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Every morning Aya goes for a $9$-kilometer-long walk and stops at a coffee shop afterwards.
When she walks at a constant speed of $s$ kilometers per hour, the walk takes her 4 hours, including $t$ minutes spent in the coffee shop.
When she walks $s+2$ kilometers per hour, the walk takes her 2 hours and 24 minutes, including $t$ minutes spent in the coffee shop.
Suppose Aya walks at $s+rac{1}{2}$ kilometers per hour.
Find the number of minutes the walk takes her, including the $t$ minutes spent in the coffee shop.

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<｜User｜>
Every morning Aya goes for a $9$-kilometer-long walk and stops at a coffee shop afterwards.
When she walks at a constant speed of $s$ kilometers per hour, the walk takes her 4 hours, including $t$ minutes spent in the coffee shop.
When she walks $s+2$ kilometers per hour, the walk takes her 2 hours and 24 minutes, including $t$ minutes spent in the coffee shop.
Suppose Aya walks at $s+rac{1}{2}$ kilometers per hour.
Find the number of minutes the walk takes her, including the $t$ minutes spent in the coffee shop.<｜Assistant｜><think>
Okay, so I have this problem where Aya walks a certain distance and stops at a coffee shop. The problem gives me two scenarios with different walking speeds and the corresponding time it takes her, including the time she spends in the coffee shop. I need to find out how long it would take her if she walks at a slightly slower speed.

Let me try to break this down step by step.

First, let's parse the information given:

1. **First Scenario:**
   - Walking speed: \( s \) km/h
   - Total time taken: 4 hours (including \( t \) minutes in the coffee shop)
   - So, the actual walking time is \( 4 - \frac{t}{60} \) hours because \( t \) minutes is converted to hours by dividing by 60.

2. **Second Scenario:**
   - Walking speed: \( s + 2 \) km/h
   - Total time taken: 2 hours and 24 minutes, which is \( 2 + \frac{24}{60} = 2.4 \) hours
   - Again, the actual walking time is \( 2.4 - \frac{t}{60} \) hours.

3. **Third Scenario:**
   - Walking speed: \( s + \frac{1}{2} \) km/h
   - We need to find the total time taken, which is the actual walking time plus \( t \) minutes.

So, the key here is to find \( s \) and \( t \) first, and then calculate the time at the new speed.

Let me denote the walking time in the first scenario as \( t_1 \) and in the second scenario as \( t_2 \).

From the first scenario:
\[ t_1 = 4 - \frac{t}{60} \]

From the second scenario:
\[ t_2 = 2.4 - \frac{t}{60} \]

Since distance is constant, I can set up equations based on the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]

So, for the first scenario:
\[ 9 = s \times t_1 \]
\[ 9 = s \times \left(4 - \frac{t}{60}\right) \]

For the second scenario:
\[ 9 = (s + 2) \times t_2 \]
\[ 9 = (s + 2) \times \left(2.4 - \frac{t}{60}\right) \]

Now, I have two equations:
1. \( 9 = s \times \left(4 - \frac{t}{60}\right) \)
2. \( 9 = (s + 2) \times \left(2.4 - \frac{t}{60}\right) \)

I can solve these two equations simultaneously to find \( s \) and \( t \).

Let me denote \( t_1 = 4 - \frac{t}{60} \) and \( t_2 = 2.4 - \frac{t}{60} \).

From the first equation:
\[ s = \frac{9}{t_1} \]

From the second equation:
\[ s + 2 = \frac{9}{t_2} \]
\[ s = \frac{9}{t_2} - 2 \]

So, I can set the two expressions for \( s \) equal to each other:
\[ \frac{9}{t_1} = \frac{9}{t_2} - 2 \]

Now, substitute \( t_1 \) and \( t_2 \) with their expressions in terms of \( t \):
\[ \frac{9}{4 - \frac{t}{60}} = \frac{9}{2.4 - \frac{t}{60}} - 2 \]

This equation can be solved for \( t \). Let me simplify it step by step.

First, let me denote \( x = \frac{t}{60} \) to make the equation less cluttered.

So, \( t = 60x \).

Substituting back:
\[ \frac{9}{4 - x} = \frac{9}{2.4 - x} - 2 \]

Now, let's solve for \( x \).

Multiply both sides by \( (4 - x)(2.4 - x) \) to eliminate denominators:
\[ 9(2.4 - x) = 9(4 - x) - 2(4 - x)(2.4 - x) \]

Let me compute each term:

Left side:
\[ 9(2.4 - x) = 21.6 - 9x \]

Right side:
First term: \( 9(4 - x) = 36 - 9x \)
Second term: \( -2(4 - x)(2.4 - x) \)
Let me compute \( (4 - x)(2.4 - x) \):
\[ (4)(2.4) - 4x - 2.4x + x^2 = 9.6 - 6.4x + x^2 \]
So, the second term becomes:
\[ -2(9.6 - 6.4x + x^2) = -19.2 + 12.8x - 2x^2 \]

Putting it all together:
Right side = \( 36 - 9x - 19.2 + 12.8x - 2x^2 \)
Simplify:
\( 36 - 19.2 = 16.8 \)
\( -9x + 12.8x = 3.8x \)
So, right side = \( 16.8 + 3.8x - 2x^2 \)

Now, set left side equal to right side:
\[ 21.6 - 9x = 16.8 + 3.8x - 2x^2 \]

Bring all terms to one side:
\[ 21.6 - 9x - 16.8 - 3.8x + 2x^2 = 0 \]
Simplify:
\( 21.6 - 16.8 = 4.8 \)
\( -9x - 3.8x = -12.8x \)
So, equation becomes:
\[ 2x^2 - 12.8x + 4.8 = 0 \]

This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where:
- \( a = 2 \)
- \( b = -12.8 \)
- \( c = 4.8 \)

Let me solve for \( x \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Compute discriminant \( D \):
\[ D = (-12.8)^2 - 4 \times 2 \times 4.8 \]
\[ D = 163.84 - 38.4 \]
\[ D = 125.44 \]

Square root of D:
\[ \sqrt{125.44} = 11.2 \]

So, solutions:
\[ x = \frac{12.8 \pm 11.2}{4} \]

Compute both possibilities:

1. \( x = \frac{12.8 + 11.2}{4} = \frac{24}{4} = 6 \)
2. \( x = \frac{12.8 - 11.2}{4} = \frac{1.6}{4} = 0.4 \)

Now, \( x = \frac{t}{60} \), so:

1. If \( x = 6 \), then \( t = 6 \times 60 = 360 \) minutes, which is 6 hours. But that seems too long for a coffee shop walk, so likely the correct solution is \( x = 0.4 \).

2. If \( x = 0.4 \), then \( t = 0.4 \times 60 = 24 \) minutes.

So, \( t = 24 \) minutes.

Now, let's find \( s \).

From the first scenario:
\[ s = \frac{9}{t_1} \]
But \( t_1 = 4 - \frac{t}{60} = 4 - 0.4 = 3.6 \) hours.

So,
\[ s = \frac{9}{3.6} = 2.5 \) km/h.

Let me verify with the second scenario to ensure consistency.

Second scenario:
\[ s + 2 = 2.5 + 2 = 4.5 \) km/h
Walking time \( t_2 = 2.4 - \frac{t}{60} = 2.4 - 0.4 = 2 \) hours.

So,
\[ 4.5 \times 2 = 9 \) km, which matches the given distance.

Great, so \( s = 2.5 \) km/h and \( t = 24 \) minutes.

Now, the third scenario is when Aya walks at \( s + \frac{1}{2} = 2.5 + 0.5 = 3 \) km/h.

We need to find the total time taken, which is the actual walking time plus \( t \) minutes.

First, compute the walking time at 3 km/h:
\[ \text{Walking time} = \frac{9}{3} = 3 \) hours.

Convert this to minutes:
\[ 3 \times 60 = 180 \) minutes.

Then, add the coffee shop time \( t = 24 \) minutes:
\[ 180 + 24 = 204 \) minutes.

So, the total time is 204 minutes.

Wait, let me double-check the calculations to ensure I didn't make any mistakes.

First, \( s = 2.5 \) km/h, \( t = 24 \) minutes.

Walking time at 3 km/h:
\[ \frac{9}{3} = 3 \) hours = 180 minutes.

Adding coffee shop time:
180 + 24 = 204 minutes.

Yes, that seems correct.

Alternatively, I can check using the equations I set up earlier.

From the first scenario:
\[ t_1 = 4 - \frac{24}{60} = 4 - 0.4 = 3.6 \) hours.

Walking time at 2.5 km/h:
\[ 3.6 \) hours.

Which is 3 hours and 36 minutes, which is 216 minutes.

Wait, but 3.6 hours is 3 hours and 36 minutes, which is 216 minutes.

But the total time including coffee shop is 216 + 24 = 240 minutes, which is 4 hours. That matches the first scenario.

Similarly, at 4.5 km/h:
Walking time is 2 hours, which is 120 minutes.

Total time including coffee shop is 120 + 24 = 144 minutes, which is 2 hours and 24 minutes. That matches the second scenario.

So, all calculations are consistent.

Therefore, at 3 km/h, the total time is 204 minutes.

**Final Answer**
The walk takes Aya 204 minutes, including the coffee shop time.
</think>

Aya walks a 9-kilometer distance and stops at a coffee shop. We need to find the time it takes her to walk at a speed of \( s + \frac{1}{2} \) kilometers per hour, including the time spent in the coffee shop.

1. **First Scenario:**
   - Walking speed: \( s \) km/h
   - Total time: 4 hours (including \( t \) minutes in the coffee shop)
   - Actual walking time: \( 4 - \frac{t}{60} \) hours
   - Equation: \( 9 = s \times \left(4 - \frac{t}{60}\right) \)

2. **Second Scenario:**
   - Walking speed: \( s + 2 \) km/h
   - Total time: 2 hours and 24 minutes (including \( t \) minutes in the coffee shop)
   - Actual walking time: \( 2.4 - \frac{t}{60} \) hours
   - Equation: \( 9 = (s + 2) \times \left(2.4 - \frac{t}{60}\right) \)

3. **Solving the Equations:**
   - Let \( x = \frac{t}{60} \)
   - From the first equation: \( s = \frac{9}{4 - x} \)
   - From the second equation: \( s = \frac{9}{2.4 - x} - 2 \)
   - Setting the expressions for \( s \) equal: \( \frac{9}{4 - x} = \frac{9}{2.4 - x} - 2 \)
   - Solving for \( x \) gives \( x = 0.4 \) (since \( x = 6 \) is too long)
   - Therefore, \( t = 24 \) minutes and \( s = 2.5 \) km/h

4. **Third Scenario:**
   - Walking speed: \( s + \frac{1}{2} = 3 \) km/h
   - Walking time: \( \frac{9}{3} = 3 \) hours = 180 minutes
   - Total time including coffee shop: \( 180 + 24 = 204 \) minutes

The final answer is \(\boxed{204}\) minutes.