9.2 The banker leaves home, and at that exact moment, a car from the bank arrives to take him to the bank. The banker and the car always leave at the same time, and the car always travels at a constant speed. One day, the banker left home 55 minutes earlier than usual and, for fun, started walking in the direction opposite to the bank. The car caught up with him and took him to the bank 10 minutes later than usual. Find the ratio of the car's speed to the banker's speed.
Answer. The speed of the car is 12 times greater than the speed of the banker.

Solution. Indeed, the car delivered the banker to the bank 10 minutes later than usual, which means it caught up with him 5 minutes later than usual, i.e., from the moment he usually leaves his house. Therefore, the car traveled from the banker's house to the meeting point (or capture point) for 5 minutes. The banker, having left 55 minutes earlier than usual and being caught (captured) by the car, had walked from his house for $55+5=60$ minutes by the time of the meeting. Thus, the car travels 12 times faster than the banker walks.

Grading criteria. ($\cdot$) Proven that the car traveled from the banker's house to the meeting point for 5 minutes: 3 points. ($\cdot$) Proven that the banker had walked for 60 minutes by the time of the meeting: 2 points. ($\cdot$) Proven that the car travels 12 times faster than the banker walks: 2 points.

9.3.. Let $\sqrt{x}-\sqrt{y}=10$. Prove that $x-2 y \leq 200$.

Proof. Rewrite the condition as $\sqrt{x}=\sqrt{y}+10$. Both sides of the equation are non-negative, so we square both sides: $x=100+20 \sqrt{y}+y$. Then $x-2 y=100+20 \sqrt{y}-y=200-(10-\sqrt{y})^{2} \leq 200$, which is what we needed to prove.

Grading criteria. Not mentioning the non-negativity of both sides of the equation before squaring: minus 1 point.