Problematics | The length of a cricket run

If you have been watching the IPL, or if you have ever watched any cricket match for that matter, you might have wondered what distance a batter travels when running a run. It is not the full length of the pitch at 22 yards, for the run is completed between one crease and the other, a distance of 58ft instead of 66ft. Even that distance need not be travelled in its entirety because the run is completed the moment the batter grounds his bat across the crease. It is another matter that the momentum will carry the batter too across the crease, which means he or she will indeed run 58ft or more. But there is a catch here too: this full length is run only when the batter is taking a single.

When the two batters run two, things happen differently. Each batter runs towards the opposite end, stretches the bat across the crease, and then turns back to their original end. The first run, as we can see, is completed after running some distance less than 58ft. It is reasonable to infer that a taller cricketer can stretch their bat earlier, thereby running a shorter distance than a shorter batter would need to.

It is in such circumstances that the following puzzle is set. As a matter of fact, it is a different version of another puzzle that had appeared here several months ago, so longtime readers will find it familiar. Nevertheless, I found it fun to take this exercise into a cricket field, and hope you will enjoy solving it even if you have already solved the previous version.

#Puzzle 88.1

Let's meet these two cricketers who are batting together for the first time. The two are the same height, which means that they travel the same distance across the pitch when running a run, and stretch their bat from exactly the same distance as they approach the opposite crease.

One of the batters, however, runs faster than the other. We meet them as they run two, bringing into play all the factors that we discussed above.

It is the non-striker who is the faster runner. The two set off for the first run, and cross each other at a point that is 23ft from the striker's crease. The non-striker completes the first run first, stretching her arm across so that her bat crosses the crease, and turns back in the same motion without losing time, momentum or speed.

Seconds later, the striker completes the first run in similar fashion, stretching her bat across after having run exactly the same distance as the non-striker had travelled for the first run. She too turns around for the second run without any loss of time or speed.

During the second run, the two cross at another point, which is 16ft from the non-striker's original crease. Needless to say, the non-striker completes the second run first, but that is not the point here.

What distance did each batter travel in order to complete the first run?

#Puzzle 88.2

These are anagrams of the titles of five novels. Once you have unscrambled them, there is a sixth anagram.

Take one letter from the same position in each title (after solving), and rearrange those five letters to form the title of yet another novel.

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 87.1

Dear Kabir,

The couple leaves office a little after 5:27pm and reaches home a little after 7:05pm.

At 5pm, the hour hand is 150° ahead of the minute hand. Say, the hands overlap t minutes after 5pm. In this time, the angular distance traversed by the minute hand (@360°/hr = 6°/min ) is 6t degrees from the 12 o'clock position while the distance traversed by the hour hand (@360°/12hr = 0.5°/minute) is 0.5t degrees from the 5 o'clock position.

As the hands overlap,

6t = 0.5t + 150, or t = 300/11 minutes.

So, the clock time at this moment is 300/11 minutes (= 27 + 3/11 minutes) past 5pm.

The couple reach home in about one-and-a-half hours. Now, at 7pm, the minute hand is 150° ahead of the hour hand. Say, the hands are opposite each other t minutes after 7pm. In that time, the minute hand travels 6t degrees from the 12 o'clock position, and the hour hand travels 0.5t degrees from 7 o'clock position.

As the hands are now opposite each other,

150° + 6t – 0.5t = 180°, or t = 60/11 minutes

So, the clock time at this moment is 60/11 minutes (= 5 + 5/11 minutes) past 7pm.

— Yadvendra Somra, Sonipat

#Puzzle 87.2

Hi Kabir,

20 years ago, since the mother was 6 times older than her daughter, the difference between their ages was 5 times the daughter’s age. At present, since the mother is twice as old as the daughter, the age difference now is equal to the daughter’s present age. This implies that the daughter’s present age is 5 times her age 20 years back.

From this we can infer that in 20 years, the daughter has added 4 times her then age. This means that her then age was 20/4 = 5 years, and her present age is 25 years. Accordingly, the mother is 50 now and was 30 then.

— Professor Anshul Kumar, Delhi

In their effort to avoid algebra, some readers have used hit and trial, but I am not complaining.

Solved both puzzles: Yadvendra Somra (Sonipat), Prof Anshul Kumar (Delhi), Dr Sunita Gupta (Delhi), Kanwarjit Singh (Chief Commissioner of Income-Tax, retired), Ajay Ashok (Mumbai), Akshay Bakhai (Mumbai), Shishir Gupta (Indore), YK Munjal (Delhi), Dr Vivek Jain (Baroda)

Solved #Puzzle 87.2: Jaikumar Inder Bhatia & Disha Bhatia (Ulhasnagar, Thane)

Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com