Question

Directions: Each group of questions is based on a set of conditions. Read the passage below and answer the question that follows.

A retail electronics chain has six new mobile phone models, T, V, W, X, Y and Z. Each model comes with at least one of three options: digital camera, music player and office document viewer. No model has any other option. The following conditions apply:

• V has both a digital camera and an office document viewer.
• W has a digital camera and a music player.
• W and Y have no options in common.
• X has more options than W.
• V and Z have exactly one option in common.
• T has fewer options than Z.

If exactly four of the six mobile phones have music player, and exactly four of the six mobile phones have digital camera, then each of the following must be true EXCEPT:

• A. T and V have no options in common.
• B. T and Y have no options in common.
• C. T and Z have exactly one option in common.
• D. W and Z have exactly one option in common.

Hint:

First show V can never carry all three options (else Z would need one option and T could not exist). With camera fixed on V, W, X and music fixed on W, X only, the two count rules force T = {music} and Z = {camera, music}. Then just read off which pair shares two options instead of one.

Answer 1

The Correct Option is D

Instead of building the model piece by piece and only then checking the choices, work the other way here: pin the skeleton with the base rules, then let the two given counts (four cameras, four music players) squeeze the model down to a single arrangement, and read the five statements off that one arrangement.

Start with the two rules that never change across this whole family of questions. W is told to carry camera and music player as a floor. X must have strictly more options than W, and three is the largest number of options any model can hold. If W ever carried all three options, no model could beat it, since three is the ceiling, so W is capped at exactly two options: camera and music player, nothing else. That immediately pushes X above two options, and the only number above two is three, so X must carry all three: camera, music player and document viewer.

Now use the rule that W and Y share no options at all. W's two options are camera and music player, so Y cannot have either of those. Every model needs at least one option, and the only option left for Y is the document viewer, so Y = {document viewer}, fully fixed.

From the base rules alone: W can only be {camera, music}, since X must beat W's option count and three is the ceiling, so W holding all three would leave nothing for X to beat it with. That forces X to {camera, music, viewer}, all three. W and Y share nothing, and W holds camera and music, so Y is squeezed down to just {document viewer}.

Next, notice V cannot be the full set {camera, music, viewer}. If it were, then "V and Z share exactly one option" would really mean "Z itself has exactly one option" (because the overlap of a full-option model with anything equals that other model). But then T would need fewer than one option, and every model needs at least one, so that path is a dead end. Hence V is exactly {camera, document viewer}.

Now bring in this question's two counts. Camera is already sitting on V, W, X (three models); the fourth camera-carrier must come from T or Z (never Y, since Y is only {viewer}). Music is already sitting on W, X only (two models, since V does not have it); the two missing music-carriers must be T and Z both.

1. Try Z without camera, i.e. Z = {music, viewer}: then T would have to supply the missing camera-carrier, so T needs both camera and music, at least two options. But T must have fewer options than Z, and Z has only two here, so T can have at most one option. A model cannot pack two required features into one option slot, so this attempt fails.
2. Try Z with camera, i.e. Z = {camera, music}: Z now already supplies the fourth camera-carrier, so T does not need camera. T still needs music, so T = {music}, one option, which is indeed fewer than Z's two options. This attempt works and is forced.

So the model is pinned uniquely: T = {music}, V = {camera, viewer}, W = {camera, music}, X = {camera, music, viewer}, Y = {viewer}, Z = {camera, music}. Reading the five statements straight off this table: T-V share nothing, T-Y share nothing, T-Z share music only (one option), Y-Z share nothing, but W-Z share both camera and music, two options, not one.

Four of the five statements match the model exactly, and only the W-Z statement claims "exactly one" when the true overlap is two options. That is the statement that does not have to be true.