# Thread: Upper and Lower bounds

1. ## 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?

2. 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?
$\displaystyle \displaystyle \frac{990}{2.6} < n < \frac{1010}{2.4}$

3. 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

4. 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.

5. 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?

6. Correct, only that if it's 1005, it'll be rounded to 1010.

EDIT: Didn't read the initial question well, corrected now.

7. Originally Posted by 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?
Yes, if 1000 is the result after rounding to the nearest 10, then a value from above 995
to below 1005 has been approximated as 1000.

So divide 1005 by 2.4 will give a maximum value,
the result of which needs to be rounded up to the next natural number