Eternity Async Tutoring

[Eternity]

n * (p + 100)% or n * (p/100 +1).

n = 30, p = 10

30 * (110)%
30* 1.1
33

or

30 * (10/100 + 1)
30 * 1.1
33

[Eternity]

Did some ratio problems. Did ~10 for ~20 minutes. Got none wrong.

I did six speed/rates problems from the textbook. I got all of them right.

@Elliot

Equations involve finding two different things that are equal. If they’re unequal but related in some clear way, you can make them equal, e.g. if something is triple the other thing, then add a division by 3 to get them equal.

This was helpful in working with the problems. I realized partially why I probably had trouble at the time I initially did this was that it was too hard to do mental math with and I was too lazy to write it out. Only issue I ran into was the second problem I referenced.

I did do a practice problem similar before re-attempting this one and I think I understand what’s going on in the problem, but I had trouble setting up an equation here. Can I make an equation here? I guess I’d try getting their positions equal. I conceptually understand that from how it was explained in the textbook. I don’t know how to express it as an equation.

Here’s my working for the other problem I referenced.

[Elliot]

In general, I convert percents to decimals or fractions when doing calculations.

n * (p + 100)% or n * (p/100 +1)

I know what you meant here, which is correct, but this is actually incorrect. The correct formula is n * (p + 1)

I specified that “p is the percent to increase”. So, for example, p may be “50%”, not 50. If p is 50%, it converts to the decimal version 0.5, so you don’t need to add or divide by 100.

The problem in the book was IMO weird by saying p%, meaning that p itself is just an integer like 50, not a percent.

@Eternity

Previously, from what I remember, I was confused as to where the p/100 came from (now I understand its just p%)

Yeah % just means divide by 100.

Let’s try a similar problem.

Suppose you start at 100 and increase by p. p is a percentage, like 30% (= 0.3), not an integer like 30. They’re writing p% specifically because p is not a percentage, which is IMO confusing when done with a variable (but normal with numbers like writing 30% isn’t confusing).

Can you write an expression for this? Just the initial expression without solving it.

If you have trouble with variables, one thing you can do is use example numbers. Like if you want to increase 100 by 30% (aka 0.3), what’s the expression for that?

@Eternity

Can I make an equation here? I guess I’d try getting their positions equal.

There are simpler equations to go for first. For example, can you make an equation that involves the position of just one dog?

[Eternity]

@Elliot

I specified that “p is the percent to increase”. So, for example, p may be “50%”, not 50. If p is 50%, it converts to the decimal version 0.5, so you don’t need to add or divide by 100.

I think I’ve been treating percents as this special thing in my head. So while I could convert between 50% and 0.5 and tell you that they are equivalent, in my head them being equivalent didn’t translate to them being the same thing.

@Elliot

Suppose you start at 100 and increase by p. p is a percentage, like 30% (= 0.3), not an integer like 30. They’re writing p% specifically because p is not a percentage, which is IMO confusing when done with a variable (but normal with numbers like writing 30% isn’t confusing).

Can you write an expression for this? Just the initial expression without solving it.

If you have trouble with variables, one thing you can do is use example numbers. Like if you want to increase 100 by 30% (aka 0.3), what’s the expression for that?

100 * (0.3 + 1) or 100 * 1.3
100 * (p + 1)

I feel like I’m missing what you’re asking here because I just wrote the correct formula you gave me except that instead of n, there’s a value there.

@Elliot

There are simpler equations to go for first. For example, can you make an equation that involves the position of just one dog?

15ft/s is Dog A, 12ft/s is Dog B
Hmm. I think if this was just a straight line they were running the position would just be their distance from the beginning. So for Dog A his position would be given by y = 15x. x here is the time he spent running and y is the total distance he’s ran.

If you wanted to know how much distance he had left to run on a 300 feet track. y = 300 - 15x. At 0 seconds he has 300 feet left, at 1 second he has 285 left, etc.

Dog B would just by y = 12x.

Does this change much for a circular track? I guess to answer your question though are those ok equations for the positions of the dog?

[Elliot]

@Eternity

100 * (0.3 + 1) or 100 * 1.3
100 * (p + 1)

I feel like I’m missing what you’re asking here because I just wrote the correct formula you gave me except that instead of n, there’s a value there.

That’s correct.

So how do you start at 100, increase by p, then decrease by p?

@Eternity

So for Dog A his position would be given by y = 15x. x here is the time he spent running and y is the total distance he’s ran.

yes. i’d recommend using meaningful variable names. like d_a = 15t. d stands for distance and t for time. you can also use a word or even multiple words as a variable name.

after you have some equations, you can look for a solution or for connections between the equations (like a way to do substitution, like you see the same thing in two different equations then substituting might help). you can also brainstorm more equations, e.g. one to deal with position on the track rather than total distance run.

you can also think more about what’s going on in the problem. like, will the solution happen when both dogs have run less than one lap? if so, you wouldn’t need to worry about position on the track, instead of distance, since they’ll be the same thing. or another thing to think about is how do you know, in terms of something else, when the solution is reached?

[Eternity]

@Elliot

So how do you start at 100, increase by p, then decrease by p?

Hmm are you asking to increase 100 then decrease 100 by p. Or are you asking how would I decrease the increase from p. Nvm, that’s the same thing.

• Start at 100
  • 100
• Increase by p, p is a percentage
  • 100 * (p + 1)
• Decrease by p
  • If an increase in p is with p+1, then a decrease would be -p+1 or 1-p. The same p, but now used to decrease.
  • (100 * (p+1)) * (1-p)

Working:

@Elliot

you can also think more about what’s going on in the problem. like, will the solution happen when both dogs have run less than one lap? if so, you wouldn’t need to worry about position on the track, instead of distance, since they’ll be the same thing. or another thing to think about is how do you know, in terms of something else, when the solution is reached?

Here’s my working for the dog problem:

I’m still having trouble thinking of how to get workable equations for positions and stuff. Going through all this work has made it more intuitive to do what the textbook told me to do for these problems: take the distance of the track and divide it by the difference in the two speeds, that should give you the time when they will meet up again. It makes sense, especially seeing one dog get closer over time.

[Elliot]

@Eternity

(100 * (p+1)) * (1-p)

Yes although you don’t need one set of parentheses.

Does that make sense and make how to do the problem clear?

@Eternity

I’m still having trouble thinking of how to get workable equations for positions and stuff. Going through all this work has made it more intuitive to do what the textbook told me to do for these problems: take the distance of the track and divide it by the difference in the two speeds, that should give you the time when they will meet up again. It makes sense, especially seeing one dog get closer over time.

d_a = 15t

d_b = 12t

So how many equations is this and how many variables are there? Do you know how many equations you (usually) need to solve for N variables?

[Eternity]

@Elliot

Does that make sense and make how to do the problem clear?

Mhm. Much less confusing then the textbook one.

@Elliot

da = 15t

db = 12t

So how many equations is this and how many variables are there? Do you know how many equations you (usually) need to solve for N variables?

Two equations. Three variables? da, db, and t. One equation per variable?

[Elliot]

@Eternity

Two equations. Three variables? da, db, and t. One equation per variable?

Yes. So you can’t solve it yet. You need one more equation.

So list the information in the problem, then check if you’re using it all. Usually with these problems you have to use all the info they give you, whereas in real life you just keep gathering info until you have enough to get a solution.

[Eternity]

I came up with da - db = 3t and solved for t. I also put into an equation that I’m looking for when the difference in their position is 300. da - db= 300.

[Elliot]

@Eternity

I also put into an equation that I’m looking for when the difference in their position is 300. da - db= 300.

Great. That uses the information from the problem (300) that you hadn’t used yet (the other ones are 12 and 15).

@Eternity

I came up with da - db = 3t and solved for t.

That’s true but it’s not enough to solve the problem without using the 300 somehow. New equations basically don’t count towards how many you have if you can derive from equations that are already on your list. That’s because they’re different forms of what you already know (which is useful) instead of new things.

So try that out. Start with:

d_a = 15t
d_b = 12t

And derive:

d_a - d_b = 3t

After that let me know if the problem is clear now, and if so you can try more word problems.

[Eternity]

@Elliot

So try that out. Start with:

da = 15t
db = 12t

And derive:

da - db = 3t

Just to clarify, are you saying to come up with equation da - db = 3t without just directly subtracting and saying da - db = 3t?

@Elliot

That’s true but it’s not enough to solve the problem without using the 300 somehow.

I’m not supposed to use the 300?

[Elliot]

@Eternity

@Elliot

That’s true but it’s not enough to solve the problem without using the 300 somehow.

I’m not supposed to use the 300?

It’s a double negative. You need the 300.

@Eternity

Just to clarify, are you saying to come up with equation da - db = 3t without just directly subtracting and saying da - db = 3t?

Just subtracting is correct. So the point is it’s implied by the other two equations rather than being new information.

You can do more word problems unles you have more questions.

[Eternity]

@Elliot

It’s a double negative. You need the 300.

Oh I see now.

@Elliot

You can do more word problems unles you have more questions.

Ok. Looking through my textbook seems like I got this problem wrong initially and just skipped it after. So I’ll do this one.

Outside of that I apparently didn’t do the review problems for the whole chapter for chapters 5, 7, 8 and 9. So I’ll just do these.
Chapter 5 - Equations and Inequalities
Chapter 7 - Ratios, Conversions, and Rates
Chapter 8 - Percents
Chapter 9 - Square Roots

Unless there’s a specific area you want me to work on. The reviews have a mix of regular problems and word problems.

[Elliot]

What do you think about tutoring so far?

[Eternity]

@Eternity

Ok. Looking through my textbook seems like I got this problem wrong initially and just skipped it after. So I’ll do this one.

Ok I was able to do it this time. I did remember the reason why I got it wrong from last time I did this was getting confused on completely clearing the tunnel. I think most problems dealing with traveling through talk deal with when you get the first part out like the nose of a train, not when the whole train gets out. Here’s my working:

I kinda have some expressions? But I mostly just thought my way through it. In the chapter 5 review some problems I could make an expression for, but most of them I just thought my way through it.

Here’s the whole review as a PDF if you wanted to see the kind of problems I did (I have the working for all of them if you wanted to it for some particular problem):
Chapter 5 Review Problems.pdf (2.6 MB)
Chapter 5 Challenge Problems.pdf (2.9 MB)

I did 37 problems( some have multiple parts) for ~2 hours and 15 minutes (I skipped the last part of the last challenge problem for now (5.70). I’m confident I can do it, I’m just a lil tired and I wanted to share what I’ve done so far).

What I got wrong:
I got part of this problem wrong (just part/problem j):

After distributing the 3 and getting -6x. I treated it as positive 6x when simplifying. I wrote it correctly, I think I was just tired and missed the -.

This is the first word problem in the review I got wrong:

My working:

I looked at the explanation and I kinda understood what I did wrong. Is there something you see particularly wrong in how I set it up?

I didn’t want clutter this reply with even more stuff, unless you want to see my working for all these at once. These are just what I got wrong. No working:

.

33/37 = 92% correct. Though not entirely accurate. I skipped the last part of a problem and missed one part of a 10 part problem and still included it. I guess 32/37 or 31/37?

[Elliot]

For 5.46 and 5.55 can you show work, especially the initial problem setup where you write down what you know and some equations? I want to know if you got that wrong or made a calculation error.

[Eternity]

Mmm. I like it so far. The primary thing I like is the sense of direction I have? When I tried self-studying philosophy and tried to do stuff by myself like working on pre-requisites my big issue was integrating what I’m learning with philosophy. I would think to myself, “What’s the point of learning stuff like arithmetic and grammar besides being generally educated?” I mean I knew general education is good but tying it into philosophy I didn’t know how to do. I felt unsure if the things I was learning were helping in my goal to learn/do philosophy.

So far I’ve liked the explanations and help you’ve given. I think I liked the help with the percentage increase and decrease problem the most.

Mmm. Nothing else comes to mind. Its good so far.

[Eternity]

Here’s the work I did for the problems. I didn’t revise anything after seeing I got it wrong, I just moved onto other problems.

Oh yeah. I gave up on this one. The way I went about it and the work I did wasn’t making sense. I’m trying to get better about moving on with stuff so I just dropped it for the time being and went to other problems.

[Elliot]

Try doing those two again but focus on being more methodical and setting up the problem correctly at the start. Start by writing out pieces of information you have, then create equations. Getting the equations right is the key. Then after you think it’s set up correctly and ready to be solved, you can do the calculations using your equations. (Try to do future problems more this way too.)