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Some Fun With Maths


Aethelwulf

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I was having some fun with some people on facebook and thought about bringing the idea here.

 

Basically, I was posting some math problems, the idea is of course to solve it.

 

It went like this -

 

Let's see how good your maths is. I have created a problem and solved it myself so I know the answer.

 

simplify terms (involving fractions) and solve for [math]k[/math]

 

[math]\frac{kr}{m} - S + (\frac{5}{ab} - \frac{3}{b^2})= T[/math]

 

values of [math]b, r[/math] and [math]a[/math] are

 

[math]b = 2[/math]

 

[math]r = 20[/math]

 

[math]a = \sqrt{\frac{\frac{4 \cdot 154}{22}}{7}}[/math]

 

 

 

I won't lie, some of it might be tricky. Some people particularly had trouble finding the value of [math]a[/math].

 

Enjoy!

Edited by Aethelwulf
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Ok, thought since you gave numbers you wanted some numeric value for k...otherwise I agree completely with belovelife and think it is just basic algebra...

 

 

There is some basic algebra involved - but I also gave a numerical task. The previous poster was very close, the actual answer is

 

[math]k = Tm+Sm -1[/math]

 

The numbers simplify, so I am not sure where the previous poster got 1/2 from... try it again.

Edited by Aethelwulf
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the sqrt is 2

then brackets give 5/4 -3/4=2/4=1/2

 

 

That's not right. the value of [math]a[/math] is what gave other people problems as well. Also, that is not how you simplfy what is inside the paranthesis.

 

The way you divide a number by a fraction is like this

 

[math]6 \frac{2}{\frac{4}{8}} = 6\frac{2}{\frac{8}{4}} = \frac{6 \cdot 2 \cdot 8}{4} = 24[/math]

 

So when you divide a number by a fraction, you have to flip the fraction around. So let's go back to our example

 

[math]a = \sqrt{\frac{4 \cdot 154}{\frac{22}{7}}}[/math]

 

swap the fraction round

 

[math]a = \sqrt{\frac{4 \cdot 154}{\frac{7}{22}}}[/math]

 

calculate

 

[math]4 \cdot 154 \cdot 7 / 22 = 196[/math]

 

Take the square root of this and it is 14.

Edited by Aethelwulf
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How do you simplify what is in the brackets?

 

That's easy

 

It is in paranthesis so you solve this part first [math](\frac{5}{ab} - \frac{3}{b^2}) = \frac{5b - 3a}{ab^2}[/math] now plug in the values. Then simplify by plugging in the value for [math]r[/math] when you get the chance.

Edited by Aethelwulf
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i think the original problem could have been visually represented different, i thought the same thing as sanctus

 

 

the original representation of (a)

 

i should say

 

 

You both wouldn't have needed to think anything if it had been calculated right. The rule for dividing whole numbers by fractions is well-known.

 

http://www.cimt.plymouth.ac.uk/resources/help/miscon5.pdf

 

An example from this site is

 

[math]3/1/4[/math]

 

The answer they give is 12. Use the rule I gave, remember to swap the fraction about

 

[math]3/4/1[/math]

 

is

 

[math]3 \cdot 4/1 = 12[/math]

 

It is perfectly consistent and nothing wrong with any kind of representation. I did warn you it could be tricky.

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So what is your answer?

 

you should end up with

 

[math]kr + \frac{36}{56} = m(T + S)[/math]

 

right? Plug in the value for [math]r[/math]

 

you should end up with

 

[math]k + \frac{56}{56} = m(T + S)[/math]

 

(I noticed in the answer I gave I forgot to plug in the value of 1 (I'll change it now), so what you have for you final answer is

 

[math]k = m(T + S) -1[/math]

Edited by Aethelwulf
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wulf: I do not agree. You say that [math]\frac{x}{\frac{y}{z}}\equiv \frac{x}{\frac{z}{y}}[/math]. You should see right away that this is wrong.

 

What you want to say I guess is [math]\frac{x}{\frac{y}{z}}\equiv \frac{x}{1}\cdot{\frac{z}{y}}[/math], so you have the swapping around you mean.

 

In general you have the rule [math]\frac{\frac{x}{k}}{\frac{y}{z}}\equiv \frac{x}{k}\cdot \frac{z}{y}[/math].

 

Applying this to your a (squared): x=4*154=616, k=22, y=7, z=1 hence it follows: [math]\frac{616}{22}\cdot \frac{1}{7}=28\cdot\frac{1}{7}=4 [/math].

 

Hence a=2.

 

But re-reading through your posts, you are not coherent on where you put the main fraction!! And this changes everything, as written in the opening post you have:

[math]a^2=\frac{\frac{4\cdot 154}{22}}{7}[/math] and then belovelife and I are right.

 

Where you say that we are wrong you actually develop:

[math]\frac{4\cdot 154}{\frac{22}{7}}[/math] which is different and leads to a =14 as you say.

 

So it is your typo in OP that started all this discussion ;-)

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How where you right? You are still doing it wrong.

 

Belovelife has calculated

 

4*154 then divides that by 22 and then by 7 - that does indeed give you 4 but that isn't the correct operation for dividing whole numbers by fractions. Let us for a moment, say we where doing it the other way around, say we where dividing a fraction by a whole number, that is still not what you get.

 

Dividing a fraction by a whole number, let's say we had

 

2/5/4 = 2/ 5 *4 = 2/20 = 1/10

 

So in my example

 

616/22/7 = 616/ 22*7 = 616/154 = 308/77

 

If you even tried both dividing a whole number by a fraction and a fraction by a whole number, one would be able to work out which one was probably intended.

Edited by Aethelwulf
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you should end up with

 

[math]kr + \frac{36}{56} = m(T + S)[/math]

 

right?

I don't think so. Starting with post #1's

 

[math]\frac{kr}{m} - S + (\frac{5}{ab} - \frac{3}{b^2})= T[/math]

 

I get

 

[math]k = \frac{m(T +S -\frac{5}{ab} +\frac{3}{b^2} )}{r} [/math]

 

Substituting the givens [imath]a=14, b=2, r=20[/imath] and simplifying, that's

 

[math]k = \frac{m(7T +7S +4)}{40}[/math]

 

Evaluating the original equation with sample values for the m, T, S, and k calculated with my result checks out. Using your doesn't. I think you've erred, Aetherwolf.

Edited by CraigD
Replaces incorrect given a=2 with correct a=14
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