This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Let G be the set of traders who sell Gari, R be the set of traders who sell Rice, and M be the set of traders who sell Maize. The total number of traders is n(U) = 100. Each trader sells at least one item, so n(G R M) = 100. Given information: • n(G) = 70 • n(R only) = 10 (traders who sell Rice but not Gari or Maize) • n(M only) = 15 (traders who sell Maize but not Gari or Rice) • n(G R) = 50 (traders who sell both Gari and Rice) • n(R M) = 45 (traders who sell both Rice and Maize) • n(M) = 65 Let x = n(G R M) be the number of traders who sell all three items. We can represent the number of traders in each disjoint region of the Venn diagram: • Traders selling only Gari: n(G only) • Traders selling only Rice: n(R only) = 10 • Traders selling only Maize: n(M only) = 15 • Traders selling Gari and Rice only (not Maize): n(G R only) = n(G R) - n(G R M) = 50 - x • Traders selling Gari and Maize only (not Rice): n(G M only) • Traders selling Rice and Maize only (not Gari): n(R M only) = n(R M) - n(G R M) = 45 - x • Traders selling Gari, Rice, and Maize: n(G R M) = x Step 1: Find n(G M only). We know n(M) = 65. This is the sum of traders selling only Maize, Maize and Gari only, Maize and Rice only, and all three. n(M) = n(M only) + n(G M only) + n(R M only) + n(G R M) 65 = 15 + n(G M only) + (45 - x) + x 65 = 15 + n(G M only) + 45 65 = 60 + n(G M only) n(G M only) = 65 - 60 = 5 Step 2: Find n(G only). We know n(G) = 70. This is the sum of traders selling only Gari, Gari and Rice only, Gari and Maize only, and all three. n(G) = n(G only) + n(G R only) + n(G M only) + n(G R M) 70 = n(G only) + (50 - x) + 5 + x 70 = n(G only) + 55 n(G only) = 70 - 55 = 15 Step 3: Find the number of traders who sell all three items (x). The total number of traders is n(G R M) = 100. This is the sum of all disjoint regions: n(G R M) = n(G only) + n(R only) + n(M only) + n(G R only) + n(G M only) + n(R M only) + n(G R M) 100 = 15 + 10 + 15 + (50 - x) + 5 + (45 - x) + x 100 = (15 + 10 + 15 + 50 + 5 + 45) - x - x + x 100 = 140 - x x = 140 - 100 x = 40 Now we can answer the specific questions: a) All three items. This is x. x = 40 The number of traders who sell all three items is 40. b) Gari and rice but not maize. This is n(G R only). n(G R only) = 50 - x = 50 - 40 = 10 The number of traders who sell Gari and Rice but not Maize is 10. c) Gari and maize but not rice. This is n(G M only). n(G M only) = 5 The number of traders who sell Gari and Maize but not Rice is 5. That's 2 down. 3 left today — send the next one.