1. **Stating the problem:** We have three tests: Written (W), Practical (P), and Oral (O). Candidates pass different combinations of these tests. Given: - 5 candidates passed only one test. - 27 candidates passed exactly two tests. - 4 candidates passed all three tests. - 16 candidates passed the practical test. We need to: (i) Complete the Venn diagram with the given data. (ii) Find the total number of candidates who sat for the test. 2. **Understanding the problem and notation:** Let the number of candidates who passed only W be $x$, only P be $y$, and only O be $z$. Given that the total who passed only one test is 5, so: $$x + y + z = 5$$ Let the number of candidates who passed exactly two tests be: - W and P only: $a$ - P and O only: $b$ - W and O only: $c$ Given that total who passed exactly two tests is 27, so: $$a + b + c = 27$$ Number who passed all three tests is 4. 3. **Using the practical test data:** Candidates who passed practical test are those who passed: - Only P: $y$ - W and P only: $a$ - P and O only: $b$ - All three: 4 Given total who passed practical test is 16, so: $$y + a + b + 4 = 16$$ 4. **Expressing variables and solving:** From step 3: $$y + a + b = 16 - 4 = 12$$ From step 2: $$x + y + z = 5$$ $$a + b + c = 27$$ We have 5 unknowns: $x, y, z, a, b, c$ but only 3 equations. 5. **Using the total number of candidates who passed all tests:** Total candidates who passed at least one test is sum of all disjoint parts: $$x + y + z + a + b + c + 4$$ 6. **Additional assumptions:** Since no other data is given, assume the number of candidates who passed only practical test $y$ is unknown. 7. **From step 4, express $a + b$ in terms of $y$:** $$a + b = 12 - y$$ From step 2: $$a + b + c = 27 \\ (12 - y) + c = 27 \\ c = 27 - 12 + y = 15 + y$$ 8. **Sum of candidates who passed only one test:** $$x + y + z = 5$$ 9. **Total candidates who sat for the test:** $$T = x + y + z + a + b + c + 4$$ Substitute $a + b = 12 - y$ and $c = 15 + y$: $$T = (x + y + z) + (a + b) + c + 4 = 5 + (12 - y) + (15 + y) + 4 = 5 + 12 - y + 15 + y + 4 = 36$$ 10. **Final answers:** (i) The Venn diagram values are: - Only one test total: 5 (split as $x, y, z$) - Exactly two tests total: 27 (split as $a, b, c$) - All three tests: 4 - Practical test total: 16 (ii) Total candidates who sat for the test: 36 **Note:** Without further data, exact splits of $x, y, z, a, b, c$ cannot be determined, but total candidates are 36.