Limits of Accuracy

• The length of a room is 10 metres to the nearest metre. In this case, the degree of accuracy is one metre.
• Connie weighs 65 kg to the nearest 5 kg. In this case, the degree of accuracy is 5 kg.
• The length of a piece of string is 10.2 cm to the nearest 0.1 cm. In this case, the degree of accuracy is 0.1 cm.
• Sebastian earned £110 yesterday to the nearest £10. In this case, the degree of accuracy is £10

A bag of oranges weighs 3 kg to the nearest kilogram. What are the upper and lower bounds for the weights of the oranges?

Solution:

Suppose the bag of oranges weighed 2.8 kg. Well, if we rounded the weight to the nearest kilogram, we would say that the oranges weighed 3 kg. Thus, the actual weight of the oranges could be 2.8 kg. Similarly, the weight could be 2.6 kg or 2.55 kg.

The question is, what is the smallest weight the oranges could be that would get rounded up to 3 kg? Here, the degree of accuracy is 1 kg, so we half this to get 0.5 kg and subtract this from 3 kg to get 2.5 kg. Thus, the lower bound is 2.5 kg.

The upper bound is a little more confusing. If we take half of the degree of accuracy and add it to 3 kg, we get 3.5 kg, but surely that would round up to 4kg...

What is the largest possible value that would round down to 3 kg? The oranges could weigh 3.4 kg because that would round to 3 kg. Similarly, the could oranges weigh 3.49 kg, because that also would round down to 3 kg. If we said that the oranges weighed 3.4999 kg, that would round down to 3 kg, but if we said the oranges weighed 3.5 kg, that would round up to 4 kg.

The answer is that there is no largest value. The largest value would be 3.49 recurring but it is not possible for a bag of oranges to weigh 3.49 recurring kilograms. Therefore, despite the fact that 3.5 kg rounds up to 4 kg, we would say 3.5 kg is the upper bound because there is no single largest value less than 3.5 kg.

Therefore, in this case, the lower bound is 2.5 kg and the upper bound is 3.5 kg.

The mass of the average apple is 175 g to the nearest gram. What are the least and greatest possible masses for the apple?

Solution:

Here, the degree of accuracy is 1 gram. If we half 1 gram, we get 0.5 grams. To get the lower bound, we subtract 0.5g from 175 g to get 174.5 g. To get the upper bound we add 0.5 g to 175 g to get 175.5 g.

If we defined the mass of the apple to be m, we could say$174.5\le m<175.5$. This is called the error interval. We use the < inequality for the upper bound to say that the mass of the apple can't actually equal 175.5 g, but it can be anything less than 175.5 g.

Tony went on a 40-mile bike ride to the nearest 10 miles. He travelled at a constant speed of 15 mph to the nearest mph. What is the minimum time he completed his bike ride in?

Solution:

Recall that$speed=\frac{dis\tan ce}{time}$. Rearranging this, we obtain that,$time=\frac{dis\tan ce}{speed}$. In order to get the smallest possible time, we need to divide the smallest possible distance by the largest possible speed. The minimum distance is 35 miles since 40 has been rounded to the nearest 10. Similarly, the maximum speed is 15.5 mph. Thus, the minimum time is$\frac{35}{15.5}=2.26$hours.