在年度最佳步兵单位竞赛中，3 个顶尖单位——A 单位、B 单位和 C 单位入围最终冠军的争夺。组织者决定为每个 NAPFA 金奖、射击优秀徽章和陆军精锐徽章分别授予 $x$、$y$ 和 $z$ 分，并宣布总分最高的单位为冠军。为了强调各单位培养更多神枪手，组织者决定将每个射击优秀徽章授予的分数设为 NAPFA 金奖分数与陆军精锐徽章分数之和。 各单位获得的 NAPFA 金奖、射击优秀徽章和陆军精锐徽章数量汇总于下表：

	NAPFA 金奖	射击优秀徽章	陆军精锐徽章
A 单位	24	27	118
B 单位	29	21	124
C 单位	26	25	121

计算得分后得知，获得 NAPFA 金奖最多的单位总得分为 295 分，落后获得射击优秀徽章最多的单位 3 分。 通过求解包含 $x$、$y$ 和 $z$ 的方程组，确定该竞赛的最终冠军。

1. Formulate an equation relating the points for each type of badge.

From the problem, the points for each Shooting Marksmanship Badge equals the sum of points for an NAPFA Gold Award and an Army Proficiency Badge:
$y = x + z$

2. Formulate an equation for the unit with the most NAPFA Gold awards.

From the table, Unit B has the most NAPFA Gold awards (29).
We are given that its total score is 295 points.
$29x + 21y + 124z = 295$

3. Formulate an equation for the unit with the most Marksmanship Badges.

From the table, Unit A has the most Marksmanship Badges (27).
Unit B falls behind Unit A by 3 points, meaning Unit A scores $295 + 3 = 298$ points.
$24x + 27y + 118z = 298$

4. Solve the system of equations for $x$, $y$, and $z$.

Substitute $y = x + z$ into the equations for Unit B and Unit A:

For Unit B:
$29x + 21(x + z) + 124z = 295$
$50x + 145z = 295 \quad \text{--- (1)}$

For Unit A:
$24x + 27(x + z) + 118z = 298$
$51x + 145z = 298 \quad \text{--- (2)}$

Subtracting (1) from (2) eliminates $z$:
$(51x - 50x) + (145z - 145z) = 298 - 295$
$x = 3$

Substitute $x = 3$ into (1):
$50(3) + 145z = 295$
$150 + 145z = 295$
$145z = 145 \implies z = 1$

Substitute $x$ and $z$ back to find $y$:
$y = 3 + 1 = 4$

5. Calculate the total score for Unit C and identify the champion.

Calculate the total points for Unit C using its tally from the table:
$\text{Score for Unit C} = 26x + 25y + 121z$
$= 26(3) + 25(4) + 121(1)$
$= 78 + 100 + 121 = 299$

Comparing the total scores:

• Unit A: 298 points
• Unit B: 295 points
• Unit C: 299 points

Since $299 > 298 > 295$, Unit C has the highest total points.