Mammoth Memory

Upper and lower boundary and difficult examples

Example 1

The formula for the height a ball can reach when thrown was given as:

`h=(v^2)/(2g)`

Where:

`v= velocity`

`g=gravity`

`h=height`

 

If the velocity is `13.6\ m//s` correct to 1 decimal place and gravity is `9.8\ m//s^2` correct to 1 decimal place. What is the lower boundary also correct to 1 D.P.?

 

NOTE:

The lower boundary of a division would be:

`Lower\ boundary=(Small)/(Big)=(Lower\ boundary\ A)/(Upper\ boundary\ B)`

 

First find the lower boundary of `V=13.6\ m//s` correct to 1 D.P.

`13.6\ m//s` to one decimal place

`13.ul6` underline the digit (the `1^(st)` decimal)

`13.ul6\0` look next door

`color(red)0`     `5` or more raise the score so `13.55` would raise to `13.6`

 

Now find the upper boundary of `g=9.8\ m//s^2` correct to 1 D.P.

`9.8\ m//s^2` to one decimal place

`9.ul8` underline the digit (the `1^(st)` decimal)

`9.ul8\0` look next door

`color(red)0`     `5` or more raise the score so `9.8499` would ignore and stay at `9.8`

This is simplified to `9.85`

 

So the lower boundary of

`H=(V^2)/(2g)`

is

`H=13.55^2/(2times9.85)`

`H=9.319923`

But rounded to 1 D.P. `H=9.3\ metres`

 

Example 2

You are borrowing a friends car and you need to know how expensive the fuel consumption is i.e. miles/gallon (mpg) of the car. Your friend left a note saying that the car travelled 1500miles and she used 31 gallons of fuel.
The mileage was correct to the nearest 100 miles and the gallons used was correct to the nearest gallon.
What are the upper and lower boundaries of the fuel consumption miles/gallon (mpg)?

 

First find the formula used i.e.

`1500\ mi\l\es=1\ gallon`

`x\ mi\l\es=1\ gallon`

(see our notes on % percentages to understand how this works)

Put a division line in

`1500/x=31/1`

Put `x` as the subject

`x=(1500times1)/31`

`x=1500/31=48.39` correct to 2 D.P.

`48.39` miles/gallon would be the answer if there were no boundaries.

But because there are boundaries it is:

`(1500\ t\o\ n\e\arest\ 100\ mi\l\es)/(31\ t\o\ n\e\arest\ gallon)`

 

Find the upper and lower boundary of the 1500 mile to nearest 100 miles.

`1500` to nearest 100 miles

`1ul5\00` underline the digit (the 100's unit)

`1ul5\color(red)0\0` look next door

`0`        `5` or more raise the score so `1450` would raise to `1500`

`0`        `4` or less just ignore `1549...` it would ignore and stay at `1500`

This is simplified to `1550`

 

So the upper and lower boundary of `1500` is `1550` and `1450`.

 

Now find the upper and lower boundaries of `31` gallons to the nearest gallon.

`31`      to nearest 1 gallon

`3ul1`      underline the digit (the 1's unit)

`3ul1.\color(red)0` look next door

`0`        `5` or more raise the score so `30.5` would raise to `31`

`0`        `4` or less just ignore `31.499...` it would ignore and stay at `31`

This is simplified to `31.5`

 

So the upper and lower boundary of `31` is `31.5` and `30.5`

 

Summary

The upper and lower boundary of `1500` is `1550` and `1450`

The upper and lower boundary of `31` is `31.5` and `30.5`

 

NOTE:

The upper boundary of division is `(Big)/(Small)=(Upper\ boundary\ A)/(Lower\ boundary\ B)`

So the upper boundary is `1550/30.5=50.82\ mi\l\es//gallon` correct to 2 D.P.

 

The lower boundary of division is `(Small)/(Big)=(Lower\ boundary\ A)/(Upper\ boundary\ B)`

So the lower boundary is `1450/31.5=46.03\ mi\l\es//gallon` correct to 2 D.P.

 

Answer: The Upper and lower boundaries are `50.82` and `46.03\ mi\l\es//gallon`

 

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