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Jim is twice as old as Stephanie, who, four years ago, was

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Jim is twice as old as Stephanie, who, four years ago, was [#permalink] New post 27 Apr 2012, 10:26
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Jim is twice as old as Stephanie, who, four years ago, was three times as old as Kate. If, five years from now, the sum of their ages will be 51, how old is Stephanie ?

A. 6
B. 10
C. 14
D. 20
E. 24

I need an help to translate words into math.

So. J= 2 S

S-4=3 (k-4)

S+5........here is the part that I do not catch. Some one can suggest me something in the right direction ??

thanks
[Reveal] Spoiler: OA

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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink] New post 27 Apr 2012, 10:37
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carcass wrote:
Jim is twice as old as Stephanie, who, four years ago, was three times as old as Kate. If, five years from now, the sum of their ages will be 51, how old is Stephanie ?

A. 6
B. 10
C. 14
D. 20
E. 24

I need an help to translate words into math.

So. J= 2 S

S-4=3 (k-4)

S+5........here is the part that I do not catch. Some one can suggest me something in the right direction ??

thanks


Jim is twice as old as Stephanie --> J=2S;

Stephanie four years ago, was three times as old as Kate --> S-4=3(K-4) --> K=(S+8)/3 (it would be better if it were "Stephanie four years ago, was three times as old as Kate was four years ago");

Five years from now, the sum of their ages will be 51 --> (J+5)+(S+5)+(K+5)=51 --> (2S+5)+(S+5)+((S+8)/3+5)=51 --> S=10.

Answer: B.
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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink] New post 27 Apr 2012, 16:14
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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink] New post 27 Apr 2012, 17:39
carcass wrote:
Was quite evident (not for me at the moment) that we solved for S and search for J and K (the other two variables).....and the rest is clear :)
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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink] New post 28 Apr 2012, 07:43
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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink] New post 02 Jul 2013, 00:16
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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DS questions on Arithmetic: search.php?search_id=tag&tag_id=30
PS questions on Arithmetic: search.php?search_id=tag&tag_id=51

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Jim is twice as old as Stephanie, who, four years ago, was [#permalink] New post 23 Nov 2014, 22:07
Best way is backsolving:

1) take C (14y.o.), so mean that S=14, J=28, their sum itself is 42+10 years from now is 52, it is over the 51,
so eliminate C,D,E
2) take B (10y.o.), meaning S=10,J=20, so 20+10+10=40 and for K=10-4/3=2+9=11, finally 40+11=51. It is correct

B

Last edited by Temurkhon on 24 Nov 2014, 00:05, edited 1 time in total.
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Jim is twice as old as Stephanie, who, four years ago, was [#permalink] New post 24 Nov 2014, 00:01
Jim ......... Stephanie ..................... Kate

.................. (a-4) ........................ \(\frac{1}{3} (a-4)\) ...................... (4 Years ago)


2a .............. a.............................................. (Current ages)


2a+5 ............ a+5 ........................ \(\frac{1}{3} (a-4) + 4 + 5\) .................. (Ages after 5 years)

Given that sum of ages post 5 years is 51

\(2a+5 + a+5 + \frac{1}{3} (a-4) + 9 = 51\)

a = 10

Answer = B
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Jim is twice as old as Stephanie, who, four years ago, was   [#permalink] 24 Nov 2014, 00:01
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