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:
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:
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 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:
Show 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.
Updated On: Jul 13, 2026
- Exactly three of the six mobile phones have digital camera.
- Exactly four of the six mobile phones have digital camera.
- Exactly three of the six mobile phones have document viewer.
- Exactly four of the six mobile phones have document viewer.
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Pin down W, X and Y first.
W is told to have a digital camera and a music player. If W also carried the document viewer, W would hold all three options, and no model could then have "more options than W" since three is the maximum. But the rules say X has more options than W, so W cannot hold all three. W must be exactly {camera, music player}, two options only.
Since X must beat W's two options, and three is the most any model can carry, X must carry all three: {camera, music player, document viewer}.
W and Y share no option at all. Because W has camera and music player, Y can have neither of those. Every model needs at least one option, so Y is forced down to just {document viewer}.
Step 2: Pin down V using the "no two alike" rule for this question.
V always carries camera and document viewer. If V also carried the music player, V would equal X's full set of three, which this question forbids (no two models may match exactly). So V is exactly {camera, document viewer}, two options.
Step 3: Work out what is left for T and Z.
The sets used so far are {camera, music}=W, {camera, music, viewer}=X, {viewer}=Y, {camera, viewer}=V. Since every model's set must be different, T and Z can only be chosen from the three sets not yet used: {camera}, {music}, {music, viewer}.
V and Z must share exactly one option. Check each candidate against V={camera, viewer}: {camera} shares camera (one option, works); {music} shares nothing (fails); {music, viewer} shares viewer (one option, works). So Z is {camera} or {music, viewer}.
T must carry fewer options than Z. If Z={camera} (just one option), T would need fewer than one, which is impossible since every model needs at least one option. So Z cannot be {camera}. That leaves Z={music, viewer}, two options, and T must have exactly one option, taken from whichever of {camera} or {music} remains unused.
Step 4: Count each option under both remaining layouts.
So the only freedom left is T={camera} or T={music}, with W={camera, music}, X={camera, music, viewer}, Y={viewer}, V={camera, viewer}, Z={music, viewer} fixed in both cases.
Document viewer is carried by V, X, Y and Z in both layouts, no matter what T is. That is always exactly four phones with a document viewer, never three.
Digital camera: if T={camera}, camera-carriers are T, V, W, X, four phones. If T={music}, camera-carriers are just V, W, X, three phones. So both "exactly three have camera" and "exactly four have camera" are each true for one of the two layouts.
Music player: if T={music}, music-carriers are T, W, X, Z, four phones. If T={camera}, music-carriers are W, X, Z, three phones. So "exactly four have music player" is true for one layout.
Final Answer:
Statements about camera counts (three or four) and document viewer (four) and music player (four) can each happen for one of the two allowed layouts. Only "exactly three of the six phones have a document viewer" never happens, since the document viewer count is locked at four regardless of which layout is chosen. That is the statement that could NOT be true. \[ \boxed{\text{Exactly three phones have a document viewer, this can never be true}} \]
W is told to have a digital camera and a music player. If W also carried the document viewer, W would hold all three options, and no model could then have "more options than W" since three is the maximum. But the rules say X has more options than W, so W cannot hold all three. W must be exactly {camera, music player}, two options only.
Since X must beat W's two options, and three is the most any model can carry, X must carry all three: {camera, music player, document viewer}.
W and Y share no option at all. Because W has camera and music player, Y can have neither of those. Every model needs at least one option, so Y is forced down to just {document viewer}.
Step 2: Pin down V using the "no two alike" rule for this question.
V always carries camera and document viewer. If V also carried the music player, V would equal X's full set of three, which this question forbids (no two models may match exactly). So V is exactly {camera, document viewer}, two options.
Step 3: Work out what is left for T and Z.
The sets used so far are {camera, music}=W, {camera, music, viewer}=X, {viewer}=Y, {camera, viewer}=V. Since every model's set must be different, T and Z can only be chosen from the three sets not yet used: {camera}, {music}, {music, viewer}.
V and Z must share exactly one option. Check each candidate against V={camera, viewer}: {camera} shares camera (one option, works); {music} shares nothing (fails); {music, viewer} shares viewer (one option, works). So Z is {camera} or {music, viewer}.
T must carry fewer options than Z. If Z={camera} (just one option), T would need fewer than one, which is impossible since every model needs at least one option. So Z cannot be {camera}. That leaves Z={music, viewer}, two options, and T must have exactly one option, taken from whichever of {camera} or {music} remains unused.
Step 4: Count each option under both remaining layouts.
So the only freedom left is T={camera} or T={music}, with W={camera, music}, X={camera, music, viewer}, Y={viewer}, V={camera, viewer}, Z={music, viewer} fixed in both cases.
Document viewer is carried by V, X, Y and Z in both layouts, no matter what T is. That is always exactly four phones with a document viewer, never three.
Digital camera: if T={camera}, camera-carriers are T, V, W, X, four phones. If T={music}, camera-carriers are just V, W, X, three phones. So both "exactly three have camera" and "exactly four have camera" are each true for one of the two layouts.
Music player: if T={music}, music-carriers are T, W, X, Z, four phones. If T={camera}, music-carriers are W, X, Z, three phones. So "exactly four have music player" is true for one layout.
Final Answer:
Statements about camera counts (three or four) and document viewer (four) and music player (four) can each happen for one of the two allowed layouts. Only "exactly three of the six phones have a document viewer" never happens, since the document viewer count is locked at four regardless of which layout is chosen. That is the statement that could NOT be true. \[ \boxed{\text{Exactly three phones have a document viewer, this can never be true}} \]
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Table 1
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