# Problem 1.

B-1

How many times is the second of the numbers $\frac{x}{2}, 2 x-3, \frac{18}{x}+1$ greater than the first, if it is known that it is as many times less than the third? Round the answer to two decimal places.
Answer: $\frac{52}{25}=2.08$

Solution. From the condition $\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}$ we get $a_{1} a_{3}=a_{2}^{2}$ or $\frac{x}{2} \cdot\left(\frac{18}{x}+1\right)=(2 x-3)^{2}$. From this, $8 x^{2}-25 x=0$. For $x=0$, the domain is violated, and for $x=\frac{25}{8}$, we get $a_{1}=\frac{25}{16}, a_{2}=\frac{13}{4}$ and $\frac{a_{2}}{a_{1}}=\frac{52}{25}$.

## B-2

How many times is the second of the numbers $\frac{x}{2}, 3 x-2, \frac{8}{x}+1$ greater than the first, if it is known that it is as many times less than the third? Round the answer to two decimal places.

Answer: 3.12

## B-3

How many times is the second of the numbers $\frac{x}{2}, 2 x-5, \frac{50}{x}+3$ greater than the first, if it is known that it is as many times less than the third? Round the answer to two decimal places.

Answer: $\frac{92}{43} \approx 2.14$

## B-4

How many times is the second of the numbers $\frac{x}{2}, 3 x-5, \frac{50}{x}+3$ greater than the first, if it is known that it is as many times less than the third? Round the answer to two decimal places.

Answer: $\frac{22}{7} \approx 3.14$

## Lomonosov Olympiad in Mathematics

Preliminary Stage 2020/21 Academic Year for 9th Grade

## B-1

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