Upper and lower boundary division
If you are looking to divide one quantity by another and you need to know the possible limits of the upper boundary and lower boundary use logic.
NOTE:
1. Upper boundary=Upper boundary ALower boundary B
If you have the largest possible amount and divide it by the smallest possible amount this will provide you with the biggest divisible number possible.
Upper boundary=BigSmall=Largest number
2. Lower boundary=Lower boundary AUpper boundary B
If you have the smallest possible amount and divide it by the largest possible amount this will provide you with the smallest divisible number possible.
Lower boundary=SmallBig=Smallest number
Example 1
3.3m and 4.2m have been rounded to one decimal point.
What are the maximum and minimum values of 3.3÷4.2m?
First find the upper and lower boundary of 3.3m which would be:
3.3m to one decimal place.
3.3̲ underline the digit (1st decimal place).
3.30̲ look next door
0 5 or more raises the score. So 3.25 would raise to 3.3.
0 four or less just ignore 3.349.9.... it would ignore and stay at 3.3.
This is simplified to 3.35
So the upper and lower boundary of 3.3m to 1 D.P. is 3.35m and 3.25m.
The upper and lower boundary of4.2m which would be:
4.2m to one decimal place.
4.2̲ underline the digit (1st decimal place).
4.20̲ look next door
0 5 or more raises the score. So 4.15 would raise to 4.2.
0 four or less just ignore 4.249.9.... it would ignore and stay at 4.2.
This is simplified to 4.25
So the upper and lower boundary of 4.2m to 1 D.P. is 4.25m and 4.15m.
Summary
Upper and lower boundary of 3.3m is 3.35m and 3.25m
Upper and lower boundary of 4.2m is 4.25m and 4.15m
To work out the upper boundary in division, we use the biggest difference between the two.
So the upper boundary of 3.34.2 is BigSmall=3.354.15=0.81 to 2 D.P.
To work out the lower boundary in division, we use the smallest difference between the two.
So the lower boundary of 3.34.2 is SmallBig=3.254.25=0.76 to 2 D.P.
Answer: The minimum and maximum values of 3.3÷4.2m is 0.81 and 0.76 to 2 D.P.
Example 2
The average fuel consumed by a car is given by the formula:
fuel consumed=(d) distance travelled (kilometres)(f) fuel used (litres)
If d=143 correct to 3 significant figures
f=13.2 litres correct to 3 significant figures
Work out the lower boundary and upper boundary of fuel consumed.
First find the upper and lower boundary of 143 which would be:
143 correct to 3 significant figures.
143̲ underline the digit (3rd significant figure).
143.0̲ look next door
0 5 or more raises the score. So 142.5 would raise to 143.
0 four or less just ignore 143.49.9..... it would ignore and stay at 143.
This is simplified to 143.5
So the upper and lower boundary of 143 to 3 significant figures is 143.5 and 142.5.
The upper and lower boundary of 13.2 to 3 significant figures would be:
13.2 to 3 significant figures.
13.2̲ underline the digit (3rd significant figure).
13.20̲ look next door
0 5 or more raises the score. So 13.15 would raise to 13.2.
0 four or less just ignore 13.249.9...... it would ignore and stay at 13.2.
This is simplified to 13.25
So the upper and lower boundary of 13.2 to 3 significant figures is 13.25 and 13.15.
Summary
Upper and lower boundary of 143 is 143.5m and 142.5m
Upper and lower boundary of 13.2 is 13.25m and 13.15m
i. To work out the upper boundary in division, you use the biggest difference between the two.
df=BigSmall=143.513.15=10.91 km/l to 2 D.P.
ii. To work out the lower boundary in division, you use the smallest difference between the two.
df=SmallBig=142.513.25=10.75 km/l to 2 D.P.
Answer:
The upper boundary of fuel consumed is 10.91 km/l to 2 D.P.
The lower boundary of fuel consumed is 0.75 km/l to 2 D.P.



