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Limits of Accuracy - GCSE Maths Exam Questions & Answers

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Limits of Accuracy - GCSE Mathematics - Revision and Examples

1. Limits of Accuracy

Although rounding is practical, accuracy is sacrificed. So, we will learn about Limits of Accuracy. The limit of accuracy of a measurement is the range of possible values a measurement can actually be when a measurement is given. 

1.1 Limits of Accuracy - Upper and Lower Limits

Think about weighing grain sacks and rounding the weight to the nearest 10 kg (this is the degree of accuracy). A sack that is tagged as weighing 50 kg may actually weigh 45 kg or 54 kg, but it will still be labelled as 50 kg in both cases.

The lower limit of accuracy is the lowest possible value that is rounded up to the estimated value. The upper limit of accuracy is the lowest value that would round up to the next estimated value.

The lower limit of accuracy in the grain sack example is 45 kg. Because 55 kg is the lowest amount that would be rounded up to the next estimated value, this is the upper limit of accuracy (60kg).

We next write the error interval, which illustrates how far the actual value could vary from the estimated value, using these upper and lower accuracy limitations. The inequality notation is used to write this.

We might have a bag of grains stamped as weighing 50 kg. We can display the potential error as

45 kg  ≤ Weight < 55 kg

to demonstrate the possible error (i.e., what the bag's true weight may be). This is also known as error interval. 

The ≤ sign means that the weight can be greater than or equal to 45 kg. The < sign means that the weight must be less than 55 kg. If it were 55 kg, it would be rounded up to the next multiple of 10 (60 kg).

Note: A quick way to calculate upper and lower bands is to halve the limit of accuracy specified, then add this to the rounded value for the upper limit and subtract it from the rounded value for the lower limit.

Example: The heights of tulips are marked to the nearest 5 cm. What are the limits of accuracy for a tulip marked as 30 cm?

Solution:

The degree of accuracy here is 5 cm, so halving gives us 2.5 cm.

The estimated value is 30 cm.

Lower limit: 30 cm - 2.5 cm = 27.5 cm

Upper limit: 30 cm + 2.5 cm = 32.5 cm

Example: A piece of paper is 50 cm long and 15 cm wide, both measured to the nearest 5 cm. What are the upper and lower limits for the area of this piece of paper?

Solution

The degree of accuracy is 5 cm. Halving this gives us 2.5 cm which can be used to obtain the lower and upper bounds for the length and width.

Length:

Lower bound = 47.5 cm

Upper bound = 52.5 cm

Width:

Lower bound = 12.5 cm

Upper bound = 17.5 cm

The smallest possible area (lower bound) = 47.5 × 12.5 = 593.75 cm²

The largest possible area (upper bound) = 52.5 × 17.5 = 918.75 cm²

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