Weight of Object - Answer/ Explaination

[Guest]

A piece of metal has 1/m part + a oz. is brass, and m/n part - b oz. is copper. Determine weight of the metal piece.

Answer provided with problem:

mn (b - a)
-------------
n + m^2 - mn

Please explain the steps to solve. Thanks.

[Guest]

Let [tex]w[/tex] be the total weight of the metal piece (in oz).
The weight (in oz) of the brass part is [tex]\frac{1}{m}w+a[/tex],
the weight (in oz) of the copper part is [tex]\frac{m}{n}w-b[/tex],
So the total weight [tex]w = \left(\frac{1}{m}w+a \right)+\left(\frac{m}{n}w-b\right)[/tex]
(assuming the metal piece only consists of brass and copper).
Rearranging gives
[tex]b-a = \frac{1}{m}w+\frac{m}{n}w-w[/tex] (group [tex]w[/tex] terms on one side, the other terms on the other side)
[tex]= \left(\frac{1}{m}+\frac{m}{n}-1\right)w[/tex] (factor out the [tex]w[/tex])
[tex]= \left(\frac{n+m^2-mn}{mn}\right)w[/tex] (simplify the fraction)
So
[tex]\frac{mn(b-a)}{n+m^2-mn}=w[/tex].

Hope this helped,

R. Baber.

[Guest]

That helped some.

I am not totally clear on the following:

(1/m + m/n -1) w (factor out the w)

( n + m^2 - mn / mn) w (simplify the fraction)

Then the answer.

Thanks.

[Guest]

I am using the fact that multiplication is distributive over addition, i.e. [tex](a+b)\times c = a\times c + b\times c[/tex].
For example according to the formula [tex](2+3)\times 5 = 2\times 5 + 3\times 5[/tex], it is easy to check this is true,
the left hand side evaluates to [tex]5\times 5[/tex], and the right hand side is [tex]10+15[/tex], both equal [tex]25[/tex].

This rule generalizes, so we can also say [tex](a+b+c)\times d = a\times d + b\times d+c\times d[/tex].

We can apply this rule to [tex]\frac{1}{m}w+\frac{m}{n}w-w[/tex] which looks very much like the right hand side of the distributive law (every term has a [tex]w[/tex], just like all the terms in the distributive law has a [tex]d[/tex]). In fact
[tex]\frac{1}{m}w+\frac{m}{n}w-w = \left(\frac{1}{m}\right)\times w+\left(\frac{m}{n}\right)\times w+(-1)\times w[/tex],
which by the distributive law we know is the same as
[tex]\left(\left(\frac{1}{m}\right)+\left(\frac{m}{n}\right)+(-1)\right)\times w[/tex],
which we can write in a more compact way (by removing brackets and getting rid of the multiplication sign which is implied) as
[tex]\left(\frac{1}{m}+\frac{m}{n}-1\right) w[/tex].

Whenever we have fractions that we want to add such as [tex]\frac{2}{3}+\frac{1}{4}[/tex], the way we do it is by first finding a common denominator (a number which all the denominators will divide into, in this case a number which [tex]3[/tex] and [tex]4[/tex] divides into). An easy way to get a common denominator is to multiply all the denominators together, so in our case we would take [tex]12[/tex] as the common denominator. Next multiply the expression by the common denominator, so in our case we get [tex]12\times\left(\frac{2}{3}+\frac{1}{4}\right)[/tex] which by the distributive law we know is the same as [tex]12\times\frac{2}{3}+12\times\frac{1}{4} = 8+3 = 11[/tex]. We've multiplied the expression by 12, so to keep things balanced/equal we should divide by 12, which means we get [tex]11/12[/tex] which is the sum of our two fractions.

We can do the same thing to [tex]\frac{1}{m}+\frac{m}{n}-1 = \frac{1}{m}+\frac{m}{n}+\frac{-1}{1}[/tex]. The common denominator is [tex]m\times n \times 1 = mn[/tex]. Multiplying the expression by the common denominator gets us
[tex]mn\times\frac{1}{m}+mn\times\frac{m}{n}+mn\times\frac{-1}{1} = n+m^2-mn[/tex]
dividing by the common denominator gives
[tex]\frac{n+m^2-mn}{mn}[/tex]

Hope this helped,

R. Baber.

[Guest]

That helped significantly. Thanks.

One more question:


Is the 1 subtracted ( 1/m + n/m - 1) due to the copper part being subtracted ?

[Guest]

No.

It comes from rearranging
[tex]w = \left(\frac{1}{m}w+a\right)+\left(\frac{m}{n}w-b\right)[/tex]

R. Baber.

[Guest]

I don't understand.

[Guest]

I'm afraid there is nothing I can do to make it easier for you to understand. You need to go back to basics and practice how to rearrange and manipulate equations.

R. Baber.

[Guest]

I looked up the topics you mentioned, but did not locate a similar type problem. I don't understand, so I am not able to solve these type of problems. My apology for wasting your time.

[Guest]

No need to apologize, it's not a waste of my time.

If you have very basic gaps in your knowledge, there is absolutely no point in attempting these types of questions. The fact that you are confused at nearly every step of the solution clearly indicates you have very large gaps in your knowledge of the underlying principles, and need to take a step back or two. You can not skip ahead when it comes to mathematics, at all (I can't emphasize this enough). You have to spend a long time learning the boring basics. Simply reading a topic is not enough, you need to do lots and lots of exercises. You need to be getting a score of about 95% on a topic before you move on to the next topic which builds on it (and if you're not getting this score you will need to take another step back and revisit even simpler topics).

If you are unwilling to do this then you should give up. There are plenty of people who refuse to do the groundwork, think they understand what's going on, and turn into "mathematical cranks" (there's one who regularly posts on this site). You need to have the ability to realize that maybe you don't understand the basics and be willing to swallow your pride and go back and revisit those basic topics.

Below is a site that has lots of resources:
http://www.mathcentre.ac.uk/students/topics/
In particular
http://www.mathcentre.ac.uk/resources/u ... 2009-1.pdf
teaches how to do basic rearranging of equations.

R. Baber.

[Guest]

R. Baber,

Thanks for the info.

I viewed the examples in the links, but I still don't understand why the 1 is subtracted in my problem; I didn't see a similar example.

[Guest]

I think I figured it out:

Is the 1 subtracted when rearranging due to being moved to the other side ?

[Guest]

Yes. The [tex]w[/tex] gets moved to the other side to become a [tex]-w = -1\times w[/tex], which is where the [tex]-1[/tex] comes from.

R. Baber.

[Guest]

Ok, thanks. I will go over the examples provided in more detail.