Count the Physicists in the room

Fun problem!So I'm trying to solve this logically and without the use of mathematics.If 99% of the people out of 100 are physicists , that adds up to a total of 99 physicists now with a little attention it can be easily seen that $\frac{98}{100} = \frac {49}{50}$ also another thing to pay attention to is the fact that when the physicists leave the room the "PEOPLE IN THE ROOM" are also reduced as well as the physicists.Now from the fraction above if 50 physicists leave the room , 98% of the rest are physicists!

let x represent the number of physicst that need to leave

so we get:

99- x \ 100-x = .98

99-x = 98 - .98x

1 = .02x

x = 50

so it is 50 physicist

Could only have achieved an even percentage if every person counted for 2%, as there was only one non-physicist. Thus, bring the total down to 50.

Well 98% is same as $\frac{98}{100}$ which can be simplified to $\frac{49}{50}$ . Now, when 50 Physicist leaves, it's now $\frac{99-50}{100-50}$ , which makes $\frac{49}{50}$ .

I solved it by looking at the number of people who are not physicists. In the first case with 99 percent in a 100 person room there is one person who is not a physicist. So in the 98 percent room there is still that one non-physicist. So then the percentage of non-physicists in the room is 0.02 percent. I then set it as an equation as 1/n = 0.02 where n is the total number of people in the room. Solving for this gave me 50 so therefore 50 physicist have to leave the room to get to exactly 98 percent.

Solved by proper algebra:

99 and 1 for 99/ 100

98 and 1 for 98/ 99

97 and 1 for 97/ 98

...

Therefore we are looking for n/ (n + 1) = 98/ 100 = 49/ 50

50 n = 49 n + 49

n = 49

For n/ (n + 1) = 49/ 50, 99 - 49 = 50 are physicists ought to leave.

let us say there are x physicists left in the room then x+1 total person would be there.. as in the starting one person was not physicist. now % of physicist is 98 therefore we can say that.. x/(x+1)=.98 ==> this gives x=49
therefore if 49 physicist are still there then 50 must have left the room...

Let n be the number of physicists leaving the room.

Since we want to know when the number of physicists in the room is equal to 98%, we want the equation:

$\frac{99-n}{100-n}=\frac{98}{100}$

Cross multiplying yields:

$9800-98n=9900-100n$

$2n=100$

$n=50$

Checking: If 50 people leave, there are now 50 remaining and 49 of which are physicists. $\frac{49}{50}=$ 98%

We start with 99 physicists and 1 normie in the room. The normal person represents 1% of the people in the room, as did each of the 99 physicists, but when some of the physicists left, the percentage of physicists changed to 98%, meaning that 2% of the people in the room are normal... but the number of normal people didn't change. Now, a single normal person represents 2%, doubling the percentage he represents. This also happens to each of the remaining physicists. Since each physicist now represents 2% of the people in there, the number of physicists in the room must be the total percentage of physicists in the room (98) divided by the percentage a single person represents (2). 98/2 is 49. There were 99 physicists before, which means 99 - 49 = 50 physicists must've left the room to achieve this ratio!

One must keep in mind that the number of non-physicists does not change in the process. Also, we have 1% of non-physicists in the beginning, and 2% of non-physicists at the end. Thus, we have :

$\frac{1}{100}\times100 = \frac{2}{100}\times\left(100 - x\right)$ , where $x$ is the number of physicists that need to leave.

Which leads to : $1 = 2 - \frac{2}{100}\times x$ $\Leftrightarrow \frac{2}{100}\times x = 1$ $\Leftrightarrow \color{#20A900} {x = 50}$

$\frac{99-x}{100-x}=\frac{98}{100} \implies 100(99-x)=98(100-x) \implies 9900-100x=9800-98x \implies 100-2x=0 \implies 100-0=2x=100 \implies x=\boxed{\large{50}}$

This is clearly a theoretical physicist problem since the non-physicist would almost certainly try to sneak off at the same time.

Let x represent the total people staying in the room, then 1/x = 2% so 1/x=2/100 so 2x =100 and x= 50. So 50 physicists have to leave the room.

To solve this, you need to find the percent for x physicists... I used 100x/(x+1) = 98, as 100*ratio=% and x/(x+1) is the physicists to the total body. (1 more person) Simplifying gives 100x = 98x + 98, and further 2x = 98. Finally, x = 49!

Now, take the difference to find how many must leave... 98 - 49 = 49! (this is not 49 factorial... just an exclamation mark)

number of physicist = $(0.99)(100)=99$

we let $x$ be the number of physicist that must leave to make the percentage of physicist to exactly be $98$ %

let's set up our equation,

$\frac{99-x}{100-x}$ $=0.98$

solving for, we obtain

$x=50$

Since the original number of people is 100, 99 % of physicists means that there are 99 physicists.

Now, Let the amount of physicists who left be X.

Then ( 99 - X ) / ( 100 - X ) = 98 / 100

( 99 - X ) / ( 100 - X ) = 49 / 50

99 - X = 98 - ( 49 X / 50 )

99 - 98 = X - 49 X / 50

1 = X / 50

X = 50

Here $x$ is equal to the number of physicists that will have to leave the room to bring down the percentage to 98%.

$99-x = \frac {98}{100} (100-x).$

$99-x = 98-\frac {98}{100} x.$

$99 = 98+\frac {x}{50}.$

$\frac {x}{50} = 1.$

$\boxed{x = 50}.$

We can think of the solution in terms of non physicists. Question can be put forth as 2 percent of what (x) is 1. X=50. Hence we have 49 physicists and 1 non physicist. So 50 of them shd leave.

There are 100 people in a room, 99% are physicists. Clearly there are 99 physicists:

$100 * 0.99 = 99$

It's temping to say only one person has to leave, but when that person leaves the total number of people in the room changes from 100 to 99, which changes the percentage left.

Let $x =$ the number of people that need to leave

The total people left in the room will be $100 - x$

The percentage of phycists we want is 98%, or $0.98$

And the resulting number of phycists will be $99 - x$

So we can adapt that first formula into:

$(100-x) * 0.98 = 99 - x$

And then solve for $x$ .

$98 - 0.98 = 99 - x$

$-0.98 + x = 99 - 98$

$0.02x = 1$

$x = 1 \div 0.02$

$x = 50$