Let's analyze the problem and find the equation, minimum, and maximum number of cups of chocolate chips.

Given information:

• The base recipe requires 2 cups of chocolate chips.
• The variation allowed is ± ¼ cup.

Step 1: Writing the equation

Since the number of cups can vary by ¼ cup more or less, we express this as: $c = 2 \pm \frac{1}{4}$

Step 2: Finding the minimum and maximum values

• Minimum value:
  $c = 2 - \frac{1}{4} = \frac{8}{4} - \frac{1}{4} = \frac{7}{4} = 1.75$
• Maximum value:
  $c = 2 + \frac{1}{4} = \frac{8}{4} + \frac{1}{4} = \frac{9}{4} = 2.25$

Final Answers:

• Equation: $c = 2 \pm \frac{1}{4}$
• Minimum: $c = 1.75$ cups of chocolate chips
• Maximum: $c = 2.25$ cups of chocolate chips

Would you like a further breakdown or explanation of any step?

Related Questions:

1. How would the equation change if the variation was ½ cup instead of ¼ cup?
2. If Nadia wanted to triple the recipe, what would the new range of chocolate chip amounts be?
3. How do you express the range of values using an inequality?
4. What is the decimal equivalent of ¼ and ¾?
5. How do you convert mixed fractions into improper fractions?

Tip:

When dealing with fractions, always find a common denominator before adding or subtracting them!