Preptest 53 game 2

I watched the last video on the game that had a similar set of rules that needed the "butterfly" configuration but to be honest it still confuses me. Can you explain it more in depth or perhaps do a quick video on it? Check out the first video on the list here. You must Log in or Sign Up to post comments. Permalink Submitted by laurlaur on Jun 19 Permalink Submitted by majorgeneraldave on Jun 21 Anyway, they suck, is what I'm saying.

Permalink Submitted by afromkin on Oct 31 You must be logged in to post a comment. You can get a free account here. Enroll now and get started in less than a minute. Take PrepTest. Review Results. Questions by Type Tool Available as online tables and downloadable booklet. About Mike Kim. Study Guide. Student Resources.

Study Schedules. Logic Game Diagrams. Readiness Checklist. Instructional Blog. Logical Reasoning Reading Comprehension Main Page. Information for Tutors. Amazon Page. Diagramming Solutions for PrepTest See More Diagrams. Game 1 Option 1. Step 1 Per the given scenario, we can write out the five elements to be placed - T, W, X, Y, and Z, and we can lay out positions in three groups - F, P, and S - with an indication that each group gets at least one and up to three assignments.

Step 2 Per the first rule, we can notate that X signs with F. Step 3 Per the second rule, we can notate that Y does not sign with F.

Step 4 Per the third rule, we can notate that Z and Y must be grouped together. Step 5 Per the fourth rule, we can notate that if T is assigned to S, W will be assigned to S as well. Game 1 Option 2. Step 3 Per the second and third rules, we can split our diagram into two frames, one in which Z and Y are assigned to P frame 1 , and another in which Z and Y are assigned to S frame 2.

Step 4 In frame 1, T and W are our only remaining options for the minimum one element assigned to S. Step 5 For frame 1, we can notate that T is the only element not yet assigned. Step 6 For frame 2, T cannot be assigned to S, because if it were, per the fourth rule, W would have to be as well, and if that were the case we would have no one left to assign to P. Step 7 For frame 2, W and T are our only remaining options for the minimum one element assigned to P.

Game 2. Step 1 Per the given scenario and rules, we can write out the six elements to be placed - G, J, L, M, P, and V - and the six positions to be filled, in order. Step 2 Per the first rule, we can notate that we must have the order P - M - L.

Step 3 Per the second rule, we can split our diagram into two frames, one in which G is before both L and J frame 1 , and another in which G is after both L and J frame 2. Step 4 Per the third rule, we can split each of our diagrams up further. Game 3. Step 1 Per the given scenario, we can write out the seven elements to be placed - S, T, V, W, X, Y, and Z, note the two types of subsets, confess c and not confess cross out c , and lay out the seven positions to be assigned, in order. Step 2 Per the first rule, we can place T into the third position.

Step 3 Per the second rule, we can notate that the element in the fourth position did not confess. Step 4 Per the third rule, we can notate that W must come before S. Step 5 Per the fourth rule, we can notate that Z must come before both X and V. Step 6 Per the fifth rule, we can notate that no one confessed after W was questioned. Step 7 Per the sixth rule, we can notate that two of the final three who were questioned confessed. Step 8 Having notated the sixth rule, the only way to satisfy the third condition that S follows W and the fifth condition that no one confessed after W is to have someone confess in 5, W confess in 6, and S not confess in 7.

Step 9 We can notate that Y is not directly restricted by any of the given rules. Game 4 Option 1. Step 1 Per the given scenario, we can write out the six elements to be placed - M, N, O, P, S, and T, as well as the subsets schools they could belong to - f, g, and h.