# 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\ A)/(Lower\ 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=(Big)/(Small)=Lar\g\e\st\ n\u\mber`

2. `Lower\ boundary=(Lower\ boundary\ A)/(Upper\ 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=(Small)/(Big)=Smal\l\e\st\ n\u\mber`

**Example 1**

`3.3m` and `4.2m` have been rounded to one decimal point.

What are the maximum and minimum values of `3.3div4.2m`?

First find the upper and lower boundary of `3.3m` which would be:

`3.3m` to one decimal place.

`3.ul3` underline the digit (`1^(st)` decimal place).

`3.ul30` look next door

`0` `5` or more raises the score. So `3.25` would raise to `3.3`.

`0` four or less just ignore `3.349dot9....` 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 of`4.2m` which would be:

`4.2m` to one decimal place.

`4.ul2` underline the digit (`1^(st)` decimal place).

`4.ul20` look next door

`0` `5` or more raises the score. So `4.15` would raise to `4.2`.

`0` four or less just ignore `4.249dot9....` 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`

The upper boundary in division we use the biggest difference between the two.

So the upper boundary of `(3.3)/(4.2)` is `(Big)/(Small)=(3.35)/(4.15)=0.81` to 2 D.P.

The lower boundary in division we use the smallest difference between the two.

So the lower boundary of `(3.3)/(4.2)` is `(Small)/(Big)=(3.25)/(4.25)=0.76` to 2 D.P.

** Answer:** The minimum and maximum values of `3.3div4.2m` 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\ cons\u\m\ed=((d)\ dist\a\nce\ travel\l\e\d\ (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.

`14ul3` underline the digit (`3^(rd)` significant figure).

`14ul3.0` look next door

`0` `5` or more raises the score. So `142.5` would raise to `143`.

`0` four or less just ignore `143.49dot9.....` 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.ul2` underline the digit (`3^(rd)` significant figure).

`13.ul20` look next door

`0` `5` or more raises the score. So `13.15` would raise to `13.2`.

`0` four or less just ignore `13.249dot9......` 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. The upper boundary in division you use the biggest difference between the two so.

`(d)/(f)=(Big)/(Small)=(143.5)/(13.15)=10.91\ km//l` to 2 D.P.

ii. The lower boundary in division you use the smallest difference between the two.

`(d)/(f)=(Small)/(Big)=(142.5)/(13.25)=10.75\ km//l` to 2 D.P.

** **

# 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\ A)/(Lower\ 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=(Big)/(Small)=Lar\g\e\st\ n\u\mber`

2. `Lower\ boundary=(Lower\ boundary\ A)/(Upper\ 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=(Small)/(Big)=Smal\l\e\st\ n\u\mber`

**Example 1**

`3.3m` and `4.2m` have been rounded to one decimal point.

What are the maximum and minimum values of `3.3div4.2m`?

First find the upper and lower boundary of `3.3m` which would be:

`3.3m` to one decimal place.

`3.ul3` underline the digit (`1^(st)` decimal place).

`3.ul30` look next door

`0` `5` or more raises the score. So `3.25` would raise to `3.3`.

`0` four or less just ignore `3.349dot9....` 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 `3.5m` and `3.25m`.

The upper and lower boundary of`4.2m` which would be:

`4.2m` to one decimal place.

`4.ul2` underline the digit (`1^(st)` decimal place).

`4.ul20` look next door

`0` `5` or more raises the score. So `4.15` would raise to `4.2`.

`0` four or less just ignore `4.249dot9....` 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`

The upper boundary in division we use the biggest difference between the two.

So the upper boundary of `(3.3)/(4.2)` is `(Big)/(Small)=(3.35)/(4.15)=0.81` to 2 D.P.

The lower boundary in division we use the smallest difference between the two.

So the lower boundary of `(3.3)/(4.2)` is `(Small)/(Big)=(3.25)/(4.25)=0.76` to 2 D.P.

** Answer:** The minimum and maximum values of `3.3div4.2m` 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\ cons\u\m\ed=((d)\ dist\a\nce\ travel\l\e\d\ (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.

`14ul3` underline the digit (`3^(rd)` significant figure).

`14ul3.0` look next door

`0` `5` or more raises the score. So `142.5` would raise to `143`.

`0` four or less just ignore `143.49dot9.....` 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.ul2` underline the digit (`3^(rd)` significant figure).

`13.ul20` look next door

`0` `5` or more raises the score. So `13.15` would raise to `4.2`.

`0` four or less just ignore `13.249dot9......` 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. The upper boundary in division you use the biggest difference between the two so.

`(d)/(f)=(Big)/(Small)=(143.5)/(13.15)=10.91\ km//l` to 2 D.P.

ii. The lower boundary in division you use the smallest difference between the two.

`(d)/(f)=(Small)/(Big)=(142.5)/(13.25)=10.75\ km//l` to 2 D.P.

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