# Thread: Charles Lutwidge Dodgson

1. ## Charles Lutwidge Dodgson

Here is a problem to try and solve:

Jack and Jill walked along a level road, up the hill, back (along the same path) down te hill, and back along the same level road to home. They started out at 3pm and arrived home at 9pm. Their speed was four miles an hour on the level, three miles an hour uphill, and six miles an hour down hill.

a.) How far did they walk in all (level, up, down, level)?

b.) You can't figure out from the given information exactly when they reached the top of the hill. How closely can you approximate when they arrived there? (e.g. can you give an interval of an hour containing the time they arrived at the top? An interval of a half hour?)

2. x = hours walked level; so 6-x = hours up and down

Miles walked
= 4x + 3[2(6-x)/3] + 6(6-x)/3
= 4x + 12 - 2x + 12 - 2x
= 24

3. Hello, matgrl!

Here's another approach to part (a).

Jack and Jill walked along a level road and up a hill.
Then walked back along the same path:
. . down the hill and along the same level road to home.
They started out at 3 pm and arrived home at 9 pm.
Their speed was 4 mph on the level, 3 mph uphill, and 6 mph downhill.

a.) How far did they walk in all (level, up, down, level)?
Code:
                        *
*
* y
*
* * * * * *
x

They walked $\,x$ miles (level) at 4 mph.
. . This took: . $\dfrac{x}{4}$ hours.

Then they walked $\,y$ miles (uphill) at 3 mph.
. . This took: . $\dfrac{y}{3}$ hours.

They walked $\,y$ miles (downhill) at 6 mph.
. . This took: . $\dfrac{y}{6}$ hours.

Then they walked $\,x$ miles (level) at 4 mph.
. . This took: . $\dfrac{x}{4}$ hours.

The entire trip took 6 hours: . $\displaystyle \frac{x}{4} + \frac{y}{3} + \frac{y}{6} + \frac{x}{4} \:=\:6$

Multiply by 12: . $3x + 4y + 2y + 3x \:=\:72$

. . $6x+6y \:=\:72 \quad\Rightarrow\quad x + y \:=\:12$

They walked 12 miles one way.

The entire walk was 24 miles.

4. Hi

Can someone please explain why this is the case?
What I don't understand is that the above equations use (include) the times/speeds for descent (return journey) so why is the result one way?
I appologide in advance for being so dense

5. First do you see that Soroban carefully stated what his variables meant: "x is miles walked on the level", "y is miles walked down hill". He then has an equation that has four parts- timed walked on the level toward the mountain, time walked uphill, time walked down hill, and time walked on the level away from the mountain, using x in both "level" walks, and y in both uphill and downhill walks. The whole route is "level toward mountain" (x), "uphill" (y), "downhill" (y again), and "level away from mountain" (x again). The whole route is x+ y+ x+ y= 2x+ 2y= 2(x+ y) so that x+ y is the first "level toward mountain" plus "uphill" (which is the same as "downhill" plus "level away from mountain". We know that x+ y is one way rather that roundtrip because we know what "x" and "y" separately meant.

(Yes, times/speeds for descent are included in the equation but the distance (y) is the same both ways.)

As I recall, in the Reverand Dodgson's (AKA Lewis Carroll) original puzzle, it was a knight and squire. Why the change to "Jack" and "Jill"?

6. Wow this is wonderful, thank you so much for exaplaining! I have one question...how did you know to multiply by 12?

7. Originally Posted by matgrl
I have one question...how did you know to multiply by 12?
That was simply to get rid of fractions:
we had x/4 + y/3 + y/6 + x/4 = 6
notice that 12 is lowest number in which 3,4 and 6 divide evenly;
multiply each term by 12 (like, x/4 times 12 = 3x):
3x + 4y + 2y + 3x = 72

Got it?

8. How would you do part b of this question? I know the answer but don't understand how to do it. I have if we assume either all flat or uphill that it could be 3 hour or 4 hours. Which would end up bieing 6/7 pm. How can we do this? Please help...

### jack up and down a hill math question

Click on a term to search for related topics.