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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, which includes \( t \) minutes in the coffee shop.
   - Therefore, the actual walking time is \( 4 - \frac{t}{60} \) hours (since \( t \) minutes is \( \frac{t}{60} \) hours).

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 includes \( t \) minutes in the coffee shop. So, the actual walking time will be \( T - \frac{t}{60} \) hours, where \( T \) is the total time in hours.

Since the distance walked is the same in both scenarios, I can set up equations based on the distance formula \( \text{distance} = \text{speed} \times \text{time} \).

Let me denote the distance as \( D \).

**First Scenario:**
\( D = s \times \left(4 - \frac{t}{60}\right) \)

**Second Scenario:**
\( D = (s + 2) \times \left(2.4 - \frac{t}{60}\right) \)

Since both expressions equal \( D \), I can set them equal to each other:

\( s \times \left(4 - \frac{t}{60}\right) = (s + 2) \times \left(2.4 - \frac{t}{60}\right) \)

Now, I can solve this equation for \( s \) in terms of \( t \).

Let me expand both sides:

Left side: \( 4s - \frac{s t}{60} \)

Right side: \( (s + 2) \times 2.4 - (s + 2) \times \frac{t}{60} \)

Expanding further:

Right side: \( 2.4s + 4.8 - \frac{2.4 t}{60} - \frac{2 t}{60} \)

Simplify the terms:

Right side: \( 2.4s + 4.8 - \frac{4.4 t}{60} \)

Now, set left side equal to right side:

\( 4s - \frac{s t}{60} = 2.4s + 4.8 - \frac{4.4 t}{60} \)

Let me bring all terms to one side to solve for \( s \):

\( 4s - 2.4s - \frac{s t}{60} + \frac{4.4 t}{60} - 4.8 = 0 \)

Simplify:

\( 1.6s - \frac{s t}{60} + \frac{4.4 t}{60} - 4.8 = 0 \)

Factor out \( s \) from the first two terms:

\( s \left(1.6 - \frac{t}{60}\right) + \frac{4.4 t}{60} - 4.8 = 0 \)

Now, solve for \( s \):

\( s \left(1.6 - \frac{t}{60}\right) = 4.8 - \frac{4.4 t}{60} \)

So,

\( s = \frac{4.8 - \frac{4.4 t}{60}}{1.6 - \frac{t}{60}} \)

Simplify numerator and denominator:

Multiply numerator and denominator by 60 to eliminate fractions:

Numerator: \( 4.8 \times 60 - 4.4 t = 288 - 4.4 t \)

Denominator: \( 1.6 \times 60 - t = 96 - t \)

So,

\( s = \frac{288 - 4.4 t}{96 - t} \)

Hmm, that seems a bit messy. Maybe I made a calculation error. Let me double-check.

Wait, 4.8 times 60 is indeed 288, and 4.4 times 60 is 264, so numerator is 288 - 264t/60? Wait, no, I think I messed up the multiplication.

Wait, no, actually, when I multiplied numerator and denominator by 60, it's:

Numerator: 4.8 * 60 = 288, and (4.4 t)/60 * 60 = 4.4 t, so numerator is 288 - 4.4 t.

Denominator: 1.6 * 60 = 96, and t * 60 = 60 t, so denominator is 96 - 60 t.

Wait, no, that's not right. Wait, the denominator is 1.6 - t/60, so multiplying numerator and denominator by 60:

Numerator: 4.8 * 60 = 288, and (4.4 t)/60 * 60 = 4.4 t, so numerator is 288 - 4.4 t.

Denominator: 1.6 * 60 = 96, and (t)/60 * 60 = t, so denominator is 96 - t.

Wait, that seems correct.

So, \( s = \frac{288 - 4.4 t}{96 - t} \)

Hmm, maybe I can simplify this fraction.

Let me see if 288 and 96 have a common factor. 288 ÷ 96 = 3, so maybe factor out 96 from numerator and denominator.

Wait, 288 = 3 * 96, and 4.4 t = 4.4 t, and 96 - t is as is.

Alternatively, perhaps I can write 4.4 as 22/5, so 4.4 t = (22/5) t.

So,

\( s = \frac{288 - \frac{22}{5} t}{96 - t} \)

To make it easier, multiply numerator and denominator by 5 to eliminate the fraction:

Numerator: 288 * 5 = 1440, and (22/5 t) * 5 = 22 t, so numerator is 1440 - 22 t.

Denominator: 96 * 5 = 480, and t * 5 = 5 t, so denominator is 480 - 5 t.

So,

\( s = \frac{1440 - 22 t}{480 - 5 t} \)

Hmm, maybe I can factor numerator and denominator:

Numerator: 1440 - 22 t = 2*(720 - 11 t)

Denominator: 480 - 5 t = 5*(96 - t)

Wait, 480 is 5*96, so 480 - 5 t = 5*(96 - t)

Similarly, 1440 is 2*720, and 720 is 15*48, but maybe that's not helpful.

Alternatively, perhaps factor numerator and denominator differently.

Wait, 1440 - 22 t = 2*(720 - 11 t)

Denominator: 480 - 5 t = 5*(96 - t)

Hmm, not sure if that helps.

Alternatively, perhaps perform polynomial division or see if numerator is a multiple of denominator.

Let me see:

Divide 1440 - 22 t by 480 - 5 t.

Let me write it as:

(1440 - 22 t) ÷ (480 - 5 t)

Let me factor out a negative sign from denominator:

= (1440 - 22 t) ÷ (-5 t + 480)

= -(1440 - 22 t) ÷ (5 t - 480)

= -(1440 - 22 t) ÷ (5 t - 480)

Hmm, not sure if that helps.

Alternatively, perhaps express numerator as multiple of denominator plus remainder.

Let me set:

1440 - 22 t = A*(480 - 5 t) + B

Find A and B such that this holds.

Expanding:

1440 - 22 t = 480 A - 5 A t + B

Comparing coefficients:

For t terms: -22 = -5 A => A = 22/5 = 4.4

For constant terms: 1440 = 480 A + B

We have A = 4.4, so:

1440 = 480*4.4 + B

Calculate 480*4.4:

480*4 = 1920

480*0.4 = 192

So total is 1920 + 192 = 2112

Thus, 1440 = 2112 + B => B = 1440 - 2112 = -672

So,

1440 - 22 t = 4.4*(480 - 5 t) - 672

Therefore,

\( s = \frac{1440 - 22 t}{480 - 5 t} = \frac{4.4*(480 - 5 t) - 672}{480 - 5 t} = 4.4 - \frac{672}{480 - 5 t} \)

Hmm, that seems a bit more manageable.

So,

\( s = 4.4 - \frac{672}{480 - 5 t} \)

But I'm not sure if this is helpful yet.

Wait, maybe I can find another equation or approach.

Alternatively, perhaps I can express \( s \) in terms of \( t \) and then plug it into the third scenario.

Wait, in the third scenario, the walking speed is \( s + 0.5 \) km/h, and the total time is \( T \) hours, which includes \( t \) minutes in the coffee shop.

So, the actual walking time is \( T - \frac{t}{60} \) hours.

Therefore, the distance \( D \) is:

\( D = (s + 0.5) \times \left(T - \frac{t}{60}\right) \)

But we also have \( D = s \times \left(4 - \frac{t}{60}\right) \) from the first scenario.

So,

\( s \times \left(4 - \frac{t}{60}\right) = (s + 0.5) \times \left(T - \frac{t}{60}\right) \)

But I don't know \( T \) yet. Hmm.

Wait, maybe I can find \( T \) from the second scenario.

In the second scenario, the walking speed is \( s + 2 \) km/h, and the total time is 2.4 hours, which includes \( t \) minutes in the coffee shop.

So, the actual walking time is \( 2.4 - \frac{t}{60} \) hours.

Therefore, the distance \( D \) is:

\( D = (s + 2) \times \left(2.4 - \frac{t}{60}\right) \)

But from the first scenario, \( D = s \times \left(4 - \frac{t}{60}\right) \)

So, equating the two expressions for \( D \):

\( s \times \left(4 - \frac{t}{60}\right) = (s + 2) \times \left(2.4 - \frac{t}{60}\right) \)

Wait, this is the same equation I had earlier, leading to \( s = \frac{288 - 4.4 t}{96 - t} \)

So, perhaps I can use this expression for \( s \) in the third scenario.

So, in the third scenario, the walking speed is \( s + 0.5 \), and the total time is \( T = 2.4 + \frac{t}{60} \) hours, because the actual walking time is \( 2.4 - \frac{t}{60} \), so total time is walking time plus coffee shop time.

Wait, no, the total time \( T \) is the actual walking time plus the coffee shop time.

So, \( T = \left(2.4 - \frac{t}{60}\right) + \frac{t}{60} = 2.4 \) hours.

Wait, that can't be right because the total time would just be 2.4 hours regardless of \( t \). That doesn't make sense.

Wait, no, actually, the total time \( T \) is the actual walking time plus the coffee shop time.

So, if the actual walking time is \( 2.4 - \frac{t}{60} \), then the total time \( T \) is \( 2.4 - \frac{t}{60} + \frac{t}{60} = 2.4 \) hours.

Wait, that would mean \( T = 2.4 \) hours regardless of \( t \), which doesn't make sense because \( t \) is the time spent in the coffee shop.

Wait, perhaps I'm misunderstanding the problem.

Wait, the problem says: "including \( t \) minutes spent in the coffee shop."

So, the total time is the actual walking time plus \( t \) minutes.

So, if the actual walking time is \( 2.4 - \frac{t}{60} \) hours, then the total time \( T \) is \( 2.4 - \frac{t}{60} + \frac{t}{60} = 2.4 \) hours.

Wait, that can't be right because \( t \) is the time spent in the coffee shop, so the total time should be more than 2.4 hours.

Wait, perhaps I made a mistake here.

Wait, let me clarify:

In the second scenario, the total time is 2.4 hours, which includes \( t \) minutes in the coffee shop.

So, the actual walking time is \( 2.4 - \frac{t}{60} \) hours.

Therefore, the distance \( D \) is:

\( D = (s + 2) \times \left(2.4 - \frac{t}{60}\right) \)

Similarly, in the first scenario, the total time is 4 hours, which includes \( t \) minutes in the coffee shop.

So, the actual walking time is \( 4 - \frac{t}{60} \) hours.

Therefore, the distance \( D \) is:

\( D = s \times \left(4 - \frac{t}{60}\right) \)

Since both expressions equal \( D \), we can set them equal:

\( s \times \left(4 - \frac{t}{60}\right) = (s + 2) \times \left(2.4 - \frac{t}{60}\right) \)

Which is the same equation as before, leading to \( s = \frac{288 - 4.4 t}{96 - t} \)

Now, in the third scenario, the walking speed is \( s + 0.5 \) km/h, and the total time is \( T \) hours, which includes \( t \) minutes in the coffee shop.

So, the actual walking time is \( T - \frac{t}{60} \) hours.

Therefore, the distance \( D \) is:

\( D = (s + 0.5) \times \left(T - \frac{t}{60}\right) \)

But we also have \( D = s \times \left(4 - \frac{t}{60}\right) \) from the first scenario.

So, equating the two expressions for \( D \):

\( s \times \left(4 - \frac{t}{60}\right) = (s + 0.5) \times \left(T - \frac{t}{60}\right) \)

But