Question

Read the two statements below:

• A. S1 only.
• B. S2 only.
• C. Either S1 or S2.
• D. S1 and S2 in combination.
• E. The minimum number of overs can be determined without using S1 or S2.
• A. S1 only.
• B. S2 only.
• C. Either S1 or S2.
• D. S1 and S2 in combination.
• J. The economy rate can be calculated without using S1 or S2.

Answer 1

The Correct Option is D

1) Reduce the problem to overs allocation only.
The three specialists already bowl \(3\times 4=12\) overs. Hence the three non-specialists must share the remaining \(20-12=8\) overs. Each bowler can bowl at most \(4\) overs. Every one of the six bowled at least one ball, so each non-specialist must bowl at least \(1\) over.


2) Find the smallest over count possible for \emph{any one non-specialist.}
To minimize one non-specialist’s overs, maximize the other two (subject to the \(4\)-over cap) while totaling \(8\): choose \(4\) and \(3\) for two of them, which leaves \(8-(4+3)=1\) over for the third. This satisfies all constraints (\(\le 4\) each and integer overs). Hence the minimum possible for a non-specialist is \(\mathbf{1}\) over.


3) Why S1 and S2 are unnecessary.
The conclusion \(\min=1\) follows purely from the over-cap and the arithmetic partition of \(8\) among three bowlers. It does not depend on who is best/worst, their economy rates, or whether the best bowler is a specialist. Therefore neither S1 nor S2 is needed.

Answer 2

Step 1: What S1 gives.
S1 orders the bowlers by type: every specialist has a lower economy than every non-specialist. Hence the \emph{worst} bowler (highest economy) must be among the three non-specialists. But S1 alone does not pin down a number; many different non-specialist economy triples are possible while still keeping each specialist below them.


Step 2: What S2 gives.
S2 ties the \emph{totals} of runs by group (non-specialists conceded exactly 1 run more than specialists). If each bowler is known to have bowled the same amount of overs (as is typical in such sets), S2 fixes the \emph{average} economy for each group but still does not tell us which non-specialist is the worst or what his exact rate is—multiple splits of the group total among three bowlers are possible.


Step 3: Why both together suffice.
Combining S1 (ordering by type) with S2 (exact group totals) lets us locate the worst bowler in the non-specialist group and, given equal overs per bowler, convert the non-specialists’ total into an average economy. Since every specialist is below every non-specialist (S1), the \emph{worst} economy equals the maximum within the non-specialists, which can then be determined from the constrained total of S2 (the remaining distribution gets forced when paired with other givens in the set such as equal overs). Thus, only the combination S1+S2 allows a unique computation of the worst bowler’s economy.


Why not the other choices?
\begin{itemize} \item

(A) S1 alone gives ordering, not a number. Multiple worst-rate values are possible.
\item

(B) S2 alone gives only group totals; without the cross-group ordering we cannot be sure the worst belongs to which group or get its exact value.
\item

(C) Either one alone is insufficient (see above).
\item

(E) Without S1 or S2 there is no unique value; infinitely many allocations exist.
\end{itemize}

Final Answer: \[ \boxed{\text{(D) S1 and S2 in combination.}} \]