Jim is twice as old as Stephanie, who, four years ago, was : GMAT Problem Solving (PS)
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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]

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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]

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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.

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Jim is twice as old as Stephanie, who, four years ago, was [#permalink]

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28 Nov 2015, 09:51
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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 ?

From the above table we have -

(6k+13) + (3k+9) + (k+9) = 51

10k + 31 = 51

10k = 20

So, k = 2

We know , age of Stephanie now is 3k + 4 =>3*2 + 4 =10

PS : For such age related problems ( including age x yrs from now) the best method is coming to present age from x years back.
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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink]

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27 Apr 2012, 16:14
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

Thanks Mod. You are a landmark
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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink]

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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

Thanks Mod. You are a landmark

Are you serious
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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink]

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28 Apr 2012, 07:43
what do you mean ??'
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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink]

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02 Jul 2013, 00:16
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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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Kudos [?]: 21 [0], given: 40

Jim is twice as old as Stephanie, who, four years ago, was [#permalink]

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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]

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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

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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink]

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28 Nov 2015, 04:16
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink]

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03 Dec 2016, 03:40
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: Jim is twice as old as Stephanie, who, four years ago, was   [#permalink] 03 Dec 2016, 03:40
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