shymexx:

Lmao... Can you provide the intersection point, bro?

1. For the specific choice I gave, yes. But of course, where exactly the point is will depend on his position.
2. Keep in mind that the problem doesn't ask you to provide the intersection point (i.e., where on the hill this event happens). It just asks, "Does there exist such a time?"

ekt_bear:
1. For the specific choice I gave, yes. But of course, where exactly the point is will depend on his position.
2. Keep in mind that the problem doesn't ask you to provide the intersection point (i.e., where on the hill this event happens). It just asks, "Does there exist such a time?"

Obviously, such a time would exist since he has to go through the same path on both days... However, the time he takes to walk the path on both days and the distance between his house and the hill aren't given, then the most likely time has to be the entry/exit point... You also have to put the "constant" into consideration...

If:
x(t) = t^2, y(t) = 1-t

x(t)=y(t)

t^2=1-t
t^2+t-1=0

t= (-1-sqrt(1+4))/2 impossible since time cannot be negative
t= (-1+sqrt(1+4))/2 = 0.618 ?

Sybellah:
If:
x(t) = t^2, y(t) = 1-t

x(t)=y(t)

t^2=1-t
t^2+t-1=0

t= (-1-sqrt(1+4))/2 impossible since time cannot be negative
t= (-1+sqrt(1+4))/2 = 0.618 ?

Yes ma'am.

You clearly remember your high school mathematics well

Hmm. I'm now intrigued more by you...

ekt_bear and sybellah

Use the same method to solve this:

Some friends on a hike leave at 9am and return at 2pm. They spend one-quarter of the total distance walking uphill, one-half walking on level ground, and one-quarter walking downhill. if their speed is 4mph on level ground, 2 mph uphill, and 6 mph downhill,approximately how far did they walk?
A. 16.9 miles B. 17.2 miles C. 10.6 miles D. 15.3 E. None of these.

None, I believe.

If you let s1 = 4, s2=2, s3=6, and D denote the unknown distance they travel, we need for:

(2pm-9am) = 5 = D*(0.25/s1 + 0.5/s2 + 0.25/s3).

Plugging in and solving, this gives D=14.117. So unless I've made a dumb algebra mistake, then the answer is none.

@ekt_bear

Na wa for this your mathematical conundrum o! We even need buy graph paper to take answer am. ℓ☺ℓ

Ekt

The answer is B... I was just trying to show you and sybellah that the method you guys used is flawed... That was why I wanted you guys to use the same method to solve a similar puzzle albeit with more information... This is maths, you don't just use "undefined terms" to solve a puzzle - that would amount to guess work than anything else, no matter how you present it... The 0.618 answer for your puzzle is BOGUS!

Solution to my question:

time = 5
total distance = x

0.25x = walking uphill
0.50x = walking on level ground
0.25x = walking downhill

walking uphill = 2mph
walking on level ground = 4mph
walking downhill = 6mph

0.25x/2 + 0.50x/4 + 0.25/6 = 5 ... note "/" denotes distance x by velocity/speed

(1.50x + 1.50x + 0.50x)/12 = 5

3.50x/12 = 5

3.50x = 60

x = 60/3.50 = 17.20miles = B

shymexx: Ekt

The answer is B... I was just trying to show you and sybellah that the method you guys used is flawed... That was why I wanted you guys to use the same method to solve a similar puzzle albeit with more information... This is maths, you don't just use "undefined terms" to solve a puzzle - that would amount to guess work than anything else, no matter how you present it... The 0.618 answer for your puzzle is BOGUS!

Solution to my question:

time = 5
total distance = x

0.25x = walking uphill
0.50x = walking on level ground
0.25x = walking downhill

walking uphill = 2mph
walking on level ground = 4mph
walking downhill = 6mph

0.25x/2 + 0.50x/4 + 0.25/6 = 5 ... note "/" denotes distance x by velocity/speed

(1.50x + 1.50x + 0.50x)/12 = 5

3.50x/12 = 5

3.50x = 60

x = 60/3.50 = 17.20miles = B

hum huum ok I understand

ekt_bear: None, I believe.

If you let s1 = 4, s2=2, s3=6, and D denote the unknown distance they travel, we need for:

(2pm-9am) = 5 = D*(0.25/s1 + 0.5/s2 + 0.25/s3).

Plugging in and solving, this gives D=14.117. So unless I've made a dumb algebra mistake, then the answer is none.

but this is the same method Shymmex used right?

5= if u multiply the d by every term in the parenthesis...u probably made some algebra mistake

let say x=d

5= d*(0.25/s1) + d*(0.5/s2) + d*(0.25/s3)
5= d*(0.25/4) + d*(0.5/2) + d*(0.25/6)

put it at the same denom 12....u find a similar answer

Sybellah:

hum huum ok I understand

More reason why the ans to ekt_bear's question has to be 1(the point of exit/entry)...

Anything outside that is bogus!!

You don't even need algebra for it, your cognitive skills alone should answer the question...

Ah, you are right.

I didn't read this sentence carefully:
"hey spend one-quarter of the total distance walking uphill, one-half walking on level ground, and one-quarter walking downhill. if their speed is 4mph on level ground, 2 mph uphill, and 6 mph downhill,approximately how far did they walk?"


I assumed that you were using the same ordering when reporting the numbers...didn't notice that you changed it.

nono guys wait i think its the same calculations u made, but it's structured differently

Sybellah:

but this is the same method Shymmex used right?

5= if u multiply the d by every term in the parenthesis...u probably made some algebra mistake

let say x=d

5= d*(0.25/s1) + d*(0.5/s2) + d*(0.25/s3)
5= d*(0.25/4) + d*(0.5/2) + d*(0.25/6)

put it at the same denom 12....u find a similar answer

He stated uphill, level ground, then downhill, then stated level ground, uphill, downhill in the second part. So if you assume the same order like I did and don't read very carefully, you'll use wrong #s.

Sybellah:
but this is the same method Shymmex used right?

5= if u multiply the d by every term in the parenthesis...u probably made some algebra mistake

Why should I multiply D by every term in the bracket?

shymexx:

Why should I multiply D by every term in the bracket?

lol, mad face of math teacher ready to slap

Ekt

Do you now agree that I got the answer to your puzzle?

kk i see now

shymexx: Ekt

Do you now agree that I got the answer to your puzzle?

no, your answer of 9am is flat out wrong. it doesn't even make any sense, unless the ijebu man has some sort of teleportation device

no but here it was a reading mistakes, but the method was the same

the mistakes were at this level

(2pm-9am) = 5 = D*(0.25/s1 + 0.5/s2 + 0.25/s3).

5= d*(0.25/4) + d*(0.5/2) + d*(0.25/6)

correcting this would be

0.25*d/2 + 0.50*d/4 + 0.25*d/6 = 5

which is the same as this

0.25x/2 + 0.50x/4 + 0.25/6 = 5

no?

ekt_bear:

no, your answer of 9am is flat out wrong. it doesn't even make any sense, unless the ijebu man has some sort of teleportation device

The distance to and from the hill to his house is irrelevant, since it's not defined...

Only the constant(9am) is relevant - that's how mathematical tricks work..

The answer is either 0((his house) or 1(point of entry/exit at the hill)..

And since "9am" is the constant in both "0" and "1", then that should be the answer...

Sybellah: no but here it was a reading mistakes, but the method was the same

the mistakes were at this level

5= d*(0.25/4) + d*(0.5/2) + d*(0.25/6)

correcting this would be

0.25*d/2 + 0.50*d/4 + 0.25*d/6 = 5

which is the same as this

0.25x/2 + 0.50x/4 + 0.25/6 = 5

no?

You are correct.

Sybellah: no but here it was a reading mistakes, but the method was the same

the mistakes were at this level

5= d*(0.25/4) + d*(0.5/2) + d*(0.25/6)

correcting this would be

0.25*d/2 + 0.50*d/4 + 0.25*d/6 = 5

which is the same as this

0.25x/2 + 0.50x/4 + 0.25/6 = 5

no?

Yep..

shymexx:

The distance to and from the hill to his house is irrelevant, since it's not defined...

Only the constant(9am) is relevant - that's how mathematical tricks work..

The answer is either 0((his house) or 1(point of entry/exit at the hill)..

And since "9am" is the constant in both "0" and "1", then that should be the answer...

Bros.

Suppose the hill is 1000 miles high. He left his house 9am on Monday. He leaves the top of the hill 9am Tuesday to go back home. Are you really saying that at 9am on Tuesday he is back home? 0 seconds after he decided to leave the top of the hill?

This Ijebu man, unless he has a teleportation device or strong juju will not be able to do that

ekt_bear:
Bros.

Suppose the hill is 1000 miles high. He left his house 9am on Monday. He leaves the top of the hill 9am Tuesday to go back home. Are you really saying that at 9am on Tuesday he is back home? 0 seconds after he decided to leave the top of the hill?

This Ijebu man, unless he has a teleportation device or strong juju will not be able to do that

I understand where you're coming from, however, the distance wasn't defined in the puzzle, was it?

So, you can't really give it a value based on assumptions... It could be 1000 miles high, or 0.0001 miles high, but we'll never know...

Well, the hill doesn't have a height of zero.

If you weren't sure whether values of 0 for the height of the hill are alowed, just ask.

And assuming that the hill has height greater than zero, then 9am won't be the answer (again, also assuming no juju or teleportation devices).

The only thing you need to know is that the hill doesn't have height zero.

anyway i got to get some work done. talk to you guys later

kk ttyl

@ekt_bear
Hmmm, seems we shall have to start a Nairaland College of Riddle Solving. LMAO

ekt_bear please answer your original question without assuming there are two men intersecting.

Q:

There was an Ijebu man who lived near a hill.

Every Monday morning @ 9am, he would walk a path from his home to the top of this hill, meditate and pray.

Then Tuesday morning @ 9am, he would walk the same path from the top of the hill back to his home.

Is there any time at which he crossed the same spot on both Monday and Tuesday?

1. The two men ONLY came into place because I transformed from the original problem into one of two different men. Do you not get how the two men problem is equivalent to the original? This is the key to the whole matter. So clarify if it doesn't make sense to you.

2. Regarding the intersection of the two men, given that you are told in the original problem that the man uses the same path, necessarily then in the two man problem they must eventually run into each other.

ekt_bear: 1. The two men ONLY came into place because I transformed from the original problem into one of two different men. Do you not get how the two men problem is equivalent to the original? This is the key to the whole matter. So clarify if it doesn't make sense to you.

2. Regarding the intersection of the two men, given that you are told in the original problem that the man uses the same path, necessarily then in the two man problem they must eventually run into each other.

oh dear,
I think I'll just drop it as I can't seem to get it across to you that asking for a point of intersection is different from your original question.
I'll take it as a miscommunication.