Severin works in a factory that produces steel bolts. One type of steel bolt should have a mass of  g; however, there is some variation, so each bolt may be heavier or lighter by as much as . Severin puts together a box of these bolts and notices that the total mass of the bolts in the box is exactly  kg.

Determine the minimum and maximum number of bolts that could be in the box.

Let's break down the problem step by step.

### Variables:
- The ideal mass of one bolt is \( m \) grams.
- Each bolt may vary by as much as \( v \) grams, so the mass of each bolt could range from \( m - v \) to \( m + v \).
- The total mass of the bolts in the box is \( T = 3 \) kg, which is equal to \( 3000 \) grams (since \( 1 \) kg = \( 1000 \) grams).

### Objective:
We are to find the **minimum** and **maximum** number of bolts in the box. To do this, we need to compute two scenarios:
1. **Maximum number of bolts:** This happens when each bolt is as light as possible, i.e., each bolt has a mass of \( m - v \).
2. **Minimum number of bolts:** This happens when each bolt is as heavy as possible, i.e., each bolt has a mass of \( m + v \).

### Maximum number of bolts:
The mass of the lightest possible bolt is \( m - v \). Let \( N_{\text{max}} \) be the maximum number of bolts. Then the total mass of the bolts is:
\[
N_{\text{max}} \times (m - v) = 3000
\]
Solving for \( N_{\text{max}} \):
\[
N_{\text{max}} = \frac{3000}{m - v}
\]

### Minimum number of bolts:
The mass of the heaviest possible bolt is \( m + v \). Let \( N_{\text{min}} \) be the minimum number of bolts. Then the total mass of the bolts is:
\[
N_{\text{min}} \times (m + v) = 3000
\]
Solving for \( N_{\text{min}} \):
\[
N_{\text{min}} = \frac{3000}{m + v}
\]

### Conclusion:
We can compute the exact values for \( N_{\text{min}} \) and \( N_{\text{max}} \) once the values of \( m \) and \( v \) are provided.

Do you have the values for \( m \) and \( v \) to continue the solution?

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**Related questions:**
1. What is the standard deviation of the mass of these bolts?
2. If the mass variation is larger, how will that affect the number of bolts in the box?
3. Can the mass distribution of bolts affect production accuracy?
4. How does Severin ensure the bolts in the box meet mass quality standards?
5. What is the probability that a random bolt weighs exactly \( m \) grams?

**Tip:** Always convert units early in a problem to avoid mistakes later on.

Severin works in a factory that produces steel bolts. One type of steel bolt should have a mass of g; however, there is some variation, so each bolt may be heavier or lighter by as much as . Severin puts together a box of these bolts and notices that the total mass of the bolts in the box is exactly 3 kg. Determine the minimum and maximum number of bolts that could be in the box.

Learn how to calculate the minimum and maximum number of steel bolts that could fit in a box, given mass variations. This problem involves algebra, unit conversion, and problem-solving skills. Ideal for students in Grades 9-11.

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