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We say that a bus is overcrowded if at some moment there is at least 50 passengers in it. Two inspectors are monitoring the number of passengers in 10 buses. One of them computed what is the percentage of the overcrowded buses, the second one - what percentage of all passengers constituted the passengers in the overcrowded buses. It is known that the number of the overcrowded busus is in set $\{1, 2, \dots, 9 \}$. Which of the inspectors got the greater number?

Lekt $k$ be the number of the overcrowded buses. Then the percentage of the overcrowded buses is $10k$. The number of passengers in the overcrowded buses $\ge 50k$ and in the other buses $< 50(10-k)$. Could be those estimations be useful in the solution?

mrnobody
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1 Answers1

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Take a guess, what do you think the answer is? If you guessed that the greater number of passengers in crowded buses contributed to the greater percentage of passengers in a overcrowded bus, you are correct. Let's try to prove it.

We are thus trying to show that:

$\frac{\text{Passengers in crowded bus}}{\text{Total passengers}}>\frac{k}{10}$

Equivalently, multiplying out, $10\times\text{Passengers in crowded bus}>k\times\text{Total passengers}$.

$\Leftrightarrow 10\times\text{Passengers in crowded bus}>k\times(\text{Passengers in crowded bus + Passengers in not-crowded bus})$

$\Leftrightarrow (10-k)\times\text{Passengers in crowded bus}>k\times\text{Passengers in not-crowded bus}$

As $k\neq 0, 10$, we can rearrange to form:

$\Leftrightarrow \frac{\text{Passengers in crowded bus}}{k}>\frac{\text{Passengers in not-crowded bus}}{10-k}$

Applying the inequalities you found,

$\frac{\text{Passengers in crowded bus}}{k}\geq 50>\frac{\text{Passengers in not-crowded bus}}{10-k}$,

so we are done.

Element118
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