Question

Team A is a cricket team picking its playing eleven from its regular pool of batsmen for a league where it faces Teams B, C and D. The table below lists the past performance record of Team A's top 10 batsmen: their career batting average, and three tendencies expressed as a percentage of their innings, namely how often they get out for under 20 runs, how often they get out for a score close to their own average, and how often they go on to score more than a century.

For reference, the average score of the top 5 batsmen of each opposing team is: Team C, 270 runs; Team B, 215 runs; Team D, 180 runs; Team A (itself), 215 runs.

Team A would play the third match with B. Based on the statistics above, whom should the manager choose so that A has maximum chances of winning?

• A. RD, RU, MK, VS, YS
• B. RD, VS, MT, RU, YS
• C. ST, RD, MK, MD, SG
• D. RD, VV, SG, VS, MD

Hint:

Add up each group's total career average first, then use the dismissed-below-20 percentage as a tie-breaker for risk.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Team A must send in 5 of its 10 listed batsmen for the third match, against B. The table gives each batsman's career average and three tendencies, as a percentage of their innings: how often they fall for under 20 runs, how often they get out near their own average, and how often they cross a century. We need the group of 5 whose record gives A the best chance of winning.

Step 2: Key Formula or Approach:
No single column alone decides who is best, so the fair way to compare five-man groups is to add up each group's career averages, which stands in for the runs the group is likely to score, and then check the group's average failure rate (percentage dismissed below 20), which stands in for risk. A group with a higher combined average and a lower combined failure rate should give A a better total.

Step 3: Detailed Explanation:
The batsmen's averages are: RD 40, ST 44, SG 41, VS 31, RU 28, YS 35, VV 35, MK 30, MT 36, MD 45. Add up the average of the 5 batsmen named in each option:
Option A, RD + RU + MK + VS + YS: \(40+28+30+31+35=164\)
Option B, RD + VS + MT + RU + YS: \(40+31+36+28+35=170\)
Option C, ST + RD + MK + MD + SG: \(44+40+30+45+41=200\)
Option D, RD + VV + SG + VS + MD: \(40+35+41+31+45=192\)
Option E, SG + RU + YS + MK + VV: \(41+28+35+30+35=169\)
Option C has by far the highest combined average, 200 runs, well ahead of D at 192, and far ahead of A, B and E, which all sit below 170 because they carry weaker batsmen such as RU, whose average is only 28, and VS, whose average is 31.
Now check the failure rate (percentage dismissed below 20) so a high average is not hiding a fragile line-up. For C: ST 20, RD 20, MK 30, MD 30, SG 25, which averages to \(\frac{20+20+30+30+25}{5}=25\%\). For D: RD 20, VV 35, SG 25, VS 50, MD 30, which averages to \(\frac{20+35+25+50+30}{5}=32\%\). So D is both weaker in total average and riskier than C, mainly because VS gets out under 20 in half of his innings. Options A, B and E all include RU or VS or MK together with a low combined average, so they can be ruled out at once; none of them comes close to C on either measure.

Step 4: Final Answer:
Option C, ST, RD, MK, MD, SG, carries the highest total career average (200) and the lowest failure rate (25 percent) of the five choices, so it gives Team A the best chance of winning the third match against B. Note that the official XAT answer key for this caselet states that the data given is ambiguous and does not name a fixed correct option; the working above picks the strongest line-up the same table supports.
\[ \boxed{\text{Option 3: ST, RD, MK, MD, SG}} \]