Question

Player S is from the weight group:

• A. 1
• B. 9
• C. 5
• D. 10
• A. 1 and 3
• B. Both from group 7
• C. 4 and 10
• D. 5 and 9
• A. 68 kg
• B. 66 kg
• C. 69 kg
• D. Cannot be determined
• A. 10
• B. 8
• C. 1
• D. Cannot be determined
• A. 5
• B. 7
• C. 10
• D. None of the above

Answer 1

The Correct Option is B

Step 1: Convert averages to totals. Selecting one player from each of the 10 groups gives an average of \(68\) kg. \[ \text{Total of 10 players} = 10 \times 68 = 680\ \text{kg}. \] Step 2: Remove \(S\) and compute the remaining total. After \(S\) leaves, \(9\) players average \(66.5\) kg. \[ \text{Total of remaining 9} = 9 \times 66.5 = 598.5\ \text{kg}. \] Step 3: Weight of \(S\). \[ w_S = 680 - 598.5 = 81.5\ \text{kg}. \] Weight group 9 covers \(80\text{–}84\) kg, which contains \(81.5\). \[ \boxed{9} \]

Answer 2

This question can be solved using a quick average-shift shortcut instead of computing both totals separately. When one player leaves a group of \(n\) and the average falls, the weight of the player who left equals the old average plus \((n-1)\) times the drop in average.

Here, the group started with \(n = 10\) players averaging 68 kg, and after S left, the remaining 9 averaged 66.5 kg, a drop of \(68 - 66.5 = 1.5\) kg. Applying the shortcut: \[ w_S = 68 + (10 - 1) \times 1.5 = 68 + 13.5 = 81.5 \text{ kg}. \]

Now checking which weight group contains 81.5 kg:

1. Group 1 (48–52 kg): This range is far below 81.5 kg, so S cannot belong here.
2. Group 9 (80–84 kg): 81.5 kg falls within this range, matching S's weight.
3. Group 5 (64–68 kg): This range is well below 81.5 kg and does not fit.
4. Group 10 (84–88 kg): 81.5 kg is below the lower bound of this range, so it does not fit either.

S's weight of 81.5 kg places them in the group covering 80–84 kg.

Therefore, the correct answer is Group 9.

Answer 3

Step 1: Total needed with two newcomers. After \(S\) left, the remaining total was \(598.5\) kg (from Q146). With two new players the team has \(11\) players with average \(68\) kg: \[ \text{New total} = 11 \times 68 = 748\ \text{kg}. \] \[ \Rightarrow \text{Sum of the two newcomers} = 748 - 598.5 = 149.5\ \text{kg}. \] Step 2: Check which pair of groups \textbf{cannot achieve a sum of \(149.5\) kg.} Use interval sums (all in kg): \[ \begin{aligned} \text{G1+G3: } &[48,52]+[56,60]=[104,112](\not\ni 149.5)
\text{G7+G7: } &[72,76]+[72,76]=[144,152](\ni 149.5)
\text{G4+G10: } &[60,64]+[84,88]=[144,152](\ni 149.5)
\text{G5+G9: } &[64,68]+[80,84]=[144,152](\ni 149.5)
\end{aligned} \] Only \(G1+G3\) cannot total \(149.5\) kg. \[ \boxed{\text{1 and 3}} \]

Answer 4

Since two newcomers with a combined weight of 149.5 kg join to raise the average to 68 kg, together they must average \( \frac{149.5}{2} = 74.75 \) kg. The question is which pair of groups can, and cannot, produce a combined weight of 149.5 kg from their weight ranges.

1. 1 and 3: Group 1 covers 48–52 kg and Group 3 covers 56–60 kg. Even taking the heaviest possible player from each, the combined maximum is only \( 52 + 60 = 112 \) kg, far short of the 149.5 kg required, so this pairing is impossible.
2. Both from group 7: Group 7 covers 72–76 kg, so two players from this group can together weigh anywhere from \( 72+72=144 \) kg to \( 76+76=152 \) kg, a range that comfortably includes 149.5 kg.
3. 4 and 10: Group 4 covers 60–64 kg and Group 10 covers 84–88 kg, giving a combined range of 144 to 152 kg, which also includes 149.5 kg.
4. 5 and 9: Group 5 covers 64–68 kg and Group 9 covers 80–84 kg, giving a combined range of 144 to 152 kg, which again includes 149.5 kg.

Groups 1 and 3 are the only pair whose combined weight can never reach 149.5 kg, no matter which individuals within those ranges are chosen.

Therefore, the correct answer is 1 and 3.

Answer 5

Step 0: Recall the facts established in earlier parts.

With one player chosen from each of the 10 groups, the average weight was \(68\) kg.
\(\Rightarrow\) Total of the original 10 \(=10\times 68=680\) kg.
Player \(S\) left; the remaining \(9\) averaged \(66.5\) kg.
\(\Rightarrow\) Total of the remaining 9 \(=9\times 66.5=598.5\) kg.
So \(S\)'s weight \(=680-598.5=81.5\) kg (lies in Group 9).
Two newcomers then joined and the new \(11\)-player average became \(68\) kg.
\(\Rightarrow\) Total of those 11 \(=11\times 68=748\) kg.
Hence, sum of the two newcomers \(=748-598.5=149.5\) kg.

Step 1: What does “all players taken together” mean here?
It refers to the original 10 (before \(S\) left) plus the two newcomers.
Therefore, the size of the combined group is \(10+2=12\) players.

Step 2: Compute the combined total for these 12 players.
\[ \text{Combined total} = \underbrace{\text{(original 10 total)}}_{680} + \underbrace{\text{(two newcomers total)}}_{149.5} = 680 + 149.5 = 829.5\ \text{kg}. \]

Step 3: Divide by the combined headcount to get the average.
\[ \bar{w}_{12} = \frac{829.5}{12} = 69.125\ \text{kg}. \] Given the answer choices are integers (to the nearest kilogram), this rounds to \(\boxed{69\ \text{kg}}\).

(Cross-check via deviations method)
Take \(68\) kg as a convenient reference.

The original 10 average is \(68\) kg \(\Rightarrow\) total deviation from \(68\) is \(0\).
The two newcomers together weigh \(149.5\) kg \(\Rightarrow\) their deviation from \(68\) each (over two people) is: \[ 149.5 - 2\times 68 = 149.5 - 136 = 13.5\ \text{kg}. \] Over \(12\) players, the average deviation from \(68\) is \(\dfrac{13.5}{12} = 1.125\) kg.
So new overall average \(=68+1.125=69.125\) kg \(\Rightarrow\) \(69\) kg (rounded).

(Sanity bounds)
Each newcomer is from some group interval. From the earlier part, we only need their sum \(=149.5\) kg; hence the 12-player average must exceed \(68\) (since we added a positive deviation of \(13.5\) kg across 12 players). Thus the only plausible integer option \(> 68\) is \(69\) kg, matching our computation.

Final Answer:
\[ \boxed{69\ \text{kg}} \]

Answer 6

Rather than summing raw totals, this can be solved using the alligation (weighted-average) method, treating the original 10 players and the two newcomers as two groups being merged.

The original 10 players average 68 kg. The two newcomers together weigh 149.5 kg (established earlier in this set), so they average \( \frac{149.5}{2} = 74.75 \) kg. When two groups of sizes 10 and 2 with averages 68 kg and 74.75 kg are combined, the combined average splits the gap between them in the inverse ratio of the group sizes: \[ \text{ratio} = 2 : 10 = 1 : 5. \]

The gap between the two averages is \( 74.75 - 68 = 6.75 \) kg. Since the combined average lies closer to the larger group's average (68 kg), it sits \( \frac{1}{1+5} = \frac{1}{6} \) of the way from 68 kg toward 74.75 kg: \[ \text{combined average} = 68 + \frac{1}{6} \times 6.75 = 68 + 1.125 = 69.125 \text{ kg}. \]

Rounded to the nearest kilogram, this is 69 kg. Because both subgroup totals are fixed by the data, this combined average is pinned down exactly, unlike the identity of the single heaviest group elsewhere in this set.

Therefore, the correct answer is 69 kg.

Answer 7

What does “contributes most” mean here?
With one player from each group, the overall average of the original 10 is fixed at \(68\) kg.
Since each group contributes exactly one player, the only way a group can “contribute more” to pulling the average upward is by having the heaviest individual (i.e., the largest positive deviation above \(68\)).
But we are not given the exact chosen weight from each interval—only the ranges:

\[ \begin{aligned} &\text{G1 }[48,52],\ \text{G2 }[52,56],\ \text{G3 }[56,60],\ \text{G4 }[60,64],\ \text{G5 }[64,68],\\ &\text{G6 }[68,72],\ \text{G7 }[72,76],\ \text{G8 }[76,80],\ \text{G9 }[80,84],\ \text{G10 }[84,88]. \end{aligned} \]
The total of the 10 players must be \(680\) kg (average \(68\) kg), but there are infinitely many ways to choose one value from each interval to sum to \(680\).
Therefore, the identity of the group with the heaviest player (hence “largest contribution”) can vary across valid selections.

Constructive proof of indeterminacy (two valid selections with different top contributors):

Baseline fact: Choosing the midpoint of each interval gives the required total \(680\) (since \(50+54+\cdots+86=680\)).

Case A (Group 10 is clearly the heaviest):
Take midpoints for all groups except adjust G10 to its maximum and compensate by lowering G1 by the same amount:
\[ \begin{aligned} &\text{G1}=48,\ \text{G2}=54,\ \text{G3}=58,\ \text{G4}=62,\ \text{G5}=66,\\ &\text{G6}=70,\ \text{G7}=74,\ \text{G8}=78,\ \text{G9}=82,\ \text{G10}=88. \end{aligned} \]
Sum \(=680\) kg, average \(=68\) kg.
Here the heaviest player is \(\mathbf{88}\) kg from \(\mathbf{G10}\) \(\Rightarrow\) G10 “contributes most”.

Case B (G10 is not uniquely the heaviest):
Pick midpoints, then set \(\text{G9}=84\) (its max, +2) and \(\text{G10}=84\) (its min, −2). Net effect on the total is \(0\), so the sum remains \(680\):
\[ \begin{aligned} &\text{G1}=50,\ \text{G2}=54,\ \text{G3}=58,\ \text{G4}=62,\ \text{G5}=66,\\ &\text{G6}=70,\ \text{G7}=74,\ \text{G8}=78,\ \text{G9}=84,\ \text{G10}=84. \end{aligned} \]
Now the maximum weight is \(84\) kg and \(\mathbf{G9}\) ties \(\mathbf{G10}\).
Thus, depending on feasible choices within intervals, “the group that contributes most” can be G10, G9, or even a tie.

Because the problem gives only ranges (not exact chosen weights), the identity of the top contributor is not uniquely determined.

\[ \boxed{\text{Cannot be determined}} \]

Answer 8

This question asks which single weight group is guaranteed to contribute the heaviest player once one player is picked from each of the 10 groups, given only that their total is 680 kg (for an average of 68 kg). Since only the weight range of each group is known, not the exact weight chosen, it helps to check whether the total constraint forces any one group to always hold the top spot.

1. Group 10: If Group 10's player is picked near the top of its 84–88 kg range while other groups are picked near their midpoints, Group 10 easily holds the highest weight. But this is just one valid selection among many satisfying the 680 kg total, not something the data forces.
2. Group 8: A selection could push Group 8's player toward the top of its 76–80 kg range while lowering another group's player by the same amount elsewhere to keep the total at 680 kg. Even so, Group 8's maximum of 80 kg still cannot beat groups whose ranges reach higher, so it is never forced to be the top contributor.
3. Group 1: Group 1's range (48–52 kg) is the lowest of all ten, so it can never hold the single heaviest player in any valid selection; this option can be ruled out entirely.
4. Cannot be determined: Because the passage fixes only the total (680 kg) and not the individual weights, and Group 9's range (80–84 kg) touches Group 10's range (84–88 kg) at exactly 84 kg, weight can be shifted between these two groups while keeping the total fixed. Depending on which valid values are chosen, either Group 9 or Group 10 can end up holding the heaviest player, so no single group is guaranteed to contribute it in every valid case.

Since different valid selections of one player per group (all consistent with the 680 kg total) can make different groups hold the top weight, the group that “contributes most” cannot be pinned down from the given information.

Therefore, the correct answer is Cannot be determined.

Answer 9

Step 1: Fix the required sum of the two newcomers. From Q147: after \(S\) left, the remaining \(9\) players totaled \(598.5\) kg; when two newcomers joined, the new \(11\)-player average became \(68\) kg, so \[ \text{New 11-player total} = 11\times 68 = 748\ \text{kg}. \] Hence the sum of the two newcomers is \[ \text{Sum(newcomers)} = 748 - 598.5 = 149.5\ \text{kg}. \] Step 2: One newcomer is from Group 4 \([60,64]\). Find the needed weight of the other. Let the Group-4 player weigh \(w_4 \in [60,64]\). Then the other newcomer must weigh \[ w_{\text{other}} = 149.5 - w_4 \in [149.5-64,\ 149.5-60] = [85.5,\ 89.5]\ \text{kg}. \] Step 3: Match \(w_{\text{other}}\) to feasible group ranges. Group ranges (kg): \[ \begin{aligned} &\text{G5 }[64,68], \text{G7 }[72,76], \text{G9 }[80,84], \text{G10 }[84,88]. \end{aligned} \] We require \(w_{\text{other}} \in [85.5,89.5]\).
G5 tops at \(68\) \(\Rightarrow\) \([64,68]\) does not reach \(85.5\). \(\Rightarrow\) impossible.
G7 tops at \(76\) \(\Rightarrow\) \([72,76]\) does not reach \(85.5\). \(\Rightarrow\) impossible.
G9 tops at \(84\) \(\Rightarrow\) \([80,84]\) does not reach \(85.5\). \(\Rightarrow\) impossible.
G10 is \([84,88]\). Intersection with \([85.5,89.5]\) is \([85.5,88]\), which is non-empty. \(\Rightarrow\) feasible.
Step 4: Existence check via interval algebra (robust). We can also solve for values of \(w_4\) that make \(w_{\text{other}}\) fall inside G10: \[ w_{\text{other}} \in [84,88] \;\Longleftrightarrow\; 149.5 - w_4 \in [84,88] \;\Longleftrightarrow\; w_4 \in [149.5-88,\ 149.5-84] = [61.5,\ 65.5]. \] Intersecting with G4 \([60,64]\) gives \([61.5,64]\), which is non-empty. So choose, for example, \(w_4=62\) kg (valid in G4), then \(w_{\text{other}}=149.5-62=87.5\) kg, which lies in G10 \([84,88]\). Thus a consistent assignment exists \(\Rightarrow\) the other player must be from Group 10. \[ \boxed{10} \]

Answer 10

The two newcomers together weigh 149.5 kg. One is known to be from Group 4 (60–64 kg); the question asks which group the other newcomer could be from, testing each option by checking whether some valid weight in Group 4 combined with some valid weight in that option can total 149.5 kg.

1. Group 5 (64–68 kg): The highest possible combined weight is \( 64 + 68 = 132 \) kg (using each group's maximum), which falls short of 149.5 kg, so this group cannot supply the other newcomer.
2. Group 7 (72–76 kg): The highest possible combined weight is \( 64 + 76 = 140 \) kg, still short of 149.5 kg, ruling this group out as well.
3. Group 10 (84–88 kg): Taking the Group 4 player near the middle of their range, say 62 kg, the Group 10 player would need to weigh \( 149.5 - 62 = 87.5 \) kg, which lies comfortably inside Group 10's 84–88 kg range, so this pairing is achievable.
4. None of the above: Since Group 10 has already been shown to work, this option does not apply.

Only Group 10 can supply a weight that, combined with a valid Group 4 weight, reaches the required 149.5 kg total.

Therefore, the correct answer is 10.