CAT 2021 Slot 2 DILR Set 4: Games & Tournaments (6 Questions) with Solutions

Clue 1: the only 10s are Barat's for o9 and Elise's for o10, so Barat holds o9 and Elise holds o10.

Barat's partner object must have an even value of at most 6, so that his bundle is even and at most 16. Only o7 (6) fits, since o8 would give 18. Barat = {o7, o9} = 16.

Elise's partner must also have an even value of at most 6. That leaves o1 or o5 (both 6), so Elise = 16.

Amar values Barat's bundle at $9+8=17$. At 16 he would envy a 16-person, which clue 6 forbids, so Amar is not at 16. Disha is odd, so Charles is the third person at 16.

Charles reaches 16 only with two 8s from o1, o2, o3, o8. Clue 2 bars two of o1 to o3, and clue 3 bars o1 with o8. That leaves {o8, o2} or {o8, o3}.

Barat values o2 and o8 at $9+8=17>16$, so he would envy. Clue 6 forbids this, so Charles = {o3, o8}.

Any person at 16 who holds o8 must be Charles, and the 16-group is Barat, Elise and Charles. Amar cannot hold o8 because that would put him at 16 with o2 or o3, which step 4 rules out.

Clue 1 gives Barat o9 and Elise o10 (the only 10s). Barat's partner must have an even value of at most 6, which leaves only o7. Barat = {o7, o9} = 16.

Elise's partner is o1 or o5 (both 6), so Elise = 16.

Amar values Barat's bundle at $9+8=17$, so he cannot be at 16. Disha is odd, so Charles is the third 16.

Charles needs two 8s from o1, o2, o3, o8. Clues 2 and 3 leave {o8, o2} or {o8, o3}, and o2 would make Barat value Charles's bundle at $9+8=17$. So Charles = {o3, o8}.

The remaining split gives Amar {o2, o6} = 12 and Disha {o1, o4} = 13, with Elise {o5, o10}. Elise's other option, o1, makes Amar envy Disha, which clue 6 forbids.

The 16-people are Barat, Elise and Charles. Clue 6 says only people below 16 can envy, so none of the three envies anyone.

Amar (12) values Barat's bundle at $9+8=17>12$, so Amar envies Barat. Disha (13) also envies, because she values Barat's bundle at $9+6=15$.

Clue 1: Barat holds o9 and Elise holds o10. Barat's partner must have an even value of at most 6, which leaves only o7. Barat = {o7, o9} = 16.

Elise's partner is o1 or o5 (both 6), so Elise = 16.

Amar values Barat's bundle at $9+8=17$, so he is not at 16. Disha is odd, so Charles is the third 16, and Charles = {o3, o8} (two 8s, with o2 excluded because Barat would value o2 and o8 at 17).

If Elise held o1, then o2, o4, o5, o6 would go to Amar and Disha. Disha odd forces o2 with one other object. Amar would then envy Disha every time, for example $9+3=12>10$ in one case and $9+7=16>6$ in another. Clue 6 forbids this, so Elise = {o5, o10}.

The objects left are o1, o2, o4, o6. Clue 2 splits o1 and o2 between Amar and Disha. Amar with o1 would have $4+3=7$, which is odd, so Amar holds o2 and Disha holds o1.

If Disha held o6, Amar would hold o4 and Disha would value Amar's bundle at $8+5=13>11$. That is an envy between two people below 16, so Disha = {o1, o4} and Amar = {o2, o6}.

Amar's value is $9+3=12$.

Clue 1 gives Barat o9 and Elise o10. Barat's partner must have an even value of at most 6, so it is o7. Barat = {o7, o9} = 16.

Elise's partner is o1 or o5, so Elise = 16.

Amar values Barat's bundle at $9+8=17$, so he is below 16. Disha is odd, so Charles is the third 16, with two 8s. Clues 2 and 3, plus Barat's $9+8=17$ for o2 and o8, give Charles = {o3, o8}.

If Elise held o1, Disha (odd) would need o2, and Amar would envy Disha in every split of o4, o5, o6. Clue 6 forbids this, so Elise = {o5, o10}.

The objects left are o1, o2, o4, o6. Clue 2 splits o1 and o2 between Amar and Disha. Amar with o1 would be at 7, which is odd, so Amar has o2 and Disha has o1.

Each of them takes one of o4 and o6. Suppose Disha takes o6 (value 11) and Amar takes o4. Disha values Amar's bundle at $8+5=13>11$, so she would envy Amar, who is also below 16. Clue 6 forbids that.

So Disha takes o4, giving Disha {o1, o4} = 13, and Amar has {o2, o6} = 12.

Clue 1 gives Barat o9 and Elise o10. Barat's partner must have an even value of at most 6, which leaves only o7. Barat = {o7, o9} = 16.

Elise's partner needs an even value of at most 6: o1 or o5 (both 6). Elise = 16 either way.

Amar values Barat's bundle at $9+8=17$, so he is below 16. Disha is odd, so Charles is the third 16. Clues 2 and 3 leave {o8, o2} or {o8, o3}, and Barat would value o2 and o8 at $9+8=17$. So Charles = {o3, o8}.

Test Elise = {o1, o10}. Then o2, o4, o5, o6 go to Amar and Disha. Disha is odd, so she holds o2 plus one more ($8+5$ or $8+3$).

In every case Amar would envy Disha. Both are below 16, and clue 6 allows envy only toward 16-people. So Elise's partner is not o1.

Elise = {o5, o10}. The rest then gives Amar {o2, o6}, Disha {o1, o4}.

Clue 1 gives Barat o9 and Elise o10. Barat's partner must have an even value of at most 6, which leaves only o7. Barat = {o7, o9} = 16.

Elise's partner is o1 or o5 (both 6). Elise = 16.

Amar values Barat's bundle at $9+8=17$, so he is not at 16. Disha is odd, so Charles is the third 16. Clues 2 and 3 limit him to {o8, o2} or {o8, o3}, and Barat would value o2 and o8 at 17. So Charles = {o3, o8}, which excludes o1 for Charles.

So o1 is with Elise, Amar or Disha. If Elise held o1, Disha (odd) would need o2 plus one of o4, o5, o6. Amar would then value Disha's bundle above his own every time ($12>10$, $16>6$, $12>10$), and clue 6 forbids envy between two people below 16. Elise = {o5, o10}.

The objects left are o1, o2, o4, o6. Clue 2 splits o1 and o2 between Amar and Disha. Amar with o1 would total $4+3=7$, which is odd, but only Disha is odd. So Amar holds o2 and Disha holds o1.

Disha's partner is o4. With o6 she would have 11 and would value Amar's bundle at $8+5=13$, which is an envy between two people below 16.