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 is playing its first match with team C. Based on the statistics above, whom should the manager choose so that the team has maximum chances of winning?

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Total the career averages of each five-man group first, since Team C's own batsmen average 270 and demand Team A's strongest possible line-up.
Updated On: Jul 10, 2026
  • RD, ST, SG, MD, YS
  • VS, YS, RU, MD, MT
  • RD, ST, SG, VS, MD
  • YS, RU, VS, MK, MD
Show Solution

The Correct Option is A

Solution and Explanation

Using the same composite reliability score as before, score = average minus half the percentage dismissed below 20, plus the century percentage, each batsman's score works out to:

  • 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: $45 - 0.5(30) + 10 = 40$

Add up the score of each option's named batsmen to see which group Team A should send in against Team C, whose own top 5 average a demanding 270 runs:

  1. RD, ST, SG, MD, YS (option A): $33+44+38.5+40+25=180.5$, the highest total because it packs in three of the four strongest scorers, ST, MD and SG, together with RD and YS, neither of whom is a weak link.
  2. VS, YS, RU, MD, MT (option B): $21+25+12.5+40+23.5=122$, well behind because RU's score of only 12.5 is the lowest of any batsman on the table.
  3. RD, ST, SG, VS, MD (option C): $33+44+38.5+21+40=176.5$, close to A but a little lower, since VS's score of 21 is well short of what YS would have added.
  4. YS, RU, VS, MK, MD (option D): $25+12.5+21+20+40=118.5$, one of the weakest groups because it carries RU, VS and MK together, three of the lower-scoring batsmen.
  5. ST, VS, RU, MD, SG (option E): $44+21+12.5+40+38.5=156$, held back again by RU's poor score.

Option A comes out on top at 180.5, just ahead of option C at 176.5, and well clear of the other three, which all fall below 160. This matches the average-and-risk check: A's edge comes from having YS in the side instead of the more fragile VS.

Let's summarize:

  • The strongest five-man core for Team A, on this table, is ST, MD, SG and RD, since all four carry the highest composite scores.
  • The fifth slot should go to YS rather than VS, because YS's lower chance of a cheap dismissal outweighs VS's slightly higher century rate.

So against Team C's strong batting line-up, Team A should field RD, ST, SG, MD, YS, option A, again keeping in mind the official key flags this caselet as having ambiguous, unresolved data.

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