The series opener between England and Pakistan is now into its third day of play. This post looks at developments in that match so far.
THE PAKISTAN INNINGS
A weather hit opening day ended with Pakistan two down, Babar Azam already past 50 and Shan Masood not far short. England bowled well on the second morning but did not get full benefit for their efforts in that department as they were badly let down by Jos Buttler who had an absolute nightmare behind the stumps. Post lunch England bowled poorly, and Masood cashed in, being well supported by Shadab Khan. Masood eventually reached 156 before his resistance was ended. Pakistan tallied 326 in total for their innings, a score that looks very good on this pitch.
England were soon 12-3 in reply, with both openers and Stokes out cheaply. Root batted a long time but did not score many, and Buttler was just able to survive to the close after Root’s dismissal. At the end of day 2 England were 92-4, with Pope who had looked a class above anyone else in the order approaching a 50. This morning Pakistan bowled superbly and England did well to get through the opening session for the loss of only one wicket – Pope got an absolute beauty. Woakes was hit by a bouncer but resisted through to lunch in company with Buttler. Early in the afternoon session Buttler has been bowled by leg spinner Yasir Shah for 38 to make it 159-6. Bess will be next man in. Taking into account Buttler’s errors with the gauntlets a generously inclined assessor would now say that he is only in a double-figure rather than a triple-figure deficit for the match. Stokes’ unfitness for bowling means that England have little batting left – Woakes is more bowler than batter (though his record in England specifically is excellent), Bess can handle a bat, but against an attack equipped with serious pace and quality wrist spin (more difficult to handle than finger spin) little can be hoped for, much less expected, from Broad, Archer and Anderson. This Pakistan team look to be made of sterner stuff than the West Indies – Masood’s ton was his third in as many tests, while Azam’s innings was a magnificent performance, and his record suggests that he deserves to be bracketed with Kohli, Smith and Williamson and placed ahead of the current version of Root as a batter. The pace bowling, with a left arm quick in Shaheen Afridi, a right arm quick in the person of 17 year old Naseem Shah, and an excellent exemplar of the steady medium-fast bowler in Mohammad Abbas looks superb. Yasir Shah with his leg spin and the second leg spinner Shadab Khan whose bowling has not yet been called on are likely to play an ever increasing role as the match goes on, and Yasir Shah has already accounted for a couple of wickets, Root yesterday as Pakistan’s keeper demonstrated that it is perfectly possible to make dismissals off a spin bowler on this pitch and Buttler today, bowled through the gate, once again failing to navigate his personal ‘Bermuda triangle’ which is located between 21 and 50. Ben Foakes has a first class batting average of 38 (having played just over 100 matches at that level – a very impressive record for someone for whom batting is the second string of the bow) and is also the best pure keeper in the country, and various young keepers are beginning to establish themselves at county level and would also be more deserving of the test gauntlets than Buttler, though my own feeling is that Foakes deserves an extended run as England’s acknowledged no1 test keeper before a youngster is blooded. Yasir Shah has nabbed a third wicket, that of Bess, while I was writing this. Archer has been sent in at no9, ahead of Broad and Anderson, and England need something major from Woakes backed by the tail – with the pitch already helping the bowlers quite a bit anything approaching a major deficit will be insuperable, and at the moment that is exactly what England will be facing.
A SOLUTION AND A NEW PROBLEM
I offered this problem from brilliant up in my previous post:
No multi-choice here (this is much too easy for that), but a bonus challenge: part 1) if there was a third square of the same size but divided into 49 smaller squares shaded in similar fashion which would have the largest shaded area, and part 2)what is the general rule relating the number of squares into which the big square is divided and the proportion of it that ends up shaded?
The first shape contains nine squares of which five are shaded, while the second contains 25 squares of which 13 are shaded. 5/9 = 0.55…, while 13/25 = 0.52, so the first shape has a greater shaded area. The 7X7 square would have an even smaller proportion of its area shaded – 25/49 = 0.51. The general rule is that the greater the number of squares the shape is divided into the closer the shaded area approaches to half the total area, while always remaining just above that limit.
Here is another problem from brilliant:
Yasir Shah has just collected his fourth wicket, that of Woakes to make it 170-8, and England are definitely in the mire.
My usual sign off…