2 batsmen each on 94 runs answer: Here is the solution to the popular WhatsApp puzzle

The ongoing coronavirus crisis has prompted the Indian public to stay indoors until the lockdown is in effect. With no live coverage of international cricket and Indian Premier League 2020 (IPL 2020) due to their postponement, cricket lovers in the country are spending their time in various other activities like playing indoor games, spending time on social media, resorting to video streaming platforms and more. Some people have also been sending various quizzes and puzzles on social media platforms like WhatsApp which one might find challenging and entertaining to solve. Among such quizzes, one of the most popular ones circulating on social media is the ‘2 batsmen each on 94 runs’ puzzle.  

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2 batsmen each on 94 runs no extras WhatsApp puzzle  

The famous '2 batsmen each on 94 runs’ puzzle is aimed at posing an interesting challenge for those who possess a sharp cricketing aptitude. The purpose of the puzzle is to mathematically determine (in cricketing terms) how two batsmen can reach their respective centuries with no extras and just three balls on offer. Here, we present you the 2 batsmen each on 94 runs puzzle answer along with a detailed explanation of the 2 batsmen each on 94 runs no extras puzzle.

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2 batsmen each on 94 runs math problem

The 2 batsmen each on 94 runs no extras puzzle is as follows. Seven runs are needed to win in the last 3 balls and both batsmen make 100 not out. Can it be done from just the last two deliveries?

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2 batsmen each on 94 runs answer

To start off, we will be considering the on-strike batsman as P1 and batsman at the non-striker's end as P2.

Ball 1: P1 hits the ball towards the boundary. Batsman runs 3 and there are 4 overthrows.

Umpires have a check and found it 1 run short.

Now, since they ran for three runs, the P2 batsman will take strike for the last ball. As one run was short and overthrow runs are counted in the batsman’s favour, P1 will have reached his century courtesy of 2 runs + 4 overthrows, i.e. 6 runs to take him from 94 to 100.

Ball 2: With still a run to get for a win and six for P2’s century, P2 hits a six to bring up his respective century and in doing so, seals the match for his side and solves the 2 batsmen each on 94 runs math problem.

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