Question:hard

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?

Show Hint

Add up each group's total career average first, then use the dismissed-below-20 percentage as a tie-breaker for risk.
Updated On: Jul 10, 2026
  • RD, RU, MK, VS, YS
  • RD, VS, MT, RU, YS
  • ST, RD, MK, MD, SG
  • RD, VV, SG, VS, MD
Show Solution

The Correct Option is C

Solution and Explanation

A different way to compare the five candidate groups is to give each of the 10 batsmen a single reliability score that blends all three tendency columns into one number, instead of reading the average and the failure rate side by side. Use the formula: score = average minus half the percentage dismissed below 20, plus the century percentage. This rewards a high average and a habit of scoring centuries, and penalizes a high chance of a cheap dismissal.

Working the formula out for all 10 batsmen:

  • RD: $40 - 0.5(20) + 3 = 33$
  • ST: $44 - 0.5(20) + 10 = 44$
  • SG: $41 - 0.5(25) + 10 = 38.5$
  • VS: $31 - 0.5(50) + 15 = 21$
  • RU: $28 - 0.5(55) + 12 = 12.5$
  • YS: $35 - 0.5(40) + 10 = 25$
  • VV: $35 - 0.5(35) + 5 = 22.5$
  • MK: $30 - 0.5(30) + 5 = 20$
  • MT: $36 - 0.5(45) + 10 = 23.5$
  • MD: $45 - 0.5(30) + 10 = 40$

Now add up the score of each option's 5 named batsmen:

  1. RD, RU, MK, VS, YS (option A): $33+12.5+20+21+25=111.5$, dragged down by RU and MK, whose scores are both under 20.
  2. RD, VS, MT, RU, YS (option B): $33+21+23.5+12.5+25=115$, still weak because RU's low score of 12.5 pulls the group down.
  3. ST, RD, MK, MD, SG (option C): $44+33+20+40+38.5=175.5$, the highest of all five groups because it packs in ST, MD and SG, the three strongest scorers on this measure, and only carries one weak link, MK.
  4. RD, VV, SG, VS, MD (option D): $33+22.5+38.5+21+40=155$, clearly behind C because VS's score of 21 is much lower than MK's contribution would have been, and VV adds less than ST would.
  5. SG, RU, YS, MK, VV (option E): $38.5+12.5+25+20+22.5=118.5$, again weak since RU's poor score of 12.5 outweighs SG's strong one.

The group with the highest total score is option C, ST, RD, MK, MD, SG, at 175.5, comfortably ahead of option D at 155 and far ahead of options A, B and E, all below 120. This agrees with the simple average-and-risk check: packing in the two best all-round batsmen, ST and MD, alongside RD and SG, outweighs the single weak link MK contributes.

Let's summarize:

  • A composite score built from average, failure rate and century rate ranks ST, MD and SG as Team A's strongest batsmen on this table.
  • Any group missing two or more of ST, RD, MD and SG scores far lower, because it has to replace them with weaker batsmen like RU or VS.

So the manager should send in ST, RD, MK, MD, SG, option C, for the best chance of beating Team B, keeping in mind that the official key marks this caselet's data as ambiguous with no single fixed answer.

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