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Bounds

Bounds

What are bounds?

  • Bounds are the smallest (lower bound, LB) and largest (upper bound, UB) numbers that a rounded number can lie between
    • It simply means how low or high the number could have been before it was rounded
  • The bounds for a number, x , can be written as  LB less or equal than x less than UB
    • Note that the lower bound is included in the range of values x could have taken but the upper bound is not

How do we find bounds when a number has been rounded?

  • The basic rule is “Half Up, Half Down”
    • UPPER BOUND – To find the upper bound add on half the degree of accuracy
    • LOWER BOUND – To find the lower bound take off half the degree of accuracy
    • ERROR INTERVAL: LB ≤ x < UB
  • Note that it is tempting to think that the Upper Bound should end in a 9, or 99, etc. but if you look at the Error Interval – LB ≤ x < UB – it does NOT INCLUDE the Upper Bound so all is well
    •  the upper bound is the cut off point for the greatest value that the number could have been rounded from but will not actually round to the number itself
  • For example, the error interval for the number 1230, rounded to 3 significant figures will be 1225 ≤ x < 1235
    • The degree of accuracy is 10 (rounding to 3 s.f. here requires rounding to the nearest ten)
    • Half the degree of accuracy is 5

    

How do we find bounds when a number has been truncated?

  • Remember that  truncating a number means to round it down
    • This means that the UPPER BOUND is found by adding 1 to the digit in the place value that the number was truncated to
    • The LOWER BOUND is always the number the value was truncated to
  • For example, the error interval for the number 1230, truncated to 3 significant figures will be 1230 ≤  x < 1240

Worked example

The length of a road, l , is given as  l equals 3.6 space km , correct to 1 decimal place.

Find the lower and upper bounds for  l.

The degree of accuracy is 1 decimal place, or 0.1 km so the true value could be up to 0.05 km above or below this

Upper bound:

3.6 + 0.05 = 3.65 km

Lower bound:

3.6 – 0.05 =  3.55 km

Upper bound: 3.65 km
Lower bound: 3.55 km

We could also write this as an error interval of  3.55 space less or equal than space l space less than space 3.65 , although this is not asked for in this question

calculation?

  • If you are adding two numbers together: T =a + b
    • The upper bound of T can be found by adding together the upper bound of a and the upper bound of b
    • The lower bound of T can be found by adding together the lower bound of a and the lower bound of b
  • If you are subtracting a number from another number: T =a – b
    • The upper bound of T can be found by using the upper bound of a and subtracting the lower bound of b
    • The lower bound of T can be found by using the lower bound of a and subtracting the upper bound of b
  • If you are multiplying two numbers together: T =a × b
    • The upper bound of T can be found by multiplying together the upper bound of a and the upper bound of b
    • The lower bound of T can be found by multiplying together the lower bound of a and the lower bound of b
  • If you are dividing a number by another number: T =a ÷ b
    • The upper bound of T can be found by using the upper bound of a and dividing it by the lower bound of b
    • The lower bound of T can be found by using the lower bound of a and dividing it by the upper bound of b

How can bounds help with calculations?

  • You can use bounds to calculate the level of accuracy of a calculation
  • This can be used to decide how to round your answer
    • e.g. If the lower bound of an value is 8.33217… and the upper bound is 8.33198…
    • The true value is between 8.33217… and 8.33198…
    • Both bounds round to 8.332 to 4sf
    • To 5sf they differ (first is 8.3322 and second is 8.3320)
    • Therefore you know the answer is definitely rounds to 8.332 to 4 significant figures

Worked example

(a) A room measures 4 m by 7 m, where each measurement is made to the nearest metre

Find the upper and lower bounds for the area of the room

Find the bounds for each dimension, you could write these as error intervals, or just write down the upper and lower bounds
As they have been rounded to the nearest metre, the true values could be up to 0.5 m bigger or smaller

3.5 space less or equal than space 4 space less than space 4.5
  6.5 space less or equal than space 7 space less than 7.5

Calculating the lower bound of the area, using the two smallest measurements

3.5 × 6.5 =

Lower Bound = 22.75 m 2

Calculating the upper bound of the area, using the two largest measurements

4.5 × 7.5 =

Upper Bound = 33.75 m 2

(b) David is trying to work out how many slabs he needs to buy in order to lay a garden path.

Slabs are 50 cm long, measured to the nearest 10 cm.

The length of the path is 6 m, measured to the nearest 10 cm.

Find the maximum number of slabs David will need to buy.

Find the bounds for each measurement, you could write these as error intervals, or just write down the upper and lower bounds
As they have been rounded to the nearest 10 cm, the true values could be up to 5 cm bigger or smaller

We have a mixture of centimetres and metres, so it is useful to change them both to metres for later calculations

Length of the slabs: or in metres:  45 space less or equal than space 50 space less than space 55 space cm
0.45 space less or equal than space 0.5 space less than space 0.55 space straight m

Length of the path:  5.95 space less or equal than space 6 space less than space 6.05 space straight m

The maximum number of slabs needed will be when the path is as long as possible (6.05 m), and the slabs are as short as possible (0.45 m)

Max number of slabs =  fraction numerator 6.05 over denominator 0.45 end fraction equals 13.444...

Assuming we can only purchase whole slabs

The maximum number of slabs to be bought is 14

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