+0

# help algebra

0
98
13

Dr. Worm leaves his house at exactly 7:20 a.m. every morning. When he averages 45 miles per hour, he arrives at his workplace five minutes late. When he averages 63 miles per hour, he arrives five minutes early. What speed should Dr. Worm average to arrive at his workplace precisely on time?

Jun 13, 2021

#1
+114044
+1

This was a little harder than it looked

it is in miles per hour so you must change 5 minutes into hours

You have   45=d/(t+5minutes)      and      63=d/(t-5minutes)

solve simultaneously to get d and t

Then speed = distance/time

Jun 13, 2021
#2
+287
+2

The answer is the harmonic mean of 45 and 63.

Jun 13, 2021
#3
+114044
+1

Sounds likely, would you care to prove it?

Melody  Jun 14, 2021
#6
+287
+2

Let $d$ be the distance.
Let $t_1$ be the time it takes to cover this distance at the rate $r_1$.
Let $t_2$ be the time it takes to cover this distance at the rate $r_2$.

Define $t_a$ to be the A.M. of the two times, that is, $t_a := (t_1+t_2)/2$.
Define $r_a$ to be the H.M. of the two rates, that is, $r_a := 2/((1/r_1)+(1/r_2))$.

Theorem. If one wants to cover distance $d$ in $t_a$ time, one should go at
the rate $r_a$.    Conversely, if one covers the distance at the rate $r_a$, one
will use up $t_a$ time.

Proof.   It's enough to show that $t_a\cdot r_a = d$.   We have
\begin{eqnarray*}
t_a\cdot r_a
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2}{\dfrac{1}{r_1} + \dfrac{1}{r_2}} \\
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2}{\dfrac{t_1}{d} + \dfrac{t_2}{d}} \\
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2}{\dfrac{t_1+t_2}{d}} \\
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2d}{t_1+t_2} \\
&=& d.
\end{eqnarray*}

The theorem extends to any positive integer $n$.

Let $d$ be the distance.
Let $t_1$ be the time it takes to cover this distance at the rate $r_1$.
Let $t_2$ be the time it takes to cover this distance at the rate $r_2$.
...
Let $t_n$ be the time it takes to cover this distance at the rate $r_n$.

Define $t_a$ to be the A.M. of the $n$ times.
Define $r_a$ to be the H.M. of the $n$ rates.

Theorem. If one wants to cover distance $d$ in $t_a$ time, one should go at
the rate $r_a$.    Conversely, if one covers the distance at the rate $r_a$, one
will use up $t_a$ time.

With this theorem, one can pose some seemingly more complicated questions like this one

Question: Dr. Worm leaves his house at exactly 7:20 a.m. every morning.
When he averages 30 miles per hour, he arrives at his workplace ten minutes late.
When he averages 40 miles per hour, he arrives at his workplace five minutes late.
When he averages 60 miles per hour, he arrives fifteen minutes early.
What speed should Dr. Worm average to arrive at his workplace precisely on time?

Bginner  Jun 19, 2021
edited by Bginner  Jun 20, 2021
edited by Bginner  Jun 20, 2021
#7
+114044
0

Thanks very much Bginner.

You have given me homework.

Fair enough, I will take a look a little later :)

oh, you should give yourself a point when you answer questions.  That is what I think anyway :)

Melody  Jun 20, 2021
#8
+114044
0

Attn:  Bginner

Please, what did I do wrong?

Question: Dr. Worm leaves his house at exactly 7:20 a.m. every morning.
When he averages 30 miles per hour, he arrives at his workplace ten minutes late.
When he averages 40 miles per hour, he arrives at his workplace five minutes late.
When he averages 60 miles per hour, he arrives fifteen minutes early.
What speed should Dr. Worm average to arrive at his workplace precisely on time?

I combined what you told me with what is in this youtube clip (the clip made a lot of sense to me)

BUT I GOT THE WRONG ANSWER

t$$30m/h \qquad\qquad t_A+10/60 \\ 40m/h \qquad\qquad t_A+5/60 \\ 60m/h \qquad \qquad t_A -15/60\\ \text{Let the distance travelled be d km}\\ \text{Arithmetic mean of the times is desired arrival time }t_A\\ \text{Harmonic mean of the rates is }r_A\\ r_A=\frac{\text{distance travelled}}{\text{time taken}}\\ \text{the total distance travelled to work on the 3 days is 3d}\\ speed=\frac{distance}{time}\\ time=\frac{distance}{speed}\\~\\ r_A=\frac{3d}{\frac{d}{30}+\frac{d}{40}+\frac{d}{60}}\\ r_A=\frac{3d}{0.075d}\\ r_A=40$$

Which is obviously wrong.

LaTex

\text{Let the distance travelled be d  km}\\
\text{Arithmetic mean of the times is desired arrival time }t_A\\
\text{Harmonic mean of the rates is }r_A\\
r_A=\frac{\text{distance travelled}}{\text{time taken}}\\
\text{the total distance travelled to work on the 3 days is 3d}\\
speed=\frac{distance}{time}\\
time=\frac{distance}{speed}\\~\\

r_A=\frac{3d}{\frac{d}{30}+\frac{d}{40}+\frac{d}{60}}\\
r_A=\frac{3d}{0.075d}\\
r_A=40

Melody  Jun 21, 2021
edited by Melody  Jun 21, 2021
#10
+287
0

Sorry, I made up those numbers and as it turned out those numbers are impossible.   They could not have occurred from a real data.  My bad.

Bginner  Jun 22, 2021
#11
+114044
0

ok.

Thanks for letting me know.

Melody  Jun 22, 2021
#12
+287
+1

There's nothing wrong with the theorem I showed to you.   Well, I proved it, didn't I :)
So it could be wrong only if my proof were invalid!

Now let me explain how I concluded that this problem could be solved by simply
taking the HM of the two given speeds.

Let x be the time it takes in minutes to arrive on time.
To arrive five minutes late means the journey takes x+5 minutes.
To arrive five minutes early means the journey takes x-5 minutes.

We are given that if r1 = 45, t1 = x+5 and
if r2 = 63, t2 = x-5.`

So ta = ((x+5)+(x-5))/2 = x minutes, exactly the the AM of t1 and t2.
(It's immaterial whether you convert the times to hour or not.
Were t1 and t2 to be counted in hours, its AM would still be the on-time time.
Why?  Because the late amount and the early amount equals, so they cancel out
when you calculate the AM.  The problem poser intended it that way.)

So according to the theorem, ra should be the HM of r1 and r2.

Now back to the case of n = 3.
Were there to be a distance d, and speeds r1, r2, r3, and desired speed x
such that (t1 + t2 + t3) / 3 = x, then the AM of r1, r2, r3 would be x and
ra, the HM of the three speeds, would be the answer as well.
However, these numbers r1, r2, r3, t1, t2, t3 must have come from reality.
In a hurry I simply picked arbitrary numbers and used them in my example.
That's bad.  These numbers aren't arbitrary.  They're dependent on one another.
So that's why you get funny result.

Bginner  Jun 22, 2021
#13
+114044
0

Yes, I got that.  thanks :)

The better question would be......

Given.

Question: Dr. Worm leaves his house at exactly 7:20 a.m. every morning.
When he averages 30 miles per hour, he arrives at his workplace ten minutes late.
When he averages 40 miles per hour, he arrives at his workplace five minutes late.
When he averages 60 miles per hour, he arrives fifteen minutes early.

Given the simplest version of this data. (no breakfast breaks, no breakdowns etc)

Prove that this set of data cannot be valid.

Melody  Jun 22, 2021
#4
+2156
+1

"To find the harmonic mean of a set of n numbers, add the reciprocals of the numbers in the set, divide the sum by n, then take the reciprocal of the result." -www.mathworlds.com

So, to find the harmonic mean of 45 you'd do: 1/((1/45+1/63)/2)

Jun 14, 2021
#5
+114044
0

That is great Cal.  I worked that out to be 52.5miles per hour.

Which is the same as I worked out my long method.

I should try and remember that. :)

Melody  Jun 14, 2021