There are 74 students in the Trevel Club. They discovered that 40 members have visited country A, 31 have visited country B, 25 have been to country C, 14 have visited countries A and B, 10 have been only to country C, and 20 have been only, to country A. Some club members have not been to any of the three foreign countries, and, curiously, an equal number have been to all three countries. (a) How many students have been to all three countries? (b) How many students have been only to country B? (e) How many students have been to all three countries?

Let \[ x = \text{number of students who visited all three countries}, \] and note that the problem states that the number of students who visited none of the countries is also \(x\). Define the following: - Only A: \(A_o = 20\) - Only B: \(B_o = y\) (unknown) - Only C: \(C_o = 10\) - Only A and B (but not C): let it be \(A\!B_o\) - Only A and C (but not B): let it be \(A\!C_o\) - Only B and C (but not A): let it be \(B\!C_o\) We are given: \[ |A| = 40,\quad |B| = 31, \quad |C| = 25, \] \[ |A\cap B| = 14, \] which includes those who visited A and B only plus those who visited all three. Hence, \[ A\!B_o = 14 - x. \] We use the fact that the total visitors in each country is the sum of those who visited only that country, those who visited that country and one other (but not the third), and those who visited all three. ### Step 1. Count for Country A For country A, the total is: \[ |A| = A_o + A\!B_o + A\!C_o + x. \] Substitute the known values: \[ 40 = 20 + (14 - x) + A\!C_o + x. \] The \(-x\) and \(+x\) cancel: \[ 40 = 20 + 14 + A\!C_o \quad \Longrightarrow \quad A\!C_o = 40 - 34 = 6. \] ### Step 2. Count for Country C For country C, the total is: \[ |C| = C_o + A\!C_o + B\!C_o + x. \] Substitute the known values: \[ 25 = 10 + 6 + B\!C_o + x. \] Simplify: \[ 25 = 16 + B\!C_o + x \quad \Longrightarrow \quad B\!C_o + x = 9. \] Thus, \[ B\!C_o = 9 - x. \] ### Step 3. Count for Country B For country B, the total is: \[ |B| = B_o + A\!B_o + B\!C_o + x. \] Substitute the known values and expressions: \[ 31 = y + (14 - x) + (9 - x) + x. \] Simplify the intersection terms: \[ 31 = y + 14 + 9 - x - x + x = y + 23 - x. \] Solve for \(y\): \[ y = 31 - 23 + x = 8 + x. \] Thus, the number of students who have been only to country \(B\) is: \[ B_o = 8 + x. \] ### Step 4. Use the Total Students and the Complement The inclusion–exclusion formula for three sets gives the total number who visited at least one country: \[ |A\cup B\cup C| = |A| + |B| + |C| - (|A\cap B| + |A\cap C| + |B\cap C|) + x. \] We know: - \(|A| + |B| + |C| = 40 + 31 + 25 = 96.\) - \(|A\cap B| = 14.\) - \(|A\cap C| = A\!C_o + x = 6 + x.\) - \(|B\cap C| = B\!C_o + x = (9 - x) + x = 9.\) Then: \[ |A\cup B\cup C| = 96 - \big(14 + (6+x) + 9\big) + x. \] Simplify the parentheses: \[ 14+6+x+9 = 29+x. \] So: \[ |A\cup B\cup C| = 96 - (29+x) + x = 96 - 29 = 67. \] Out of 74 students, the number who did not visit any country is: \[ d = 74 - 67 = 7. \] But the problem states \(d = x\). Therefore, \[ x = 7. \] ### Step 5. Answer the Questions (a) The number of students who visited all three countries is: \[ x = 7. \] (b) The number of students who have been only to country \(B\) is: \[ B_o = 8 + x = 8 + 7 = 15. \] (e) This repeats the question regarding the number of students who visited all three countries, so the answer remains: \[ 7. \]