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Epoxy Resin worksheet

Grades 6–8 · CCSS 7.G.B.6, 8.G.C.9, 6.RP.A.3

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Units: Metric (cm, mm, mL)
- Project: Rectangular coating or pour
- Length: 85 cm
- Width (for a river table, the average gap width): 45 cm
- Thickness or pour depth: 3 mm
- Mix ratio: resin (part A): 1
- Mix ratio: hardener (part B): 1
- Extra for drips, mixing cups and edges: 9%
Find: Mixed epoxy needed
Answer
2.
Problem 2.
- Units: US (inches, fl oz)
- Project: Rectangular coating or pour
- Length: 28 in
- Width (for a river table, the average gap width): 18 in
- Thickness or pour depth: 0.11 in
- Mix ratio: resin (part A): 1
- Mix ratio: hardener (part B): 1
- Extra for drips, mixing cups and edges: 11%
Find: Mixed epoxy needed
Answer
3.
Problem 3.
- Units: Metric (cm, mm, mL)
- Project: Round coating or pour
- Diameter: 36 cm
- Thickness or pour depth: 4 mm
- Mix ratio: resin (part A): 1
- Mix ratio: hardener (part B): 2
- Extra for drips, mixing cups and edges: 11%
Find: Mixed epoxy needed
Answer
4.
Problem 4.
- Units: Metric (cm, mm, mL)
- Project: Rectangular coating or pour
- Length: 48 cm
- Width (for a river table, the average gap width): 43 cm
- Thickness or pour depth: 4 mm
- Mix ratio: resin (part A): 1
- Mix ratio: hardener (part B): 1
- Extra for drips, mixing cups and edges: 7%
Find: Mixed epoxy needed
Answer
5.
Problem 5.
- Units: Metric (cm, mm, mL)
- Project: Rectangular coating or pour
- Length: 55 cm
- Width (for a river table, the average gap width): 50 cm
- Thickness or pour depth: 2 mm
- Mix ratio: resin (part A): 2
- Mix ratio: hardener (part B): 2
- Extra for drips, mixing cups and edges: 7%
Find: Mixed epoxy needed
Answer
6.
Problem 6.
- Units: Metric (cm, mm, mL)
- Project: Rectangular coating or pour
- Length: 75 cm
- Width (for a river table, the average gap width): 40 cm
- Thickness or pour depth: 2 mm
- Mix ratio: resin (part A): 1
- Mix ratio: hardener (part B): 1
- Extra for drips, mixing cups and edges: 9%
Find: Mixed epoxy needed
Answer
7.
Problem 7.
- Units: Metric (cm, mm, mL)
- Project: Round coating or pour
- Diameter: 28 cm
- Thickness or pour depth: 3 mm
- Mix ratio: resin (part A): 1
- Mix ratio: hardener (part B): 2
- Extra for drips, mixing cups and edges: 8%
Find: Mixed epoxy needed
Answer
8.
Problem 8.
- Units: Metric (cm, mm, mL)
- Project: Round coating or pour
- Diameter: 24 cm
- Thickness or pour depth: 3 mm
- Mix ratio: resin (part A): 1
- Mix ratio: hardener (part B): 1
- Extra for drips, mixing cups and edges: 11%
Find: Mixed epoxy needed
Answer
9.
Problem 9.
- Units: US (inches, fl oz)
- Project: River table gap
- Length: 27 in
- Width (for a river table, the average gap width): 16 in
- Thickness or pour depth: 0.11 in
- Mix ratio: resin (part A): 1
- Mix ratio: hardener (part B): 1
- Extra for drips, mixing cups and edges: 7%
Find: Mixed epoxy needed
Answer
10.
Problem 10.
- Units: US (inches, fl oz)
- Project: River table gap
- Length: 30 in
- Width (for a river table, the average gap width): 18 in
- Thickness or pour depth: 0.12 in
- Mix ratio: resin (part A): 1
- Mix ratio: hardener (part B): 1
- Extra for drips, mixing cups and edges: 12%
Find: Mixed epoxy needed
Answer

Answer key: Epoxy Resin worksheet

1. Mixed epoxy needed: 1,251 mL
1. Find the surface area: length times width
  $85 \times 45$ $= 3,825 cm²$
2. Multiply the area by the depth to get the volume (mm ÷ 10 gives cm, and 1 cm³ = 1 mL)
  $3,825 \times 0.3$ $= 1,147.5 mL$
3. Add 9% extra for drips and the cup
  $1,147.5 \times 1.09$ $= 1,250.775 mL$
4. Resin (part A) is 1 of every 2 parts of the mix
  $\frac{1,250.775 \times 1}{1 + 1}$ $= 625.3875 mL$
5. Hardener (part B) = the rest of the mix
  $\frac{1,250.775 \times 1}{1 + 1}$ $= 625.3875 mL$
2. Mixed epoxy needed: 34.1 fl oz
1. Find the surface area: length times width
  $28 \times 18$ $= 504 in²$
2. Multiply the area by the depth to get the volume to fill
  $504 \times 0.11$ $= 55.44 in³$
3. Change to fluid ounces (1 fl oz = 1.8046875 in³) and add 11% extra for drips and the cup
  $\frac{55.44}{1.8046875} \times 1.11$ $= 34.0992 fl oz$
4. Resin (part A) is 1 of every 2 parts of the mix
  $\frac{34.0992 \times 1}{1 + 1}$ $= 17.0496 fl oz$
5. Hardener (part B) = the rest of the mix
  $\frac{34.0992 \times 1}{1 + 1}$ $= 17.0496 fl oz$
3. Mixed epoxy needed: 452 mL
1. Find the area of the circle: π (about 3.14) times half the diameter, squared
  $π \times 18^{2}$ $= 1,017.876 cm²$
2. Multiply the area by the depth to get the volume (mm ÷ 10 gives cm, and 1 cm³ = 1 mL)
  $1,017.876 \times 0.4$ $= 407.1504 mL$
3. Add 11% extra for drips and the cup
  $407.1504 \times 1.11$ $= 451.937 mL$
4. Resin (part A) is 1 of every 3 parts of the mix
  $\frac{451.937 \times 1}{1 + 2}$ $= 150.6457 mL$
5. Hardener (part B) = the rest of the mix
  $\frac{451.937 \times 2}{1 + 2}$ $= 301.2913 mL$
4. Mixed epoxy needed: 883 mL
1. Find the surface area: length times width
  $48 \times 43$ $= 2,064 cm²$
2. Multiply the area by the depth to get the volume (mm ÷ 10 gives cm, and 1 cm³ = 1 mL)
  $2,064 \times 0.4$ $= 825.6 mL$
3. Add 7% extra for drips and the cup
  $825.6 \times 1.07$ $= 883.392 mL$
4. Resin (part A) is 1 of every 2 parts of the mix
  $\frac{883.392 \times 1}{1 + 1}$ $= 441.696 mL$
5. Hardener (part B) = the rest of the mix
  $\frac{883.392 \times 1}{1 + 1}$ $= 441.696 mL$
5. Mixed epoxy needed: 589 mL
1. Find the surface area: length times width
  $55 \times 50$ $= 2,750 cm²$
2. Multiply the area by the depth to get the volume (mm ÷ 10 gives cm, and 1 cm³ = 1 mL)
  $2,750 \times 0.2$ $= 550 mL$
3. Add 7% extra for drips and the cup
  $550 \times 1.07$ $= 588.5 mL$
4. Resin (part A) is 2 of every 4 parts of the mix
  $\frac{588.5 \times 2}{2 + 2}$ $= 294.25 mL$
5. Hardener (part B) = the rest of the mix
  $\frac{588.5 \times 2}{2 + 2}$ $= 294.25 mL$
6. Mixed epoxy needed: 654 mL
1. Find the surface area: length times width
  $75 \times 40$ $= 3,000 cm²$
2. Multiply the area by the depth to get the volume (mm ÷ 10 gives cm, and 1 cm³ = 1 mL)
  $3,000 \times 0.2$ $= 600 mL$
3. Add 9% extra for drips and the cup
  $600 \times 1.09$ $= 654 mL$
4. Resin (part A) is 1 of every 2 parts of the mix
  $\frac{654 \times 1}{1 + 1}$ $= 327 mL$
5. Hardener (part B) = the rest of the mix
  $\frac{654 \times 1}{1 + 1}$ $= 327 mL$
7. Mixed epoxy needed: 200 mL
1. Find the area of the circle: π (about 3.14) times half the diameter, squared
  $π \times 14^{2}$ $= 615.7522 cm²$
2. Multiply the area by the depth to get the volume (mm ÷ 10 gives cm, and 1 cm³ = 1 mL)
  $615.7522 \times 0.3$ $= 184.7256 mL$
3. Add 8% extra for drips and the cup
  $184.7256 \times 1.08$ $= 199.5037 mL$
4. Resin (part A) is 1 of every 3 parts of the mix
  $\frac{199.5037 \times 1}{1 + 2}$ $= 66.5012 mL$
5. Hardener (part B) = the rest of the mix
  $\frac{199.5037 \times 2}{1 + 2}$ $= 133.0025 mL$
8. Mixed epoxy needed: 151 mL
1. Find the area of the circle: π (about 3.14) times half the diameter, squared
  $π \times 12^{2}$ $= 452.3893 cm²$
2. Multiply the area by the depth to get the volume (mm ÷ 10 gives cm, and 1 cm³ = 1 mL)
  $452.3893 \times 0.3$ $= 135.7168 mL$
3. Add 11% extra for drips and the cup
  $135.7168 \times 1.11$ $= 150.6457 mL$
4. Resin (part A) is 1 of every 2 parts of the mix
  $\frac{150.6457 \times 1}{1 + 1}$ $= 75.3228 mL$
5. Hardener (part B) = the rest of the mix
  $\frac{150.6457 \times 1}{1 + 1}$ $= 75.3228 mL$
9. Mixed epoxy needed: 28.17 fl oz
1. Find the gap's area: its length times its average width
  $27 \times 16$ $= 432 in²$
2. Multiply the area by the depth to get the volume to fill
  $432 \times 0.11$ $= 47.52 in³$
3. Change to fluid ounces (1 fl oz = 1.8046875 in³) and add 7% extra for drips and the cup
  $\frac{47.52}{1.8046875} \times 1.07$ $= 28.1746 fl oz$
4. Resin (part A) is 1 of every 2 parts of the mix
  $\frac{28.1746 \times 1}{1 + 1}$ $= 14.0873 fl oz$
5. Hardener (part B) = the rest of the mix
  $\frac{28.1746 \times 1}{1 + 1}$ $= 14.0873 fl oz$
10. Mixed epoxy needed: 40.22 fl oz
1. Find the gap's area: its length times its average width
  $30 \times 18$ $= 540 in²$
2. Multiply the area by the depth to get the volume to fill
  $540 \times 0.12$ $= 64.8 in³$
3. Change to fluid ounces (1 fl oz = 1.8046875 in³) and add 12% extra for drips and the cup
  $\frac{64.8}{1.8046875} \times 1.12$ $= 40.2153 fl oz$
4. Resin (part A) is 1 of every 2 parts of the mix
  $\frac{40.2153 \times 1}{1 + 1}$ $= 20.1076 fl oz$
5. Hardener (part B) = the rest of the mix
  $\frac{40.2153 \times 1}{1 + 1}$ $= 20.1076 fl oz$