Explanation:
Okay, let me try to figure out this problem. So, there are 360 students in total. Some take calculus, physics, and chemistry. Some don't take any of them. We need to find out how many students take physics. Let me jot down the given information first:

Total students = 360.

Number of students taking all three subjects (calculus, physics, chemistry) = 15.

Number of students taking none of the subjects = 15.

Number of students taking calculus = 180.

Twice as many students take chemistry as take physics. So, if we let the number of students taking physics be P, then the number taking chemistry would be 2P.

Number of students taking both calculus and chemistry = 75.

Number of students taking both physics and chemistry = 75.

Only 30 take both physics and calculus. Wait, "only" here might mean that those 30 are the ones taking both physics and calculus but not chemistry? Or does it mean that the intersection of physics and calculus is 30? Hmm, the wording says "Only 30 take both physics and calculus." The word "only" might be important here. If it's only 30 taking both physics and calculus, then that would mean that there are no students taking all three subjects in that intersection? Wait, but we already know that 15 students take all three. So if we have students taking both physics and calculus, some of them might be taking all three. Therefore, "only 30 take both physics and calculus" would mean that the number of students taking both physics and calculus but not chemistry is 30? Or maybe the total number of students taking both physics and calculus is 30? This is a bit confusing. Let me think.

In set theory, usually, when they say "the number of students taking both A and B," that includes those who might also be taking other subjects. However, if it's specified as "only" taking both A and B, then it would exclude those taking all three. However, the problem says "Only 30 take both physics and calculus." The wording here is crucial. If we have 15 students taking all three, then the total number of students taking both physics and calculus would be those who take only physics and calculus, plus those who take all three. If the problem says "only 30 take both physics and calculus," that may mean that the only ones taking both are those 30, so the total number taking both physics and calculus is 30, and those 30 are not taking chemistry. However, we also know that 15 take all three. So that would mean that the total number of students taking both physics and calculus is 30 + 15 = 45? But that contradicts the given information. Wait, let me check.

Wait, the problem says: "Only 30 take both physics and calculus." If "only" modifies "take both physics and calculus," that might mean that 30 is the number of students taking exactly both physics and calculus (not chemistry). However, there are 15 students taking all three. Therefore, the total number of students taking both physics and calculus would be 30 (exactly both) plus 15 (all three) = 45. However, in that case, the problem might have said "Only 30 take both physics and calculus and not chemistry." But since it says "Only 30 take both physics and calculus," it's ambiguous. However, in standard set notation problems, if they say "the number of students taking both physics and calculus is X," that includes those taking all three. However, if they specify "only" taking both, then that would be exclusive. But the problem says "Only 30 take both physics and calculus," which might mean that the number of students taking both physics and calculus is 30. But then, given that 15 take all three, that would imply that the number taking both physics and calculus (including those taking all three) is 30. But that would mean that the number taking both physics and calculus and not chemistry is 30 - 15 = 15. But I need to clarify this.

Alternatively, maybe the problem is using "only" to mean that the intersection of physics and calculus is 30, and there are no students taking all three? But that contradicts the given information that 15 take all three. Therefore, probably, the problem is that the number of students taking both physics and calculus is 30, and this includes those taking all three. Therefore, the number of students taking both physics and calculus is 30, and the 15 taking all three are included in that count. Wait, but then if 15 take all three, and 30 take both physics and calculus, that would mean that 30 - 15 = 15 take only physics and calculus. However, the problem says "Only 30 take both physics and calculus." If "only" is intended to mean that there are no students taking all three in that intersection, but that conflicts with the information given that 15 take all three. Therefore, I think the problem is using standard set terminology where the number of students taking both physics and calculus is 30, which includes those taking all three. So, the 15 taking all three are part of that 30. So, perhaps the problem says "Only 30 take both physics and calculus," meaning that it's the only number for that intersection, so total number of students taking both physics and calculus is 30, which includes those taking all three. Therefore, if that is the case, then the total number taking both physics and calculus is 30, which includes those taking all three. But then how does that reconcile with the other numbers? Let me note that.

Let me try to draw a Venn diagram with three circles: C for calculus, P for physics, and Ch for chemistry. The total number of students is 360, and 15 don't take any of them, so the number of students taking at least one subject is 360 - 15 = 345.

We need to find the number of students taking physics, which is the total in the P circle. To find that, perhaps we can use the principle of inclusion-exclusion.

Let me recall the inclusion-exclusion formula for three sets:

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

We know |C ∪ P ∪ Ch| = 345, |C| = 180, |C ∩ P| = ?, |C ∩ Ch| = 75, |P ∩ Ch| = 75, |C ∩ P ∩ Ch| = 15. Also, we have that twice as many take chemistry as take physics, so |Ch| = 2|P|. Also, the problem says "Only 30 take both physics and calculus." This is probably referring to |C ∩ P| = 30? But if that's the case, then we can plug in the numbers. Wait, but if |C ∩ P| is 30, but we also know that 15 take all three. Therefore, the number of students taking both C and P and not Ch is 30 - 15 = 15. But maybe that's okay. Let me try.

So, if we assume that |C ∩ P| = 30, then:

345 = 180 + |P| + |Ch| - 30 - 75 - 75 + 15

Wait, let me plug in the values:

|C| = 180

|Ch| = 2|P|

|C ∩ P| = 30

|C ∩ Ch| = 75

|P ∩ Ch| = 75

|C ∩ P ∩ Ch| = 15

Therefore, substituting into inclusion-exclusion formula:

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

Let me compute that:

First, combine terms:

180 + 3|P| - 30 - 75 - 75 + 15

Calculate constants:

180 - 30 - 75 -75 +15

Compute step by step:

180 - 30 = 150

150 - 75 = 75

75 - 75 = 0

0 + 15 = 15

So, total becomes 15 + 3|P| = 345

Therefore, 3|P| = 345 - 15 = 330

Therefore, |P| = 330 / 3 = 110

So, number of students taking physics would be 110?

Wait, but let me check if this is consistent with other given numbers. Let me verify if that makes sense.

But hold on, if |P| = 110, then |Ch| = 2*110 = 220.

Let me check if the inclusion-exclusion works. Let me recalculate:

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

= 180 + 110 + 220 - 30 - 75 - 75 + 15

Let me compute:

180 + 110 = 290; 290 + 220 = 510

Then subtract 30: 510 - 30 = 480

Subtract 75: 480 - 75 = 405

Subtract 75: 405 - 75 = 330

Add 15: 330 + 15 = 345. Which matches the total number of students taking at least one subject. So that seems okay. But let me check if the number of students taking both physics and chemistry is 75. If |P ∩ Ch| = 75, and |C ∩ P ∩ Ch| =15, then the number of students taking both physics and chemistry but not calculus is 75 - 15 = 60. Similarly, the number taking both calculus and chemistry but not physics is |C ∩ Ch| - |C ∩ P ∩ Ch| = 75 - 15 = 60. The number taking both calculus and physics but not chemistry is |C ∩ P| - |C ∩ P ∩ Ch| = 30 - 15 = 15. Then, the number taking only calculus is |C| - (students taking both C and P or C and Ch or all three). Wait, maybe we can check if the numbers make sense.

But also, we can check the number of students taking only calculus. Let me try to compute that. The total number taking calculus is 180. Which includes:

- Only calculus

- Calculus and physics only

- Calculus and chemistry only

- All three

So, |C| = Only C + (C and P only) + (C and Ch only) + All three

We have:

Only C = ?

(C and P only) = 15 (as calculated above)

(C and Ch only) = 60 (as calculated above)

All three = 15

Therefore:

Only C + 15 + 60 + 15 = 180 => Only C + 90 = 180 => Only C = 90

Similarly, total number taking physics is |P| = 110. Which includes:

- Only P

- P and C only (15)

- P and Ch only (60)

- All three (15)

Therefore, Only P + 15 + 60 + 15 = 110 => Only P + 90 = 110 => Only P = 20

Similarly, total number taking chemistry is |Ch| = 220. Which includes:

- Only Ch

- Ch and C only (60)

- Ch and P only (60)

- All three (15)

Therefore, Only Ch + 60 + 60 + 15 = 220 => Only Ch + 135 = 220 => Only Ch = 85

So, now, total number of students taking at least one subject is:

Only C (90) + Only P (20) + Only Ch (85) + (C and P only) (15) + (C and Ch only) (60) + (P and Ch only) (60) + All three (15) =

Let me add them up:

90 + 20 = 110

110 + 85 = 195

195 + 15 = 210

210 + 60 = 270

270 + 60 = 330

330 + 15 = 345. Which matches. So that works. And total students are 360, with 15 not taking any, so that also checks out.

But let me check if the numbers given in the problem are consistent with this. The problem says "75 take both calculus and chemistry" which is |C ∩ Ch| = 75, which we have as 60 (only C and Ch) + 15 (all three) = 75. Correct. Similarly, "75 take both physics and chemistry" which is |P ∩ Ch| = 75, which is 60 (only P and Ch) +15 (all three) = 75. Correct. Also, "Only 30 take both physics and calculus." Wait, in our calculation, (C and P only) is 15, and all three is 15, so total taking both physics and calculus is 15 + 15 = 30. Therefore, "Only 30 take both physics and calculus" meaning that the total number taking both is 30, which includes those taking all three. So the wording might have meant that the total number of students taking both physics and calculus is 30, which includes the ones taking all three. Therefore, that makes sense. So the answer would be |P| = 110. Therefore, the number of students taking physics is 110.

But let me check again if all the given numbers fit. Let me see:

- Number taking all three: 15. Correct.

- Number not taking any: 15. Correct.

- Number taking calculus: 180. Correct, as we calculated.

- Twice as many take chemistry as take physics: |Ch|=220, |P|=110. 220 is twice 110. Correct.

- 75 take both calculus and chemistry. Correct, as 60 + 15 = 75.

- 75 take both physics and chemistry. Correct, as 60 +15 = 75.

- Only 30 take both physics and calculus. Wait, but total number taking both physics and calculus is 30, which includes those taking all three. Therefore, the "only" here might be confusing. But if the problem says "Only 30 take both physics and calculus," that might mean that there are no students taking all three in that category, but that contradicts the given information. However, as we saw, if we interpret that |C ∩ P| = 30, then the number of students taking both is 30, which includes the 15 taking all three. Therefore, the only students taking both physics and calculus are those in the intersection, which includes those taking all three. Therefore, the wording is a bit confusing, but the calculation checks out.

Alternatively, maybe "Only 30 take both physics and calculus" means that 30 take both physics and calculus but not chemistry. In that case, the total number taking both physics and calculus would be 30 + 15 = 45. Let me check if that would change the result.

Suppose that |C ∩ P| - |C ∩ P ∩ Ch| = 30, meaning that the number taking both physics and calculus but not chemistry is 30, so total |C ∩ P| = 30 + 15 = 45. Then, let's redo the inclusion-exclusion formula.

So, let's redo the earlier steps with |C ∩ P| = 45.

So, using inclusion-exclusion:

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

Which is:

345 = 180 + |P| + 2|P| - 45 - 75 - 75 + 15

Which simplifies to:

345 = 180 + 3|P| - 45 - 75 - 75 +15

Calculating constants:

180 -45 = 135

135 -75 = 60

60 -75 = -15

-15 +15 = 0

So:

345 = 0 + 3|P| => 3|P| = 345 => |P| = 115

So then, |P| would be 115, |Ch| = 230.

But let's check if that works with other data.

So, then:

If |P| = 115, |Ch| = 230.

Now, let's check the numbers again.

Total taking at least one subject is 345.

Using inclusion-exclusion:

180 + 115 + 230 - 45 - 75 - 75 +15 = Let's compute:

180 + 115 = 295; 295 + 230 = 525

Subtract 45: 525 - 45 = 480

Subtract 75: 480 -75 = 405

Subtract 75: 405 -75 = 330

Add 15: 330 +15 = 345. Correct.

Now, let's check the numbers:

Number taking both physics and chemistry is 75, which is given. Since |P ∩ Ch| = 75. Which includes those taking all three. So, number taking both physics and chemistry but not calculus is 75 - 15 = 60.

Number taking both calculus and chemistry is 75, which includes those taking all three. So, number taking both calculus and chemistry but not physics is 75 -15 = 60.

Number taking both calculus and physics is 45, which includes those taking all three. So, number taking both calculus and physics but not chemistry is 45 -15 = 30. Which would fit with the statement that "Only 30 take both physics and calculus." If "Only 30 take both physics and calculus" means that 30 take both physics and calculus but not chemistry, then this is consistent.

But then, the problem says "Only 30 take both physics and calculus," which in this case would mean that 30 take both physics and calculus but not chemistry. However, in the previous interpretation, it was the total taking both physics and calculus. So this is ambiguous. However, the problem says "Only 30 take both physics and calculus," which might be intended to mean that the intersection of physics and calculus is 30, but since 15 take all three, that would imply that the number taking both physics and calculus but not chemistry is 15. However, that conflicts with the given numbers. Wait, no, in this case, if "Only 30 take both physics and calculus," meaning only 30 take both (and not any other subject), then the total taking both would be 30 + 15 = 45, as we considered in this second scenario.

But then which interpretation is correct? The problem is ambiguous, but given that the answer is different in the two scenarios, we need to figure out which one is correct based on standard problem conventions.

In standard set problems, when they say "Only X take both A and B," it usually means that X is the number of students taking exactly both A and B, not including those taking all three. However, if they say "the number of students taking both A and B is X," that includes those taking all three. Therefore, if the problem says "Only 30 take both physics and calculus," it's more likely that they mean the number of students taking exactly both physics and calculus is 30, so that the total number taking both physics and calculus (including those taking all three) is 30 + 15 = 45. Therefore, in this case, the answer would be |P| = 115. But earlier, when we thought that |C ∩ P| = 30, we had |P| = 110. Therefore, there are two possibilities based on interpretation.

But let's see if the problem gives any other clues. The problem also says "Only 30 take both physics and calculus," and "Twice as many students take chemistry as take physics." If we consider the second interpretation where |C ∩ P| = 45 (so exactly both physics and calculus is 30), leading to |P| = 115, then:

Number taking chemistry is 230. Let's check if the number taking both physics and chemistry is 75. Which includes 60 taking both physics and chemistry but not calculus, plus 15 taking all three. So 75. Okay. Similarly, number taking both calculus and chemistry is 75, which is 60 + 15. Okay.

But also, the number taking calculus is 180. Let me check that. In this scenario, the number of students taking calculus is:

Only C + (C and P only) + (C and Ch only) + All three.

Earlier, we calculated:

If |C ∩ P| = 45, then (C and P only) = 45 - 15 = 30.

Similarly, (C and Ch only) = 75 -15 = 60.

All three =15.

So then, Only C = |C| - (C and P only) - (C and Ch only) - All three = 180 - 30 -60 -15 = 180 - 105 = 75.

So Only C is 75. Then, total number taking physics is 115. Which includes Only P + (P and C only) + (P and Ch only) + All three.

Only P = ?

(P and C only) = 30.

(P and Ch only) = 60.

All three = 15.

So, Only P = 115 - 30 -60 -15 = 115 - 105 = 10.

Similarly, total taking chemistry is 230, which includes Only Ch + (Ch and C only) + (Ch and P only) + All three = Only Ch + 60 + 60 + 15 = Only Ch + 135 = 230 => Only Ch = 95.

Then, total students taking at least one subject is:

Only C (75) + Only P (10) + Only Ch (95) + (C and P only) (30) + (C and Ch only) (60) + (P and Ch only) (60) + All three (15) =

75 +10 = 85; 85 +95=180; 180 +30=210; 210 +60=270; 270 +60=330; 330 +15=345. Correct.

So that also works. But then, in this case, the number of students taking only physics is 10, which seems low, but maybe okay.

But here's the problem: the question says "Only 30 take both physics and calculus." If "Only" is meant to specify that there are no other subjects, then "Only 30 take both physics and calculus" would mean that the number taking exactly both physics and calculus is 30, which is consistent with the second interpretation leading to |P| = 115. However, if the problem had said "30 take both physics and calculus," without the "Only," then we could have interpreted it as the total. However, given that the problem says "Only 30 take both physics and calculus," that supports the interpretation that exactly both is 30, so total taking both is 30 + 15 = 45. Therefore, leading to |P| = 115. But then why does the first interpretation lead to |P| = 110?

Because the initial assumption was that "Only 30 take both physics and calculus" meant total both is 30, which would conflict with the fact that 15 take all three. However, since the problem says "Only 30 take both physics and calculus," perhaps that 30 is the total, so including those taking all three. But the word "only" might be confusing here.

Alternatively, perhaps there's a mistake in the problem's wording? Or perhaps the problem is from a source where "Only 30 take both physics and calculus" is meant to say that the intersection of physics and calculus is 30, and since 15 take all three, that is allowed.

This is a bit of a conundrum. However, given that when we assume that |C ∩ P| = 30 (total), then we get |P| = 110, and that works with the rest of the numbers (as checked earlier). However, if we assume that the number taking exactly both physics and calculus is 30, we get |P| = 115, which also works.

Therefore, how can we determine which one is correct? Let me check if there's another given number that can help. The problem says "75 take both calculus and chemistry, and 75 take both physics and chemistry." If we take the second interpretation where |C ∩ P| = 45 (total taking both physics and calculus), then the number taking both physics and calculus is 45, which is more than 30. But the problem says "Only 30 take both physics and calculus," so maybe that implies that the intersection is 30, not more. Therefore, the first interpretation where |C ∩ P| = 30 is correct, even though it includes the 15 taking all three. Hence, the answer would be |P| = 110.

But why is the problem saying "Only 30 take both physics and calculus"? Maybe it's trying to say that the number taking exactly both is 30, hence total taking both is 30 + 15 = 45. However, the problem says "Only 30 take both physics and calculus," which is a bit unclear. But if we take "Only" as in "only those 30 take both, and no more," which would mean that there are no students taking both physics and calculus and another subject, but that contradicts the 15 taking all three. Therefore, that can't be. Therefore, the only way to interpret it is that the total number taking both physics and calculus is 30, which includes those taking all three. Therefore, the first calculation of |P| = 110 is correct.

Alternatively, perhaps the problem has a typo, and the correct number is that "Only 30 take both physics and calculus and not chemistry," which would mean that exactly both is 30. However, given the problem as stated, the more straightforward interpretation is that "Only 30 take both physics and calculus" means that the total number taking both is 30, hence the answer is 110. However, since both interpretations lead to different answers, and the problem may have intended the first one.

But let me check once again. If |C ∩ P| = 30, then the number of students taking both physics and calculus is 30, which includes those taking all three. Therefore, the number taking all three is 15, so the number taking both physics and calculus but not chemistry is 30 - 15 = 15. Therefore, the statement "Only 30 take both physics and calculus" is conflicting with the fact that 15 take all three, but since it's allowed in the intersection, it's not conflicting. Therefore, the answer is 110. Therefore, I think that the answer is 110. Hence, the number of students taking physics is 110.

Alternatively, maybe I need to check with another approach. Let me try to use a table or the formula for two-set intersections.

Wait, another way: Let's denote:

Let me use variables:

Let’s define:

- a = number of students taking only calculus.

- b = number of students taking only physics.

- c = number of students taking only chemistry.

- d = number of students taking calculus and physics but not chemistry.

- e = number of students taking calculus and chemistry but not physics.

- f = number of students taking physics and chemistry but not calculus.

- g = number of students taking all three.

Given:

Total students = 360.

Students taking none = 15 => Students taking at least one = 345.

Given:

g = 15.

Number taking calculus = 180. Calculus includes a, d, e, g. So:

a + d + e + g = 180.

Number taking physics = b + d + f + g = ?

Number taking chemistry = c + e + f + g = 2*(number taking physics).

Number taking both calculus and chemistry = e + g = 75.

Number taking both physics and chemistry = f + g = 75.

Number taking both physics and calculus = d + g = 30 (if we take the first interpretation), or d + g = 45 (if we take the second interpretation). But the problem says "Only 30 take both physics and calculus," which may be d = 30? Or d + g = 30?

If "Only 30 take both physics and calculus," and "take both physics and calculus" means taking exactly both, then d = 30. Hence, d = 30. If it means taking both physics and calculus (including those taking all three), then d + g = 30. But since g =15, then d = 15. However, given that d + g = 30, then d = 15. But then the number taking only physics and calculus is 15. However, this would mean that the number taking both physics and calculus is 15 + 15 = 30, which is the total. But this would match the first interpretation. However, if the problem says "Only 30 take both physics and calculus," this could mean that d = 30, i.e., only those taking both physics and calculus (not chemistry). But then d = 30, and g is separate. But g is 15. So then the total taking both physics and calculus would be d + g = 30 + 15 = 45. Therefore, the problem's wording is conflicting.

Alternatively, maybe the problem is using "take both physics and calculus" to mean exactly both, so d = 30, and then the total taking physics and calculus (including all three) is 30 + 15 = 45. However, the problem says "Only 30 take both physics and calculus," which might mean that the number taking both physics and calculus (exactly) is 30, hence d = 30.

So let's try this approach. Assume that "Only 30 take both physics and calculus" means exactly both, so d = 30. Then:

From given:

Number taking both calculus and chemistry = e + g = 75 => e = 75 - g = 75 - 15 = 60.

Similarly, number taking both physics and chemistry = f + g = 75 => f = 75 - 15 = 60.

Number taking both physics and calculus = d + g = 30 + 15 = 45. But the problem says "Only 30 take both physics and calculus," which might mean that this total is 30, hence we would have d + g = 30 => d = 30 - g = 30 -15=15. So if we have d = 15.

Wait, this is very confusing. Let me clarify:

If the problem says "Only 30 take both physics and calculus," and it's meant to say that the number of students taking both physics and calculus (and possibly other subjects) is 30, then:

d + g = 30.

But since g = 15, then d = 15. Hence, the number taking both physics and calculus but not chemistry is d = 15.

Alternatively, if the problem says that the number taking exactly both physics and calculus is 30, then d = 30.

But which one is correct? The problem is not very clear. However, given that it's a math problem, likely the "Only" is intended to mean that the intersection of physics and calculus is 30, hence d + g = 30. But since g = 15, then d = 15. Let's test this.

So, if d + g = 30 => d = 15.

Then, we have:

From calculus: a + d + e + g = 180 => a + 15 + 60 + 15 = 180 => a + 90 = 180 => a = 90.

From chemistry: c + e + f + g = 2*(number taking physics). Let's denote number taking physics as P = b + d + f + g = b + 15 + 60 + 15 = b + 90.

Therefore, chemistry is c + 60 + 60 + 15 = c + 135 = 2*(b + 90).

Total students taking at least one subject is a + b + c + d + e + f + g = 90 + b + c + 15 + 60 + 60 + 15 = 90 + b + c + 150 = b + c + 240 = 345 => b + c = 345 - 240 = 105.

From above, c + 135 = 2*(b + 90). Let's write:

c + 135 = 2b + 180 => c = 2b + 180 - 135 => c = 2b + 45.

But we also have b + c = 105. Substitute c:

b + (2b + 45) = 105 => 3b + 45 = 105 => 3b = 60 => b = 20.

Therefore, c = 2*20 +45 = 40 + 45 = 85.

Therefore, number taking physics is P = b + 90 = 20 + 90 = 110. So, same as before.

So, this gives P = 110. Here, "Only 30 take both physics and calculus" is interpreted as the total number taking both is 30, which includes d + g = 15 + 15 = 30. Therefore, this is consistent. Thus, the answer is 110.

Therefore, this approach leads to |P| = 110, and the numbers check out. Therefore, the answer is 110.

Thus, despite the confusion in wording, interpreting "Only 30 take both physics and calculus" as the total number taking both (including those taking all three) gives us a consistent answer. Therefore, the number of students taking physics is 110.

**Final Answer**
The number of students who take physics is \boxed{110}.

To determine the number of students who take **physics**, we use the given information and apply **principles of set theory**, specifically the **inclusion-exclusion principle**, to account for overlapping groupings.

---

### **Step 1: Define Total Groups and Key Parameters**

Total number of students: $ 360 $
Students taking **none** of the three subjects: $ 15 $
So, the number of students taking **at least one** of the subjects:

$$
360 - 15 = 345
$$

We are given:

- 15 students take **all three** subjects (i.e., $ |C \cap P \cap Ch| = 15 $)
- 180 students take **calculus** ($ |C| = 180 $)
- **Twice as many** students take chemistry as take physics:
  $$
  |Ch| = 2 \cdot |P|
  $$
- 75 students take **both calculus and chemistry** ($ |C \cap Ch| = 75 $)
- 75 students take **both physics and chemistry** ($ |P \cap Ch| = 75 $)
- Only 30 take **both physics and calculus**: this is interpreted as the total number of students taking both physics and calculus (including those who also take chemistry), so:
  $$
  |C \cap P| = 30
  $$

---

### **Step 2: Apply the Inclusion-Exclusion Principle**

The principle for three sets is:

$$
|C \cup P \cup Ch| = |C| + |P| + |Ch| - |C \cap P| - |C \cap Ch| - |P \cap Ch| + |C \cap P \cap Ch|
$$

Substitute the known values:

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

Simplify:

$$
345 = 3|P| - 15
$$

Solving:

$$
3|P| = 360 \quad \Rightarrow \quad |P| = \frac{360}{3} = 110
$$

---

### **Step 3: Verify Consistency with Other Information**

Let’s check whether all the values are consistent:

- $ |C| = 180 $
- $ |P| = 110 $
- $ |Ch| = 2 \cdot 110 = 220 $
- $ |C \cap P| = 30 $
- $ |C \cap Ch| = 75 $
- $ |P \cap Ch| = 75 $
- $ |C \cap P \cap Ch| = 15 $

Plug into the inclusion-exclusion formula:

$$
|C \cup P \cup Ch| = 180 + 110 + 220 - 30 - 75 - 75 + 15 = 345
$$

This matches the total number of students taking at least one course. Therefore, the values are internally consistent.

---

### **Final Result**

The number of students who take **physics** is:

$$
\boxed{110}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, there are 360 students in total. Some take calculus, physics, and chemistry. Some don't take any of them. We need to find out how many students take physics. Let me jot down the given information first:

Total students = 360.

Number of students taking all three subjects (calculus, physics, chemistry) = 15.

Number of students taking none of the subjects = 15.

Number of students taking calculus = 180.

Twice as many students take chemistry as take physics. So, if we let the number of students taking physics be P, then the number taking chemistry would be 2P.

Number of students taking both calculus and chemistry = 75.

Number of students taking both physics and chemistry = 75.

Only 30 take both physics and calculus. Wait, "only" here might mean that those 30 are the ones taking both physics and calculus but not chemistry? Or does it mean that the intersection of physics and calculus is 30? Hmm, the wording says "Only 30 take both physics and calculus." The word "only" might be important here. If it's only 30 taking both physics and calculus, then that would mean that there are no students taking all three subjects in that intersection? Wait, but we already know that 15 students take all three. So if we have students taking both physics and calculus, some of them might be taking all three. Therefore, "only 30 take both physics and calculus" would mean that the number of students taking both physics and calculus but not chemistry is 30? Or maybe the total number of students taking both physics and calculus is 30? This is a bit confusing. Let me think.

In set theory, usually, when they say "the number of students taking both A and B," that includes those who might also be taking other subjects. However, if it's specified as "only" taking both A and B, then it would exclude those taking all three. However, the problem says "Only 30 take both physics and calculus." The wording here is crucial. If we have 15 students taking all three, then the total number of students taking both physics and calculus would be those who take only physics and calculus, plus those who take all three. If the problem says "only 30 take both physics and calculus," that may mean that the only ones taking both are those 30, so the total number taking both physics and calculus is 30, and those 30 are not taking chemistry. However, we also know that 15 take all three. So that would mean that the total number of students taking both physics and calculus is 30 + 15 = 45? But that contradicts the given information. Wait, let me check.

Wait, the problem says: "Only 30 take both physics and calculus." If "only" modifies "take both physics and calculus," that might mean that 30 is the number of students taking exactly both physics and calculus (not chemistry). However, there are 15 students taking all three. Therefore, the total number of students taking both physics and calculus would be 30 (exactly both) plus 15 (all three) = 45. However, in that case, the problem might have said "Only 30 take both physics and calculus and not chemistry." But since it says "Only 30 take both physics and calculus," it's ambiguous. However, in standard set notation problems, if they say "the number of students taking both physics and calculus is X," that includes those taking all three. However, if they specify "only" taking both, then that would be exclusive. But the problem says "Only 30 take both physics and calculus," which might mean that the number of students taking both physics and calculus is 30. But then, given that 15 take all three, that would imply that the number taking both physics and calculus (including those taking all three) is 30. But that would mean that the number taking both physics and calculus and not chemistry is 30 - 15 = 15. But I need to clarify this.

Alternatively, maybe the problem is using "only" to mean that the intersection of physics and calculus is 30, and there are no students taking all three? But that contradicts the given information that 15 take all three. Therefore, probably, the problem is that the number of students taking both physics and calculus is 30, and this includes those taking all three. Therefore, the number of students taking both physics and calculus is 30, and the 15 taking all three are included in that count. Wait, but then if 15 take all three, and 30 take both physics and calculus, that would mean that 30 - 15 = 15 take only physics and calculus. However, the problem says "Only 30 take both physics and calculus." If "only" is intended to mean that there are no students taking all three in that intersection, but that conflicts with the information given that 15 take all three. Therefore, I think the problem is using standard set terminology where the number of students taking both physics and calculus is 30, which includes those taking all three. So, the 15 taking all three are part of that 30. So, perhaps the problem says "Only 30 take both physics and calculus," meaning that it's the only number for that intersection, so total number of students taking both physics and calculus is 30, which includes those taking all three. Therefore, if that is the case, then the total number taking both physics and calculus is 30, which includes those taking all three. But then how does that reconcile with the other numbers? Let me note that.

Let me try to draw a Venn diagram with three circles: C for calculus, P for physics, and Ch for chemistry. The total number of students is 360, and 15 don't take any of them, so the number of students taking at least one subject is 360 - 15 = 345.

We need to find the number of students taking physics, which is the total in the P circle. To find that, perhaps we can use the principle of inclusion-exclusion.

Let me recall the inclusion-exclusion formula for three sets:

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

We know |C ∪ P ∪ Ch| = 345, |C| = 180, |C ∩ P| = ?, |C ∩ Ch| = 75, |P ∩ Ch| = 75, |C ∩ P ∩ Ch| = 15. Also, we have that twice as many take chemistry as take physics, so |Ch| = 2|P|. Also, the problem says "Only 30 take both physics and calculus." This is probably referring to |C ∩ P| = 30? But if that's the case, then we can plug in the numbers. Wait, but if |C ∩ P| is 30, but we also know that 15 take all three. Therefore, the number of students taking both C and P and not Ch is 30 - 15 = 15. But maybe that's okay. Let me try.

So, if we assume that |C ∩ P| = 30, then:

345 = 180 + |P| + |Ch| - 30 - 75 - 75 + 15

Wait, let me plug in the values:

|C| = 180

|Ch| = 2|P|

|C ∩ P| = 30

|C ∩ Ch| = 75

|P ∩ Ch| = 75

|C ∩ P ∩ Ch| = 15

Therefore, substituting into inclusion-exclusion formula:

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

Let me compute that:

First, combine terms:

180 + 3|P| - 30 - 75 - 75 + 15

Calculate constants:

180 - 30 - 75 -75 +15

Compute step by step:

180 - 30 = 150

150 - 75 = 75

75 - 75 = 0

0 + 15 = 15

So, total becomes 15 + 3|P| = 345

Therefore, 3|P| = 345 - 15 = 330

Therefore, |P| = 330 / 3 = 110

So, number of students taking physics would be 110?

Wait, but let me check if this is consistent with other given numbers. Let me verify if that makes sense.

But hold on, if |P| = 110, then |Ch| = 2*110 = 220.

Let me check if the inclusion-exclusion works. Let me recalculate:

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

= 180 + 110 + 220 - 30 - 75 - 75 + 15

Let me compute:

180 + 110 = 290; 290 + 220 = 510

Then subtract 30: 510 - 30 = 480

Subtract 75: 480 - 75 = 405

Subtract 75: 405 - 75 = 330

Add 15: 330 + 15 = 345. Which matches the total number of students taking at least one subject. So that seems okay. But let me check if the number of students taking both physics and chemistry is 75. If |P ∩ Ch| = 75, and |C ∩ P ∩ Ch| =15, then the number of students taking both physics and chemistry but not calculus is 75 - 15 = 60. Similarly, the number taking both calculus and chemistry but not physics is |C ∩ Ch| - |C ∩ P ∩ Ch| = 75 - 15 = 60. The number taking both calculus and physics but not chemistry is |C ∩ P| - |C ∩ P ∩ Ch| = 30 - 15 = 15. Then, the number taking only calculus is |C| - (students taking both C and P or C and Ch or all three). Wait, maybe we can check if the numbers make sense.

But also, we can check the number of students taking only calculus. Let me try to compute that. The total number taking calculus is 180. Which includes:

- Only calculus

- Calculus and physics only

- Calculus and chemistry only

- All three

So, |C| = Only C + (C and P only) + (C and Ch only) + All three

We have:

Only C = ?

(C and P only) = 15 (as calculated above)

(C and Ch only) = 60 (as calculated above)

All three = 15

Therefore:

Only C + 15 + 60 + 15 = 180 => Only C + 90 = 180 => Only C = 90

Similarly, total number taking physics is |P| = 110. Which includes:

- Only P

- P and C only (15)

- P and Ch only (60)

- All three (15)

Therefore, Only P + 15 + 60 + 15 = 110 => Only P + 90 = 110 => Only P = 20

Similarly, total number taking chemistry is |Ch| = 220. Which includes:

- Only Ch

- Ch and C only (60)

- Ch and P only (60)

- All three (15)

Therefore, Only Ch + 60 + 60 + 15 = 220 => Only Ch + 135 = 220 => Only Ch = 85

So, now, total number of students taking at least one subject is:

Only C (90) + Only P (20) + Only Ch (85) + (C and P only) (15) + (C and Ch only) (60) + (P and Ch only) (60) + All three (15) =

Let me add them up:

90 + 20 = 110

110 + 85 = 195

195 + 15 = 210

210 + 60 = 270

270 + 60 = 330

330 + 15 = 345. Which matches. So that works. And total students are 360, with 15 not taking any, so that also checks out.

But let me check if the numbers given in the problem are consistent with this. The problem says "75 take both calculus and chemistry" which is |C ∩ Ch| = 75, which we have as 60 (only C and Ch) + 15 (all three) = 75. Correct. Similarly, "75 take both physics and chemistry" which is |P ∩ Ch| = 75, which is 60 (only P and Ch) +15 (all three) = 75. Correct. Also, "Only 30 take both physics and calculus." Wait, in our calculation, (C and P only) is 15, and all three is 15, so total taking both physics and calculus is 15 + 15 = 30. Therefore, "Only 30 take both physics and calculus" meaning that the total number taking both is 30, which includes those taking all three. So the wording might have meant that the total number of students taking both physics and calculus is 30, which includes the ones taking all three. Therefore, that makes sense. So the answer would be |P| = 110. Therefore, the number of students taking physics is 110.

But let me check again if all the given numbers fit. Let me see:

- Number taking all three: 15. Correct.

- Number not taking any: 15. Correct.

- Number taking calculus: 180. Correct, as we calculated.

- Twice as many take chemistry as take physics: |Ch|=220, |P|=110. 220 is twice 110. Correct.

- 75 take both calculus and chemistry. Correct, as 60 + 15 = 75.

- 75 take both physics and chemistry. Correct, as 60 +15 = 75.

- Only 30 take both physics and calculus. Wait, but total number taking both physics and calculus is 30, which includes those taking all three. Therefore, the "only" here might be confusing. But if the problem says "Only 30 take both physics and calculus," that might mean that there are no students taking all three in that category, but that contradicts the given information. However, as we saw, if we interpret that |C ∩ P| = 30, then the number of students taking both is 30, which includes the 15 taking all three. Therefore, the only students taking both physics and calculus are those in the intersection, which includes those taking all three. Therefore, the wording is a bit confusing, but the calculation checks out.

Alternatively, maybe "Only 30 take both physics and calculus" means that 30 take both physics and calculus but not chemistry. In that case, the total number taking both physics and calculus would be 30 + 15 = 45. Let me check if that would change the result.

Suppose that |C ∩ P| - |C ∩ P ∩ Ch| = 30, meaning that the number taking both physics and calculus but not chemistry is 30, so total |C ∩ P| = 30 + 15 = 45. Then, let's redo the inclusion-exclusion formula.

So, let's redo the earlier steps with |C ∩ P| = 45.

So, using inclusion-exclusion:

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

Which is:

345 = 180 + |P| + 2|P| - 45 - 75 - 75 + 15

Which simplifies to:

345 = 180 + 3|P| - 45 - 75 - 75 +15

Calculating constants:

180 -45 = 135

135 -75 = 60

60 -75 = -15

-15 +15 = 0

So:

345 = 0 + 3|P| => 3|P| = 345 => |P| = 115

So then, |P| would be 115, |Ch| = 230.

But let's check if that works with other data.

So, then:

If |P| = 115, |Ch| = 230.

Now, let's check the numbers again.

Total taking at least one subject is 345.

Using inclusion-exclusion:

180 + 115 + 230 - 45 - 75 - 75 +15 = Let's compute:

180 + 115 = 295; 295 + 230 = 525

Subtract 45: 525 - 45 = 480

Subtract 75: 480 -75 = 405

Subtract 75: 405 -75 = 330

Add 15: 330 +15 = 345. Correct.

Now, let's check the numbers:

Number taking both physics and chemistry is 75, which is given. Since |P ∩ Ch| = 75. Which includes those taking all three. So, number taking both physics and chemistry but not calculus is 75 - 15 = 60.

Number taking both calculus and chemistry is 75, which includes those taking all three. So, number taking both calculus and chemistry but not physics is 75 -15 = 60.

Number taking both calculus and physics is 45, which includes those taking all three. So, number taking both calculus and physics but not chemistry is 45 -15 = 30. Which would fit with the statement that "Only 30 take both physics and calculus." If "Only 30 take both physics and calculus" means that 30 take both physics and calculus but not chemistry, then this is consistent.

But then, the problem says "Only 30 take both physics and calculus," which in this case would mean that 30 take both physics and calculus but not chemistry. However, in the previous interpretation, it was the total taking both physics and calculus. So this is ambiguous. However, the problem says "Only 30 take both physics and calculus," which might be intended to mean that the intersection of physics and calculus is 30, but since 15 take all three, that would imply that the number taking both physics and calculus but not chemistry is 15. However, that conflicts with the given numbers. Wait, no, in this case, if "Only 30 take both physics and calculus," meaning only 30 take both (and not any other subject), then the total taking both would be 30 + 15 = 45, as we considered in this second scenario.

But then which interpretation is correct? The problem is ambiguous, but given that the answer is different in the two scenarios, we need to figure out which one is correct based on standard problem conventions.

In standard set problems, when they say "Only X take both A and B," it usually means that X is the number of students taking exactly both A and B, not including those taking all three. However, if they say "the number of students taking both A and B is X," that includes those taking all three. Therefore, if the problem says "Only 30 take both physics and calculus," it's more likely that they mean the number of students taking exactly both physics and calculus is 30, so that the total number taking both physics and calculus (including those taking all three) is 30 + 15 = 45. Therefore, in this case, the answer would be |P| = 115. But earlier, when we thought that |C ∩ P| = 30, we had |P| = 110. Therefore, there are two possibilities based on interpretation.

But let's see if the problem gives any other clues. The problem also says "Only 30 take both physics and calculus," and "Twice as many students take chemistry as take physics." If we consider the second interpretation where |C ∩ P| = 45 (so exactly both physics and calculus is 30), leading to |P| = 115, then:

Number taking chemistry is 230. Let's check if the number taking both physics and chemistry is 75. Which includes 60 taking both physics and chemistry but not calculus, plus 15 taking all three. So 75. Okay. Similarly, number taking both calculus and chemistry is 75, which is 60 + 15. Okay.

But also, the number taking calculus is 180. Let me check that. In this scenario, the number of students taking calculus is:

Only C + (C and P only) + (C and Ch only) + All three.

Earlier, we calculated:

If |C ∩ P| = 45, then (C and P only) = 45 - 15 = 30.

Similarly, (C and Ch only) = 75 -15 = 60.

All three =15.

So then, Only C = |C| - (C and P only) - (C and Ch only) - All three = 180 - 30 -60 -15 = 180 - 105 = 75.

So Only C is 75. Then, total number taking physics is 115. Which includes Only P + (P and C only) + (P and Ch only) + All three.

Only P = ?

(P and C only) = 30.

(P and Ch only) = 60.

All three = 15.

So, Only P = 115 - 30 -60 -15 = 115 - 105 = 10.

Similarly, total taking chemistry is 230, which includes Only Ch + (Ch and C only) + (Ch and P only) + All three = Only Ch + 60 + 60 + 15 = Only Ch + 135 = 230 => Only Ch = 95.

Then, total students taking at least one subject is:

Only C (75) + Only P (10) + Only Ch (95) + (C and P only) (30) + (C and Ch only) (60) + (P and Ch only) (60) + All three (15) =

75 +10 = 85; 85 +95=180; 180 +30=210; 210 +60=270; 270 +60=330; 330 +15=345. Correct.

So that also works. But then, in this case, the number of students taking only physics is 10, which seems low, but maybe okay.

But here's the problem: the question says "Only 30 take both physics and calculus." If "Only" is meant to specify that there are no other subjects, then "Only 30 take both physics and calculus" would mean that the number taking exactly both physics and calculus is 30, which is consistent with the second interpretation leading to |P| = 115. However, if the problem had said "30 take both physics and calculus," without the "Only," then we could have interpreted it as the total. However, given that the problem says "Only 30 take both physics and calculus," that supports the interpretation that exactly both is 30, so total taking both is 30 + 15 = 45. Therefore, leading to |P| = 115. But then why does the first interpretation lead to |P| = 110?

Because the initial assumption was that "Only 30 take both physics and calculus" meant total both is 30, which would conflict with the fact that 15 take all three. However, since the problem says "Only 30 take both physics and calculus," perhaps that 30 is the total, so including those taking all three. But the word "only" might be confusing here.

Alternatively, perhaps there's a mistake in the problem's wording? Or perhaps the problem is from a source where "Only 30 take both physics and calculus" is meant to say that the intersection of physics and calculus is 30, and since 15 take all three, that is allowed.

This is a bit of a conundrum. However, given that when we assume that |C ∩ P| = 30 (total), then we get |P| = 110, and that works with the rest of the numbers (as checked earlier). However, if we assume that the number taking exactly both physics and calculus is 30, we get |P| = 115, which also works.

Therefore, how can we determine which one is correct? Let me check if there's another given number that can help. The problem says "75 take both calculus and chemistry, and 75 take both physics and chemistry." If we take the second interpretation where |C ∩ P| = 45 (total taking both physics and calculus), then the number taking both physics and calculus is 45, which is more than 30. But the problem says "Only 30 take both physics and calculus," so maybe that implies that the intersection is 30, not more. Therefore, the first interpretation where |C ∩ P| = 30 is correct, even though it includes the 15 taking all three. Hence, the answer would be |P| = 110.

But why is the problem saying "Only 30 take both physics and calculus"? Maybe it's trying to say that the number taking exactly both is 30, hence total taking both is 30 + 15 = 45. However, the problem says "Only 30 take both physics and calculus," which is a bit unclear. But if we take "Only" as in "only those 30 take both, and no more," which would mean that there are no students taking both physics and calculus and another subject, but that contradicts the 15 taking all three. Therefore, that can't be. Therefore, the only way to interpret it is that the total number taking both physics and calculus is 30, which includes those taking all three. Therefore, the first calculation of |P| = 110 is correct.

Alternatively, perhaps the problem has a typo, and the correct number is that "Only 30 take both physics and calculus and not chemistry," which would mean that exactly both is 30. However, given the problem as stated, the more straightforward interpretation is that "Only 30 take both physics and calculus" means that the total number taking both is 30, hence the answer is 110. However, since both interpretations lead to different answers, and the problem may have intended the first one.

But let me check once again. If |C ∩ P| = 30, then the number of students taking both physics and calculus is 30, which includes those taking all three. Therefore, the number taking all three is 15, so the number taking both physics and calculus but not chemistry is 30 - 15 = 15. Therefore, the statement "Only 30 take both physics and calculus" is conflicting with the fact that 15 take all three, but since it's allowed in the intersection, it's not conflicting. Therefore, the answer is 110. Therefore, I think that the answer is 110. Hence, the number of students taking physics is 110.

Alternatively, maybe I need to check with another approach. Let me try to use a table or the formula for two-set intersections.

Wait, another way: Let's denote:

Let me use variables:

Let’s define:

- a = number of students taking only calculus.

- b = number of students taking only physics.

- c = number of students taking only chemistry.

- d = number of students taking calculus and physics but not chemistry.

- e = number of students taking calculus and chemistry but not physics.

- f = number of students taking physics and chemistry but not calculus.

- g = number of students taking all three.

Given:

Total students = 360.

Students taking none = 15 => Students taking at least one = 345.

Given:

g = 15.

Number taking calculus = 180. Calculus includes a, d, e, g. So:

a + d + e + g = 180.

Number taking physics = b + d + f + g = ?

Number taking chemistry = c + e + f + g = 2*(number taking physics).

Number taking both calculus and chemistry = e + g = 75.

Number taking both physics and chemistry = f + g = 75.

Number taking both physics and calculus = d + g = 30 (if we take the first interpretation), or d + g = 45 (if we take the second interpretation). But the problem says "Only 30 take both physics and calculus," which may be d = 30? Or d + g = 30?

If "Only 30 take both physics and calculus," and "take both physics and calculus" means taking exactly both, then d = 30. Hence, d = 30. If it means taking both physics and calculus (including those taking all three), then d + g = 30. But since g =15, then d = 15. However, given that d + g = 30, then d = 15. But then the number taking only physics and calculus is 15. However, this would mean that the number taking both physics and calculus is 15 + 15 = 30, which is the total. But this would match the first interpretation. However, if the problem says "Only 30 take both physics and calculus," this could mean that d = 30, i.e., only those taking both physics and calculus (not chemistry). But then d = 30, and g is separate. But g is 15. So then the total taking both physics and calculus would be d + g = 30 + 15 = 45. Therefore, the problem's wording is conflicting.

Alternatively, maybe the problem is using "take both physics and calculus" to mean exactly both, so d = 30, and then the total taking physics and calculus (including all three) is 30 + 15 = 45. However, the problem says "Only 30 take both physics and calculus," which might mean that the number taking both physics and calculus (exactly) is 30, hence d = 30.

So let's try this approach. Assume that "Only 30 take both physics and calculus" means exactly both, so d = 30. Then:

From given:

Number taking both calculus and chemistry = e + g = 75 => e = 75 - g = 75 - 15 = 60.

Similarly, number taking both physics and chemistry = f + g = 75 => f = 75 - 15 = 60.

Number taking both physics and calculus = d + g = 30 + 15 = 45. But the problem says "Only 30 take both physics and calculus," which might mean that this total is 30, hence we would have d + g = 30 => d = 30 - g = 30 -15=15. So if we have d = 15.

Wait, this is very confusing. Let me clarify:

If the problem says "Only 30 take both physics and calculus," and it's meant to say that the number of students taking both physics and calculus (and possibly other subjects) is 30, then:

d + g = 30.

But since g = 15, then d = 15. Hence, the number taking both physics and calculus but not chemistry is d = 15.

Alternatively, if the problem says that the number taking exactly both physics and calculus is 30, then d = 30.

But which one is correct? The problem is not very clear. However, given that it's a math problem, likely the "Only" is intended to mean that the intersection of physics and calculus is 30, hence d + g = 30. But since g = 15, then d = 15. Let's test this.

So, if d + g = 30 => d = 15.

Then, we have:

From calculus: a + d + e + g = 180 => a + 15 + 60 + 15 = 180 => a + 90 = 180 => a = 90.

From chemistry: c + e + f + g = 2*(number taking physics). Let's denote number taking physics as P = b + d + f + g = b + 15 + 60 + 15 = b + 90.

Therefore, chemistry is c + 60 + 60 + 15 = c + 135 = 2*(b + 90).

Total students taking at least one subject is a + b + c + d + e + f + g = 90 + b + c + 15 + 60 + 60 + 15 = 90 + b + c + 150 = b + c + 240 = 345 => b + c = 345 - 240 = 105.

From above, c + 135 = 2*(b + 90). Let's write:

c + 135 = 2b + 180 => c = 2b + 180 - 135 => c = 2b + 45.

But we also have b + c = 105. Substitute c:

b + (2b + 45) = 105 => 3b + 45 = 105 => 3b = 60 => b = 20.

Therefore, c = 2*20 +45 = 40 + 45 = 85.

Therefore, number taking physics is P = b + 90 = 20 + 90 = 110. So, same as before.

So, this gives P = 110. Here, "Only 30 take both physics and calculus" is interpreted as the total number taking both is 30, which includes d + g = 15 + 15 = 30. Therefore, this is consistent. Thus, the answer is 110.

Therefore, this approach leads to |P| = 110, and the numbers check out. Therefore, the answer is 110.

Thus, despite the confusion in wording, interpreting "Only 30 take both physics and calculus" as the total number taking both (including those taking all three) gives us a consistent answer. Therefore, the number of students taking physics is 110.

**Final Answer**
The number of students who take physics is \boxed{110}.

To determine the number of students who take **physics**, we use the given information and apply **principles of set theory**, specifically the **inclusion-exclusion principle**, to account for overlapping groupings.

---

### **Step 1: Define Total Groups and Key Parameters**

Total number of students: $ 360 $
Students taking **none** of the three subjects: $ 15 $
So, the number of students taking **at least one** of the subjects:

$$
360 - 15 = 345
$$

We are given:

- 15 students take **all three** subjects (i.e., $ |C \cap P \cap Ch| = 15 $)
- 180 students take **calculus** ($ |C| = 180 $)
- **Twice as many** students take chemistry as take physics:
  $$
  |Ch| = 2 \cdot |P|
  $$
- 75 students take **both calculus and chemistry** ($ |C \cap Ch| = 75 $)
- 75 students take **both physics and chemistry** ($ |P \cap Ch| = 75 $)
- Only 30 take **both physics and calculus**: this is interpreted as the total number of students taking both physics and calculus (including those who also take chemistry), so:
  $$
  |C \cap P| = 30
  $$

---

### **Step 2: Apply the Inclusion-Exclusion Principle**

The principle for three sets is:

$$
|C \cup P \cup Ch| = |C| + |P| + |Ch| - |C \cap P| - |C \cap Ch| - |P \cap Ch| + |C \cap P \cap Ch|
$$

Substitute the known values:

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

Simplify:

$$
345 = 3|P| - 15
$$

Solving:

$$
3|P| = 360 \quad \Rightarrow \quad |P| = \frac{360}{3} = 110
$$

---

### **Step 3: Verify Consistency with Other Information**

Let’s check whether all the values are consistent:

- $ |C| = 180 $
- $ |P| = 110 $
- $ |Ch| = 2 \cdot 110 = 220 $
- $ |C \cap P| = 30 $
- $ |C \cap Ch| = 75 $
- $ |P \cap Ch| = 75 $
- $ |C \cap P \cap Ch| = 15 $

Plug into the inclusion-exclusion formula:

$$
|C \cup P \cup Ch| = 180 + 110 + 220 - 30 - 75 - 75 + 15 = 345
$$

This matches the total number of students taking at least one course. Therefore, the values are internally consistent.

---

### **Final Result**

The number of students who take **physics** is:

$$
\boxed{110}
$$

Solution:
Let $x$ be the number of students taking physics, so the number in chemistry is $2x$.  There are 15 students taking all three, and 30 students in both physics and calculus, meaning there are $30 - 15 = 15$ students in just physics and calculus.  Similarly there are $60$ students in just chemistry and calculus, and $60$ in physics and chemistry.  Since there are $x$ students in physics and $15 + 15 + 60 = 90$ students taking physics along with other classes, $x - 90$ students are just taking physics.  Similarly, there are $2x - 135$ students taking just chemistry and $90$ students taking just calculus.  Knowing that there are 15 students not taking any of them, the sum of these eight categories is 360, the total number of people at the school:   \[
(x - 90) + (2x - 135) + 90 + 60 + 15 + 60 + 15 + 15 = 360.
\] We solve for $x$ and find that the number of physics students is $x = \boxed{110}$.


Explanation:
Okay, so Sam has this 2-gallon bucket of milk, and he's carrying it 3 miles to his house. But there's a leak in the bucket. The problem says that for each mile he walks, there's 2/3 as much milk in the bucket as at the beginning of the mile. Hmm, let me think about how to model this.

First, let me make sure I understand the problem correctly. The bucket starts with 2 gallons. Every mile he walks, the amount of milk decreases to 2/3 of what it was at the start of that mile. So, if he walks one mile, then after that mile, the milk is 2/3 of the original 2 gallons. Then, if he walks another mile, the milk would be 2/3 of that amount, and so on for each mile.

Wait, but does this mean that after each mile, the milk is multiplied by 2/3? Let me check. It says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, if he starts a mile with some amount of milk, after walking that mile, the amount is 2/3 of that starting amount. So yes, it seems like each mile, the milk is multiplied by 2/3. Therefore, after each mile, the milk is (2/3) times what it was at the start of that mile.

Therefore, for the first mile, starting with 2 gallons, after the first mile, he has 2 * (2/3) gallons. Then, for the second mile, he starts with that amount, so after the second mile, it's 2 * (2/3) * (2/3) = 2*(2/3)^2. Then, after the third mile, it's 2*(2/3)^3. Since he's walking 3 miles, that would mean when he gets home, the amount of milk is 2*(2/3)^3.

Wait, is that right? Let me check again. So, starting amount is 2 gallons. After first mile, it's 2*(2/3). After second mile, it's that amount times (2/3) again, so 2*(2/3)^2. After third mile, 2*(2/3)^3. So yes, that seems correct.

But let me think again. Let me take an example. Suppose he walks 1 mile. Then the milk is 2*(2/3) = 4/3 gallons. Then he walks another mile, so that 4/3 gallons becomes (4/3)*(2/3) = 8/9 gallons. Then, another mile, so 8/9*(2/3) = 16/27 gallons. So after 3 miles, he has 16/27 gallons. Let me compute 2*(2/3)^3. 2*(8/27) = 16/27. Yep, same result.

Therefore, is the answer 16/27 gallons? Wait, but let me check again if my interpretation is correct. The problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, does that mean that during the mile, the milk is leaking so that at the end of the mile, it's 2/3 of what it was at the beginning? So, if he starts the mile with, say, M gallons, then after walking the mile, it's (2/3)*M. So yes, that is the same as multiplying by 2/3 each mile.

Alternatively, could it mean that the milk decreases by 2/3 per mile? Like, if it's decreasing by 2/3, then it would be M - (2/3)M = (1/3)M. But the wording is "there is 2/3 as much milk... as at the beginning of the mile." So, "as much as" so that would mean the amount is 2/3 of the beginning. So, that is, the amount is multiplied by 2/3. So, that's the correct interpretation.

Therefore, the total amount after 3 miles is 2*(2/3)^3 = 16/27. Is there a different way to model this?

Alternatively, maybe we need to consider the leakage rate. Suppose that the leak causes the milk to decrease exponentially. The problem says that for each mile, the amount is 2/3 of what it was at the beginning of the mile. Therefore, each mile, it's multiplied by 2/3. So, after n miles, it's 2*(2/3)^n.

Therefore, for 3 miles, n=3, so 2*(2/3)^3 = 16/27. So, 16/27 is approximately 0.59259 gallons. So, that's about 0.59 gallons. But the question is asking for the exact value, so 16/27.

But let me think again. Maybe there's another interpretation. Suppose that the leak is constant, and the rate depends on the distance. Wait, but the problem says that for each mile he walks, there is 2/3 as much milk as at the beginning of the mile. So, this is not a rate; it's a proportion. So, regardless of the distance walked, every mile, the milk is reduced to 2/3 of its initial value for that mile.

Alternatively, maybe the problem is that the rate of leakage depends on the distance? But the problem is not stating that. Let me think again.

Alternatively, maybe it's a geometric series? Wait, if he walks mile by mile, each mile decreasing the milk by 2/3, so first mile, he loses 1 - 2/3 = 1/3 of the initial amount. Then, the second mile, he loses 1/3 of the remaining amount. Wait, but that would be a different interpretation. Wait, but the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, if at the beginning of the mile, there is some amount, then after the mile, it's 2/3 of that. Therefore, the amount lost during the mile is 1/3 of the amount at the beginning of the mile.

Therefore, if that is the case, then the total amount would be 2*(2/3)^3. But maybe someone could think that the amount lost is 2/3 per mile, but that would be different. But no, the problem says "there is 2/3 as much milk... as at the beginning of the mile," so it's not the amount lost, but the remaining amount.

Alternatively, maybe the problem is that for each mile, the milk decreases by a factor of 2/3. Which is the same thing. So, if you start with M, after mile 1, you have (2/3)M. After mile 2, (2/3)^2 M. After mile 3, (2/3)^3 M. So, 2*(2/3)^3 = 16/27. So, that seems to be the answer.

Alternatively, maybe the problem is a trick question? For example, if the bucket is leaking continuously, so that over each mile, the milk decreases by some rate. But since the problem is stated as "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile," it's more of a discrete process: each mile, it's multiplied by 2/3. Therefore, the total is 2*(2/3)^3. Therefore, 16/27.

Alternatively, let me think of it as a geometric sequence. The amount of milk after each mile is:

After 0 miles: 2 gallons

After 1 mile: 2*(2/3)

After 2 miles: 2*(2/3)^2

After 3 miles: 2*(2/3)^3

So, yes, that's straightforward. Therefore, the answer should be 16/27 gallons.

Wait, 2*(2/3)^3. Let me compute that again:

(2/3)^3 is 8/27. Multiply by 2: 16/27. Yep. So, 16/27 is the exact value. So, the answer is 16/27.

But let me check if there's another way to model this. For example, suppose that the leakage is continuous, so that over each mile, the milk decreases exponentially. But the problem states that after each mile, the milk is 2/3 of the beginning of the mile. So, that suggests that the reduction is happening over the mile, but the result after the mile is 2/3 of the initial. Therefore, it's not necessarily a continuous exponential decay, but rather a discrete multiplier each mile.

Therefore, the answer is 16/27. So, converting that to a decimal, it's approximately 0.59259 gallons. Since the question asks for the exact value, 16/27 is the correct answer.

Alternatively, if someone thought that the leak was 2/3 gallon per mile, then the answer would be different. But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile," which is a ratio, not a fixed amount. Therefore, the answer is 16/27.

Therefore, I think that is the correct answer.

**Final Answer**
The amount of milk in the bucket when Sam gets home is \boxed{\dfrac{16}{27}} gallons.

To determine how much milk remains in Sam’s bucket when he reaches his house, we need to analyze the situation carefully.

Sam is carrying a **2-gallon bucket** of milk, and he walks **3 miles** to his house. However, the bucket has a leak: **for each mile he walks**, the amount of milk in the bucket becomes **$\frac{2}{3}$** of what it was at the **beginning of that mile**.

---

### Step-by-Step Breakdown:

Let’s assume the initial amount of milk is:

$$
M_0 = 2 \text{ gallons}
$$

At the start of the first mile:
- The amount of milk is $ M_0 = 2 $ gallons.

At the end of the first mile:
- The amount of milk becomes $ M_1 = \frac{2}{3} \times M_0 = \frac{2}{3} \times 2 = \frac{4}{3} $ gallons.

At the start of the second mile:
- The amount of milk is $ M_1 = \frac{4}{3} $ gallons.

At the end of the second mile:
- The amount of milk becomes $ M_2 = \frac{2}{3} \times M_1 = \frac{2}{3} \times \frac{4}{3} = \frac{8}{9} $ gallons.

At the start of the third mile:
- The amount of milk is $ M_2 = \frac{8}{9} $ gallons.

At the end of the third mile:
- The amount of milk becomes $ M_3 = \frac{2}{3} \times M_2 = \frac{2}{3} \times \frac{8}{9} = \frac{16}{27} $ gallons.

---

### General Formula

This pattern shows that after $ n $ miles, the amount of milk remaining is:

$$
M_n = 2 \times \left( \frac{2}{3} \right)^n
$$

Since Sam walks **3 miles**, we substitute $ n = 3 $:

$$
M_3 = 2 \times \left( \frac{2}{3} \right)^3 = 2 \times \frac{8}{27} = \frac{16}{27}
$$

---

### Final Answer

$$
\boxed{\dfrac{16}{27}}
$$

So, when Sam gets home, there will be **$\dfrac{16}{27}$ gallons** of milk left in the bucket.

[Assistant]
Explanation:
Okay, so Sam has this 2-gallon bucket of milk, and he's carrying it 3 miles to his house. But there's a leak in the bucket. The problem says that for each mile he walks, there's 2/3 as much milk in the bucket as at the beginning of the mile. Hmm, let me think about how to model this.

First, let me make sure I understand the problem correctly. The bucket starts with 2 gallons. Every mile he walks, the amount of milk decreases to 2/3 of what it was at the start of that mile. So, if he walks one mile, then after that mile, the milk is 2/3 of the original 2 gallons. Then, if he walks another mile, the milk would be 2/3 of that amount, and so on for each mile.

Wait, but does this mean that after each mile, the milk is multiplied by 2/3? Let me check. It says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, if he starts a mile with some amount of milk, after walking that mile, the amount is 2/3 of that starting amount. So yes, it seems like each mile, the milk is multiplied by 2/3. Therefore, after each mile, the milk is (2/3) times what it was at the start of that mile.

Therefore, for the first mile, starting with 2 gallons, after the first mile, he has 2 * (2/3) gallons. Then, for the second mile, he starts with that amount, so after the second mile, it's 2 * (2/3) * (2/3) = 2*(2/3)^2. Then, after the third mile, it's 2*(2/3)^3. Since he's walking 3 miles, that would mean when he gets home, the amount of milk is 2*(2/3)^3.

Wait, is that right? Let me check again. So, starting amount is 2 gallons. After first mile, it's 2*(2/3). After second mile, it's that amount times (2/3) again, so 2*(2/3)^2. After third mile, 2*(2/3)^3. So yes, that seems correct.

But let me think again. Let me take an example. Suppose he walks 1 mile. Then the milk is 2*(2/3) = 4/3 gallons. Then he walks another mile, so that 4/3 gallons becomes (4/3)*(2/3) = 8/9 gallons. Then, another mile, so 8/9*(2/3) = 16/27 gallons. So after 3 miles, he has 16/27 gallons. Let me compute 2*(2/3)^3. 2*(8/27) = 16/27. Yep, same result.

Therefore, is the answer 16/27 gallons? Wait, but let me check again if my interpretation is correct. The problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, does that mean that during the mile, the milk is leaking so that at the end of the mile, it's 2/3 of what it was at the beginning? So, if he starts the mile with, say, M gallons, then after walking the mile, it's (2/3)*M. So yes, that is the same as multiplying by 2/3 each mile.

Alternatively, could it mean that the milk decreases by 2/3 per mile? Like, if it's decreasing by 2/3, then it would be M - (2/3)M = (1/3)M. But the wording is "there is 2/3 as much milk... as at the beginning of the mile." So, "as much as" so that would mean the amount is 2/3 of the beginning. So, that is, the amount is multiplied by 2/3. So, that's the correct interpretation.

Therefore, the total amount after 3 miles is 2*(2/3)^3 = 16/27. Is there a different way to model this?

Alternatively, maybe we need to consider the leakage rate. Suppose that the leak causes the milk to decrease exponentially. The problem says that for each mile, the amount is 2/3 of what it was at the beginning of the mile. Therefore, each mile, it's multiplied by 2/3. So, after n miles, it's 2*(2/3)^n.

Therefore, for 3 miles, n=3, so 2*(2/3)^3 = 16/27. So, 16/27 is approximately 0.59259 gallons. So, that's about 0.59 gallons. But the question is asking for the exact value, so 16/27.

But let me think again. Maybe there's another interpretation. Suppose that the leak is constant, and the rate depends on the distance. Wait, but the problem says that for each mile he walks, there is 2/3 as much milk as at the beginning of the mile. So, this is not a rate; it's a proportion. So, regardless of the distance walked, every mile, the milk is reduced to 2/3 of its initial value for that mile.

Alternatively, maybe the problem is that the rate of leakage depends on the distance? But the problem is not stating that. Let me think again.

Alternatively, maybe it's a geometric series? Wait, if he walks mile by mile, each mile decreasing the milk by 2/3, so first mile, he loses 1 - 2/3 = 1/3 of the initial amount. Then, the second mile, he loses 1/3 of the remaining amount. Wait, but that would be a different interpretation. Wait, but the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, if at the beginning of the mile, there is some amount, then after the mile, it's 2/3 of that. Therefore, the amount lost during the mile is 1/3 of the amount at the beginning of the mile.

Therefore, if that is the case, then the total amount would be 2*(2/3)^3. But maybe someone could think that the amount lost is 2/3 per mile, but that would be different. But no, the problem says "there is 2/3 as much milk... as at the beginning of the mile," so it's not the amount lost, but the remaining amount.

Alternatively, maybe the problem is that for each mile, the milk decreases by a factor of 2/3. Which is the same thing. So, if you start with M, after mile 1, you have (2/3)M. After mile 2, (2/3)^2 M. After mile 3, (2/3)^3 M. So, 2*(2/3)^3 = 16/27. So, that seems to be the answer.

Alternatively, maybe the problem is a trick question? For example, if the bucket is leaking continuously, so that over each mile, the milk decreases by some rate. But since the problem is stated as "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile," it's more of a discrete process: each mile, it's multiplied by 2/3. Therefore, the total is 2*(2/3)^3. Therefore, 16/27.

Alternatively, let me think of it as a geometric sequence. The amount of milk after each mile is:

After 0 miles: 2 gallons

After 1 mile: 2*(2/3)

After 2 miles: 2*(2/3)^2

After 3 miles: 2*(2/3)^3

So, yes, that's straightforward. Therefore, the answer should be 16/27 gallons.

Wait, 2*(2/3)^3. Let me compute that again:

(2/3)^3 is 8/27. Multiply by 2: 16/27. Yep. So, 16/27 is the exact value. So, the answer is 16/27.

But let me check if there's another way to model this. For example, suppose that the leakage is continuous, so that over each mile, the milk decreases exponentially. But the problem states that after each mile, the milk is 2/3 of the beginning of the mile. So, that suggests that the reduction is happening over the mile, but the result after the mile is 2/3 of the initial. Therefore, it's not necessarily a continuous exponential decay, but rather a discrete multiplier each mile.

Therefore, the answer is 16/27. So, converting that to a decimal, it's approximately 0.59259 gallons. Since the question asks for the exact value, 16/27 is the correct answer.

Alternatively, if someone thought that the leak was 2/3 gallon per mile, then the answer would be different. But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile," which is a ratio, not a fixed amount. Therefore, the answer is 16/27.

Therefore, I think that is the correct answer.

**Final Answer**
The amount of milk in the bucket when Sam gets home is \boxed{\dfrac{16}{27}} gallons.

To determine how much milk remains in Sam’s bucket when he reaches his house, we need to analyze the situation carefully.

Sam is carrying a **2-gallon bucket** of milk, and he walks **3 miles** to his house. However, the bucket has a leak: **for each mile he walks**, the amount of milk in the bucket becomes **$\frac{2}{3}$** of what it was at the **beginning of that mile**.

---

### Step-by-Step Breakdown:

Let’s assume the initial amount of milk is:

$$
M_0 = 2 \text{ gallons}
$$

At the start of the first mile:
- The amount of milk is $ M_0 = 2 $ gallons.

At the end of the first mile:
- The amount of milk becomes $ M_1 = \frac{2}{3} \times M_0 = \frac{2}{3} \times 2 = \frac{4}{3} $ gallons.

At the start of the second mile:
- The amount of milk is $ M_1 = \frac{4}{3} $ gallons.

At the end of the second mile:
- The amount of milk becomes $ M_2 = \frac{2}{3} \times M_1 = \frac{2}{3} \times \frac{4}{3} = \frac{8}{9} $ gallons.

At the start of the third mile:
- The amount of milk is $ M_2 = \frac{8}{9} $ gallons.

At the end of the third mile:
- The amount of milk becomes $ M_3 = \frac{2}{3} \times M_2 = \frac{2}{3} \times \frac{8}{9} = \frac{16}{27} $ gallons.

---

### General Formula

This pattern shows that after $ n $ miles, the amount of milk remaining is:

$$
M_n = 2 \times \left( \frac{2}{3} \right)^n
$$

Since Sam walks **3 miles**, we substitute $ n = 3 $:

$$
M_3 = 2 \times \left( \frac{2}{3} \right)^3 = 2 \times \frac{8}{27} = \frac{16}{27}
$$

---

### Final Answer

$$
\boxed{\dfrac{16}{27}}
$$

So, when Sam gets home, there will be **$\dfrac{16}{27}$ gallons** of milk left in the bucket.

Solution: At the end of the first mile, there will be $\frac{2}{3}$ of the initial milk in the bucket. Each additional mile multiplies this amount by $\frac{2}{3}$. Thus, when he arrives at home at the end of the third mile, there will be $\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \left(\frac{2}{3}\right)^{3}$ as much milk in the bucket. Since he had 2 gallons initially, the amount in the bucket when he gets home is $2 \cdot \left(\frac{2}{3}\right)^{3}$. Because $\left(\frac{a}{b}\right)^{n} = \frac{a^{n}}{b^{n}}$, this expression is equivalent to $2 \cdot \frac{2^{3}}{3^{3}}$. Because $n^{a} \cdot n^{b} = n^{a+b}$, this equals $\frac{2^{4}}{3^{3}}$. Multiplying the exponents out, we get $\boxed{\frac{16}{27}}$ gallons.