Question

Alex could arrive on time for his appointment if he leaves now and he drives 50 mph. However alex leaves the house 15 minutes early driving 40 mph and arrives on time. How far from Alex’s house is the appointment. If the appointment is X miles away from Alex house the equation is X/50= the appointment was __miles away from Alex house

Please help

Answer 1

the appointment was 1 mile away from Alex's house for every minute of driving time at 50 mph.

Explanation:

Let's call the distance from Alex's house to the appointment "d".

If Alex drives 50 mph, he will cover this distance in d/50 hours.

If he leaves 15 minutes (or 0.25 hours) early and drives 40 mph, he will cover the same distance in (d/40) + 0.25 hours.

Since both of these times are the same (since Alex arrives on time), we can set them equal to each other and solve for d:

d/50 = (d/40) + 0.25

Multiplying both sides by 200 (the least common multiple of 50 and 40) to get rid of the denominators, we get:

4d = 5d + 50

Subtracting 4d from both sides, we get:

d = 50

Therefore, the appointment is 50 miles away from Alex's house.

As you mentioned, the equation X/50 = the appointment was __ miles away from Alex house can be used to solve the problem. Plugging in X = 50 gives:

50/50 = 1

So the appointment was 1 mile away from Alex's house for every minute of driving time at 50 mph.