Problem Setup

Let the 7 friends' scores be $s_1, s_2, s_3, s_4, s_5, s_6, s_7$ in ascending order.

• Maximum score ($s_7$) = 25 (P1).
• Minimum score ($s_1$) = 0 (P7).
• All scores are distinct integers.
• Median score ($s_4$) = 15.
• Average score = 12.57.
• Two players (P2, P3) scored more than median but less than maximum: $15 < \text{score} < 25$.
• Their combined score: $P2_{score} + P3_{score} < 40$.
• One score $x$ is double the sum of two other non-zero scores $y, z$: $x = 2 \times (y + z)$, where $x, y, z \in S \setminus \{0\}$ and $y \neq z$.

The ordered scores are: $0 < s_2 < s_3 < 15 < s_5 < s_6 < 25$.

Score Calculation

The total sum of the 7 scores is calculated from the average:

$ \text{Sum} = \text{Average} \times 7 = 12.57 \times 7 = 87.99 $

Since scores are integers, the sum is 88.

$ s_1 + s_2 + s_3 + s_4 + s_5 + s_6 + s_7 = 88 $

Substituting the known values:

$ 0 + s_2 + s_3 + 15 + s_5 + s_6 + 25 = 88 $

$ s_2 + s_3 + s_5 + s_6 = 88 - 15 - 25 = 48 $

The scores $s_5$ and $s_6$ must correspond to P2 and P3, as they are the only scores strictly between the median (15) and maximum (25). Thus, $15 < s_5 < s_6 < 25$ and $s_5 + s_6 < 40$.

Analyzing Potential Scores (Second Highest Score, $s_6$)

We test the given options for $s_6$.

Option B: Second Highest Score is 20

Assume $s_6 = 20$.

We need $15 < s_5 < 20$ and $s_5 + 20 < 40$, which means $s_5 < 20$. Possible values for $s_5$ are 16, 17, 18, 19.

Let's test $s_5 = 18$. This satisfies $15 < 18 < 20$ and $s_5 + s_6 = 18 + 20 = 38 < 40$.

Now, find $s_2 + s_3$: $s_2 + s_3 = 48 - (s_5 + s_6) = 48 - 38 = 10$.

We need $0 < s_2 < s_3 < 15$ such that $s_2 + s_3 = 10$. Possible pairs $(s_2, s_3)$ are (1, 9), (2, 8), (3, 7), (4, 6).

Consider the set of scores $S = \{0, 2, 8, 15, 18, 20, 25\}$ (using $s_2=2, s_3=8$).

• Check distinct scores: Yes.
• Check order: $0 < 2 < 8 < 15 < 18 < 20 < 25$. Yes.
• Check sum: $0+2+8+15+18+20+25 = 88$. Average $88/7 \approx 12.57$. Yes.
• Check P2/P3 condition: Scores > 15 and < 25 are 18 and 20. Their sum is $18+20 = 38 < 40$. Yes.
• Check "double sum" condition: Non-zero scores are $\{2, 8, 15, 18, 20, 25\}$. Does any score $x$ equal $2(y+z)$? Yes, $20 = 2 \times (2+8)$. Here $x=20, y=2, z=8$.

All conditions are met. Thus, 20 is a possible value for the second highest score.

Option C: Second Highest Score is 21

Assume $s_6 = 21$.

We need $15 < s_5 < 21$ and $s_5 + 21 < 40$, which means $s_5 < 19$. Possible values for $s_5$ are 16, 17, 18.

Let's test $s_5 = 18$. This satisfies $15 < 18 < 21$ and $s_5 + s_6 = 18 + 21 = 39 < 40$.

Now, find $s_2 + s_3$: $s_2 + s_3 = 48 - (s_5 + s_6) = 48 - 39 = 9$.

We need $0 < s_2 < s_3 < 15$ such that $s_2 + s_3 = 9$. Possible pairs $(s_2, s_3)$ are (1, 8), (2, 7), (3, 6), (4, 5).

Consider the set of scores $S = \{0, 1, 8, 15, 18, 21, 25\}$ (using $s_2=1, s_3=8$).

• Check distinct scores: Yes.
• Check order: $0 < 1 < 8 < 15 < 18 < 21 < 25$. Yes.
• Check sum: $0+1+8+15+18+21+25 = 88$. Average $88/7 \approx 12.57$. Yes.
• Check P2/P3 condition: Scores > 15 and < 25 are 18 and 21. Their sum is $18+21 = 39 < 40$. Yes.
• Check "double sum" condition: Non-zero scores are $\{1, 8, 15, 18, 21, 25\}$. Does any score $x$ equal $2(y+z)$? Yes, $18 = 2 \times (1+8)$. Here $x=18, y=1, z=8$.

All conditions are met. Thus, 21 is a possible value for the second highest score.

Option A: Second Highest Score is 18

If $s_6 = 18$, possible $s_5$ values satisfying $15 < s_5 < 18$ and $s_5 + 18 < 40$ are $s_5 \in \{16, 17\}$.

If $s_5=17$, $s_5+s_6=35$. Then $s_2+s_3 = 48-35=13$. Possible $(s_2, s_3)$ are (4,9), (5,8), (6,7). Example set: $\{0, 4, 9, 15, 17, 18, 25\}$. Non-zero scores $\{4, 9, 15, 17, 18, 25\}$. We need $x=2(y+z)$. Possible sums $y+z$: $4+9=13 \implies x=26$ (no); $4+15=19 \implies x=38$ (no); $9+15=24 \implies x=48$ (no); $15+17=32 \implies x=64$ (no); $17+18=35 \implies x=70$ (no); $18+25=43 \implies x=86$ (no). It can be shown that no combination satisfies the double sum condition for $s_6=18$.

Option D: Second Highest Score is 22

If $s_6 = 22$, possible $s_5$ values satisfying $15 < s_5 < 22$ and $s_5 + 22 < 40$ are $s_5 \in \{16, 17\}$. ($s_5=18$ gives sum 40, which is not allowed).

If $s_5=17$, $s_5+s_6=39$. Then $s_2+s_3 = 48-39=9$. Possible $(s_2, s_3)$ are (1,8), (2,7), (3,6), (4,5). Example set: $\{0, 1, 8, 15, 17, 22, 25\}$. Non-zero scores $\{1, 8, 15, 17, 22, 25\}$. Check $x=2(y+z)$. Sums $y+z$: $1+8=9 \implies x=18$ (not in set); $1+15=16 \implies x=32$ (not in set); $8+15=23 \implies x=46$ (not in set); $15+17=32 \implies x=64$ (not in set); $17+22=39 \implies x=78$ (not in set); $22+25=47 \implies x=94$ (not in set). It can be shown that no combination satisfies the double sum condition for $s_6=22$.

Conclusion

The possible values for the second highest score are 20 and 21.