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Borrowing from the many experiences of Henry Ernest Dudeney (1847-1930), Martin Crowe starts off with a pitcher of K gallons (K=10), and a jar of J gallons (J=?). After he removes the first jarful of wine, there is K-J gallons of wine (Ponting’s magic potion) left in the K-gallon pitcher. When he removes the second gallon of the "mixture", he is removing J*(K-J)/K gallons of wine. In the end, he has K/2 gallons of wine, so K - J - J * (K - J)/K = K/2; 10 - J - J * (10 - J)/10 = 5; J^2/10 - 2*J + 5 = 0 Solving, we obtain J = 10 +/- 5*sqrt(2); Since 10+5*sqrt(2) is larger than the capacity of our original pitcher, the solution is: 10-5*sqrt(2) [approximately equal to 2.929 gallons] . The first removal leaves 10 - 10 + 5*sqrt(2) = 5*sqrt(2) gallons of wine in the pitcher, and the second removal leaves 10 - 10 + 5*sqrt(2) - (10 - 5*sqrt(2))(10 - 10 + 5*sqrt(2))/10 = 5*sqrt(2) - (10 - 5*sqrt(2))(sqrt(2)/2) = 5*sqrt(2) - 5*sqrt(2) + 5*(sqrt(2)^2)/2 = 5 gallons. Vijayesh Narayan Singh of the Panjab University Department of Mathematics says that the capacity of the jar is equal to { 10 - 5*(sqr root of 2)} litres = 2.9289 litres. Let us suppose the quantity of the jar is ‘x’ litres. Then, firstly, he took ‘x’ litres of wine from 10-litre wine pitcher and then filled it with water fully. Then, the percentage of the wine left in the 10-litred mixture, is = { (10-x)/10} * 100 = 10(10 - x) and the water percentage is (x/10)*100 = 10x. After this, another ‘x’ litres of mixture is taken out and the remaining is filled with water; so when the mixture was out, the wine left in the mixture is 10(10 - x) % of (10 - x) = { (10 - x)^2} /10 ...(1) The water left in the mixture is 10x % of (10 - x) = x(10 - x) / 10 ...(2) which is added to another ‘x’ litres of water. Now the quantities of water and wine are equal in the pitcher; so by equating the equations (1) and (2), one gets the value of x. Also, by equating any of these equations with the value 5, one can get the solution, since the total quantity of the pitcher is 10. By solving this, one gets the quadratic equation x^2 - 20x + 50 = 0 which has two solutions. One is x = 10 - 5*(sq root of 2) and another is x = 10 + 5*(sq root of 2). The second solution is not possible because the jar’s quantity cannot be greater than the pitcher’s quantity. "If only playing cricket like
Ricky Ponting was just as simple..." says Dr R. Chawla. Dr Tarsem
Lal, U.K. Gupta, Ravneet Kotwal, Madhur Sharma, Navdeep, Amarjeet Kaur,
Harvinder Pal Singh, Arshdeep Singh, Divya Sood, Rakesh Satija,
Manjinder Arora, Aman Bali, Dr Vikas Handa, Puneet Sandhu and Sudhir
Dhamija also have that magic potion. (Write at The Tribune or
adityarishi99@yahoo.co.in) |
