'Plausible Estimation' Estimating for Amazing Facts Tasks - Set #3 (solutions) Malcolm Swan Mathematics Education University of Nottingham Malcolm.Swan@nottingham.ac.uk Jim Ridgway School of Education University of Durham Jim.Ridgway@durham.ac.uk The aim of this assessment is to provide the opportunity for you to: · develop a chain of reasoning that will enable you to estimate quantities to an appropriate degree of accuracy · choose suitable units for your estimate · communicate the assumptions upon which your estimate is based. ____________________________________________________ 1. High stack Suppose you have a very large sheet of paper. You tear it in half and put one half on top of the other. You now have a stack of two sheets. You now tear the whole stack in half and place one half on top of the other to make a new stack. You repeat this process, tearing 50 times. (Yes I know its impossible - just imagine you could). How high would the stack be? 50 feet? 100 feet? A mile? or more...? Make a sensible estimate, based on careful reasoning. Assumptions · A ream of paper (500 sheets) is about 2 inches thick Reasoning Each sheet is therefore approx 1/250th of an inch thick. After 50 tears, the stack will contain 250 sheets of paper. This is equal to approximately 1.1259 x 1015 sheets or 4.5 x 1012 inches = 71million miles. Answer: 71 million miles! 2. The swimming pool and the glass. How long would it take you to empty an olympic size swimming pool with a glass? Assumptions · Dimensions of swimming pool = 50 meters x 20 meters · Depth of water = 6 feet (2 meters) · The glass holds half a pint. · You empty the pool at one glass per second. Reasoning This gives volume of pool as 2000 cubic meters or 2 x 106 liters. One half pint glass has a volume of about 25 x 10-5 cubic meters. Thus approximately 8 million glasses will empty the pool. This would take 2,215 hours or about 90 days. Answer: 92 days or about 3 months, working day and night! 3. The briefcase of cents Suppose you filled a briefcase with one cent coins. How much would the money be worth? Assumptions · A cent coin has a diameter of 20 mm (0.75 inches) and a thickness of 1.5 mm · The briefcase is approx 100mm x 350mm x 500mm Reasoning The volume of the coin is therefore given by pr2h = 3.14 x (20/2)2 x 1.5 Å 470 mm3 The volume of the case is approx 500 x 100 x 350 = 17,500,000 mm3 Thus the number of coins that will fit in (assuming no gaps) will be about 17,500,000 Ö 470 = 37,000 (approx). There would be some gaps between coins, so Answer: The money will be worth around $350. (Note that the total weight of the case would be almost 100 kilograms - far too heavy!)