Befuddling word problem

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Applz

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Peter Parker leaves his aunt's house at exactly 7:00 a.m. every morning to get to school. When he averages 40 miles per hour, he arrives at his school exactly 3 minutes late. When he averages 60 miles per hour, he arrives three minutes early.
Find x using the information you found above. How many miles is it from the house to school?

This is the first problem in a set of two I was stuck at the beginning i review my notes and cant understand how to solve it.

Peter Parker leaves his aunt's house at exactly 7:00 a.m. every morning to get to school. When he averages 40 miles per hour, he arrives at his school exactly 3 minutes late. When he averages 60 miles per hour, he arrives three minutes early.
At what average speed, in miles per hour, should Peter Parker travel to arrive at his school precisely on time? (just enter the number, not mph)

Here is the second part if someone could help me that would be great!
 
boo, distance = rate x time

that's one of those equations that you as a non caveman should just know.

We don't know how far away the school is from his house so we'll call it [MATH]D[/MATH] miles
We don't know how long the trip should take yet so we'll call it [MATH]T[/MATH] minutes

If he travels at 40 mph he's 3 minutes late.
We need to convert miles per hour to miles per minute to keep the units consistent.
40 mph is 2/3 mile per minute.

[MATH]\dfrac{D}{\frac 2 3} = T+3[/MATH]
Now you figure out and post the second equation you can get from the given info.
 
Applz said:
Peter Parker leaves his aunt's house at exactly 7:00 a.m. every morning to get to school. When he averages 40 miles per hour, he arrives at his school exactly 3 minutes late. When he averages 60 miles per hour, he arrives three minutes early.
Find x using the information you found above. How many miles is it from the house to school?

This is the first problem in a set of two I was stuck at the beginning i review my notes and cant understand how to solve it.

Peter Parker leaves his aunt's house at exactly 7:00 a.m. every morning to get to school. When he averages 40 miles per hour, he arrives at his school exactly 3 minutes late. When he averages 60 miles per hour, he arrives three minutes early.
At what average speed, in miles per hour, should Peter Parker travel to arrive at his school precisely on time? (just enter the number, not mph)

Here is the second part if someone could help me that would be great!
Click to expand...
Peter Parker leaves his aunt's house at exactly 7:00 a.m. every morning to get to school. When he averages 40 miles per hour, he arrives at his school exactly 3 minutes late. When he averages 60 miles per hour, he arrives three minutes early.
Find x using the information you found above. How many miles is it from the house to school?

What do you need to FIND in this problem?

Applz said:
Peter Parker leaves his aunt's house at exactly 7:00 a.m. every morning to get to school. When he averages 40 miles per hour, he arrives at his school exactly 3 minutes late. When he averages 60 miles per hour, he arrives three minutes early.
At what average speed, in miles per hour, should Peter Parker travel to arrive at his school precisely on time? (just enter the number, not mph)
Click to expand...
What do you need to FIND in this problem?
 
Applz said:
Peter Parker leaves his aunt's house at exactly 7:00 a.m. every morning to get to school. When he averages 40 miles per hour, he arrives at his school exactly 3 minutes late. When he averages 60 miles per hour, he arrives three minutes early.
Find x using the information you found above. How many miles is it from the house to school?

This is the first problem in a set of two I was stuck at the beginning i review my notes and cant understand how to solve it.

Peter Parker leaves his aunt's house at exactly 7:00 a.m. every morning to get to school. When he averages 40 miles per hour, he arrives at his school exactly 3 minutes late. When he averages 60 miles per hour, he arrives three minutes early.
At what average speed, in miles per hour, should Peter Parker travel to arrive at his school precisely on time? (just enter the number, not mph)

Here is the second part if someone could help me that would be great!
Click to expand...
Tedious but straight forward.

Let D be the distance he travels to school, in miles, and T the times school starts. If he starts at 7 A.M. the time it should take him to be on time is T- 7 hours. Travelling at v miles per hour, v(T- 7)= D.

At 40 miles per hour he is three minutes late. Since 3 minutes is 3/60= 1/20 hour, it must take him T- 7+ 1/20 hours. D= 40(T- 7+ 1/20) .

At 60 miles per hour he is three minutes early so it takes him T- 7- 1/20 hours.
D= 60(T- 7- 1/20).

D= 40(T- 7+ 1/20)= 60(T- 7- 1/20).
Divide both sides by 20:
2(T- 7+ 1/20)= 3(T- 7- 1/20).
2T- 14+ 2/20= 3T- 21- 3/20.
T= 21- 14+ 2/20+ 3/20= 7+ 5/20.
T= 7+ 1/4. School starts at 7:15.
In order to be exactly on time he must take 15 minutes to go to school.

T= 28/4+ 1/4= 29/4.
So D= 40(29/4- 7+ 1/20)= 40(1/4+ 1/20)= 10+ 2. He goes 12 miles to school,

Check:
At 40 mph, he will take 12/40= 3/10 hour or 18 minutes arriving at 7:18, 3 minutes late.
At 60 mph, he will take 12/60 hour or 12 minutes arriving at 7:12, 3 minutes early.


How fast should he drive to go 12 miles in 15 min= 1/4 hour?

(That is, in fact, the harmonic mean of 40 and 60 mph.)
Harmonic mean - Wikipedia
 
Last edited:
HallsofIvy said:
Tedious but straight forward.

Let D be the distance he travels to school, in miles, and T the times school starts. If he starts at 7 A.M. the time it should take him to be on time is T- 7 hours. Travelling at v miles per hour, v(T- 7)= D.

At 40 miles per hour he is three minutes late. Since 3 minutes is 3/60= 1/20 hour, it must take him T- 7+ 1/20 hours. D= 40(T- 7+ 1/20) .

At 60 miles per hour he is three minutes early so it takes him T- 7- 1/20 hours.
D= 60(T- 7- 1/20).

D= 40(T- 7+ 1/20)= 60(T- 7- 1/20).
Divide both sides by 60:
(2/3)(T- 7+ 1/20)= T- 7- 1/20.
(2/3)T- 14/3+ 1/30= T- 7- 1/20.
-(1/3)T= 14/3- 7- 1/20- 1/10= 14/3- 21/3- 1/20- 3/30
-(1/3)T= -7/3- 4/30= -7/3- 5/60= -7/3- 1/12.
T= 7+ 1/4. School starts at 7:15.
In order to be exactly on time he must take 15 minutes to go to school.

T= 28/4+ 1/4= 29/4.
Also D= 40(29/4- 7+ 1/20)= 40(1/4+ 1/20)= 10+ 2. He goes 12 miles to school,

Check:
At 40 mph, he will take 12/40= 3/10 hour or 18 minutes arriving at 7:18, 3 minutes late.
At 60 mph, he will take 12/60 he will take 12 minutes arriving at 7:12, 3 minutes early.


How fast should he drive to go 12 miles in 15 min= 1/4 hour?
Click to expand...
Prof. Ivey,

Please do not give out full answer especially before the original-poster showed any significant effort
 
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