# Upper and Lower bounds

• Nov 20th 2010, 11:32 AM
Natasha1
Upper and Lower bounds
Hi, i would just need someone to explain to me how to do this question in a very straight forward easy way.

The volume of oil in a tank is 1000 litres, correct to the nearest 10 litres. The oil is poured into tins of volume 2.5 litres, correct to one decimal place.

Calculate the upper bound of the number of tins which will be required?
(Worried)
• Nov 20th 2010, 01:59 PM
skeeter
Quote:

Originally Posted by Natasha1
Hi, i would just need someone to explain to me how to do this question in a very straight forward easy way.

The volume of oil in a tank is 1000 litres, correct to the nearest 10 litres. The oil is poured into tins of volume 2.5 litres, correct to one decimal place.

Calculate the upper bound of the number of tins which will be required?

$\displaystyle \frac{990}{2.6} < n < \frac{1010}{2.4}$
• Nov 20th 2010, 02:01 PM
Natasha1
Thanks. Is the answer to the question Calculate the upper bound of the number of tins which will be required? just n < 1010 / 2.4
• Nov 20th 2010, 03:30 PM
Quote:

Originally Posted by Natasha1
Thanks. Is the answer to the question Calculate the upper bound of the number of tins which will be required? just n < 1010 / 2.4

Yes, the maximum possible amount of oil divided by the minimum amount per tin of 2.4 litres gives the max number of tins.
That will cause the maximum amount of tins to be required.
You will have to round up to the nearest whole number of tins to contain all the oil.
• Nov 21st 2010, 04:20 AM
Natasha1
Yes but to the nearest 10 surely 1000's upper bound is 1004.9999999999999 as soon as it's 1005 it rounds to 1005. So why is it not 1005 divided by 2.4?
• Nov 21st 2010, 05:08 AM
Unknown008
Correct, only that if it's 1005, it'll be rounded to 1010.

EDIT: Didn't read the initial question well, corrected now.
• Nov 21st 2010, 05:13 AM