To solve the problem of determining the mass shown on each triple beam balance, we need to read the values from the three beams (the 100g beam, the 10g beam, and the 1g beam) and add them together. Let's go through each balance step by step.
Balance 1:
1.
100g Beam: The pointer is at the 200g mark.
- Value: \( 200 \, \text{g} \)
2.
10g Beam: The pointer is at the 50g mark.
- Value: \( 50 \, \text{g} \)
3.
1g Beam (Rider): The rider is at the 4.6g mark.
- Value: \( 4.6 \, \text{g} \)
Adding these values:
\[
200 \, \text{g} + 50 \, \text{g} + 4.6 \, \text{g} = 254.6 \, \text{g}
\]
Answer for Balance 1: \( \boxed{254.6 \, \text{g}} \)
---
Balance 2:
1.
100g Beam: The pointer is at the 300g mark.
- Value: \( 300 \, \text{g} \)
2.
10g Beam: The pointer is at the 70g mark.
- Value: \( 70 \, \text{g} \)
3.
1g Beam (Rider): The rider is at the 8.4g mark.
- Value: \( 8.4 \, \text{g} \)
Adding these values:
\[
300 \, \text{g} + 70 \, \text{g} + 8.4 \, \text{g} = 378.4 \, \text{g}
\]
Answer for Balance 2: \( \boxed{378.4 \, \text{g}} \)
---
Balance 3:
1.
100g Beam: The pointer is at the 100g mark.
- Value: \( 100 \, \text{g} \)
2.
10g Beam: The pointer is at the 30g mark.
- Value: \( 30 \, \text{g} \)
3.
1g Beam (Rider): The rider is at the 5.2g mark.
- Value: \( 5.2 \, \text{g} \)
Adding these values:
\[
100 \, \text{g} + 30 \, \text{g} + 5.2 \, \text{g} = 135.2 \, \text{g}
\]
Answer for Balance 3: \( \boxed{135.2 \, \text{g}} \)
---
Final Answers:
1. \( \boxed{254.6 \, \text{g}} \)
2. \( \boxed{378.4 \, \text{g}} \)
3. \( \boxed{135.2 \, \text{g}} \)
These are the masses shown on each triple beam balance. Each value includes the contributions from the 100g beam, the 10g beam, and the 1g beam (rider).
Parent Tip: Review the logic above to help your child master the concept of reading a triple beam balance worksheet.
