If Julian walks from his house to his school at a speed of 3 miles per hour he will be five minutes late for class. If he cycles to school at a speed of 10 miles per hour, he will be 16 minutes early for class. Find the distance from his house to his school while only using ONE variable to solve the problem. Please help as quickly as possible, this is due today.

To solve this problem, we must first determine the difference between the times taken to reach school when walking and when cycling. In other words, we have to figure out how much time is saved by cycling rather than walking.

Let's assume that Julian's class starts at 10:30 am.

• When walking, he is 5 minutes late for class. Therefore, he would reach school at 10:35 am.

• When cycling, he is 16 minutes early for class. Hence, he would reach school at 10:14 am.

Therefore, the difference is 35 - 14 = 21 minutes

This means that Julian arrives 21 minutes earlier when cycling than he would if he had walked.

Next, let's assume the distance between Julian's house and school is x miles.

Since [tex]\mathrm{time \ taken = \frac{distance}{speed}}[/tex], the time taken when walking and cycling is x/3 hours and x/10 hours respectively.

We know that the difference between these times is 21 minutes, i.e. [tex]\frac{21}{60}[/tex] hours.

Therefore,

[tex]\frac{x}{3} - \frac{x}{10} = \frac{21}{60}[/tex]

⇒ [tex]\frac{10x}{30} - \frac{3x}{30} = \frac{21}{60}[/tex] [Making the denominators of the fractions equal]

⇒ [tex]\frac{10x-3x}{30} = \frac{21}{60}[/tex]

⇒ [tex]7x = \frac{21}{60} \times 30[/tex] [Multiplying both sides of the equation by 30]

⇒ [tex]7x = 10.5[/tex]

⇒ [tex]x = \bf 1.5[/tex]

Therefore, the distance between Julian's house and his school is 1.5 miles.