does this look promising

[allegansveritatem]

Here is the problem:


Here is how I am trying to set it up:


I am aware that this is not exactly the way to go--I have already tried it. But is it in the ballpark at all? Obviously I am treating this as a hypotenuse situation. For one thing, the problem that came before was solved hypotneusely, so to speak. The idea governing the above equation is: If distance = rate over time, then letting x = rate of cycle and x+10= rate of car, what we have is the distance each covered divided by the time it took squared will = the hypotenuse divided by time (squared). Ain't quite right but does it even make sense?

 

[Steven G]

Here is the problem:
View attachment 10968

Here is how I am trying to set it up:
View attachment 10969

I am aware that this is not exactly the way to go--I have already tried it. But is it in the ballpark at all? Obviously I am treating this as a hypotenuse situation. For one thing, the problem that came before was solved hypotneusely, so to speak. The idea governing the above equation is: If distance = rate over time, then letting x = rate of cycle and x+10= rate of car, what we have is the distance each covered divided by the time it took squared will = the hypotenuse divided by time (squared). Ain't quite right but does it even make sense?

How much further will the car go than the motorcycle in 3 hours? The answer to that question is the whole problem.

 

[tkhunny]

Here is the problem:
View attachment 10968

Here is how I am trying to set it up:
View attachment 10969

I am aware that this is not exactly the way to go--I have already tried it. But is it in the ballpark at all? Obviously I am treating this as a hypotenuse situation. For one thing, the problem that came before was solved hypotneusely, so to speak. The idea governing the above equation is: If distance = rate over time, then letting x = rate of cycle and x+10= rate of car, what we have is the distance each covered divided by the time it took squared will = the hypotenuse divided by time (squared). Ain't quite right but does it even make sense?

\(\displaystyle Distance = Rate * Time \ne \dfrac{Rate}{Time}\)

Checking Units will often save you.

x mph / 3 hr = \(\displaystyle \dfrac{x}{3}\cdot\dfrac{mile}{hour^{2}}\) -- What even is that?

 

[allegansveritatem]

\(\displaystyle Distance = Rate * Time \ne \dfrac{Rate}{Time}\)

Checking Units will often save you.

x mph / 3 hr = \(\displaystyle \dfrac{x}{3}\cdot\dfrac{mile}{hour^{2}}\) -- What even is that?

You're right. I screwed this up royally. I had a sense I was on the edge of no return while setting this up and now I see that I meant rate TIMES time and wrote rate divided by time. What an idiot! Would that make any more sense? I mean, if instead of squared fractions I substituted squared products? I am not able to follow up on this at the moment but will come back shortly.

 

[Steven G]

You're right. I screwed this up royally. I had a sense I was on the edge of no return while setting this up and now I see that I meant rate TIMES time and wrote rate divided by time. What an idiot! Would that make any more sense? I mean, if instead of squared fractions I substituted squared products? I am not able to follow up on this at the moment but will come back shortly.

Reread my last post. You can not do this problem without knowing the answer to my question.

 

[Denis]

Wonder why that problem is so weirdly written.
Seems to suggest that the 180 miles apart
is an hypotenuse: roads meeting at right angle...

If so, then car is 144 miles from intersection,
motorcycle is 108 miles from intersection,
so no "nice" solution!

 

[allegansveritatem]

How much further will the car go than the motorcycle in 3 hours? The answer to that question is the whole problem.

so this is strictly a motion problem? 180/x +180/x+10 =3 (?)

 

[JeffM]

Here is the problem:
View attachment 10968

Here is how I am trying to set it up:
View attachment 10969

I am aware that this is not exactly the way to go--I have already tried it. But is it in the ballpark at all? Obviously I am treating this as a hypotenuse situation. For one thing, the problem that came before was solved hypotneusely, so to speak. The idea governing the above equation is: If distance = rate over time, then letting x = rate of cycle and x+10= rate of car, what we have is the distance each covered divided by the time it took squared will = the hypotenuse divided by time (squared). Ain't quite right but does it even make sense?

I am going to attack this differently from Jomo, but you end up in the same place.

I like to start by asking myself what I do not know. I do that before I try to set up equations. I want to know what I am looking for.

You don't know distances from the intersection and you don't know speeds.

That is four numbers you do not know. So assign a symbol to each unknown number.

\(\displaystyle a = \text {distance of cycle from intersection.}\)

\(\displaystyle b= \text {speed of motorcycle.}\)

\(\displaystyle u = \text {distance of car from intersection.}\)

\(\displaystyle v = \text {speed of car.}\)

Notice that this involves no algebra. It involves thinking about what numbers are relevant to solving the problem.

Solving for four unknowns requires four equations.

\(\displaystyle a^2 + u^2 = 180^2.\)

The distance between 2 points on perpendicular lines is determined by the Pythagorean Theorem, not rates of speed. Do not blindly copy patterns or apply formulas. Think a bit first.

\(\displaystyle b = v - 10.\)

\(\displaystyle 3b = a.\)

\(\displaystyle 3v = u.\)

Not one of those equations is hard to find. Trying to find a single equation that embodies all those relationships is not so obvious.

And what is the general method for solving a system of simultaneous equations? SUBSTITUTION.

Now it is just mechanics. The thinking is done.

I admit that jomo's way involves less mechanics and is more elegant, but first you have to have jomo's flash of insight. Once jomo asked his question, I saw his simple approach, but I did not see it without his hint. Neverthesless, despite my lack of insight, I can solve the problem by being systematic in my thinking.

EDIT: As you do the substitutions, you will see what jomo was talking about. He just got there on the express while I travel on the local.

 

[Harry_the_cat]

Wonder why that problem is so weirdly written.
Seems to suggest that the 180 miles apart
is an hypotenuse: roads meeting at right angle...

I think that is the correct interpretation.

Draw a right angled triangle (with the right angle in the bottom left corner). This is where the roads meet (at right angles).

Let the car travel from N to S towards the intersection, and the motorbike travel from E to W toward the intersection.

Let the motorbike travel d miles at x m/h. Therefore the car travels \(\displaystyle \sqrt{180^2 - d^2}\) miles at (x+10) m/h.

Time = 3 hours = \(\displaystyle \frac{distance}{speed}\) in both cases.

This gives 2 equations with two unknowns (d and x).

 

[Denis]

I think that is the correct interpretation.

Draw a right angled triangle (with the right angle in the bottom left corner). This is where the roads meet (at right angles).

Let the car travel from N to S towards the intersection, and the motorbike travel from E to W toward the intersection.

Let the motorbike travel d miles at x m/h. Therefore the car travels \(\displaystyle \sqrt{180^2 - d^2}\) miles at (x+10) m/h.

Time = 3 hours = \(\displaystyle \frac{distance}{speed}\) in both cases.

This gives 2 equations with two unknowns (d and x).

BUT that means a "messy" solution:
108-144-180 being the only integer sided right triangle with hypotenuse = 180...

 

[Harry_the_cat]

BUT that means a "messy" solution:
108-144-180 being the only integer sided right triangle with hypotenuse = 180...

Why should it be restricted to integers?

 

[Denis]

Why should it be restricted to integers?

Of course it doesn't have to...but these problems usually are...right?

If we let m = motorcycle speed, then:
motorcycle travels 3m miles and car travels 3(m + 10) = 3m + 30 miles.
So we have:
(3m)^2 + (3m + 30)^2 = 180^2
Solving: m = 5(sqrt(71) - 1) ...YIKES!!

But if the 180 was somehow intended to mean the sum of
the distance travelled by both, then:
3m + 3(m + 10) = 180, so m = 25

Interesting to note that IF the car was 12mph faster than motorcycle
(instead of 10mph) then we'd have:
(3m)^2 + (3m + 36)^2 = 180^2
solving: m = 36, so car = 48 .... so we all live happily forever after!!
So perhaps the 10 is a typo...12 was intended?????

 

[Harry_the_cat]

Of course it doesn't have to...but these problems usually are...right?

No, not these days when everyone has a calculator.

I remember the days when all problems involving circumference and area of circles had a radius of 7 or 14 or 21 and you used approx. of 22/7 for pi.

 

[Deleted member 4993]

\(\displaystyle Distance = Rate * Time \ne \dfrac{Rate}{Time}\)

Checking Units will often save you.

x mph / 3 hr = \(\displaystyle \dfrac{x}{3}\cdot\dfrac{mile}{hour^{2}}\) -- What even is that?

Acceleration....

 

[allegansveritatem]

I am going to attack this differently from Jomo, but you end up in the same place.

I like to start by asking myself what I do not know. I do that before I try to set up equations. I want to know what I am looking for.

You don't know distances from the intersection and you don't know speeds.

That is four numbers you do not know. So assign a symbol to each unknown number.

\(\displaystyle a = \text {distance of cycle from intersection.}\)

\(\displaystyle b= \text {speed of motorcycle.}\)

\(\displaystyle u = \text {distance of car from intersection.}\)

\(\displaystyle v = \text {speed of car.}\)

Notice that this involves no algebra. It involves thinking about what numbers are relevant to solving the problem.

Solving for four unknowns requires four equations.

\(\displaystyle a^2 + u^2 = 180^2.\)

The distance between 2 points on perpendicular lines is determined by the Pythagorean Theorem, not rates of speed. Do not blindly copy patterns or apply formulas. Think a bit first.

\(\displaystyle b = v - 10.\)

\(\displaystyle 3b = a.\)

\(\displaystyle 3v = u.\)

Not one of those equations is hard to find. Trying to find a single equation that embodies all those relationships is not so obvious.

And what is the general method for solving a system of simultaneous equations? SUBSTITUTION.

Now it is just mechanics. The thinking is done.

I admit that jomo's way involves less mechanics and is more elegant, but first you have to have jomo's flash of insight. Once jomo asked his question, I saw his simple approach, but I did not see it without his hint. Neverthesless, despite my lack of insight, I can solve the problem by being systematic in my thinking.

EDIT: As you do the substitutions, you will see what jomo was talking about. He just got there on the express while I travel on the local.

I think this is the key mto these word problems: Think them through. Understand them thoroughly. Of course, you can't prove it by me....

 

[allegansveritatem]

So, I got to work on this again today and came up with this:


This I think will fill the bill...I worked within the constraints of pleasing the puzzle's author. I am pretty sure he wanted that hypotenuse to get a part in the drama that has been gripping so many for so long...I guess I should speak for myself on the score of who got gripped and by what. Anyway...if this is wrong, how is it wrong?

 

[Denis]

Anyway...if this is wrong, how is it wrong?

It's correct...now, go away