In the glorious tradition of this column’s constant teetering on the tripwire of correctly loopholed illegality, may I present a way out for laggards. Suppose you have some immediate work, report or article to submit but aren’t done yet. Simply send the incomplete stuff to this site which then carefully corrupts it so that when the end-reader opens it he or she can’t. They think you delivered on time but they had a technical glitch. Meaning you get some extra time to send it again. Cool, no? (Want more info?)
Now to less important things as usual. A vessel contains a mixture of wine and water. Had there been a litre more of wine and a litre less of water the ratio of wine to water would have been 7:8. But had there been a litre more of water and a litre less of wine the ratio would have been 2:3. What was the original mixture in litres?
Throughput
(The older gen problem was: “A cylindrical marshmallow has a diameter of 5 and height of 3. A dessert is made with 24 marshmallows consisting of two stacked 3 x 4 arrays. Chocolate is poured fill the gaps between the stacks. “What is the volume of chocolate?”)
The volume can be calculated as the product of the sum of the areas of the six gaps and the height of two layers of marshmallows. If we draw squares, with tangential lines, enclosing the circles representing the bases of the marshmallows, we can see that each gap is made up of four segments, each segment, one quarter of the difference between the area of the square and that of the circle. That means, a gap is equal to the difference between the area of the square and that of the circle. Volume = Area of the base × height = (5^2 - (pi*5^2)/4)*6*6 = 193.142 units. -- Balagopalan Nair K, balagopalannair@gmail.com
The volume of chocolate is 193.14 cubic units. Each of the six spaces has an area equal to the difference between the area of a 5*5 square and the area of a circle of 2.5 unit radius. Six times this difference gives the surface area of the chocolate region and that surface area multiplied by 6, the height, gives the volume of the chocolate as 193.1416529. -- Indira Vaseekhar, youmaycontactme@yahoo.com
(The other puzzle was: “How can you show that in any year the same days of the month in March and November fall on the same day of the week?”)
There are exactly 245 days from 1st March to 31st October. 245 divided by 7 is exactly 35, which means completion of 35 weeks. Hence November 1 falls on the same day as March 1. The occurrence is irrespective of leap years as February does not come into the picture. -- Sanath Kumar T S, sanathkumarts1958@gmail.com
Another small detail. Since a year consists of 365 days (not a leap year) if first of January this year is a Sunday, the first of Jan next year would be a Monday. Hence by remembering the index numbers for all the months and with a base year as reference, anyone can easily calculate on which weekday any day of any year falls (of course discounting for leap years). – A V Ramana Rao, raoavr@gmail.com
(Among others who also got it correct are the following five: Narayanan P S, narayananpsn@gmail.com; Krishna D V, krishp84@gmail.com; Ganesh Ram Palanisamy, 1969ganram@gmail.com; Dr P Gnanaseharan, gnanam.chithrabanu@gmail.com; Binata Mohanta, binatamohanta@rediffmail.com.)
(The third problem was: “A slice of uniform thickness of one centimetre is cut from a sphere of diameter 10 centimetres. What is the curved area of the slice?”)
Since the diameter of the sphere is 10 cm, it’s radius R is 5 cm and the curved surface area of the slice is 2pi*5 = 10pi = 31.42 square cm. As the curved area is 2piR*1 (1 being the thickness) we can see that it depends only on the radius of the sphere and the thickness of the slice and is independent of how far from the centre of the sphere the slice is cut. -- Balagopalan Nair K, balagopalannair@gmail.com
But Google This Now
1. What do the following words have in common: Bedecked, Icebox, Choice, Kidded?
2. A circular fort of diameter 6 kms has two gates only: one in the north and the other in the south. A big tree stands 2 kms north of the north gate. What distance must be covered by a soldier walking east from the south gate to see the tree without obstruction?
— Sharma is a scriptwriter
and former editor of Science Today magazine.(mukul.mindsport@gmail.com)
