Analytical Reasoning Test

Z is last, so first place goes to W, X or Y. W did not finish first, leaving X or Y, and since X finished ahead of Y, Y cannot be first, so X finished first. Placing Y first ignores that X is ahead of Y. Placing W first ignores that W did not finish first. Concluding it cannot be determined overlooks that the three clues together fix first place exactly.

G is in seat 3, and F is immediately to its left, so F is in seat 2. J must be left of F, and the only seat left of seat 2 is seat 1, so J is in seat 1, leaving H in seat 4. Placing F in seat 1 ignores that F is pinned to seat 2. Placing H in seat 1 ignores that J must sit to the left of F. Concluding it cannot be determined overlooks that the clues fix every seat.

P and Q share a group, and T is barred from it, so T is in the other group. R and S are split, so exactly one of them joins P and Q, filling that group of three. That leaves T, the other of R and S, and U together, so U must be with T. Placing T with P is barred directly; placing T with Q is barred too, because Q sits with P, so joining Q would put T with P. Concluding it cannot be determined misses that U is forced, even though which of R or S joins T is not.

The clues chain into Eastfield smaller than Riverside smaller than Northgate, with Westbank largest above them all, so Eastfield is smallest. Choosing Riverside stops at it being below Northgate but ignores that Eastfield is smaller still. Choosing Northgate reverses the comparison, since Northgate is second largest. Concluding it cannot be determined overlooks that the chain orders all four sites.

Stock takes Thursday, so Safety, Systems and Service fill Monday to Wednesday. Systems immediately before Service means the pair is either Monday and Tuesday or Tuesday and Wednesday; the second leaves Safety on Monday, which is barred, so Systems is Monday, Service Tuesday and Safety Wednesday. Monday breaks the “not Monday” clue; Tuesday is Service’s day; Thursday is Stock’s day.

If the manual were in C, box B would be empty; but box B is not empty, so by the contrapositive the manual is not in C. It is not in A either, so it must be in B. Choosing box C ignores that a full box B rules C out. Choosing box A breaks the “not in A” clue. Concluding it cannot be determined misses that eliminating A and C leaves B as the only possibility.

The novel is at the bottom. The dictionary sits on the thesaurus, and the atlas is above the dictionary with the guide directly above the atlas, giving from the top: guide, atlas, dictionary, thesaurus, novel. So the guide is on top. Choosing the atlas forgets the guide sits above it. Choosing the dictionary starts the pile there, ignoring the atlas and guide above it. Choosing the thesaurus inverts the “directly on top of” relation.

Bea has timing. Ade cannot take notes, the chair or timing, so Ade takes welcome. Cai takes welcome only if Ade takes the chair; Ade does not, and welcome is gone, so Cai cannot take welcome. Chair and notes remain for Cai and Dot, and since Dot will not take the chair, Dot takes notes and Cai takes the chair. Choosing welcome ignores the “only if” condition; choosing notes would force Dot into the chair, which is barred; timing is Bea’s role.

Test each winner against the “exactly one true” rule. If Ada won, Ben’s statement and Cara’s are both true; if Ben won, Ada’s and Cara’s are both true; if Cara won, only Ada’s statement is true and the other two are false, which is exactly one, so Cara won. A winner of Ada or Ben each produces two true statements, breaking the rule. Concluding it cannot be determined misses that only one winner satisfies the count.

Every mug is blue, so a green item cannot be a mug, and as each item is only a mug or a plate, the green item must be a plate. Claiming it is a mug contradicts “all mugs are blue”. Claiming it is blue contradicts the stated colour of the item. Concluding it cannot be determined overlooks that ruling out “mug” leaves “plate” as the only category available.