Question

the data table shows the mass and volumes of boxes a, b and c.

box a
mass = 36 g
volume = 6 cm³

box b
mass = 81 g
volume = 9 cm³

box c
mass = 48 g
volume = 8 cm³

density = \\(\frac{mass}{volume}\\)

1. which box is represented below?

diagram of a rectangular prism with dimensions (e.g., 3 cm, 2 cm, 1 cm) indicated by arrows

1. what is the density of box c? ______ don’t forget units.

1. which box has the greatest density? box a, b, or c? ______ show your work:

Explanation:

Question 1

Step1: Calculate volume of the box in the diagram

The box is a rectangular prism, volume \( V = l \times w \times h \). Given \( l = 3\,\text{cm} \), \( w = 2\,\text{cm} \), \( h = 1\,\text{cm} \), so \( V = 3\times2\times1 = 6\,\text{cm}^3 \)? Wait, no, wait the table: Box A: mass 36g, volume 6cm³; Box B: 81g, 9cm³; Box C: 48g, 8cm³. Wait the diagram: length 3, width 2, height 1? Wait no, maybe I misread. Wait the diagram's dimensions: 3cm, 2cm, 1cm? Wait volume is \( 3\times2\times1 = 6\,\text{cm}^3 \)? But Box A has volume 6cm³, mass 36g. Wait no, wait the diagram: maybe the box in the diagram is Box A? Wait no, let's check the table. Box A: mass 36g, volume 6cm³. The diagram's volume: 321=6 cm³. So the box in the diagram is Box A? Wait no, wait the question is "Which box is represented below?" (the diagram). So calculate the volume of the diagram's box: \( V = 3 \times 2 \times 1 = 6\,\text{cm}^3 \). Now check the table: Box A has volume 6 cm³, Box B 9, Box C 8. So the diagram's box has volume 6 cm³, so it's Box A? Wait no, wait the diagram's dimensions: 3cm, 2cm, 1cm? Wait 321=6. Box A's volume is 6. So the box in the diagram is Box A? Wait maybe I made a mistake. Wait the diagram: let's see, the arrows: 3cm, 2cm, 1cm. So volume is length × width × height = 3×2×1 = 6 cm³. Box A has volume 6 cm³, so the box in the diagram is Box A.

Step2: Confirm with mass? Wait no, the diagram doesn't show mass. So based on volume, the diagram's box has volume 6 cm³, which matches Box A's volume (6 cm³). So the box represented below is Box A.

Step1: Recall density formula

Density \(
ho = \frac{\text{Mass}}{\text{Volume}} \).

Step2: Identify mass and volume of Box C

From the table, Box C has mass \( m = 48\,\text{g} \) and volume \( V = 8\,\text{cm}^3 \).

Step3: Calculate density

\(
ho = \frac{48\,\text{g}}{8\,\text{cm}^3} = 6\,\text{g/cm}^3 \).

Step1: Calculate density of Box A

Density formula \(
ho = \frac{\text{Mass}}{\text{Volume}} \). For Box A: \( m = 36\,\text{g} \), \( V = 6\,\text{cm}^3 \). So \(
ho_A = \frac{36}{6} = 6\,\text{g/cm}^3 \)? Wait no, 36/6=6? Wait no, 36 divided by 6 is 6? Wait 6*6=36. Wait no, wait 36/6=6? Wait no, 36 divided by 6 is 6? Wait yes. Wait Box B: mass 81g, volume 9cm³. \(
ho_B = \frac{81}{9} = 9\,\text{g/cm}^3 \). Box C: we calculated \(
ho_C = 6\,\text{g/cm}^3 \) (from question 2). Wait wait, no, wait Box A: 36g / 6cm³ = 6 g/cm³. Box B: 81g / 9cm³ = 9 g/cm³. Box C: 48g / 8cm³ = 6 g/cm³. So comparing \(
ho_A = 6 \), \(
ho_B = 9 \), \(
ho_C = 6 \). So Box B has the greatest density.

Step2: Verify calculations

• Box A: \( \frac{36}{6} = 6\,\text{g/cm}^3 \)
• Box B: \( \frac{81}{9} = 9\,\text{g/cm}^3 \)
• Box C: \( \frac{48}{8} = 6\,\text{g/cm}^3 \)

Since \( 9 > 6 \), Box B has the greatest density.

Answer:

A (Box A)

Question 2