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CAT 2021 Slot 2 · DILR · Games and Tournaments · medium

CAT 2021 Slot 2 DILR: Chess Round-Robin with Rating Transfers

Five players – A, B, C, D and E – took part in a round-robin chess tournament in which every player played every other player exactly once. Before the tournament their ratings were:

| Player | A | B | C | D | E |
|---|---|---|---|---|---|
| Pre-tournament rating | 2400 | 2350 | 2300 | 2250 | 2200 |

After each game, rating points were transferred between the two players as follows. Let g be the difference between the two players' pre-tournament ratings (so g is always a multiple of 50), and let k = g/25.

- If the higher-rated player won, the loser transferred (10 − k) points to the winner.
- If the lower-rated player won, the loser transferred (10 + k) points to the winner.
- If the game was drawn, the higher-rated player transferred k points to the lower-rated player.

No other points were gained or lost. The table below gives each player's record in the tournament (W = wins, D = draws, L = losses).

| Player | W | D | L |
|---|---|---|---|
| A | 3 | 0 | 1 |
| B | 2 | 1 | 1 |
| C | 0 | 1 | 3 |
| D | 1 | 0 | 3 |
| E | 2 | 2 | 0 |

Question 1

Single correct option

Which players did E defeat?
  1. A. A and B
  2. B. A and D
  3. C. C and D
  4. D. B and D
Show answer and solution

Answer: (B) A and D

Deducing the results. Total draws = (0 + 1 + 1 + 0 + 2)/2 = 2 games. E drew twice and the only other players with a draw are B and C, so the draws are B–E and C–E. E lost nothing, so E beat A and D.

A lost only once (to E), so A beat B, C and D. C won nothing and drew only with E, so C lost to A, B and D. D's one win must then be over C, so D lost to B. B: beat C and D, lost to A, drew with E (2-1-1 ✓).

| Game | Result | k | Points moved |
|---|---|---|---|
| A–B | A won | 2 | 8 to A |
| A–C | A won | 4 | 6 to A |
| A–D | A won | 6 | 4 to A |
| A–E | E won (upset) | 8 | 18 to E |
| B–C | B won | 2 | 8 to B |
| B–D | B won | 4 | 6 to B |
| B–E | draw | 6 | 6 to E |
| C–D | D won (upset) | 2 | 12 to D |
| C–E | draw | 4 | 4 to E |
| D–E | E won (upset) | 2 | 12 to E |

Final ratings: A 2400, B 2350, C 2270, D 2240, E 2240.

So E defeated A and D and drew with B and C.

Question 2

Numerical answer

What was C's rating at the end of the tournament?
Show answer and solution

Answer: 2270

C lost to A (higher-rated won, k = 4: C gives 6), lost to B (k = 2: gives 8), lost to D (lower-rated D won, k = 2: C gives 10 + 2 = 12) and drew with E (C is higher-rated, k = 4: C gives 4).

Total loss = 6 + 8 + 12 + 4 = 30, so C ends at 2300 − 30 = 2270.

Deducing the results. Total draws = (0 + 1 + 1 + 0 + 2)/2 = 2 games. E drew twice and the only other players with a draw are B and C, so the draws are B–E and C–E. E lost nothing, so E beat A and D.

A lost only once (to E), so A beat B, C and D. C won nothing and drew only with E, so C lost to A, B and D. D's one win must then be over C, so D lost to B. B: beat C and D, lost to A, drew with E (2-1-1 ✓).

| Game | Result | k | Points moved |
|---|---|---|---|
| A–B | A won | 2 | 8 to A |
| A–C | A won | 4 | 6 to A |
| A–D | A won | 6 | 4 to A |
| A–E | E won (upset) | 8 | 18 to E |
| B–C | B won | 2 | 8 to B |
| B–D | B won | 4 | 6 to B |
| B–E | draw | 6 | 6 to E |
| C–D | D won (upset) | 2 | 12 to D |
| C–E | draw | 4 | 4 to E |
| D–E | E won (upset) | 2 | 12 to E |

Final ratings: A 2400, B 2350, C 2270, D 2240, E 2240.

Question 3

Numerical answer

What was the largest number of rating points transferred in any single game of the tournament?
Show answer and solution

Answer: 18

The biggest transfers come from upsets across large rating gaps. The upsets were E over A (g = 200, k = 8: 10 + 8 = 18), D over C (k = 2: 12) and E over D (k = 2: 12). No win by a higher-rated player can move more than 10 − 2 = 8 points and no draw more than 8.

Maximum = 18 (E's win over A).

Deducing the results. Total draws = (0 + 1 + 1 + 0 + 2)/2 = 2 games. E drew twice and the only other players with a draw are B and C, so the draws are B–E and C–E. E lost nothing, so E beat A and D.

A lost only once (to E), so A beat B, C and D. C won nothing and drew only with E, so C lost to A, B and D. D's one win must then be over C, so D lost to B. B: beat C and D, lost to A, drew with E (2-1-1 ✓).

| Game | Result | k | Points moved |
|---|---|---|---|
| A–B | A won | 2 | 8 to A |
| A–C | A won | 4 | 6 to A |
| A–D | A won | 6 | 4 to A |
| A–E | E won (upset) | 8 | 18 to E |
| B–C | B won | 2 | 8 to B |
| B–D | B won | 4 | 6 to B |
| B–E | draw | 6 | 6 to E |
| C–D | D won (upset) | 2 | 12 to D |
| C–E | draw | 4 | 4 to E |
| D–E | E won (upset) | 2 | 12 to E |

Final ratings: A 2400, B 2350, C 2270, D 2240, E 2240.

Question 4

Single correct option

Which of the following players finished the tournament with exactly the same rating they started with?
  1. A. A only
  2. B. B only
  3. C. A and B
  4. D. B and E
Show answer and solution

Answer: (C) A and B

A: +8 (vs B) + 6 (vs C) + 4 (vs D) − 18 (vs E) = 0.
B: −8 (vs A) + 8 (vs C) + 6 (vs D) − 6 (draw with E) = 0.
C: −30, D: −4 − 6 + 12 − 12 = −10, E: +18 + 6 + 4 + 12 = +40.

So A and B both finish exactly where they started. (Check: net changes 0 + 0 − 30 − 10 + 40 = 0 ✓.)

Deducing the results. Total draws = (0 + 1 + 1 + 0 + 2)/2 = 2 games. E drew twice and the only other players with a draw are B and C, so the draws are B–E and C–E. E lost nothing, so E beat A and D.

A lost only once (to E), so A beat B, C and D. C won nothing and drew only with E, so C lost to A, B and D. D's one win must then be over C, so D lost to B. B: beat C and D, lost to A, drew with E (2-1-1 ✓).

| Game | Result | k | Points moved |
|---|---|---|---|
| A–B | A won | 2 | 8 to A |
| A–C | A won | 4 | 6 to A |
| A–D | A won | 6 | 4 to A |
| A–E | E won (upset) | 8 | 18 to E |
| B–C | B won | 2 | 8 to B |
| B–D | B won | 4 | 6 to B |
| B–E | draw | 6 | 6 to E |
| C–D | D won (upset) | 2 | 12 to D |
| C–E | draw | 4 | 4 to E |
| D–E | E won (upset) | 2 | 12 to E |

Final ratings: A 2400, B 2350, C 2270, D 2240, E 2240.

Question 5

Single correct option

If E had won the game against B instead of drawing it (all other results unchanged), what would E's final rating have been?
  1. A. 2246
  2. B. 2250
  3. C. 2256
  4. D. 2234
Show answer and solution

Answer: (B) 2250

For B–E, g = 150 so k = 6. In the actual draw E gained k = 6. Had E (the lower-rated player) won, E would have gained 10 + 6 = 16.

E's actual final rating = 2200 + 18 + 6 + 4 + 12 = 2240. Replacing the +6 by +16: 2240 − 6 + 16 = 2250.

(2256 adds 16 without removing the 6; 2246 uses the higher-rated winner's 10 − k.)

Deducing the results. Total draws = (0 + 1 + 1 + 0 + 2)/2 = 2 games. E drew twice and the only other players with a draw are B and C, so the draws are B–E and C–E. E lost nothing, so E beat A and D.

A lost only once (to E), so A beat B, C and D. C won nothing and drew only with E, so C lost to A, B and D. D's one win must then be over C, so D lost to B. B: beat C and D, lost to A, drew with E (2-1-1 ✓).

| Game | Result | k | Points moved |
|---|---|---|---|
| A–B | A won | 2 | 8 to A |
| A–C | A won | 4 | 6 to A |
| A–D | A won | 6 | 4 to A |
| A–E | E won (upset) | 8 | 18 to E |
| B–C | B won | 2 | 8 to B |
| B–D | B won | 4 | 6 to B |
| B–E | draw | 6 | 6 to E |
| C–D | D won (upset) | 2 | 12 to D |
| C–E | draw | 4 | 4 to E |
| D–E | E won (upset) | 2 | 12 to E |

Final ratings: A 2400, B 2350, C 2270, D 2240, E 2240.
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Original question written in the CAT 2021 Slot 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake