A person reaches her office 15 minutes early at 30km/h. At 25 km/h, she is 1 minute early. How many minutes early or late will she be at 20 km/h?

Let's denote the following:
- \( d \) as the distance to the office (in km).
- \( t \) as the time she should ideally take to reach the office (in hours).

Given:
1. At 30 km/h, she is 15 minutes early.
2. At 25 km/h, she is 1 minute early.

We need to find how many minutes early or late she will be at 20 km/h.

First, let's set up the equations based on the given information.

1. At 30 km/h, she takes \( t - \frac{15}{60} \) hours to reach the office:
   \[
   \frac{d}{30} = t - \frac{15}{60}
   \]
   Simplifying, we get:
   \[
   \frac{d}{30} = t - \frac{1}{4}
   \]

2. At 25 km/h, she takes \( t - \frac{1}{60} \) hours to reach the office:
   \[
   \frac{d}{25} = t - \frac{1}{60}
   \]

We now have two equations:
\[
\frac{d}{30} = t - \frac{1}{4}
\]
\[
\frac{d}{25} = t - \frac{1}{60}
\]

We solve these two equations simultaneously. 

First, express \( t \) in terms of \( d \) from the first equation:
\[
t = \frac{d}{30} + \frac{1}{4}
\]

Substitute this expression for \( t \) in the second equation:
\[
\frac{d}{25} = \left( \frac{d}{30} + \frac{1}{4} \right) - \frac{1}{60}
\]

Simplify the equation:
\[
\frac{d}{25} = \frac{d}{30} + \frac{1}{4} - \frac{1}{60}
\]

Find a common denominator for the fractions on the right:
\[
\frac{d}{25} = \frac{d}{30} + \frac{15}{60} - \frac{1}{60}
\]
\[
\frac{d}{25} = \frac{d}{30} + \frac{14}{60}
\]
\[
\frac{d}{25} = \frac{d}{30} + \frac{7}{30}
\]
\[
\frac{d}{25} = \frac{d + 7}{30}
\]

Cross-multiplying to solve for \( d \):
\[
30d = 25(d + 7)
\]
\[
30d = 25d + 175
\]
\[
5d = 175
\]
\[
d = 35 \text{ km}
\]

Now substitute \( d = 35 \) back into the expression for \( t \):
\[
t = \frac{35}{30} + \frac{1}{4}
\]
\[
t = \frac{7}{6} + \frac{1}{4}
\]
\[
t = \frac{14}{12} + \frac{3}{12}
\]
\[
t = \frac{17}{12} \text{ hours}
\]
\[
t = 1 \frac{5}{12} \text{ hours}
\]

This means she should take \( 1 \frac{5}{12} \) hours or \( 1 \times 60 + \frac{5}{12} \times 60 \) minutes:
\[
1 \frac{5}{12} \text{ hours} = 1 \times 60 + 25 = 85 \text{ minutes}
\]

Now, calculate the time taken at 20 km/h:
\[
\frac{35}{20} = 1.75 \text{ hours}
\]
\[
1.75 \text{ hours} = 1 \times 60 + 45 = 105 \text{ minutes}
\]

She should take 85 minutes to reach on time. At 20 km/h, she takes 105 minutes, which means she will be late by:
\[
105 - 85 = 20 \text{ minutes}
\]

Therefore, she will be 20 minutes late at 20 km/h.

Would you like any further details or have any questions?

Here are 5 more questions you might want to ask:
1. How do you calculate the speed needed to arrive exactly on time?
2. Can you solve another problem involving distance, speed, and time?
3. What is the formula for average speed?
4. How do you convert between different units of speed?
5. How do you handle problems involving acceleration?

Tip: When solving problems involving distance, speed, and time, remember that units must be consistent throughout your calculations.

Learn how to solve distance, speed, and time problems with detailed step-by-step solutions. This problem involves calculating arrival time using different speeds and distances. Suitable for students in Grades 7-9.

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