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Here is an easy riddle. Recommended for those who haven't studied too much of maths.

There are 100 people in a room.

$99\%$ of the people in the room are literate.

Some people are removed.

Now, $98\%$ of the people in the room are literate.

How many people were removed?

And don't tell me that someone was educated sitting in the room... :)

P.S.

If your first answer was $1$, you're wrong.

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  • 1
    $\begingroup$ I think you mean at least how many people were removed. $\endgroup$ – Rohcana Sep 15 '15 at 19:18
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The amount of people removed was

$50$, and all of them were literate.

Proof:

With $99\%$ of people in the room of $100$ being literate, there must be $99$ literate people and $1$ illiterate person. $98\%$ can only be achieved by two fractions: $98/100$ and $49/50$. Since we must remove at least one person to decrease the percentage from $99\%$, the only possible fraction left is $49/50$ to achieve $98\%$ literacy. Therefore, we must remove $50$ literate people from the room.

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  • $\begingroup$ does that means the room is left with 1 illiterate and 49 literate? $\endgroup$ – Alex Sep 15 '15 at 16:58
  • $\begingroup$ @Alex Yes................ $\endgroup$ – ghosts_in_the_code Sep 15 '15 at 16:59
  • $\begingroup$ right i am dumb.... i was thinking how 49 people in a 50 group makes up 98% of the population $\endgroup$ – Alex Sep 15 '15 at 17:18

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