Math 3 Quiz: Density And Volume Modeling
6 questions · exam conditions
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Density And Volume ModelingQuestion 1 of 6

A truncated cone (frustum) has a top radius of 4 cm, bottom radius of 8 cm, and height of 12 cm. A liquid with density 1.5 g/cm³ fills the frustum completely. If the frustum is inverted so that the smaller radius is at the bottom, what is the new height of the liquid?

10.8 cm representing liquid redistribution after inversion
11.2 cm representing liquid redistribution after inversion
12.0 cm representing liquid redistribution after inversion
12.6 cm representing liquid redistribution after inversion
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Math 3 Quiz

Math 3 Quiz: Density And Volume Modeling

Practice Density And Volume Modeling in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Density And Volume Modeling, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A truncated cone (frustum) has a top radius of 4 cm, bottom radius of 8 cm, and height of 12 cm. A liquid with density 1.5 g/cm³ fills the frustum completely. If the frustum is inverted so that the smaller radius is at the bottom, what is the new height of the liquid?

  1. 10.8 cm representing liquid redistribution after inversion
  2. 11.2 cm representing liquid redistribution after inversion
  3. 12.0 cm representing liquid redistribution after inversion (correct answer)
  4. 12.6 cm representing liquid redistribution after inversion
Explanation: The volume of a frustum is V=πh3(r12+r1r2+r22)V = \frac{\pi h}{3}(r_1^2 + r_1 r_2 + r_2^2) where r1r_1 and r2r_2 are the radii of the two ends. Original volume: V=π×123(42+4×8+82)=4π(16+32+64)=4π×112=448πV = \frac{\pi \times 12}{3}(4^2 + 4 \times 8 + 8^2) = 4\pi(16 + 32 + 64) = 4\pi \times 112 = 448\pi cm³. When inverted, the frustum has the same shape and volume, so the liquid will still occupy the same volume 448π448\pi cm³. Since the container shape is identical (just flipped), the liquid height remains 12 cm - it completely fills the frustum regardless of orientation. Choice A assumes some liquid compression. Choice B accounts for surface tension effects incorrectly. Choice D assumes liquid expansion.

Question 2

A cylindrical oil storage tank has a radius of 25 feet and height of 40 feet. Due to a leak, oil flows out at a rate proportional to the pressure at the bottom, which depends on the height of oil in the tank. When the tank is 75% full, oil flows out at 15 cubic feet per minute. What is the rate of change of the oil surface level at this moment?

  1. −3125π-\frac{3}{125\pi} feet per minute due to leak pressure
  2. −6125π-\frac{6}{125\pi} feet per minute due to leak pressure
  3. −15625π-\frac{15}{625\pi} feet per minute due to leak pressure (correct answer)
  4. −12625π-\frac{12}{625\pi} feet per minute due to leak pressure
Explanation: The cross-sectional area of the cylindrical tank is A=πr2=π(252)=625πA = \pi r^2 = \pi (25^2) = 625\pi square feet. The rate of volume change due to the leak is dVdt=−15\frac{dV}{dt} = -15 cubic feet per minute (negative because oil is flowing out). For a cylinder, dVdt=A×dhdt\frac{dV}{dt} = A \times \frac{dh}{dt}, where hh is the height of oil. Therefore: −15=625π×dhdt-15 = 625\pi \times \frac{dh}{dt}. Solving for the rate of surface level change: dhdt=−15625π\frac{dh}{dt} = \frac{-15}{625\pi} feet per minute. Choice A incorrectly uses radius instead of area. Choice B doubles the flow rate incorrectly. Choice D uses wrong area calculation.

Question 3

A cylindrical water tank has a radius of 8 feet and a height of 12 feet. Water flows into the tank at a rate of 15 cubic feet per minute. If the tank starts empty, what is the rate at which the water level is rising when the tank is half full?

  1. 1564π\frac{15}{64\pi} feet per minute (correct answer)
  2. 15128π\frac{15}{128\pi} feet per minute
  3. 1532π\frac{15}{32\pi} feet per minute
  4. 15256π\frac{15}{256\pi} feet per minute
Explanation: The volume of water at height h is V = πr²h = π(8)²h = 64πh. Taking the derivative with respect to time: dV/dt = 64π(dh/dt). Since water flows in at 15 ft³/min, dV/dt = 15. Therefore: 15 = 64π(dh/dt), so dh/dt = 15/(64π) feet per minute. This rate is constant regardless of how full the tank is, since the cross-sectional area remains constant. Choice B uses the total volume formula incorrectly. Choice C uses radius instead of radius squared. Choice D doubles the error from choice B.

Question 4

A pharmaceutical company produces tablets with an active ingredient that has a density of 2.3 g/cm³. Each tablet is cylindrical with a diameter of 8 mm and a thickness that varies depending on the dosage required.

If a tablet needs to contain exactly 184 mg of the active ingredient, and the tablet is made of 100% active ingredient, what should be the thickness of the tablet?

  1. 2.0 mm
  2. 1.6 mm (correct answer)
  3. 1.8 mm
  4. 2.2 mm
Explanation: First, convert the required mass: 184 mg = 0.184 g. The required volume is V = mass/density = 0.184/2.3 = 0.08 cm³. The tablet has radius 4 mm = 0.4 cm. For a cylinder, V = πr²h, so h = V/(πr²) = 0.08/(π × 0.16) = 0.08/(0.16π) = 0.5/π ≈ 0.159 cm = 1.59 mm ≈ 1.6 mm. Choice A would give excess mass. Choice C would give about 15% excess mass. Choice D would give about 38% excess mass.

Question 5

A hemispherical bowl with radius 12 cm is being filled with honey at a constant rate of 15 cm³/min. The density of honey is 1.4 g/cm³. When the honey is 8 cm deep (measured vertically from the bottom), at what rate is the mass of honey in the bowl increasing?

  1. 21 g/min (correct answer)
  2. 15 g/min
  3. 18 g/min
  4. 24 g/min
Explanation: Since honey is being added at a constant volume rate of 15 cm³/min, and the density is constant at 1.4 g/cm³, the rate of mass increase is simply dm/dt = density × dV/dt = 1.4 × 15 = 21 g/min. This rate is independent of the current depth because the volume addition rate is constant. The depth information is a distractor. Choice B gives just the volume rate without applying density. Choice C uses an incorrect density of 1.2. Choice D uses an incorrect density of 1.6.

Question 6

A conical sand pile has a height that is always twice its base radius. Sand is being added to the pile at a rate of 12 cubic meters per minute. The density of the sand is 1,600 kg/m³. At what rate is the mass of the pile increasing when the radius is 4 meters?

  1. 19,200 kg per minute (correct answer)
  2. 7,200 kg per minute
  3. 14,400 kg per minute
  4. 28,800 kg per minute
Explanation: Since mass = density × volume, and density is constant at 1,600 kg/m³, the rate of mass increase is dm/dt = ρ(dV/dt) = 1,600 × 12 = 19,200 kg/min. This rate is independent of the current radius because we're told sand is added at a constant volume rate. Choice B incorrectly uses 1,600 × 12/2.67 assuming some radius dependency. Choice C uses 1,600 × 9 as if the volume rate were 9 instead of 12. Choice D uses 1,600 × 18 as if the volume rate were doubled.