<Question>
There are 360 people in my school.  15 take calculus, physics, and chemistry, and 15 don't take any of them.  180 take calculus.  Twice as many students take chemistry as take physics.  75 take both calculus and chemistry, and 75 take both physics and chemistry.  Only 30 take both physics and calculus.  How many students take physics?
</Question>

<think>
<Plan_1>
Okay, let me try to figure out this problem step by step. It's a bit complicated with all the overlapping subjects, but maybe if I break it down, it'll make more sense.

First, let's list out all the information given:

- Total number of people in the school: 360.
- 15 take all three subjects: calculus, physics, and chemistry.
- 15 don't take any of the three subjects.
- 180 take calculus.
- Twice as many students take chemistry as take physics.
- 75 take both calculus and chemistry.
- 75 take both physics and chemistry.
- 30 take both physics and calculus.
- We need to find how many students take physics.

Hmm, okay. So, we have some numbers and relationships between the numbers of students taking different combinations of subjects. This sounds like a problem that can be solved using set theory and Venn diagrams. Let me recall the principle of inclusion-exclusion for three sets.

The formula for the union of three sets is:

Total = C + P + Ch - (C∩P) - (C∩Ch) - (P∩Ch) + (C∩P∩Ch)

Where:
- C is the number of students taking calculus,
- P is the number taking physics,
- Ch is the number taking chemistry,
- C∩P is those taking both calculus and physics,
- C∩Ch is those taking both calculus and chemistry,
- P∩Ch is those taking both physics and chemistry,
- C∩P∩Ch is those taking all three.

But we also know that 15 don't take any subjects, so the number taking at least one subject is 360 - 15 = 345. So the total in the union is 345.

Let me note down the known values:

<prompt_1.1> List the known values from the problem statement. </prompt_1.1>
-----

But let me check if this is consistent with the other given numbers. For example, let's check if the numbers for the intersections make sense.

<prompt_1.2> Verify the consistency of the solution with the given numbers. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
- Total (taking at least one subject): 345
- C = 180
- Ch = 2P (since twice as many take chemistry as physics)
- C∩Ch = 75
- P∩Ch = 75
- C∩P = 30
- C∩P∩Ch = 15 (Wait, no. Wait, the problem says 15 take calculus, physics, and chemistry. So C∩P∩Ch = 15. But hold on, usually in these problems, the numbers given for intersections might already include those taking all three. Wait, but in standard set theory, when they say 75 take both calculus and chemistry, that would be C∩Ch, which includes those who might be taking all three. Similarly, 75 take both physics and chemistry, which would also include the 15 who are taking all three. So, actually, the intersections given (75, 75, 30) are the total numbers for those two-subject overlaps, including those who might be taking a third subject. Therefore, the numbers 75, 75, 30 are the counts for each pair, regardless of whether they are also in the third set. But the 15 who take all three would be part of each of those pairwise intersections. So, for example, the 75 who take both calculus and chemistry include the 15 who take all three. Similarly, the 75 who take both physics and chemistry include the 15. So, to apply the inclusion-exclusion formula correctly, we need to use the given pairwise intersections as they are, which already include the triple overlap.
</execution_1.1>

<execution_1.2>
We know that 75 take both calculus and chemistry. According to our values, C∩Ch = 75. From our calculation, yes, that's given. Similarly, P∩Ch = 75, and C∩P = 30. All these are given, so they fit into the formula.

Another check: Let's compute the number of students taking only physics. If P = 120, then:

Only physics would be P - (C∩P) - (P∩Ch) + (C∩P∩Ch) ?

Wait, no. Wait, the formula for only physics is P - (number taking both physics and calculus) - (number taking both physics and chemistry) + (number taking all three). Wait, no, actually, when you subtract the intersections, you've subtracted the triple overlap twice, so you need to add it back once.

Wait, maybe it's clearer to use the inclusion-exclusion step by step.

Alternatively, let's compute the number of students in each region of the Venn diagram.

But maybe that's overcomplicating. Alternatively, let's check if the total numbers add up.

Wait, let's compute the total number of students taking at least one subject. According to inclusion-exclusion:

C + P + Ch - C∩P - C∩Ch - P∩Ch + C∩P∩Ch = 180 + 120 + 240 - 30 - 75 - 75 + 15.

Let's compute that:

180 + 120 = 300; 300 + 240 = 540.

Subtract 30: 540 - 30 = 510; 510 - 75 = 435; 435 - 75 = 360.

Then add 15: 360 + 15 = 375.

Wait, but we were told that 345 students take at least one subject. But according to this calculation, it's 375. That's a problem. There's a discrepancy here. So something's wrong.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
So, plugging into the inclusion-exclusion formula:

345 = 180 + P + 2P - 30 - 75 - 75 + 15

Wait, let me write that out again:

345 = C + P + Ch - (C∩P) - (C∩Ch) - (P∩Ch) + (C∩P∩Ch)

Substituting the known values:

<prompt_2.1> Solve the equation to find the number of students taking physics. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
C = 180

P = P (unknown)

Ch = 2P

(C∩P) = 30

(C∩Ch) = 75

(P∩Ch) = 75

(C∩P∩Ch) = 15

So substituting:

345 = 180 + P + 2P - 30 - 75 - 75 + 15

Combine like terms:

On the right side:

180 + P + 2P = 180 + 3P

Then subtract 30 + 75 + 75 = 180

Then add 15.

So:

345 = 180 + 3P - 180 + 15

Simplify:

180 - 180 cancels out, so:

345 = 3P + 15

Subtract 15 from both sides:

330 = 3P

Divide both sides by 3:

P = 110

Wait, so according to this, the number of students taking physics is 110? But let me check if this makes sense because when we did the Venn diagram earlier, the number came out to 120.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Hmm, maybe I made a mistake in the interpretation of the problem. Let me double-check the given data.

<prompt_3.1> Re-evaluate the interpretation of the problem and verify the setup of the equation. </prompt_3.1>
-----

Based on execution_1.2:
So clearly, there's an error in my calculation somewhere.

Let me go back.

<prompt_3.2> Identify and correct the error in the inclusion-exclusion calculation. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Wait, the problem says "Twice as many students take chemistry as take physics". So Ch = 2P. That seems correct. "75 take both calculus and chemistry", that's C∩Ch = 75. "75 take both physics and chemistry", so P∩Ch = 75. "30 take both physics and calculus", so C∩P = 30. "15 take all three subjects", so C∩P∩Ch = 15. All right, that seems right.

Wait, but in the inclusion-exclusion formula, the pairwise intersections already include the triple overlap. So when we subtract the pairwise intersections, we have to subtract them once. But when we add back the triple overlap because it was subtracted three times. Wait, let me recall the formula:

|C ∪ P ∪ Ch| = |C| + |P| + |Ch| - |C∩P| - |C∩Ch| - |P∩Ch| + |C∩P∩Ch|

Yes, that's correct. So the formula already accounts for the overlaps correctly. So substituting the values as I did before should be correct.
</execution_3.1>

<execution_3.2>
Wait, I think I messed up the inclusion-exclusion formula. Let me check the formula again.

The correct formula for three sets is:

|C ∪ P ∪ Ch| = |C| + |P| + |Ch| - |C∩P| - |C∩Ch| - |P∩Ch| + |C∩P∩Ch|

Yes, that's correct. So substituting the numbers:

345 = 180 + 120 + 240 - 30 - 75 - 75 + 15

Compute step by step:

180 + 120 = 300

300 + 240 = 540

540 - 30 = 510

510 - 75 = 435

435 - 75 = 360

360 + 15 = 375

But 375 ≠ 345. There's a discrepancy here. So where did I go wrong?

Ah! Wait a minute. The problem states that 15 take none of the subjects. Therefore, the total taking at least one subject is 360 - 15 = 345. But according to the inclusion-exclusion calculation, it's 375. That's a problem. So there's an inconsistency here.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
Let me check if my interpretation of the pairwise intersections is wrong. Maybe the numbers given for the pairwise intersections (75 for C∩Ch, 75 for P∩Ch, 30 for C∩P) are not including the triple overlap? But in standard set theory, when they give the pairwise intersections, they include those in all three sets. For example, if 15 take all three, then the number taking both calculus and chemistry would be the 15 plus those taking only calculus and chemistry. So if the problem states that 75 take both calculus and chemistry, that would already include the 15. Therefore, the given numbers for the pairwise intersections are already accounting for the triple overlap. Therefore, the inclusion-exclusion formula should be applied directly with those numbers.

But then why is there a discrepancy?

<prompt_4.1> Reassess the inclusion-exclusion calculation and consider alternative approaches. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Let me check my arithmetic again.

Ch = 2P. If P = 120, then Ch = 240.

So substituting into the inclusion-exclusion formula:

|C ∪ P ∪ Ch| = 180 + 120 + 240 - 30 - 75 - 75 + 15

Calculating step by step:

180 + 120 = 300

300 + 240 = 540

540 - 30 = 510

510 - 75 = 435

435 - 75 = 360

360 + 15 = 375

Yes, that's 375. But the number of students taking at least one subject is 345. So 375 vs. 345. That's a difference of 30. What's going on here?
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Maybe the problem is that the triple overlap is 15, but when we calculated P, we didn't consider that properly? Wait, no. The formula already includes the triple overlap once. Let me check the formula again.

<prompt_5.1> Re-evaluate the calculations and consider if there is an error in the problem setup. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
The formula is:

Total = C + P + Ch - (C∩P) - (C∩Ch) - (P∩Ch) + (C∩P∩Ch)

So, if the triple overlap is 15, that's added back after subtracting the pairwise overlaps. So if the numbers given for the pairwise overlaps already include the triple overlap, then the formula is correct. So in that case, why the discrepancy?

Alternatively, maybe the problem is that the numbers given for the pairwise overlaps do not include the triple overlap? But that contradicts standard set theory. If the pairwise overlaps are given as only those taking exactly two subjects, then the formula would be different. But the problem states "75 take both calculus and chemistry", which would include those taking all three. Similarly for the others. So the given numbers are the total for each pair, including those in all three.

Alternatively, perhaps there's an error in the problem's numbers? But that's unlikely. Maybe I made a miscalculation.

Wait, let's try solving the equation again.

We had:

345 = 180 + P + 2P - 30 - 75 - 75 + 15

Simplify:

345 = 180 + 3P - 180 + 15

Because 30 + 75 + 75 = 180, so subtracting 180, then adding 15.

So 180 - 180 = 0, so 3P + 15 = 345

Thus, 3P = 330 => P = 110.

But when we plug back in, we get 375. So that's a problem. Therefore, either the problem is inconsistent, or there's a miscalculation on my part.

Wait, let's try solving for P again.

Given:

Total in at least one subject = 345

Inclusion-exclusion formula:

345 = 180 + P + 2P - 30 - 75 - 75 + 15

Combine like terms:

180 + P + 2P = 180 + 3P

Then subtract 30 + 75 + 75 = 180

Then add 15.

So:

345 = 180 + 3P - 180 + 15

Simplify:

345 = 3P + 15

Therefore, 3P = 345 - 15 = 330 => P = 110.

So according to the equation, P = 110. But when plugging back in, we get 375 instead of 345. Therefore, there's a contradiction here.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
This suggests that there's an error in the problem's data, or perhaps in the interpretation of the problem. Wait, let me check the problem statement again.

<prompt_6.1> Perform a final check of the problem's data and calculations. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
"Twice as many students take chemistry as take physics". So Ch = 2P.

"75 take both calculus and chemistry", so C∩Ch = 75.

"75 take both physics and chemistry", so P∩Ch = 75.

"30 take both physics and calculus", so C∩P = 30.

"15 take all three subjects", so C∩P∩Ch = 15.

Wait, but in that case, when we use the inclusion-exclusion formula, we have to make sure that the pairwise intersections include the triple overlap. But in the formula, we subtract the pairwise intersections, which already include the triple overlap, then add back the triple overlap. So the formula is correct as applied.

But then why the discrepancy? Let's see.

If P = 110, then Ch = 220.

Then:

|C ∪ P ∪ Ch| = 180 + 110 + 220 - 30 - 75 - 75 + 15

Calculate:

180 + 110 = 290; 290 + 220 = 510

510 - 30 = 480; 480 - 75 = 405; 405 - 75 = 330

330 + 15 = 345.

Wait, wait a minute! Wait, 180 + 110 + 220 is 510. Then subtract 30 gives 480, subtract 75 gives 405, subtract another 75 gives 330, then add 15 gives 345. So that's correct. Then 345 is the number taking at least one subject, which matches the given 345. So my previous calculation was wrong. Let me check again.

Yes, 180 + 110 + 220 = 510. Then subtract 30: 480. Subtract 75: 405. Subtract 75: 330. Add 15: 345. So that's correct. Then 345 is the number taking at least one subject, which is correct because 360 - 15 = 345. So that works out. So my previous mistake was miscalculating the sum. So P = 110 is correct.

But then why did I get a different result earlier? Because I thought 180 + 120 + 240, but if P is 110, then Ch is 220. So the correct calculation is:

180 (Calculus) + 110 (Physics) + 220 (Chemistry) - 30 (Calculus and Physics) - 75 (Calculus and Chemistry) - 75 (Physics and Chemistry) + 15 (All three) = 180 + 110 + 220 - 30 - 75 - 75 + 15

Which is 510 - 180 + 15 = 345. Yes, correct.

So earlier, I mistakenly used P = 120, which was my initial wrong conclusion. But solving the equation gives P = 110, which does satisfy the total number of students taking at least one subject as 345. Therefore, the correct answer should be 110.

But wait, this contradicts the Venn diagram approach where I initially got 120. Let me check that again.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
If P = 110, then Ch = 220. Let's see how the numbers break down in the Venn diagram.

<prompt_7.1> Re-evaluate the Venn diagram approach and confirm the correct number of students taking physics. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
The regions would be:

Only Calculus: C - (C∩P) - (C∩Ch) + (C∩P∩Ch) = 180 - 30 - 75 + 15 = 180 - 105 + 15 = 90

Only Physics: P - (C∩P) - (P∩Ch) + (C∩P∩Ch) = 110 - 30 - 75 + 15 = 110 - 105 + 15 = 20

Only Chemistry: Ch - (C∩Ch) - (P∩Ch) + (C∩P∩Ch) = 220 - 75 - 75 + 15 = 220 - 150 + 15 = 85

Calculus and Physics only: 30 - 15 = 15 (since 15 are in all three)

Calculus and Chemistry only: 75 - 15 = 60

Physics and Chemistry only: 75 - 15 = 60

All three: 15

Now, let's sum all these up:

90 (Calculus only) + 20 (Physics only) + 85 (Chemistry only) + 15 (Calculus and Physics) + 60 (Calculus and Chemistry) + 60 (Physics and Chemistry) + 15 (All three) =

90 + 20 = 110

110 + 85 = 195

195 + 15 = 210

210 + 60 = 270

270 + 60 = 330

330 + 15 = 345.

Yes, that adds up to 345, which matches the required number. So that works out. Therefore, the number of students taking Physics is 110.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
But wait, earlier when I thought P was 120, that led to a discrepancy, but solving the equation correctly gives P = 110, which works. So that must be the answer. But let me just check once again to make sure I didn't make any mistakes.

<prompt_8.1> Verify the solution by checking all calculations and logical consistency. </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
Let me verify the arithmetic:

If P = 110, then Ch = 220.

Inclusion-exclusion:

180 + 110 + 220 = 510

Subtract (30 + 75 + 75) = 180

510 - 180 = 330

Add 15: 330 + 15 = 345.

Yes, correct.

Therefore, despite my initial confusion and miscalculations, the correct answer is 110. So the number of students taking Physics is 110.
</execution_8.1>
</Execution_8>

<Final_answer>
Therefore, the number of students who take physics is $\boxed{110}$.
</Final_answer>
</think>