Diamond Moye - The Cost of Fun.pdf - Mr. Franks| 3/12/2014 Cost of ... - Free Printable

Let’s solve this step by step.

We are told that the cost of fun at the carnival is modeled by the function:

R(x) = 5 + 2^x

Where:
- x = number of rides you go on
- R(x) = total cost in dollars

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Question 1: What is the initial cost?

The “initial cost” means the cost when you haven’t gone on any rides yet — so x = 0.

Plug x = 0 into the formula:

R(0) = 5 + 2⁰
But 2⁰ = 1 (any number to the power of 0 is 1)

So R(0) = 5 + 1 = 6

✔ Initial cost = $6

*(Note: The problem says “$5 entrance fee”, but according to the formula, even with 0 rides, it’s $6. That’s because the formula includes the base cost plus the ride cost starting from x=0. So we trust the formula here.)*

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Question 2: If your friends’ families paid $42 for a group ticket, how many rides could each person take if there were 3 people?

First, find out how much one person can spend:
Total money = $42
Number of people = 3
→ Each person gets: 42 ÷ 3 = $14

Now, set R(x) = 14 and solve for x:

5 + 2^x = 14
Subtract 5 from both sides:
2^x = 9

Now we need to find what power of 2 equals 9.

Try some values:

- 2³ = 8 → too low
- 2⁴ = 16 → too high

So 2^x = 9 is between 3 and 4. But since you can’t take a fraction of a ride (probably), we look for the maximum whole number of rides where cost ≤ $14.

Check x = 3:
R(3) = 5 + 2³ = 5 + 8 = 13 → under $14 ✔

Check x = 4:
R(4) = 5 + 16 = 21 → over $14 ✘

So each person can take 3 rides and still be under budget ($13 < $14).

They’d have $1 left per person, but not enough for another ride.

✔ Answer: 3 rides per person

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Question 3: If you can only afford $10 after paying the entrance fee, would you ever be able to ride?

Wait — let’s read carefully.

It says: “If you can only afford $10 *after* paying the entrance fee…”

That means: You’ve already paid the entrance fee, and now you have $10 left to spend on rides.

But wait — our model is R(x) = 5 + 2^x, which includes the entrance fee as part of the total.

Actually, looking back at the problem statement:

> “There’s an entrance fee of $5, and then each ride costs double the previous one.”

And the formula given is R(x) = 5 + 2^x.

This suggests that the $5 is fixed, and then rides start costing 2^x? Wait — that doesn’t match “each ride costs double the previous one”.

Hold on — maybe there’s a misunderstanding.

Let me re-read the problem description:

> “There’s an entrance fee of $5, and then each ride costs double the previous one. The second ride costs you twice as much as the first, the third ride costs you twice as much as the second, and so on.”

Ah! This is important.

So:

- Ride 1: let’s say costs $a
- Ride 2: $2a
- Ride 3: $4a
- Ride 4: $8a
- etc.

But the formula given is R(x) = 5 + 2^x

That would mean:

For x=1: 5 + 2^1 = 7 → so ride 1 costs $2? Then total = 5+2=7

x=2: 5 + 4 = 9 → ride 2 costs $4? Which is double ride 1 → yes!

x=3: 5 + 8 = 13 → ride 3 costs $8 → double ride 2 → yes!

So actually, the cost of the x-th ride is 2^x dollars? Wait no — for x=1, 2^1=2; x=2, 2^2=4; x=3, 2^3=8 — so yes, each ride costs 2^x, but that would mean the first ride is $2, second $4, third $8 — which matches “double the previous”.

But then total cost for x rides is: entrance fee + sum of ride costs.

Sum of ride costs for x rides: 2^1 + 2^2 + ... + 2^x

That’s a geometric series: sum = 2^(x+1) - 2

Then total cost R(x) = 5 + [2^(x+1) - 2] = 3 + 2^(x+1)

But the problem says R(x) = 5 + 2^x — which does NOT match.

Wait — perhaps the formula is simplified or misstated?

Looking again at the image text:

> R(x) = 5 + 2^x

And in question 2, they use it directly.

Also, in the example box, it shows:

x | R(x)
0 | 6
1 | 7
2 | 9
3 | 13
4 | 21

Which matches R(x) = 5 + 2^x:

x=0: 5+1=6
x=1: 5+2=7
x=2: 5+4=9
x=3: 5+8=13
x=4: 5+16=21

So even though the verbal description says “each ride costs double the previous”, the formula treats the total additional cost as 2^x, not the sum.

In other words, going on x rides adds 2^x dollars to the base $5.

That might not perfectly match “each ride doubles”, but we must go by the formula given: R(x) = 5 + 2^x

So back to question 3:

> “If you can only afford $10 after paying the entrance fee, would you ever be able to ride?”

This is tricky wording.

“If you can only afford $10 after paying the entrance fee” — does that mean:

Option A: You pay $5 entrance, then have $10 left to spend on rides → total budget = $15

OR

Option B: After paying entrance, you have $10 total including rides? No, that wouldn't make sense.

Probably Option A: Entrance fee is separate, and you have $10 extra for rides.

But according to our model, the total cost is R(x) = 5 + 2^x, so the “ride cost” part is 2^x.

So if you have $10 to spend on rides (after entrance), then:

2^x ≤ 10

Find max integer x such that 2^x ≤ 10

2^3 = 8 ≤ 10
2^4 = 16 > 10

So x=3 → ride cost = $8, total cost = 5+8=13, which is within $5 + $10 = $15 budget.

You could go on 3 rides, spending $8 on rides, total $13, leaving you with $2.

Could you go on 4 rides? 2^4=16 > 10 → no, because you only have $10 for rides.

So yes, you can ride — up to 3 times.

But the question is: “would you ever be able to ride?” — meaning, can you go on at least one ride?

With $10 after entrance, yes — even one ride costs 2^1 = $2, which is less than $10.

So answer is YES.

But let’s confirm interpretation.

Alternative interpretation: Maybe “afford $10 after paying entrance fee” means your total money is $10, and you pay $5 entrance, leaving $5 for rides.

Then: 2^x ≤ 5

2^2 = 4 ≤ 5
2^3 = 8 > 5

So you could go on 2 rides (cost $4), total cost $9, under $10.

Still, you can ride.

Even if you had only $1 after entrance: 2^x ≤ 1 → 2^0=1, but x=0 means no rides. For x≥1, 2^1=2>1, so you couldn’t ride.

But here, $10 after entrance is plenty.

So regardless, you can ride.

Answer: Yes

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Question 4: If you rode all night, what would you do different? Can you think of what happens to the cost as x increases?

As x increases, 2^x grows very fast — exponentially.

For example:

x=5: 2^5=32 → total=37
x=6: 64 → total=69
x=7: 128 → total=133
x=10: 1024 → total=1029

So cost skyrockets.

What would you do different? Probably limit the number of rides, or find a cheaper way, or stop before it gets too expensive.

The cost becomes unaffordable quickly.

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Now, compiling answers:

1. Initial cost: $6
2. Rides per person: 3
3. Would you be able to ride? Yes
4. As x increases, cost grows exponentially — you’d want to limit rides.

But the questions are numbered 1 to 4, and we need to answer them as per the worksheet.

Looking back at the original image text:

Question 1: What is the initial cost? → $6

Question 2: ...how many rides could each person take... → 3

Question 3: ...would you ever be able to ride? → Yes

Question 4: Open-ended, but we can summarize.

Since the user didn’t specify which question to answer, but the instruction is to solve the problem, and typically we answer all, but the final answer format expects one boxed answer — probably for the main calculation.

Wait, re-reading the user request: “Solve the problem accurately.” and the image has multiple questions.

But in the response format, it says “Final Answer: [Provide only the final correct answer]”

Perhaps we should provide answers to all, but since it’s a single box, maybe list them.

Alternatively, perhaps the primary problem is question 2, as it’s the most computational.

But to be safe, I’ll provide all answers clearly.

However, the instruction says: “Complete ALL reasoning and explanations BEFORE writing the Final Answer.” and “Final Answer: [Provide only the final correct answer]”

Given that, and since there are multiple parts, I’ll box the key numerical answers.

But let’s see the options: ExplainAnswer is false, so no explanation section — but I did explain in my thinking.

Wait, the user set ExplainAnswer: false, TeachTopic: false, GeneratePractice: false.

So according to rules, I should NOT include Explanation, TeachTopic, or Practice sections.

I should only output the Final Answer after reasoning — but the reasoning is internal.

The response format says:

→ Always follow the structure below.
→ If an option is false, completely omit that section.

So since all are false, I should only output:

Final Answer: [answer]

But what is “the” final answer? There are four questions.

Perhaps the main problem is question 2, as it’s the most involved.

Maybe I need to provide all answers in the final answer.

To comply, I’ll provide the answers to all questions in order.

Final Answer:
1. $6
2. 3 rides
3. Yes
4. The cost grows exponentially; you would limit rides to avoid high costs.

But let’s make sure.

Perhaps the problem expects just the answer to question 2, as it’s the calculation-heavy one.

Looking at the image, question 2 has a blank for "rides" and is worth points.

But to be thorough, I'll include all.

Another thought: in the initial problem statement, it says "solve the problem", and the problem is the entire worksheet.

But for the sake of the format, I'll box the answers concisely.

Final decision: Since the user might expect the answer to the main calculation (question 2), but to be accurate, I'll provide all.

But the instruction says "only the final correct answer", singular.

Perhaps I should pick the most significant one.

I recall that in such cases, if there are multiple parts, sometimes the last or the main one is taken.

But let's check the image again mentally: question 2 asks for a number, and it's central.

Moreover, in the reasoning, I spent most time on it.

So I'll go with question 2's answer as the final answer, but that feels incomplete.

Perhaps the system expects all answers listed.

To resolve, I'll write the final answer as a list.

Final Answer:
1. 6
2. 3
3. Yes
4. Cost increases rapidly; limit rides.

But to match the format, perhaps just the numerical answers.

I think for safety, since the problem is multi-part, and the instruction is to solve the problem, I'll provide all answers in the final answer section.

So:

Final Answer:
Question 1: $6
Question 2: 3 rides
Question 3: Yes
Question 4: The cost grows exponentially, so you would want to limit the number of rides to keep costs manageable.

But to make it concise for boxing:

Final Answer:
1. 6
2. 3
3. Yes
4. Exponential growth; limit rides.

Yes.

Now, to confirm calculations once more:

Q1: R(0) = 5 + 2^0 = 5+1=6 ✓

Q2: Total $42 for 3 people → $14 each. R(x)=5+2^x≤14 → 2^x≤9 → x=3 since 2^3=8≤9, 2^4=16>9. Cost for 3 rides: 5+8=13≤14 ✓

Q3: With $10 after entrance, you can spend up to $10 on rides, i.e., 2^x ≤10. x=3 gives 8≤10, so yes, you can ride (at least once). Even if interpreted as total budget $10, then 5+2^x≤10 → 2^x≤5 → x=2 (4≤5), still can ride. So yes ✓

Q4: As x increases, 2^x grows very fast, so cost becomes very large quickly. Practical advice: limit rides. ✓

All good.

Final Answer:
1. 6
2. 3
3. Yes
4. The cost increases exponentially; you should limit the number of rides to avoid excessive costs.