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.

Suppose no two mobile phone models have exactly the same set of options as one another. In that case, each of the following could be true EXCEPT:

• A. Exactly three of the six mobile phones have digital camera.
• B. Exactly four of the six mobile phones have digital camera.
• C. Exactly three of the six mobile phones have document viewer.
• D. Exactly four of the six mobile phones have document viewer.

Hint:

First pin W to exactly {camera, music} and X to all three options, since X must beat W and three is the maximum. Then work out Y = {document viewer} and V = {camera, viewer}, and count each option across the two remaining layouts for T and Z.

Answer 1

The Correct Option is C

A faster way to attack this kind of "EXCEPT" question is to lock the option counts for camera, music player and document viewer first, and only then check the five statements one by one, instead of building the whole model from scratch before looking at the answer choices.

Start from the two rules that pin single models directly. W has camera and music player as a floor, and X must beat W's option count, and the ceiling for any model is three options total. If W held all three options, nothing could beat it, so W is stuck at exactly two: camera and music player. That pushes X to the only value bigger than two, which is three, so X gets all three options.

W and Y share nothing, and W's two options are camera and music player, so Y is barred from both. Since Y needs at least one option and camera/music are both off limits, Y can only be the document viewer.

Now use this question's special rule: no two of the six models share the exact same option set. V already must include camera and document viewer; if it also took music player it would duplicate X's set, which is banned here, so V is fixed at exactly {camera, document viewer}.

Four sets are now used up: {camera, music}, {camera, music, viewer}, {viewer}, {camera, viewer}. Only {camera}, {music}, and {music, viewer} remain for T and Z, and they must stay distinct from each other and from the used sets.

1. Testing Z = {camera}: V and Z would share just camera, one option, which fits the "exactly one in common" rule for V and Z. But then T would need fewer than one option, and nothing can have zero options, so this branch is dead.
2. Testing Z = {music, viewer}: V and Z share only viewer, again exactly one option, which fits. T then needs fewer than two options, so T gets exactly one, either {camera} or {music}, whichever is still free. This branch survives.

So the puzzle settles into two live pictures, differing only in whether T is {camera} or {music}, while W, X, Y, V, Z stay fixed as above. Now check each answer choice against both pictures:
Document viewer belongs to V, X, Y, Z in both pictures - always four phones, never three.
Camera belongs to V, W, X always, plus T in the {camera} picture (four total) but not in the {music} picture (three total). Both three and four are achievable.
Music player belongs to W, X, Z always, plus T in the {music} picture (four total) but not in the {camera} picture (three total). Four is achievable.

Every statement about camera and music counts turns out to be achievable in one picture or the other, but the document-viewer count never drops to three, it is locked at four in every valid arrangement. That is the one statement that can never be true.