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:
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:
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 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:
Show 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.
Updated On: Jul 13, 2026
- T and V have no options in common.
- T and Y have no options in common.
- T and Z have exactly one option in common.
- W and Z have exactly one option in common.
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Rebuild the fixed skeleton from the base rules.
X must have more options than W, and three is the highest number of options any model can carry. So W cannot hold all three options, otherwise nothing could beat it. W is stuck at exactly two options: camera and music player. That pushes X above two, so X must carry all three options: camera, music player and document viewer.
W and Y share nothing. Since W has camera and music, Y is barred from both, so Y is left with only the document viewer.
Step 2: Show that V cannot carry all three options.
V always has camera and document viewer, and might or might not have music player. Suppose V did carry all three. Then V would be the full set of options, so whatever Z is, V and Z's overlap would just equal Z itself. The rule says this overlap must be exactly one option, so Z would have to be a single-option model. But then T needs fewer options than Z, that is fewer than one, which is impossible since every model needs at least one option. So V cannot carry music player. V is fixed at exactly {camera, document viewer}.
Step 3: Use this question's counts to fix T and Z.
Camera is already guaranteed on V, W and X, three models. We need the total to be four, and Y never has camera, so exactly one of T, Z must carry a camera.
Music player is already guaranteed on W and X only (V has been ruled out above), two models. We need the total to be four, and Y never has music, so both T and Z together must supply two more music carriers, meaning BOTH T and Z carry music player.
Step 4: Pin Z using the V-Z overlap rule.
Z must include music player (Step 3). Check which music-containing sets overlap V={camera, viewer} in exactly one option: {music} alone overlaps zero, fails; {camera, music} overlaps just camera, one option, works; {music, viewer} overlaps just viewer, one option, works; {camera, music, viewer} overlaps two options, fails. So Z is {camera, music} or {music, viewer}.
Step 5: Use the T-versus-Z size rule to choose between the two Z options.
If Z = {music, viewer} (no camera), then since exactly one of T, Z needs camera, it would have to be T. T would then need both camera and music, two options at least, but T must have fewer options than Z's two options, that is at most one option. A model cannot carry two required options in only one slot, so this branch collapses.
So Z = {camera, music}, which already supplies the one camera-carrier needed among T and Z, meaning T must NOT carry camera. T still needs music (Step 3), so T = {music} exactly, one option. Checking T against Z: one option is fewer than two, which satisfies the rule.
Step 6: The layout is now completely fixed. Check each statement against it.
T = {music}, V = {camera, viewer}, W = {camera, music}, X = {camera, music, viewer}, Y = {viewer}, Z = {camera, music}.
T and V: {music} vs {camera, viewer}, no shared option, so "no options in common" holds.
T and Y: {music} vs {viewer}, no shared option, holds.
T and Z: {music} vs {camera, music}, share only music, exactly one option, holds.
W and Z: {camera, music} vs {camera, music}, share BOTH camera and music, that is two options in common, not one. This breaks the claim of "exactly one option in common."
Y and Z: {viewer} vs {camera, music}, no shared option, holds.
Final Answer:
Every statement checks out except the one about W and Z, which actually share two options, not one. That is the statement that is not forced to be true. \[ \boxed{\text{W and Z share two options, not one}} \]
X must have more options than W, and three is the highest number of options any model can carry. So W cannot hold all three options, otherwise nothing could beat it. W is stuck at exactly two options: camera and music player. That pushes X above two, so X must carry all three options: camera, music player and document viewer.
W and Y share nothing. Since W has camera and music, Y is barred from both, so Y is left with only the document viewer.
Step 2: Show that V cannot carry all three options.
V always has camera and document viewer, and might or might not have music player. Suppose V did carry all three. Then V would be the full set of options, so whatever Z is, V and Z's overlap would just equal Z itself. The rule says this overlap must be exactly one option, so Z would have to be a single-option model. But then T needs fewer options than Z, that is fewer than one, which is impossible since every model needs at least one option. So V cannot carry music player. V is fixed at exactly {camera, document viewer}.
Step 3: Use this question's counts to fix T and Z.
Camera is already guaranteed on V, W and X, three models. We need the total to be four, and Y never has camera, so exactly one of T, Z must carry a camera.
Music player is already guaranteed on W and X only (V has been ruled out above), two models. We need the total to be four, and Y never has music, so both T and Z together must supply two more music carriers, meaning BOTH T and Z carry music player.
Step 4: Pin Z using the V-Z overlap rule.
Z must include music player (Step 3). Check which music-containing sets overlap V={camera, viewer} in exactly one option: {music} alone overlaps zero, fails; {camera, music} overlaps just camera, one option, works; {music, viewer} overlaps just viewer, one option, works; {camera, music, viewer} overlaps two options, fails. So Z is {camera, music} or {music, viewer}.
Step 5: Use the T-versus-Z size rule to choose between the two Z options.
If Z = {music, viewer} (no camera), then since exactly one of T, Z needs camera, it would have to be T. T would then need both camera and music, two options at least, but T must have fewer options than Z's two options, that is at most one option. A model cannot carry two required options in only one slot, so this branch collapses.
So Z = {camera, music}, which already supplies the one camera-carrier needed among T and Z, meaning T must NOT carry camera. T still needs music (Step 3), so T = {music} exactly, one option. Checking T against Z: one option is fewer than two, which satisfies the rule.
Step 6: The layout is now completely fixed. Check each statement against it.
T = {music}, V = {camera, viewer}, W = {camera, music}, X = {camera, music, viewer}, Y = {viewer}, Z = {camera, music}.
T and V: {music} vs {camera, viewer}, no shared option, so "no options in common" holds.
T and Y: {music} vs {viewer}, no shared option, holds.
T and Z: {music} vs {camera, music}, share only music, exactly one option, holds.
W and Z: {camera, music} vs {camera, music}, share BOTH camera and music, that is two options in common, not one. This breaks the claim of "exactly one option in common."
Y and Z: {viewer} vs {camera, music}, no shared option, holds.
Final Answer:
Every statement checks out except the one about W and Z, which actually share two options, not one. That is the statement that is not forced to be true. \[ \boxed{\text{W and Z share two options, not one}} \]
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