Question

Directions for Questions 102 to 107: A retail electronics showroom chain has six new mobile phone models, T, V, W, X, Y and Z, each equipped with at least one of three options: digital camera, music player and office document viewer. No phone has any other option. The following conditions apply:

• V features 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.

Q102. For exactly how many of the six mobile phones is it possible to determine exactly which option each one has?

• A. Two
• B. Three
• C. Four
• D. Five

Hint:

Work out Y first from the no-overlap rule with W, then use the three-option ceiling to pin down W, X and V.

Answer 1

The Correct Option is C

Step 1: Pin down Y.
W has a digital camera and a music player, so W's option set includes at least {camera, music}. Since W and Y share no options at all, Y cannot have a camera or a music player. Every phone must have at least one option, so Y is left with only the document viewer. Y = {document viewer}, fully fixed.

Step 2: Pin down W and X together.
W has at least {camera, music}; it might also carry the document viewer, giving it up to 3 options. X must have strictly more options than W. Since 3 is the maximum any phone can carry, if W already had all 3 options, nothing could beat it, which breaks the "X has more" rule. So W cannot have all three; W must be exactly {camera, music}, 2 options. That forces X to have more than 2, and since 3 is the ceiling, X must have all three options: {camera, music, document viewer}.

Step 3: Pin down V.
V is given at least {camera, document viewer}. Suppose V also had music, making it all three options. Then "V and Z share exactly one option" would force Z to be a strict subset of V's options with size 1, and T (which must have fewer options than Z) would need fewer than 1 option, which is impossible since every phone needs at least one. So V cannot have all three; V is exactly {camera, document viewer}, 2 options.

Step 4: Check Z and T.
V and Z share exactly one option. Since V = {camera, document viewer}, Z shares either the camera or the document viewer, but not both. If that single shared item were Z's only option, Z would have 1 option, forcing T (with fewer options than Z) to have 0, impossible. So Z must carry the shared item plus the music player, giving Z exactly 2 options: either {camera, music} or {document viewer, music}. Either way, T only needs 1 option, and T could independently be camera, music, or document viewer alone. Both Z's exact identity and T's exact identity stay open; neither is pinned to one unique answer.

Final Answer:
V, W, X and Y each come out to one single, forced option set, four phones in total. Z and T each have more than one option set that still fits every rule, so they are not fully determined.
\[ \boxed{\text{Four phones (V, W, X, Y) can be determined exactly.}} \]