

A larger total population can be spread over a larger area at a lower average density.
Problem 3 Two liquids for the same tank

An empty tank can hold 7.2 L to its fill line. Two different liquid batches each have mass 6.3 kg. Batch A has a density of 1.05 grams per cubic centimeter. Batch B has a density of 0.9 grams per cubic centimeter.

Test each batch separately in the empty tank. Does it fit below the fill line? If it fits, how much room is left up to that line?
1. Hint 1
  Since 1 L is 1,000 cubic centimeters and 1 kg is 1,000 g, the numerical density in kilograms per liter matches the numerical density in grams per cubic centimeter.
2. Hint 2
  Divide each mass by its own density to find how many liters that batch occupies. Compare each volume with 7.2 L.
Answer

Batch A fits with 1.2 L left. Batch B fits with 0.2 L left.

Full solution

Batch A.
At 1.05 kilograms per liter, the volume in liters is
$V_A=\frac{6.3}{1.05}.$
That is 6 L, leaving 1.2 L.

Batch B.
At 0.9 kilograms per liter, the volume in liters is
$V_B=\frac{6.3}{0.9}.$
That is 7 L, leaving 0.2 L.

With the same mass in each batch, the less dense liquid takes up more space.
That agrees with batch B leaving less room.

Answer

Batch A fits with 1.2 L left. Batch B fits with 0.2 L left.

Key idea

Capacity is a volume. Convert each mass into the volume it occupies before comparing it with capacity.
Problem 4 Map a river study area

On a paper map, a trapezoidal study area has parallel sides of 2.4 cm and 3.6 cm. The perpendicular distance between them is 1.8 cm. A scale bar 1.6 cm long on the map represents 280 m along the ground.

Find the actual study area in square meters. A straight sampling line elsewhere is 490 m long on the ground. Find its length on the map.

Use the labeled map measurements. Screen size can vary.

Text description of this figure

The map trapezoid has parallel sides 2.4 and 3.6 centimeters, with perpendicular separation 1.8 centimeters. A 1.6-centimeter bar on that printed map represents 280 meters.
1. Hint 1
  Use the scale bar to find how many ground meters one map centimeter represents.
2. Hint 2
  Find the trapezoid’s map area and apply the square of the length conversion. For the line, use the length conversion in reverse.
Answer

Study area 165,375 square meters. Sampling line 2.8 cm on the map.

Full solution

The scale represents 175 actual meters per map centimeter because 280 divided by 1.6 is 175.

Study area.
Average the two parallel side lengths and multiply by the perpendicular distance.
The map area in square centimeters is
$A_{\mathrm{map}}=\frac{2.4+3.6}{2}\cdot1.8.$
It is 5.4 square centimeters.

In square meters, the actual area is
$A_{\mathrm{actual}}=5.4(175)^2.$
This is 165,375 square meters.

Sampling line.
The actual length is given, so divide by 175 actual meters per map centimeter.
The result is 2.8 map centimeters, longer than the 1.6 cm bar as expected.

Answer

Study area 165,375 square meters. Sampling line 2.8 cm on the map.

Key idea

Use the squared conversion for the footprint and the length conversion for the line.
Problem 5 Which measurements give capacity?

A model cooler has rectangular outside dimensions 20 cm by 16 cm by 15 cm. Its rectangular cavity measures 16 cm by 12 cm by 10 cm up to the rim. Two full-size coolers have outside dimensions twice those of the model.

Cooler A. Its inside lengths are also twice the model’s. Predict its capacity to the rim in liters.

Cooler B. Its wall and floor thicknesses are unknown. Decide whether its capacity to the rim must equal Cooler A’s. Give one possible rectangular inside that shows why.

Top view. Use the stated heights to find volumes.

Text description of this figure

A top-view schematic shows the 20-by-16-centimeter outside rectangle and the 16-by-12-centimeter cavity within it. Depths are given in the problem.
1. Hint 1
  Capacity uses cavity dimensions. Only Cooler A specifies an inside length factor.
2. Hint 2
  The outside dimensions for Cooler B are 40 cm by 32 cm by 30 cm. A narrower cavity than Cooler A’s can fit within that shell.
Answer

Cooler A has capacity 15.36 L. Cooler B need not match. For example, an inside of 30 cm by 22 cm by 18 cm would hold 11.88 L.

Full solution

Cooler A.
The model cavity volume is
$V_{\mathrm{model}}=16\cdot12\cdot10.$
That is 1,920 cubic centimeters.
The actual capacity in cubic centimeters is
$V_A=1{,}920(2^3).$
This is 15,360 cubic centimeters, or 15.36 L.

Cooler B.
A rectangular cavity 30 cm by 22 cm by 18 cm fits within the stated outside dimensions of 40 cm by 32 cm by 30 cm, with positive room for walls and a floor.
It holds 11,880 cubic centimeters, or 11.88 L.

The two full-size coolers can share outside dimensions and have different capacities because the space used by the walls and floor can differ.

Answer

Cooler A has capacity 15.36 L. Cooler B need not match. For example, an inside of 30 cm by 22 cm by 18 cm would hold 11.88 L.

Key idea

An outside scale predicts an inside volume only with information about the corresponding cavity.
Problem 6 An average across two zones

Two nonoverlapping zones make up a district. Zone A covers 2 square kilometers and Zone B covers 6 square kilometers. The district has 1,600 residents.

A separate count bounds Zone B at no more than 175 residents per square kilometer. Find the district average and the least possible average in Zone A. Give a population for each zone that attains the least average.

The widths show the ratio of the areas. The zone populations are left unlabeled.

Text description of this figure

One long rectangle shows Zone A, 2 square kilometers, and Zone B, 6 square kilometers. Together the two adjacent areas make up the district.
1. Hint 1
  Find the greatest population Zone B can have under its density limit.
2. Hint 2
  Subtract that count from the district total. Check that both zone counts are possible.
Answer

The district averages 200 residents per square kilometer. Zone A averages at least 275 per square kilometer, attained with 550 residents in A and 1,050 in B.

Full solution

District.
The two areas total 8 square kilometers, so $1{,}600\div8=200$ residents per square kilometer.

Zone B.
Its greatest allowed count is $175(6)=1{,}050$ residents.

Zone A.
At least $1{,}600-1{,}050=550$ residents must live in its 2 square kilometers.
Its average is at least $550\div2=275$ residents per square kilometer.

The counts 550 and 1,050 add to 1,600 and give Zone B its allowed upper density of 175 per square kilometer.
They attain the lower bound for Zone A.

Answer

The district averages 200 residents per square kilometer. Zone A averages at least 275 per square kilometer, attained with 550 residents in A and 1,050 in B.

Key idea

An upper bound on one region’s count can force a lower bound on another region’s density.
Problem 7 One prediction in two unit systems

A miniature solid occupies 54 cubic centimeters. The actual solid has similar shape at a model-to-actual scale of 1:20 and is made of material with density 0.9 grams per cubic centimeter.

Route A. Use cubic centimeters and grams, then report the actual mass in kilograms.

Route B. Use cubic meters and kilograms throughout the mass calculation. Explain why the two routes agree for any positive actual volume and the same density expressed in the two unit systems.
1. Hint 1
  Both routes start with an actual volume from the cube of the length factor.
2. Hint 2
  In the second route, each cubic meter contains one million cubic centimeters. A mass in grams becomes kilograms when divided by 1,000. Convert the density as well as the volume.
Answer

Both routes give 388.8 kg. The combined conversion factors in Route B divide the gram-based mass number by 1,000, just as Route A does.

Full solution

The actual volume in cubic centimeters is
$V_{\mathrm{actual}}=54(20^3).$
That is 432,000 cubic centimeters.

Route A.
Multiplying by 0.9 grams per cubic centimeter gives 388,800 grams.
Dividing by 1,000 gives 388.8 kg.

Route B.
One cubic meter contains 1,000,000 cubic centimeters, so this volume is 0.432 cubic meters.
At 0.9 grams per cubic centimeter, one cubic meter has a mass of 900,000 grams, or 900 kg.
Multiply 0.432 cubic meters by 900 kilograms per cubic meter to get 388.8 kg.

Why the routes agree.
For any actual volume with numerical measure $V$ in cubic centimeters and density number $d$ in grams per cubic centimeter, Route A gives $Vd/1{,}000$ kilograms.

Route B divides the volume number by one million and multiplies the density number by one thousand.
In kilograms,
$\frac{V}{10^6}(10^3d)=\frac{Vd}{10^3}.$
This matches Route A.

Answer

Both routes give 388.8 kg. The combined conversion factors in Route B divide the gram-based mass number by 1,000, just as Route A does.

Key idea

Converting both the volume and the density preserves the mass when the units are used consistently.
Problem 8 Survey a reserve from a plan

A reserve consists of two nonoverlapping zones. On its printed map, 1 cm represents 0.5 km.

East. The map area is 18 square centimeters. The average density there is 48 nests per square kilometer.

West. The map area is 10 square centimeters. Surveyors recorded 150 nests there.

Which zone has the higher average nesting density? Find the average across the entire reserve and explain which zone’s rate it is closer to.

Schematic label panels. Read the areas from the labels.

Text description of this figure

The map measurements for East and West are labeled beside their separate nesting data. East has 18 map square centimeters and 48 nests per actual square kilometer. West has 10 map square centimeters and 150 nests.
1. Hint 1
  Convert both areas using the square of 0.5. The east count is its actual area times its density.
2. Hint 2
  Compute the total number of nests and the total actual area before dividing to find the reserve average.
Answer

West has the higher rate at 60 nests per square kilometer. The reserve averages $52\frac27$ nests per square kilometer, closer to East’s 48.

Full solution

East.
Each map square centimeter represents 0.25 square kilometers.
East covers 4.5 square kilometers.
At 48 nests per square kilometer, East has 216 nests.

West.
West covers 2.5 square kilometers.
Divide 150 nests by 2.5 square kilometers to get 60 nests per square kilometer.
West has the higher rate.

Reserve.
Divide the 366 nests by the 7 square kilometers to get $52\frac27$ nests per square kilometer.
That is closer to 48 than to 60 because East has the larger share of the total area.

Answer

West has the higher rate at 60 nests per square kilometer. The reserve averages $52\frac27$ nests per square kilometer, closer to East’s 48.

Key idea

Combining two area densities weights each rate by its actual area.
Problem 9 The same mold with a hollow center

A model for a solid resin casting occupies 112 cubic centimeters. Each actual length is three times its model length. The resin has a mass density of 1.25 grams per cubic centimeter.

Solid casting. The full actual shape is made of resin.

Hollow casting. The same outer shape surrounds an empty cavity. The corresponding cavity in the model would occupy 28 cubic centimeters.

Find both modeled masses and the amount of mass saved by the cavity. Compare the hollow casting’s average density using its outer volume with the density of the resin itself.

Schematic sections show where the cavity is. Use the stated volumes for calculations.

Text description of this figure

Schematic sections show the same outer rectangle for each casting. The hollow casting has a smaller, closed empty rectangle completely inside it.
1. Hint 1
  The outer volume and the cavity volume both use the cube of the same scale factor.
2. Hint 2
  Only the part outside the cavity contains resin. For average density of the whole hollow casting, use the entire outer volume.
Answer

Solid casting 3,780 g. Hollow casting 2,835 g. Mass saved 945 g. The hollow casting averages 0.9375 grams per cubic centimeter using its outer volume, below the resin’s 1.25.

Full solution

Both outer shapes have a volume of 3,024 cubic centimeters because $112(3^3)$ gives the actual outer volume.

Solid casting.
Multiply 3,024 cubic centimeters by 1.25 grams per cubic centimeter to get 3,780 g.

Hollow casting.
The cavity accounts for a quarter of the model volume.
Since both volumes use the same factor, it also accounts for a quarter of the actual outer volume.
The resin mass is three quarters of 3,780 g, or 2,835 g.

Comparison.
The hollow casting saves 945 g.
Its average density using all 3,024 cubic centimeters is 0.9375 grams per cubic centimeter.
The resin itself is still 1.25 grams per cubic centimeter.

Answer

Solid casting 3,780 g. Hollow casting 2,835 g. Mass saved 945 g. The hollow casting averages 0.9375 grams per cubic centimeter using its outer volume, below the resin’s 1.25.

Key idea

The denominator for an object’s average density can include space its material does not occupy.
Problem 10 Check the fill mark on a planter

A school planter’s scale model has a rectangular inside base 5 cm by 4 cm and a rim 2.5 cm above the floor. All actual interior lengths are 10 times the corresponding model lengths. A cart can carry at most 35.2 kg of soil in the planter after allowing for the empty planter’s mass. The model assumes a bulk soil density of 0.88 kilograms per liter.

Rim. Would a full planter meet the soil mass limit in the model?

Fill mark. If it would not, find the maximum actual fill depth allowed by the model.

Check. A 5 L sample of the soil prepared the same way has measured mass 4.35 kg. Use this independent measurement to check the estimate at the fill mark and explain how far that check goes.

All figure dimensions are from the model.

Text description of this figure

The planter model has a rectangular inside base 5 centimeters by 4 centimeters. The side view indicates that its rim is 2.5 centimeters above the inside floor.
1. Hint 1
  Use the scale to find the actual inside base and rim depth. Convert its rim volume to liters before using the density.
2. Hint 2
  The mass limit gives a volume limit at the assumed density. Divide that volume in cubic centimeters by the actual base area.
3. Hint 3
  Compare the number of 5 L batches in the allowed fill with what one measured batch weighs.
Answer

The rim would contain 44 kg of soil, over the limit. The model allows a 20 cm actual fill depth. The measured sample suggests about 34.8 kg at that depth if it represents the load, close to the model’s 35.2 kg.

Full solution

The actual inside base is 50 cm by 40 cm and the rim depth is 25 cm.

Rim.
The rim volume is 50,000 cubic centimeters, or 50 L.
At 0.88 kilograms per liter, a full planter contains 44 kg of soil.
That exceeds 35.2 kg.

Fill mark.
The mass limit divided by the assumed density gives a volume of 40 L, or 40,000 cubic centimeters.
Divide by the actual base area of 2,000 square centimeters.
The maximum fill depth in this model is 20 cm, below the rim.

Check.
Forty liters is eight batches of 5 L.
Multiplying the measured 4.35 kg per batch by eight gives 34.8 kg, close to the modeled 35.2 kg.
This one sample supports the estimate if it represents the soil in the bed.
The measured mass of the full load would settle whether it meets the limit.

The modeled limit uses four fifths of the actual rim depth.

Answer

The rim would contain 44 kg of soil, over the limit. The model allows a 20 cm actual fill depth. The measured sample suggests about 34.8 kg at that depth if it represents the load, close to the model’s 35.2 kg.

Key idea

Scaling predicts the available space. The soil density and the mass limit determine how much of that space to fill.

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