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

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 3

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 4

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 5

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. \[ \boxed{10} \]