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Old 11-18-2004, 12:22 AM
skate_mate Australia skate_mate is offline
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Algebraic Ages [Maths]

Here's another assignment that I've been given; and this one really is hard. I think I've found the answers, but I'm not quite sure how I've gone about it.

Quote:
The combined ages of Marie and her sister Anne are 44 years, and Marie is twice as old as Anne was when Marie was half as old as Anne will be when Anne is three times as old as Marie was when Marie was three times as old as Anne. Deduce Marie's age and enter it in the space provided on this form.
I'm not quite sure, but I got 27.5 for Marie's age and 16.5 for Anne's age. I worked it out by seperating that long sentence into a few different equations and then substituting them into each other until I came out with those two answers. I'll write up the working soon, but until then could anyone verify this for me?

EDIT:

Alrighty, I think the key in that problem is to focus on the will be's and the when's.

Algebraicly:

Let M be Marie's curent age.
Let A be Anne's current age.
I'm unsure how to say "M is directly proportional to A", because they're sisters in algebraic terms :-S.
I started off having the new age of the people known as "M1, M2, etc" but because that got too confusing I put it in a time scenario, e.g

|<---y---M (current age) ---x--->| (and then another one coming out called "f")

This how how I wrote the equation.. I don't know if its very good :-S.

M + A = 44

M/2 = (A-x)
(M-x)*2=(A+y)
(A+y)/3=(M-f)
(M-f)/3=(A-f)

I couldn't work it out algebraicly, only by substitution; trial and error.
First, I tried '22' as Marie's age.

22 + A = 44
A=44

22/2=22-x
11=22-x
11=x

(22-11)*2=(22+y)
22=22+y
0=y

(22+0)/3=(22-f)
7.333=(22-f)
14.666=f

(22-14.666)/3=(22-14.666)
2.444 does not equal 7.334
Here, you can see that this is obviously incorrect.

From there, you can see that the gap between the sister's ages needs to be more then 0. After a bit more trial and error I came up with 27.5.

27.5+A=44
A=16.5

27.5/2=16.5-x
13.75=16.5-x
2.75=x

(27.5-2.75)*2=16.5+y
24.75*2=16.5+y
49.5=16.5+y
33=y

(16.5+33)/3=27.5-f
16.5=27.5-f
11=f

(27.5-11)/3=16.5-11
5.5=5.5

=)

Now I know I've gotten the question correct, I just don't know how to explain it properly in mathematical terminology, help please :-S

EDIT#2: I've uploaded the file, its a bit neater then the above writing. Can someone check it for me, please?
Attached Files
File Type: doc JMA356 - Marie and Anne.doc (28.5 KB, 5 views)
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Last Edited by skate_mate; 11-18-2004 at 02:29 AM. Reason: Reply With Quote
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Old 11-21-2004, 04:05 PM
Czardas Czardas is offline
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Re: Algebraic Ages [Maths]

Ok, a bit late but perhaps still helpful?
A quite fun problem if I may say so.

This is how I solved the problem without the use of iteration:

First, let's set a few variables. I'll let both sisters' ages be linear functions of time (on the form y=kx+m as I'm sure you are familiar with) and I let s be the time when we wish to know Marie's age. I also used the variable u as a number of years before s. Note that both s and u are integrers (part of Z).

aM(t) will indicate Marie's age (I wrote it as a index M but that's a bit hard here) and aA(t) will indicate Anne's age at any time t.

It is then easy to realize that
aM(t) = t + aM(0) (1a) and
aA(t) = t + aA(0) (1b)

From the text we now read
aM(0) = 3 aA(0) (2) (from the last part of the last sentence)
aM(s) + aA(s) = 44 (3) (the very first words)

Combine (3) with (1a) and (1b) to get
2 s + aM(0) + aA(0) = 44
and further with (2) to get
2 s + 4aA(0) = 44 (4)

More reading (from the complex sentence) tells us
aM(s) = 2 aA(s-u) (5)
aM(s-u) = 1/2 * 3 aM(0) (6)

Combine (6) with (2) to get
aM(s-u) = 9/2 aA(0) (7)

But aM(s-u) & aA(s-u) are
aM(s-u) = s - u + aM(0) (8)
aA(s-u) = s - u + aA(0) (9)

Combine (2), (8) & (6)
s - u = 3/2 aA(0) (10)

Combine (1a), (5) & (9)
s - 2 u = aA(0) (11)

We now have all we need to solve the equation
(4): 2 s + 4aA(0) = 44
(10): s - u = 3/2 aA(0)
(11): s - 2 u = aA(0)
Three equations and three unknowns. I suggest that you eliminate u first to get that s=11. That fact leads to that aA(0)=5.5 and thus aM(0)=16.5

Add 11 to 16.5 to get aM(s)=27.5
Attached Files
File Type: pdf annemarie.pdf (15.8 KB, 1 views)
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Last Edited by Czardas; 11-21-2004 at 04:15 PM. Reason: Reply With Quote
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Old 11-21-2004, 04:21 PM
Link_Of_Hyru1e Link_Of_Hyru1e is offline
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Re: Algebraic Ages [Maths]

i hate that stupid stuff. i never got it all that well... i passed somehow tho... O.o
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Old 12-03-2004, 09:36 PM
skate_mate Australia skate_mate is offline
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Re: Algebraic Ages [Maths]

CZardas; thank you very, very much for the hard work you put into that post . may I ask what grade you're in (if you are indeed still at school)?

Quote:
(on the form y=kx+m as I'm sure you are familiar with)
Actually I'm not familiar with it. We've always used "y=mx+c" (which I guess is the same with with different names for the variables :-P).
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