29(1105). Two cars set off simultaneously from $A$ to $B$ and from $B$ to $A$, and they met after 3 hours. The first car arrived in $B$ 1.1 hours later than the second car arrived in $A$. How many times greater is the speed of the second car compared to the speed of the first car?
Solution. Let the speed of the first car be $v_{1}$ kilometers per hour, and the speed of the second car be $v_{2}$ kilometers per hour. The first car traveled $3 v_{1}$ kilometers before the meeting, and the second car traveled $3 v_{2}$ kilometers. After the meeting, the first car traveled for $\frac{3 v_{2}}{v_{1}}$ hours, and the second car for $\frac{3 v_{1}}{v_{2}}$ hours.

According to the condition, we have:

$$
\frac{3 v_{2}}{v_{1}}-\frac{3 v_{1}}{v_{2}}=\frac{11}{10}
$$

To solve this equation, let $\frac{v_{2}}{v_{1}}=x$. Then $\frac{v_{1}}{v_{2}}=\frac{1}{x}, 3 x-\frac{3}{x}=\frac{11}{10}$, or $30 x^{2}-11 x-30=0$. The condition of the problem is satisfied only by the positive root $x=\frac{6}{5}, \frac{v_{2}}{v_{1}}=\frac{6}{5}$.

Answer: 1.2 times.

Remark 1. The question to the problem is better formulated as: “Find the ratio of the speeds of the cars.”

Remark 2. After problem 29(1105), it is advisable to solve problem 1005 from [2].

30(1106). The ore mined in a quarry can be hauled out by the first truck 3 hours faster than by the second truck. If one-third of the ore is hauled out by the first truck and the remaining part by the second truck, it will take $7 \frac{1}{3}$ hours longer than if both trucks worked simultaneously. How many hours can each truck haul out the ore?

Solution. Let the first truck haul out the ore in $t$ hours, then the second truck will take $(t+3)$ hours. In 1 hour, the first truck can haul $\frac{1}{t}$ part, and the second truck $\frac{1}{t+3}$ part of the ore mined in the quarry. When working simultaneously, in 1 hour the trucks can haul $\frac{2 t+3}{t(t+3)}\left(\frac{1}{t}+\frac{1}{t+3}=\frac{2 t+3}{t(t+3)}\right)$ part of the ore, spending $\frac{t(t+3)}{2 t+3}$ hours on the entire work $\left(1: \frac{2 t+3}{t(t+3)}=\frac{t(t+3)}{2 t+3}\right)$.

The first truck can haul $\frac{1}{3}$ part of the ore in $\frac{1}{3}$ hours, and the remaining part $\left(\frac{2}{3}\right)$ the second truck can haul in $\frac{2 t+6}{3}$ hours $\left(\frac{2}{3}: \frac{1}{t+3}=\frac{2 t+6}{3}\right)$.

According to the condition of the problem, we form the equation:

$$
\frac{t}{3}+\frac{2 t+6}{3}=\frac{t(t+3)}{2 t+3}+\frac{22}{3}
$$

from which $3 t^{2}-32 t-48=0, \quad t_{1}=12, \quad t_{2}=-\frac{4}{3}$. The condition of the problem is satisfied only by the positive root of the equation: $t=12$, $t+3=15$.

Answer: in 12 hours and 15 hours.

Remark. It is useful to check the solution according to the condition of the problem.

31(1107). Two mechanics received a task. For its completion, the first mechanic would need 7 hours more than the second. After both mechanics completed half of the task, the work had to be finished by the second mechanic alone, and the task was completed 4.5 hours later than if they had completed the entire task together. How many hours could each mechanic complete the task?

Solution. Let the first mechanic need $t$ hours to complete the task, and the second $(t-7)$ hours. In 1 hour, the first mechanic can complete $\frac{1}{t}$ part of the task, and the second $\frac{1}{t-7}$ part. When working simultaneously, in 1 hour both mechanics can complete $\frac{2 t-7}{t(t-7)}$ part of the task $\left(\frac{1}{t}+\frac{1}{t-7}=\frac{2 t-7}{t(t-7)}\right)$, spending $\frac{t(t-7)}{2 t-7}$ hours on the entire work $\left(1: \frac{2 t-7}{t(t-7)}=\frac{t(t-7)}{2 t-7}\right)$.

For half of the task, both mechanics working together spent $\frac{t(t-7)}{2(2 t-7)}$ hours, and the second mechanic spent $\frac{t-7}{2}$ hours on the remaining half of the task.

According to the condition of the problem, we form the equation:

$$
\frac{t(t-7)}{2(2 t-7)}+\frac{t-7}{2}=\frac{t(t-7)}{2 t-7}+\frac{9}{2}
$$

from which

$$
t^{2}-32 t+112=0, t_{1}=28, t_{2}=4
$$

The root $t_{2}=4$ does not satisfy the condition of the problem, as $t_{2}-7<0$. Therefore, the first mechanic can complete the task in 28 hours, and the second in 21 hours $(28-7=21)$.

Answer: in 28 hours and 21 hours.

Remark. It is advisable to check the solution according to the condition of the problem.

32(1108). A two-digit number is given. The number of its units is 3 less than the number of tens. The product of this number and the number written with the same digits in reverse order is 574. Find the given number.

Solution. Let the number of units of the desired number be $x$, and the number of tens be $x+3$. Then the desired number is $10(x+3)+x$. According to the condition, we have:

$$
(10(x+3)+x)(10 x+x+3)=574 \text {, or } x^{2}+3 x-4=0 \text {, }
$$

from which $x=1$. (The negative root $x=-4$ does not satisfy the condition of the problem.) The desired number is $41(10 \cdot 4+1=41)$.

By checking, we easily see that the problem is solved correctly: $41 \cdot 14=574$.

Answer: 41