## Task B-1.3.

A ship traveling along a river has covered $24 \mathrm{~km}$ upstream and $28 \mathrm{~km}$ downstream. For this journey, it took half an hour less than for traveling $30 \mathrm{~km}$ upstream and $21 \mathrm{~km}$ downstream, or half an hour more than for traveling $15 \mathrm{~km}$ upstream and $42 \mathrm{~km}$ downstream, assuming that both the ship and the river move uniformly.

Determine the speed of the ship in still water and the speed of the river.

Solution:
## Solution.

Let $t$ be the time required for the boat to travel $24 \mathrm{~km}$ upstream and $28 \mathrm{~km}$ downstream, $v_{R}$ the speed of the river, and $v_{B}$ the speed of the boat. When the boat is traveling upstream, its speed is $v_{B}-v_{R}$, and when it is traveling downstream, its speed is $v_{B}+v_{R}$.

Since $t=\frac{s}{v}$, from the given data, we obtain the following system of equations:

$\left\{\begin{array}{l}t=\frac{24}{v_{B}-v_{R}}+\frac{28}{v_{B}+v_{R}} \\ t+0.5=\frac{30}{v_{B}-v_{R}}+\frac{21}{v_{B}+v_{R}} \\ t-0.5=\frac{15}{v_{B}-v_{R}}+\frac{42}{v_{B}+v_{R}}\end{array}\right.$

By introducing new variables $x=\frac{3}{v_{B}-v_{R}}, y=\frac{7}{v_{B}+v_{R}}$, the system transforms into:

$\left\{\begin{array}{l}t=8 x+4 y \\ t+0.5=10 x+3 y \\ t-0.5=5 x+6 y\end{array}\right.$

Substituting $t$ from the first equation into the remaining two, we get:

$\left\{\begin{array}{l}8 x+4 y+0.5=10 x+3 y \\ 8 x+4 y-0.5=5 x+6 y\end{array}\right.$

$\left\{\begin{array}{l}2 x-y=0.5 \\ 3 x-2 y=0.5\end{array}\right.$

The solution to the last system is (0.5, 0.5). Then we have:

$\frac{3}{v_{B}-v_{R}}=0.5$, hence, $v_{B}-v_{R}=6 \mathrm{~and}$

$\frac{7}{v_{B}+v_{R}}=0.5$, hence, $v_{B}+v_{R}=14$.

The speed of the river is $v_{R}=4 \mathrm{~km} / \mathrm{h}$, and the speed of the boat is $v_{B}=10 \mathrm{~km} / \mathrm{h}$.

## Note:

By substituting $x=\frac{1}{v_{B}-v_{R}}, y=\frac{1}{v_{B}+v_{R}} \mathrm{~and}$ following the same procedure, the initial system transforms into the system $\left\{\begin{array}{l}6 x-7 y=0.5 \\ 9 x-14 y=0.5\end{array}\right.$

The solution to this system is $\left(\frac{1}{6}, \frac{1}{14}\right)$.

Answer: v_{R}=4\mathrm{~}/\mathrm{},v_{B}=10\mathrm{~}/\mathrm{}

<think>
Okay, so I need to find the speed of the ship in still water and the speed of the river. Let me start by recalling that when a ship is moving upstream, its effective speed is the speed of the ship minus the speed of the river. Conversely, when moving downstream, its effective speed is the ship's speed plus the river's speed.

Let me denote the speed of the ship in still water as \( v \) (in km/h) and the speed of the river as \( r \) (also in km/h). Then, the upstream speed would be \( v - r \), and the downstream speed would be \( v + r \).

The problem mentions three different journeys:

1. 24 km upstream and 28 km downstream, taking half an hour less than the second journey.
2. 30 km upstream and 21 km downstream, which takes half an hour more than the first journey.
3. 15 km upstream and 42 km downstream, which takes half an hour less than the first journey.

Wait, actually, the problem states: "For this journey, it took half an hour less than for traveling 30 km upstream and 21 km downstream, or half an hour more than for traveling 15 km upstream and 42 km downstream..."

Hmm, let me parse that again. The first journey (24 km upstream, 28 km downstream) took half an hour less than the journey with 30 km upstream and 21 km downstream. Alternatively, it took half an hour more than the journey with 15 km upstream and 42 km downstream. So, there are two comparisons here:

- Time for 24 up + 28 down = Time for 30 up + 21 down - 0.5 hours
- Time for 24 up + 28 down = Time for 15 up + 42 down + 0.5 hours

Therefore, we can set up two equations based on these time differences.

First, let me express the time taken for each journey. Time is equal to distance divided by speed. So, for the first journey (24 km up, 28 km down):

Time1 = \( \frac{24}{v - r} + \frac{28}{v + r} \)

For the second journey (30 km up, 21 km down):

Time2 = \( \frac{30}{v - r} + \frac{21}{v + r} \)

According to the problem, Time1 = Time2 - 0.5

Similarly, for the third journey (15 km up, 42 km down):

Time3 = \( \frac{15}{v - r} + \frac{42}{v + r} \)

And here, Time1 = Time3 + 0.5

So, we have two equations:

1. \( \frac{24}{v - r} + \frac{28}{v + r} = \frac{30}{v - r} + \frac{21}{v + r} - 0.5 \)
2. \( \frac{24}{v - r} + \frac{28}{v + r} = \frac{15}{v - r} + \frac{42}{v + r} + 0.5 \)

Okay, so now I need to solve these two equations for \( v \) and \( r \).

Let me first work on the first equation:

\( \frac{24}{v - r} + \frac{28}{v + r} = \frac{30}{v - r} + \frac{21}{v + r} - 0.5 \)

Let me subtract \( \frac{24}{v - r} + \frac{28}{v + r} \) from both sides, but maybe it's better to bring all terms to the left side:

\( \frac{24}{v - r} + \frac{28}{v + r} - \frac{30}{v - r} - \frac{21}{v + r} + 0.5 = 0 \)

Simplify the terms:

For the upstream terms: \( 24 - 30 = -6 \), so \( \frac{-6}{v - r} \)

For the downstream terms: \( 28 - 21 = 7 \), so \( \frac{7}{v + r} \)

So, the equation becomes:

\( \frac{-6}{v - r} + \frac{7}{v + r} + 0.5 = 0 \)

Similarly, for the second equation:

\( \frac{24}{v - r} + \frac{28}{v + r} - \frac{15}{v - r} - \frac{42}{v + r} - 0.5 = 0 \)

Simplify:

Upstream terms: \( 24 - 15 = 9 \), so \( \frac{9}{v - r} \)

Downstream terms: \( 28 - 42 = -14 \), so \( \frac{-14}{v + r} \)

Thus:

\( \frac{9}{v - r} - \frac{14}{v + r} - 0.5 = 0 \)

Now, we have two equations:

1. \( -\frac{6}{v - r} + \frac{7}{v + r} + 0.5 = 0 \)
2. \( \frac{9}{v - r} - \frac{14}{v + r} - 0.5 = 0 \)

Let me denote \( x = \frac{1}{v - r} \) and \( y = \frac{1}{v + r} \). Then, the equations become:

1. \( -6x + 7y + 0.5 = 0 \) ⇒ \( -6x + 7y = -0.5 \)
2. \( 9x - 14y - 0.5 = 0 \) ⇒ \( 9x - 14y = 0.5 \)

Now, we have a system of linear equations:

Equation (1): \( -6x + 7y = -0.5 \)

Equation (2): \( 9x - 14y = 0.5 \)

Let me solve this system using the elimination method. Let's multiply Equation (1) by 2 to make the coefficients of y terms opposites:

Equation (1) multiplied by 2: \( -12x + 14y = -1 \)

Now, add this to Equation (2):

\( (-12x + 14y) + (9x -14y) = -1 + 0.5 \)

Simplify:

\( -3x = -0.5 \)

Thus, \( x = (-0.5)/(-3) = 0.5/3 ≈ 0.1667 \)

But exact value is \( x = 1/6 \)

Now, substitute \( x = 1/6 \) back into Equation (1):

\( -6*(1/6) + 7y = -0.5 \)

Simplify:

\( -1 + 7y = -0.5 \)

Add 1 to both sides:

7y = 0.5

Thus, \( y = 0.5/7 = 1/14 ≈ 0.0714 \)

So, \( x = 1/6 \) and \( y = 1/14 \)

Recall that \( x = \frac{1}{v - r} = 1/6 \) ⇒ \( v - r = 6 \)

Similarly, \( y = \frac{1}{v + r} = 1/14 \) ⇒ \( v + r = 14 \)

Therefore, we have two equations:

1. \( v - r = 6 \)
2. \( v + r = 14 \)

Adding these equations:

\( (v - r) + (v + r) = 6 + 14 \)

Simplify:

\( 2v = 20 \) ⇒ \( v = 10 \)

Substituting back into \( v + r = 14 \):

\( 10 + r = 14 \) ⇒ \( r = 4 \)

So, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.

Let me verify these results by plugging them back into the original time equations.

First journey: 24/(10-4) + 28/(10+4) = 24/6 + 28/14 = 4 + 2 = 6 hours.

Second journey: 30/(10-4) + 21/(10+4) = 30/6 + 21/14 = 5 + 1.5 = 6.5 hours. So, 6.5 - 0.5 = 6 hours, which matches the first journey. Good.

Third journey: 15/(10-4) + 42/(10+4) = 15/6 + 42/14 = 2.5 + 3 = 5.5 hours. Then, 5.5 + 0.5 = 6 hours, which also matches the first journey. Perfect.

Therefore, the solution is correct.

**Final Answer**
The speed of the ship in still water is \boxed{10} km/h and the speed of the river is \boxed{4} km/h.
</think>

Let \( v \) be the speed of the ship in still water (in km/h) and \( r \) be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \( v - r \) and \( v + r \), respectively.

We are given three journeys with the following time relationships:
1. The journey of 24 km upstream and 28 km downstream takes half an hour less than the journey of 30 km upstream and 21 km downstream.
2. The journey of 24 km upstream and 28 km downstream takes half an hour more than the journey of 15 km upstream and 42 km downstream.

We set up the equations for the times taken for each journey:
1. For the first journey (24 km upstream, 28 km downstream):
   \[
   \frac{24}{v - r} + \frac{28}{v + r}
   \]
2. For the second journey (30 km upstream, 21 km downstream):
   \[
   \frac{30}{v - r} + \frac{21}{v + r}
   \]
3. For the third journey (15 km upstream, 42 km downstream):
   \[
   \frac{15}{v - r} + \frac{42}{v + r}
   \]

The time relationships give us the following equations:
1. \[
   \frac{24}{v - r} + \frac{28}{v + r} = \frac{30}{v - r} + \frac{21}{v + r} - 0.5
   \]
2. \[
   \frac{24}{v - r} + \frac{28}{v + r} = \frac{15}{v - r} + \frac{42}{v + r} + 0.5
   \]

Introducing variables \( x = \frac{1}{v - r} \) and \( y = \frac{1}{v + r} \), we transform the equations into:
1. \[
   -6x + 7y = -0.5
   \]
2. \[
   9x - 14y = 0.5
   \]

Solving this system of linear equations:
1. Multiply the first equation by 2:
   \[
   -12x + 14y = -1
   \]
2. Add the modified first equation to the second equation:
   \[
   -3x = -0.5 \implies x = \frac{1}{6}
   \]
3. Substitute \( x = \frac{1}{6} \) into the first equation:
   \[
   -6 \left( \frac{1}{6} \right) + 7y = -0.5 \implies -1 + 7y = -0.5 \implies 7y = 0.5 \implies y = \frac{1}{14}
   \]

Converting back to \( v \) and \( r \):
1. \( v - r = 6 \)
2. \( v + r = 14 \)

Solving these equations:
1. Adding the two equations:
   \[
   2v = 20 \implies v = 10
   \]
2. Substituting \( v = 10 \) into \( v + r = 14 \):
   \[
   10 + r = 14 \implies r = 4
   \]

The speed of the ship in still water is \(\boxed{10}\) km/h and the speed of the river is \(\boxed{4}\) km/h.

<think>
Okay, so I need to solve this problem about a ship traveling upstream and downstream. The question is asking for the speed of the ship in still water and the speed of the river. Let me think through this step by step.

First, let me parse the problem again. The ship covers different distances upstream and downstream, and the time taken for these journeys is compared. Specifically, it says that traveling 24 km upstream and 28 km downstream took half an hour less than traveling 30 km upstream and 21 km downstream. Additionally, that same journey (24 km up, 28 km down) took half an hour more than traveling 15 km upstream and 42 km downstream.

All right, so the key here is that the time differences are half an hour in each case. Since the ship and river are moving uniformly, their speeds are constant. Let me denote the speed of the ship in still water as \( v \) (km/h) and the speed of the river as \( r \) (km/h). Then, when the ship is going upstream, its effective speed is \( v - r \), and downstream it's \( v + r \).

Time is distance divided by speed, so the time taken for each part of the journey (upstream and downstream) can be calculated by dividing the distance by the respective effective speed. The total time for a journey would be the sum of the time upstream and the time downstream.

Let me write down the equations based on the problem statements. Let's consider the first comparison:

1. The journey of 24 km upstream and 28 km downstream took half an hour less than the journey of 30 km upstream and 21 km downstream.

Let me denote the time for 24 km up and 28 km down as \( T_1 \), and the time for 30 km up and 21 km down as \( T_2 \). According to the problem, \( T_1 = T_2 - 0.5 \) hours.

Similarly, the second comparison:

2. The same journey (24 km up, 28 km down) took half an hour more than the journey of 15 km upstream and 42 km downstream.

Let me denote the time for 15 km up and 42 km down as \( T_3 \). Then, \( T_1 = T_3 + 0.5 \) hours.

Therefore, we have two equations:

\[
\frac{24}{v - r} + \frac{28}{v + r} = \frac{30}{v - r} + \frac{21}{v + r} - 0.5
\]

and

\[
\frac{24}{v - r} + \frac{28}{v + r} = \frac{15}{v - r} + \frac{42}{v + r} + 0.5
\]

Hmm, these are the two equations we need to solve. Let me simplify these equations.

Starting with the first equation:

Left side: \( \frac{24}{v - r} + \frac{28}{v + r} \)

Right side: \( \frac{30}{v - r} + \frac{21}{v + r} - 0.5 \)

Let me subtract the left side from the right side to set the equation to zero:

\( \frac{30}{v - r} + \frac{21}{v + r} - 0.5 - \left( \frac{24}{v - r} + \frac{28}{v + r} \right) = 0 \)

Simplify term by term:

\( \frac{30 - 24}{v - r} + \frac{21 - 28}{v + r} - 0.5 = 0 \)

Which simplifies to:

\( \frac{6}{v - r} - \frac{7}{v + r} - 0.5 = 0 \)

Similarly, for the second equation:

Left side: \( \frac{24}{v - r} + \frac{28}{v + r} \)

Right side: \( \frac{15}{v - r} + \frac{42}{v + r} + 0.5 \)

Subtract left side from right side:

\( \frac{15}{v - r} + \frac{42}{v + r} + 0.5 - \left( \frac{24}{v - r} + \frac{28}{v + r} \right) = 0 \)

Simplify term by term:

\( \frac{15 - 24}{v - r} + \frac{42 - 28}{v + r} + 0.5 = 0 \)

Which simplifies to:

\( -\frac{9}{v - r} + \frac{14}{v + r} + 0.5 = 0 \)

So now we have two equations:

1. \( \frac{6}{v - r} - \frac{7}{v + r} = 0.5 \)
2. \( -\frac{9}{v - r} + \frac{14}{v + r} = -0.5 \)

Let me write them as:

Equation (1): \( \frac{6}{v - r} - \frac{7}{v + r} = 0.5 \)

Equation (2): \( -\frac{9}{v - r} + \frac{14}{v + r} = -0.5 \)

Hmm, these are two equations with two variables, \( \frac{1}{v - r} \) and \( \frac{1}{v + r} \). Let me denote \( x = \frac{1}{v - r} \) and \( y = \frac{1}{v + r} \). Then, the equations become:

Equation (1): \( 6x - 7y = 0.5 \)

Equation (2): \( -9x + 14y = -0.5 \)

Now, I can solve this system of linear equations for x and y. Let me write them again:

1. \( 6x - 7y = 0.5 \)
2. \( -9x + 14y = -0.5 \)

Let me use the elimination method. If I multiply equation (1) by 2, I get:

Equation (1a): \( 12x - 14y = 1 \)

Now, equation (2) is \( -9x + 14y = -0.5 \)

If I add equation (1a) and equation (2), the y terms will cancel out:

\( (12x - 14y) + (-9x + 14y) = 1 + (-0.5) \)

Simplify:

\( 3x = 0.5 \)

Therefore, \( x = 0.5 / 3 = 1/6 \)

Now, substitute x = 1/6 back into equation (1):

\( 6*(1/6) -7y = 0.5 \)

Simplify:

\( 1 - 7y = 0.5 \)

Subtract 1:

\( -7y = -0.5 \)

Divide:

\( y = (-0.5)/(-7) = 0.5/7 = 1/14 \)

So, we have x = 1/6 and y = 1/14. Recall that:

\( x = \frac{1}{v - r} = 1/6 \)

\( y = \frac{1}{v + r} = 1/14 \)

Therefore, we can write:

\( v - r = 6 \)

\( v + r = 14 \)

Now, solving these two equations:

Add the two equations:

\( (v - r) + (v + r) = 6 + 14 \)

Which simplifies to:

\( 2v = 20 \)

Therefore, \( v = 10 \) km/h

Substituting back into \( v - r = 6 \):

\( 10 - r = 6 \)

Thus, \( r = 10 - 6 = 4 \) km/h

So, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.

Let me verify these results to ensure there are no mistakes.

First, check the first journey: 24 km upstream and 28 km downstream.

Upstream speed: 10 - 4 = 6 km/h. Time upstream: 24 / 6 = 4 hours.

Downstream speed: 10 + 4 = 14 km/h. Time downstream: 28 / 14 = 2 hours.

Total time: 4 + 2 = 6 hours.

Second journey: 30 km upstream and 21 km downstream.

Time upstream: 30 / 6 = 5 hours.

Time downstream: 21 / 14 = 1.5 hours.

Total time: 5 + 1.5 = 6.5 hours.

Difference between the two journeys: 6.5 - 6 = 0.5 hours, which is half an hour. That's correct.

Third journey: 15 km upstream and 42 km downstream.

Time upstream: 15 / 6 = 2.5 hours.

Time downstream: 42 / 14 = 3 hours.

Total time: 2.5 + 3 = 5.5 hours.

Difference between the first journey (6 hours) and this one: 6 - 5.5 = 0.5 hours, which is half an hour more. That also checks out.

Therefore, the solution seems correct. The speed of the ship in still water is 10 km/h and the speed of the river is 4 km/h.

**Final Answer**
The speed of the ship in still water is \boxed{10} km/h and the speed of the river is \boxed{4} km/h.
</think>

Let \( v \) be the speed of the ship in still water (in km/h) and \( r \) be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \( v - r \) and \( v + r \) respectively.

We are given two comparisons involving the time taken for different journeys. The first comparison states that traveling 24 km upstream and 28 km downstream took half an hour less than traveling 30 km upstream and 21 km downstream. The second comparison states that the same journey (24 km upstream and 28 km downstream) took half an hour more than traveling 15 km upstream and 42 km downstream.

We set up the following equations based on the time differences:

1. For the first comparison:
   \[
   \frac{24}{v - r} + \frac{28}{v + r} = \frac{30}{v - r} + \frac{21}{v + r} - 0.5
   \]
   Simplifying this, we get:
   \[
   \frac{6}{v - r} - \frac{7}{v + r} = 0.5
   \]

2. For the second comparison:
   \[
   \frac{24}{v - r} + \frac{28}{v + r} = \frac{15}{v - r} + \frac{42}{v + r} + 0.5
   \]
   Simplifying this, we get:
   \[
   -\frac{9}{v - r} + \frac{14}{v + r} = -0.5
   \]

Introducing \( x = \frac{1}{v - r} \) and \( y = \frac{1}{v + r} \), we convert these equations into a system of linear equations:
1. \( 6x - 7y = 0.5 \)
2. \( -9x + 14y = -0.5 \)

Solving this system, we multiply the first equation by 2:
1. \( 12x - 14y = 1 \)

Adding this to the second equation:
\[
3x = 0.5 \implies x = \frac{1}{6}
\]

Substituting \( x = \frac{1}{6} \) back into the first equation:
\[
6 \left( \frac{1}{6} \right) - 7y = 0.5 \implies 1 - 7y = 0.5 \implies y = \frac{1}{14}
\]

Thus, we have:
\[
v - r = 6 \quad \text{and} \quad v + r = 14
\]

Solving these equations:
\[
2v = 20 \implies v = 10 \quad \text{and} \quad r = 10 - 6 = 4
\]

The speed of the ship in still water is \boxed{10} km/h and the speed of the river is \boxed{4} km/h.

[User]
## Task B-1.3.

A ship traveling along a river has covered $24 \mathrm{~km}$ upstream and $28 \mathrm{~km}$ downstream. For this journey, it took half an hour less than for traveling $30 \mathrm{~km}$ upstream and $21 \mathrm{~km}$ downstream, or half an hour more than for traveling $15 \mathrm{~km}$ upstream and $42 \mathrm{~km}$ downstream, assuming that both the ship and the river move uniformly.

Determine the speed of the ship in still water and the speed of the river.

[Assistant]
<think>
Okay, so I need to find the speed of the ship in still water and the speed of the river. Let me start by recalling that when a ship is moving upstream, its effective speed is the speed of the ship minus the speed of the river. Conversely, when moving downstream, its effective speed is the ship's speed plus the river's speed.

Let me denote the speed of the ship in still water as \( v \) (in km/h) and the speed of the river as \( r \) (also in km/h). Then, the upstream speed would be \( v - r \), and the downstream speed would be \( v + r \).

The problem mentions three different journeys:

1. 24 km upstream and 28 km downstream, taking half an hour less than the second journey.
2. 30 km upstream and 21 km downstream, which takes half an hour more than the first journey.
3. 15 km upstream and 42 km downstream, which takes half an hour less than the first journey.

Wait, actually, the problem states: "For this journey, it took half an hour less than for traveling 30 km upstream and 21 km downstream, or half an hour more than for traveling 15 km upstream and 42 km downstream..."

Hmm, let me parse that again. The first journey (24 km upstream, 28 km downstream) took half an hour less than the journey with 30 km upstream and 21 km downstream. Alternatively, it took half an hour more than the journey with 15 km upstream and 42 km downstream. So, there are two comparisons here:

- Time for 24 up + 28 down = Time for 30 up + 21 down - 0.5 hours
- Time for 24 up + 28 down = Time for 15 up + 42 down + 0.5 hours

Therefore, we can set up two equations based on these time differences.

First, let me express the time taken for each journey. Time is equal to distance divided by speed. So, for the first journey (24 km up, 28 km down):

Time1 = \( \frac{24}{v - r} + \frac{28}{v + r} \)

For the second journey (30 km up, 21 km down):

Time2 = \( \frac{30}{v - r} + \frac{21}{v + r} \)

According to the problem, Time1 = Time2 - 0.5

Similarly, for the third journey (15 km up, 42 km down):

Time3 = \( \frac{15}{v - r} + \frac{42}{v + r} \)

And here, Time1 = Time3 + 0.5

So, we have two equations:

1. \( \frac{24}{v - r} + \frac{28}{v + r} = \frac{30}{v - r} + \frac{21}{v + r} - 0.5 \)
2. \( \frac{24}{v - r} + \frac{28}{v + r} = \frac{15}{v - r} + \frac{42}{v + r} + 0.5 \)

Okay, so now I need to solve these two equations for \( v \) and \( r \).

Let me first work on the first equation:

\( \frac{24}{v - r} + \frac{28}{v + r} = \frac{30}{v - r} + \frac{21}{v + r} - 0.5 \)

Let me subtract \( \frac{24}{v - r} + \frac{28}{v + r} \) from both sides, but maybe it's better to bring all terms to the left side:

\( \frac{24}{v - r} + \frac{28}{v + r} - \frac{30}{v - r} - \frac{21}{v + r} + 0.5 = 0 \)

Simplify the terms:

For the upstream terms: \( 24 - 30 = -6 \), so \( \frac{-6}{v - r} \)

For the downstream terms: \( 28 - 21 = 7 \), so \( \frac{7}{v + r} \)

So, the equation becomes:

\( \frac{-6}{v - r} + \frac{7}{v + r} + 0.5 = 0 \)

Similarly, for the second equation:

\( \frac{24}{v - r} + \frac{28}{v + r} - \frac{15}{v - r} - \frac{42}{v + r} - 0.5 = 0 \)

Simplify:

Upstream terms: \( 24 - 15 = 9 \), so \( \frac{9}{v - r} \)

Downstream terms: \( 28 - 42 = -14 \), so \( \frac{-14}{v + r} \)

Thus:

\( \frac{9}{v - r} - \frac{14}{v + r} - 0.5 = 0 \)

Now, we have two equations:

1. \( -\frac{6}{v - r} + \frac{7}{v + r} + 0.5 = 0 \)
2. \( \frac{9}{v - r} - \frac{14}{v + r} - 0.5 = 0 \)

Let me denote \( x = \frac{1}{v - r} \) and \( y = \frac{1}{v + r} \). Then, the equations become:

1. \( -6x + 7y + 0.5 = 0 \) ⇒ \( -6x + 7y = -0.5 \)
2. \( 9x - 14y - 0.5 = 0 \) ⇒ \( 9x - 14y = 0.5 \)

Now, we have a system of linear equations:

Equation (1): \( -6x + 7y = -0.5 \)

Equation (2): \( 9x - 14y = 0.5 \)

Let me solve this system using the elimination method. Let's multiply Equation (1) by 2 to make the coefficients of y terms opposites:

Equation (1) multiplied by 2: \( -12x + 14y = -1 \)

Now, add this to Equation (2):

\( (-12x + 14y) + (9x -14y) = -1 + 0.5 \)

Simplify:

\( -3x = -0.5 \)

Thus, \( x = (-0.5)/(-3) = 0.5/3 ≈ 0.1667 \)

But exact value is \( x = 1/6 \)

Now, substitute \( x = 1/6 \) back into Equation (1):

\( -6*(1/6) + 7y = -0.5 \)

Simplify:

\( -1 + 7y = -0.5 \)

Add 1 to both sides:

7y = 0.5

Thus, \( y = 0.5/7 = 1/14 ≈ 0.0714 \)

So, \( x = 1/6 \) and \( y = 1/14 \)

Recall that \( x = \frac{1}{v - r} = 1/6 \) ⇒ \( v - r = 6 \)

Similarly, \( y = \frac{1}{v + r} = 1/14 \) ⇒ \( v + r = 14 \)

Therefore, we have two equations:

1. \( v - r = 6 \)
2. \( v + r = 14 \)

Adding these equations:

\( (v - r) + (v + r) = 6 + 14 \)

Simplify:

\( 2v = 20 \) ⇒ \( v = 10 \)

Substituting back into \( v + r = 14 \):

\( 10 + r = 14 \) ⇒ \( r = 4 \)

So, the speed of the ship in still water is 10 km/h, and the speed of the river is 4 km/h.

Let me verify these results by plugging them back into the original time equations.

First journey: 24/(10-4) + 28/(10+4) = 24/6 + 28/14 = 4 + 2 = 6 hours.

Second journey: 30/(10-4) + 21/(10+4) = 30/6 + 21/14 = 5 + 1.5 = 6.5 hours. So, 6.5 - 0.5 = 6 hours, which matches the first journey. Good.

Third journey: 15/(10-4) + 42/(10+4) = 15/6 + 42/14 = 2.5 + 3 = 5.5 hours. Then, 5.5 + 0.5 = 6 hours, which also matches the first journey. Perfect.

Therefore, the solution is correct.

**Final Answer**
The speed of the ship in still water is \boxed{10} km/h and the speed of the river is \boxed{4} km/h.
</think>

Let \( v \) be the speed of the ship in still water (in km/h) and \( r \) be the speed of the river (in km/h). The effective speeds of the ship upstream and downstream are \( v - r \) and \( v + r \), respectively.

We are given three journeys with the following time relationships:
1. The journey of 24 km upstream and 28 km downstream takes half an hour less than the journey of 30 km upstream and 21 km downstream.
2. The journey of 24 km upstream and 28 km downstream takes half an hour more than the journey of 15 km upstream and 42 km downstream.

We set up the equations for the times taken for each journey:
1. For the first journey (24 km upstream, 28 km downstream):
   \[
   \frac{24}{v - r} + \frac{28}{v + r}
   \]
2. For the second journey (30 km upstream, 21 km downstream):
   \[
   \frac{30}{v - r} + \frac{21}{v + r}
   \]
3. For the third journey (15 km upstream, 42 km downstream):
   \[
   \frac{15}{v - r} + \frac{42}{v + r}
   \]

The time relationships give us the following equations:
1. \[
   \frac{24}{v - r} + \frac{28}{v + r} = \frac{30}{v - r} + \frac{21}{v + r} - 0.5
   \]
2. \[
   \frac{24}{v - r} + \frac{28}{v + r} = \frac{15}{v - r} + \frac{42}{v + r} + 0.5
   \]

Introducing variables \( x = \frac{1}{v - r} \) and \( y = \frac{1}{v + r} \), we transform the equations into:
1. \[
   -6x + 7y = -0.5
   \]
2. \[
   9x - 14y = 0.5
   \]

Solving this system of linear equations:
1. Multiply the first equation by 2:
   \[
   -12x + 14y = -1
   \]
2. Add the modified first equation to the second equation:
   \[
   -3x = -0.5 \implies x = \frac{1}{6}
   \]
3. Substitute \( x = \frac{1}{6} \) into the first equation:
   \[
   -6 \left( \frac{1}{6} \right) + 7y = -0.5 \implies -1 + 7y = -0.5 \implies 7y = 0.5 \implies y = \frac{1}{14}
   \]

Converting back to \( v \) and \( r \):
1. \( v - r = 6 \)
2. \( v + r = 14 \)

Solving these equations:
1. Adding the two equations:
   \[
   2v = 20 \implies v = 10
   \]
2. Substituting \( v = 10 \) into \( v + r = 14 \):
   \[
   10 + r = 14 \implies r = 4
   \]

The speed of the ship in still water is \(\boxed{10}\) km/h and the speed of the river is \(\boxed{4}\) km/h.


3. (6 points) A construction company was building a tunnel. When $\frac{1}{3}$ of the tunnel was completed at the original speed, they started using new equipment, which increased the construction speed by $20 \%$ and reduced the working hours to $80 \%$ of the original. As a result, it took a total of 185 days to complete the tunnel. If they had not used the new equipment and continued at the original speed, it would have taken $\qquad$ days to complete the tunnel.

Solution:
3. (6 points) A construction company builds a tunnel. When $\frac{1}{3}$ of the tunnel is completed at the original speed, new equipment is used, increasing the construction speed by $20 \%$ and reducing the daily working hours to $80 \%$ of the original. As a result, the tunnel is completed in 185 days. If the new equipment was not used and the original speed was maintained, it would take $\qquad$ 180 days.

【Solution】Solution: $\left(1-\frac{1}{3}\right) \div[(1+20 \%) \times 80 \%]$
$$
\begin{array}{l}
=\frac{2}{3} \div[120 \% \times 80 \%], \\
=\frac{2}{3} \div \frac{24}{25} \\
=\frac{25}{36} ; \\
185 \div\left(\frac{1}{3}+\frac{25}{36}\right) \\
=185 \div \frac{37}{36}, \\
=180 \text { (days). }
\end{array}
$$

Answer: If the original speed was maintained, it would take 180 days.
Therefore, the answer is: 180.

Answer: 180

<think>
Okay, let me try to figure out this problem. So, a construction company was building a tunnel. When 1/3 of the tunnel was completed at the original speed, they started using new equipment. This new equipment increased their construction speed by 20% and reduced their working hours to 80% of the original. The total time taken with the new equipment was 185 days. We need to find out how many days it would have taken if they hadn't used the new equipment and continued at the original speed.

Hmm. Let me break this down. Let's denote the total length of the tunnel as 1 unit for simplicity. So, they completed 1/3 of the tunnel at the original speed. Then, the remaining 2/3 was completed with the new equipment, which made them 20% faster and working 80% of the original hours per day.

Wait, the problem says "reduced the working hours to 80% of the original." So, their daily working hours are now 80% of what they used to be. But their speed increased by 20%. So, the rate at which they construct the tunnel per day is affected both by the speed increase and the reduction in working hours.

First, let's define some variables. Let's let the original construction speed be "v" units per day. Wait, but actually, maybe it's better to think in terms of time. Let me think. Suppose the original rate is such that without any changes, they would take T days to complete the tunnel. So, total work is 1 tunnel, original time is T days. Therefore, original rate is 1/T tunnels per day.

But when they started, they completed 1/3 of the tunnel at the original speed, then switched to the new equipment for the remaining 2/3. The total time taken was 185 days. We need to find T.

Wait, but maybe we need to relate the original time to the actual time taken with the changes. Let me approach step by step.

First, original total time would be T days. But they didn't complete the tunnel in T days; instead, they changed their equipment partway. Let's think about the two phases: Phase 1 is building 1/3 of the tunnel at original speed, and Phase 2 is building 2/3 at the new speed with reduced working hours.

Let me define the original rate as R (tunnel per day). Then, the original time to complete the whole tunnel would be T = 1/R days.

In Phase 1, they built 1/3 of the tunnel at rate R. So, the time taken for Phase 1 is (1/3)/R = 1/(3R) days.

In Phase 2, they have 2/3 of the tunnel left. Their speed increases by 20%, so the new speed is R + 20% of R = 1.2R. However, their working hours are reduced to 80% of the original. If the original working hours per day are H, then the new working hours are 0.8H. But how does this affect the construction rate?

Wait, speed is in tunnel per day, which is dependent on how much they work each day. If their speed was originally R tunnels per day with H hours per day, then their rate per hour is R/H tunnels per hour. When they increase their speed by 20%, that would be 1.2*(R/H) tunnels per hour. But they also reduce their working hours to 0.8H. Therefore, the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R tunnels per day. Wait, is that correct?

Wait, maybe. Let me clarify. The original rate is R tunnels per day, which is based on working H hours per day. So, per hour, they complete R/H tunnels. If their speed (efficiency) increases by 20%, then their per hour rate becomes 1.2*(R/H). However, they are now working only 0.8H hours per day, so the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R. So, their daily rate actually decreases to 0.96R? Wait, that seems counterintuitive. If they work faster but fewer hours, the net effect is a decrease in daily rate? Hmm.

Wait, but 20% faster in speed, but 20% less time working. Let me check the math. If original rate is R per day (H hours). Then per hour rate is R/H. After 20% increase, per hour rate is 1.2R/H. Then, working 0.8H hours per day, so daily rate is 1.2R/H * 0.8H = 0.96R. So yes, the daily rate becomes 0.96R. So even though they are working faster, the reduction in hours leads to a lower daily rate. Therefore, the time taken for Phase 2 is (2/3)/0.96R days.

But total time is 185 days, so:

Time for Phase 1 + Time for Phase 2 = 185

Which is:

1/(3R) + (2/3)/(0.96R) = 185

We can solve for R, then T = 1/R is the original total time.

Let me compute that.

First, let's write the equation:

1/(3R) + (2/3)/(0.96R) = 185

Combine the terms:

[1/3 + (2/3)/0.96] / R = 185

Calculate the numerator:

1/3 + (2/3)/0.96

First compute (2/3)/0.96:

2/3 divided by 0.96 = (2/3)*(100/96) = (2/3)*(25/24) = (50/72) = 25/36 ≈ 0.6944

Then add 1/3: 1/3 ≈ 0.3333, so total ≈ 0.6944 + 0.3333 ≈ 1.0277

So numerator ≈ 1.0277

Therefore, 1.0277 / R = 185

So R ≈ 1.0277 / 185 ≈ 0.005555 tunnels per day

Therefore, original time T = 1/R ≈ 1 / 0.005555 ≈ 180 days

Wait, so is T 180 days? Let me check the exact calculation instead of approximating.

Let me compute 1/3 + (2/3)/0.96 exactly.

First, (2/3)/0.96:

0.96 is 24/25, so dividing by 24/25 is multiplying by 25/24.

Thus, (2/3)*(25/24) = (50/72) = 25/36.

Then 1/3 is 12/36, so 12/36 + 25/36 = 37/36.

Therefore, numerator is 37/36.

So the equation is (37/36)/R = 185

Hence, R = 37/(36*185) = 37/(6660) = 1/180 per day.

Therefore, T = 1/R = 180 days.

So the original time would be 180 days.

Wait, so the answer is 180 days? Let me confirm.

If original time is 180 days, then original rate is 1/180 per day.

Phase 1: 1/3 of the tunnel at 1/180 per day. Time = (1/3)/(1/180) = 60 days.

Phase 2: 2/3 of the tunnel at the new rate. The new rate is 0.96*(1/180) = 0.96/180 = 0.005333... per day.

Time for Phase 2: (2/3)/0.005333... = (2/3)/(0.005333...) ≈ (2/3)/ (8/1500) ) Because 0.005333 is 8/1500.

Wait, 0.005333... is 8/1500? Let me check. 8 divided by 1500 is 0.005333..., yes.

So (2/3) divided by (8/1500) = (2/3)*(1500/8) = (3000/24) = 125 days.

Wait, 60 days + 125 days = 185 days, which matches the given total time. Therefore, the original time T is indeed 180 days. So the answer is 180.

Therefore, if they had not used the new equipment, it would have taken 180 days.

**Final Answer**
\boxed{180}
</think>

Let \( T \) be the original number of days required to complete the tunnel. The original construction speed is \( \frac{1}{T} \) tunnels per day.

When \( \frac{1}{3} \) of the tunnel was completed at the original speed, the time taken for this phase is:
\[
\frac{1/3}{1/T} = \frac{T}{3} \text{ days}
\]

For the remaining \( \frac{2}{3} \) of the tunnel, the construction speed increased by 20% and the working hours were reduced to 80% of the original. The new construction speed is \( 1.2 \times \frac{1}{T} \) per hour, but due to reduced working hours, the effective daily rate becomes:
\[
1.2 \times \frac{1}{T} \times 0.8 = \frac{0.96}{T} \text{ tunnels per day}
\]

The time required to complete the remaining \( \frac{2}{3} \) of the tunnel at the new rate is:
\[
\frac{2/3}{0.96/T} = \frac{2T}{3 \times 0.96} = \frac{2T}{2.88} = \frac{25T}{36} \text{ days}
\]

The total time taken with the new equipment is 185 days:
\[
\frac{T}{3} + \frac{25T}{36} = 185
\]

Combining the terms:
\[
\frac{12T}{36} + \frac{25T}{36} = \frac{37T}{36} = 185
\]

Solving for \( T \):
\[
T = 185 \times \frac{36}{37} = 180
\]

Thus, if they had not used the new equipment and continued at the original speed, it would have taken \(\boxed{180}\) days to complete the tunnel.

<think>
Alright, let's try to solve this problem step by step. So, we have a construction company building a tunnel. They completed the first 1/3 of the tunnel at the original speed, then started using new equipment which increased their speed by 20% and reduced their working hours to 80% of the original. The total time taken was 185 days. We need to find out how many days it would have taken if they didn't use the new equipment and continued at the original speed.

Okay, let's break this down. Let's denote the original speed as S (maybe in terms of tunnel length per day), and the original working hours per day as H. Then, with the new equipment, their speed becomes S + 20% of S, which is 1.2S. Their working hours become 80% of H, which is 0.8H.

Wait, but how does working hours per day affect the total days? Hmm. If they reduce the working hours per day, then even if their speed is higher, the amount they can do each day would be speed multiplied by hours. So maybe we need to consider the daily progress before and after the equipment change.

Let me think. Let's assume that originally, without any new equipment, the total tunnel length is L. Then, the total work is L. The original speed is S, so the time to complete the tunnel would be T = L / S days. But they might have a certain number of hours per day, so maybe S is already accounting for hours per day? Wait, the problem mentions "reduced the working hours to 80% of the original". So working hours per day is a factor here.

Wait, perhaps it's better to model this in terms of man-hours or something? Let's see. Let me try to structure it.

Let me denote:

- Original speed: S (tunnel length per hour)

- Original working hours per day: H

So originally, the amount of tunnel they can build in one day is S * H.

After the new equipment:

- Speed increases by 20%, so new speed is 1.2S

- Working hours per day decrease to 80% of original, so new working hours per day is 0.8H

Therefore, the new daily progress is 1.2S * 0.8H = 0.96SH

Wait, that's interesting. So even though speed increased by 20%, the working hours decreased to 80%, so the total daily progress is 0.96 times the original daily progress. So actually, the daily progress decreased by 4%?

But that seems counterintuitive. Let me check that math again.

Original daily progress: S * H

New daily progress: (1.2S) * (0.8H) = 1.2 * 0.8 * S * H = 0.96SH. Yes, so 0.96 times original daily progress. So actually, despite increasing speed by 20%, the reduction in working hours led to a 4% decrease in daily progress.

Hmm, so their daily rate of construction actually decreased slightly when they changed the equipment? But they still managed to complete the tunnel in 185 days. But if they had continued at the original speed and hours, how long would it have taken?

Wait, maybe my approach is missing something. Let's see. The problem says "increased the construction speed by 20% and reduced the working hours to 80% of the original". So speed per hour is increased by 20%, but they are working fewer hours each day.

Alternatively, maybe "construction speed" is already in terms of tunnel length per day? Then, increasing the speed by 20% would mean 1.2 times the original daily speed, but then they reduced the working hours to 80%. Wait, that might be a different interpretation.

Wait, the problem is a bit ambiguous. The construction speed could be per hour or per day. The problem says "increased the construction speed by 20%". If the original speed was, say, x tunnel length per hour, then a 20% increase would make it 1.2x per hour. Then, if they reduce working hours to 80% of original, the daily progress becomes 1.2x * 0.8H = 0.96xH, which is again 96% of the original daily progress. So the same as before.

Alternatively, if the original speed was given as tunnel length per day, and then they increased that speed by 20%, but then reduced the working hours. Hmm. Wait, the problem says "increased the construction speed by 20%". So "construction speed" is probably the rate at which they construct, which could be per hour. Then, when they reduce the working hours, their daily progress is (speed per hour)*(hours per day).

So in any case, the daily progress becomes 0.96 times the original daily progress. So even though they work faster per hour, they work fewer hours each day, resulting in a slight decrease in daily progress.

But according to the problem, after they started using the new equipment, they managed to complete the tunnel in a total of 185 days. But how does that compare to the original plan?

Wait, let's structure the problem.

First, they build 1/3 of the tunnel at the original speed. Then, they build the remaining 2/3 with the new equipment. The total time is 185 days. If they hadn't used the new equipment, they would have continued at the original speed for the entire tunnel, and that hypothetical total time is what we need to find.

So let's denote:

Let T be the original total time needed to build the tunnel without any new equipment. Then, the original daily progress is L / T, where L is the total length.

Wait, maybe that's a better way. Let me assume the total length of the tunnel is 1 unit for simplicity. Then, original total time is T days. So original daily progress is 1/T per day.

But wait, when they start using the new equipment after 1/3 of the tunnel is built, they change their daily progress. Let's figure out the time taken for the first 1/3 and the time taken for the remaining 2/3.

But how is the daily progress affected?

If originally, the entire tunnel takes T days, then the original daily progress is 1/T per day. So the first 1/3 of the tunnel would take (1/3) / (1/T) = T/3 days.

After that, they switch to the new equipment. With the new equipment, their daily progress is 0.96 * (1/T) per day, as we calculated before? Wait, but no. Wait, if the original daily progress is 1/T per day (since they complete 1 tunnel in T days), then with the new equipment, their daily progress is 0.96*(1/T). Therefore, the remaining 2/3 of the tunnel would take (2/3) / (0.96/T) = (2/3)*(T/0.96) = (2T)/(3*0.96) days.

Therefore, the total time is T/3 + (2T)/(3*0.96) = 185 days.

So solving for T:

T/3 + (2T)/(3*0.96) = 185

Let me compute this equation.

First, factor out T/3:

T/3 [1 + 2/0.96] = 185

Compute 2/0.96: 2 divided by 0.96 is equal to 2.083333...

So, 1 + 2.083333... = 3.083333...

Therefore, T/3 * 3.083333... = 185

Multiply both sides by 3:

T * 3.083333... = 555

Therefore, T = 555 / 3.083333...

Compute 3.083333... as a fraction. 3.083333... is 3 + 0.083333..., and 0.083333... is 1/12. So 3 + 1/12 = 37/12.

Therefore, T = 555 / (37/12) = 555 * (12/37) = (555/37)*12

Compute 555 divided by 37:

37*15 = 555, so 555/37 = 15

Therefore, T = 15*12 = 180 days.

Wait, so the original total time would have been 180 days, and according to the problem, with the new equipment, it took 185 days. That seems like using the new equipment actually made them slower? Because 185 is more than 180. But the problem states that "as a result, it took a total of 185 days". Wait, maybe I messed up the daily progress.

Wait, hold on. If the original total time is T days, then when they switch to the new equipment after 1/3 of the tunnel is done, which took T/3 days, and then the remaining 2/3 is done at a slower daily rate (0.96 of original daily rate), which would take longer, so total time becomes longer than T? But that contradicts the problem statement, which probably implies that using the new equipment should have made them faster, but maybe not? Wait, maybe not. Wait, let's check.

Wait, according to the calculation, when they use the new equipment, their daily progress is 96% of the original, so they actually progress slower each day, so the remaining 2/3 takes longer. So even though they increased speed by 20%, the reduction in working hours caused their daily progress to decrease. Therefore, the total time increased from 180 days to 185 days. But the question is, if they had NOT used the new equipment, how long would it have taken? The answer is 180 days.

But let's check again if my reasoning is correct.

Let me verify with numbers. Suppose the original total time is 180 days.

First 1/3 of the tunnel: 180 / 3 = 60 days.

Remaining 2/3 at original speed: 180 - 60 = 120 days.

But with the new equipment, daily progress is 0.96 of original. So time taken for remaining 2/3 is 120 / 0.96 = 125 days.

Total time: 60 + 125 = 185 days. Which matches the problem's statement.

So that makes sense. Therefore, the original total time is 180 days. Therefore, the answer is 180.

But let me think again. The problem says "increased the construction speed by 20% and reduced the working hours to 80% of the original". So construction speed (per hour) increased by 20%, but working hours per day decreased to 80%. So the daily progress is 1.2 * 0.8 = 0.96 times original. So 96% of original daily progress. Therefore, the remaining 2/3 would take (2/3)/0.96 of the original time. Original time for 2/3 is (2/3)*T, so new time is (2/3)*T /0.96 = (2T)/(3*0.96). Then total time is T/3 + (2T)/(3*0.96) = 185.

Solving for T gives 180. Therefore, the answer is 180 days. That seems correct.

Alternatively, maybe my initial interpretation of speed is wrong. Let's try another approach.

Suppose the original speed is S (length per day). Then, the total length L = S * T, where T is the original total time.

They complete 1/3 of the tunnel at speed S, which takes (L/3)/S = T/3 days.

Then, they use new equipment, which increases speed by 20%, so new speed is 1.2S. But working hours are reduced to 80% of original. Wait, if the original speed S is length per day, which is dependent on working hours per day. So if they reduce working hours per day to 80%, but speed per hour increases by 20%, then the new speed in terms of length per day would be 1.2 * original speed per hour * 0.8 * original hours per day.

If original speed per hour is s, then original speed per day is s * H, where H is original hours per day. Then, new speed per day is (1.2s) * (0.8H) = 0.96sH = 0.96S. Therefore, new speed per day is 0.96S.

Therefore, remaining 2/3 of the tunnel would take (2L/3)/(0.96S) = (2/3)*(L/S)/0.96 = (2/3)*T /0.96 = (2T)/(3*0.96) days.

So total time is T/3 + (2T)/(3*0.96) = 185.

Which leads to T = 180. So same result.

Therefore, regardless of whether we model speed as per day or per hour, we get the same result. Therefore, the original total time would have been 180 days.

Therefore, the answer is 180, boxed.

**Final Answer**
\boxed{180}
</think>

Let \( T \) be the original total time needed to complete the tunnel without using the new equipment. The tunnel is divided into two parts: the first \( \frac{1}{3} \) completed at the original speed, and the remaining \( \frac{2}{3} \) completed with the new equipment.

1. **Time to complete the first \( \frac{1}{3} \) of the tunnel:**
   \[
   \frac{T}{3} \text{ days}
   \]

2. **Effect of new equipment:**
   - Construction speed increases by 20%, so the new speed is \( 1.2 \) times the original speed.
   - Working hours are reduced to 80% of the original, so the new working hours are \( 0.8 \) times the original.

   The new daily progress is \( 1.2 \times 0.8 = 0.96 \) times the original daily progress. Therefore, the time to complete the remaining \( \frac{2}{3} \) of the tunnel with the new equipment is:
   \[
   \frac{\frac{2}{3}T}{0.96} = \frac{2T}{3 \times 0.96}
   \]

3. **Total time with new equipment:**
   \[
   \frac{T}{3} + \frac{2T}{3 \times 0.96} = 185
   \]

4. **Solving for \( T \):**
   \[
   \frac{T}{3} + \frac{2T}{2.88} = 185
   \]
   \[
   \frac{T}{3} + \frac{5T}{7.2} = 185
   \]
   Converting to a common denominator:
   \[
   \frac{2.4T}{7.2} + \frac{5T}{7.2} = 185
   \]
   \[
   \frac{7.4T}{7.2} = 185
   \]
   \[
   T = 185 \times \frac{7.2}{7.4}
   \]
   \[
   T = 180
   \]

Thus, if they had not used the new equipment and continued at the original speed, it would have taken \(\boxed{180}\) days to complete the tunnel.

[User] 3. (6 points) A construction company was building a tunnel. When $\frac{1}{3}$ of the tunnel was completed at the original speed, they started using new equipment, which increased the construction speed by $20 \%$ and reduced the working hours to $80 \%$ of the original. As a result, it took a total of 185 days to complete the tunnel. If they had not used the new equipment and continued at the original speed, it would have taken $\qquad$ days to complete the tunnel.

[Assistant]
<think>
Okay, let me try to figure out this problem. So, a construction company was building a tunnel. When 1/3 of the tunnel was completed at the original speed, they started using new equipment. This new equipment increased their construction speed by 20% and reduced their working hours to 80% of the original. The total time taken with the new equipment was 185 days. We need to find out how many days it would have taken if they hadn't used the new equipment and continued at the original speed.

Hmm. Let me break this down. Let's denote the total length of the tunnel as 1 unit for simplicity. So, they completed 1/3 of the tunnel at the original speed. Then, the remaining 2/3 was completed with the new equipment, which made them 20% faster and working 80% of the original hours per day.

Wait, the problem says "reduced the working hours to 80% of the original." So, their daily working hours are now 80% of what they used to be. But their speed increased by 20%. So, the rate at which they construct the tunnel per day is affected both by the speed increase and the reduction in working hours.

First, let's define some variables. Let's let the original construction speed be "v" units per day. Wait, but actually, maybe it's better to think in terms of time. Let me think. Suppose the original rate is such that without any changes, they would take T days to complete the tunnel. So, total work is 1 tunnel, original time is T days. Therefore, original rate is 1/T tunnels per day.

But when they started, they completed 1/3 of the tunnel at the original speed, then switched to the new equipment for the remaining 2/3. The total time taken was 185 days. We need to find T.

Wait, but maybe we need to relate the original time to the actual time taken with the changes. Let me approach step by step.

First, original total time would be T days. But they didn't complete the tunnel in T days; instead, they changed their equipment partway. Let's think about the two phases: Phase 1 is building 1/3 of the tunnel at original speed, and Phase 2 is building 2/3 at the new speed with reduced working hours.

Let me define the original rate as R (tunnel per day). Then, the original time to complete the whole tunnel would be T = 1/R days.

In Phase 1, they built 1/3 of the tunnel at rate R. So, the time taken for Phase 1 is (1/3)/R = 1/(3R) days.

In Phase 2, they have 2/3 of the tunnel left. Their speed increases by 20%, so the new speed is R + 20% of R = 1.2R. However, their working hours are reduced to 80% of the original. If the original working hours per day are H, then the new working hours are 0.8H. But how does this affect the construction rate?

Wait, speed is in tunnel per day, which is dependent on how much they work each day. If their speed was originally R tunnels per day with H hours per day, then their rate per hour is R/H tunnels per hour. When they increase their speed by 20%, that would be 1.2*(R/H) tunnels per hour. But they also reduce their working hours to 0.8H. Therefore, the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R tunnels per day. Wait, is that correct?

Wait, maybe. Let me clarify. The original rate is R tunnels per day, which is based on working H hours per day. So, per hour, they complete R/H tunnels. If their speed (efficiency) increases by 20%, then their per hour rate becomes 1.2*(R/H). However, they are now working only 0.8H hours per day, so the new daily rate is 1.2*(R/H) * 0.8H = 1.2*0.8*R = 0.96R. So, their daily rate actually decreases to 0.96R? Wait, that seems counterintuitive. If they work faster but fewer hours, the net effect is a decrease in daily rate? Hmm.

Wait, but 20% faster in speed, but 20% less time working. Let me check the math. If original rate is R per day (H hours). Then per hour rate is R/H. After 20% increase, per hour rate is 1.2R/H. Then, working 0.8H hours per day, so daily rate is 1.2R/H * 0.8H = 0.96R. So yes, the daily rate becomes 0.96R. So even though they are working faster, the reduction in hours leads to a lower daily rate. Therefore, the time taken for Phase 2 is (2/3)/0.96R days.

But total time is 185 days, so:

Time for Phase 1 + Time for Phase 2 = 185

Which is:

1/(3R) + (2/3)/(0.96R) = 185

We can solve for R, then T = 1/R is the original total time.

Let me compute that.

First, let's write the equation:

1/(3R) + (2/3)/(0.96R) = 185

Combine the terms:

[1/3 + (2/3)/0.96] / R = 185

Calculate the numerator:

1/3 + (2/3)/0.96

First compute (2/3)/0.96:

2/3 divided by 0.96 = (2/3)*(100/96) = (2/3)*(25/24) = (50/72) = 25/36 ≈ 0.6944

Then add 1/3: 1/3 ≈ 0.3333, so total ≈ 0.6944 + 0.3333 ≈ 1.0277

So numerator ≈ 1.0277

Therefore, 1.0277 / R = 185

So R ≈ 1.0277 / 185 ≈ 0.005555 tunnels per day

Therefore, original time T = 1/R ≈ 1 / 0.005555 ≈ 180 days

Wait, so is T 180 days? Let me check the exact calculation instead of approximating.

Let me compute 1/3 + (2/3)/0.96 exactly.

First, (2/3)/0.96:

0.96 is 24/25, so dividing by 24/25 is multiplying by 25/24.

Thus, (2/3)*(25/24) = (50/72) = 25/36.

Then 1/3 is 12/36, so 12/36 + 25/36 = 37/36.

Therefore, numerator is 37/36.

So the equation is (37/36)/R = 185

Hence, R = 37/(36*185) = 37/(6660) = 1/180 per day.

Therefore, T = 1/R = 180 days.

So the original time would be 180 days.

Wait, so the answer is 180 days? Let me confirm.

If original time is 180 days, then original rate is 1/180 per day.

Phase 1: 1/3 of the tunnel at 1/180 per day. Time = (1/3)/(1/180) = 60 days.

Phase 2: 2/3 of the tunnel at the new rate. The new rate is 0.96*(1/180) = 0.96/180 = 0.005333... per day.

Time for Phase 2: (2/3)/0.005333... = (2/3)/(0.005333...) ≈ (2/3)/ (8/1500) ) Because 0.005333 is 8/1500.

Wait, 0.005333... is 8/1500? Let me check. 8 divided by 1500 is 0.005333..., yes.

So (2/3) divided by (8/1500) = (2/3)*(1500/8) = (3000/24) = 125 days.

Wait, 60 days + 125 days = 185 days, which matches the given total time. Therefore, the original time T is indeed 180 days. So the answer is 180.

Therefore, if they had not used the new equipment, it would have taken 180 days.

**Final Answer**
\boxed{180}
</think>

Let \( T \) be the original number of days required to complete the tunnel. The original construction speed is \( \frac{1}{T} \) tunnels per day.

When \( \frac{1}{3} \) of the tunnel was completed at the original speed, the time taken for this phase is:
\[
\frac{1/3}{1/T} = \frac{T}{3} \text{ days}
\]

For the remaining \( \frac{2}{3} \) of the tunnel, the construction speed increased by 20% and the working hours were reduced to 80% of the original. The new construction speed is \( 1.2 \times \frac{1}{T} \) per hour, but due to reduced working hours, the effective daily rate becomes:
\[
1.2 \times \frac{1}{T} \times 0.8 = \frac{0.96}{T} \text{ tunnels per day}
\]

The time required to complete the remaining \( \frac{2}{3} \) of the tunnel at the new rate is:
\[
\frac{2/3}{0.96/T} = \frac{2T}{3 \times 0.96} = \frac{2T}{2.88} = \frac{25T}{36} \text{ days}
\]

The total time taken with the new equipment is 185 days:
\[
\frac{T}{3} + \frac{25T}{36} = 185
\]

Combining the terms:
\[
\frac{12T}{36} + \frac{25T}{36} = \frac{37T}{36} = 185
\]

Solving for \( T \):
\[
T = 185 \times \frac{36}{37} = 180
\]

Thus, if they had not used the new equipment and continued at the original speed, it would have taken \(\boxed{180}\) days to complete the tunnel.