A student was simply solving a routine word problem of Class 10 when he noticed something strange. A tiny shortcut appeared to give the answer instantly. The class was impressed. Then someone asked the question that turns school mathematics into mathematical investigation. The Original Question : A man bought some pens for ₹5760. If each pen had cost ₹8 less, he would have received 10 more pens for the same amount. Find: The original cost of each pen The number of pens bought originally Most students solved it algebraically. Let the original cost per pen be p. Then the original number of pens is: 5760/p If the price decreases by ₹8, the new price becomes: p-8 And the new number of pens becomes: 5760/(p-8) Since he gets 10 more pens: (5760/p) +10 = 5760/(p-8) 5760 ( p − 8 ) + 10 p ( p − 8 ) = 5760 p . 5760p-46080+10p^2-80p=5760p. 10 p^ 2 − 80 p − 46080 = 0. p^ 2 − 8 p − 4608 = 0. Solving gives: p = 72 Original cost = ₹72 Original number of pens = 5760/72 = 80 At this stage, it ...

  This is a narrative of a classroom episode in which students could see how mathematics can help someone avoid some common mistakes and take a right decision. A simple question involving ratios and simultaneous linear equations could do this with right learning strategy. Students were given a question to work on, and the discussion build on that as given below. A class was working on the following problem: A vendor makes two types of milkshakes using milk and syrup. The ratio of these two ingredients in two types of Shakes is given as - ShakeX, milk : syrup = 2 : 3   - ShakeO, milk : syrup = 3 : 4   He has 23 liters of milk and 31 liters of syrup.   How much of each shake should he prepare so that everything is used? The discussion went like:  We assume he prepares (x) liters of ShakeX and (y) liters of ShakeO,  It was noted: - Each shake consumes some milk and some syrup   - Total milk used must become 23 liters   - Total syrup used must become ...