Upper and Lower Bounds A number has been

Upper and Lower Bounds

A number has been rounded to the nearest whole number, what could it have been? If it was rounded up, it could have been- 4. 9 4. 87 5 4. 61 What’s the smallest number it could have been? If it was rounded down, it could have been- 5. 1 5. 34 5. 48 4. 5 What’s the biggest number it could have been? 5. 4999999999999 9999999. . .

If 5 has been rounded to the nearest whole number we say �The lower bound is 4. 5 (the smallest it could have been) �The upper bound is 5. 5 (it’s easier than writing 4. 499999999999. . . , even though technically it would round to give 6!)

Try these. . These numbers have been rounded to the nearest whole number, write the upper and lower bounds: a) 7 b) 11 c) 15 6. 5 and 7. 5 10. 5 and 11. 5 d) 17 14. 5 and 15. 5 e) 23 16. 5 and 17. 5 f) 100 22. 5 and 23. 5 99. 5 and 100. 5

A general rule � To find the upper bound we have to add half of the unit we rounded to for example: �If we rounded to the nearest 10, we add on 5 (half of 10) �If we rounded to the nearest 1 (whole number), we add on 0. 5 (half of 1) �If we rounded to the nearest 0. 1 (1 dp), we add on 0. 05 (half of 0. 1) �If we rounded to the nearest 0. 01 (2 dp), we add on 0. 005 (half of 0. 01) �And if we want the lower bound, we subtract have of the unit

Limits 1) These numbers have been rounded to the nearest 10, write down the largest and smallest values they could be: 1) 50 2) 80 3) 110 2) These numbers have been rounded to the nearest whole number, write down the upper and lower limits: 1) 3 2) 17 3) 23 4) 100 5) -3 3) These lengths have been rounded to the nearest 10 th of a cm, write the upper and lower limits: 1) 12. 5 cm 2) 21. 7 cm 3) 35. 8 cm 4) 52. 1 cm 5) 80. 4 cm Answers 4) A field is 100 m wide and 120 m long, both lengths have been 1. rounded to the nearest metre. 1. 45 and 54. 9999999. . 2. 75 and 84. 999. . a) Find the perimeter and area of the field if these 3. 105 and 114. 9999. . . measurements are accurate 2. b) Find the largest and smallest possible perimeter 1. 2. 5 and 3. 4999. . . c) Find the largest 2. and smallest possible area. 16. 5 and 17. 4999. . . 3. 22. 5 and 23. 49999. . . 4. 99. 5 and 100. 4999. . 5) A rectangle has it’s area rounded to the nearest 5. -3. 499999 and whole -2. 5 number, 3. it becomes 40 cm 2. One side of the rectangle is exactly 10 cm; find 1. 12. 45 and 12. 55 the maximum and minimum lengths theand other 21. 75 length could have. 2. 21. 65 3. 35. 75 and 38. 5 4. 52. 05 and 52. 15 6) Two lengths of wood are stuck together and their combined 5. 80. 35 and 80. 45 4. length is rounded to the nearest mm and itand is 14. 9 cm, length 2 a) p=44 om area=one 12000 m largest= is rounded to the nearest mmb) and is 7. 1 cm. 442 m Find the smallest=438 m minimum c) largest= 12110. 25 m 2 and maximum length of the othersmallest=11890. 25 m length. 2 5. smallest- 3. 95 cm largest- 4. 05 cm 6. largest- 7. 9 cm smallest- 7. 7 cm