Year 8 Bounds Dr J Frost jfrosttiffin kingston

Year 8: Bounds Dr J Frost (jfrost@tiffin. kingston. sch. uk) Last modified: 20 th June 2014

Starter Write down the number 24. 98063 to the following number of significant figures: 1 sf 2 sf 3 sf 4 sf 5 sf ? 25. 0 (the. 0 is? required!) 24. 98 ? 24. 981 ? 20 The length of a squirrel is measured as 14 cm correct to the nearest cm. What is the smallest possible length the squirrel could be? What is the largest possible length? Smallest possible length: ? 13. 5 cm Largest possible length: ? 14. 5 cm*

More Squirreling For a length of 14 cm measured, the range of possible values was 13. 5 cm to 14. 5 cm. Lower Bound: Upper Bound: 13. 5 cm 14. 5 cm Question 1: How would we represent the range of possible values on a number line? ? 12 cm 13 cm 14 cm 15 cm The value of 14. 5 cm isn’t include, because this would have rounded to 15 cm.

Activity In pairs, work on the handout. Value 7 cm 5. 3 m 200 km 8. 04 mm 3 m 830 litres 100 miles Accuracy Nearest cm Nearest 0. 1 m Nearest 10 km Nearest hundredth of a mm Nearest cm 2 sf 3 sf 1 sf Lower Bound 6. 5 cm 5. 25 m ? 195 km ? 8. 035 mm ? 2. 995 m? 825 litres ? 829. 5 litres ? 95 m ? Upper Bound 7. 5 cm 5. 35 m ? 205 km ? 8. 045 mm ? 3. 005 m ? 835 litres ? 830. 5 litres ? 150 m ? Can you think of an easy way of getting the bounds given the accuracy?

Easily getting bounds To find bounds, just add or subtract half the accuracy! Value: 250 m Correct to: nearest 10 m Lower Bound = 250 – (10/2) ? = 245 m Upper Bound = 250 + (10/2) ? = 255 m Value: 423 mm Correct to: tenth of a mm Lower Bound = 423 – (0. 1/2)? = 423. 05 mm Upper Bound = 423 + (0. 1/2)? = 422. 95 mm

Exercises 1 ? ? ? 2 ? 3 4 ? ? N ?