Some Tough Ones

[MathCat]

Hello out there,

I am a long time lurker of this site, but this is my fist post.

When I was in primary school, I never particularly cared for mathematics. I used to think it was boring and was not given any context for the work I was doing. Years later, in college, I would read the work of Rene Descartes and it really revolutionized how I thought about mathematics, geometry, and the world around me. I might be a bit of an eccentric, but I spend a lot of my free time studying maths and get a lot of enjoyment out of it. This enterprise wound up leading me into finding this website and spending a lot of time on it. I started studying from the ground up on all the stuff that I missed out early on in my academic career and, three years later, I am now employed full time as a mathematics tutor.

Recently, I've been trying to get a better grasp of Algebra and Geometry. I'm kind of embarrassed because I can tutor higher maths but these basic questions are currently tripping me up. I was wondering if anyone on this site would be willing to lend me a hand since these guys are tripping me up. It might be kind of challenging, but that's why I broke down and got an account.

Here's the first one:

What's the perimeter of this shape?
My original impression was to divide it into two rectangles. My issue with doing it this way is that I cannot seem to divide the image into enough meaningful parts to get a perimeter. So while I can create defined auxiliary lines to separate the image into two rectangles (lines a and b), I am left with either a rectangle simply composed of 2L or simply composed of 2W.

 

My second thought was to try to turn it into one big rectangle, but I don't think I can do that meaningfully either since there's no way (I can think of) to do so yet control for the extra space I'm adding to this shape. So, for example, if I find the perimeter of one big rectangle L = 20 and W = 30, I don't really know how much space to subtract from the smaller missing piece.

 

The only other thing I can think of is dividing it into three rectangles and dividing those three into six triangles. That would be a huge thorn in my side but I'm wondering if that's possible.

Please note that the image is not drawn to scale.

Here's the second question:

 

"A chemist takes a solution containing 40% water and mixes it with a solution containing 50% water in order to make 200L of a solution containing 44% water. How much of the 50% solution is used?"

I can solve this question with the following equation:

 

.40(200-X) (.50)X = (.44) 200


As indicated, I know how to answer this one. My question is why does this equation actually work? What is this piece of information saying? The way that I read it is "If you take the value of two hundred minus the amount of fifty percent solution and multiply it by forty percent and multiply this number by the value of fifty percent times the amount of fifty percent solution, you will get the value of forty-four percent times two-hundred."

 

To me, when I read that equation, I don't feel like it meaningfully interprets the problem from English to math. Can someone lend me a hand on this one?

The rest of the stuff that I'm currently working on is combining like terms, simplifying expressions, FOIL, and Cartesian Planes. I feel confident in my ability to handle that stuff because the information is kind of plainly written in my books. The ones that I posted I cannot seem to find anywhere online and that's why I figured I'd post them here and ask for help. I DID READ THE RULES BEFORE POSTING, and If this isn't the most appropriate subforum I apologize in advance, but I felt that it would be best placed here.

Thanks,

MathCat

Edited May 2, 2014 by MathCat

[Endy0816]

lol, first one makes for an enjoyable logic problem.

 

Perimeter = Side A + Side B + ...

 

The trick is that you don't need to know the missing information to find the answer. So the perimeter is the same as if you were dealing with just a rectangle with those same dimensions(20x30).

 

20 + 30 + 20 + 30 = 100

 

To put it another way the sum of the sides you don't know is equal to the sides you do know.

 

20 + 30 + x + y + z + a = 100

Edited May 2, 2014 by Endy0816

[MathCat]

 

 

The trick is that you don't need to know the missing information to find the answer. So the perimeter is the same as if you were dealing with just a rectangle with those same dimensions(20x30)

 

 

Thanks for this, and I understand your reasoning. If we define the perimeter of the distance around a two dimensional shape, it still is a little odd to me why the perimeter of a full rectangle is equivalent to a rectangle with a large piece missing from it.

 

Anybody have any thoughts on the second one?

[studiot]

 

When I was in primary school, I never particularly cared for mathematics. I used to think it was boring and was not given any context for the work I was doing.

 

No offence meant, but I wonder if you missed out on basic geometry, ratio and proportion as a result.

 

It is basic geometry to prove that you can move the two indented sides out to complete the large rectangle and find the perimeter = 2(20+30)

 

It is basic ratio & proportion to combine ingredients to form compound % of a whole without using equations (although they will also work).

 

I suggest you get hold of a elementary algebra/arithmetic book such as Hall & Knight or Lockwood and Down and study it.

[MathCat]

 

 

No offence meant, but I wonder if you missed out on basic geometry, ratio and proportion as a result.

It is basic geometry to prove that you can move the two indented sides out to complete the large rectangle and find the perimeter = 2(20+30)

It is basic ratio & proportion to combine ingredients to form compound % of a whole without using equations (although they will also work).

I suggest you get hold of a elementary algebra/arithmetic book such as Hall & Knight or Lockwood and Down and study it.

 

No offense taken. It is true that I did not get very good education growing up and that's why I'm humbly asking for help on this forum. To clarify, these questions aren't for any kind of test or class. I simply am studying mathematics because I find it enjoyable.

 

I will however return the remark. No offense intended, but I get the impression that you are much older or had access to good educational facilities growing up. My experience as a university tutor for 4+ years is that most college algebra classes don't even go over this kind of material anymore since it isn't necessary to fulfilling many higher math curriculums. To clarify, I did mean universIty tutor (as in, not community college). In fact, where I grew up, Geometry wasn't even offered until High School, and the idea of an elementary school geometry or algebra book is foreign to me. Furthermore, I can see how my post openly illustrates a lower strata on algebraic understanding, but I don't see how my post demonstrates any misunderstanding of arithmetic.

 

I am currently working through two college algebra texts, a survey of geometry text, and two software packages from Hawkes Learning System. I started about six days ago.

[Endy0816]

 

 

Thanks for this, and I understand your reasoning. If we define the perimeter of the distance around a two dimensional shape, it still is a little odd to me why the perimeter of a full rectangle is equivalent to a rectangle with a large piece missing from it.

 

 

I think you are thinking of area. That would be different. Not enough information to find the Area, but via a bit of mental rearrangement you can find the perimeter.

 

 

.40(200-X) (.50)X = (.44) 200

 

so you could say:

 

40%(A) x 50%(B) = 44%(200)

 

A+B=200

 

so

 

200-B=A

 

Plugging back in:

 

40%(200-B) x 50%(B) = 44%(200)

 

which is the equivalent of what you had to start with.

 

Note: I'm mostly working towards the middle here. Never was all that good at adding compounds.

Edited May 2, 2014 by Endy0816

[studiot]

 

My experience as a university tutor for 4+ years is that most college algebra classes don't even go over this kind of material anymore since it isn't necessary to fulfilling many higher math curriculums.

 

Yes I agree that is so, which is why the books I recommended are not modern.

 

 

My experience as a university tutor for 4+ years is that most college algebra classes don't even go over this kind of material anymore since it isn't necessary to fulfilling many higher math curriculums. To clarify, I did mean universIty tutor (as in, not community college). In fact, where I grew up, Geometry wasn't even offered until High School, and the idea of an elementary school geometry or algebra book is foreign to me. Furthermore, I can see how my post openly illustrates a lower strata on algebraic understanding, but I don't see how my post demonstrates any misunderstanding of arithmetic.

 

 

 

You are right the system I grew up in was not directly comparable to the one you describe.

Neither the terminology nor the organisation was the same.

 

The material I am referring to used to be taught to the last year of primary school and the first two years of secondary school, ie between the ages of 10 and 13 approximately.

Since that school system was tiered, similar material was also taught to older pupils, in lower tiered schools, who were operating at lower level and destined to become secretaries, shop assistants and the like.

 

Above all it was practical.

 

There were even books entitled Practical Maths or more likely Practical Arithmetic or similar.

 

These books contained many little tricks and quite a lot of understanding that more formal 'High School' studies were built on.

 

It is good that you are taking steps to recover lost ground, but I still recommend using the old ways. You will not miss out a good coverage that way. Furthermore the old ways used much repetition, which you should not need, plus you could avoid spending a year on logarithms as they are no longer required.

 

I really am trying to help, but not just with a couple of specifics, but by pointing the way to efficiently cover the entire area.

 

[MathCat]

Thank you Endy0819 for your response as I found your insight very useful on the second problem.

Thank you Studiot for sharing some of you oldschool wisdom. I agree that the current curriculum is piecemeal and lacks context, but was not aware of these older books / methods. I will look into these texts.

[nellythedog]

 

No offense taken. It is true that I did not get very good education growing up and that's why I'm humbly asking for help on this forum. To clarify, these questions aren't for any kind of test or class. I simply am studying mathematics because I find it enjoyable.

 

I will however return the remark. No offense intended, but I get the impression that you are much older or had access to good educational facilities growing up. My experience as a university tutor for 4+ years is that most college algebra classes don't even go over this kind of material anymore since it isn't necessary to fulfilling many higher math curriculums. To clarify, I did mean universIty tutor (as in, not community college). In fact, where I grew up, Geometry wasn't even offered until High School, and the idea of an elementary school geometry or algebra book is foreign to me. Furthermore, I can see how my post openly illustrates a lower strata on algebraic understanding, but I don't see how my post demonstrates any misunderstanding of arithmetic.

 

I am currently working through two college algebra texts, a survey of geometry text, and two software packages from Hawkes Learning System. I started about six days ago.

Good luck with algebra. I didn't understand it at school either.

[John]

While the given solution to the first problem is probably what's intended, I'm going to be "that guy" and note that from the diagram, we can't assume we're dealing with right angles at all, and thus we can't be sure we simply have a rectangle with one corner "flipped."

In a geometry class, just based on the illustration given, the correct answer (though it might make a teacher roll his eyes) is that there isn't enough information to solve the problem.

Edit: As a fun example, and the one my geometry textbook uses as part of its section on reasoning from diagrams, consider the fallacy of the isosceles triangle.

Edited May 2, 2014 by John

[studiot]

Here is the derivation of your chemist's equation.

 

Let X be the number of litres of solution B (50% water) added.

 

This therefore contains 0.5X litres of water.

 

Further this means that (200-X) litres of solution A (40% water) are used since 200L are made in all.

 

This contains 0.4(200-X) litres of water

 

Now 200 litres of final mixture contain 200 x 0.44 litres of water ie 88 litres.

 

Thus adding the contributions from solutions A and B

 

0.5X + .4(200-X) = 88

 

removing decimals

 

5X + 800 - 4X = 880

 

X = 80 litres