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Let's break down the problem to solve it:

Henry’s father wants to buy him a six-month movie pass for $30. With the pass, the price of popcorn is reduced to $1.00. Without the pass, popcorn costs $3.50. We need to find how many times Henry must attend for the pass to be worth the $30.

1. **Cost without the pass**: $3.50 per popcorn.
2. **Cost with the pass**: $1.00 per popcorn.
3. **Savings per popcorn**: $3.50 - $1.00 = $2.50 saved for each visit.

Now, if Henry saves $2.50 per visit, how many visits does it take for those savings to equal or exceed the $30 cost of the pass?

To find the number of visits $ x $, we solve the equation:
\[
2.50 \times x = 30
\]
Solving for $ x $:
\[
x = \frac{30}{2.50} = 12
\]

So, Henry needs to attend **12 times** for the pass to be beneficial.

The correct answer is **D. 12 times**.

Would you like any more details or have questions on this? Here are some follow-up questions:

1. What if the pass cost $40 instead of $30?
2. How would the problem change if the normal popcorn price was $4 instead of $3.50?
3. Can we generalize this approach to other scenarios with different savings?
4. What if the discount was a percentage instead of a fixed amount?
5. How can we model this scenario with a graph for better visualization?

**Tip**: Whenever dealing with savings problems like this, setting up a simple equation relating savings per unit to the cost will help solve them efficiently.

Henry’s father is thinking of buying his son a six-month movie pass for $30. With the pass, popcorn costs $1.00. If popcorn is normally $3.50 each, how many times must Henry attend in order for it to benefit his father to buy the pass?

Learn how to calculate the number of times Henry must attend the movies to make the six-month movie pass worth it. Using algebraic concepts, we'll explore how savings per popcorn visit lead to a simple equation. Suitable for Grades 5-7.

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