Percentage "MORE"

[niravrph]

Hi,

If I know that 65% of people will drop out of a study but I want to end up with 450 people at the end of the study --- how many people must I start out with?

How do I calculate this?

THANKS!

 

[soroban]

Hello, niravrph!

If I know that 65% of people will drop out of a study,
but I want to end up with 450 people at the end of the study.
How many people must I start out with?

Let \(\displaystyle x\) = number of people (originally).

Only 35% of them will finish the study.

We want 35% of \(\displaystyle x\) to equal 450.

There is our equation . . . \(\displaystyle 0.35x \,=\,450\)

. . \(\displaystyle x \:=\:\dfrac{450}{0.35} \:=\:1285.71428\)


You must start with 1286 people.

 

[niravrph]

Ahhh... perfect. Thank you!!

 

[mmm4444bot]

Soroban, why could the answer not be 1,285 people or 1,287 people?

 

[soroban]

Hello, mmm4444bot!

You're mostly right . . . and thanks for the heads-up.

Soroban, why could the answer not be 1,285 people or 1,287 people?

I assume we are dealing with "whole" people . . . so we "drop decimals".

\(\displaystyle 35\%\text{ of }1285 \:=\:449.75\quad\Rightarrow\quad \text{only 449 people}\)


However:

. . \(\displaystyle \begin {array}{ccccc}35\%\text{ of }1287 \:=\: 450.45 & \Rightarrow & \text{450 people} \\ 35\%\text{ of }1288 \:=\:450.80 & \Rightarrow & \text{450 people} \end{array}\)


The answer is: .\(\displaystyle 1286, 1287,\text{ or }1288\)

 

[mmm4444bot]

I assume we are dealing with "whole" people . . . so we "drop decimals".

I had considered this, but also I had noticed that you rounded up your calculation for x, so I was not certain.

Seems like there's more than one possible answer regardless of rounding scheme.